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Topic: Rubi 4 is now available
Replies: 25   Last Post: Jun 22, 2013 4:14 PM

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did

Posts: 77
Registered: 9/14/05
Re: Rubi 4 is now available
Posted: May 20, 2013 3:55 AM
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What Rubi 4 on MMA 9 gives:

> INT(ASIN(x)*LN(x), x)
unable

> INT(x*ASIN(x)/SQRT(1 - x^2), x)
x - Sqrt[1 - x^2] ArcSin[x]

> INT(ASIN(SQRT(x + 1) - SQRT(x)), x)
unable

> INT(LN(1 + x*SQRT(1 + x^2)), x)
-2 x - Sqrt[1/10 (1 + Sqrt[5])] ArcTan[Sqrt[2/(1 + Sqrt[5])] x] +
2 Sqrt[1/5 (2 + Sqrt[5])] ArcTan[Sqrt[2/(1 + Sqrt[5])] x] +
Sqrt[2/(5 (-1 + Sqrt[5]))]
ArcTan[Sqrt[2/(-1 + Sqrt[5])] Sqrt[1 + x^2]] +
Sqrt[2/5 (-1 + Sqrt[5])]
ArcTan[Sqrt[2/(-1 + Sqrt[5])] Sqrt[1 + x^2]] +
2 Sqrt[1/5 (-2 + Sqrt[5])] ArcTanh[Sqrt[2/(-1 + Sqrt[5])] x] +
Sqrt[1/10 (-1 + Sqrt[5])] ArcTanh[Sqrt[2/(-1 + Sqrt[5])] x] +
Sqrt[2/(5 (1 + Sqrt[5]))]
ArcTanh[Sqrt[2/(1 + Sqrt[5])] Sqrt[1 + x^2]] -
Sqrt[2/5 (1 + Sqrt[5])]
ArcTanh[Sqrt[2/(1 + Sqrt[5])] Sqrt[1 + x^2]] +
x Log[1 + x Sqrt[1 + x^2]]

> INT(COS(x)^2/SQRT(COS(x)^4 + COS(x)^2 + 1), x)
(EllipticPi[1/2 (3 - I Sqrt[3]),
ArcSin[Sqrt[-((2 I)/(3 I + Sqrt[3]))] Tan[x]], (3 I + Sqrt[3])/(
3 I - Sqrt[3])] (3 + 3 Tan[x]^2 + Tan[x]^4))/(3 Sqrt[-((2 I)/(
3 I + Sqrt[3]))] (1 + Tan[x]^2) Sqrt[
1 + (2 Tan[x]^2)/(3 - I Sqrt[3])] Sqrt[
1 + (2 Tan[x]^2)/(3 + I Sqrt[3])] Sqrt[(
3 + 3 Tan[x]^2 + Tan[x]^4)/(1 + Tan[x]^2)^2])

> INT(TAN(x)*SQRT(1 + TAN(x)^4), x)
-(1/2) ArcSinh[Tan[x]^2] -
ArcTanh[(1 - Tan[x]^2)/(Sqrt[2] Sqrt[1 + Tan[x]^4])]/Sqrt[2] +
1/2 Sqrt[1 + Tan[x]^4]

> INT(TAN(x)/SQRT(SEC(x)^3 + 1), x)
-(2/3) ArcTanh[Sqrt[1 + Sec[x]^3]]

> INT(SQRT(TAN(x)^2 + 2*TAN(x) + 2), x)
ArcSinh[1 + Tan[x]] -
1/2 I Sqrt[1 - 2 I]
ArcTanh[((4 - 2 I) + (2 - 2 I) Tan[x])/(
2 Sqrt[1 - 2 I] Sqrt[2 + 2 Tan[x] + Tan[x]^2])] +
1/2 I Sqrt[1 + 2 I]
ArcTanh[((4 + 2 I) + (2 + 2 I) Tan[x])/(
2 Sqrt[1 + 2 I] Sqrt[2 + 2 Tan[x] + Tan[x]^2])]

> INT(SIN(x)*ATAN(SQRT(SEC(x) - 1)), x)
1/2 ArcTan[Sqrt[-1 + Sec[x]]] - ArcTan[Sqrt[-1 + Sec[x]]] Cos[x] +
1/2 Cos[x] Sqrt[-1 + Sec[x]]

> INT(x^3*EXP(ASIN(x))/SQRT(1 - x^2), x)
3/10 E^ArcSin[x] x + 1/10 E^ArcSin[x] x^3 -
3/10 E^ArcSin[x] Sqrt[1 - x^2] - 3/10 E^ArcSin[x] x^2 Sqrt[1 - x^2]

Every results were obtained instantaneously.

Did



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