3.939 \(\int \frac {e^{\tanh ^{-1}(a x)} x^5}{(1-a^2 x^2)^{5/2}} \, dx\)

Optimal. Leaf size=76 \[ -\frac {1}{a^6 (1-a x)}+\frac {1}{8 a^6 (a x+1)}+\frac {1}{8 a^6 (1-a x)^2}-\frac {23 \log (1-a x)}{16 a^6}+\frac {7 \log (a x+1)}{16 a^6}-\frac {x}{a^5} \]

[Out]

-x/a^5+1/8/a^6/(-a*x+1)^2-1/a^6/(-a*x+1)+1/8/a^6/(a*x+1)-23/16*ln(-a*x+1)/a^6+7/16*ln(a*x+1)/a^6

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Rubi [A]  time = 0.13, antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {6150, 88} \[ -\frac {x}{a^5}-\frac {1}{a^6 (1-a x)}+\frac {1}{8 a^6 (a x+1)}+\frac {1}{8 a^6 (1-a x)^2}-\frac {23 \log (1-a x)}{16 a^6}+\frac {7 \log (a x+1)}{16 a^6} \]

Antiderivative was successfully verified.

[In]

Int[(E^ArcTanh[a*x]*x^5)/(1 - a^2*x^2)^(5/2),x]

[Out]

-(x/a^5) + 1/(8*a^6*(1 - a*x)^2) - 1/(a^6*(1 - a*x)) + 1/(8*a^6*(1 + a*x)) - (23*Log[1 - a*x])/(16*a^6) + (7*L
og[1 + a*x])/(16*a^6)

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rule 6150

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[c^p, Int[x^m*(1 -
a*x)^(p - n/2)*(1 + a*x)^(p + n/2), x], x] /; FreeQ[{a, c, d, m, n, p}, x] && EqQ[a^2*c + d, 0] && (IntegerQ[p
] || GtQ[c, 0])

Rubi steps

\begin {align*} \int \frac {e^{\tanh ^{-1}(a x)} x^5}{\left (1-a^2 x^2\right )^{5/2}} \, dx &=\int \frac {x^5}{(1-a x)^3 (1+a x)^2} \, dx\\ &=\int \left (-\frac {1}{a^5}-\frac {1}{4 a^5 (-1+a x)^3}-\frac {1}{a^5 (-1+a x)^2}-\frac {23}{16 a^5 (-1+a x)}-\frac {1}{8 a^5 (1+a x)^2}+\frac {7}{16 a^5 (1+a x)}\right ) \, dx\\ &=-\frac {x}{a^5}+\frac {1}{8 a^6 (1-a x)^2}-\frac {1}{a^6 (1-a x)}+\frac {1}{8 a^6 (1+a x)}-\frac {23 \log (1-a x)}{16 a^6}+\frac {7 \log (1+a x)}{16 a^6}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 55, normalized size = 0.72 \[ \frac {2 \left (-8 a x+\frac {8}{a x-1}+\frac {1}{a x+1}+\frac {1}{(a x-1)^2}\right )-23 \log (1-a x)+7 \log (a x+1)}{16 a^6} \]

Antiderivative was successfully verified.

[In]

Integrate[(E^ArcTanh[a*x]*x^5)/(1 - a^2*x^2)^(5/2),x]

[Out]

(2*(-8*a*x + (-1 + a*x)^(-2) + 8/(-1 + a*x) + (1 + a*x)^(-1)) - 23*Log[1 - a*x] + 7*Log[1 + a*x])/(16*a^6)

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fricas [A]  time = 0.54, size = 117, normalized size = 1.54 \[ -\frac {16 \, a^{4} x^{4} - 16 \, a^{3} x^{3} - 34 \, a^{2} x^{2} + 18 \, a x - 7 \, {\left (a^{3} x^{3} - a^{2} x^{2} - a x + 1\right )} \log \left (a x + 1\right ) + 23 \, {\left (a^{3} x^{3} - a^{2} x^{2} - a x + 1\right )} \log \left (a x - 1\right ) + 12}{16 \, {\left (a^{9} x^{3} - a^{8} x^{2} - a^{7} x + a^{6}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)/(-a^2*x^2+1)^3*x^5,x, algorithm="fricas")

[Out]

-1/16*(16*a^4*x^4 - 16*a^3*x^3 - 34*a^2*x^2 + 18*a*x - 7*(a^3*x^3 - a^2*x^2 - a*x + 1)*log(a*x + 1) + 23*(a^3*
x^3 - a^2*x^2 - a*x + 1)*log(a*x - 1) + 12)/(a^9*x^3 - a^8*x^2 - a^7*x + a^6)

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giac [A]  time = 0.17, size = 64, normalized size = 0.84 \[ -\frac {x}{a^{5}} + \frac {7 \, \log \left ({\left | a x + 1 \right |}\right )}{16 \, a^{6}} - \frac {23 \, \log \left ({\left | a x - 1 \right |}\right )}{16 \, a^{6}} + \frac {9 \, a^{2} x^{2} - a x - 6}{8 \, {\left (a x + 1\right )} {\left (a x - 1\right )}^{2} a^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)/(-a^2*x^2+1)^3*x^5,x, algorithm="giac")

[Out]

-x/a^5 + 7/16*log(abs(a*x + 1))/a^6 - 23/16*log(abs(a*x - 1))/a^6 + 1/8*(9*a^2*x^2 - a*x - 6)/((a*x + 1)*(a*x
- 1)^2*a^6)

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maple [A]  time = 0.04, size = 65, normalized size = 0.86 \[ -\frac {x}{a^{5}}-\frac {23 \ln \left (a x -1\right )}{16 a^{6}}+\frac {1}{8 a^{6} \left (a x -1\right )^{2}}+\frac {1}{a^{6} \left (a x -1\right )}+\frac {1}{8 a^{6} \left (a x +1\right )}+\frac {7 \ln \left (a x +1\right )}{16 a^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x+1)/(-a^2*x^2+1)^3*x^5,x)

[Out]

-x/a^5-23/16/a^6*ln(a*x-1)+1/8/a^6/(a*x-1)^2+1/a^6/(a*x-1)+1/8/a^6/(a*x+1)+7/16*ln(a*x+1)/a^6

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maxima [A]  time = 0.33, size = 72, normalized size = 0.95 \[ \frac {9 \, a^{2} x^{2} - a x - 6}{8 \, {\left (a^{9} x^{3} - a^{8} x^{2} - a^{7} x + a^{6}\right )}} - \frac {x}{a^{5}} + \frac {7 \, \log \left (a x + 1\right )}{16 \, a^{6}} - \frac {23 \, \log \left (a x - 1\right )}{16 \, a^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)/(-a^2*x^2+1)^3*x^5,x, algorithm="maxima")

[Out]

1/8*(9*a^2*x^2 - a*x - 6)/(a^9*x^3 - a^8*x^2 - a^7*x + a^6) - x/a^5 + 7/16*log(a*x + 1)/a^6 - 23/16*log(a*x -
1)/a^6

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mupad [B]  time = 0.93, size = 73, normalized size = 0.96 \[ \frac {7\,\ln \left (a\,x+1\right )}{16\,a^6}-\frac {23\,\ln \left (a\,x-1\right )}{16\,a^6}-\frac {x}{a^5}+\frac {\frac {x}{8}-\frac {9\,a\,x^2}{8}+\frac {3}{4\,a}}{-a^8\,x^3+a^7\,x^2+a^6\,x-a^5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(x^5*(a*x + 1))/(a^2*x^2 - 1)^3,x)

[Out]

(7*log(a*x + 1))/(16*a^6) - (23*log(a*x - 1))/(16*a^6) - x/a^5 + (x/8 - (9*a*x^2)/8 + 3/(4*a))/(a^6*x - a^5 +
a^7*x^2 - a^8*x^3)

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sympy [A]  time = 0.43, size = 71, normalized size = 0.93 \[ - \frac {- 9 a^{2} x^{2} + a x + 6}{8 a^{9} x^{3} - 8 a^{8} x^{2} - 8 a^{7} x + 8 a^{6}} - \frac {x}{a^{5}} - \frac {\frac {23 \log {\left (x - \frac {1}{a} \right )}}{16} - \frac {7 \log {\left (x + \frac {1}{a} \right )}}{16}}{a^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)/(-a**2*x**2+1)**3*x**5,x)

[Out]

-(-9*a**2*x**2 + a*x + 6)/(8*a**9*x**3 - 8*a**8*x**2 - 8*a**7*x + 8*a**6) - x/a**5 - (23*log(x - 1/a)/16 - 7*l
og(x + 1/a)/16)/a**6

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