3.940 \(\int \frac {e^{\tanh ^{-1}(a x)} x^4}{(1-a^2 x^2)^{5/2}} \, dx\)

Optimal. Leaf size=72 \[ -\frac {3}{4 a^5 (1-a x)}-\frac {1}{8 a^5 (a x+1)}+\frac {1}{8 a^5 (1-a x)^2}-\frac {11 \log (1-a x)}{16 a^5}-\frac {5 \log (a x+1)}{16 a^5} \]

[Out]

1/8/a^5/(-a*x+1)^2-3/4/a^5/(-a*x+1)-1/8/a^5/(a*x+1)-11/16*ln(-a*x+1)/a^5-5/16*ln(a*x+1)/a^5

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Rubi [A]  time = 0.12, antiderivative size = 72, 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} \[ -\frac {3}{4 a^5 (1-a x)}-\frac {1}{8 a^5 (a x+1)}+\frac {1}{8 a^5 (1-a x)^2}-\frac {11 \log (1-a x)}{16 a^5}-\frac {5 \log (a x+1)}{16 a^5} \]

Antiderivative was successfully verified.

[In]

Int[(E^ArcTanh[a*x]*x^4)/(1 - a^2*x^2)^(5/2),x]

[Out]

1/(8*a^5*(1 - a*x)^2) - 3/(4*a^5*(1 - a*x)) - 1/(8*a^5*(1 + a*x)) - (11*Log[1 - a*x])/(16*a^5) - (5*Log[1 + a*
x])/(16*a^5)

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^4}{\left (1-a^2 x^2\right )^{5/2}} \, dx &=\int \frac {x^4}{(1-a x)^3 (1+a x)^2} \, dx\\ &=\int \left (-\frac {1}{4 a^4 (-1+a x)^3}-\frac {3}{4 a^4 (-1+a x)^2}-\frac {11}{16 a^4 (-1+a x)}+\frac {1}{8 a^4 (1+a x)^2}-\frac {5}{16 a^4 (1+a x)}\right ) \, dx\\ &=\frac {1}{8 a^5 (1-a x)^2}-\frac {3}{4 a^5 (1-a x)}-\frac {1}{8 a^5 (1+a x)}-\frac {11 \log (1-a x)}{16 a^5}-\frac {5 \log (1+a x)}{16 a^5}\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 55, normalized size = 0.76 \[ \frac {\frac {2 \left (5 a^2 x^2+3 a x-6\right )}{(a x-1)^2 (a x+1)}-11 \log (1-a x)-5 \log (a x+1)}{16 a^5} \]

Antiderivative was successfully verified.

[In]

Integrate[(E^ArcTanh[a*x]*x^4)/(1 - a^2*x^2)^(5/2),x]

[Out]

((2*(-6 + 3*a*x + 5*a^2*x^2))/((-1 + a*x)^2*(1 + a*x)) - 11*Log[1 - a*x] - 5*Log[1 + a*x])/(16*a^5)

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fricas [A]  time = 0.50, size = 101, normalized size = 1.40 \[ \frac {10 \, a^{2} x^{2} + 6 \, a x - 5 \, {\left (a^{3} x^{3} - a^{2} x^{2} - a x + 1\right )} \log \left (a x + 1\right ) - 11 \, {\left (a^{3} x^{3} - a^{2} x^{2} - a x + 1\right )} \log \left (a x - 1\right ) - 12}{16 \, {\left (a^{8} x^{3} - a^{7} x^{2} - a^{6} x + a^{5}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)/(-a^2*x^2+1)^3*x^4,x, algorithm="fricas")

[Out]

1/16*(10*a^2*x^2 + 6*a*x - 5*(a^3*x^3 - a^2*x^2 - a*x + 1)*log(a*x + 1) - 11*(a^3*x^3 - a^2*x^2 - a*x + 1)*log
(a*x - 1) - 12)/(a^8*x^3 - a^7*x^2 - a^6*x + a^5)

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giac [A]  time = 0.27, size = 58, normalized size = 0.81 \[ -\frac {5 \, \log \left ({\left | a x + 1 \right |}\right )}{16 \, a^{5}} - \frac {11 \, \log \left ({\left | a x - 1 \right |}\right )}{16 \, a^{5}} + \frac {5 \, a^{2} x^{2} + 3 \, a x - 6}{8 \, {\left (a x + 1\right )} {\left (a x - 1\right )}^{2} a^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)/(-a^2*x^2+1)^3*x^4,x, algorithm="giac")

[Out]

-5/16*log(abs(a*x + 1))/a^5 - 11/16*log(abs(a*x - 1))/a^5 + 1/8*(5*a^2*x^2 + 3*a*x - 6)/((a*x + 1)*(a*x - 1)^2
*a^5)

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maple [A]  time = 0.04, size = 60, normalized size = 0.83 \[ \frac {1}{8 a^{5} \left (a x -1\right )^{2}}+\frac {3}{4 a^{5} \left (a x -1\right )}-\frac {11 \ln \left (a x -1\right )}{16 a^{5}}-\frac {1}{8 a^{5} \left (a x +1\right )}-\frac {5 \ln \left (a x +1\right )}{16 a^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x+1)/(-a^2*x^2+1)^3*x^4,x)

[Out]

1/8/a^5/(a*x-1)^2+3/4/a^5/(a*x-1)-11/16/a^5*ln(a*x-1)-1/8/a^5/(a*x+1)-5/16*ln(a*x+1)/a^5

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maxima [A]  time = 0.32, size = 66, normalized size = 0.92 \[ \frac {5 \, a^{2} x^{2} + 3 \, a x - 6}{8 \, {\left (a^{8} x^{3} - a^{7} x^{2} - a^{6} x + a^{5}\right )}} - \frac {5 \, \log \left (a x + 1\right )}{16 \, a^{5}} - \frac {11 \, \log \left (a x - 1\right )}{16 \, a^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)/(-a^2*x^2+1)^3*x^4,x, algorithm="maxima")

[Out]

1/8*(5*a^2*x^2 + 3*a*x - 6)/(a^8*x^3 - a^7*x^2 - a^6*x + a^5) - 5/16*log(a*x + 1)/a^5 - 11/16*log(a*x - 1)/a^5

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mupad [B]  time = 0.18, size = 67, normalized size = 0.93 \[ -\frac {11\,\ln \left (a\,x-1\right )}{16\,a^5}-\frac {5\,\ln \left (a\,x+1\right )}{16\,a^5}-\frac {\frac {3\,x}{8\,a^4}-\frac {3}{4\,a^5}+\frac {5\,x^2}{8\,a^3}}{-a^3\,x^3+a^2\,x^2+a\,x-1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(x^4*(a*x + 1))/(a^2*x^2 - 1)^3,x)

[Out]

- (11*log(a*x - 1))/(16*a^5) - (5*log(a*x + 1))/(16*a^5) - ((3*x)/(8*a^4) - 3/(4*a^5) + (5*x^2)/(8*a^3))/(a*x
+ a^2*x^2 - a^3*x^3 - 1)

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sympy [A]  time = 0.38, size = 68, normalized size = 0.94 \[ - \frac {- 5 a^{2} x^{2} - 3 a x + 6}{8 a^{8} x^{3} - 8 a^{7} x^{2} - 8 a^{6} x + 8 a^{5}} - \frac {\frac {11 \log {\left (x - \frac {1}{a} \right )}}{16} + \frac {5 \log {\left (x + \frac {1}{a} \right )}}{16}}{a^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)/(-a**2*x**2+1)**3*x**4,x)

[Out]

-(-5*a**2*x**2 - 3*a*x + 6)/(8*a**8*x**3 - 8*a**7*x**2 - 8*a**6*x + 8*a**5) - (11*log(x - 1/a)/16 + 5*log(x +
1/a)/16)/a**5

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