3.929 \(\int \frac {e^{\tanh ^{-1}(a x)} x^4}{(1-a^2 x^2)^{3/2}} \, dx\)

Optimal. Leaf size=58 \[ \frac {1}{2 a^5 (1-a x)}+\frac {7 \log (1-a x)}{4 a^5}+\frac {\log (a x+1)}{4 a^5}+\frac {x}{a^4}+\frac {x^2}{2 a^3} \]

[Out]

x/a^4+1/2*x^2/a^3+1/2/a^5/(-a*x+1)+7/4*ln(-a*x+1)/a^5+1/4*ln(a*x+1)/a^5

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Rubi [A]  time = 0.12, antiderivative size = 58, 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} \[ \frac {x^2}{2 a^3}+\frac {x}{a^4}+\frac {1}{2 a^5 (1-a x)}+\frac {7 \log (1-a x)}{4 a^5}+\frac {\log (a x+1)}{4 a^5} \]

Antiderivative was successfully verified.

[In]

Int[(E^ArcTanh[a*x]*x^4)/(1 - a^2*x^2)^(3/2),x]

[Out]

x/a^4 + x^2/(2*a^3) + 1/(2*a^5*(1 - a*x)) + (7*Log[1 - a*x])/(4*a^5) + Log[1 + a*x]/(4*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 )^{3/2}} \, dx &=\int \frac {x^4}{(1-a x)^2 (1+a x)} \, dx\\ &=\int \left (\frac {1}{a^4}+\frac {x}{a^3}+\frac {1}{2 a^4 (-1+a x)^2}+\frac {7}{4 a^4 (-1+a x)}+\frac {1}{4 a^4 (1+a x)}\right ) \, dx\\ &=\frac {x}{a^4}+\frac {x^2}{2 a^3}+\frac {1}{2 a^5 (1-a x)}+\frac {7 \log (1-a x)}{4 a^5}+\frac {\log (1+a x)}{4 a^5}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 45, normalized size = 0.78 \[ \frac {2 \left (a^2 x^2+2 a x+\frac {1}{1-a x}\right )+7 \log (1-a x)+\log (a x+1)}{4 a^5} \]

Antiderivative was successfully verified.

[In]

Integrate[(E^ArcTanh[a*x]*x^4)/(1 - a^2*x^2)^(3/2),x]

[Out]

(2*(2*a*x + a^2*x^2 + (1 - a*x)^(-1)) + 7*Log[1 - a*x] + Log[1 + a*x])/(4*a^5)

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fricas [A]  time = 0.61, size = 62, normalized size = 1.07 \[ \frac {2 \, a^{3} x^{3} + 2 \, a^{2} x^{2} - 4 \, a x + {\left (a x - 1\right )} \log \left (a x + 1\right ) + 7 \, {\left (a x - 1\right )} \log \left (a x - 1\right ) - 2}{4 \, {\left (a^{6} x - a^{5}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.15, size = 56, normalized size = 0.97 \[ \frac {\log \left ({\left | a x + 1 \right |}\right )}{4 \, a^{5}} + \frac {7 \, \log \left ({\left | a x - 1 \right |}\right )}{4 \, a^{5}} + \frac {a^{3} x^{2} + 2 \, a^{2} x}{2 \, a^{6}} - \frac {1}{2 \, {\left (a x - 1\right )} a^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*log(abs(a*x + 1))/a^5 + 7/4*log(abs(a*x - 1))/a^5 + 1/2*(a^3*x^2 + 2*a^2*x)/a^6 - 1/2/((a*x - 1)*a^5)

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maple [A]  time = 0.03, size = 49, normalized size = 0.84 \[ \frac {x^{2}}{2 a^{3}}+\frac {x}{a^{4}}-\frac {1}{2 a^{5} \left (a x -1\right )}+\frac {7 \ln \left (a x -1\right )}{4 a^{5}}+\frac {\ln \left (a x +1\right )}{4 a^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x^2/a^3+x/a^4-1/2/a^5/(a*x-1)+7/4/a^5*ln(a*x-1)+1/4*ln(a*x+1)/a^5

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maxima [A]  time = 0.31, size = 52, normalized size = 0.90 \[ -\frac {1}{2 \, {\left (a^{6} x - a^{5}\right )}} + \frac {a x^{2} + 2 \, x}{2 \, a^{4}} + \frac {\log \left (a x + 1\right )}{4 \, a^{5}} + \frac {7 \, \log \left (a x - 1\right )}{4 \, a^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/(a^6*x - a^5) + 1/2*(a*x^2 + 2*x)/a^4 + 1/4*log(a*x + 1)/a^5 + 7/4*log(a*x - 1)/a^5

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mupad [B]  time = 0.91, size = 54, normalized size = 0.93 \[ \frac {7\,\ln \left (a\,x-1\right )}{4\,a^5}+\frac {\ln \left (a\,x+1\right )}{4\,a^5}-\frac {1}{2\,a\,\left (a^5\,x-a^4\right )}+\frac {x}{a^4}+\frac {x^2}{2\,a^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(7*log(a*x - 1))/(4*a^5) + log(a*x + 1)/(4*a^5) - 1/(2*a*(a^5*x - a^4)) + x/a^4 + x^2/(2*a^3)

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sympy [A]  time = 0.26, size = 48, normalized size = 0.83 \[ - \frac {1}{2 a^{6} x - 2 a^{5}} + \frac {x^{2}}{2 a^{3}} + \frac {x}{a^{4}} + \frac {\frac {7 \log {\left (x - \frac {1}{a} \right )}}{4} + \frac {\log {\left (x + \frac {1}{a} \right )}}{4}}{a^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/(2*a**6*x - 2*a**5) + x**2/(2*a**3) + x/a**4 + (7*log(x - 1/a)/4 + log(x + 1/a)/4)/a**5

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