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

Optimal. Leaf size=71 \[ \frac{3 a}{4 (1-a x)}-\frac{a}{8 (a x+1)}+\frac{a}{8 (1-a x)^2}+a \log (x)-\frac{23}{16} a \log (1-a x)+\frac{7}{16} a \log (a x+1)-\frac{1}{x} \]

[Out]

-x^(-1) + a/(8*(1 - a*x)^2) + (3*a)/(4*(1 - a*x)) - a/(8*(1 + a*x)) + a*Log[x] - (23*a*Log[1 - a*x])/16 + (7*a
*Log[1 + a*x])/16

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Rubi [A]  time = 0.126846, antiderivative size = 71, normalized size of antiderivative = 1., 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} \[ \frac{3 a}{4 (1-a x)}-\frac{a}{8 (a x+1)}+\frac{a}{8 (1-a x)^2}+a \log (x)-\frac{23}{16} a \log (1-a x)+\frac{7}{16} a \log (a x+1)-\frac{1}{x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-x^(-1) + a/(8*(1 - a*x)^2) + (3*a)/(4*(1 - a*x)) - a/(8*(1 + a*x)) + a*Log[x] - (23*a*Log[1 - a*x])/16 + (7*a
*Log[1 + a*x])/16

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])

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]))

Rubi steps

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

Mathematica [A]  time = 0.0601992, size = 65, normalized size = 0.92 \[ \frac{1}{16} \left (\frac{12 a}{1-a x}-\frac{2 a}{a x+1}+\frac{2 a}{(a x-1)^2}+16 a \log (x)-23 a \log (1-a x)+7 a \log (a x+1)-\frac{16}{x}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-16/x + (12*a)/(1 - a*x) + (2*a)/(-1 + a*x)^2 - (2*a)/(1 + a*x) + 16*a*Log[x] - 23*a*Log[1 - a*x] + 7*a*Log[1
+ a*x])/16

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Maple [A]  time = 0.038, size = 59, normalized size = 0.8 \begin{align*} -{x}^{-1}+a\ln \left ( x \right ) -{\frac{a}{8\,ax+8}}+{\frac{7\,a\ln \left ( ax+1 \right ) }{16}}+{\frac{a}{8\, \left ( ax-1 \right ) ^{2}}}-{\frac{3\,a}{4\,ax-4}}-{\frac{23\,a\ln \left ( ax-1 \right ) }{16}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/x+a*ln(x)-1/8*a/(a*x+1)+7/16*a*ln(a*x+1)+1/8*a/(a*x-1)^2-3/4*a/(a*x-1)-23/16*a*ln(a*x-1)

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Maxima [A]  time = 0.95228, size = 97, normalized size = 1.37 \begin{align*} \frac{7}{16} \, a \log \left (a x + 1\right ) - \frac{23}{16} \, a \log \left (a x - 1\right ) + a \log \left (x\right ) - \frac{15 \, a^{3} x^{3} - 11 \, a^{2} x^{2} - 14 \, a x + 8}{8 \,{\left (a^{3} x^{4} - a^{2} x^{3} - a x^{2} + x\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

7/16*a*log(a*x + 1) - 23/16*a*log(a*x - 1) + a*log(x) - 1/8*(15*a^3*x^3 - 11*a^2*x^2 - 14*a*x + 8)/(a^3*x^4 -
a^2*x^3 - a*x^2 + x)

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Fricas [B]  time = 2.03365, size = 316, normalized size = 4.45 \begin{align*} -\frac{30 \, a^{3} x^{3} - 22 \, a^{2} x^{2} - 28 \, a x - 7 \,{\left (a^{4} x^{4} - a^{3} x^{3} - a^{2} x^{2} + a x\right )} \log \left (a x + 1\right ) + 23 \,{\left (a^{4} x^{4} - a^{3} x^{3} - a^{2} x^{2} + a x\right )} \log \left (a x - 1\right ) - 16 \,{\left (a^{4} x^{4} - a^{3} x^{3} - a^{2} x^{2} + a x\right )} \log \left (x\right ) + 16}{16 \,{\left (a^{3} x^{4} - a^{2} x^{3} - a x^{2} + x\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*(30*a^3*x^3 - 22*a^2*x^2 - 28*a*x - 7*(a^4*x^4 - a^3*x^3 - a^2*x^2 + a*x)*log(a*x + 1) + 23*(a^4*x^4 - a
^3*x^3 - a^2*x^2 + a*x)*log(a*x - 1) - 16*(a^4*x^4 - a^3*x^3 - a^2*x^2 + a*x)*log(x) + 16)/(a^3*x^4 - a^2*x^3
- a*x^2 + x)

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Sympy [A]  time = 0.772041, size = 78, normalized size = 1.1 \begin{align*} a \log{\left (x \right )} - \frac{23 a \log{\left (x - \frac{1}{a} \right )}}{16} + \frac{7 a \log{\left (x + \frac{1}{a} \right )}}{16} - \frac{15 a^{3} x^{3} - 11 a^{2} x^{2} - 14 a x + 8}{8 a^{3} x^{4} - 8 a^{2} x^{3} - 8 a x^{2} + 8 x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*log(x) - 23*a*log(x - 1/a)/16 + 7*a*log(x + 1/a)/16 - (15*a**3*x**3 - 11*a**2*x**2 - 14*a*x + 8)/(8*a**3*x**
4 - 8*a**2*x**3 - 8*a*x**2 + 8*x)

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Giac [A]  time = 1.20685, size = 90, normalized size = 1.27 \begin{align*} \frac{7}{16} \, a \log \left ({\left | a x + 1 \right |}\right ) - \frac{23}{16} \, a \log \left ({\left | a x - 1 \right |}\right ) + a \log \left ({\left | x \right |}\right ) - \frac{15 \, a^{3} x^{3} - 11 \, a^{2} x^{2} - 14 \, a x + 8}{8 \,{\left (a x + 1\right )}{\left (a x - 1\right )}^{2} x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

7/16*a*log(abs(a*x + 1)) - 23/16*a*log(abs(a*x - 1)) + a*log(abs(x)) - 1/8*(15*a^3*x^3 - 11*a^2*x^2 - 14*a*x +
8)/((a*x + 1)*(a*x - 1)^2*x)