3.1063 \(\int \frac{e^{2 \tanh ^{-1}(a x)}}{x (c-a^2 c x^2)^2} \, dx\)

Optimal. Leaf size=64 \[ \frac{3}{4 c^2 (1-a x)}+\frac{1}{4 c^2 (1-a x)^2}-\frac{7 \log (1-a x)}{8 c^2}-\frac{\log (a x+1)}{8 c^2}+\frac{\log (x)}{c^2} \]

[Out]

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

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Rubi [A]  time = 0.0992315, antiderivative size = 64, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.08, Rules used = {6150, 72} \[ \frac{3}{4 c^2 (1-a x)}+\frac{1}{4 c^2 (1-a x)^2}-\frac{7 \log (1-a x)}{8 c^2}-\frac{\log (a x+1)}{8 c^2}+\frac{\log (x)}{c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 72

Int[((e_.) + (f_.)*(x_))^(p_.)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Int[ExpandIntegrand[(
e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[p]

Rubi steps

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

Mathematica [A]  time = 0.0381694, size = 48, normalized size = 0.75 \[ \frac{\frac{6}{1-a x}+\frac{2}{(a x-1)^2}-7 \log (1-a x)-\log (a x+1)+8 \log (x)}{8 c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.038, size = 54, normalized size = 0.8 \begin{align*}{\frac{\ln \left ( x \right ) }{{c}^{2}}}-{\frac{\ln \left ( ax+1 \right ) }{8\,{c}^{2}}}+{\frac{1}{4\,{c}^{2} \left ( ax-1 \right ) ^{2}}}-{\frac{3}{4\,{c}^{2} \left ( ax-1 \right ) }}-{\frac{7\,\ln \left ( ax-1 \right ) }{8\,{c}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.956038, size = 81, normalized size = 1.27 \begin{align*} -\frac{3 \, a x - 4}{4 \,{\left (a^{2} c^{2} x^{2} - 2 \, a c^{2} x + c^{2}\right )}} - \frac{\log \left (a x + 1\right )}{8 \, c^{2}} - \frac{7 \, \log \left (a x - 1\right )}{8 \, c^{2}} + \frac{\log \left (x\right )}{c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(3*a*x - 4)/(a^2*c^2*x^2 - 2*a*c^2*x + c^2) - 1/8*log(a*x + 1)/c^2 - 7/8*log(a*x - 1)/c^2 + log(x)/c^2

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Fricas [A]  time = 2.40164, size = 215, normalized size = 3.36 \begin{align*} -\frac{6 \, a x +{\left (a^{2} x^{2} - 2 \, a x + 1\right )} \log \left (a x + 1\right ) + 7 \,{\left (a^{2} x^{2} - 2 \, a x + 1\right )} \log \left (a x - 1\right ) - 8 \,{\left (a^{2} x^{2} - 2 \, a x + 1\right )} \log \left (x\right ) - 8}{8 \,{\left (a^{2} c^{2} x^{2} - 2 \, a c^{2} x + c^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(6*a*x + (a^2*x^2 - 2*a*x + 1)*log(a*x + 1) + 7*(a^2*x^2 - 2*a*x + 1)*log(a*x - 1) - 8*(a^2*x^2 - 2*a*x +
1)*log(x) - 8)/(a^2*c^2*x^2 - 2*a*c^2*x + c^2)

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Sympy [A]  time = 0.66909, size = 58, normalized size = 0.91 \begin{align*} - \frac{3 a x - 4}{4 a^{2} c^{2} x^{2} - 8 a c^{2} x + 4 c^{2}} - \frac{- \log{\left (x \right )} + \frac{7 \log{\left (x - \frac{1}{a} \right )}}{8} + \frac{\log{\left (x + \frac{1}{a} \right )}}{8}}{c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(3*a*x - 4)/(4*a**2*c**2*x**2 - 8*a*c**2*x + 4*c**2) - (-log(x) + 7*log(x - 1/a)/8 + log(x + 1/a)/8)/c**2

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Giac [A]  time = 1.1363, size = 68, normalized size = 1.06 \begin{align*} -\frac{\log \left ({\left | a x + 1 \right |}\right )}{8 \, c^{2}} - \frac{7 \, \log \left ({\left | a x - 1 \right |}\right )}{8 \, c^{2}} + \frac{\log \left ({\left | x \right |}\right )}{c^{2}} - \frac{3 \, a x - 4}{4 \,{\left (a x - 1\right )}^{2} c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*log(abs(a*x + 1))/c^2 - 7/8*log(abs(a*x - 1))/c^2 + log(abs(x))/c^2 - 1/4*(3*a*x - 4)/((a*x - 1)^2*c^2)