3.1065 \(\int \frac{e^{2 \tanh ^{-1}(a x)}}{x^3 (c-a^2 c x^2)^2} \, dx\)

Optimal. Leaf size=99 \[ \frac{7 a^2}{4 c^2 (1-a x)}+\frac{a^2}{4 c^2 (1-a x)^2}+\frac{4 a^2 \log (x)}{c^2}-\frac{31 a^2 \log (1-a x)}{8 c^2}-\frac{a^2 \log (a x+1)}{8 c^2}-\frac{2 a}{c^2 x}-\frac{1}{2 c^2 x^2} \]

[Out]

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

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Rubi [A]  time = 0.122351, antiderivative size = 99, 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, 88} \[ \frac{7 a^2}{4 c^2 (1-a x)}+\frac{a^2}{4 c^2 (1-a x)^2}+\frac{4 a^2 \log (x)}{c^2}-\frac{31 a^2 \log (1-a x)}{8 c^2}-\frac{a^2 \log (a x+1)}{8 c^2}-\frac{2 a}{c^2 x}-\frac{1}{2 c^2 x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/(2*c^2*x^2) - (2*a)/(c^2*x) + a^2/(4*c^2*(1 - a*x)^2) + (7*a^2)/(4*c^2*(1 - a*x)) + (4*a^2*Log[x])/c^2 - (3
1*a^2*Log[1 - a*x])/(8*c^2) - (a^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 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^{2 \tanh ^{-1}(a x)}}{x^3 \left (c-a^2 c x^2\right )^2} \, dx &=\frac{\int \frac{1}{x^3 (1-a x)^3 (1+a x)} \, dx}{c^2}\\ &=\frac{\int \left (\frac{1}{x^3}+\frac{2 a}{x^2}+\frac{4 a^2}{x}-\frac{a^3}{2 (-1+a x)^3}+\frac{7 a^3}{4 (-1+a x)^2}-\frac{31 a^3}{8 (-1+a x)}-\frac{a^3}{8 (1+a x)}\right ) \, dx}{c^2}\\ &=-\frac{1}{2 c^2 x^2}-\frac{2 a}{c^2 x}+\frac{a^2}{4 c^2 (1-a x)^2}+\frac{7 a^2}{4 c^2 (1-a x)}+\frac{4 a^2 \log (x)}{c^2}-\frac{31 a^2 \log (1-a x)}{8 c^2}-\frac{a^2 \log (1+a x)}{8 c^2}\\ \end{align*}

Mathematica [A]  time = 0.0957903, size = 72, normalized size = 0.73 \[ -\frac{\frac{14 a^2}{a x-1}-\frac{2 a^2}{(a x-1)^2}-32 a^2 \log (x)+31 a^2 \log (1-a x)+a^2 \log (a x+1)+\frac{16 a}{x}+\frac{4}{x^2}}{8 c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(4/x^2 + (16*a)/x - (2*a^2)/(-1 + a*x)^2 + (14*a^2)/(-1 + a*x) - 32*a^2*Log[x] + 31*a^2*Log[1 - a*x] + a^2*Lo
g[1 + a*x])/(8*c^2)

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Maple [A]  time = 0.039, size = 87, normalized size = 0.9 \begin{align*} -{\frac{1}{2\,{c}^{2}{x}^{2}}}-2\,{\frac{a}{x{c}^{2}}}+4\,{\frac{{a}^{2}\ln \left ( x \right ) }{{c}^{2}}}-{\frac{{a}^{2}\ln \left ( ax+1 \right ) }{8\,{c}^{2}}}+{\frac{{a}^{2}}{4\,{c}^{2} \left ( ax-1 \right ) ^{2}}}-{\frac{7\,{a}^{2}}{4\,{c}^{2} \left ( ax-1 \right ) }}-{\frac{31\,{a}^{2}\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^3/(-a^2*c*x^2+c)^2,x)

[Out]

-1/2/c^2/x^2-2*a/x/c^2+4*a^2*ln(x)/c^2-1/8*a^2*ln(a*x+1)/c^2+1/4/c^2*a^2/(a*x-1)^2-7/4/c^2*a^2/(a*x-1)-31/8/c^
2*a^2*ln(a*x-1)

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Maxima [A]  time = 0.95721, size = 124, normalized size = 1.25 \begin{align*} -\frac{a^{2} \log \left (a x + 1\right )}{8 \, c^{2}} - \frac{31 \, a^{2} \log \left (a x - 1\right )}{8 \, c^{2}} + \frac{4 \, a^{2} \log \left (x\right )}{c^{2}} - \frac{15 \, a^{3} x^{3} - 22 \, a^{2} x^{2} + 4 \, a x + 2}{4 \,{\left (a^{2} c^{2} x^{4} - 2 \, a c^{2} x^{3} + c^{2} x^{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^3/(-a^2*c*x^2+c)^2,x, algorithm="maxima")

[Out]

-1/8*a^2*log(a*x + 1)/c^2 - 31/8*a^2*log(a*x - 1)/c^2 + 4*a^2*log(x)/c^2 - 1/4*(15*a^3*x^3 - 22*a^2*x^2 + 4*a*
x + 2)/(a^2*c^2*x^4 - 2*a*c^2*x^3 + c^2*x^2)

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Fricas [A]  time = 2.34618, size = 301, normalized size = 3.04 \begin{align*} -\frac{30 \, a^{3} x^{3} - 44 \, a^{2} x^{2} + 8 \, a x +{\left (a^{4} x^{4} - 2 \, a^{3} x^{3} + a^{2} x^{2}\right )} \log \left (a x + 1\right ) + 31 \,{\left (a^{4} x^{4} - 2 \, a^{3} x^{3} + a^{2} x^{2}\right )} \log \left (a x - 1\right ) - 32 \,{\left (a^{4} x^{4} - 2 \, a^{3} x^{3} + a^{2} x^{2}\right )} \log \left (x\right ) + 4}{8 \,{\left (a^{2} c^{2} x^{4} - 2 \, a c^{2} x^{3} + c^{2} x^{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^3/(-a^2*c*x^2+c)^2,x, algorithm="fricas")

[Out]

-1/8*(30*a^3*x^3 - 44*a^2*x^2 + 8*a*x + (a^4*x^4 - 2*a^3*x^3 + a^2*x^2)*log(a*x + 1) + 31*(a^4*x^4 - 2*a^3*x^3
+ a^2*x^2)*log(a*x - 1) - 32*(a^4*x^4 - 2*a^3*x^3 + a^2*x^2)*log(x) + 4)/(a^2*c^2*x^4 - 2*a*c^2*x^3 + c^2*x^2
)

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Sympy [A]  time = 0.884441, size = 92, normalized size = 0.93 \begin{align*} - \frac{15 a^{3} x^{3} - 22 a^{2} x^{2} + 4 a x + 2}{4 a^{2} c^{2} x^{4} - 8 a c^{2} x^{3} + 4 c^{2} x^{2}} - \frac{- 4 a^{2} \log{\left (x \right )} + \frac{31 a^{2} \log{\left (x - \frac{1}{a} \right )}}{8} + \frac{a^{2} \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**3/(-a**2*c*x**2+c)**2,x)

[Out]

-(15*a**3*x**3 - 22*a**2*x**2 + 4*a*x + 2)/(4*a**2*c**2*x**4 - 8*a*c**2*x**3 + 4*c**2*x**2) - (-4*a**2*log(x)
+ 31*a**2*log(x - 1/a)/8 + a**2*log(x + 1/a)/8)/c**2

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Giac [A]  time = 1.13579, size = 107, normalized size = 1.08 \begin{align*} -\frac{a^{2} \log \left ({\left | a x + 1 \right |}\right )}{8 \, c^{2}} - \frac{31 \, a^{2} \log \left ({\left | a x - 1 \right |}\right )}{8 \, c^{2}} + \frac{4 \, a^{2} \log \left ({\left | x \right |}\right )}{c^{2}} - \frac{15 \, a^{3} x^{3} - 22 \, a^{2} x^{2} + 4 \, a x + 2}{4 \,{\left (a x - 1\right )}^{2} c^{2} x^{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^3/(-a^2*c*x^2+c)^2,x, algorithm="giac")

[Out]

-1/8*a^2*log(abs(a*x + 1))/c^2 - 31/8*a^2*log(abs(a*x - 1))/c^2 + 4*a^2*log(abs(x))/c^2 - 1/4*(15*a^3*x^3 - 22
*a^2*x^2 + 4*a*x + 2)/((a*x - 1)^2*c^2*x^2)