3.1059 \(\int \frac{e^{2 \tanh ^{-1}(a x)} x^3}{(c-a^2 c x^2)^2} \, dx\)

Optimal. Leaf size=70 \[ -\frac{5}{4 a^4 c^2 (1-a x)}+\frac{1}{4 a^4 c^2 (1-a x)^2}-\frac{7 \log (1-a x)}{8 a^4 c^2}-\frac{\log (a x+1)}{8 a^4 c^2} \]

[Out]

1/(4*a^4*c^2*(1 - a*x)^2) - 5/(4*a^4*c^2*(1 - a*x)) - (7*Log[1 - a*x])/(8*a^4*c^2) - Log[1 + a*x]/(8*a^4*c^2)

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Rubi [A]  time = 0.106701, antiderivative size = 70, 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} \[ -\frac{5}{4 a^4 c^2 (1-a x)}+\frac{1}{4 a^4 c^2 (1-a x)^2}-\frac{7 \log (1-a x)}{8 a^4 c^2}-\frac{\log (a x+1)}{8 a^4 c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

1/(4*a^4*c^2*(1 - a*x)^2) - 5/(4*a^4*c^2*(1 - a*x)) - (7*Log[1 - a*x])/(8*a^4*c^2) - Log[1 + a*x]/(8*a^4*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{x^3}{(1-a x)^3 (1+a x)} \, dx}{c^2}\\ &=\frac{\int \left (-\frac{1}{2 a^3 (-1+a x)^3}-\frac{5}{4 a^3 (-1+a x)^2}-\frac{7}{8 a^3 (-1+a x)}-\frac{1}{8 a^3 (1+a x)}\right ) \, dx}{c^2}\\ &=\frac{1}{4 a^4 c^2 (1-a x)^2}-\frac{5}{4 a^4 c^2 (1-a x)}-\frac{7 \log (1-a x)}{8 a^4 c^2}-\frac{\log (1+a x)}{8 a^4 c^2}\\ \end{align*}

Mathematica [A]  time = 0.0358648, size = 53, normalized size = 0.76 \[ -\frac{-10 a x+7 (a x-1)^2 \log (1-a x)+(a x-1)^2 \log (a x+1)+8}{8 a^4 c^2 (a x-1)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(8 - 10*a*x + 7*(-1 + a*x)^2*Log[1 - a*x] + (-1 + a*x)^2*Log[1 + a*x])/(8*a^4*c^2*(-1 + a*x)^2)

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Maple [A]  time = 0.033, size = 60, normalized size = 0.9 \begin{align*} -{\frac{\ln \left ( ax+1 \right ) }{8\,{a}^{4}{c}^{2}}}+{\frac{1}{4\,{a}^{4}{c}^{2} \left ( ax-1 \right ) ^{2}}}+{\frac{5}{4\,{a}^{4}{c}^{2} \left ( ax-1 \right ) }}-{\frac{7\,\ln \left ( ax-1 \right ) }{8\,{a}^{4}{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]

-1/8*ln(a*x+1)/a^4/c^2+1/4/c^2/a^4/(a*x-1)^2+5/4/c^2/a^4/(a*x-1)-7/8/c^2/a^4*ln(a*x-1)

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Maxima [A]  time = 0.958823, size = 89, normalized size = 1.27 \begin{align*} \frac{5 \, a x - 4}{4 \,{\left (a^{6} c^{2} x^{2} - 2 \, a^{5} c^{2} x + a^{4} c^{2}\right )}} - \frac{\log \left (a x + 1\right )}{8 \, a^{4} c^{2}} - \frac{7 \, \log \left (a x - 1\right )}{8 \, a^{4} 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, algorithm="maxima")

[Out]

1/4*(5*a*x - 4)/(a^6*c^2*x^2 - 2*a^5*c^2*x + a^4*c^2) - 1/8*log(a*x + 1)/(a^4*c^2) - 7/8*log(a*x - 1)/(a^4*c^2
)

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Fricas [A]  time = 2.22161, size = 178, normalized size = 2.54 \begin{align*} \frac{10 \, 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}{8 \,{\left (a^{6} c^{2} x^{2} - 2 \, a^{5} c^{2} x + a^{4} 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^3/(-a^2*c*x^2+c)^2,x, algorithm="fricas")

[Out]

1/8*(10*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^6*c^2*x^2 - 2*
a^5*c^2*x + a^4*c^2)

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Sympy [A]  time = 0.537136, size = 61, normalized size = 0.87 \begin{align*} \frac{5 a x - 4}{4 a^{6} c^{2} x^{2} - 8 a^{5} c^{2} x + 4 a^{4} c^{2}} - \frac{\frac{7 \log{\left (x - \frac{1}{a} \right )}}{8} + \frac{\log{\left (x + \frac{1}{a} \right )}}{8}}{a^{4} 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]

(5*a*x - 4)/(4*a**6*c**2*x**2 - 8*a**5*c**2*x + 4*a**4*c**2) - (7*log(x - 1/a)/8 + log(x + 1/a)/8)/(a**4*c**2)

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Giac [A]  time = 1.141, size = 70, normalized size = 1. \begin{align*} -\frac{\log \left ({\left | a x + 1 \right |}\right )}{8 \, a^{4} c^{2}} - \frac{7 \, \log \left ({\left | a x - 1 \right |}\right )}{8 \, a^{4} c^{2}} + \frac{5 \, a x - 4}{4 \,{\left (a x - 1\right )}^{2} a^{4} 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, algorithm="giac")

[Out]

-1/8*log(abs(a*x + 1))/(a^4*c^2) - 7/8*log(abs(a*x - 1))/(a^4*c^2) + 1/4*(5*a*x - 4)/((a*x - 1)^2*a^4*c^2)