3.1073 \(\int \frac{e^{2 \tanh ^{-1}(a x)}}{x (c-a^2 c x^2)^3} \, dx\)

Optimal. Leaf size=93 \[ \frac{11}{16 c^3 (1-a x)}+\frac{1}{16 c^3 (a x+1)}+\frac{1}{4 c^3 (1-a x)^2}+\frac{1}{12 c^3 (1-a x)^3}-\frac{13 \log (1-a x)}{16 c^3}-\frac{3 \log (a x+1)}{16 c^3}+\frac{\log (x)}{c^3} \]

[Out]

1/(12*c^3*(1 - a*x)^3) + 1/(4*c^3*(1 - a*x)^2) + 11/(16*c^3*(1 - a*x)) + 1/(16*c^3*(1 + a*x)) + Log[x]/c^3 - (
13*Log[1 - a*x])/(16*c^3) - (3*Log[1 + a*x])/(16*c^3)

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Rubi [A]  time = 0.118672, antiderivative size = 93, 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} \[ \frac{11}{16 c^3 (1-a x)}+\frac{1}{16 c^3 (a x+1)}+\frac{1}{4 c^3 (1-a x)^2}+\frac{1}{12 c^3 (1-a x)^3}-\frac{13 \log (1-a x)}{16 c^3}-\frac{3 \log (a x+1)}{16 c^3}+\frac{\log (x)}{c^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

1/(12*c^3*(1 - a*x)^3) + 1/(4*c^3*(1 - a*x)^2) + 11/(16*c^3*(1 - a*x)) + 1/(16*c^3*(1 + a*x)) + Log[x]/c^3 - (
13*Log[1 - a*x])/(16*c^3) - (3*Log[1 + a*x])/(16*c^3)

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 \left (c-a^2 c x^2\right )^3} \, dx &=\frac{\int \frac{1}{x (1-a x)^4 (1+a x)^2} \, dx}{c^3}\\ &=\frac{\int \left (\frac{1}{x}+\frac{a}{4 (-1+a x)^4}-\frac{a}{2 (-1+a x)^3}+\frac{11 a}{16 (-1+a x)^2}-\frac{13 a}{16 (-1+a x)}-\frac{a}{16 (1+a x)^2}-\frac{3 a}{16 (1+a x)}\right ) \, dx}{c^3}\\ &=\frac{1}{12 c^3 (1-a x)^3}+\frac{1}{4 c^3 (1-a x)^2}+\frac{11}{16 c^3 (1-a x)}+\frac{1}{16 c^3 (1+a x)}+\frac{\log (x)}{c^3}-\frac{13 \log (1-a x)}{16 c^3}-\frac{3 \log (1+a x)}{16 c^3}\\ \end{align*}

Mathematica [A]  time = 0.0670385, size = 66, normalized size = 0.71 \[ \frac{\frac{33}{1-a x}+\frac{3}{a x+1}+\frac{12}{(a x-1)^2}-\frac{4}{(a x-1)^3}-39 \log (1-a x)-9 \log (a x+1)+48 \log (x)}{48 c^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(33/(1 - a*x) - 4/(-1 + a*x)^3 + 12/(-1 + a*x)^2 + 3/(1 + a*x) + 48*Log[x] - 39*Log[1 - a*x] - 9*Log[1 + a*x])
/(48*c^3)

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Maple [A]  time = 0.039, size = 78, normalized size = 0.8 \begin{align*}{\frac{\ln \left ( x \right ) }{{c}^{3}}}+{\frac{1}{16\,{c}^{3} \left ( ax+1 \right ) }}-{\frac{3\,\ln \left ( ax+1 \right ) }{16\,{c}^{3}}}-{\frac{1}{12\,{c}^{3} \left ( ax-1 \right ) ^{3}}}+{\frac{1}{4\,{c}^{3} \left ( ax-1 \right ) ^{2}}}-{\frac{11}{16\,{c}^{3} \left ( ax-1 \right ) }}-{\frac{13\,\ln \left ( ax-1 \right ) }{16\,{c}^{3}}} \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)^3,x)

[Out]

ln(x)/c^3+1/16/c^3/(a*x+1)-3/16*ln(a*x+1)/c^3-1/12/c^3/(a*x-1)^3+1/4/c^3/(a*x-1)^2-11/16/c^3/(a*x-1)-13/16/c^3
*ln(a*x-1)

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Maxima [A]  time = 0.979564, size = 120, normalized size = 1.29 \begin{align*} -\frac{15 \, a^{3} x^{3} - 18 \, a^{2} x^{2} - 19 \, a x + 26}{24 \,{\left (a^{4} c^{3} x^{4} - 2 \, a^{3} c^{3} x^{3} + 2 \, a c^{3} x - c^{3}\right )}} - \frac{3 \, \log \left (a x + 1\right )}{16 \, c^{3}} - \frac{13 \, \log \left (a x - 1\right )}{16 \, c^{3}} + \frac{\log \left (x\right )}{c^{3}} \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)^3,x, algorithm="maxima")

[Out]

-1/24*(15*a^3*x^3 - 18*a^2*x^2 - 19*a*x + 26)/(a^4*c^3*x^4 - 2*a^3*c^3*x^3 + 2*a*c^3*x - c^3) - 3/16*log(a*x +
1)/c^3 - 13/16*log(a*x - 1)/c^3 + log(x)/c^3

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Fricas [A]  time = 2.38349, size = 329, normalized size = 3.54 \begin{align*} -\frac{30 \, a^{3} x^{3} - 36 \, a^{2} x^{2} - 38 \, a x + 9 \,{\left (a^{4} x^{4} - 2 \, a^{3} x^{3} + 2 \, a x - 1\right )} \log \left (a x + 1\right ) + 39 \,{\left (a^{4} x^{4} - 2 \, a^{3} x^{3} + 2 \, a x - 1\right )} \log \left (a x - 1\right ) - 48 \,{\left (a^{4} x^{4} - 2 \, a^{3} x^{3} + 2 \, a x - 1\right )} \log \left (x\right ) + 52}{48 \,{\left (a^{4} c^{3} x^{4} - 2 \, a^{3} c^{3} x^{3} + 2 \, a c^{3} x - c^{3}\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)^3,x, algorithm="fricas")

[Out]

-1/48*(30*a^3*x^3 - 36*a^2*x^2 - 38*a*x + 9*(a^4*x^4 - 2*a^3*x^3 + 2*a*x - 1)*log(a*x + 1) + 39*(a^4*x^4 - 2*a
^3*x^3 + 2*a*x - 1)*log(a*x - 1) - 48*(a^4*x^4 - 2*a^3*x^3 + 2*a*x - 1)*log(x) + 52)/(a^4*c^3*x^4 - 2*a^3*c^3*
x^3 + 2*a*c^3*x - c^3)

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Sympy [A]  time = 0.924615, size = 87, normalized size = 0.94 \begin{align*} - \frac{15 a^{3} x^{3} - 18 a^{2} x^{2} - 19 a x + 26}{24 a^{4} c^{3} x^{4} - 48 a^{3} c^{3} x^{3} + 48 a c^{3} x - 24 c^{3}} + \frac{\log{\left (x \right )} - \frac{13 \log{\left (x - \frac{1}{a} \right )}}{16} - \frac{3 \log{\left (x + \frac{1}{a} \right )}}{16}}{c^{3}} \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)**3,x)

[Out]

-(15*a**3*x**3 - 18*a**2*x**2 - 19*a*x + 26)/(24*a**4*c**3*x**4 - 48*a**3*c**3*x**3 + 48*a*c**3*x - 24*c**3) +
(log(x) - 13*log(x - 1/a)/16 - 3*log(x + 1/a)/16)/c**3

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Giac [A]  time = 1.13504, size = 99, normalized size = 1.06 \begin{align*} -\frac{3 \, \log \left ({\left | a x + 1 \right |}\right )}{16 \, c^{3}} - \frac{13 \, \log \left ({\left | a x - 1 \right |}\right )}{16 \, c^{3}} + \frac{\log \left ({\left | x \right |}\right )}{c^{3}} - \frac{15 \, a^{3} x^{3} - 18 \, a^{2} x^{2} - 19 \, a x + 26}{24 \,{\left (a x + 1\right )}{\left (a x - 1\right )}^{3} c^{3}} \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)^3,x, algorithm="giac")

[Out]

-3/16*log(abs(a*x + 1))/c^3 - 13/16*log(abs(a*x - 1))/c^3 + log(abs(x))/c^3 - 1/24*(15*a^3*x^3 - 18*a^2*x^2 -
19*a*x + 26)/((a*x + 1)*(a*x - 1)^3*c^3)