3.1066 \(\int \frac{e^{2 \tanh ^{-1}(a x)}}{x^4 (c-a^2 c x^2)^2} \, dx\)

Optimal. Leaf size=110 \[ \frac{9 a^3}{4 c^2 (1-a x)}+\frac{a^3}{4 c^2 (1-a x)^2}-\frac{4 a^2}{c^2 x}+\frac{6 a^3 \log (x)}{c^2}-\frac{49 a^3 \log (1-a x)}{8 c^2}+\frac{a^3 \log (a x+1)}{8 c^2}-\frac{a}{c^2 x^2}-\frac{1}{3 c^2 x^3} \]

[Out]

-1/(3*c^2*x^3) - a/(c^2*x^2) - (4*a^2)/(c^2*x) + a^3/(4*c^2*(1 - a*x)^2) + (9*a^3)/(4*c^2*(1 - a*x)) + (6*a^3*
Log[x])/c^2 - (49*a^3*Log[1 - a*x])/(8*c^2) + (a^3*Log[1 + a*x])/(8*c^2)

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Rubi [A]  time = 0.133771, antiderivative size = 110, 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} \[ \frac{9 a^3}{4 c^2 (1-a x)}+\frac{a^3}{4 c^2 (1-a x)^2}-\frac{4 a^2}{c^2 x}+\frac{6 a^3 \log (x)}{c^2}-\frac{49 a^3 \log (1-a x)}{8 c^2}+\frac{a^3 \log (a x+1)}{8 c^2}-\frac{a}{c^2 x^2}-\frac{1}{3 c^2 x^3} \]

Antiderivative was successfully verified.

[In]

Int[E^(2*ArcTanh[a*x])/(x^4*(c - a^2*c*x^2)^2),x]

[Out]

-1/(3*c^2*x^3) - a/(c^2*x^2) - (4*a^2)/(c^2*x) + a^3/(4*c^2*(1 - a*x)^2) + (9*a^3)/(4*c^2*(1 - a*x)) + (6*a^3*
Log[x])/c^2 - (49*a^3*Log[1 - a*x])/(8*c^2) + (a^3*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^4 \left (c-a^2 c x^2\right )^2} \, dx &=\frac{\int \frac{1}{x^4 (1-a x)^3 (1+a x)} \, dx}{c^2}\\ &=\frac{\int \left (\frac{1}{x^4}+\frac{2 a}{x^3}+\frac{4 a^2}{x^2}+\frac{6 a^3}{x}-\frac{a^4}{2 (-1+a x)^3}+\frac{9 a^4}{4 (-1+a x)^2}-\frac{49 a^4}{8 (-1+a x)}+\frac{a^4}{8 (1+a x)}\right ) \, dx}{c^2}\\ &=-\frac{1}{3 c^2 x^3}-\frac{a}{c^2 x^2}-\frac{4 a^2}{c^2 x}+\frac{a^3}{4 c^2 (1-a x)^2}+\frac{9 a^3}{4 c^2 (1-a x)}+\frac{6 a^3 \log (x)}{c^2}-\frac{49 a^3 \log (1-a x)}{8 c^2}+\frac{a^3 \log (1+a x)}{8 c^2}\\ \end{align*}

Mathematica [A]  time = 0.0786887, size = 87, normalized size = 0.79 \[ \frac{\frac{9 a^3}{4-4 a x}+\frac{a^3}{4 (a x-1)^2}-\frac{4 a^2}{x}+6 a^3 \log (x)-\frac{49}{8} a^3 \log (1-a x)+\frac{1}{8} a^3 \log (a x+1)-\frac{a}{x^2}-\frac{1}{3 x^3}}{c^2} \]

Antiderivative was successfully verified.

[In]

Integrate[E^(2*ArcTanh[a*x])/(x^4*(c - a^2*c*x^2)^2),x]

[Out]

(-1/(3*x^3) - a/x^2 - (4*a^2)/x + (9*a^3)/(4 - 4*a*x) + a^3/(4*(-1 + a*x)^2) + 6*a^3*Log[x] - (49*a^3*Log[1 -
a*x])/8 + (a^3*Log[1 + a*x])/8)/c^2

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Maple [A]  time = 0.038, size = 98, normalized size = 0.9 \begin{align*} -{\frac{1}{3\,{c}^{2}{x}^{3}}}-{\frac{a}{{c}^{2}{x}^{2}}}-4\,{\frac{{a}^{2}}{x{c}^{2}}}+6\,{\frac{{a}^{3}\ln \left ( x \right ) }{{c}^{2}}}+{\frac{{a}^{3}\ln \left ( ax+1 \right ) }{8\,{c}^{2}}}+{\frac{{a}^{3}}{4\,{c}^{2} \left ( ax-1 \right ) ^{2}}}-{\frac{9\,{a}^{3}}{4\,{c}^{2} \left ( ax-1 \right ) }}-{\frac{49\,{a}^{3}\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^4/(-a^2*c*x^2+c)^2,x)

[Out]

-1/3/c^2/x^3-a/c^2/x^2-4*a^2/x/c^2+6*a^3*ln(x)/c^2+1/8*a^3*ln(a*x+1)/c^2+1/4/c^2*a^3/(a*x-1)^2-9/4/c^2*a^3/(a*
x-1)-49/8/c^2*a^3*ln(a*x-1)

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Maxima [A]  time = 0.96654, size = 135, normalized size = 1.23 \begin{align*} \frac{a^{3} \log \left (a x + 1\right )}{8 \, c^{2}} - \frac{49 \, a^{3} \log \left (a x - 1\right )}{8 \, c^{2}} + \frac{6 \, a^{3} \log \left (x\right )}{c^{2}} - \frac{75 \, a^{4} x^{4} - 114 \, a^{3} x^{3} + 28 \, a^{2} x^{2} + 4 \, a x + 4}{12 \,{\left (a^{2} c^{2} x^{5} - 2 \, a c^{2} x^{4} + c^{2} x^{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^4/(-a^2*c*x^2+c)^2,x, algorithm="maxima")

[Out]

1/8*a^3*log(a*x + 1)/c^2 - 49/8*a^3*log(a*x - 1)/c^2 + 6*a^3*log(x)/c^2 - 1/12*(75*a^4*x^4 - 114*a^3*x^3 + 28*
a^2*x^2 + 4*a*x + 4)/(a^2*c^2*x^5 - 2*a*c^2*x^4 + c^2*x^3)

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Fricas [A]  time = 2.35848, size = 328, normalized size = 2.98 \begin{align*} -\frac{150 \, a^{4} x^{4} - 228 \, a^{3} x^{3} + 56 \, a^{2} x^{2} + 8 \, a x - 3 \,{\left (a^{5} x^{5} - 2 \, a^{4} x^{4} + a^{3} x^{3}\right )} \log \left (a x + 1\right ) + 147 \,{\left (a^{5} x^{5} - 2 \, a^{4} x^{4} + a^{3} x^{3}\right )} \log \left (a x - 1\right ) - 144 \,{\left (a^{5} x^{5} - 2 \, a^{4} x^{4} + a^{3} x^{3}\right )} \log \left (x\right ) + 8}{24 \,{\left (a^{2} c^{2} x^{5} - 2 \, a c^{2} x^{4} + c^{2} x^{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^4/(-a^2*c*x^2+c)^2,x, algorithm="fricas")

[Out]

-1/24*(150*a^4*x^4 - 228*a^3*x^3 + 56*a^2*x^2 + 8*a*x - 3*(a^5*x^5 - 2*a^4*x^4 + a^3*x^3)*log(a*x + 1) + 147*(
a^5*x^5 - 2*a^4*x^4 + a^3*x^3)*log(a*x - 1) - 144*(a^5*x^5 - 2*a^4*x^4 + a^3*x^3)*log(x) + 8)/(a^2*c^2*x^5 - 2
*a*c^2*x^4 + c^2*x^3)

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Sympy [A]  time = 1.00502, size = 100, normalized size = 0.91 \begin{align*} - \frac{75 a^{4} x^{4} - 114 a^{3} x^{3} + 28 a^{2} x^{2} + 4 a x + 4}{12 a^{2} c^{2} x^{5} - 24 a c^{2} x^{4} + 12 c^{2} x^{3}} - \frac{- 6 a^{3} \log{\left (x \right )} + \frac{49 a^{3} \log{\left (x - \frac{1}{a} \right )}}{8} - \frac{a^{3} \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**4/(-a**2*c*x**2+c)**2,x)

[Out]

-(75*a**4*x**4 - 114*a**3*x**3 + 28*a**2*x**2 + 4*a*x + 4)/(12*a**2*c**2*x**5 - 24*a*c**2*x**4 + 12*c**2*x**3)
- (-6*a**3*log(x) + 49*a**3*log(x - 1/a)/8 - a**3*log(x + 1/a)/8)/c**2

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Giac [A]  time = 1.12835, size = 117, normalized size = 1.06 \begin{align*} \frac{a^{3} \log \left ({\left | a x + 1 \right |}\right )}{8 \, c^{2}} - \frac{49 \, a^{3} \log \left ({\left | a x - 1 \right |}\right )}{8 \, c^{2}} + \frac{6 \, a^{3} \log \left ({\left | x \right |}\right )}{c^{2}} - \frac{75 \, a^{4} x^{4} - 114 \, a^{3} x^{3} + 28 \, a^{2} x^{2} + 4 \, a x + 4}{12 \,{\left (a x - 1\right )}^{2} c^{2} x^{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^4/(-a^2*c*x^2+c)^2,x, algorithm="giac")

[Out]

1/8*a^3*log(abs(a*x + 1))/c^2 - 49/8*a^3*log(abs(a*x - 1))/c^2 + 6*a^3*log(abs(x))/c^2 - 1/12*(75*a^4*x^4 - 11
4*a^3*x^3 + 28*a^2*x^2 + 4*a*x + 4)/((a*x - 1)^2*c^2*x^3)