3.32 \(\int \frac{e^{4 \coth ^{-1}(a x)}}{x^4} \, dx\)

Optimal. Leaf size=54 \[ \frac{4 a^3}{1-a x}-\frac{8 a^2}{x}+12 a^3 \log (x)-12 a^3 \log (1-a x)-\frac{2 a}{x^2}-\frac{1}{3 x^3} \]

[Out]

-1/(3*x^3) - (2*a)/x^2 - (8*a^2)/x + (4*a^3)/(1 - a*x) + 12*a^3*Log[x] - 12*a^3*Log[1 - a*x]

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Rubi [A]  time = 0.0612882, antiderivative size = 54, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {6167, 6126, 88} \[ \frac{4 a^3}{1-a x}-\frac{8 a^2}{x}+12 a^3 \log (x)-12 a^3 \log (1-a x)-\frac{2 a}{x^2}-\frac{1}{3 x^3} \]

Antiderivative was successfully verified.

[In]

Int[E^(4*ArcCoth[a*x])/x^4,x]

[Out]

-1/(3*x^3) - (2*a)/x^2 - (8*a^2)/x + (4*a^3)/(1 - a*x) + 12*a^3*Log[x] - 12*a^3*Log[1 - a*x]

Rule 6167

Int[E^(ArcCoth[(a_.)*(x_)]*(n_))*(u_.), x_Symbol] :> Dist[(-1)^(n/2), Int[u*E^(n*ArcTanh[a*x]), x], x] /; Free
Q[a, x] && IntegerQ[n/2]

Rule 6126

Int[E^(ArcTanh[(a_.)*(x_)]*(n_))*(x_)^(m_.), x_Symbol] :> Int[(x^m*(1 + a*x)^(n/2))/(1 - a*x)^(n/2), x] /; Fre
eQ[{a, m, n}, x] &&  !IntegerQ[(n - 1)/2]

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^{4 \coth ^{-1}(a x)}}{x^4} \, dx &=\int \frac{e^{4 \tanh ^{-1}(a x)}}{x^4} \, dx\\ &=\int \frac{(1+a x)^2}{x^4 (1-a x)^2} \, dx\\ &=\int \left (\frac{1}{x^4}+\frac{4 a}{x^3}+\frac{8 a^2}{x^2}+\frac{12 a^3}{x}+\frac{4 a^4}{(-1+a x)^2}-\frac{12 a^4}{-1+a x}\right ) \, dx\\ &=-\frac{1}{3 x^3}-\frac{2 a}{x^2}-\frac{8 a^2}{x}+\frac{4 a^3}{1-a x}+12 a^3 \log (x)-12 a^3 \log (1-a x)\\ \end{align*}

Mathematica [A]  time = 0.0424887, size = 54, normalized size = 1. \[ \frac{4 a^3}{1-a x}-\frac{8 a^2}{x}+12 a^3 \log (x)-12 a^3 \log (1-a x)-\frac{2 a}{x^2}-\frac{1}{3 x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[E^(4*ArcCoth[a*x])/x^4,x]

[Out]

-1/(3*x^3) - (2*a)/x^2 - (8*a^2)/x + (4*a^3)/(1 - a*x) + 12*a^3*Log[x] - 12*a^3*Log[1 - a*x]

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Maple [A]  time = 0.049, size = 51, normalized size = 0.9 \begin{align*} -{\frac{1}{3\,{x}^{3}}}-2\,{\frac{a}{{x}^{2}}}-8\,{\frac{{a}^{2}}{x}}+12\,{a}^{3}\ln \left ( x \right ) -4\,{\frac{{a}^{3}}{ax-1}}-12\,{a}^{3}\ln \left ( ax-1 \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a*x-1)^2*(a*x+1)^2/x^4,x)

[Out]

-1/3/x^3-2*a/x^2-8*a^2/x+12*a^3*ln(x)-4*a^3/(a*x-1)-12*a^3*ln(a*x-1)

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Maxima [A]  time = 1.0022, size = 76, normalized size = 1.41 \begin{align*} -12 \, a^{3} \log \left (a x - 1\right ) + 12 \, a^{3} \log \left (x\right ) - \frac{36 \, a^{3} x^{3} - 18 \, a^{2} x^{2} - 5 \, a x - 1}{3 \,{\left (a x^{4} - x^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x-1)^2*(a*x+1)^2/x^4,x, algorithm="maxima")

[Out]

-12*a^3*log(a*x - 1) + 12*a^3*log(x) - 1/3*(36*a^3*x^3 - 18*a^2*x^2 - 5*a*x - 1)/(a*x^4 - x^3)

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Fricas [A]  time = 1.89311, size = 173, normalized size = 3.2 \begin{align*} -\frac{36 \, a^{3} x^{3} - 18 \, a^{2} x^{2} - 5 \, a x + 36 \,{\left (a^{4} x^{4} - a^{3} x^{3}\right )} \log \left (a x - 1\right ) - 36 \,{\left (a^{4} x^{4} - a^{3} x^{3}\right )} \log \left (x\right ) - 1}{3 \,{\left (a x^{4} - x^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x-1)^2*(a*x+1)^2/x^4,x, algorithm="fricas")

[Out]

-1/3*(36*a^3*x^3 - 18*a^2*x^2 - 5*a*x + 36*(a^4*x^4 - a^3*x^3)*log(a*x - 1) - 36*(a^4*x^4 - a^3*x^3)*log(x) -
1)/(a*x^4 - x^3)

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Sympy [A]  time = 0.462664, size = 49, normalized size = 0.91 \begin{align*} 12 a^{3} \left (\log{\left (x \right )} - \log{\left (x - \frac{1}{a} \right )}\right ) - \frac{36 a^{3} x^{3} - 18 a^{2} x^{2} - 5 a x - 1}{3 a x^{4} - 3 x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x-1)**2*(a*x+1)**2/x**4,x)

[Out]

12*a**3*(log(x) - log(x - 1/a)) - (36*a**3*x**3 - 18*a**2*x**2 - 5*a*x - 1)/(3*a*x**4 - 3*x**3)

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Giac [A]  time = 1.1818, size = 100, normalized size = 1.85 \begin{align*} 12 \, a^{3} \log \left ({\left | -\frac{1}{a x - 1} - 1 \right |}\right ) - \frac{4 \, a^{3}}{a x - 1} + \frac{31 \, a^{3} + \frac{69 \, a^{3}}{a x - 1} + \frac{39 \, a^{3}}{{\left (a x - 1\right )}^{2}}}{3 \,{\left (\frac{1}{a x - 1} + 1\right )}^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x-1)^2*(a*x+1)^2/x^4,x, algorithm="giac")

[Out]

12*a^3*log(abs(-1/(a*x - 1) - 1)) - 4*a^3/(a*x - 1) + 1/3*(31*a^3 + 69*a^3/(a*x - 1) + 39*a^3/(a*x - 1)^2)/(1/
(a*x - 1) + 1)^3