3.804 \(\int \frac{e^{4 \coth ^{-1}(a x)}}{(c-\frac{c}{a^2 x^2})^3} \, dx\)

Optimal. Leaf size=111 \[ \frac{111}{16 a c^3 (1-a x)}-\frac{49}{16 a c^3 (1-a x)^2}+\frac{11}{12 a c^3 (1-a x)^3}-\frac{1}{8 a c^3 (1-a x)^4}+\frac{129 \log (1-a x)}{32 a c^3}-\frac{\log (a x+1)}{32 a c^3}+\frac{x}{c^3} \]

[Out]

x/c^3 - 1/(8*a*c^3*(1 - a*x)^4) + 11/(12*a*c^3*(1 - a*x)^3) - 49/(16*a*c^3*(1 - a*x)^2) + 111/(16*a*c^3*(1 - a
*x)) + (129*Log[1 - a*x])/(32*a*c^3) - Log[1 + a*x]/(32*a*c^3)

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Rubi [A]  time = 0.19596, antiderivative size = 111, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {6167, 6157, 6150, 88} \[ \frac{111}{16 a c^3 (1-a x)}-\frac{49}{16 a c^3 (1-a x)^2}+\frac{11}{12 a c^3 (1-a x)^3}-\frac{1}{8 a c^3 (1-a x)^4}+\frac{129 \log (1-a x)}{32 a c^3}-\frac{\log (a x+1)}{32 a c^3}+\frac{x}{c^3} \]

Antiderivative was successfully verified.

[In]

Int[E^(4*ArcCoth[a*x])/(c - c/(a^2*x^2))^3,x]

[Out]

x/c^3 - 1/(8*a*c^3*(1 - a*x)^4) + 11/(12*a*c^3*(1 - a*x)^3) - 49/(16*a*c^3*(1 - a*x)^2) + 111/(16*a*c^3*(1 - a
*x)) + (129*Log[1 - a*x])/(32*a*c^3) - Log[1 + a*x]/(32*a*c^3)

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 6157

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)/(x_)^2)^(p_.), x_Symbol] :> Dist[d^p, Int[(u*(1 - a^2*x^
2)^p*E^(n*ArcTanh[a*x]))/x^(2*p), x], x] /; FreeQ[{a, c, d, n}, x] && EqQ[c + a^2*d, 0] && IntegerQ[p]

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^{4 \coth ^{-1}(a x)}}{\left (c-\frac{c}{a^2 x^2}\right )^3} \, dx &=\int \frac{e^{4 \tanh ^{-1}(a x)}}{\left (c-\frac{c}{a^2 x^2}\right )^3} \, dx\\ &=-\frac{a^6 \int \frac{e^{4 \tanh ^{-1}(a x)} x^6}{\left (1-a^2 x^2\right )^3} \, dx}{c^3}\\ &=-\frac{a^6 \int \frac{x^6}{(1-a x)^5 (1+a x)} \, dx}{c^3}\\ &=-\frac{a^6 \int \left (-\frac{1}{a^6}-\frac{1}{2 a^6 (-1+a x)^5}-\frac{11}{4 a^6 (-1+a x)^4}-\frac{49}{8 a^6 (-1+a x)^3}-\frac{111}{16 a^6 (-1+a x)^2}-\frac{129}{32 a^6 (-1+a x)}+\frac{1}{32 a^6 (1+a x)}\right ) \, dx}{c^3}\\ &=\frac{x}{c^3}-\frac{1}{8 a c^3 (1-a x)^4}+\frac{11}{12 a c^3 (1-a x)^3}-\frac{49}{16 a c^3 (1-a x)^2}+\frac{111}{16 a c^3 (1-a x)}+\frac{129 \log (1-a x)}{32 a c^3}-\frac{\log (1+a x)}{32 a c^3}\\ \end{align*}

Mathematica [A]  time = 0.0627582, size = 89, normalized size = 0.8 \[ \frac{2 \left (48 a^5 x^5-192 a^4 x^4-45 a^3 x^3+660 a^2 x^2-701 a x+224\right )+387 (a x-1)^4 \log (1-a x)-3 (a x-1)^4 \log (a x+1)}{96 a c^3 (a x-1)^4} \]

Antiderivative was successfully verified.

[In]

Integrate[E^(4*ArcCoth[a*x])/(c - c/(a^2*x^2))^3,x]

[Out]

(2*(224 - 701*a*x + 660*a^2*x^2 - 45*a^3*x^3 - 192*a^4*x^4 + 48*a^5*x^5) + 387*(-1 + a*x)^4*Log[1 - a*x] - 3*(
-1 + a*x)^4*Log[1 + a*x])/(96*a*c^3*(-1 + a*x)^4)

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Maple [A]  time = 0.058, size = 95, normalized size = 0.9 \begin{align*}{\frac{x}{{c}^{3}}}-{\frac{\ln \left ( ax+1 \right ) }{32\,a{c}^{3}}}-{\frac{1}{8\,a{c}^{3} \left ( ax-1 \right ) ^{4}}}-{\frac{11}{12\,a{c}^{3} \left ( ax-1 \right ) ^{3}}}-{\frac{49}{16\,a{c}^{3} \left ( ax-1 \right ) ^{2}}}-{\frac{111}{16\,a{c}^{3} \left ( ax-1 \right ) }}+{\frac{129\,\ln \left ( ax-1 \right ) }{32\,a{c}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a*x-1)^2*(a*x+1)^2/(c-c/a^2/x^2)^3,x)

[Out]

x/c^3-1/32*ln(a*x+1)/a/c^3-1/8/a/c^3/(a*x-1)^4-11/12/a/c^3/(a*x-1)^3-49/16/a/c^3/(a*x-1)^2-111/16/a/c^3/(a*x-1
)+129/32/a/c^3*ln(a*x-1)

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Maxima [A]  time = 1.0878, size = 144, normalized size = 1.3 \begin{align*} -\frac{333 \, a^{3} x^{3} - 852 \, a^{2} x^{2} + 749 \, a x - 224}{48 \,{\left (a^{5} c^{3} x^{4} - 4 \, a^{4} c^{3} x^{3} + 6 \, a^{3} c^{3} x^{2} - 4 \, a^{2} c^{3} x + a c^{3}\right )}} + \frac{x}{c^{3}} - \frac{\log \left (a x + 1\right )}{32 \, a c^{3}} + \frac{129 \, \log \left (a x - 1\right )}{32 \, a c^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x-1)^2*(a*x+1)^2/(c-c/a^2/x^2)^3,x, algorithm="maxima")

[Out]

-1/48*(333*a^3*x^3 - 852*a^2*x^2 + 749*a*x - 224)/(a^5*c^3*x^4 - 4*a^4*c^3*x^3 + 6*a^3*c^3*x^2 - 4*a^2*c^3*x +
a*c^3) + x/c^3 - 1/32*log(a*x + 1)/(a*c^3) + 129/32*log(a*x - 1)/(a*c^3)

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Fricas [A]  time = 1.5235, size = 370, normalized size = 3.33 \begin{align*} \frac{96 \, a^{5} x^{5} - 384 \, a^{4} x^{4} - 90 \, a^{3} x^{3} + 1320 \, a^{2} x^{2} - 1402 \, a x - 3 \,{\left (a^{4} x^{4} - 4 \, a^{3} x^{3} + 6 \, a^{2} x^{2} - 4 \, a x + 1\right )} \log \left (a x + 1\right ) + 387 \,{\left (a^{4} x^{4} - 4 \, a^{3} x^{3} + 6 \, a^{2} x^{2} - 4 \, a x + 1\right )} \log \left (a x - 1\right ) + 448}{96 \,{\left (a^{5} c^{3} x^{4} - 4 \, a^{4} c^{3} x^{3} + 6 \, a^{3} c^{3} x^{2} - 4 \, a^{2} c^{3} x + a c^{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/(c-c/a^2/x^2)^3,x, algorithm="fricas")

[Out]

1/96*(96*a^5*x^5 - 384*a^4*x^4 - 90*a^3*x^3 + 1320*a^2*x^2 - 1402*a*x - 3*(a^4*x^4 - 4*a^3*x^3 + 6*a^2*x^2 - 4
*a*x + 1)*log(a*x + 1) + 387*(a^4*x^4 - 4*a^3*x^3 + 6*a^2*x^2 - 4*a*x + 1)*log(a*x - 1) + 448)/(a^5*c^3*x^4 -
4*a^4*c^3*x^3 + 6*a^3*c^3*x^2 - 4*a^2*c^3*x + a*c^3)

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Sympy [A]  time = 0.923373, size = 114, normalized size = 1.03 \begin{align*} a^{6} \left (- \frac{333 a^{3} x^{3} - 852 a^{2} x^{2} + 749 a x - 224}{48 a^{11} c^{3} x^{4} - 192 a^{10} c^{3} x^{3} + 288 a^{9} c^{3} x^{2} - 192 a^{8} c^{3} x + 48 a^{7} c^{3}} + \frac{x}{a^{6} c^{3}} + \frac{\frac{129 \log{\left (x - \frac{1}{a} \right )}}{32} - \frac{\log{\left (x + \frac{1}{a} \right )}}{32}}{a^{7} c^{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/(c-c/a**2/x**2)**3,x)

[Out]

a**6*(-(333*a**3*x**3 - 852*a**2*x**2 + 749*a*x - 224)/(48*a**11*c**3*x**4 - 192*a**10*c**3*x**3 + 288*a**9*c*
*3*x**2 - 192*a**8*c**3*x + 48*a**7*c**3) + x/(a**6*c**3) + (129*log(x - 1/a)/32 - log(x + 1/a)/32)/(a**7*c**3
))

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Giac [A]  time = 1.12352, size = 176, normalized size = 1.59 \begin{align*} \frac{a x - 1}{a c^{3}} - \frac{4 \, \log \left (\frac{{\left | a x - 1 \right |}}{{\left (a x - 1\right )}^{2}{\left | a \right |}}\right )}{a c^{3}} - \frac{\log \left ({\left | -\frac{2}{a x - 1} - 1 \right |}\right )}{32 \, a c^{3}} - \frac{\frac{333 \, a^{11} c^{9}}{a x - 1} + \frac{147 \, a^{11} c^{9}}{{\left (a x - 1\right )}^{2}} + \frac{44 \, a^{11} c^{9}}{{\left (a x - 1\right )}^{3}} + \frac{6 \, a^{11} c^{9}}{{\left (a x - 1\right )}^{4}}}{48 \, a^{12} c^{12}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x-1)^2*(a*x+1)^2/(c-c/a^2/x^2)^3,x, algorithm="giac")

[Out]

(a*x - 1)/(a*c^3) - 4*log(abs(a*x - 1)/((a*x - 1)^2*abs(a)))/(a*c^3) - 1/32*log(abs(-2/(a*x - 1) - 1))/(a*c^3)
- 1/48*(333*a^11*c^9/(a*x - 1) + 147*a^11*c^9/(a*x - 1)^2 + 44*a^11*c^9/(a*x - 1)^3 + 6*a^11*c^9/(a*x - 1)^4)
/(a^12*c^12)