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

Optimal. Leaf size=109 \[ -\frac{1}{16 a c^3 (1-a x)}+\frac{39}{16 a c^3 (a x+1)}-\frac{5}{8 a c^3 (a x+1)^2}+\frac{1}{12 a c^3 (a x+1)^3}-\frac{\log (1-a x)}{4 a c^3}+\frac{9 \log (a x+1)}{4 a c^3}-\frac{x}{c^3} \]

[Out]

-(x/c^3) - 1/(16*a*c^3*(1 - a*x)) + 1/(12*a*c^3*(1 + a*x)^3) - 5/(8*a*c^3*(1 + a*x)^2) + 39/(16*a*c^3*(1 + a*x
)) - Log[1 - a*x]/(4*a*c^3) + (9*Log[1 + a*x])/(4*a*c^3)

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Rubi [A]  time = 0.16565, antiderivative size = 109, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {6157, 6150, 88} \[ -\frac{1}{16 a c^3 (1-a x)}+\frac{39}{16 a c^3 (a x+1)}-\frac{5}{8 a c^3 (a x+1)^2}+\frac{1}{12 a c^3 (a x+1)^3}-\frac{\log (1-a x)}{4 a c^3}+\frac{9 \log (a x+1)}{4 a c^3}-\frac{x}{c^3} \]

Antiderivative was successfully verified.

[In]

Int[1/(E^(2*ArcTanh[a*x])*(c - c/(a^2*x^2))^3),x]

[Out]

-(x/c^3) - 1/(16*a*c^3*(1 - a*x)) + 1/(12*a*c^3*(1 + a*x)^3) - 5/(8*a*c^3*(1 + a*x)^2) + 39/(16*a*c^3*(1 + a*x
)) - Log[1 - a*x]/(4*a*c^3) + (9*Log[1 + a*x])/(4*a*c^3)

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^{-2 \tanh ^{-1}(a x)}}{\left (c-\frac{c}{a^2 x^2}\right )^3} \, dx &=-\frac{a^6 \int \frac{e^{-2 \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)^2 (1+a x)^4} \, dx}{c^3}\\ &=-\frac{a^6 \int \left (\frac{1}{a^6}+\frac{1}{16 a^6 (-1+a x)^2}+\frac{1}{4 a^6 (-1+a x)}+\frac{1}{4 a^6 (1+a x)^4}-\frac{5}{4 a^6 (1+a x)^3}+\frac{39}{16 a^6 (1+a x)^2}-\frac{9}{4 a^6 (1+a x)}\right ) \, dx}{c^3}\\ &=-\frac{x}{c^3}-\frac{1}{16 a c^3 (1-a x)}+\frac{1}{12 a c^3 (1+a x)^3}-\frac{5}{8 a c^3 (1+a x)^2}+\frac{39}{16 a c^3 (1+a x)}-\frac{\log (1-a x)}{4 a c^3}+\frac{9 \log (1+a x)}{4 a c^3}\\ \end{align*}

Mathematica [A]  time = 0.0623581, size = 104, normalized size = 0.95 \[ \frac{-2 \left (6 a^5 x^5+12 a^4 x^4-15 a^3 x^3-24 a^2 x^2+7 a x+11\right )-3 (a x-1) (a x+1)^3 \log (1-a x)+27 (a x-1) (a x+1)^3 \log (a x+1)}{12 a (a x-1) (a c x+c)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(E^(2*ArcTanh[a*x])*(c - c/(a^2*x^2))^3),x]

[Out]

(-2*(11 + 7*a*x - 24*a^2*x^2 - 15*a^3*x^3 + 12*a^4*x^4 + 6*a^5*x^5) - 3*(-1 + a*x)*(1 + a*x)^3*Log[1 - a*x] +
27*(-1 + a*x)*(1 + a*x)^3*Log[1 + a*x])/(12*a*(-1 + a*x)*(c + a*c*x)^3)

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Maple [A]  time = 0.041, size = 96, normalized size = 0.9 \begin{align*} -{\frac{x}{{c}^{3}}}+{\frac{1}{12\,a{c}^{3} \left ( ax+1 \right ) ^{3}}}-{\frac{5}{8\,a{c}^{3} \left ( ax+1 \right ) ^{2}}}+{\frac{39}{16\,a{c}^{3} \left ( ax+1 \right ) }}+{\frac{9\,\ln \left ( ax+1 \right ) }{4\,a{c}^{3}}}+{\frac{1}{16\,a{c}^{3} \left ( ax-1 \right ) }}-{\frac{\ln \left ( ax-1 \right ) }{4\,a{c}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x/c^3+1/12/a/c^3/(a*x+1)^3-5/8/a/c^3/(a*x+1)^2+39/16/a/c^3/(a*x+1)+9/4*ln(a*x+1)/a/c^3+1/16/a/c^3/(a*x-1)-1/4
/a/c^3*ln(a*x-1)

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Maxima [A]  time = 0.972072, size = 132, normalized size = 1.21 \begin{align*} \frac{15 \, a^{3} x^{3} + 12 \, a^{2} x^{2} - 13 \, a x - 11}{6 \,{\left (a^{5} c^{3} x^{4} + 2 \, a^{4} c^{3} x^{3} - 2 \, a^{2} c^{3} x - a c^{3}\right )}} - \frac{x}{c^{3}} + \frac{9 \, \log \left (a x + 1\right )}{4 \, a c^{3}} - \frac{\log \left (a x - 1\right )}{4 \, a c^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(15*a^3*x^3 + 12*a^2*x^2 - 13*a*x - 11)/(a^5*c^3*x^4 + 2*a^4*c^3*x^3 - 2*a^2*c^3*x - a*c^3) - x/c^3 + 9/4*
log(a*x + 1)/(a*c^3) - 1/4*log(a*x - 1)/(a*c^3)

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Fricas [A]  time = 1.86269, size = 308, normalized size = 2.83 \begin{align*} -\frac{12 \, a^{5} x^{5} + 24 \, a^{4} x^{4} - 30 \, a^{3} x^{3} - 48 \, a^{2} x^{2} + 14 \, a x - 27 \,{\left (a^{4} x^{4} + 2 \, a^{3} x^{3} - 2 \, a x - 1\right )} \log \left (a x + 1\right ) + 3 \,{\left (a^{4} x^{4} + 2 \, a^{3} x^{3} - 2 \, a x - 1\right )} \log \left (a x - 1\right ) + 22}{12 \,{\left (a^{5} c^{3} x^{4} + 2 \, a^{4} c^{3} x^{3} - 2 \, 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^2*x^2+1)/(c-c/a^2/x^2)^3,x, algorithm="fricas")

[Out]

-1/12*(12*a^5*x^5 + 24*a^4*x^4 - 30*a^3*x^3 - 48*a^2*x^2 + 14*a*x - 27*(a^4*x^4 + 2*a^3*x^3 - 2*a*x - 1)*log(a
*x + 1) + 3*(a^4*x^4 + 2*a^3*x^3 - 2*a*x - 1)*log(a*x - 1) + 22)/(a^5*c^3*x^4 + 2*a^4*c^3*x^3 - 2*a^2*c^3*x -
a*c^3)

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Sympy [A]  time = 3.70591, size = 104, normalized size = 0.95 \begin{align*} - a^{6} \left (- \frac{15 a^{3} x^{3} + 12 a^{2} x^{2} - 13 a x - 11}{6 a^{11} c^{3} x^{4} + 12 a^{10} c^{3} x^{3} - 12 a^{8} c^{3} x - 6 a^{7} c^{3}} + \frac{x}{a^{6} c^{3}} + \frac{\frac{\log{\left (x - \frac{1}{a} \right )}}{4} - \frac{9 \log{\left (x + \frac{1}{a} \right )}}{4}}{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**2*x**2+1)/(c-c/a**2/x**2)**3,x)

[Out]

-a**6*(-(15*a**3*x**3 + 12*a**2*x**2 - 13*a*x - 11)/(6*a**11*c**3*x**4 + 12*a**10*c**3*x**3 - 12*a**8*c**3*x -
6*a**7*c**3) + x/(a**6*c**3) + (log(x - 1/a)/4 - 9*log(x + 1/a)/4)/(a**7*c**3))

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Giac [A]  time = 1.19834, size = 189, normalized size = 1.73 \begin{align*} -\frac{{\left (a x + 1\right )}{\left (\frac{65}{a x + 1} - 32\right )}}{32 \, a c^{3}{\left (\frac{2}{a x + 1} - 1\right )}} - \frac{2 \, \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 )}{4 \, a c^{3}} + \frac{\frac{117 \, a^{11} c^{6}}{a x + 1} - \frac{30 \, a^{11} c^{6}}{{\left (a x + 1\right )}^{2}} + \frac{4 \, a^{11} c^{6}}{{\left (a x + 1\right )}^{3}}}{48 \, a^{12} c^{9}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/32*(a*x + 1)*(65/(a*x + 1) - 32)/(a*c^3*(2/(a*x + 1) - 1)) - 2*log(abs(a*x + 1)/((a*x + 1)^2*abs(a)))/(a*c^
3) - 1/4*log(abs(-2/(a*x + 1) + 1))/(a*c^3) + 1/48*(117*a^11*c^6/(a*x + 1) - 30*a^11*c^6/(a*x + 1)^2 + 4*a^11*
c^6/(a*x + 1)^3)/(a^12*c^9)