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

Optimal. Leaf size=86 \[ -\frac{3}{16 a c^3 (1-a x)}+\frac{1}{16 a c^3 (a x+1)}-\frac{1}{8 a c^3 (1-a x)^2}-\frac{1}{12 a c^3 (1-a x)^3}-\frac{\tanh ^{-1}(a x)}{4 a c^3} \]

[Out]

-1/(12*a*c^3*(1 - a*x)^3) - 1/(8*a*c^3*(1 - a*x)^2) - 3/(16*a*c^3*(1 - a*x)) + 1/(16*a*c^3*(1 + a*x)) - ArcTan
h[a*x]/(4*a*c^3)

________________________________________________________________________________________

Rubi [A]  time = 0.100349, antiderivative size = 86, 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, 6140, 44, 207} \[ -\frac{3}{16 a c^3 (1-a x)}+\frac{1}{16 a c^3 (a x+1)}-\frac{1}{8 a c^3 (1-a x)^2}-\frac{1}{12 a c^3 (1-a x)^3}-\frac{\tanh ^{-1}(a x)}{4 a c^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/(12*a*c^3*(1 - a*x)^3) - 1/(8*a*c^3*(1 - a*x)^2) - 3/(16*a*c^3*(1 - a*x)) + 1/(16*a*c^3*(1 + a*x)) - ArcTan
h[a*x]/(4*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 6140

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

Rule 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rule 207

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rubi steps

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

Mathematica [A]  time = 0.0367401, size = 63, normalized size = 0.73 \[ \frac{3 a^3 x^3-6 a^2 x^2+a x-3 (a x-1)^3 (a x+1) \tanh ^{-1}(a x)+4}{12 a c^3 (a x-1)^3 (a x+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4 + a*x - 6*a^2*x^2 + 3*a^3*x^3 - 3*(-1 + a*x)^3*(1 + a*x)*ArcTanh[a*x])/(12*a*c^3*(-1 + a*x)^3*(1 + a*x))

________________________________________________________________________________________

Maple [A]  time = 0.05, size = 90, normalized size = 1.1 \begin{align*}{\frac{1}{16\,a{c}^{3} \left ( ax+1 \right ) }}-{\frac{\ln \left ( ax+1 \right ) }{8\,a{c}^{3}}}+{\frac{1}{12\,a{c}^{3} \left ( ax-1 \right ) ^{3}}}-{\frac{1}{8\,a{c}^{3} \left ( ax-1 \right ) ^{2}}}+{\frac{3}{16\,a{c}^{3} \left ( ax-1 \right ) }}+{\frac{\ln \left ( ax-1 \right ) }{8\,a{c}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16/a/c^3/(a*x+1)-1/8*ln(a*x+1)/a/c^3+1/12/a/c^3/(a*x-1)^3-1/8/a/c^3/(a*x-1)^2+3/16/a/c^3/(a*x-1)+1/8/a/c^3*l
n(a*x-1)

________________________________________________________________________________________

Maxima [A]  time = 1.06068, size = 123, normalized size = 1.43 \begin{align*} \frac{3 \, a^{3} x^{3} - 6 \, a^{2} x^{2} + a x + 4}{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 )}} - \frac{\log \left (a x + 1\right )}{8 \, a c^{3}} + \frac{\log \left (a x - 1\right )}{8 \, a c^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 1.5362, size = 266, normalized size = 3.09 \begin{align*} \frac{6 \, a^{3} x^{3} - 12 \, a^{2} x^{2} + 2 \, a x - 3 \,{\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 ) + 8}{24 \,{\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)*(a*x+1)/(-a^2*c*x^2+c)^3,x, algorithm="fricas")

[Out]

1/24*(6*a^3*x^3 - 12*a^2*x^2 + 2*a*x - 3*(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) + 8)/(a^5*c^3*x^4 - 2*a^4*c^3*x^3 + 2*a^2*c^3*x - a*c^3)

________________________________________________________________________________________

Sympy [A]  time = 0.675573, size = 83, normalized size = 0.97 \begin{align*} \frac{3 a^{3} x^{3} - 6 a^{2} x^{2} + a x + 4}{12 a^{5} c^{3} x^{4} - 24 a^{4} c^{3} x^{3} + 24 a^{2} c^{3} x - 12 a c^{3}} - \frac{- \frac{\log{\left (x - \frac{1}{a} \right )}}{8} + \frac{\log{\left (x + \frac{1}{a} \right )}}{8}}{a c^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [A]  time = 1.11658, size = 100, normalized size = 1.16 \begin{align*} -\frac{\log \left ({\left | a x + 1 \right |}\right )}{8 \, a c^{3}} + \frac{\log \left ({\left | a x - 1 \right |}\right )}{8 \, a c^{3}} + \frac{3 \, a^{3} x^{3} - 6 \, a^{2} x^{2} + a x + 4}{12 \,{\left (a x + 1\right )}{\left (a x - 1\right )}^{3} a c^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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