3.1062 \(\int \frac{e^{2 \tanh ^{-1}(a x)}}{(c-a^2 c x^2)^2} \, dx\)

Optimal. Leaf size=51 \[ \frac{1}{4 a c^2 (1-a x)}+\frac{1}{4 a c^2 (1-a x)^2}+\frac{\tanh ^{-1}(a x)}{4 a c^2} \]

[Out]

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

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Rubi [A]  time = 0.052897, antiderivative size = 51, 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 = {6140, 44, 207} \[ \frac{1}{4 a c^2 (1-a x)}+\frac{1}{4 a c^2 (1-a x)^2}+\frac{\tanh ^{-1}(a x)}{4 a c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

1/(4*a*c^2*(1 - a*x)^2) + 1/(4*a*c^2*(1 - a*x)) + ArcTanh[a*x]/(4*a*c^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 \tanh ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^2} \, dx &=\frac{\int \frac{1}{(1-a x)^3 (1+a x)} \, dx}{c^2}\\ &=\frac{\int \left (-\frac{1}{2 (-1+a x)^3}+\frac{1}{4 (-1+a x)^2}-\frac{1}{4 \left (-1+a^2 x^2\right )}\right ) \, dx}{c^2}\\ &=\frac{1}{4 a c^2 (1-a x)^2}+\frac{1}{4 a c^2 (1-a x)}-\frac{\int \frac{1}{-1+a^2 x^2} \, dx}{4 c^2}\\ &=\frac{1}{4 a c^2 (1-a x)^2}+\frac{1}{4 a c^2 (1-a x)}+\frac{\tanh ^{-1}(a x)}{4 a c^2}\\ \end{align*}

Mathematica [A]  time = 0.0194994, size = 35, normalized size = 0.69 \[ \frac{-a x+(a x-1)^2 \tanh ^{-1}(a x)+2}{4 a c^2 (a x-1)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.032, size = 60, normalized size = 1.2 \begin{align*}{\frac{\ln \left ( ax+1 \right ) }{8\,a{c}^{2}}}+{\frac{1}{4\,a{c}^{2} \left ( ax-1 \right ) ^{2}}}-{\frac{1}{4\,a{c}^{2} \left ( ax-1 \right ) }}-{\frac{\ln \left ( ax-1 \right ) }{8\,a{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)/(-a^2*c*x^2+c)^2,x)

[Out]

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

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Maxima [A]  time = 0.972146, size = 85, normalized size = 1.67 \begin{align*} -\frac{a x - 2}{4 \,{\left (a^{3} c^{2} x^{2} - 2 \, a^{2} c^{2} x + a c^{2}\right )}} + \frac{\log \left (a x + 1\right )}{8 \, a c^{2}} - \frac{\log \left (a x - 1\right )}{8 \, a 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)/(-a^2*c*x^2+c)^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.28199, size = 173, normalized size = 3.39 \begin{align*} -\frac{2 \, a x -{\left (a^{2} x^{2} - 2 \, a x + 1\right )} \log \left (a x + 1\right ) +{\left (a^{2} x^{2} - 2 \, a x + 1\right )} \log \left (a x - 1\right ) - 4}{8 \,{\left (a^{3} c^{2} x^{2} - 2 \, a^{2} c^{2} x + a c^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.474745, size = 56, normalized size = 1.1 \begin{align*} - \frac{a x - 2}{4 a^{3} c^{2} x^{2} - 8 a^{2} c^{2} x + 4 a c^{2}} - \frac{\frac{\log{\left (x - \frac{1}{a} \right )}}{8} - \frac{\log{\left (x + \frac{1}{a} \right )}}{8}}{a 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)/(-a**2*c*x**2+c)**2,x)

[Out]

-(a*x - 2)/(4*a**3*c**2*x**2 - 8*a**2*c**2*x + 4*a*c**2) - (log(x - 1/a)/8 - log(x + 1/a)/8)/(a*c**2)

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Giac [A]  time = 1.14465, size = 69, normalized size = 1.35 \begin{align*} \frac{\log \left ({\left | a x + 1 \right |}\right )}{8 \, a c^{2}} - \frac{\log \left ({\left | a x - 1 \right |}\right )}{8 \, a c^{2}} - \frac{a x - 2}{4 \,{\left (a x - 1\right )}^{2} a 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)/(-a^2*c*x^2+c)^2,x, algorithm="giac")

[Out]

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