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

Optimal. Leaf size=49 \[ \frac{1}{4 a c^2 (a x+1)}+\frac{1}{4 a c^2 (a x+1)^2}-\frac{\tanh ^{-1}(a x)}{4 a c^2} \]

[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 = 0.0764191, antiderivative size = 49, 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} \[ \frac{1}{4 a c^2 (a x+1)}+\frac{1}{4 a c^2 (a x+1)^2}-\frac{\tanh ^{-1}(a x)}{4 a c^2} \]

Antiderivative was successfully verified.

[In]

Int[1/(E^(2*ArcCoth[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 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 )^2} \, dx &=-\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) (1+a x)^3} \, 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 = 0.0265241, size = 33, normalized size = 0.67 \[ \frac{a x+(a x+1)^2 \left (-\tanh ^{-1}(a x)\right )+2}{4 a (a c x+c)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.05, size = 60, normalized size = 1.2 \begin{align*}{\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}}}+{\frac{\ln \left ( ax-1 \right ) }{8\,a{c}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.11156, size = 85, normalized size = 1.73 \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)/(a*x+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 = 1.50556, size = 171, normalized size = 3.49 \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)/(a*x+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 = 0.463682, size = 54, 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)/(a*x+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 = 1.10729, size = 69, normalized size = 1.41 \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)/(a*x+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)