\(\int \frac {e^{-\coth ^{-1}(a x)}}{(c-a^2 c x^2)^2} \, dx\) [595]

Optimal result

Integrand size = 22, antiderivative size = 55 \[ \int \frac {e^{-\coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^2} \, dx=-\frac {2 e^{-\coth ^{-1}(a x)}}{3 a c^2}+\frac {e^{-\coth ^{-1}(a x)} (1+2 a x)}{3 a c^2 \left (1-a^2 x^2\right )} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {6320, 6318} \[ \int \frac {e^{-\coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^2} \, dx=\frac {(2 a x+1) e^{-\coth ^{-1}(a x)}}{3 a c^2 \left (1-a^2 x^2\right )}-\frac {2 e^{-\coth ^{-1}(a x)}}{3 a c^2} \]

[In]

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

[Out]

-2/(3*a*c^2*E^ArcCoth[a*x]) + (1 + 2*a*x)/(3*a*c^2*E^ArcCoth[a*x]*(1 - a^2*x^2))

Rule 6318

Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))/((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[E^(n*ArcCoth[a*x])/(a*c*n), x] /; F
reeQ[{a, c, d, n}, x] && EqQ[a^2*c + d, 0] &&  !IntegerQ[n/2]

Rule 6320

Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*((c_) + (d_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(n + 2*a*(p + 1)*x)*(c + d*x^2
)^(p + 1)*(E^(n*ArcCoth[a*x])/(a*c*(n^2 - 4*(p + 1)^2))), x] - Dist[2*(p + 1)*((2*p + 3)/(c*(n^2 - 4*(p + 1)^2
))), Int[(c + d*x^2)^(p + 1)*E^(n*ArcCoth[a*x]), x], x] /; FreeQ[{a, c, d, n}, x] && EqQ[a^2*c + d, 0] &&  !In
tegerQ[n/2] && LtQ[p, -1] && NeQ[p, -3/2] && NeQ[n^2 - 4*(p + 1)^2, 0] && (IntegerQ[p] ||  !IntegerQ[n])

Rubi steps \begin{align*} \text {integral}& = \frac {e^{-\coth ^{-1}(a x)} (1+2 a x)}{3 a c^2 \left (1-a^2 x^2\right )}+\frac {2 \int \frac {e^{-\coth ^{-1}(a x)}}{c-a^2 c x^2} \, dx}{3 c} \\ & = -\frac {2 e^{-\coth ^{-1}(a x)}}{3 a c^2}+\frac {e^{-\coth ^{-1}(a x)} (1+2 a x)}{3 a c^2 \left (1-a^2 x^2\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.30 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.87 \[ \int \frac {e^{-\coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^2} \, dx=-\frac {\sqrt {1-\frac {1}{a^2 x^2}} x \left (-1+2 a x+2 a^2 x^2\right )}{3 (-1+a x) (c+a c x)^2} \]

[In]

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

[Out]

-1/3*(Sqrt[1 - 1/(a^2*x^2)]*x*(-1 + 2*a*x + 2*a^2*x^2))/((-1 + a*x)*(c + a*c*x)^2)

Maple [A] (verified)

Time = 0.50 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.89

method	result	size
gosper	\(-\frac {\sqrt {\frac {a x -1}{a x +1}}\, \left (2 a^{2} x^{2}+2 a x -1\right )}{3 \left (a^{2} x^{2}-1\right ) a \,c^{2}}\)	\(49\)
default	\(-\frac {\sqrt {\frac {a x -1}{a x +1}}\, \left (2 a^{2} x^{2}+2 a x -1\right )}{3 c^{2} \left (a x +1\right ) \left (a x -1\right ) a}\)	\(52\)
trager	\(-\frac {\left (2 a^{2} x^{2}+2 a x -1\right ) \sqrt {-\frac {-a x +1}{a x +1}}}{3 a \,c^{2} \left (a x -1\right ) \left (a x +1\right )}\)	\(54\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.91 \[ \int \frac {e^{-\coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^2} \, dx=-\frac {{\left (2 \, a^{2} x^{2} + 2 \, a x - 1\right )} \sqrt {\frac {a x - 1}{a x + 1}}}{3 \, {\left (a^{3} c^{2} x^{2} - a c^{2}\right )}} \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \frac {e^{-\coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^2} \, dx=\frac {\int \frac {\sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}}}{a^{4} x^{4} - 2 a^{2} x^{2} + 1}\, dx}{c^{2}} \]

[In]

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

[Out]

Integral(sqrt(a*x/(a*x + 1) - 1/(a*x + 1))/(a**4*x**4 - 2*a**2*x**2 + 1), x)/c**2

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.22 \[ \int \frac {e^{-\coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^2} \, dx=\frac {1}{12} \, a {\left (\frac {\left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}} - 6 \, \sqrt {\frac {a x - 1}{a x + 1}}}{a^{2} c^{2}} - \frac {3}{a^{2} c^{2} \sqrt {\frac {a x - 1}{a x + 1}}}\right )} \]

[In]

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

[Out]

1/12*a*((((a*x - 1)/(a*x + 1))^(3/2) - 6*sqrt((a*x - 1)/(a*x + 1)))/(a^2*c^2) - 3/(a^2*c^2*sqrt((a*x - 1)/(a*x
+ 1))))

Giac [F]

\[ \int \frac {e^{-\coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^2} \, dx=\int { \frac {\sqrt {\frac {a x - 1}{a x + 1}}}{{\left (a^{2} c x^{2} - c\right )}^{2}} \,d x } \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-\coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^2} \, dx=-\frac {\frac {6\,\left (a\,x-1\right )}{a\,x+1}-\frac {{\left (a\,x-1\right )}^2}{{\left (a\,x+1\right )}^2}+3}{12\,a\,c^2\,\sqrt {\frac {a\,x-1}{a\,x+1}}} \]

[In]

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

[Out]

-((6*(a*x - 1))/(a*x + 1) - (a*x - 1)^2/(a*x + 1)^2 + 3)/(12*a*c^2*((a*x - 1)/(a*x + 1))^(1/2))