\(\int \frac {e^{\coth ^{-1}(a x)}}{(c-a^2 c x^2)^2} \, dx\) [561]

Optimal result

Integrand size = 20, antiderivative size = 51 \[ \int \frac {e^{\coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^2} \, dx=\frac {2 e^{\coth ^{-1}(a x)}}{3 a c^2}-\frac {e^{\coth ^{-1}(a x)} (1-2 a x)}{3 a c^2 \left (1-a^2 x^2\right )} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {6320, 6318} \[ \int \frac {e^{\coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^2} \, dx=\frac {2 e^{\coth ^{-1}(a x)}}{3 a c^2}-\frac {(1-2 a x) e^{\coth ^{-1}(a x)}}{3 a c^2 \left (1-a^2 x^2\right )} \]

[In]

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

[Out]

(2*E^ArcCoth[a*x])/(3*a*c^2) - (E^ArcCoth[a*x]*(1 - 2*a*x))/(3*a*c^2*(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 = 50, normalized size of antiderivative = 0.98 \[ \int \frac {e^{\coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^2} \, dx=\frac {\sqrt {1-\frac {1}{a^2 x^2}} x \left (-1-2 a x+2 a^2 x^2\right )}{3 c^2 (-1+a x)^2 (1+a x)} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.47 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.92

method	result	size
trager	\(\frac {\left (2 a^{2} x^{2}-2 a x -1\right ) \sqrt {-\frac {-a x +1}{a x +1}}}{3 a \,c^{2} \left (a x -1\right )^{2}}\)	\(47\)
gosper	\(\frac {2 a^{2} x^{2}-2 a x -1}{3 \left (a^{2} x^{2}-1\right ) c^{2} \sqrt {\frac {a x -1}{a x +1}}\, a}\)	\(49\)
default	\(\frac {2 a^{2} x^{2}-2 a x -1}{3 \sqrt {\frac {a x -1}{a x +1}}\, c^{2} \left (a x -1\right ) a \left (a x +1\right )}\)	\(52\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.14 \[ \int \frac {e^{\coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^2} \, dx=\frac {{\left (2 \, a^{2} x^{2} - 2 \, a x - 1\right )} \sqrt {\frac {a x - 1}{a x + 1}}}{3 \, {\left (a^{3} c^{2} x^{2} - 2 \, a^{2} c^{2} x + a c^{2}\right )}} \]

[In]

integrate(1/((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 - 2*a^2*c^2*x + a*c^2)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.27 \[ \int \frac {e^{\coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^2} \, dx=\frac {1}{12} \, a {\left (\frac {3 \, \sqrt {\frac {a x - 1}{a x + 1}}}{a^{2} c^{2}} + \frac {\frac {6 \, {\left (a x - 1\right )}}{a x + 1} - 1}{a^{2} c^{2} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}}}\right )} \]

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.98 \[ \int \frac {e^{\coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^2} \, dx=\frac {-2\,a^2\,x^2+2\,a\,x+1}{\left (3\,a\,c^2-3\,a^3\,c^2\,x^2\right )\,\sqrt {\frac {a\,x-1}{a\,x+1}}} \]

[In]

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

[Out]

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