\(\int \frac {e^{\coth ^{-1}(a x)}}{(c-a^2 c x^2)^3} \, dx\) [562]

Optimal result

Integrand size = 20, antiderivative size = 85 \[ \int \frac {e^{\coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^3} \, dx=\frac {8 e^{\coth ^{-1}(a x)}}{15 a c^3}-\frac {e^{\coth ^{-1}(a x)} (1-4 a x)}{15 a c^3 \left (1-a^2 x^2\right )^2}-\frac {4 e^{\coth ^{-1}(a x)} (1-2 a x)}{15 a c^3 \left (1-a^2 x^2\right )} \]

[Out]

8/15/((a*x-1)/(a*x+1))^(1/2)/a/c^3-1/15/((a*x-1)/(a*x+1))^(1/2)*(-4*a*x+1)/a/c^3/(-a^2*x^2+1)^2-4/15/((a*x-1)/
(a*x+1))^(1/2)*(-2*a*x+1)/a/c^3/(-a^2*x^2+1)

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {6320, 6318} \[ \int \frac {e^{\coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^3} \, dx=-\frac {(1-4 a x) e^{\coth ^{-1}(a x)}}{15 a c^3 \left (1-a^2 x^2\right )^2}-\frac {4 (1-2 a x) e^{\coth ^{-1}(a x)}}{15 a c^3 \left (1-a^2 x^2\right )}+\frac {8 e^{\coth ^{-1}(a x)}}{15 a c^3} \]

[In]

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

[Out]

(8*E^ArcCoth[a*x])/(15*a*c^3) - (E^ArcCoth[a*x]*(1 - 4*a*x))/(15*a*c^3*(1 - a^2*x^2)^2) - (4*E^ArcCoth[a*x]*(1
- 2*a*x))/(15*a*c^3*(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)} (1-4 a x)}{15 a c^3 \left (1-a^2 x^2\right )^2}+\frac {4 \int \frac {e^{\coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^2} \, dx}{5 c} \\ & = -\frac {e^{\coth ^{-1}(a x)} (1-4 a x)}{15 a c^3 \left (1-a^2 x^2\right )^2}-\frac {4 e^{\coth ^{-1}(a x)} (1-2 a x)}{15 a c^3 \left (1-a^2 x^2\right )}+\frac {8 \int \frac {e^{\coth ^{-1}(a x)}}{c-a^2 c x^2} \, dx}{15 c^2} \\ & = \frac {8 e^{\coth ^{-1}(a x)}}{15 a c^3}-\frac {e^{\coth ^{-1}(a x)} (1-4 a x)}{15 a c^3 \left (1-a^2 x^2\right )^2}-\frac {4 e^{\coth ^{-1}(a x)} (1-2 a x)}{15 a c^3 \left (1-a^2 x^2\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.38 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.78 \[ \int \frac {e^{\coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^3} \, dx=\frac {\sqrt {1-\frac {1}{a^2 x^2}} x \left (3+12 a x-12 a^2 x^2-8 a^3 x^3+8 a^4 x^4\right )}{15 c^3 (-1+a x)^3 (1+a x)^2} \]

[In]

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

[Out]

(Sqrt[1 - 1/(a^2*x^2)]*x*(3 + 12*a*x - 12*a^2*x^2 - 8*a^3*x^3 + 8*a^4*x^4))/(15*c^3*(-1 + a*x)^3*(1 + a*x)^2)

Maple [A] (verified)

Time = 0.48 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.76

method	result	size
gosper	\(\frac {8 a^{4} x^{4}-8 a^{3} x^{3}-12 a^{2} x^{2}+12 a x +3}{15 \left (a^{2} x^{2}-1\right )^{2} c^{3} \sqrt {\frac {a x -1}{a x +1}}\, a}\)	\(65\)
default	\(\frac {8 a^{4} x^{4}-8 a^{3} x^{3}-12 a^{2} x^{2}+12 a x +3}{15 \sqrt {\frac {a x -1}{a x +1}}\, c^{3} \left (a x -1\right )^{2} a \left (a x +1\right )^{2}}\)	\(68\)
trager	\(\frac {\left (8 a^{4} x^{4}-8 a^{3} x^{3}-12 a^{2} x^{2}+12 a x +3\right ) \sqrt {-\frac {-a x +1}{a x +1}}}{15 a \,c^{3} \left (a x +1\right ) \left (a x -1\right )^{3}}\)	\(70\)

[In]

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

[Out]

1/15*(8*a^4*x^4-8*a^3*x^3-12*a^2*x^2+12*a*x+3)/(a^2*x^2-1)^2/c^3/((a*x-1)/(a*x+1))^(1/2)/a

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.01 \[ \int \frac {e^{\coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^3} \, dx=\frac {{\left (8 \, a^{4} x^{4} - 8 \, a^{3} x^{3} - 12 \, a^{2} x^{2} + 12 \, a x + 3\right )} \sqrt {\frac {a x - 1}{a x + 1}}}{15 \, {\left (a^{5} c^{3} x^{4} - 2 \, a^{4} c^{3} x^{3} + 2 \, a^{2} c^{3} x - a c^{3}\right )}} \]

[In]

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

[Out]

1/15*(8*a^4*x^4 - 8*a^3*x^3 - 12*a^2*x^2 + 12*a*x + 3)*sqrt((a*x - 1)/(a*x + 1))/(a^5*c^3*x^4 - 2*a^4*c^3*x^3
+ 2*a^2*c^3*x - a*c^3)

Sympy [F]

\[ \int \frac {e^{\coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^3} \, dx=- \frac {\int \frac {1}{a^{6} x^{6} \sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}} - 3 a^{4} x^{4} \sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}} + 3 a^{2} x^{2} \sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}} - \sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}}}\, dx}{c^{3}} \]

[In]

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

[Out]

-Integral(1/(a**6*x**6*sqrt(a*x/(a*x + 1) - 1/(a*x + 1)) - 3*a**4*x**4*sqrt(a*x/(a*x + 1) - 1/(a*x + 1)) + 3*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**3

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.16 \[ \int \frac {e^{\coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^3} \, dx=-\frac {1}{240} \, a {\left (\frac {5 \, {\left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}} - 12 \, \sqrt {\frac {a x - 1}{a x + 1}}\right )}}{a^{2} c^{3}} + \frac {\frac {20 \, {\left (a x - 1\right )}}{a x + 1} - \frac {90 \, {\left (a x - 1\right )}^{2}}{{\left (a x + 1\right )}^{2}} - 3}{a^{2} c^{3} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {5}{2}}}\right )} \]

[In]

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

[Out]

-1/240*a*(5*(((a*x - 1)/(a*x + 1))^(3/2) - 12*sqrt((a*x - 1)/(a*x + 1)))/(a^2*c^3) + (20*(a*x - 1)/(a*x + 1) -
90*(a*x - 1)^2/(a*x + 1)^2 - 3)/(a^2*c^3*((a*x - 1)/(a*x + 1))^(5/2)))

Giac [F]

\[ \int \frac {e^{\coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^3} \, dx=\int { -\frac {1}{{\left (a^{2} c x^{2} - c\right )}^{3} \sqrt {\frac {a x - 1}{a x + 1}}} \,d x } \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 4.30 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.71 \[ \int \frac {e^{\coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^3} \, dx=\frac {8\,a^4\,x^4-8\,a^3\,x^3-12\,a^2\,x^2+12\,a\,x+3}{15\,a\,c^3\,{\left (a\,x+1\right )}^4\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{5/2}} \]

[In]

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

[Out]

(12*a*x - 12*a^2*x^2 - 8*a^3*x^3 + 8*a^4*x^4 + 3)/(15*a*c^3*(a*x + 1)^4*((a*x - 1)/(a*x + 1))^(5/2))