\(\int \frac {e^{-3 \coth ^{-1}(a x)}}{(c-a^2 c x^2)^{3/2}} \, dx\) [665]

Optimal result

Integrand size = 24, antiderivative size = 46 \[ \int \frac {e^{-3 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{3/2}} \, dx=-\frac {a^2 \left (1-\frac {1}{a^2 x^2}\right )^{3/2} x^3}{2 (1+a x)^2 \left (c-a^2 c x^2\right )^{3/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {6327, 6328, 32} \[ \int \frac {e^{-3 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{3/2}} \, dx=-\frac {a^2 x^3 \left (1-\frac {1}{a^2 x^2}\right )^{3/2}}{2 (a x+1)^2 \left (c-a^2 c x^2\right )^{3/2}} \]

[In]

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

[Out]

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

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rule 6327

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

Rule 6328

Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)/(x_)^2)^(p_.), x_Symbol] :> Dist[c^p/a^(2*p), Int[(u/x^(
2*p))*(-1 + a*x)^(p - n/2)*(1 + a*x)^(p + n/2), x], x] /; FreeQ[{a, c, d, n, p}, x] && EqQ[c + a^2*d, 0] &&  !
IntegerQ[n/2] && (IntegerQ[p] || GtQ[c, 0]) && IntegersQ[2*p, p + n/2]

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

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.11 \[ \int \frac {e^{-3 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{3/2}} \, dx=-\frac {\sqrt {1-\frac {1}{a^2 x^2}} x \sqrt {c-a^2 c x^2}}{2 c^2 (-1+a x) (1+a x)^3} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.52 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.85

method	result	size
gosper	\(-\frac {\left (a x +1\right ) \left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}}}{2 a \left (-a^{2} c \,x^{2}+c \right )^{\frac {3}{2}}}\)	\(39\)
default	\(-\frac {\left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}} \sqrt {-c \left (a^{2} x^{2}-1\right )}}{2 \left (a x -1\right ) \left (a^{2} x^{2}-1\right ) a \,c^{2}}\)	\(56\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.85 \[ \int \frac {e^{-3 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{3/2}} \, dx=-\frac {\sqrt {-a^{2} c}}{2 \, {\left (a^{4} c^{2} x^{2} + 2 \, a^{3} c^{2} x + a^{2} c^{2}\right )}} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 4.39 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.26 \[ \int \frac {e^{-3 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{3/2}} \, dx=\frac {\sqrt {\frac {a\,x-1}{a\,x+1}}}{2\,a^2\,c\,\left (x\,\sqrt {c-a^2\,c\,x^2}+\frac {\sqrt {c-a^2\,c\,x^2}}{a}\right )} \]

[In]

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

[Out]

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