\(\int \frac {e^{\coth ^{-1}(a x)} x}{(c-a^2 c x^2)^{3/2}} \, dx\) [694]

Optimal result

Integrand size = 23, antiderivative size = 87 \[ \int \frac {e^{\coth ^{-1}(a x)} x}{\left (c-a^2 c x^2\right )^{3/2}} \, dx=\frac {a \left (1-\frac {1}{a^2 x^2}\right )^{3/2} x^3}{2 (1-a x) \left (c-a^2 c x^2\right )^{3/2}}-\frac {a \left (1-\frac {1}{a^2 x^2}\right )^{3/2} x^3 \text {arctanh}(a x)}{2 \left (c-a^2 c x^2\right )^{3/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {6327, 6328, 78, 213} \[ \int \frac {e^{\coth ^{-1}(a x)} x}{\left (c-a^2 c x^2\right )^{3/2}} \, dx=\frac {a x^3 \left (1-\frac {1}{a^2 x^2}\right )^{3/2}}{2 (1-a x) \left (c-a^2 c x^2\right )^{3/2}}-\frac {a x^3 \left (1-\frac {1}{a^2 x^2}\right )^{3/2} \text {arctanh}(a x)}{2 \left (c-a^2 c x^2\right )^{3/2}} \]

[In]

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

[Out]

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

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 213

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[b, 2])^(-1))*ArcTanh[Rt[b, 2]*(x/Rt[-a, 2])]
, x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

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

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.60 \[ \int \frac {e^{\coth ^{-1}(a x)} x}{\left (c-a^2 c x^2\right )^{3/2}} \, dx=\frac {a \left (1-\frac {1}{a^2 x^2}\right )^{3/2} x^3 \left (\frac {1}{1-a x}-\text {arctanh}(a x)\right )}{2 \left (c-a^2 c x^2\right )^{3/2}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.54 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.97

method	result	size
default	\(-\frac {\sqrt {-c \left (a^{2} x^{2}-1\right )}\, \left (a \ln \left (a x +1\right ) x -a \ln \left (a x -1\right ) x -\ln \left (a x +1\right )+\ln \left (a x -1\right )+2\right )}{4 \sqrt {\frac {a x -1}{a x +1}}\, \left (a^{2} x^{2}-1\right ) c^{2} a^{2}}\)	\(84\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.99 \[ \int \frac {e^{\coth ^{-1}(a x)} x}{\left (c-a^2 c x^2\right )^{3/2}} \, dx=-\frac {{\left (a^{2} x - a\right )} \sqrt {-c} \log \left (\frac {a^{2} c x^{2} - 2 \, \sqrt {-a^{2} c} \sqrt {-c} x + c}{a^{2} x^{2} - 1}\right ) + 2 \, \sqrt {-a^{2} c}}{4 \, {\left (a^{4} c^{2} x - a^{3} c^{2}\right )}} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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