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

Optimal. Leaf size=87 \[ \frac{a x^3 \left (1-\frac{1}{a^2 x^2}\right )^{3/2}}{2 (1-a x) \left (c-a^2 c x^2\right )^{3/2}}-\frac{a x^3 \left (1-\frac{1}{a^2 x^2}\right )^{3/2} \tanh ^{-1}(a x)}{2 \left (c-a^2 c x^2\right )^{3/2}} \]

[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))

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Rubi [A]  time = 0.210995, antiderivative size = 87, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {6192, 6193, 77, 207} \[ \frac{a x^3 \left (1-\frac{1}{a^2 x^2}\right )^{3/2}}{2 (1-a x) \left (c-a^2 c x^2\right )^{3/2}}-\frac{a x^3 \left (1-\frac{1}{a^2 x^2}\right )^{3/2} \tanh ^{-1}(a x)}{2 \left (c-a^2 c x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[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 6192

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 6193

Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)/(x_)^2)^(p_.), x_Symbol] :> Dist[c^p/a^(2*p), Int[(u*(-1
+ a*x)^(p - n/2)*(1 + a*x)^(p + n/2))/x^(2*p), 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]

Rule 77

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 207

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

Rubi steps

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

Mathematica [A]  time = 0.0412113, size = 59, normalized size = 0.68 \[ \frac{x \sqrt{1-\frac{1}{a^2 x^2}} \left ((a x-1) \tanh ^{-1}(a x)+1\right )}{2 a c (a x-1) \sqrt{c-a^2 c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.153, size = 84, normalized size = 1. \begin{align*} -{\frac{ax\ln \left ( ax+1 \right ) -\ln \left ( ax-1 \right ) xa-\ln \left ( ax+1 \right ) +\ln \left ( ax-1 \right ) +2}{ \left ( 4\,{a}^{2}{x}^{2}-4 \right ){c}^{2}{a}^{2}}\sqrt{-c \left ({a}^{2}{x}^{2}-1 \right ) }{\frac{1}{\sqrt{{\frac{ax-1}{ax+1}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x}{{\left (-a^{2} c x^{2} + c\right )}^{\frac{3}{2}} \sqrt{\frac{a x - 1}{a x + 1}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [A]  time = 1.5346, size = 177, normalized size = 2.03 \begin{align*} -\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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x}{{\left (-a^{2} c x^{2} + c\right )}^{\frac{3}{2}} \sqrt{\frac{a x - 1}{a x + 1}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)