3.703 \(\int \frac{e^{\coth ^{-1}(a x)} x}{(c-a^2 c x^2)^{5/2}} \, dx\)

Optimal. Leaf size=137 \[ -\frac{a^3 x^5 \left (1-\frac{1}{a^2 x^2}\right )^{5/2}}{8 (a x+1) \left (c-a^2 c x^2\right )^{5/2}}-\frac{a^3 x^5 \left (1-\frac{1}{a^2 x^2}\right )^{5/2}}{8 (1-a x)^2 \left (c-a^2 c x^2\right )^{5/2}}+\frac{a^3 x^5 \left (1-\frac{1}{a^2 x^2}\right )^{5/2} \tanh ^{-1}(a x)}{8 \left (c-a^2 c x^2\right )^{5/2}} \]

[Out]

-(a^3*(1 - 1/(a^2*x^2))^(5/2)*x^5)/(8*(1 - a*x)^2*(c - a^2*c*x^2)^(5/2)) - (a^3*(1 - 1/(a^2*x^2))^(5/2)*x^5)/(
8*(1 + a*x)*(c - a^2*c*x^2)^(5/2)) + (a^3*(1 - 1/(a^2*x^2))^(5/2)*x^5*ArcTanh[a*x])/(8*(c - a^2*c*x^2)^(5/2))

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Rubi [A]  time = 0.221176, antiderivative size = 137, 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} \[ -\frac{a^3 x^5 \left (1-\frac{1}{a^2 x^2}\right )^{5/2}}{8 (a x+1) \left (c-a^2 c x^2\right )^{5/2}}-\frac{a^3 x^5 \left (1-\frac{1}{a^2 x^2}\right )^{5/2}}{8 (1-a x)^2 \left (c-a^2 c x^2\right )^{5/2}}+\frac{a^3 x^5 \left (1-\frac{1}{a^2 x^2}\right )^{5/2} \tanh ^{-1}(a x)}{8 \left (c-a^2 c x^2\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(E^ArcCoth[a*x]*x)/(c - a^2*c*x^2)^(5/2),x]

[Out]

-(a^3*(1 - 1/(a^2*x^2))^(5/2)*x^5)/(8*(1 - a*x)^2*(c - a^2*c*x^2)^(5/2)) - (a^3*(1 - 1/(a^2*x^2))^(5/2)*x^5)/(
8*(1 + a*x)*(c - a^2*c*x^2)^(5/2)) + (a^3*(1 - 1/(a^2*x^2))^(5/2)*x^5*ArcTanh[a*x])/(8*(c - a^2*c*x^2)^(5/2))

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

Mathematica [A]  time = 0.0677811, size = 60, normalized size = 0.44 \[ -\frac{a^3 x^5 \left (1-\frac{1}{a^2 x^2}\right )^{5/2} \left (\frac{1}{a x+1}+\frac{1}{(a x-1)^2}-\tanh ^{-1}(a x)\right )}{8 \left (c-a^2 c x^2\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(E^ArcCoth[a*x]*x)/(c - a^2*c*x^2)^(5/2),x]

[Out]

-(a^3*(1 - 1/(a^2*x^2))^(5/2)*x^5*((-1 + a*x)^(-2) + (1 + a*x)^(-1) - ArcTanh[a*x]))/(8*(c - a^2*c*x^2)^(5/2))

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Maple [A]  time = 0.15, size = 164, normalized size = 1.2 \begin{align*} -{\frac{{a}^{3}{x}^{3}\ln \left ( ax+1 \right ) -\ln \left ( ax-1 \right ){x}^{3}{a}^{3}-\ln \left ( ax+1 \right ){a}^{2}{x}^{2}+\ln \left ( ax-1 \right ){a}^{2}{x}^{2}-2\,{a}^{2}{x}^{2}-ax\ln \left ( ax+1 \right ) +\ln \left ( ax-1 \right ) xa+2\,ax+\ln \left ( ax+1 \right ) -\ln \left ( ax-1 \right ) -4}{ \left ( 16\,ax-16 \right ) \left ({a}^{2}{x}^{2}-1 \right ){c}^{3}{a}^{2} \left ( ax+1 \right ) }\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)^(5/2),x)

[Out]

-1/16/((a*x-1)/(a*x+1))^(1/2)/(a*x-1)*(-c*(a^2*x^2-1))^(1/2)*(a^3*x^3*ln(a*x+1)-ln(a*x-1)*x^3*a^3-ln(a*x+1)*a^
2*x^2+ln(a*x-1)*a^2*x^2-2*a^2*x^2-a*x*ln(a*x+1)+ln(a*x-1)*x*a+2*a*x+ln(a*x+1)-ln(a*x-1)-4)/(a^2*x^2-1)/c^3/a^2
/(a*x+1)

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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{5}{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)^(5/2),x, algorithm="maxima")

[Out]

integrate(x/((-a^2*c*x^2 + c)^(5/2)*sqrt((a*x - 1)/(a*x + 1))), x)

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Fricas [A]  time = 1.51764, size = 270, normalized size = 1.97 \begin{align*} -\frac{{\left (a^{4} x^{3} - a^{3} x^{2} - 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 \,{\left (a^{2} x^{2} - a x + 2\right )} \sqrt{-a^{2} c}}{16 \,{\left (a^{6} c^{3} x^{3} - a^{5} c^{3} x^{2} - a^{4} c^{3} x + a^{3} c^{3}\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)^(5/2),x, algorithm="fricas")

[Out]

-1/16*((a^4*x^3 - a^3*x^2 - 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*(a^2*x^2 - a*x + 2)*sqrt(-a^2*c))/(a^6*c^3*x^3 - a^5*c^3*x^2 - a^4*c^3*x + a^3*c^3)

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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)**(5/2),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{5}{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)^(5/2),x, algorithm="giac")

[Out]

integrate(x/((-a^2*c*x^2 + c)^(5/2)*sqrt((a*x - 1)/(a*x + 1))), x)