3.701 \(\int \frac{e^{\coth ^{-1}(a x)} x^3}{(c-a^2 c x^2)^{5/2}} \, dx\)

Optimal. Leaf size=176 \[ \frac{a x^5 \left (1-\frac{1}{a^2 x^2}\right )^{5/2}}{2 (1-a x) \left (c-a^2 c x^2\right )^{5/2}}-\frac{a x^5 \left (1-\frac{1}{a^2 x^2}\right )^{5/2}}{8 (a x+1) \left (c-a^2 c x^2\right )^{5/2}}-\frac{a x^5 \left (1-\frac{1}{a^2 x^2}\right )^{5/2}}{8 (1-a x)^2 \left (c-a^2 c x^2\right )^{5/2}}-\frac{3 a x^5 \left (1-\frac{1}{a^2 x^2}\right )^{5/2} \tanh ^{-1}(a x)}{8 \left (c-a^2 c x^2\right )^{5/2}} \]

[Out]

-(a*(1 - 1/(a^2*x^2))^(5/2)*x^5)/(8*(1 - a*x)^2*(c - a^2*c*x^2)^(5/2)) + (a*(1 - 1/(a^2*x^2))^(5/2)*x^5)/(2*(1
- a*x)*(c - a^2*c*x^2)^(5/2)) - (a*(1 - 1/(a^2*x^2))^(5/2)*x^5)/(8*(1 + a*x)*(c - a^2*c*x^2)^(5/2)) - (3*a*(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 = 0.267924, antiderivative size = 176, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used = {6192, 6193, 88, 207} \[ \frac{a x^5 \left (1-\frac{1}{a^2 x^2}\right )^{5/2}}{2 (1-a x) \left (c-a^2 c x^2\right )^{5/2}}-\frac{a x^5 \left (1-\frac{1}{a^2 x^2}\right )^{5/2}}{8 (a x+1) \left (c-a^2 c x^2\right )^{5/2}}-\frac{a x^5 \left (1-\frac{1}{a^2 x^2}\right )^{5/2}}{8 (1-a x)^2 \left (c-a^2 c x^2\right )^{5/2}}-\frac{3 a x^5 \left (1-\frac{1}{a^2 x^2}\right )^{5/2} \tanh ^{-1}(a x)}{8 \left (c-a^2 c x^2\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(E^ArcCoth[a*x]*x^3)/(c - a^2*c*x^2)^(5/2),x]

[Out]

-(a*(1 - 1/(a^2*x^2))^(5/2)*x^5)/(8*(1 - a*x)^2*(c - a^2*c*x^2)^(5/2)) + (a*(1 - 1/(a^2*x^2))^(5/2)*x^5)/(2*(1
- a*x)*(c - a^2*c*x^2)^(5/2)) - (a*(1 - 1/(a^2*x^2))^(5/2)*x^5)/(8*(1 + a*x)*(c - a^2*c*x^2)^(5/2)) - (3*a*(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 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

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^3}{\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^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 \frac{x^3}{(-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^3 (-1+a x)^3}+\frac{1}{2 a^3 (-1+a x)^2}+\frac{1}{8 a^3 (1+a x)^2}+\frac{3}{8 a^3 \left (-1+a^2 x^2\right )}\right ) \, dx}{\left (c-a^2 c x^2\right )^{5/2}}\\ &=-\frac{a \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 \left (1-\frac{1}{a^2 x^2}\right )^{5/2} x^5}{2 (1-a x) \left (c-a^2 c x^2\right )^{5/2}}-\frac{a \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 (3 a^2 \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 \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 \left (1-\frac{1}{a^2 x^2}\right )^{5/2} x^5}{2 (1-a x) \left (c-a^2 c x^2\right )^{5/2}}-\frac{a \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{3 a \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 = 0.0641193, size = 86, normalized size = 0.49 \[ -\frac{x \sqrt{1-\frac{1}{a^2 x^2}} \left (5 a^2 x^2-a x+3 (a x-1)^2 (a x+1) \tanh ^{-1}(a x)-2\right )}{8 a^3 c^2 (a x-1)^2 (a x+1) \sqrt{c-a^2 c x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(E^ArcCoth[a*x]*x^3)/(c - a^2*c*x^2)^(5/2),x]

[Out]

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

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Maple [A]  time = 0.142, size = 169, normalized size = 1. \begin{align*}{\frac{3\,{a}^{3}{x}^{3}\ln \left ( ax+1 \right ) -3\,\ln \left ( ax-1 \right ){x}^{3}{a}^{3}-3\,\ln \left ( ax+1 \right ){a}^{2}{x}^{2}+3\,\ln \left ( ax-1 \right ){a}^{2}{x}^{2}+10\,{a}^{2}{x}^{2}-3\,ax\ln \left ( ax+1 \right ) +3\,\ln \left ( ax-1 \right ) xa-2\,ax+3\,\ln \left ( ax+1 \right ) -3\,\ln \left ( ax-1 \right ) -4}{ \left ( 16\,ax-16 \right ) \left ({a}^{2}{x}^{2}-1 \right ){c}^{3}{a}^{4} \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^3/(-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)*(3*a^3*x^3*ln(a*x+1)-3*ln(a*x-1)*x^3*a^3-3*ln(a*x+
1)*a^2*x^2+3*ln(a*x-1)*a^2*x^2+10*a^2*x^2-3*a*x*ln(a*x+1)+3*ln(a*x-1)*x*a-2*a*x+3*ln(a*x+1)-3*ln(a*x-1)-4)/(a^
2*x^2-1)/c^3/a^4/(a*x+1)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{3}}{{\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^3/(-a^2*c*x^2+c)^(5/2),x, algorithm="maxima")

[Out]

integrate(x^3/((-a^2*c*x^2 + c)^(5/2)*sqrt((a*x - 1)/(a*x + 1))), x)

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Fricas [A]  time = 1.61313, size = 275, normalized size = 1.56 \begin{align*} -\frac{3 \,{\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 (5 \, a^{2} x^{2} - a x - 2\right )} \sqrt{-a^{2} c}}{16 \,{\left (a^{8} c^{3} x^{3} - a^{7} c^{3} x^{2} - a^{6} c^{3} x + a^{5} 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^3/(-a^2*c*x^2+c)^(5/2),x, algorithm="fricas")

[Out]

-1/16*(3*(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*(5*a^2*x^2 - a*x - 2)*sqrt(-a^2*c))/(a^8*c^3*x^3 - a^7*c^3*x^2 - a^6*c^3*x + a^5*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**3/(-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^{3}}{{\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^3/(-a^2*c*x^2+c)^(5/2),x, algorithm="giac")

[Out]

integrate(x^3/((-a^2*c*x^2 + c)^(5/2)*sqrt((a*x - 1)/(a*x + 1))), x)