3.700 \(\int \frac{e^{\coth ^{-1}(a x)} x^4}{(c-a^2 c x^2)^{5/2}} \, dx\)

Optimal. Leaf size=217 \[ \frac{3 x^5 \left (1-\frac{1}{a^2 x^2}\right )^{5/2}}{4 (1-a x) \left (c-a^2 c x^2\right )^{5/2}}+\frac{x^5 \left (1-\frac{1}{a^2 x^2}\right )^{5/2}}{8 (a x+1) \left (c-a^2 c x^2\right )^{5/2}}-\frac{x^5 \left (1-\frac{1}{a^2 x^2}\right )^{5/2}}{8 (1-a x)^2 \left (c-a^2 c x^2\right )^{5/2}}+\frac{11 x^5 \left (1-\frac{1}{a^2 x^2}\right )^{5/2} \log (1-a x)}{16 \left (c-a^2 c x^2\right )^{5/2}}+\frac{5 x^5 \left (1-\frac{1}{a^2 x^2}\right )^{5/2} \log (a x+1)}{16 \left (c-a^2 c x^2\right )^{5/2}} \]

[Out]

-((1 - 1/(a^2*x^2))^(5/2)*x^5)/(8*(1 - a*x)^2*(c - a^2*c*x^2)^(5/2)) + (3*(1 - 1/(a^2*x^2))^(5/2)*x^5)/(4*(1 -
a*x)*(c - a^2*c*x^2)^(5/2)) + ((1 - 1/(a^2*x^2))^(5/2)*x^5)/(8*(1 + a*x)*(c - a^2*c*x^2)^(5/2)) + (11*(1 - 1/
(a^2*x^2))^(5/2)*x^5*Log[1 - a*x])/(16*(c - a^2*c*x^2)^(5/2)) + (5*(1 - 1/(a^2*x^2))^(5/2)*x^5*Log[1 + a*x])/(
16*(c - a^2*c*x^2)^(5/2))

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Rubi [A]  time = 0.277597, antiderivative size = 217, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.12, Rules used = {6192, 6193, 88} \[ \frac{3 x^5 \left (1-\frac{1}{a^2 x^2}\right )^{5/2}}{4 (1-a x) \left (c-a^2 c x^2\right )^{5/2}}+\frac{x^5 \left (1-\frac{1}{a^2 x^2}\right )^{5/2}}{8 (a x+1) \left (c-a^2 c x^2\right )^{5/2}}-\frac{x^5 \left (1-\frac{1}{a^2 x^2}\right )^{5/2}}{8 (1-a x)^2 \left (c-a^2 c x^2\right )^{5/2}}+\frac{11 x^5 \left (1-\frac{1}{a^2 x^2}\right )^{5/2} \log (1-a x)}{16 \left (c-a^2 c x^2\right )^{5/2}}+\frac{5 x^5 \left (1-\frac{1}{a^2 x^2}\right )^{5/2} \log (a x+1)}{16 \left (c-a^2 c x^2\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(E^ArcCoth[a*x]*x^4)/(c - a^2*c*x^2)^(5/2),x]

[Out]

-((1 - 1/(a^2*x^2))^(5/2)*x^5)/(8*(1 - a*x)^2*(c - a^2*c*x^2)^(5/2)) + (3*(1 - 1/(a^2*x^2))^(5/2)*x^5)/(4*(1 -
a*x)*(c - a^2*c*x^2)^(5/2)) + ((1 - 1/(a^2*x^2))^(5/2)*x^5)/(8*(1 + a*x)*(c - a^2*c*x^2)^(5/2)) + (11*(1 - 1/
(a^2*x^2))^(5/2)*x^5*Log[1 - a*x])/(16*(c - a^2*c*x^2)^(5/2)) + (5*(1 - 1/(a^2*x^2))^(5/2)*x^5*Log[1 + a*x])/(
16*(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]))

Rubi steps

\begin{align*} \int \frac{e^{\coth ^{-1}(a x)} x^4}{\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} \, 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^4}{(-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^4 (-1+a x)^3}+\frac{3}{4 a^4 (-1+a x)^2}+\frac{11}{16 a^4 (-1+a x)}-\frac{1}{8 a^4 (1+a x)^2}+\frac{5}{16 a^4 (1+a x)}\right ) \, dx}{\left (c-a^2 c x^2\right )^{5/2}}\\ &=-\frac{\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{3 \left (1-\frac{1}{a^2 x^2}\right )^{5/2} x^5}{4 (1-a x) \left (c-a^2 c x^2\right )^{5/2}}+\frac{\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{11 \left (1-\frac{1}{a^2 x^2}\right )^{5/2} x^5 \log (1-a x)}{16 \left (c-a^2 c x^2\right )^{5/2}}+\frac{5 \left (1-\frac{1}{a^2 x^2}\right )^{5/2} x^5 \log (1+a x)}{16 \left (c-a^2 c x^2\right )^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.094583, size = 84, normalized size = 0.39 \[ \frac{x^5 \left (1-\frac{1}{a^2 x^2}\right )^{5/2} \left (-\frac{2 \left (5 a^2 x^2+3 a x-6\right )}{(a x-1)^2 (a x+1)}+11 \log (1-a x)+5 \log (a x+1)\right )}{16 \left (c-a^2 c x^2\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(E^ArcCoth[a*x]*x^4)/(c - a^2*c*x^2)^(5/2),x]

[Out]

((1 - 1/(a^2*x^2))^(5/2)*x^5*((-2*(-6 + 3*a*x + 5*a^2*x^2))/((-1 + a*x)^2*(1 + a*x)) + 11*Log[1 - a*x] + 5*Log
[1 + a*x]))/(16*(c - a^2*c*x^2)^(5/2))

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Maple [A]  time = 0.143, size = 169, normalized size = 0.8 \begin{align*} -{\frac{5\,{a}^{3}{x}^{3}\ln \left ( ax+1 \right ) +11\,\ln \left ( ax-1 \right ){x}^{3}{a}^{3}-5\,\ln \left ( ax+1 \right ){a}^{2}{x}^{2}-11\,\ln \left ( ax-1 \right ){a}^{2}{x}^{2}-10\,{a}^{2}{x}^{2}-5\,ax\ln \left ( ax+1 \right ) -11\,\ln \left ( ax-1 \right ) xa-6\,ax+5\,\ln \left ( ax+1 \right ) +11\,\ln \left ( ax-1 \right ) +12}{ \left ( 16\,ax-16 \right ) \left ({a}^{2}{x}^{2}-1 \right ){c}^{3}{a}^{5} \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^4/(-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)*(5*a^3*x^3*ln(a*x+1)+11*ln(a*x-1)*x^3*a^3-5*ln(a*
x+1)*a^2*x^2-11*ln(a*x-1)*a^2*x^2-10*a^2*x^2-5*a*x*ln(a*x+1)-11*ln(a*x-1)*x*a-6*a*x+5*ln(a*x+1)+11*ln(a*x-1)+1
2)/(a^2*x^2-1)/c^3/a^5/(a*x+1)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{4}}{{\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^4/(-a^2*c*x^2+c)^(5/2),x, algorithm="maxima")

[Out]

integrate(x^4/((-a^2*c*x^2 + c)^(5/2)*sqrt((a*x - 1)/(a*x + 1))), x)

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Fricas [A]  time = 1.54577, size = 257, normalized size = 1.18 \begin{align*} \frac{{\left (10 \, a^{2} x^{2} + 6 \, a x - 5 \,{\left (a^{3} x^{3} - a^{2} x^{2} - a x + 1\right )} \log \left (a x + 1\right ) - 11 \,{\left (a^{3} x^{3} - a^{2} x^{2} - a x + 1\right )} \log \left (a x - 1\right ) - 12\right )} \sqrt{-a^{2} c}}{16 \,{\left (a^{9} c^{3} x^{3} - a^{8} c^{3} x^{2} - a^{7} c^{3} x + a^{6} 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^4/(-a^2*c*x^2+c)^(5/2),x, algorithm="fricas")

[Out]

1/16*(10*a^2*x^2 + 6*a*x - 5*(a^3*x^3 - a^2*x^2 - a*x + 1)*log(a*x + 1) - 11*(a^3*x^3 - a^2*x^2 - a*x + 1)*log
(a*x - 1) - 12)*sqrt(-a^2*c)/(a^9*c^3*x^3 - a^8*c^3*x^2 - a^7*c^3*x + a^6*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**4/(-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^{4}}{{\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^4/(-a^2*c*x^2+c)^(5/2),x, algorithm="giac")

[Out]

integrate(x^4/((-a^2*c*x^2 + c)^(5/2)*sqrt((a*x - 1)/(a*x + 1))), x)