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

Optimal. Leaf size=271 \[ -\frac{a^5 x^5 \left (1-\frac{1}{a^2 x^2}\right )^{5/2}}{2 (1-a x) \left (c-a^2 c x^2\right )^{5/2}}-\frac{a^5 x^5 \left (1-\frac{1}{a^2 x^2}\right )^{5/2}}{8 (a x+1) \left (c-a^2 c x^2\right )^{5/2}}-\frac{a^5 x^5 \left (1-\frac{1}{a^2 x^2}\right )^{5/2}}{8 (1-a x)^2 \left (c-a^2 c x^2\right )^{5/2}}-\frac{a^5 x^5 \left (1-\frac{1}{a^2 x^2}\right )^{5/2} \log (x)}{\left (c-a^2 c x^2\right )^{5/2}}+\frac{11 a^5 x^5 \left (1-\frac{1}{a^2 x^2}\right )^{5/2} \log (1-a x)}{16 \left (c-a^2 c x^2\right )^{5/2}}+\frac{5 a^5 x^5 \left (1-\frac{1}{a^2 x^2}\right )^{5/2} \log (a x+1)}{16 \left (c-a^2 c x^2\right )^{5/2}} \]

[Out]

-(a^5*(1 - 1/(a^2*x^2))^(5/2)*x^5)/(8*(1 - a*x)^2*(c - a^2*c*x^2)^(5/2)) - (a^5*(1 - 1/(a^2*x^2))^(5/2)*x^5)/(
2*(1 - a*x)*(c - a^2*c*x^2)^(5/2)) - (a^5*(1 - 1/(a^2*x^2))^(5/2)*x^5)/(8*(1 + a*x)*(c - a^2*c*x^2)^(5/2)) - (
a^5*(1 - 1/(a^2*x^2))^(5/2)*x^5*Log[x])/(c - a^2*c*x^2)^(5/2) + (11*a^5*(1 - 1/(a^2*x^2))^(5/2)*x^5*Log[1 - a*
x])/(16*(c - a^2*c*x^2)^(5/2)) + (5*a^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 = 0.2836, antiderivative size = 271, 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} \[ -\frac{a^5 x^5 \left (1-\frac{1}{a^2 x^2}\right )^{5/2}}{2 (1-a x) \left (c-a^2 c x^2\right )^{5/2}}-\frac{a^5 x^5 \left (1-\frac{1}{a^2 x^2}\right )^{5/2}}{8 (a x+1) \left (c-a^2 c x^2\right )^{5/2}}-\frac{a^5 x^5 \left (1-\frac{1}{a^2 x^2}\right )^{5/2}}{8 (1-a x)^2 \left (c-a^2 c x^2\right )^{5/2}}-\frac{a^5 x^5 \left (1-\frac{1}{a^2 x^2}\right )^{5/2} \log (x)}{\left (c-a^2 c x^2\right )^{5/2}}+\frac{11 a^5 x^5 \left (1-\frac{1}{a^2 x^2}\right )^{5/2} \log (1-a x)}{16 \left (c-a^2 c x^2\right )^{5/2}}+\frac{5 a^5 x^5 \left (1-\frac{1}{a^2 x^2}\right )^{5/2} \log (a x+1)}{16 \left (c-a^2 c x^2\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^5*(1 - 1/(a^2*x^2))^(5/2)*x^5)/(8*(1 - a*x)^2*(c - a^2*c*x^2)^(5/2)) - (a^5*(1 - 1/(a^2*x^2))^(5/2)*x^5)/(
2*(1 - a*x)*(c - a^2*c*x^2)^(5/2)) - (a^5*(1 - 1/(a^2*x^2))^(5/2)*x^5)/(8*(1 + a*x)*(c - a^2*c*x^2)^(5/2)) - (
a^5*(1 - 1/(a^2*x^2))^(5/2)*x^5*Log[x])/(c - a^2*c*x^2)^(5/2) + (11*a^5*(1 - 1/(a^2*x^2))^(5/2)*x^5*Log[1 - a*
x])/(16*(c - a^2*c*x^2)^(5/2)) + (5*a^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 \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^6} \, 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{1}{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}{x}+\frac{a}{4 (-1+a x)^3}-\frac{a}{2 (-1+a x)^2}+\frac{11 a}{16 (-1+a x)}+\frac{a}{8 (1+a x)^2}+\frac{5 a}{16 (1+a x)}\right ) \, dx}{\left (c-a^2 c x^2\right )^{5/2}}\\ &=-\frac{a^5 \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^5 \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^5 \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^5 \left (1-\frac{1}{a^2 x^2}\right )^{5/2} x^5 \log (x)}{\left (c-a^2 c x^2\right )^{5/2}}+\frac{11 a^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}}+\frac{5 a^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 = 0.0961849, size = 88, normalized size = 0.32 \[ \frac{a^5 x^5 \left (1-\frac{1}{a^2 x^2}\right )^{5/2} \left (\frac{8}{a x-1}-\frac{2}{a x+1}-\frac{2}{(a x-1)^2}+11 \log (1-a x)+5 \log (a x+1)-16 \log (x)\right )}{16 \left (c-a^2 c x^2\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^5*(1 - 1/(a^2*x^2))^(5/2)*x^5*(-2/(-1 + a*x)^2 + 8/(-1 + a*x) - 2/(1 + a*x) - 16*Log[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 = 0.139, size = 196, normalized size = 0.7 \begin{align*}{\frac{16\,{a}^{3}\ln \left ( x \right ){x}^{3}-5\,{a}^{3}{x}^{3}\ln \left ( ax+1 \right ) -11\,\ln \left ( ax-1 \right ){x}^{3}{a}^{3}-16\,{a}^{2}\ln \left ( x \right ){x}^{2}+5\,\ln \left ( ax+1 \right ){a}^{2}{x}^{2}+11\,\ln \left ( ax-1 \right ){a}^{2}{x}^{2}-6\,{a}^{2}{x}^{2}-16\,a\ln \left ( x \right ) x+5\,ax\ln \left ( ax+1 \right ) +11\,\ln \left ( ax-1 \right ) xa-2\,ax+16\,\ln \left ( x \right ) -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} \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)*(16*a^3*ln(x)*x^3-5*a^3*x^3*ln(a*x+1)-11*ln(a*x-1)
*x^3*a^3-16*a^2*ln(x)*x^2+5*ln(a*x+1)*a^2*x^2+11*ln(a*x-1)*a^2*x^2-6*a^2*x^2-16*a*ln(x)*x+5*a*x*ln(a*x+1)+11*l
n(a*x-1)*x*a-2*a*x+16*ln(x)-5*ln(a*x+1)-11*ln(a*x-1)+12)/(a^2*x^2-1)/c^3/(a*x+1)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (-a^{2} c x^{2} + c\right )}^{\frac{5}{2}} x \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(1/((-a^2*c*x^2 + c)^(5/2)*x*sqrt((a*x - 1)/(a*x + 1))), x)

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Fricas [A]  time = 1.67067, size = 311, normalized size = 1.15 \begin{align*} -\frac{{\left (6 \, a^{2} x^{2} + 2 \, 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 ) - 16 \,{\left (a^{3} x^{3} - a^{2} x^{2} - a x + 1\right )} \log \left (x\right ) - 12\right )} \sqrt{-a^{2} c}}{16 \,{\left (a^{4} c^{3} x^{3} - a^{3} c^{3} x^{2} - a^{2} c^{3} x + a 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*(6*a^2*x^2 + 2*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) - 16*(a^3*x^3 - a^2*x^2 - a*x + 1)*log(x) - 12)*sqrt(-a^2*c)/(a^4*c^3*x^3 - a^3*c^3*x^2 - a^2*c^3*x
+ a*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{1}{{\left (-a^{2} c x^{2} + c\right )}^{\frac{5}{2}} x \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(1/((-a^2*c*x^2 + c)^(5/2)*x*sqrt((a*x - 1)/(a*x + 1))), x)