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

Optimal. Leaf size=252 \[ \frac{a^5 x^3 \left (1-\frac{1}{a^2 x^2}\right )^{3/2}}{2 (1-a x) \left (c-a^2 c x^2\right )^{3/2}}-\frac{a^4 x^2 \left (1-\frac{1}{a^2 x^2}\right )^{3/2}}{\left (c-a^2 c x^2\right )^{3/2}}-\frac{a^3 x \left (1-\frac{1}{a^2 x^2}\right )^{3/2}}{2 \left (c-a^2 c x^2\right )^{3/2}}+\frac{2 a^5 x^3 \left (1-\frac{1}{a^2 x^2}\right )^{3/2} \log (x)}{\left (c-a^2 c x^2\right )^{3/2}}-\frac{7 a^5 x^3 \left (1-\frac{1}{a^2 x^2}\right )^{3/2} \log (1-a x)}{4 \left (c-a^2 c x^2\right )^{3/2}}-\frac{a^5 x^3 \left (1-\frac{1}{a^2 x^2}\right )^{3/2} \log (a x+1)}{4 \left (c-a^2 c x^2\right )^{3/2}} \]

[Out]

-(a^3*(1 - 1/(a^2*x^2))^(3/2)*x)/(2*(c - a^2*c*x^2)^(3/2)) - (a^4*(1 - 1/(a^2*x^2))^(3/2)*x^2)/(c - a^2*c*x^2)
^(3/2) + (a^5*(1 - 1/(a^2*x^2))^(3/2)*x^3)/(2*(1 - a*x)*(c - a^2*c*x^2)^(3/2)) + (2*a^5*(1 - 1/(a^2*x^2))^(3/2
)*x^3*Log[x])/(c - a^2*c*x^2)^(3/2) - (7*a^5*(1 - 1/(a^2*x^2))^(3/2)*x^3*Log[1 - a*x])/(4*(c - a^2*c*x^2)^(3/2
)) - (a^5*(1 - 1/(a^2*x^2))^(3/2)*x^3*Log[1 + a*x])/(4*(c - a^2*c*x^2)^(3/2))

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Rubi [A]  time = 0.271751, antiderivative size = 252, 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} \[ \frac{a^5 x^3 \left (1-\frac{1}{a^2 x^2}\right )^{3/2}}{2 (1-a x) \left (c-a^2 c x^2\right )^{3/2}}-\frac{a^4 x^2 \left (1-\frac{1}{a^2 x^2}\right )^{3/2}}{\left (c-a^2 c x^2\right )^{3/2}}-\frac{a^3 x \left (1-\frac{1}{a^2 x^2}\right )^{3/2}}{2 \left (c-a^2 c x^2\right )^{3/2}}+\frac{2 a^5 x^3 \left (1-\frac{1}{a^2 x^2}\right )^{3/2} \log (x)}{\left (c-a^2 c x^2\right )^{3/2}}-\frac{7 a^5 x^3 \left (1-\frac{1}{a^2 x^2}\right )^{3/2} \log (1-a x)}{4 \left (c-a^2 c x^2\right )^{3/2}}-\frac{a^5 x^3 \left (1-\frac{1}{a^2 x^2}\right )^{3/2} \log (a x+1)}{4 \left (c-a^2 c x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^3*(1 - 1/(a^2*x^2))^(3/2)*x)/(2*(c - a^2*c*x^2)^(3/2)) - (a^4*(1 - 1/(a^2*x^2))^(3/2)*x^2)/(c - a^2*c*x^2)
^(3/2) + (a^5*(1 - 1/(a^2*x^2))^(3/2)*x^3)/(2*(1 - a*x)*(c - a^2*c*x^2)^(3/2)) + (2*a^5*(1 - 1/(a^2*x^2))^(3/2
)*x^3*Log[x])/(c - a^2*c*x^2)^(3/2) - (7*a^5*(1 - 1/(a^2*x^2))^(3/2)*x^3*Log[1 - a*x])/(4*(c - a^2*c*x^2)^(3/2
)) - (a^5*(1 - 1/(a^2*x^2))^(3/2)*x^3*Log[1 + a*x])/(4*(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 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^3 \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^6} \, 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{1}{x^3 (-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}{x^3}+\frac{a}{x^2}+\frac{2 a^2}{x}+\frac{a^3}{2 (-1+a x)^2}-\frac{7 a^3}{4 (-1+a x)}-\frac{a^3}{4 (1+a x)}\right ) \, dx}{\left (c-a^2 c x^2\right )^{3/2}}\\ &=-\frac{a^3 \left (1-\frac{1}{a^2 x^2}\right )^{3/2} x}{2 \left (c-a^2 c x^2\right )^{3/2}}-\frac{a^4 \left (1-\frac{1}{a^2 x^2}\right )^{3/2} x^2}{\left (c-a^2 c x^2\right )^{3/2}}+\frac{a^5 \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{2 a^5 \left (1-\frac{1}{a^2 x^2}\right )^{3/2} x^3 \log (x)}{\left (c-a^2 c x^2\right )^{3/2}}-\frac{7 a^5 \left (1-\frac{1}{a^2 x^2}\right )^{3/2} x^3 \log (1-a x)}{4 \left (c-a^2 c x^2\right )^{3/2}}-\frac{a^5 \left (1-\frac{1}{a^2 x^2}\right )^{3/2} x^3 \log (1+a x)}{4 \left (c-a^2 c x^2\right )^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.080199, size = 94, normalized size = 0.37 \[ \frac{a^3 x^3 \left (1-\frac{1}{a^2 x^2}\right )^{3/2} \left (\frac{2 a^2}{1-a x}+8 a^2 \log (x)-7 a^2 \log (1-a x)-a^2 \log (a x+1)-\frac{4 a}{x}-\frac{2}{x^2}\right )}{4 \left (c-a^2 c x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*(1 - 1/(a^2*x^2))^(3/2)*x^3*(-2/x^2 - (4*a)/x + (2*a^2)/(1 - a*x) + 8*a^2*Log[x] - 7*a^2*Log[1 - a*x] - a
^2*Log[1 + a*x]))/(4*(c - a^2*c*x^2)^(3/2))

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Maple [A]  time = 0.143, size = 138, normalized size = 0.6 \begin{align*}{\frac{8\,{a}^{3}\ln \left ( x \right ){x}^{3}-{a}^{3}{x}^{3}\ln \left ( ax+1 \right ) -7\,\ln \left ( ax-1 \right ){x}^{3}{a}^{3}-8\,{a}^{2}\ln \left ( x \right ){x}^{2}+\ln \left ( ax+1 \right ){a}^{2}{x}^{2}+7\,\ln \left ( ax-1 \right ){a}^{2}{x}^{2}-6\,{a}^{2}{x}^{2}+2\,ax+2}{ \left ( 4\,{a}^{2}{x}^{2}-4 \right ){c}^{2}{x}^{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^3/(-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)*(8*a^3*ln(x)*x^3-a^3*x^3*ln(a*x+1)-7*ln(a*x-1)*x^3*a^3-8*a^
2*ln(x)*x^2+ln(a*x+1)*a^2*x^2+7*ln(a*x-1)*a^2*x^2-6*a^2*x^2+2*a*x+2)/(a^2*x^2-1)/c^2/x^2

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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{3}{2}} x^{3} \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)^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.72366, size = 232, normalized size = 0.92 \begin{align*} -\frac{{\left (6 \, a^{2} x^{2} - 2 \, a x +{\left (a^{3} x^{3} - a^{2} x^{2}\right )} \log \left (a x + 1\right ) + 7 \,{\left (a^{3} x^{3} - a^{2} x^{2}\right )} \log \left (a x - 1\right ) - 8 \,{\left (a^{3} x^{3} - a^{2} x^{2}\right )} \log \left (x\right ) - 2\right )} \sqrt{-a^{2} c}}{4 \,{\left (a^{2} c^{2} x^{3} - a c^{2} x^{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^3/(-a^2*c*x^2+c)^(3/2),x, algorithm="fricas")

[Out]

-1/4*(6*a^2*x^2 - 2*a*x + (a^3*x^3 - a^2*x^2)*log(a*x + 1) + 7*(a^3*x^3 - a^2*x^2)*log(a*x - 1) - 8*(a^3*x^3 -
a^2*x^2)*log(x) - 2)*sqrt(-a^2*c)/(a^2*c^2*x^3 - a*c^2*x^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**3/(-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{1}{{\left (-a^{2} c x^{2} + c\right )}^{\frac{3}{2}} x^{3} \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)^(3/2),x, algorithm="giac")

[Out]

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