### 3.697 $$\int \frac{e^{\coth ^{-1}(a x)}}{x^2 (c-a^2 c x^2)^{3/2}} \, dx$$

Optimal. Leaf size=214 $\frac{a^4 x^3 \left (1-\frac{1}{a^2 x^2}\right )^{3/2}}{2 (1-a x) \left (c-a^2 c x^2\right )^{3/2}}-\frac{a^3 x^2 \left (1-\frac{1}{a^2 x^2}\right )^{3/2}}{\left (c-a^2 c x^2\right )^{3/2}}+\frac{a^4 x^3 \left (1-\frac{1}{a^2 x^2}\right )^{3/2} \log (x)}{\left (c-a^2 c x^2\right )^{3/2}}-\frac{5 a^4 x^3 \left (1-\frac{1}{a^2 x^2}\right )^{3/2} \log (1-a x)}{4 \left (c-a^2 c x^2\right )^{3/2}}+\frac{a^4 x^3 \left (1-\frac{1}{a^2 x^2}\right )^{3/2} \log (a x+1)}{4 \left (c-a^2 c x^2\right )^{3/2}}$

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

________________________________________________________________________________________

Rubi [A]  time = 0.265469, antiderivative size = 214, 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} $\frac{a^4 x^3 \left (1-\frac{1}{a^2 x^2}\right )^{3/2}}{2 (1-a x) \left (c-a^2 c x^2\right )^{3/2}}-\frac{a^3 x^2 \left (1-\frac{1}{a^2 x^2}\right )^{3/2}}{\left (c-a^2 c x^2\right )^{3/2}}+\frac{a^4 x^3 \left (1-\frac{1}{a^2 x^2}\right )^{3/2} \log (x)}{\left (c-a^2 c x^2\right )^{3/2}}-\frac{5 a^4 x^3 \left (1-\frac{1}{a^2 x^2}\right )^{3/2} \log (1-a x)}{4 \left (c-a^2 c x^2\right )^{3/2}}+\frac{a^4 x^3 \left (1-\frac{1}{a^2 x^2}\right )^{3/2} \log (a x+1)}{4 \left (c-a^2 c x^2\right )^{3/2}}$

Antiderivative was successfully veriﬁed.

[In]

Int[E^ArcCoth[a*x]/(x^2*(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^3)/(2*(1 - a*x)*(c
- a^2*c*x^2)^(3/2)) + (a^4*(1 - 1/(a^2*x^2))^(3/2)*x^3*Log[x])/(c - a^2*c*x^2)^(3/2) - (5*a^4*(1 - 1/(a^2*x^2
))^(3/2)*x^3*Log[1 - a*x])/(4*(c - a^2*c*x^2)^(3/2)) + (a^4*(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^2 \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^5} \, 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^2 (-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^2}+\frac{a}{x}+\frac{a^2}{2 (-1+a x)^2}-\frac{5 a^2}{4 (-1+a x)}+\frac{a^2}{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^3}{2 (1-a x) \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^3 \log (x)}{\left (c-a^2 c x^2\right )^{3/2}}-\frac{5 a^4 \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^4 \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 = 0.0651484, size = 79, normalized size = 0.37 $\frac{a^3 x^3 \left (1-\frac{1}{a^2 x^2}\right )^{3/2} \left (\frac{2 a}{1-a x}+4 a \log (x)-5 a \log (1-a x)+a \log (a x+1)-\frac{4}{x}\right )}{4 \left (c-a^2 c x^2\right )^{3/2}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.136, size = 118, normalized size = 0.6 \begin{align*}{\frac{4\,{a}^{2}\ln \left ( x \right ){x}^{2}+\ln \left ( ax+1 \right ){a}^{2}{x}^{2}-5\,\ln \left ( ax-1 \right ){a}^{2}{x}^{2}-4\,a\ln \left ( x \right ) x-ax\ln \left ( ax+1 \right ) +5\,\ln \left ( ax-1 \right ) xa-6\,ax+4}{ \left ( 4\,{a}^{2}{x}^{2}-4 \right ){c}^{2}x}\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^2/(-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)*(4*a^2*ln(x)*x^2+ln(a*x+1)*a^2*x^2-5*ln(a*x-1)*a^2*x^2-4*a*
ln(x)*x-a*x*ln(a*x+1)+5*ln(a*x-1)*x*a-6*a*x+4)/(a^2*x^2-1)/c^2/x

________________________________________________________________________________________

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^{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^2/(-a^2*c*x^2+c)^(3/2),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 1.72626, size = 197, normalized size = 0.92 \begin{align*} -\frac{\sqrt{-a^{2} c}{\left (6 \, a x -{\left (a^{2} x^{2} - a x\right )} \log \left (a x + 1\right ) + 5 \,{\left (a^{2} x^{2} - a x\right )} \log \left (a x - 1\right ) - 4 \,{\left (a^{2} x^{2} - a x\right )} \log \left (x\right ) - 4\right )}}{4 \,{\left (a^{2} c^{2} x^{2} - a c^{2} x\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^2/(-a^2*c*x^2+c)^(3/2),x, algorithm="fricas")

[Out]

-1/4*sqrt(-a^2*c)*(6*a*x - (a^2*x^2 - a*x)*log(a*x + 1) + 5*(a^2*x^2 - a*x)*log(a*x - 1) - 4*(a^2*x^2 - a*x)*l
og(x) - 4)/(a^2*c^2*x^2 - a*c^2*x)

________________________________________________________________________________________

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**2/(-a**2*c*x**2+c)**(3/2),x)

[Out]

Timed out

________________________________________________________________________________________

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^{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^2/(-a^2*c*x^2+c)^(3/2),x, algorithm="giac")

[Out]

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