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

Optimal. Leaf size=175 \[ \frac{\left (1-a^2 x^2\right )^{3/2}}{2 a^4 x^3 (1-a x) \left (c-\frac{c}{a^2 x^2}\right )^{3/2}}+\frac{\left (1-a^2 x^2\right )^{3/2}}{a^3 x^2 \left (c-\frac{c}{a^2 x^2}\right )^{3/2}}+\frac{5 \left (1-a^2 x^2\right )^{3/2} \log (1-a x)}{4 a^4 x^3 \left (c-\frac{c}{a^2 x^2}\right )^{3/2}}-\frac{\left (1-a^2 x^2\right )^{3/2} \log (a x+1)}{4 a^4 x^3 \left (c-\frac{c}{a^2 x^2}\right )^{3/2}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.184058, antiderivative size = 175, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {6160, 6150, 88} \[ \frac{\left (1-a^2 x^2\right )^{3/2}}{2 a^4 x^3 (1-a x) \left (c-\frac{c}{a^2 x^2}\right )^{3/2}}+\frac{\left (1-a^2 x^2\right )^{3/2}}{a^3 x^2 \left (c-\frac{c}{a^2 x^2}\right )^{3/2}}+\frac{5 \left (1-a^2 x^2\right )^{3/2} \log (1-a x)}{4 a^4 x^3 \left (c-\frac{c}{a^2 x^2}\right )^{3/2}}-\frac{\left (1-a^2 x^2\right )^{3/2} \log (a x+1)}{4 a^4 x^3 \left (c-\frac{c}{a^2 x^2}\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 6160

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)/(x_)^2)^(p_), x_Symbol] :> Dist[(x^(2*p)*(c + d/x^2)^p)/
(1 + (c*x^2)/d)^p, Int[(u*(1 + (c*x^2)/d)^p*E^(n*ArcTanh[a*x]))/x^(2*p), x], x] /; FreeQ[{a, c, d, n, p}, x] &
& EqQ[c + a^2*d, 0] &&  !IntegerQ[p] &&  !IntegerQ[n/2]

Rule 6150

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[c^p, Int[x^m*(1 -
a*x)^(p - n/2)*(1 + a*x)^(p + n/2), x], x] /; FreeQ[{a, c, d, m, n, p}, x] && EqQ[a^2*c + d, 0] && (IntegerQ[p
] || GtQ[c, 0])

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

Mathematica [A]  time = 0.0516773, size = 91, normalized size = 0.52 \[ -\frac{\sqrt{1-a^2 x^2} \left (a^2 x^2-1\right ) \left (\frac{x}{a^3}+\frac{1}{2 a^4 (1-a x)}+\frac{5 \log (1-a x)}{4 a^4}-\frac{\log (a x+1)}{4 a^4}\right )}{x^3 \left (c-\frac{c}{a^2 x^2}\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.174, size = 94, normalized size = 0.5 \begin{align*}{\frac{ \left ( -4\,{a}^{2}{x}^{2}+ax\ln \left ( ax+1 \right ) -5\,\ln \left ( ax-1 \right ) xa+4\,ax-\ln \left ( ax+1 \right ) +5\,\ln \left ( ax-1 \right ) +2 \right ) \left ( ax+1 \right ) }{4\,{a}^{4}{x}^{3}}\sqrt{-{a}^{2}{x}^{2}+1} \left ({\frac{c \left ({a}^{2}{x}^{2}-1 \right ) }{{a}^{2}{x}^{2}}} \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(-4*a^2*x^2+a*x*ln(a*x+1)-5*ln(a*x-1)*x*a+4*a*x-ln(a*x+1)+5*ln(a*x-1)+2)*(a*x+1)*(-a^2*x^2+1)^(1/2)/a^4/x^
3/(c*(a^2*x^2-1)/a^2/x^2)^(3/2)

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{a x + 1}{\sqrt{-a^{2} x^{2} + 1}{\left (c - \frac{c}{a^{2} x^{2}}\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)/(-a^2*x^2+1)^(1/2)/(c-c/a^2/x^2)^(3/2),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{\sqrt{-a^{2} x^{2} + 1} a^{4} x^{4} \sqrt{\frac{a^{2} c x^{2} - c}{a^{2} x^{2}}}}{a^{5} c^{2} x^{5} - a^{4} c^{2} x^{4} - 2 \, a^{3} c^{2} x^{3} + 2 \, a^{2} c^{2} x^{2} + a c^{2} x - c^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)/(-a^2*x^2+1)^(1/2)/(c-c/a^2/x^2)^(3/2),x, algorithm="fricas")

[Out]

integral(-sqrt(-a^2*x^2 + 1)*a^4*x^4*sqrt((a^2*c*x^2 - c)/(a^2*x^2))/(a^5*c^2*x^5 - a^4*c^2*x^4 - 2*a^3*c^2*x^
3 + 2*a^2*c^2*x^2 + a*c^2*x - c^2), x)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{a x + 1}{\sqrt{- \left (a x - 1\right ) \left (a x + 1\right )} \left (- c \left (-1 + \frac{1}{a x}\right ) \left (1 + \frac{1}{a x}\right )\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{a x + 1}{\sqrt{-a^{2} x^{2} + 1}{\left (c - \frac{c}{a^{2} x^{2}}\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)/(-a^2*x^2+1)^(1/2)/(c-c/a^2/x^2)^(3/2),x, algorithm="giac")

[Out]

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