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

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

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

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Rubi [A]  time = 0.185951, antiderivative size = 175, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {6160, 6150, 88} \[ \frac{\left (1-a^2 x^2\right )^{3/2}}{2 a^4 x^3 (a x+1) \left (c-\frac{c}{a^2 x^2}\right )^{3/2}}-\frac{\left (1-a^2 x^2\right )^{3/2}}{a^3 x^2 \left (c-\frac{c}{a^2 x^2}\right )^{3/2}}-\frac{\left (1-a^2 x^2\right )^{3/2} \log (1-a x)}{4 a^4 x^3 \left (c-\frac{c}{a^2 x^2}\right )^{3/2}}+\frac{5 \left (1-a^2 x^2\right )^{3/2} \log (a x+1)}{4 a^4 x^3 \left (c-\frac{c}{a^2 x^2}\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

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) (1+a x)^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 \left (-\frac{1}{a^3}-\frac{1}{4 a^3 (-1+a x)}-\frac{1}{2 a^3 (1+a x)^2}+\frac{5}{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{\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{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}\\ \end{align*}

Mathematica [A]  time = 0.0534217, size = 91, normalized size = 0.52 \[ -\frac{\sqrt{1-a^2 x^2} \left (a^2 x^2-1\right ) \left (-\frac{x}{a^3}+\frac{1}{2 a^4 (a x+1)}-\frac{\log (1-a x)}{4 a^4}+\frac{5 \log (a x+1)}{4 a^4}\right )}{x^3 \left (c-\frac{c}{a^2 x^2}\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

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Maple [A]  time = 0.168, size = 95, normalized size = 0.5 \begin{align*} -{\frac{ \left ( -4\,{a}^{2}{x}^{2}+5\,ax\ln \left ( ax+1 \right ) -\ln \left ( ax-1 \right ) xa-4\,ax+5\,\ln \left ( ax+1 \right ) -\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(1/(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+5*a*x*ln(a*x+1)-ln(a*x-1)*x*a-4*a*x+5*ln(a*x+1)-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)

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

[Out]

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

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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(1/(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)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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