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

Optimal. Leaf size=170 \[ -\frac{3 \left (1-a^2 x^2\right )^{3/2}}{a^4 x^3 (a x+1) \left (c-\frac{c}{a^2 x^2}\right )^{3/2}}+\frac{\left (1-a^2 x^2\right )^{3/2}}{a^3 x^2 \left (c-\frac{c}{a^2 x^2}\right )^{3/2}}+\frac{\left (1-a^2 x^2\right )^{3/2}}{2 a^4 x^3 (a x+1)^2 \left (c-\frac{c}{a^2 x^2}\right )^{3/2}}-\frac{3 \left (1-a^2 x^2\right )^{3/2} \log (a x+1)}{a^4 x^3 \left (c-\frac{c}{a^2 x^2}\right )^{3/2}} \]

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

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Rubi [A]  time = 0.197348, antiderivative size = 170, 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, 43} \[ -\frac{3 \left (1-a^2 x^2\right )^{3/2}}{a^4 x^3 (a x+1) \left (c-\frac{c}{a^2 x^2}\right )^{3/2}}+\frac{\left (1-a^2 x^2\right )^{3/2}}{a^3 x^2 \left (c-\frac{c}{a^2 x^2}\right )^{3/2}}+\frac{\left (1-a^2 x^2\right )^{3/2}}{2 a^4 x^3 (a x+1)^2 \left (c-\frac{c}{a^2 x^2}\right )^{3/2}}-\frac{3 \left (1-a^2 x^2\right )^{3/2} \log (a x+1)}{a^4 x^3 \left (c-\frac{c}{a^2 x^2}\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/(E^(3*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)^2) - (3*(1 - a^2*x^2)^(3/2))/(a^4*(c - c/(a^2*x^2))^(3/2)*x^3*(1 + a*x)) - (3*(1 - a^2*x^2)^(3/2)*L
og[1 + a*x])/(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 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

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

Mathematica [A]  time = 0.0557708, size = 86, normalized size = 0.51 \[ -\frac{\sqrt{1-a^2 x^2} \left (a^2 x^2-1\right ) \left (\frac{x}{a^3}-\frac{3}{a^4 (a x+1)}+\frac{1}{2 a^4 (a x+1)^2}-\frac{3 \log (a x+1)}{a^4}\right )}{x^3 \left (c-\frac{c}{a^2 x^2}\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

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Maple [A]  time = 0.177, size = 106, normalized size = 0.6 \begin{align*}{\frac{ \left ( -2\,{x}^{3}{a}^{3}+6\,\ln \left ( ax+1 \right ){a}^{2}{x}^{2}-4\,{a}^{2}{x}^{2}+12\,ax\ln \left ( ax+1 \right ) +4\,ax+6\,\ln \left ( ax+1 \right ) +5 \right ) \left ( ax-1 \right ) }{ \left ( 2\,ax+2 \right ){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)^3*(-a^2*x^2+1)^(3/2)/(c-c/a^2/x^2)^(3/2),x)

[Out]

1/2*(-2*x^3*a^3+6*ln(a*x+1)*a^2*x^2-4*a^2*x^2+12*a*x*ln(a*x+1)+4*a*x+6*ln(a*x+1)+5)*(a*x-1)*(-a^2*x^2+1)^(1/2)
/(a*x+1)/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{{\left (-a^{2} x^{2} + 1\right )}^{\frac{3}{2}}}{{\left (a x + 1\right )}^{3}{\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)^3*(-a^2*x^2+1)^(3/2)/(c-c/a^2/x^2)^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.1977, size = 994, normalized size = 5.85 \begin{align*} \left [-\frac{3 \,{\left (a^{4} x^{4} + 2 \, a^{3} x^{3} - 2 \, a x - 1\right )} \sqrt{-c} \log \left (\frac{a^{6} c x^{6} + 4 \, a^{5} c x^{5} + 5 \, a^{4} c x^{4} - 4 \, a^{2} c x^{2} - 4 \, a c x +{\left (a^{5} x^{5} + 4 \, a^{4} x^{4} + 6 \, a^{3} x^{3} + 4 \, a^{2} x^{2}\right )} \sqrt{-a^{2} x^{2} + 1} \sqrt{-c} \sqrt{\frac{a^{2} c x^{2} - c}{a^{2} x^{2}}} - 2 \, c}{a^{4} x^{4} + 2 \, a^{3} x^{3} - 2 \, a x - 1}\right ) +{\left (2 \, a^{4} x^{4} + 9 \, a^{3} x^{3} + 6 \, a^{2} x^{2}\right )} \sqrt{-a^{2} x^{2} + 1} \sqrt{\frac{a^{2} c x^{2} - c}{a^{2} x^{2}}}}{2 \,{\left (a^{5} c^{2} x^{4} + 2 \, a^{4} c^{2} x^{3} - 2 \, a^{2} c^{2} x - a c^{2}\right )}}, \frac{6 \,{\left (a^{4} x^{4} + 2 \, a^{3} x^{3} - 2 \, a x - 1\right )} \sqrt{c} \arctan \left (\frac{{\left (a^{2} x^{2} + 2 \, a x + 2\right )} \sqrt{-a^{2} x^{2} + 1} \sqrt{c} \sqrt{\frac{a^{2} c x^{2} - c}{a^{2} x^{2}}}}{a^{3} c x^{3} + 2 \, a^{2} c x^{2} - a c x - 2 \, c}\right ) -{\left (2 \, a^{4} x^{4} + 9 \, a^{3} x^{3} + 6 \, a^{2} x^{2}\right )} \sqrt{-a^{2} x^{2} + 1} \sqrt{\frac{a^{2} c x^{2} - c}{a^{2} x^{2}}}}{2 \,{\left (a^{5} c^{2} x^{4} + 2 \, a^{4} c^{2} x^{3} - 2 \, a^{2} c^{2} x - a c^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2*(3*(a^4*x^4 + 2*a^3*x^3 - 2*a*x - 1)*sqrt(-c)*log((a^6*c*x^6 + 4*a^5*c*x^5 + 5*a^4*c*x^4 - 4*a^2*c*x^2 -
4*a*c*x + (a^5*x^5 + 4*a^4*x^4 + 6*a^3*x^3 + 4*a^2*x^2)*sqrt(-a^2*x^2 + 1)*sqrt(-c)*sqrt((a^2*c*x^2 - c)/(a^2
*x^2)) - 2*c)/(a^4*x^4 + 2*a^3*x^3 - 2*a*x - 1)) + (2*a^4*x^4 + 9*a^3*x^3 + 6*a^2*x^2)*sqrt(-a^2*x^2 + 1)*sqrt
((a^2*c*x^2 - c)/(a^2*x^2)))/(a^5*c^2*x^4 + 2*a^4*c^2*x^3 - 2*a^2*c^2*x - a*c^2), 1/2*(6*(a^4*x^4 + 2*a^3*x^3
- 2*a*x - 1)*sqrt(c)*arctan((a^2*x^2 + 2*a*x + 2)*sqrt(-a^2*x^2 + 1)*sqrt(c)*sqrt((a^2*c*x^2 - c)/(a^2*x^2))/(
a^3*c*x^3 + 2*a^2*c*x^2 - a*c*x - 2*c)) - (2*a^4*x^4 + 9*a^3*x^3 + 6*a^2*x^2)*sqrt(-a^2*x^2 + 1)*sqrt((a^2*c*x
^2 - c)/(a^2*x^2)))/(a^5*c^2*x^4 + 2*a^4*c^2*x^3 - 2*a^2*c^2*x - a*c^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)**3*(-a**2*x**2+1)**(3/2)/(c-c/a**2/x**2)**(3/2),x)

[Out]

Timed out

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

[Out]

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