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

Optimal. Leaf size=146 \[ -\frac{a^3 x^4 \left (c-\frac{c}{a^2 x^2}\right )^{3/2}}{\left (1-a^2 x^2\right )^{3/2}}+\frac{3 a x^2 \left (c-\frac{c}{a^2 x^2}\right )^{3/2}}{\left (1-a^2 x^2\right )^{3/2}}-\frac{x \left (c-\frac{c}{a^2 x^2}\right )^{3/2}}{2 \left (1-a^2 x^2\right )^{3/2}}+\frac{3 a^2 x^3 \log (x) \left (c-\frac{c}{a^2 x^2}\right )^{3/2}}{\left (1-a^2 x^2\right )^{3/2}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.167182, antiderivative size = 146, 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} \[ -\frac{a^3 x^4 \left (c-\frac{c}{a^2 x^2}\right )^{3/2}}{\left (1-a^2 x^2\right )^{3/2}}+\frac{3 a x^2 \left (c-\frac{c}{a^2 x^2}\right )^{3/2}}{\left (1-a^2 x^2\right )^{3/2}}-\frac{x \left (c-\frac{c}{a^2 x^2}\right )^{3/2}}{2 \left (1-a^2 x^2\right )^{3/2}}+\frac{3 a^2 x^3 \log (x) \left (c-\frac{c}{a^2 x^2}\right )^{3/2}}{\left (1-a^2 x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.0369828, size = 64, normalized size = 0.44 \[ \frac{c \sqrt{c-\frac{c}{a^2 x^2}} \left (2 a^3 x^3-6 a^2 x^2 \log (x)-6 a x+1\right )}{2 a^2 x \sqrt{1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c*Sqrt[c - c/(a^2*x^2)]*(1 - 6*a*x + 2*a^3*x^3 - 6*a^2*x^2*Log[x]))/(2*a^2*x*Sqrt[1 - a^2*x^2])

________________________________________________________________________________________

Maple [A]  time = 0.162, size = 70, normalized size = 0.5 \begin{align*}{\frac{x \left ( -2\,{x}^{3}{a}^{3}+6\,{a}^{2}\ln \left ( x \right ){x}^{2}+6\,ax-1 \right ) }{2\, \left ({a}^{2}{x}^{2}-1 \right ) ^{2}} \left ({\frac{c \left ({a}^{2}{x}^{2}-1 \right ) }{{a}^{2}{x}^{2}}} \right ) ^{{\frac{3}{2}}}\sqrt{-{a}^{2}{x}^{2}+1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(c*(a^2*x^2-1)/a^2/x^2)^(3/2)*x/(a^2*x^2-1)^2*(-a^2*x^2+1)^(1/2)*(-2*x^3*a^3+6*a^2*ln(x)*x^2+6*a*x-1)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 2.08172, size = 779, normalized size = 5.34 \begin{align*} \left [\frac{3 \,{\left (a^{3} c x^{3} - a c x\right )} \sqrt{-c} \log \left (\frac{a^{2} c x^{6} + a^{2} c x^{2} - c x^{4} -{\left (a x^{5} - a x\right )} \sqrt{-a^{2} x^{2} + 1} \sqrt{-c} \sqrt{\frac{a^{2} c x^{2} - c}{a^{2} x^{2}}} - c}{a^{2} x^{4} - x^{2}}\right ) -{\left (2 \, a^{3} c x^{3} -{\left (2 \, a^{3} - 6 \, a + 1\right )} c x^{2} - 6 \, a c x + c\right )} \sqrt{-a^{2} x^{2} + 1} \sqrt{\frac{a^{2} c x^{2} - c}{a^{2} x^{2}}}}{2 \,{\left (a^{4} x^{3} - a^{2} x\right )}}, \frac{6 \,{\left (a^{3} c x^{3} - a c x\right )} \sqrt{c} \arctan \left (\frac{\sqrt{-a^{2} x^{2} + 1}{\left (a x^{3} + a x\right )} \sqrt{c} \sqrt{\frac{a^{2} c x^{2} - c}{a^{2} x^{2}}}}{a^{2} c x^{4} -{\left (a^{2} + 1\right )} c x^{2} + c}\right ) -{\left (2 \, a^{3} c x^{3} -{\left (2 \, a^{3} - 6 \, a + 1\right )} c x^{2} - 6 \, a c x + c\right )} \sqrt{-a^{2} x^{2} + 1} \sqrt{\frac{a^{2} c x^{2} - c}{a^{2} x^{2}}}}{2 \,{\left (a^{4} x^{3} - a^{2} x\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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