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

Optimal. Leaf size=153 \[ -\frac{a x^4 \sqrt{c-\frac{c}{a^2 x^2}}}{3 \sqrt{1-a^2 x^2}}-\frac{3 x^3 \sqrt{c-\frac{c}{a^2 x^2}}}{2 \sqrt{1-a^2 x^2}}-\frac{4 x^2 \sqrt{c-\frac{c}{a^2 x^2}}}{a \sqrt{1-a^2 x^2}}-\frac{4 x \sqrt{c-\frac{c}{a^2 x^2}} \log (1-a x)}{a^2 \sqrt{1-a^2 x^2}} \]

[Out]

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

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Rubi [A]  time = 0.2323, antiderivative size = 153, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {6160, 6150, 77} \[ -\frac{a x^4 \sqrt{c-\frac{c}{a^2 x^2}}}{3 \sqrt{1-a^2 x^2}}-\frac{3 x^3 \sqrt{c-\frac{c}{a^2 x^2}}}{2 \sqrt{1-a^2 x^2}}-\frac{4 x^2 \sqrt{c-\frac{c}{a^2 x^2}}}{a \sqrt{1-a^2 x^2}}-\frac{4 x \sqrt{c-\frac{c}{a^2 x^2}} \log (1-a x)}{a^2 \sqrt{1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*Sqrt[c - c/(a^2*x^2)]*x^2)/(a*Sqrt[1 - a^2*x^2]) - (3*Sqrt[c - c/(a^2*x^2)]*x^3)/(2*Sqrt[1 - a^2*x^2]) - (
a*Sqrt[c - c/(a^2*x^2)]*x^4)/(3*Sqrt[1 - a^2*x^2]) - (4*Sqrt[c - c/(a^2*x^2)]*x*Log[1 - a*x])/(a^2*Sqrt[1 - a^
2*x^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 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rubi steps

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

Mathematica [A]  time = 0.0424301, size = 65, normalized size = 0.42 \[ \frac{x \sqrt{c-\frac{c}{a^2 x^2}} \left (-\frac{4 \log (1-a x)}{a^2}-\frac{a x^3}{3}-\frac{4 x}{a}-\frac{3 x^2}{2}\right )}{\sqrt{1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.141, size = 78, normalized size = 0.5 \begin{align*}{\frac{x \left ( 2\,{x}^{3}{a}^{3}+9\,{a}^{2}{x}^{2}+24\,ax+24\,\ln \left ( ax-1 \right ) \right ) }{ \left ( 6\,{a}^{2}{x}^{2}-6 \right ){a}^{2}}\sqrt{{\frac{c \left ({a}^{2}{x}^{2}-1 \right ) }{{a}^{2}{x}^{2}}}}\sqrt{-{a}^{2}{x}^{2}+1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(c*(a^2*x^2-1)/a^2/x^2)^(1/2)*x*(-a^2*x^2+1)^(1/2)*(2*x^3*a^3+9*a^2*x^2+24*a*x+24*ln(a*x-1))/(a^2*x^2-1)/a
^2

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Maxima [C]  time = 1.32941, size = 232, normalized size = 1.52 \begin{align*} \frac{1}{6} \, a^{3}{\left (\frac{2 \,{\left (i \, a^{2} \sqrt{c} x^{3} + 3 i \, \sqrt{c} x\right )}}{a^{5}} - \frac{3 i \, \sqrt{c} \log \left (a x + 1\right )}{a^{6}} + \frac{3 i \, \sqrt{c} \log \left (a x - 1\right )}{a^{6}}\right )} - \frac{3}{2} \, a^{2}{\left (-\frac{i \, \sqrt{c} x^{2}}{a^{3}} - \frac{i \, \sqrt{c} \log \left (a x + 1\right )}{a^{5}} - \frac{i \, \sqrt{c} \log \left (a x - 1\right )}{a^{5}}\right )} - \frac{3}{2} \, a{\left (-\frac{2 i \, \sqrt{c} x}{a^{3}} + \frac{i \, \sqrt{c} \log \left (a x + 1\right )}{a^{4}} - \frac{i \, \sqrt{c} \log \left (a x - 1\right )}{a^{4}}\right )} + \frac{i \, \sqrt{c} \log \left (a x + 1\right )}{2 \, a^{3}} + \frac{i \, \sqrt{c} \log \left (a x - 1\right )}{2 \, a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.52007, size = 846, normalized size = 5.53 \begin{align*} \left [\frac{12 \,{\left (a^{2} x^{2} - 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} + 24 \, a^{2} x^{2}\right )} \sqrt{-a^{2} x^{2} + 1} \sqrt{\frac{a^{2} c x^{2} - c}{a^{2} x^{2}}}}{6 \,{\left (a^{5} x^{2} - a^{3}\right )}}, \frac{24 \,{\left (a^{2} x^{2} - 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} + 24 \, a^{2} x^{2}\right )} \sqrt{-a^{2} x^{2} + 1} \sqrt{\frac{a^{2} c x^{2} - c}{a^{2} x^{2}}}}{6 \,{\left (a^{5} x^{2} - a^{3}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/6*(12*(a^2*x^2 - 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 + 24*a^2*x^2)*sqrt(-a^2*x^2 + 1)*sqrt((a^2*c*x^2 - c)/(a
^2*x^2)))/(a^5*x^2 - a^3), 1/6*(24*(a^2*x^2 - 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 + 24*a^2*
x^2)*sqrt(-a^2*x^2 + 1)*sqrt((a^2*c*x^2 - c)/(a^2*x^2)))/(a^5*x^2 - a^3)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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