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

Optimal. Leaf size=152 \[ \frac{a x^4 \sqrt{c-\frac{c}{a^2 x^2}}}{3 \sqrt{1-a^2 x^2}}-\frac{3 x^3 \sqrt{c-\frac{c}{a^2 x^2}}}{2 \sqrt{1-a^2 x^2}}+\frac{4 x^2 \sqrt{c-\frac{c}{a^2 x^2}}}{a \sqrt{1-a^2 x^2}}-\frac{4 x \sqrt{c-\frac{c}{a^2 x^2}} \log (a x+1)}{a^2 \sqrt{1-a^2 x^2}} \]

[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 = 0.240186, antiderivative size = 152, 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} \[ \frac{a x^4 \sqrt{c-\frac{c}{a^2 x^2}}}{3 \sqrt{1-a^2 x^2}}-\frac{3 x^3 \sqrt{c-\frac{c}{a^2 x^2}}}{2 \sqrt{1-a^2 x^2}}+\frac{4 x^2 \sqrt{c-\frac{c}{a^2 x^2}}}{a \sqrt{1-a^2 x^2}}-\frac{4 x \sqrt{c-\frac{c}{a^2 x^2}} \log (a x+1)}{a^2 \sqrt{1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(Sqrt[c - c/(a^2*x^2)]*x^2)/E^(3*ArcTanh[a*x]),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 = 0.0462516, size = 64, normalized size = 0.42 \[ \frac{x \sqrt{c-\frac{c}{a^2 x^2}} \left (-\frac{4 \log (a x+1)}{a^2}+\frac{a x^3}{3}+\frac{4 x}{a}-\frac{3 x^2}{2}\right )}{\sqrt{1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sqrt[c - c/(a^2*x^2)]*x^2)/E^(3*ArcTanh[a*x]),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 = 0.139, 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(x^2*(c-c/a^2/x^2)^(1/2)/(a*x+1)^3*(-a^2*x^2+1)^(3/2),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 [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (-a^{2} x^{2} + 1\right )}^{\frac{3}{2}} \sqrt{c - \frac{c}{a^{2} x^{2}}} x^{2}}{{\left (a x + 1\right )}^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.45616, size = 846, normalized size = 5.57 \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(x^2*(c-c/a^2/x^2)^(1/2)/(a*x+1)^3*(-a^2*x^2+1)^(3/2),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(-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(x**2*(c-c/a**2/x**2)**(1/2)/(a*x+1)**3*(-a**2*x**2+1)**(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}} \sqrt{c - \frac{c}{a^{2} x^{2}}} x^{2}}{{\left (a x + 1\right )}^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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