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

Optimal. Leaf size=187 \[ \frac{a x^5 \sqrt{c-\frac{c}{a^2 x^2}}}{4 \sqrt{1-a^2 x^2}}-\frac{x^4 \sqrt{c-\frac{c}{a^2 x^2}}}{\sqrt{1-a^2 x^2}}+\frac{2 x^3 \sqrt{c-\frac{c}{a^2 x^2}}}{a \sqrt{1-a^2 x^2}}-\frac{4 x^2 \sqrt{c-\frac{c}{a^2 x^2}}}{a^2 \sqrt{1-a^2 x^2}}+\frac{4 x \sqrt{c-\frac{c}{a^2 x^2}} \log (a x+1)}{a^3 \sqrt{1-a^2 x^2}} \]

[Out]

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

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Rubi [A]  time = 0.265933, antiderivative size = 187, 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, 88} \[ \frac{a x^5 \sqrt{c-\frac{c}{a^2 x^2}}}{4 \sqrt{1-a^2 x^2}}-\frac{x^4 \sqrt{c-\frac{c}{a^2 x^2}}}{\sqrt{1-a^2 x^2}}+\frac{2 x^3 \sqrt{c-\frac{c}{a^2 x^2}}}{a \sqrt{1-a^2 x^2}}-\frac{4 x^2 \sqrt{c-\frac{c}{a^2 x^2}}}{a^2 \sqrt{1-a^2 x^2}}+\frac{4 x \sqrt{c-\frac{c}{a^2 x^2}} \log (a x+1)}{a^3 \sqrt{1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*Sqrt[c - c/(a^2*x^2)]*x^2)/(a^2*Sqrt[1 - a^2*x^2]) + (2*Sqrt[c - c/(a^2*x^2)]*x^3)/(a*Sqrt[1 - a^2*x^2]) -
(Sqrt[c - c/(a^2*x^2)]*x^4)/Sqrt[1 - a^2*x^2] + (a*Sqrt[c - c/(a^2*x^2)]*x^5)/(4*Sqrt[1 - a^2*x^2]) + (4*Sqrt
[c - c/(a^2*x^2)]*x*Log[1 + a*x])/(a^3*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 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 e^{-3 \tanh ^{-1}(a x)} \sqrt{c-\frac{c}{a^2 x^2}} x^3 \, dx &=\frac{\left (\sqrt{c-\frac{c}{a^2 x^2}} x\right ) \int e^{-3 \tanh ^{-1}(a x)} x^2 \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^2 (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^2}+\frac{4 x}{a}-3 x^2+a x^3+\frac{4}{a^2 (1+a x)}\right ) \, dx}{\sqrt{1-a^2 x^2}}\\ &=-\frac{4 \sqrt{c-\frac{c}{a^2 x^2}} x^2}{a^2 \sqrt{1-a^2 x^2}}+\frac{2 \sqrt{c-\frac{c}{a^2 x^2}} x^3}{a \sqrt{1-a^2 x^2}}-\frac{\sqrt{c-\frac{c}{a^2 x^2}} x^4}{\sqrt{1-a^2 x^2}}+\frac{a \sqrt{c-\frac{c}{a^2 x^2}} x^5}{4 \sqrt{1-a^2 x^2}}+\frac{4 \sqrt{c-\frac{c}{a^2 x^2}} x \log (1+a x)}{a^3 \sqrt{1-a^2 x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0494406, size = 70, normalized size = 0.37 \[ \frac{x \sqrt{c-\frac{c}{a^2 x^2}} \left (-\frac{4 x}{a^2}+\frac{4 \log (a x+1)}{a^3}+\frac{a x^4}{4}+\frac{2 x^2}{a}-x^3\right )}{\sqrt{1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.138, size = 85, normalized size = 0.5 \begin{align*} -{\frac{x \left ({x}^{4}{a}^{4}-4\,{x}^{3}{a}^{3}+8\,{a}^{2}{x}^{2}-16\,ax+16\,\ln \left ( ax+1 \right ) \right ) }{ \left ( 4\,{a}^{2}{x}^{2}-4 \right ){a}^{3}}\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^3*(c-c/a^2/x^2)^(1/2)/(a*x+1)^3*(-a^2*x^2+1)^(3/2),x)

[Out]

-1/4*(c*(a^2*x^2-1)/a^2/x^2)^(1/2)*x*(-a^2*x^2+1)^(1/2)*(x^4*a^4-4*x^3*a^3+8*a^2*x^2-16*a*x+16*ln(a*x+1))/(a^2
*x^2-1)/a^3

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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^{3}}{{\left (a x + 1\right )}^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(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^3/(a*x + 1)^3, x)

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Fricas [A]  time = 2.51729, size = 873, normalized size = 4.67 \begin{align*} \left [\frac{8 \,{\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 (a^{5} x^{5} - 4 \, a^{4} x^{4} + 8 \, a^{3} x^{3} - 16 \, a^{2} x^{2}\right )} \sqrt{-a^{2} x^{2} + 1} \sqrt{\frac{a^{2} c x^{2} - c}{a^{2} x^{2}}}}{4 \,{\left (a^{6} x^{2} - a^{4}\right )}}, -\frac{16 \,{\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 (a^{5} x^{5} - 4 \, a^{4} x^{4} + 8 \, a^{3} x^{3} - 16 \, a^{2} x^{2}\right )} \sqrt{-a^{2} x^{2} + 1} \sqrt{\frac{a^{2} c x^{2} - c}{a^{2} x^{2}}}}{4 \,{\left (a^{6} x^{2} - a^{4}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*(8*(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)) - (a^5*x^5 - 4*a^4*x^4 + 8*a^3*x^3 - 16*a^2*x^2)*sqrt(-a^2*x^2 + 1)*sqrt((a^2*c*x^
2 - c)/(a^2*x^2)))/(a^6*x^2 - a^4), -1/4*(16*(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)) + (a^5*x^5 - 4*a^4*x^4 +
8*a^3*x^3 - 16*a^2*x^2)*sqrt(-a^2*x^2 + 1)*sqrt((a^2*c*x^2 - c)/(a^2*x^2)))/(a^6*x^2 - a^4)]

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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**3*(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^{3}}{{\left (a x + 1\right )}^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(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^3/(a*x + 1)^3, x)