3.1377 \(\int \frac{e^{-3 \tanh ^{-1}(a x)} x^2}{(c-a^2 c x^2)^{11/2}} \, dx\)

Optimal. Leaf size=60 \[ \frac{(3 a x+1) \sqrt{1-a^2 x^2}}{24 a^3 c^5 (1-a x)^3 (a x+1)^6 \sqrt{c-a^2 c x^2}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.236635, antiderivative size = 60, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {6153, 6150, 81} \[ \frac{(3 a x+1) \sqrt{1-a^2 x^2}}{24 a^3 c^5 (1-a x)^3 (a x+1)^6 \sqrt{c-a^2 c x^2}} \]

Antiderivative was successfully verified.

[In]

Int[x^2/(E^(3*ArcTanh[a*x])*(c - a^2*c*x^2)^(11/2)),x]

[Out]

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

Rule 6153

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(c^IntPart[p]*(c +
d*x^2)^FracPart[p])/(1 - a^2*x^2)^FracPart[p], Int[x^m*(1 - a^2*x^2)^p*E^(n*ArcTanh[a*x]), x], x] /; FreeQ[{a,
c, d, m, n, p}, x] && EqQ[a^2*c + d, 0] &&  !(IntegerQ[p] || GtQ[c, 0]) &&  !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 81

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(c + d*
x)^(n + 1)*(e + f*x)^(p + 1)*(2*a*d*f*(n + p + 3) - b*(d*e*(n + 2) + c*f*(p + 2)) + b*d*f*(n + p + 2)*x))/(d^2
*f^2*(n + p + 2)*(n + p + 3)), x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2, 0] && NeQ[n + p + 3,
0] && EqQ[d*f*(n + p + 2)*(a^2*d*f*(n + p + 3) - b*(b*c*e + a*(d*e*(n + 1) + c*f*(p + 1)))) - b*(d*e*(n + 1)
+ c*f*(p + 1))*(a*d*f*(n + p + 4) - b*(d*e*(n + 2) + c*f*(p + 2))), 0]

Rubi steps

\begin{align*} \int \frac{e^{-3 \tanh ^{-1}(a x)} x^2}{\left (c-a^2 c x^2\right )^{11/2}} \, dx &=\frac{\sqrt{1-a^2 x^2} \int \frac{e^{-3 \tanh ^{-1}(a x)} x^2}{\left (1-a^2 x^2\right )^{11/2}} \, dx}{c^5 \sqrt{c-a^2 c x^2}}\\ &=\frac{\sqrt{1-a^2 x^2} \int \frac{x^2}{(1-a x)^4 (1+a x)^7} \, dx}{c^5 \sqrt{c-a^2 c x^2}}\\ &=\frac{(1+3 a x) \sqrt{1-a^2 x^2}}{24 a^3 c^5 (1-a x)^3 (1+a x)^6 \sqrt{c-a^2 c x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0915533, size = 59, normalized size = 0.98 \[ -\frac{(3 a x+1) \sqrt{1-a^2 x^2}}{24 a^3 c^5 (a x-1)^3 (a x+1)^6 \sqrt{c-a^2 c x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2/(E^(3*ArcTanh[a*x])*(c - a^2*c*x^2)^(11/2)),x]

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.032, size = 49, normalized size = 0.8 \begin{align*} -{\frac{ \left ( 3\,ax+1 \right ) \left ( ax-1 \right ) }{24\,{a}^{3} \left ( ax+1 \right ) ^{2}} \left ( -{a}^{2}{x}^{2}+1 \right ) ^{{\frac{3}{2}}} \left ( -{a}^{2}c{x}^{2}+c \right ) ^{-{\frac{11}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(a*x+1)^3*(-a^2*x^2+1)^(3/2)/(-a^2*c*x^2+c)^(11/2),x)

[Out]

-1/24*(a*x-1)*(3*a*x+1)*(-a^2*x^2+1)^(3/2)/(a*x+1)^2/a^3/(-a^2*c*x^2+c)^(11/2)

________________________________________________________________________________________

Maxima [A]  time = 1.1023, size = 126, normalized size = 2.1 \begin{align*} -\frac{3 \, a x + 1}{24 \,{\left (a^{12} c^{\frac{11}{2}} x^{9} + 3 \, a^{11} c^{\frac{11}{2}} x^{8} - 8 \, a^{9} c^{\frac{11}{2}} x^{6} - 6 \, a^{8} c^{\frac{11}{2}} x^{5} + 6 \, a^{7} c^{\frac{11}{2}} x^{4} + 8 \, a^{6} c^{\frac{11}{2}} x^{3} - 3 \, a^{4} c^{\frac{11}{2}} x - a^{3} c^{\frac{11}{2}}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(a*x+1)^3*(-a^2*x^2+1)^(3/2)/(-a^2*c*x^2+c)^(11/2),x, algorithm="maxima")

[Out]

-1/24*(3*a*x + 1)/(a^12*c^(11/2)*x^9 + 3*a^11*c^(11/2)*x^8 - 8*a^9*c^(11/2)*x^6 - 6*a^8*c^(11/2)*x^5 + 6*a^7*c
^(11/2)*x^4 + 8*a^6*c^(11/2)*x^3 - 3*a^4*c^(11/2)*x - a^3*c^(11/2))

________________________________________________________________________________________

Fricas [B]  time = 2.17869, size = 392, normalized size = 6.53 \begin{align*} \frac{{\left (a^{6} x^{9} + 3 \, a^{5} x^{8} - 8 \, a^{3} x^{6} - 6 \, a^{2} x^{5} + 6 \, a x^{4} + 8 \, x^{3}\right )} \sqrt{-a^{2} c x^{2} + c} \sqrt{-a^{2} x^{2} + 1}}{24 \,{\left (a^{11} c^{6} x^{11} + 3 \, a^{10} c^{6} x^{10} - a^{9} c^{6} x^{9} - 11 \, a^{8} c^{6} x^{8} - 6 \, a^{7} c^{6} x^{7} + 14 \, a^{6} c^{6} x^{6} + 14 \, a^{5} c^{6} x^{5} - 6 \, a^{4} c^{6} x^{4} - 11 \, a^{3} c^{6} x^{3} - a^{2} c^{6} x^{2} + 3 \, a c^{6} x + c^{6}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(a*x+1)^3*(-a^2*x^2+1)^(3/2)/(-a^2*c*x^2+c)^(11/2),x, algorithm="fricas")

[Out]

1/24*(a^6*x^9 + 3*a^5*x^8 - 8*a^3*x^6 - 6*a^2*x^5 + 6*a*x^4 + 8*x^3)*sqrt(-a^2*c*x^2 + c)*sqrt(-a^2*x^2 + 1)/(
a^11*c^6*x^11 + 3*a^10*c^6*x^10 - a^9*c^6*x^9 - 11*a^8*c^6*x^8 - 6*a^7*c^6*x^7 + 14*a^6*c^6*x^6 + 14*a^5*c^6*x
^5 - 6*a^4*c^6*x^4 - 11*a^3*c^6*x^3 - a^2*c^6*x^2 + 3*a*c^6*x + c^6)

________________________________________________________________________________________

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/(a*x+1)**3*(-a**2*x**2+1)**(3/2)/(-a**2*c*x**2+c)**(11/2),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}} x^{2}}{{\left (-a^{2} c x^{2} + c\right )}^{\frac{11}{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/(a*x+1)^3*(-a^2*x^2+1)^(3/2)/(-a^2*c*x^2+c)^(11/2),x, algorithm="giac")

[Out]

integrate((-a^2*x^2 + 1)^(3/2)*x^2/((-a^2*c*x^2 + c)^(11/2)*(a*x + 1)^3), x)