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

Optimal. Leaf size=57 \[ \frac{x^2 (a x+1)}{3 a c^2 \left (1-a^2 x^2\right )^{3/2}}-\frac{2}{3 a^3 c^2 \sqrt{1-a^2 x^2}} \]

[Out]

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

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Rubi [A]  time = 0.093813, antiderivative size = 57, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {6148, 796, 12, 261} \[ \frac{x^2 (a x+1)}{3 a c^2 \left (1-a^2 x^2\right )^{3/2}}-\frac{2}{3 a^3 c^2 \sqrt{1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 6148

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

Rule 796

Int[(x_)^2*((f_) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(x^2*(a*g - c*f*x)*(a + c*x^2)^(p
+ 1))/(2*a*c*(p + 1)), x] - Dist[1/(2*a*c*(p + 1)), Int[x*Simp[2*a*g - c*f*(2*p + 5)*x, x]*(a + c*x^2)^(p + 1
), x], x] /; FreeQ[{a, c, f, g}, x] && EqQ[a*g^2 + f^2*c, 0] && LtQ[p, -2]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 261

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

Rubi steps

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

Mathematica [A]  time = 0.0197114, size = 45, normalized size = 0.79 \[ \frac{-a^2 x^2-2 a x+2}{3 a^3 c^2 (a x-1) \sqrt{1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.029, size = 41, normalized size = 0.7 \begin{align*} -{\frac{{a}^{2}{x}^{2}+2\,ax-2}{ \left ( 3\,ax-3 \right ){c}^{2}{a}^{3}}{\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.62398, size = 182, normalized size = 3.19 \begin{align*} -\frac{2 \, a^{3} x^{3} - 2 \, a^{2} x^{2} - 2 \, a x -{\left (a^{2} x^{2} + 2 \, a x - 2\right )} \sqrt{-a^{2} x^{2} + 1} + 2}{3 \,{\left (a^{6} c^{2} x^{3} - a^{5} c^{2} x^{2} - a^{4} c^{2} x + a^{3} c^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{x^{2}}{a^{4} x^{4} \sqrt{- a^{2} x^{2} + 1} - 2 a^{2} x^{2} \sqrt{- a^{2} x^{2} + 1} + \sqrt{- a^{2} x^{2} + 1}}\, dx + \int \frac{a x^{3}}{a^{4} x^{4} \sqrt{- a^{2} x^{2} + 1} - 2 a^{2} x^{2} \sqrt{- a^{2} x^{2} + 1} + \sqrt{- a^{2} x^{2} + 1}}\, dx}{c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(Integral(x**2/(a**4*x**4*sqrt(-a**2*x**2 + 1) - 2*a**2*x**2*sqrt(-a**2*x**2 + 1) + sqrt(-a**2*x**2 + 1)), x)
+ Integral(a*x**3/(a**4*x**4*sqrt(-a**2*x**2 + 1) - 2*a**2*x**2*sqrt(-a**2*x**2 + 1) + sqrt(-a**2*x**2 + 1)),
x))/c**2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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