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

Optimal. Leaf size=74 \[ \frac{x^2 (a x+1)}{3 a^2 c^2 \left (1-a^2 x^2\right )^{3/2}}-\frac{3 a x+2}{3 a^4 c^2 \sqrt{1-a^2 x^2}}+\frac{\sin ^{-1}(a x)}{a^4 c^2} \]

[Out]

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

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Rubi [A]  time = 0.103928, antiderivative size = 74, 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, 819, 778, 216} \[ \frac{x^2 (a x+1)}{3 a^2 c^2 \left (1-a^2 x^2\right )^{3/2}}-\frac{3 a x+2}{3 a^4 c^2 \sqrt{1-a^2 x^2}}+\frac{\sin ^{-1}(a x)}{a^4 c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x^2*(1 + a*x))/(3*a^2*c^2*(1 - a^2*x^2)^(3/2)) - (2 + 3*a*x)/(3*a^4*c^2*Sqrt[1 - a^2*x^2]) + ArcSin[a*x]/(a^4
*c^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 819

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x)^(
m - 1)*(a + c*x^2)^(p + 1)*(a*(e*f + d*g) - (c*d*f - a*e*g)*x))/(2*a*c*(p + 1)), x] - Dist[1/(2*a*c*(p + 1)),
Int[(d + e*x)^(m - 2)*(a + c*x^2)^(p + 1)*Simp[a*e*(e*f*(m - 1) + d*g*m) - c*d^2*f*(2*p + 3) + e*(a*e*g*m - c*
d*f*(m + 2*p + 2))*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && GtQ
[m, 1] && (EqQ[d, 0] || (EqQ[m, 2] && EqQ[p, -3] && RationalQ[a, c, d, e, f, g]) ||  !ILtQ[m + 2*p + 3, 0])

Rule 778

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((a*(e*f + d*g) -
(c*d*f - a*e*g)*x)*(a + c*x^2)^(p + 1))/(2*a*c*(p + 1)), x] - Dist[(a*e*g - c*d*f*(2*p + 3))/(2*a*c*(p + 1)),
Int[(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && LtQ[p, -1]

Rule 216

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[(Rt[-b, 2]*x)/Sqrt[a]]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

Rubi steps

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

Mathematica [A]  time = 0.0395464, size = 69, normalized size = 0.93 \[ \frac{-4 a^2 x^2+3 (a x-1) \sqrt{1-a^2 x^2} \sin ^{-1}(a x)+a x+2}{3 a^4 c^2 (a x-1) \sqrt{1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.042, size = 160, normalized size = 2.2 \begin{align*}{\frac{1}{{c}^{2}{a}^{3}}\arctan \left ({x\sqrt{{a}^{2}}{\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}}} \right ){\frac{1}{\sqrt{{a}^{2}}}}}+{\frac{1}{4\,{c}^{2}{a}^{5} \left ( x+{a}^{-1} \right ) }\sqrt{-{a}^{2} \left ( x+{a}^{-1} \right ) ^{2}+2\,a \left ( x+{a}^{-1} \right ) }}+{\frac{1}{6\,{c}^{2}{a}^{6}}\sqrt{-{a}^{2} \left ( x-{a}^{-1} \right ) ^{2}-2\,a \left ( x-{a}^{-1} \right ) } \left ( x-{a}^{-1} \right ) ^{-2}}+{\frac{13}{12\,{c}^{2}{a}^{5}}\sqrt{-{a}^{2} \left ( x-{a}^{-1} \right ) ^{2}-2\,a \left ( x-{a}^{-1} \right ) } \left ( x-{a}^{-1} \right ) ^{-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^3/(-a^2*c*x^2+c)^2,x)

[Out]

1/c^2/a^3/(a^2)^(1/2)*arctan((a^2)^(1/2)*x/(-a^2*x^2+1)^(1/2))+1/4/c^2/a^5/(x+1/a)*(-a^2*(x+1/a)^2+2*a*(x+1/a)
)^(1/2)+1/6/c^2/a^6/(x-1/a)^2*(-a^2*(x-1/a)^2-2*a*(x-1/a))^(1/2)+13/12/c^2/a^5/(x-1/a)*(-a^2*(x-1/a)^2-2*a*(x-
1/a))^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} a \int \frac{x^{4}}{{\left (a^{4} c^{2} x^{4} - 2 \, a^{2} c^{2} x^{2} + c^{2}\right )} \sqrt{a x + 1} \sqrt{-a x + 1}}\,{d x} + \frac{3 \, a^{2} x^{2} - 2}{3 \,{\left (-a^{2} x^{2} + 1\right )}^{\frac{3}{2}} a^{4} 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^3/(-a^2*c*x^2+c)^2,x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 1.59463, size = 281, normalized size = 3.8 \begin{align*} -\frac{2 \, a^{3} x^{3} - 2 \, a^{2} x^{2} - 2 \, a x + 6 \,{\left (a^{3} x^{3} - a^{2} x^{2} - a x + 1\right )} \arctan \left (\frac{\sqrt{-a^{2} x^{2} + 1} - 1}{a x}\right ) -{\left (4 \, a^{2} x^{2} - a x - 2\right )} \sqrt{-a^{2} x^{2} + 1} + 2}{3 \,{\left (a^{7} c^{2} x^{3} - a^{6} c^{2} x^{2} - a^{5} c^{2} x + a^{4} 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^3/(-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 + 6*(a^3*x^3 - a^2*x^2 - a*x + 1)*arctan((sqrt(-a^2*x^2 + 1) - 1)/(a*x)) -
(4*a^2*x^2 - a*x - 2)*sqrt(-a^2*x^2 + 1) + 2)/(a^7*c^2*x^3 - a^6*c^2*x^2 - a^5*c^2*x + a^4*c^2)

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

[Out]

(Integral(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)
+ Integral(a*x**4/(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^{3}}{{\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^3/(-a^2*c*x^2+c)^2,x, algorithm="giac")

[Out]

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