3.910 \(\int \frac{e^{\tanh ^{-1}(a x)} x^4}{(c-a^2 c x^2)^3} \, dx\)

Optimal. Leaf size=81 \[ \frac{x^4 (a x+1)}{5 a c^3 \left (1-a^2 x^2\right )^{5/2}}+\frac{4}{5 a^5 c^3 \sqrt{1-a^2 x^2}}-\frac{4}{15 a^5 c^3 \left (1-a^2 x^2\right )^{3/2}} \]

[Out]

(x^4*(1 + a*x))/(5*a*c^3*(1 - a^2*x^2)^(5/2)) - 4/(15*a^5*c^3*(1 - a^2*x^2)^(3/2)) + 4/(5*a^5*c^3*Sqrt[1 - a^2
*x^2])

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Rubi [A]  time = 0.116196, antiderivative size = 81, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {6148, 805, 266, 43} \[ \frac{x^4 (a x+1)}{5 a c^3 \left (1-a^2 x^2\right )^{5/2}}+\frac{4}{5 a^5 c^3 \sqrt{1-a^2 x^2}}-\frac{4}{15 a^5 c^3 \left (1-a^2 x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(E^ArcTanh[a*x]*x^4)/(c - a^2*c*x^2)^3,x]

[Out]

(x^4*(1 + a*x))/(5*a*c^3*(1 - a^2*x^2)^(5/2)) - 4/(15*a^5*c^3*(1 - a^2*x^2)^(3/2)) + 4/(5*a^5*c^3*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 805

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

Rule 266

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

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \frac{e^{\tanh ^{-1}(a x)} x^4}{\left (c-a^2 c x^2\right )^3} \, dx &=\frac{\int \frac{x^4 (1+a x)}{\left (1-a^2 x^2\right )^{7/2}} \, dx}{c^3}\\ &=\frac{x^4 (1+a x)}{5 a c^3 \left (1-a^2 x^2\right )^{5/2}}-\frac{4 \int \frac{x^3}{\left (1-a^2 x^2\right )^{5/2}} \, dx}{5 a c^3}\\ &=\frac{x^4 (1+a x)}{5 a c^3 \left (1-a^2 x^2\right )^{5/2}}-\frac{2 \operatorname{Subst}\left (\int \frac{x}{\left (1-a^2 x\right )^{5/2}} \, dx,x,x^2\right )}{5 a c^3}\\ &=\frac{x^4 (1+a x)}{5 a c^3 \left (1-a^2 x^2\right )^{5/2}}-\frac{2 \operatorname{Subst}\left (\int \left (\frac{1}{a^2 \left (1-a^2 x\right )^{5/2}}-\frac{1}{a^2 \left (1-a^2 x\right )^{3/2}}\right ) \, dx,x,x^2\right )}{5 a c^3}\\ &=\frac{x^4 (1+a x)}{5 a c^3 \left (1-a^2 x^2\right )^{5/2}}-\frac{4}{15 a^5 c^3 \left (1-a^2 x^2\right )^{3/2}}+\frac{4}{5 a^5 c^3 \sqrt{1-a^2 x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0263161, size = 68, normalized size = 0.84 \[ \frac{3 a^4 x^4+12 a^3 x^3-12 a^2 x^2-8 a x+8}{15 a^5 c^3 (a x-1)^2 (a x+1) \sqrt{1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(E^ArcTanh[a*x]*x^4)/(c - a^2*c*x^2)^3,x]

[Out]

(8 - 8*a*x - 12*a^2*x^2 + 12*a^3*x^3 + 3*a^4*x^4)/(15*a^5*c^3*(-1 + a*x)^2*(1 + a*x)*Sqrt[1 - a^2*x^2])

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Maple [A]  time = 0.03, size = 58, normalized size = 0.7 \begin{align*} -{\frac{3\,{x}^{4}{a}^{4}+12\,{x}^{3}{a}^{3}-12\,{a}^{2}{x}^{2}-8\,ax+8}{ \left ( 15\,ax-15 \right ){c}^{3}{a}^{5}} \left ( -{a}^{2}{x}^{2}+1 \right ) ^{-{\frac{3}{2}}}} \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^4/(-a^2*c*x^2+c)^3,x)

[Out]

-1/15*(3*a^4*x^4+12*a^3*x^3-12*a^2*x^2-8*a*x+8)/(a*x-1)/c^3/(-a^2*x^2+1)^(3/2)/a^5

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

[Out]

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

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Fricas [B]  time = 1.51358, size = 300, normalized size = 3.7 \begin{align*} \frac{8 \, a^{5} x^{5} - 8 \, a^{4} x^{4} - 16 \, a^{3} x^{3} + 16 \, a^{2} x^{2} + 8 \, a x -{\left (3 \, a^{4} x^{4} + 12 \, a^{3} x^{3} - 12 \, a^{2} x^{2} - 8 \, a x + 8\right )} \sqrt{-a^{2} x^{2} + 1} - 8}{15 \,{\left (a^{10} c^{3} x^{5} - a^{9} c^{3} x^{4} - 2 \, a^{8} c^{3} x^{3} + 2 \, a^{7} c^{3} x^{2} + a^{6} c^{3} x - a^{5} c^{3}\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^4/(-a^2*c*x^2+c)^3,x, algorithm="fricas")

[Out]

1/15*(8*a^5*x^5 - 8*a^4*x^4 - 16*a^3*x^3 + 16*a^2*x^2 + 8*a*x - (3*a^4*x^4 + 12*a^3*x^3 - 12*a^2*x^2 - 8*a*x +
8)*sqrt(-a^2*x^2 + 1) - 8)/(a^10*c^3*x^5 - a^9*c^3*x^4 - 2*a^8*c^3*x^3 + 2*a^7*c^3*x^2 + a^6*c^3*x - a^5*c^3)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{x^{4}}{- a^{6} x^{6} \sqrt{- a^{2} x^{2} + 1} + 3 a^{4} x^{4} \sqrt{- a^{2} x^{2} + 1} - 3 a^{2} x^{2} \sqrt{- a^{2} x^{2} + 1} + \sqrt{- a^{2} x^{2} + 1}}\, dx + \int \frac{a x^{5}}{- a^{6} x^{6} \sqrt{- a^{2} x^{2} + 1} + 3 a^{4} x^{4} \sqrt{- a^{2} x^{2} + 1} - 3 a^{2} x^{2} \sqrt{- a^{2} x^{2} + 1} + \sqrt{- a^{2} x^{2} + 1}}\, dx}{c^{3}} \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**4/(-a**2*c*x**2+c)**3,x)

[Out]

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

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

[Out]

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