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

Optimal. Leaf size=74 \[ \frac {a x^5}{5 c^3 \left (1-a^2 x^2\right )^{5/2}}-\frac {1}{3 a^4 c^3 \left (1-a^2 x^2\right )^{3/2}}+\frac {1}{5 a^4 c^3 \left (1-a^2 x^2\right )^{5/2}} \]

[Out]

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

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Rubi [A]  time = 0.11, antiderivative size = 88, normalized size of antiderivative = 1.19, 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, 191} \[ \frac {x^2 (a x+1)}{5 a^2 c^3 \left (1-a^2 x^2\right )^{5/2}}+\frac {x}{5 a^3 c^3 \sqrt {1-a^2 x^2}}-\frac {3 a x+2}{15 a^4 c^3 \left (1-a^2 x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 191

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

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 68, normalized size = 0.92 \[ \frac {3 a^4 x^4-3 a^3 x^3+3 a^2 x^2+2 a x-2}{15 a^4 c^3 (a x-1)^2 (a x+1) \sqrt {1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.55, size = 145, normalized size = 1.96 \[ -\frac {2 \, a^{5} x^{5} - 2 \, a^{4} x^{4} - 4 \, a^{3} x^{3} + 4 \, a^{2} x^{2} + 2 \, a x + {\left (3 \, a^{4} x^{4} - 3 \, a^{3} x^{3} + 3 \, a^{2} x^{2} + 2 \, a x - 2\right )} \sqrt {-a^{2} x^{2} + 1} - 2}{15 \, {\left (a^{9} c^{3} x^{5} - a^{8} c^{3} x^{4} - 2 \, a^{7} c^{3} x^{3} + 2 \, a^{6} c^{3} x^{2} + a^{5} c^{3} x - a^{4} c^{3}\right )}} \]

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)^3,x, algorithm="fricas")

[Out]

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

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

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)^3,x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:sym2
poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [A]  time = 0.03, size = 58, normalized size = 0.78 \[ -\frac {3 x^{4} a^{4}-3 x^{3} a^{3}+3 a^{2} x^{2}+2 a x -2}{15 \left (a x -1\right ) c^{3} \left (-a^{2} x^{2}+1\right )^{\frac {3}{2}} a^{4}} \]

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)^3,x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -a \int \frac {x^{4}}{{\left (a^{6} c^{3} x^{6} - 3 \, a^{4} c^{3} x^{4} + 3 \, a^{2} c^{3} x^{2} - c^{3}\right )} \sqrt {a x + 1} \sqrt {-a x + 1}}\,{d x} + \frac {5 \, a^{2} x^{2} - 2}{15 \, {\left (-a^{2} x^{2} + 1\right )}^{\frac {5}{2}} a^{4} c^{3}} \]

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)^3,x, algorithm="maxima")

[Out]

-a*integrate(x^4/((a^6*c^3*x^6 - 3*a^4*c^3*x^4 + 3*a^2*c^3*x^2 - c^3)*sqrt(a*x + 1)*sqrt(-a*x + 1)), x) + 1/15
*(5*a^2*x^2 - 2)/((-a^2*x^2 + 1)^(5/2)*a^4*c^3)

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mupad [B]  time = 0.08, size = 287, normalized size = 3.88 \[ \frac {\sqrt {1-a^2\,x^2}}{24\,\left (a^6\,c^3\,x^2+2\,a^5\,c^3\,x+a^4\,c^3\right )}-\frac {2\,\sqrt {1-a^2\,x^2}}{15\,\left (a^6\,c^3\,x^2-2\,a^5\,c^3\,x+a^4\,c^3\right )}-\frac {\sqrt {1-a^2\,x^2}}{20\,\sqrt {-a^2}\,\left (a^2\,c^3\,\sqrt {-a^2}+3\,a^4\,c^3\,x^2\,\sqrt {-a^2}-a^5\,c^3\,x^3\,\sqrt {-a^2}-3\,a^3\,c^3\,x\,\sqrt {-a^2}\right )}+\frac {7\,\sqrt {1-a^2\,x^2}}{48\,\left (a^2\,c^3\,\sqrt {-a^2}+a^3\,c^3\,x\,\sqrt {-a^2}\right )\,\sqrt {-a^2}}-\frac {13\,\sqrt {1-a^2\,x^2}}{240\,\left (a^2\,c^3\,\sqrt {-a^2}-a^3\,c^3\,x\,\sqrt {-a^2}\right )\,\sqrt {-a^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(1 - a^2*x^2)^(1/2)/(24*(a^4*c^3 + 2*a^5*c^3*x + a^6*c^3*x^2)) - (2*(1 - a^2*x^2)^(1/2))/(15*(a^4*c^3 - 2*a^5*
c^3*x + a^6*c^3*x^2)) - (1 - a^2*x^2)^(1/2)/(20*(-a^2)^(1/2)*(a^2*c^3*(-a^2)^(1/2) + 3*a^4*c^3*x^2*(-a^2)^(1/2
) - a^5*c^3*x^3*(-a^2)^(1/2) - 3*a^3*c^3*x*(-a^2)^(1/2))) + (7*(1 - a^2*x^2)^(1/2))/(48*(a^2*c^3*(-a^2)^(1/2)
+ a^3*c^3*x*(-a^2)^(1/2))*(-a^2)^(1/2)) - (13*(1 - a^2*x^2)^(1/2))/(240*(a^2*c^3*(-a^2)^(1/2) - a^3*c^3*x*(-a^
2)^(1/2))*(-a^2)^(1/2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {x^{3}}{- a^{6} x^{6} \sqrt {- a^{2} x^{2} + 1} + 3 a^{4} x^{4} \sqrt {- a^{2} x^{2} + 1} - 3 a^{2} x^{2} \sqrt {- a^{2} x^{2} + 1} + \sqrt {- a^{2} x^{2} + 1}}\, dx + \int \frac {a x^{4}}{- a^{6} x^{6} \sqrt {- a^{2} x^{2} + 1} + 3 a^{4} x^{4} \sqrt {- a^{2} x^{2} + 1} - 3 a^{2} x^{2} \sqrt {- a^{2} x^{2} + 1} + \sqrt {- a^{2} x^{2} + 1}}\, dx}{c^{3}} \]

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)**3,x)

[Out]

(Integral(x**3/(-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**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))/c**3

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