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

Optimal. Leaf size=80 \[ -\frac {2 x}{15 a c^3 \sqrt {1-a^2 x^2}}-\frac {x}{15 a c^3 \left (1-a^2 x^2\right )^{3/2}}+\frac {a x+1}{5 a^2 c^3 \left (1-a^2 x^2\right )^{5/2}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.07, antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {6148, 778, 192, 191} \[ -\frac {2 x}{15 a c^3 \sqrt {1-a^2 x^2}}-\frac {x}{15 a c^3 \left (1-a^2 x^2\right )^{3/2}}+\frac {a x+1}{5 a^2 c^3 \left (1-a^2 x^2\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(1 + a*x)/(5*a^2*c^3*(1 - a^2*x^2)^(5/2)) - x/(15*a*c^3*(1 - a^2*x^2)^(3/2)) - (2*x)/(15*a*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 192

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

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

________________________________________________________________________________________

Mathematica [A]  time = 0.03, size = 68, normalized size = 0.85 \[ \frac {-2 a^4 x^4+2 a^3 x^3+3 a^2 x^2-3 a x+3}{15 a^2 c^3 (a x-1)^2 (a x+1) \sqrt {1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [B]  time = 0.68, size = 145, normalized size = 1.81 \[ \frac {3 \, a^{5} x^{5} - 3 \, a^{4} x^{4} - 6 \, a^{3} x^{3} + 6 \, a^{2} x^{2} + 3 \, a x + {\left (2 \, a^{4} x^{4} - 2 \, a^{3} x^{3} - 3 \, a^{2} x^{2} + 3 \, a x - 3\right )} \sqrt {-a^{2} x^{2} + 1} - 3}{15 \, {\left (a^{7} c^{3} x^{5} - a^{6} c^{3} x^{4} - 2 \, a^{5} c^{3} x^{3} + 2 \, a^{4} c^{3} x^{2} + a^{3} c^{3} x - a^{2} c^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.03, size = 58, normalized size = 0.72 \[ \frac {2 x^{4} a^{4}-2 x^{3} a^{3}-3 a^{2} x^{2}+3 a x -3}{15 \left (a x -1\right ) c^{3} \left (-a^{2} x^{2}+1\right )^{\frac {3}{2}} a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 0.07, size = 271, normalized size = 3.39 \[ \frac {\sqrt {1-a^2\,x^2}}{30\,\left (a^4\,c^3\,x^2-2\,a^3\,c^3\,x+a^2\,c^3\right )}+\frac {\sqrt {1-a^2\,x^2}}{24\,\left (a^4\,c^3\,x^2+2\,a^3\,c^3\,x+a^2\,c^3\right )}-\frac {5\,\sqrt {1-a^2\,x^2}}{48\,\left (c^3\,\sqrt {-a^2}+a\,c^3\,x\,\sqrt {-a^2}\right )\,\sqrt {-a^2}}+\frac {7\,\sqrt {1-a^2\,x^2}}{240\,\left (c^3\,\sqrt {-a^2}-a\,c^3\,x\,\sqrt {-a^2}\right )\,\sqrt {-a^2}}-\frac {\sqrt {1-a^2\,x^2}}{20\,\sqrt {-a^2}\,\left (c^3\,\sqrt {-a^2}+3\,a^2\,c^3\,x^2\,\sqrt {-a^2}-a^3\,c^3\,x^3\,\sqrt {-a^2}-3\,a\,c^3\,x\,\sqrt {-a^2}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*(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)/(30*(a^2*c^3 - 2*a^3*c^3*x + a^4*c^3*x^2)) + (1 - a^2*x^2)^(1/2)/(24*(a^2*c^3 + 2*a^3*c^3*
x + a^4*c^3*x^2)) - (5*(1 - a^2*x^2)^(1/2))/(48*(c^3*(-a^2)^(1/2) + a*c^3*x*(-a^2)^(1/2))*(-a^2)^(1/2)) + (7*(
1 - a^2*x^2)^(1/2))/(240*(c^3*(-a^2)^(1/2) - a*c^3*x*(-a^2)^(1/2))*(-a^2)^(1/2)) - (1 - a^2*x^2)^(1/2)/(20*(-a
^2)^(1/2)*(c^3*(-a^2)^(1/2) + 3*a^2*c^3*x^2*(-a^2)^(1/2) - a^3*c^3*x^3*(-a^2)^(1/2) - 3*a*c^3*x*(-a^2)^(1/2)))

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(Integral(x/(-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**2/(-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

________________________________________________________________________________________