3.1198 \(\int \frac {e^{-\tanh ^{-1}(a x)}}{(c-a^2 c x^2)^3} \, dx\)

Optimal. Leaf size=75 \[ \frac {8 x}{15 c^3 \sqrt {1-a^2 x^2}}+\frac {4 x}{15 c^3 \left (1-a^2 x^2\right )^{3/2}}-\frac {1-a x}{5 a c^3 \left (1-a^2 x^2\right )^{5/2}} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {6139, 639, 192, 191} \[ \frac {8 x}{15 c^3 \sqrt {1-a^2 x^2}}+\frac {4 x}{15 c^3 \left (1-a^2 x^2\right )^{3/2}}-\frac {1-a x}{5 a c^3 \left (1-a^2 x^2\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/(E^ArcTanh[a*x]*(c - a^2*c*x^2)^3),x]

[Out]

-(1 - a*x)/(5*a*c^3*(1 - a^2*x^2)^(5/2)) + (4*x)/(15*c^3*(1 - a^2*x^2)^(3/2)) + (8*x)/(15*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 639

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

Rule 6139

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

Rubi steps

\begin {align*} \int \frac {e^{-\tanh ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^3} \, dx &=\frac {\int \frac {1-a x}{\left (1-a^2 x^2\right )^{7/2}} \, dx}{c^3}\\ &=-\frac {1-a x}{5 a c^3 \left (1-a^2 x^2\right )^{5/2}}+\frac {4 \int \frac {1}{\left (1-a^2 x^2\right )^{5/2}} \, dx}{5 c^3}\\ &=-\frac {1-a x}{5 a c^3 \left (1-a^2 x^2\right )^{5/2}}+\frac {4 x}{15 c^3 \left (1-a^2 x^2\right )^{3/2}}+\frac {8 \int \frac {1}{\left (1-a^2 x^2\right )^{3/2}} \, dx}{15 c^3}\\ &=-\frac {1-a x}{5 a c^3 \left (1-a^2 x^2\right )^{5/2}}+\frac {4 x}{15 c^3 \left (1-a^2 x^2\right )^{3/2}}+\frac {8 x}{15 c^3 \sqrt {1-a^2 x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 59, normalized size = 0.79 \[ -\frac {8 a^4 x^4+8 a^3 x^3-12 a^2 x^2-12 a x+3}{15 a c^3 (1-a x)^{3/2} (a x+1)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(E^ArcTanh[a*x]*(c - a^2*c*x^2)^3),x]

[Out]

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

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fricas [B]  time = 0.84, size = 141, normalized size = 1.88 \[ -\frac {3 \, a^{5} x^{5} + 3 \, a^{4} x^{4} - 6 \, a^{3} x^{3} - 6 \, a^{2} x^{2} + 3 \, a x + {\left (8 \, a^{4} x^{4} + 8 \, a^{3} x^{3} - 12 \, a^{2} x^{2} - 12 \, a x + 3\right )} \sqrt {-a^{2} x^{2} + 1} + 3}{15 \, {\left (a^{6} c^{3} x^{5} + a^{5} c^{3} x^{4} - 2 \, a^{4} c^{3} x^{3} - 2 \, a^{3} c^{3} x^{2} + a^{2} c^{3} x + a c^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x+1)*(-a^2*x^2+1)^(1/2)/(-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 + (8*a^4*x^4 + 8*a^3*x^3 - 12*a^2*x^2 - 12*a*x +
3)*sqrt(-a^2*x^2 + 1) + 3)/(a^6*c^3*x^5 + a^5*c^3*x^4 - 2*a^4*c^3*x^3 - 2*a^3*c^3*x^2 + a^2*c^3*x + a*c^3)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.03, size = 58, normalized size = 0.77 \[ -\frac {8 x^{4} a^{4}+8 x^{3} a^{3}-12 a^{2} x^{2}-12 a x +3}{15 \left (-a^{2} x^{2}+1\right )^{\frac {3}{2}} \left (a x +1\right ) c^{3} a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.08, size = 64, normalized size = 0.85 \[ -\frac {\sqrt {1-a^2\,x^2}\,\left (8\,a^4\,x^4+8\,a^3\,x^3-12\,a^2\,x^2-12\,a\,x+3\right )}{15\,a\,c^3\,{\left (a\,x-1\right )}^2\,{\left (a\,x+1\right )}^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {1}{a^{5} x^{5} \sqrt {- a^{2} x^{2} + 1} + a^{4} x^{4} \sqrt {- a^{2} x^{2} + 1} - 2 a^{3} x^{3} \sqrt {- a^{2} x^{2} + 1} - 2 a^{2} x^{2} \sqrt {- a^{2} x^{2} + 1} + a x \sqrt {- a^{2} x^{2} + 1} + \sqrt {- a^{2} x^{2} + 1}}\, dx}{c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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