3.1260 \(\int \frac{e^{-3 \tanh ^{-1}(a x)}}{(c-a^2 c x^2)^3} \, dx\)

Optimal. Leaf size=116 \[ \frac{8 x}{35 c^3 \sqrt{1-a^2 x^2}}-\frac{4}{35 a c^3 (a x+1) \sqrt{1-a^2 x^2}}-\frac{4}{35 a c^3 (a x+1)^2 \sqrt{1-a^2 x^2}}-\frac{1}{7 a c^3 (a x+1)^3 \sqrt{1-a^2 x^2}} \]

[Out]

(8*x)/(35*c^3*Sqrt[1 - a^2*x^2]) - 1/(7*a*c^3*(1 + a*x)^3*Sqrt[1 - a^2*x^2]) - 4/(35*a*c^3*(1 + a*x)^2*Sqrt[1
- a^2*x^2]) - 4/(35*a*c^3*(1 + a*x)*Sqrt[1 - a^2*x^2])

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Rubi [A]  time = 0.084844, antiderivative size = 116, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {6139, 655, 659, 191} \[ \frac{8 x}{35 c^3 \sqrt{1-a^2 x^2}}-\frac{4}{35 a c^3 (a x+1) \sqrt{1-a^2 x^2}}-\frac{4}{35 a c^3 (a x+1)^2 \sqrt{1-a^2 x^2}}-\frac{1}{7 a c^3 (a x+1)^3 \sqrt{1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(8*x)/(35*c^3*Sqrt[1 - a^2*x^2]) - 1/(7*a*c^3*(1 + a*x)^3*Sqrt[1 - a^2*x^2]) - 4/(35*a*c^3*(1 + a*x)^2*Sqrt[1
- a^2*x^2]) - 4/(35*a*c^3*(1 + a*x)*Sqrt[1 - a^2*x^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]

Rule 655

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

Rule 659

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[(e*(d + e*x)^m*(a + c*x^2)^(p + 1)
)/(2*c*d*(m + p + 1)), x] + Dist[Simplify[m + 2*p + 2]/(2*d*(m + p + 1)), Int[(d + e*x)^(m + 1)*(a + c*x^2)^p,
x], x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && ILtQ[Simplify[m + 2*p + 2
], 0]

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]

Rubi steps

\begin{align*} \int \frac{e^{-3 \tanh ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^3} \, dx &=\frac{\int \frac{(1-a x)^3}{\left (1-a^2 x^2\right )^{9/2}} \, dx}{c^3}\\ &=\frac{\int \frac{1}{(1+a x)^3 \left (1-a^2 x^2\right )^{3/2}} \, dx}{c^3}\\ &=-\frac{1}{7 a c^3 (1+a x)^3 \sqrt{1-a^2 x^2}}+\frac{4 \int \frac{1}{(1+a x)^2 \left (1-a^2 x^2\right )^{3/2}} \, dx}{7 c^3}\\ &=-\frac{1}{7 a c^3 (1+a x)^3 \sqrt{1-a^2 x^2}}-\frac{4}{35 a c^3 (1+a x)^2 \sqrt{1-a^2 x^2}}+\frac{12 \int \frac{1}{(1+a x) \left (1-a^2 x^2\right )^{3/2}} \, dx}{35 c^3}\\ &=-\frac{1}{7 a c^3 (1+a x)^3 \sqrt{1-a^2 x^2}}-\frac{4}{35 a c^3 (1+a x)^2 \sqrt{1-a^2 x^2}}-\frac{4}{35 a c^3 (1+a x) \sqrt{1-a^2 x^2}}+\frac{8 \int \frac{1}{\left (1-a^2 x^2\right )^{3/2}} \, dx}{35 c^3}\\ &=\frac{8 x}{35 c^3 \sqrt{1-a^2 x^2}}-\frac{1}{7 a c^3 (1+a x)^3 \sqrt{1-a^2 x^2}}-\frac{4}{35 a c^3 (1+a x)^2 \sqrt{1-a^2 x^2}}-\frac{4}{35 a c^3 (1+a x) \sqrt{1-a^2 x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0231819, size = 59, normalized size = 0.51 \[ \frac{8 a^4 x^4+24 a^3 x^3+20 a^2 x^2-4 a x-13}{35 a c^3 \sqrt{1-a x} (a x+1)^{7/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-13 - 4*a*x + 20*a^2*x^2 + 24*a^3*x^3 + 8*a^4*x^4)/(35*a*c^3*Sqrt[1 - a*x]*(1 + a*x)^(7/2))

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Maple [A]  time = 0.03, size = 58, normalized size = 0.5 \begin{align*}{\frac{8\,{x}^{4}{a}^{4}+24\,{x}^{3}{a}^{3}+20\,{a}^{2}{x}^{2}-4\,ax-13}{35\, \left ( ax+1 \right ) ^{3}{c}^{3}a}{\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/35/(-a^2*x^2+1)^(1/2)*(8*a^4*x^4+24*a^3*x^3+20*a^2*x^2-4*a*x-13)/(a*x+1)^3/c^3/a

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.72811, size = 309, normalized size = 2.66 \begin{align*} -\frac{13 \, a^{5} x^{5} + 39 \, a^{4} x^{4} + 26 \, a^{3} x^{3} - 26 \, a^{2} x^{2} - 39 \, a x +{\left (8 \, a^{4} x^{4} + 24 \, a^{3} x^{3} + 20 \, a^{2} x^{2} - 4 \, a x - 13\right )} \sqrt{-a^{2} x^{2} + 1} - 13}{35 \,{\left (a^{6} c^{3} x^{5} + 3 \, a^{5} c^{3} x^{4} + 2 \, a^{4} c^{3} x^{3} - 2 \, a^{3} c^{3} x^{2} - 3 \, a^{2} c^{3} x - a c^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/35*(13*a^5*x^5 + 39*a^4*x^4 + 26*a^3*x^3 - 26*a^2*x^2 - 39*a*x + (8*a^4*x^4 + 24*a^3*x^3 + 20*a^2*x^2 - 4*a
*x - 13)*sqrt(-a^2*x^2 + 1) - 13)/(a^6*c^3*x^5 + 3*a^5*c^3*x^4 + 2*a^4*c^3*x^3 - 2*a^3*c^3*x^2 - 3*a^2*c^3*x -
a*c^3)

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

[Out]

Integral(1/(-a**5*x**5*sqrt(-a**2*x**2 + 1) - 3*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) + 3*a*x*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^{2} x^{2} + 1\right )}^{\frac{3}{2}}}{{\left (a^{2} c x^{2} - c\right )}^{3}{\left (a x + 1\right )}^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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