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

Optimal. Leaf size=74 \[ \frac{8 x}{15 c^3 \sqrt{1-a^2 x^2}}+\frac{4 x}{15 c^3 \left (1-a^2 x^2\right )^{3/2}}+\frac{a x+1}{5 a c^3 \left (1-a^2 x^2\right )^{5/2}} \]

[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]
)

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Rubi [A]  time = 0.0453206, antiderivative size = 74, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {6138, 639, 192, 191} \[ \frac{8 x}{15 c^3 \sqrt{1-a^2 x^2}}+\frac{4 x}{15 c^3 \left (1-a^2 x^2\right )^{3/2}}+\frac{a x+1}{5 a c^3 \left (1-a^2 x^2\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[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 6138

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] && IGtQ[(n + 1)/2, 0] &&
!IntegerQ[p - n/2]

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 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 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^{\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*}

Mathematica [A]  time = 0.021533, size = 59, normalized size = 0.8 \[ \frac{8 a^4 x^4-8 a^3 x^3-12 a^2 x^2+12 a x+3}{15 a c^3 (1-a x)^{5/2} (a x+1)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.033, size = 58, normalized size = 0.8 \begin{align*} -{\frac{8\,{x}^{4}{a}^{4}-8\,{x}^{3}{a}^{3}-12\,{a}^{2}{x}^{2}+12\,ax+3}{ \left ( 15\,ax-15 \right ){c}^{3}a} \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)/(-a^2*c*x^2+c)^3,x)

[Out]

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

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

[Out]

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

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Fricas [B]  time = 1.52217, size = 293, normalized size = 3.96 \begin{align*} \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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(Integral(a*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(1/(-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{a x + 1}{{\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)/(-a^2*c*x^2+c)^3,x, algorithm="giac")

[Out]

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