3.223 \(\int \frac{e^{-3 \tanh ^{-1}(a x)}}{(c-a c x)^4} \, dx\)

Optimal. Leaf size=55 \[ \frac{2 x}{3 c^4 \sqrt{1-a^2 x^2}}+\frac{1}{3 a c^4 (1-a x) \sqrt{1-a^2 x^2}} \]

[Out]

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

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Rubi [A]  time = 0.0445643, antiderivative size = 55, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {6127, 659, 191} \[ \frac{2 x}{3 c^4 \sqrt{1-a^2 x^2}}+\frac{1}{3 a c^4 (1-a x) \sqrt{1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Int[1/(E^(3*ArcTanh[a*x])*(c - a*c*x)^4),x]

[Out]

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

Rule 6127

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

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

Mathematica [A]  time = 0.0168939, size = 45, normalized size = 0.82 \[ \frac{2 a^2 x^2-2 a x-1}{3 a c^4 (a x-1) \sqrt{1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(E^(3*ArcTanh[a*x])*(c - a*c*x)^4),x]

[Out]

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

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Maple [A]  time = 0.031, size = 49, normalized size = 0.9 \begin{align*}{\frac{2\,{a}^{2}{x}^{2}-2\,ax-1}{3\, \left ( ax-1 \right ) ^{3}{c}^{4}a \left ( ax+1 \right ) ^{2}} \left ( -{a}^{2}{x}^{2}+1 \right ) ^{{\frac{3}{2}}}} \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*c*x+c)^4,x)

[Out]

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

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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 c x - c\right )}^{4}{\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*c*x+c)^4,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.62778, size = 173, normalized size = 3.15 \begin{align*} \frac{a^{3} x^{3} - a^{2} x^{2} - a x -{\left (2 \, a^{2} x^{2} - 2 \, a x - 1\right )} \sqrt{-a^{2} x^{2} + 1} + 1}{3 \,{\left (a^{4} c^{4} x^{3} - a^{3} c^{4} x^{2} - a^{2} c^{4} x + a c^{4}\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*c*x+c)^4,x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{\sqrt{- a^{2} x^{2} + 1}}{a^{7} x^{7} - a^{6} x^{6} - 3 a^{5} x^{5} + 3 a^{4} x^{4} + 3 a^{3} x^{3} - 3 a^{2} x^{2} - a x + 1}\, dx + \int - \frac{a^{2} x^{2} \sqrt{- a^{2} x^{2} + 1}}{a^{7} x^{7} - a^{6} x^{6} - 3 a^{5} x^{5} + 3 a^{4} x^{4} + 3 a^{3} x^{3} - 3 a^{2} x^{2} - a x + 1}\, dx}{c^{4}} \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*c*x+c)**4,x)

[Out]

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

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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 c x - c\right )}^{4}{\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*c*x+c)^4,x, algorithm="giac")

[Out]

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