3.224 \(\int \frac{e^{-3 \tanh ^{-1}(a x)}}{(c-a c x)^5} \, dx\)

Optimal. Leaf size=87 \[ \frac{2 x}{5 c^5 \sqrt{1-a^2 x^2}}+\frac{1}{5 a c^5 (1-a x) \sqrt{1-a^2 x^2}}+\frac{1}{5 a c^5 (1-a x)^2 \sqrt{1-a^2 x^2}} \]

[Out]

(2*x)/(5*c^5*Sqrt[1 - a^2*x^2]) + 1/(5*a*c^5*(1 - a*x)^2*Sqrt[1 - a^2*x^2]) + 1/(5*a*c^5*(1 - a*x)*Sqrt[1 - a^
2*x^2])

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Rubi [A]  time = 0.0615994, antiderivative size = 87, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {6127, 659, 191} \[ \frac{2 x}{5 c^5 \sqrt{1-a^2 x^2}}+\frac{1}{5 a c^5 (1-a x) \sqrt{1-a^2 x^2}}+\frac{1}{5 a c^5 (1-a x)^2 \sqrt{1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Int[1/(E^(3*ArcTanh[a*x])*(c - a*c*x)^5),x]

[Out]

(2*x)/(5*c^5*Sqrt[1 - a^2*x^2]) + 1/(5*a*c^5*(1 - a*x)^2*Sqrt[1 - a^2*x^2]) + 1/(5*a*c^5*(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)^5} \, dx &=\frac{\int \frac{1}{(c-a c x)^2 \left (1-a^2 x^2\right )^{3/2}} \, dx}{c^3}\\ &=\frac{1}{5 a c^5 (1-a x)^2 \sqrt{1-a^2 x^2}}+\frac{3 \int \frac{1}{(c-a c x) \left (1-a^2 x^2\right )^{3/2}} \, dx}{5 c^4}\\ &=\frac{1}{5 a c^5 (1-a x)^2 \sqrt{1-a^2 x^2}}+\frac{1}{5 a c^5 (1-a x) \sqrt{1-a^2 x^2}}+\frac{2 \int \frac{1}{\left (1-a^2 x^2\right )^{3/2}} \, dx}{5 c^5}\\ &=\frac{2 x}{5 c^5 \sqrt{1-a^2 x^2}}+\frac{1}{5 a c^5 (1-a x)^2 \sqrt{1-a^2 x^2}}+\frac{1}{5 a c^5 (1-a x) \sqrt{1-a^2 x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0193726, size = 52, normalized size = 0.6 \[ \frac{2 a^3 x^3-4 a^2 x^2+a x+2}{5 a c^5 (a x-1)^2 \sqrt{1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(E^(3*ArcTanh[a*x])*(c - a*c*x)^5),x]

[Out]

(2 + a*x - 4*a^2*x^2 + 2*a^3*x^3)/(5*a*c^5*(-1 + a*x)^2*Sqrt[1 - a^2*x^2])

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Maple [A]  time = 0.033, size = 56, normalized size = 0.6 \begin{align*}{\frac{2\,{x}^{3}{a}^{3}-4\,{a}^{2}{x}^{2}+ax+2}{5\, \left ( ax-1 \right ) ^{4}{c}^{5}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)^5,x)

[Out]

1/5*(-a^2*x^2+1)^(3/2)*(2*a^3*x^3-4*a^2*x^2+a*x+2)/(a*x-1)^4/c^5/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 )}^{5}{\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)^5,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.61503, size = 200, normalized size = 2.3 \begin{align*} \frac{2 \, a^{4} x^{4} - 4 \, a^{3} x^{3} + 4 \, a x -{\left (2 \, a^{3} x^{3} - 4 \, a^{2} x^{2} + a x + 2\right )} \sqrt{-a^{2} x^{2} + 1} - 2}{5 \,{\left (a^{5} c^{5} x^{4} - 2 \, a^{4} c^{5} x^{3} + 2 \, a^{2} c^{5} x - a c^{5}\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)^5,x, algorithm="fricas")

[Out]

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

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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^{8} x^{8} - 2 a^{7} x^{7} - 2 a^{6} x^{6} + 6 a^{5} x^{5} - 6 a^{3} x^{3} + 2 a^{2} x^{2} + 2 a x - 1}\, dx + \int - \frac{a^{2} x^{2} \sqrt{- a^{2} x^{2} + 1}}{a^{8} x^{8} - 2 a^{7} x^{7} - 2 a^{6} x^{6} + 6 a^{5} x^{5} - 6 a^{3} x^{3} + 2 a^{2} x^{2} + 2 a x - 1}\, dx}{c^{5}} \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)**5,x)

[Out]

-(Integral(sqrt(-a**2*x**2 + 1)/(a**8*x**8 - 2*a**7*x**7 - 2*a**6*x**6 + 6*a**5*x**5 - 6*a**3*x**3 + 2*a**2*x*
*2 + 2*a*x - 1), x) + Integral(-a**2*x**2*sqrt(-a**2*x**2 + 1)/(a**8*x**8 - 2*a**7*x**7 - 2*a**6*x**6 + 6*a**5
*x**5 - 6*a**3*x**3 + 2*a**2*x**2 + 2*a*x - 1), x))/c**5

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Giac [C]  time = 1.38634, size = 235, normalized size = 2.7 \begin{align*} \frac{1}{40} \,{\left (a{\left (\frac{5}{a^{3} c^{7} \sqrt{-\frac{2 \, c}{a c x - c} - 1}} - \frac{a^{12} c^{28}{\left (\frac{2 \, c}{a c x - c} + 1\right )}^{2} \sqrt{-\frac{2 \, c}{a c x - c} - 1} + 5 \, a^{12} c^{28}{\left (-\frac{2 \, c}{a c x - c} - 1\right )}^{\frac{3}{2}} + 15 \, a^{12} c^{28} \sqrt{-\frac{2 \, c}{a c x - c} - 1}}{a^{15} c^{35}}\right )} \mathrm{sgn}\left (\frac{1}{a c x - c}\right ) \mathrm{sgn}\left (a\right ) \mathrm{sgn}\left (c\right ) + \frac{16 i \, \mathrm{sgn}\left (\frac{1}{a c x - c}\right ) \mathrm{sgn}\left (a\right ) \mathrm{sgn}\left (c\right )}{a^{2} c^{7}}\right )} c^{2}{\left | a \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)^5,x, algorithm="giac")

[Out]

1/40*(a*(5/(a^3*c^7*sqrt(-2*c/(a*c*x - c) - 1)) - (a^12*c^28*(2*c/(a*c*x - c) + 1)^2*sqrt(-2*c/(a*c*x - c) - 1
) + 5*a^12*c^28*(-2*c/(a*c*x - c) - 1)^(3/2) + 15*a^12*c^28*sqrt(-2*c/(a*c*x - c) - 1))/(a^15*c^35))*sgn(1/(a*
c*x - c))*sgn(a)*sgn(c) + 16*I*sgn(1/(a*c*x - c))*sgn(a)*sgn(c)/(a^2*c^7))*c^2*abs(a)