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

Optimal. Leaf size=97 \[ \frac{2 \sqrt{1-a^2 x^2}}{15 a c^4 (1-a x)}+\frac{2 \sqrt{1-a^2 x^2}}{15 a c^4 (1-a x)^2}+\frac{\sqrt{1-a^2 x^2}}{5 a c^4 (1-a x)^3} \]

[Out]

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

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Rubi [A]  time = 0.0681872, antiderivative size = 97, 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, 651} \[ \frac{2 \sqrt{1-a^2 x^2}}{15 a c^4 (1-a x)}+\frac{2 \sqrt{1-a^2 x^2}}{15 a c^4 (1-a x)^2}+\frac{\sqrt{1-a^2 x^2}}{5 a c^4 (1-a x)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 651

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*(p + 1)), x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && EqQ[m + 2*p
+ 2, 0]

Rubi steps

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

Mathematica [A]  time = 0.0200372, size = 43, normalized size = 0.44 \[ \frac{\sqrt{a x+1} \left (2 a^2 x^2-6 a x+7\right )}{15 a c^4 (1-a x)^{5/2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(Sqrt[1 + a*x]*(7 - 6*a*x + 2*a^2*x^2))/(15*a*c^4*(1 - a*x)^(5/2))

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Maple [A]  time = 0.03, size = 42, normalized size = 0.4 \begin{align*} -{\frac{2\,{a}^{2}{x}^{2}-6\,ax+7}{15\, \left ( ax-1 \right ) ^{3}{c}^{4}a}\sqrt{-{a}^{2}{x}^{2}+1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/15*(-a^2*x^2+1)^(1/2)*(2*a^2*x^2-6*a*x+7)/(a*x-1)^3/c^4/a

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{-a^{2} x^{2} + 1}}{{\left (a c x - c\right )}^{4}{\left (a x + 1\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x+1)*(-a^2*x^2+1)^(1/2)/(-a*c*x+c)^4,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.70404, size = 190, normalized size = 1.96 \begin{align*} \frac{7 \, a^{3} x^{3} - 21 \, a^{2} x^{2} + 21 \, a x -{\left (2 \, a^{2} x^{2} - 6 \, a x + 7\right )} \sqrt{-a^{2} x^{2} + 1} - 7}{15 \,{\left (a^{4} c^{4} x^{3} - 3 \, a^{3} c^{4} x^{2} + 3 \, 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)*(-a^2*x^2+1)^(1/2)/(-a*c*x+c)^4,x, algorithm="fricas")

[Out]

1/15*(7*a^3*x^3 - 21*a^2*x^2 + 21*a*x - (2*a^2*x^2 - 6*a*x + 7)*sqrt(-a^2*x^2 + 1) - 7)/(a^4*c^4*x^3 - 3*a^3*c
^4*x^2 + 3*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^{5} x^{5} - 3 a^{4} x^{4} + 2 a^{3} x^{3} + 2 a^{2} x^{2} - 3 a x + 1}\, dx}{c^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.2111, size = 196, normalized size = 2.02 \begin{align*} -\frac{2 \,{\left (\frac{20 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}}{a^{2} x} - \frac{40 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}^{2}}{a^{4} x^{2}} + \frac{30 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}^{3}}{a^{6} x^{3}} - \frac{15 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}^{4}}{a^{8} x^{4}} - 7\right )}}{15 \, c^{4}{\left (\frac{\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a}{a^{2} x} - 1\right )}^{5}{\left | a \right |}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x+1)*(-a^2*x^2+1)^(1/2)/(-a*c*x+c)^4,x, algorithm="giac")

[Out]

-2/15*(20*(sqrt(-a^2*x^2 + 1)*abs(a) + a)/(a^2*x) - 40*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^2/(a^4*x^2) + 30*(sqrt(
-a^2*x^2 + 1)*abs(a) + a)^3/(a^6*x^3) - 15*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^4/(a^8*x^4) - 7)/(c^4*((sqrt(-a^2*x
^2 + 1)*abs(a) + a)/(a^2*x) - 1)^5*abs(a))