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

Optimal. Leaf size=65 \[ \frac{\left (1-a^2 x^2\right )^{3/2}}{15 a c^3 (1-a x)^3}+\frac{\left (1-a^2 x^2\right )^{3/2}}{5 a c^3 (1-a x)^4} \]

[Out]

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

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Rubi [A]  time = 0.0516873, antiderivative size = 65, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.188, Rules used = {6127, 659, 651} \[ \frac{\left (1-a^2 x^2\right )^{3/2}}{15 a c^3 (1-a x)^3}+\frac{\left (1-a^2 x^2\right )^{3/2}}{5 a c^3 (1-a x)^4} \]

Antiderivative was successfully verified.

[In]

Int[E^ArcTanh[a*x]/(c - a*c*x)^3,x]

[Out]

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

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

Mathematica [A]  time = 0.0160793, size = 35, normalized size = 0.54 \[ \frac{(4-a x) (a x+1)^{3/2}}{15 a c^3 (1-a x)^{5/2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [A]  time = 0.033, size = 40, normalized size = 0.6 \begin{align*} -{\frac{ \left ( ax-4 \right ) \left ( ax+1 \right ) ^{2}}{15\, \left ( ax-1 \right ) ^{2}{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((a*x+1)/(-a^2*x^2+1)^(1/2)/(-a*c*x+c)^3,x)

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \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*c*x+c)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.57504, size = 188, normalized size = 2.89 \begin{align*} \frac{4 \, a^{3} x^{3} - 12 \, a^{2} x^{2} + 12 \, a x +{\left (a^{2} x^{2} - 3 \, a x - 4\right )} \sqrt{-a^{2} x^{2} + 1} - 4}{15 \,{\left (a^{4} c^{3} x^{3} - 3 \, 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((a*x+1)/(-a^2*x^2+1)^(1/2)/(-a*c*x+c)^3,x, algorithm="fricas")

[Out]

1/15*(4*a^3*x^3 - 12*a^2*x^2 + 12*a*x + (a^2*x^2 - 3*a*x - 4)*sqrt(-a^2*x^2 + 1) - 4)/(a^4*c^3*x^3 - 3*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{a x}{a^{3} x^{3} \sqrt{- a^{2} x^{2} + 1} - 3 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 + \int \frac{1}{a^{3} x^{3} \sqrt{- a^{2} x^{2} + 1} - 3 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((a*x+1)/(-a**2*x**2+1)**(1/2)/(-a*c*x+c)**3,x)

[Out]

-(Integral(a*x/(a**3*x**3*sqrt(-a**2*x**2 + 1) - 3*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) + Integral(1/(a**3*x**3*sqrt(-a**2*x**2 + 1) - 3*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 [B]  time = 1.16223, size = 196, normalized size = 3.02 \begin{align*} -\frac{2 \,{\left (\frac{5 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}}{a^{2} x} - \frac{25 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}^{2}}{a^{4} x^{2}} + \frac{15 \,{\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}} - 4\right )}}{15 \, c^{3}{\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((a*x+1)/(-a^2*x^2+1)^(1/2)/(-a*c*x+c)^3,x, algorithm="giac")

[Out]

-2/15*(5*(sqrt(-a^2*x^2 + 1)*abs(a) + a)/(a^2*x) - 25*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^2/(a^4*x^2) + 15*(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) - 4)/(c^3*((sqrt(-a^2*x^
2 + 1)*abs(a) + a)/(a^2*x) - 1)^5*abs(a))