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

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

[Out]

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

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Rubi [A]  time = 0.0784652, antiderivative size = 65, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.176, Rules used = {6128, 793, 651} \[ \frac{\left (1-a^2 x^2\right )^{3/2}}{5 a^2 c^3 (1-a x)^4}-\frac{4 \left (1-a^2 x^2\right )^{3/2}}{15 a^2 c^3 (1-a x)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 6128

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

Rule 793

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d*g - e*f)*(
d + e*x)^m*(a + c*x^2)^(p + 1))/(2*c*d*(m + p + 1)), x] + Dist[(m*(g*c*d + c*e*f) + 2*e*c*f*(p + 1))/(e*(2*c*d
)*(m + p + 1)), Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && EqQ[c*d^2
+ a*e^2, 0] && ((LtQ[m, -1] &&  !IGtQ[m + p + 1, 0]) || (LtQ[m, 0] && LtQ[p, -1]) || EqQ[m + 2*p + 2, 0]) &&
NeQ[m + p + 1, 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)} x}{(c-a c x)^3} \, dx &=c \int \frac{x \sqrt{1-a^2 x^2}}{(c-a c x)^4} \, dx\\ &=\frac{\left (1-a^2 x^2\right )^{3/2}}{5 a^2 c^3 (1-a x)^4}-\frac{4 \int \frac{\sqrt{1-a^2 x^2}}{(c-a c x)^3} \, dx}{5 a}\\ &=\frac{\left (1-a^2 x^2\right )^{3/2}}{5 a^2 c^3 (1-a x)^4}-\frac{4 \left (1-a^2 x^2\right )^{3/2}}{15 a^2 c^3 (1-a x)^3}\\ \end{align*}

Mathematica [A]  time = 0.0182955, size = 35, normalized size = 0.54 \[ \frac{(a x+1)^{3/2} (4 a x-1)}{15 a^2 c^3 (1-a x)^{5/2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [A]  time = 0.034, size = 41, normalized size = 0.6 \begin{align*}{\frac{ \left ( 4\,ax-1 \right ) \left ( ax+1 \right ) ^{2}}{15\,{c}^{3} \left ( ax-1 \right ) ^{2}{a}^{2}}{\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)*x/(-a*c*x+c)^3,x)

[Out]

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

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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)*x/(-a*c*x+c)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.56598, size = 189, normalized size = 2.91 \begin{align*} -\frac{a^{3} x^{3} - 3 \, a^{2} x^{2} + 3 \, a x +{\left (4 \, a^{2} x^{2} + 3 \, a x - 1\right )} \sqrt{-a^{2} x^{2} + 1} - 1}{15 \,{\left (a^{5} c^{3} x^{3} - 3 \, a^{4} c^{3} x^{2} + 3 \, a^{3} c^{3} x - a^{2} 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)*x/(-a*c*x+c)^3,x, algorithm="fricas")

[Out]

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

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

[Out]

-(Integral(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(a*x**2/(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 = 1.21856, size = 163, normalized size = 2.51 \begin{align*} \frac{2 \,{\left (\frac{5 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}}{a^{2} x} + \frac{5 \,{\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}} - 1\right )}}{15 \, a 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)*x/(-a*c*x+c)^3,x, algorithm="giac")

[Out]

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