3.175 \(\int \frac{e^{2 \tanh ^{-1}(a x)}}{(c-a c x)^3} \, dx\)

Optimal. Leaf size=37 \[ \frac{2}{3 a c^3 (1-a x)^3}-\frac{1}{2 a c^3 (1-a x)^2} \]

[Out]

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

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Rubi [A]  time = 0.0391528, antiderivative size = 37, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {6129, 43} \[ \frac{2}{3 a c^3 (1-a x)^3}-\frac{1}{2 a c^3 (1-a x)^2} \]

Antiderivative was successfully verified.

[In]

Int[E^(2*ArcTanh[a*x])/(c - a*c*x)^3,x]

[Out]

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

Rule 6129

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

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

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

Mathematica [A]  time = 0.0137785, size = 23, normalized size = 0.62 \[ -\frac{3 a x+1}{6 a c^3 (a x-1)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[E^(2*ArcTanh[a*x])/(c - a*c*x)^3,x]

[Out]

-(1 + 3*a*x)/(6*a*c^3*(-1 + a*x)^3)

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Maple [A]  time = 0.035, size = 30, normalized size = 0.8 \begin{align*}{\frac{1}{{c}^{3}} \left ( -{\frac{1}{2\,a \left ( ax-1 \right ) ^{2}}}-{\frac{2}{3\,a \left ( ax-1 \right ) ^{3}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.948002, size = 63, normalized size = 1.7 \begin{align*} -\frac{3 \, a x + 1}{6 \,{\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)^2/(-a^2*x^2+1)/(-a*c*x+c)^3,x, algorithm="maxima")

[Out]

-1/6*(3*a*x + 1)/(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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Fricas [A]  time = 1.9254, size = 95, normalized size = 2.57 \begin{align*} -\frac{3 \, a x + 1}{6 \,{\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)^2/(-a^2*x^2+1)/(-a*c*x+c)^3,x, algorithm="fricas")

[Out]

-1/6*(3*a*x + 1)/(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 [A]  time = 0.490163, size = 48, normalized size = 1.3 \begin{align*} - \frac{3 a x + 1}{6 a^{4} c^{3} x^{3} - 18 a^{3} c^{3} x^{2} + 18 a^{2} c^{3} x - 6 a c^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(3*a*x + 1)/(6*a**4*c**3*x**3 - 18*a**3*c**3*x**2 + 18*a**2*c**3*x - 6*a*c**3)

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Giac [A]  time = 1.20623, size = 28, normalized size = 0.76 \begin{align*} -\frac{3 \, a x + 1}{6 \,{\left (a x - 1\right )}^{3} a c^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^2/(-a^2*x^2+1)/(-a*c*x+c)^3,x, algorithm="giac")

[Out]

-1/6*(3*a*x + 1)/((a*x - 1)^3*a*c^3)