3.195 \(\int \frac{e^{4 \tanh ^{-1}(a x)}}{(c-a c x)^3} \, dx\)

Optimal. Leaf size=52 \[ \frac{1}{2 a c^3 (1-a x)^2}-\frac{4}{3 a c^3 (1-a x)^3}+\frac{1}{a c^3 (1-a x)^4} \]

[Out]

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

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Rubi [A]  time = 0.0431774, antiderivative size = 52, 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} \[ \frac{1}{2 a c^3 (1-a x)^2}-\frac{4}{3 a c^3 (1-a x)^3}+\frac{1}{a c^3 (1-a x)^4} \]

Antiderivative was successfully verified.

[In]

Int[E^(4*ArcTanh[a*x])/(c - a*c*x)^3,x]

[Out]

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

Mathematica [A]  time = 0.0169185, size = 31, normalized size = 0.6 \[ \frac{3 a^2 x^2+2 a x+1}{6 a c^3 (a x-1)^4} \]

Antiderivative was successfully verified.

[In]

Integrate[E^(4*ArcTanh[a*x])/(c - a*c*x)^3,x]

[Out]

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

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Maple [A]  time = 0.032, size = 41, normalized size = 0.8 \begin{align*}{\frac{1}{{c}^{3}} \left ({\frac{1}{2\,a \left ( ax-1 \right ) ^{2}}}+{\frac{4}{3\,a \left ( ax-1 \right ) ^{3}}}+{\frac{1}{a \left ( ax-1 \right ) ^{4}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.951407, size = 88, normalized size = 1.69 \begin{align*} \frac{3 \, a^{2} x^{2} + 2 \, a x + 1}{6 \,{\left (a^{5} c^{3} x^{4} - 4 \, a^{4} c^{3} x^{3} + 6 \, a^{3} c^{3} x^{2} - 4 \, 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)^4/(-a^2*x^2+1)^2/(-a*c*x+c)^3,x, algorithm="maxima")

[Out]

1/6*(3*a^2*x^2 + 2*a*x + 1)/(a^5*c^3*x^4 - 4*a^4*c^3*x^3 + 6*a^3*c^3*x^2 - 4*a^2*c^3*x + a*c^3)

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Fricas [A]  time = 1.55492, size = 131, normalized size = 2.52 \begin{align*} \frac{3 \, a^{2} x^{2} + 2 \, a x + 1}{6 \,{\left (a^{5} c^{3} x^{4} - 4 \, a^{4} c^{3} x^{3} + 6 \, a^{3} c^{3} x^{2} - 4 \, 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)^4/(-a^2*x^2+1)^2/(-a*c*x+c)^3,x, algorithm="fricas")

[Out]

1/6*(3*a^2*x^2 + 2*a*x + 1)/(a^5*c^3*x^4 - 4*a^4*c^3*x^3 + 6*a^3*c^3*x^2 - 4*a^2*c^3*x + a*c^3)

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Sympy [A]  time = 0.5357, size = 66, normalized size = 1.27 \begin{align*} \frac{3 a^{2} x^{2} + 2 a x + 1}{6 a^{5} c^{3} x^{4} - 24 a^{4} c^{3} x^{3} + 36 a^{3} c^{3} x^{2} - 24 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)**4/(-a**2*x**2+1)**2/(-a*c*x+c)**3,x)

[Out]

(3*a**2*x**2 + 2*a*x + 1)/(6*a**5*c**3*x**4 - 24*a**4*c**3*x**3 + 36*a**3*c**3*x**2 - 24*a**2*c**3*x + 6*a*c**
3)

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Giac [A]  time = 1.17137, size = 39, normalized size = 0.75 \begin{align*} \frac{3 \, a^{2} x^{2} + 2 \, a x + 1}{6 \,{\left (a x - 1\right )}^{4} a c^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^4/(-a^2*x^2+1)^2/(-a*c*x+c)^3,x, algorithm="giac")

[Out]

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