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

Optimal. Leaf size=37 \[ \frac{c^3 (1-a x)^4}{4 a}-\frac{2 c^3 (1-a x)^3}{3 a} \]

[Out]

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

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Rubi [A]  time = 0.0347904, 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} \[ \frac{c^3 (1-a x)^4}{4 a}-\frac{2 c^3 (1-a x)^3}{3 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.0135797, size = 30, normalized size = 0.81 \[ \frac{1}{12} c^3 x \left (3 a^3 x^3-4 a^2 x^2-6 a x+12\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c^3*x*(12 - 6*a*x - 4*a^2*x^2 + 3*a^3*x^3))/12

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Maple [A]  time = 0.032, size = 29, normalized size = 0.8 \begin{align*}{c}^{3} \left ({\frac{{x}^{4}{a}^{3}}{4}}-{\frac{{x}^{3}{a}^{2}}{3}}-{\frac{a{x}^{2}}{2}}+x \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]

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

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Maxima [A]  time = 0.957431, size = 50, normalized size = 1.35 \begin{align*} \frac{1}{4} \, a^{3} c^{3} x^{4} - \frac{1}{3} \, a^{2} c^{3} x^{3} - \frac{1}{2} \, a c^{3} x^{2} + c^{3} x \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]

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

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Fricas [A]  time = 1.56118, size = 80, normalized size = 2.16 \begin{align*} \frac{1}{4} \, a^{3} c^{3} x^{4} - \frac{1}{3} \, a^{2} c^{3} x^{3} - \frac{1}{2} \, a c^{3} x^{2} + c^{3} x \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]

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

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Sympy [A]  time = 0.089358, size = 37, normalized size = 1. \begin{align*} \frac{a^{3} c^{3} x^{4}}{4} - \frac{a^{2} c^{3} x^{3}}{3} - \frac{a c^{3} x^{2}}{2} + c^{3} x \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]

a**3*c**3*x**4/4 - a**2*c**3*x**3/3 - a*c**3*x**2/2 + c**3*x

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Giac [A]  time = 1.18426, size = 50, normalized size = 1.35 \begin{align*} \frac{1}{4} \, a^{3} c^{3} x^{4} - \frac{1}{3} \, a^{2} c^{3} x^{3} - \frac{1}{2} \, a c^{3} x^{2} + c^{3} x \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]

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