3.190 \(\int e^{4 \coth ^{-1}(a x)} (c-a c x)^3 \, dx\)

Optimal. Leaf size=35 \[ \frac{2 c^3 (a x+1)^3}{3 a}-\frac{c^3 (a x+1)^4}{4 a} \]

[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 = 0.0585238, antiderivative size = 35, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {6167, 6129, 43} \[ \frac{2 c^3 (a x+1)^3}{3 a}-\frac{c^3 (a x+1)^4}{4 a} \]

Antiderivative was successfully verified.

[In]

Int[E^(4*ArcCoth[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 6167

Int[E^(ArcCoth[(a_.)*(x_)]*(n_))*(u_.), x_Symbol] :> Dist[(-1)^(n/2), Int[u*E^(n*ArcTanh[a*x]), x], x] /; Free
Q[a, x] && IntegerQ[n/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 e^{4 \coth ^{-1}(a x)} (c-a c x)^3 \, dx &=\int e^{4 \tanh ^{-1}(a x)} (c-a c x)^3 \, dx\\ &=c^3 \int (1-a x) (1+a x)^2 \, 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 = 0.0147218, size = 30, normalized size = 0.86 \[ -\frac{1}{12} c^3 x \left (3 a^3 x^3+4 a^2 x^2-6 a x-12\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[E^(4*ArcCoth[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 = 0.04, 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(1/(a*x-1)^2*(a*x+1)^2*(-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 = 1.01588, size = 50, normalized size = 1.43 \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(1/(a*x-1)^2*(a*x+1)^2*(-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 = 1.42124, size = 81, normalized size = 2.31 \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(1/(a*x-1)^2*(a*x+1)^2*(-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 = 0.177027, size = 37, normalized size = 1.06 \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(1/(a*x-1)**2*(a*x+1)**2*(-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 = 1.14638, size = 57, normalized size = 1.63 \begin{align*} -\frac{{\left (3 \, c^{3} + \frac{16 \, c^{3}}{a x - 1} + \frac{24 \, c^{3}}{{\left (a x - 1\right )}^{2}}\right )}{\left (a x - 1\right )}^{4}}{12 \, a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(3*c^3 + 16*c^3/(a*x - 1) + 24*c^3/(a*x - 1)^2)*(a*x - 1)^4/a