3.209 \(\int e^{-2 \coth ^{-1}(a x)} (c-a c x)^2 \, dx\)

Optimal. Leaf size=55 \[ -\frac{c^2 (1-a x)^3}{3 a}-\frac{c^2 (1-a x)^2}{a}-\frac{8 c^2 \log (a x+1)}{a}+4 c^2 x \]

[Out]

4*c^2*x - (c^2*(1 - a*x)^2)/a - (c^2*(1 - a*x)^3)/(3*a) - (8*c^2*Log[1 + a*x])/a

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Rubi [A]  time = 0.0582331, antiderivative size = 55, 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} \[ -\frac{c^2 (1-a x)^3}{3 a}-\frac{c^2 (1-a x)^2}{a}-\frac{8 c^2 \log (a x+1)}{a}+4 c^2 x \]

Antiderivative was successfully verified.

[In]

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

[Out]

4*c^2*x - (c^2*(1 - a*x)^2)/a - (c^2*(1 - a*x)^3)/(3*a) - (8*c^2*Log[1 + a*x])/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^{-2 \coth ^{-1}(a x)} (c-a c x)^2 \, dx &=-\int e^{-2 \tanh ^{-1}(a x)} (c-a c x)^2 \, dx\\ &=-\left (c^2 \int \frac{(1-a x)^3}{1+a x} \, dx\right )\\ &=-\left (c^2 \int \left (-4-2 (1-a x)-(1-a x)^2+\frac{8}{1+a x}\right ) \, dx\right )\\ &=4 c^2 x-\frac{c^2 (1-a x)^2}{a}-\frac{c^2 (1-a x)^3}{3 a}-\frac{8 c^2 \log (1+a x)}{a}\\ \end{align*}

Mathematica [A]  time = 0.014293, size = 39, normalized size = 0.71 \[ \frac{c^2 \left (a^3 x^3-6 a^2 x^2+21 a x-24 \log (a x+1)-4\right )}{3 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c^2*(-4 + 21*a*x - 6*a^2*x^2 + a^3*x^3 - 24*Log[1 + a*x]))/(3*a)

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Maple [A]  time = 0.04, size = 42, normalized size = 0.8 \begin{align*}{\frac{{a}^{2}{c}^{2}{x}^{3}}{3}}-2\,{c}^{2}{x}^{2}a+7\,x{c}^{2}-8\,{\frac{{c}^{2}\ln \left ( ax+1 \right ) }{a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*a^2*c^2*x^3-2*c^2*x^2*a+7*x*c^2-8*c^2*ln(a*x+1)/a

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Maxima [A]  time = 1.016, size = 55, normalized size = 1. \begin{align*} \frac{1}{3} \, a^{2} c^{2} x^{3} - 2 \, a c^{2} x^{2} + 7 \, c^{2} x - \frac{8 \, c^{2} \log \left (a x + 1\right )}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*a^2*c^2*x^3 - 2*a*c^2*x^2 + 7*c^2*x - 8*c^2*log(a*x + 1)/a

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Fricas [A]  time = 1.41193, size = 97, normalized size = 1.76 \begin{align*} \frac{a^{3} c^{2} x^{3} - 6 \, a^{2} c^{2} x^{2} + 21 \, a c^{2} x - 24 \, c^{2} \log \left (a x + 1\right )}{3 \, a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(a^3*c^2*x^3 - 6*a^2*c^2*x^2 + 21*a*c^2*x - 24*c^2*log(a*x + 1))/a

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Sympy [A]  time = 0.644037, size = 41, normalized size = 0.75 \begin{align*} \frac{a^{2} c^{2} x^{3}}{3} - 2 a c^{2} x^{2} + 7 c^{2} x - \frac{8 c^{2} \log{\left (a x + 1 \right )}}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**2*c**2*x**3/3 - 2*a*c**2*x**2 + 7*c**2*x - 8*c**2*log(a*x + 1)/a

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Giac [A]  time = 1.13615, size = 70, normalized size = 1.27 \begin{align*} -\frac{8 \, c^{2} \log \left ({\left | a x + 1 \right |}\right )}{a} + \frac{a^{5} c^{2} x^{3} - 6 \, a^{4} c^{2} x^{2} + 21 \, a^{3} c^{2} x}{3 \, a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-8*c^2*log(abs(a*x + 1))/a + 1/3*(a^5*c^2*x^3 - 6*a^4*c^2*x^2 + 21*a^3*c^2*x)/a^3