### 3.191 $$\int e^{4 \coth ^{-1}(a x)} (c-a c x)^2 \, dx$$

Optimal. Leaf size=17 $\frac{c^2 (a x+1)^3}{3 a}$

[Out]

(c^2*(1 + a*x)^3)/(3*a)

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Rubi [A]  time = 0.0464065, antiderivative size = 17, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 18, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.167, Rules used = {6167, 6129, 32} $\frac{c^2 (a x+1)^3}{3 a}$

Antiderivative was successfully veriﬁed.

[In]

Int[E^(4*ArcCoth[a*x])*(c - a*c*x)^2,x]

[Out]

(c^2*(1 + a*x)^3)/(3*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 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rubi steps

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

Mathematica [A]  time = 0.0125278, size = 21, normalized size = 1.24 $c^2 \left (\frac{a^2 x^3}{3}+a x^2+x\right )$

Antiderivative was successfully veriﬁed.

[In]

Integrate[E^(4*ArcCoth[a*x])*(c - a*c*x)^2,x]

[Out]

c^2*(x + a*x^2 + (a^2*x^3)/3)

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Maple [A]  time = 0.039, size = 16, normalized size = 0.9 \begin{align*}{\frac{{c}^{2} \left ( ax+1 \right ) ^{3}}{3\,a}} \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)^2,x)

[Out]

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

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Maxima [A]  time = 1.0172, size = 34, normalized size = 2. \begin{align*} \frac{1}{3} \, a^{2} c^{2} x^{3} + a c^{2} x^{2} + c^{2} 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)^2,x, algorithm="maxima")

[Out]

1/3*a^2*c^2*x^3 + a*c^2*x^2 + c^2*x

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Fricas [A]  time = 1.48649, size = 50, normalized size = 2.94 \begin{align*} \frac{1}{3} \, a^{2} c^{2} x^{3} + a c^{2} x^{2} + c^{2} 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)^2,x, algorithm="fricas")

[Out]

1/3*a^2*c^2*x^3 + a*c^2*x^2 + c^2*x

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Sympy [A]  time = 0.156255, size = 24, normalized size = 1.41 \begin{align*} \frac{a^{2} c^{2} x^{3}}{3} + a c^{2} x^{2} + c^{2} 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)**2,x)

[Out]

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

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Giac [B]  time = 1.17359, size = 54, normalized size = 3.18 \begin{align*} \frac{{\left (c^{2} + \frac{6 \, c^{2}}{a x - 1} + \frac{12 \, c^{2}}{{\left (a x - 1\right )}^{2}}\right )}{\left (a x - 1\right )}^{3}}{3 \, 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)^2,x, algorithm="giac")

[Out]

1/3*(c^2 + 6*c^2/(a*x - 1) + 12*c^2/(a*x - 1)^2)*(a*x - 1)^3/a