### 3.584 $$\int e^{4 \coth ^{-1}(a x)} (c-a^2 c x^2)^2 \, dx$$

Optimal. Leaf size=17 $\frac{c^2 (a x+1)^5}{5 a}$

[Out]

(c^2*(1 + a*x)^5)/(5*a)

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Rubi [A]  time = 0.0571792, antiderivative size = 17, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 22, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.136, Rules used = {6167, 6140, 32} $\frac{c^2 (a x+1)^5}{5 a}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(c^2*(1 + a*x)^5)/(5*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 6140

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*((c_) + (d_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[c^p, Int[(1 - a*x)^(p - n/2)*
(1 + a*x)^(p + n/2), x], x] /; FreeQ[{a, c, d, n, p}, x] && EqQ[a^2*c + d, 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)} \left (c-a^2 c x^2\right )^2 \, dx &=\int e^{4 \tanh ^{-1}(a x)} \left (c-a^2 c x^2\right )^2 \, dx\\ &=c^2 \int (1+a x)^4 \, dx\\ &=\frac{c^2 (1+a x)^5}{5 a}\\ \end{align*}

Mathematica [B]  time = 0.0283111, size = 37, normalized size = 2.18 $c^2 \left (\frac{a^4 x^5}{5}+a^3 x^4+2 a^2 x^3+2 a x^2+x\right )$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [A]  time = 0.04, size = 16, normalized size = 0.9 \begin{align*}{\frac{{c}^{2} \left ( ax+1 \right ) ^{5}}{5\,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^2*c*x^2+c)^2,x)

[Out]

1/5*c^2*(a*x+1)^5/a

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Maxima [B]  time = 1.08093, size = 63, normalized size = 3.71 \begin{align*} \frac{1}{5} \, a^{4} c^{2} x^{5} + a^{3} c^{2} x^{4} + 2 \, a^{2} c^{2} x^{3} + 2 \, 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^2*c*x^2+c)^2,x, algorithm="maxima")

[Out]

1/5*a^4*c^2*x^5 + a^3*c^2*x^4 + 2*a^2*c^2*x^3 + 2*a*c^2*x^2 + c^2*x

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Fricas [B]  time = 1.53714, size = 93, normalized size = 5.47 \begin{align*} \frac{1}{5} \, a^{4} c^{2} x^{5} + a^{3} c^{2} x^{4} + 2 \, a^{2} c^{2} x^{3} + 2 \, 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^2*c*x^2+c)^2,x, algorithm="fricas")

[Out]

1/5*a^4*c^2*x^5 + a^3*c^2*x^4 + 2*a^2*c^2*x^3 + 2*a*c^2*x^2 + c^2*x

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Sympy [B]  time = 0.091581, size = 48, normalized size = 2.82 \begin{align*} \frac{a^{4} c^{2} x^{5}}{5} + a^{3} c^{2} x^{4} + 2 a^{2} c^{2} x^{3} + 2 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**2*c*x**2+c)**2,x)

[Out]

a**4*c**2*x**5/5 + a**3*c**2*x**4 + 2*a**2*c**2*x**3 + 2*a*c**2*x**2 + c**2*x

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Giac [B]  time = 1.11551, size = 86, normalized size = 5.06 \begin{align*} \frac{{\left (c^{2} + \frac{10 \, c^{2}}{a x - 1} + \frac{40 \, c^{2}}{{\left (a x - 1\right )}^{2}} + \frac{80 \, c^{2}}{{\left (a x - 1\right )}^{3}} + \frac{80 \, c^{2}}{{\left (a x - 1\right )}^{4}}\right )}{\left (a x - 1\right )}^{5}}{5 \, 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^2*c*x^2+c)^2,x, algorithm="giac")

[Out]

1/5*(c^2 + 10*c^2/(a*x - 1) + 40*c^2/(a*x - 1)^2 + 80*c^2/(a*x - 1)^3 + 80*c^2/(a*x - 1)^4)*(a*x - 1)^5/a