### 3.567 $$\int e^{2 \coth ^{-1}(a x)} (c-a^2 c x^2)^2 \, dx$$

Optimal. Leaf size=35 $\frac{c^2 (a x+1)^5}{5 a}-\frac{c^2 (a x+1)^4}{2 a}$

[Out]

-(c^2*(1 + a*x)^4)/(2*a) + (c^2*(1 + a*x)^5)/(5*a)

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Rubi [A]  time = 0.0668577, antiderivative size = 35, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 22, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.136, Rules used = {6167, 6140, 43} $\frac{c^2 (a x+1)^5}{5 a}-\frac{c^2 (a x+1)^4}{2 a}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

-(c^2*(1 + a*x)^4)/(2*a) + (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 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)} \left (c-a^2 c x^2\right )^2 \, dx &=-\int e^{2 \tanh ^{-1}(a x)} \left (c-a^2 c x^2\right )^2 \, dx\\ &=-\left (c^2 \int (1-a x) (1+a x)^3 \, dx\right )\\ &=-\left (c^2 \int \left (2 (1+a x)^3-(1+a x)^4\right ) \, dx\right )\\ &=-\frac{c^2 (1+a x)^4}{2 a}+\frac{c^2 (1+a x)^5}{5 a}\\ \end{align*}

Mathematica [A]  time = 0.0162947, size = 23, normalized size = 0.66 $\frac{c^2 (a x+1)^4 (2 a x-3)}{10 a}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(c^2*(1 + a*x)^4*(-3 + 2*a*x))/(10*a)

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Maple [A]  time = 0.039, size = 31, normalized size = 0.9 \begin{align*}{c}^{2} \left ({\frac{{x}^{5}{a}^{4}}{5}}+{\frac{{x}^{4}{a}^{3}}{2}}-a{x}^{2}-x \right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.04745, size = 51, normalized size = 1.46 \begin{align*} \frac{1}{5} \, a^{4} c^{2} x^{5} + \frac{1}{2} \, a^{3} c^{2} x^{4} - 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)*(a*x+1)*(-a^2*c*x^2+c)^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.44836, size = 74, normalized size = 2.11 \begin{align*} \frac{1}{5} \, a^{4} c^{2} x^{5} + \frac{1}{2} \, a^{3} c^{2} x^{4} - 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)*(a*x+1)*(-a^2*c*x^2+c)^2,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.081442, size = 36, normalized size = 1.03 \begin{align*} \frac{a^{4} c^{2} x^{5}}{5} + \frac{a^{3} c^{2} x^{4}}{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)*(a*x+1)*(-a**2*c*x**2+c)**2,x)

[Out]

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

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Giac [A]  time = 1.12924, size = 51, normalized size = 1.46 \begin{align*} \frac{1}{5} \, a^{4} c^{2} x^{5} + \frac{1}{2} \, a^{3} c^{2} x^{4} - 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)*(a*x+1)*(-a^2*c*x^2+c)^2,x, algorithm="giac")

[Out]

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