3.1032 \(\int e^{2 \tanh ^{-1}(a x)} (c-a^2 c x^2)^2 \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.0370518, antiderivative size = 35, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.091, Rules used = {6140, 43} \[ \frac{c^2 (a x+1)^4}{2 a}-\frac{c^2 (a x+1)^5}{5 a} \]

Antiderivative was successfully verified.

[In]

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

Mathematica [A]  time = 0.0130379, 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*ArcTanh[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 = 0.023, size = 28, normalized size = 0.8 \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)^2/(-a^2*x^2+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 = 0.950569, size = 49, normalized size = 1.4 \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((a*x+1)^2/(-a^2*x^2+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 = 1.69535, size = 76, normalized size = 2.17 \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((a*x+1)^2/(-a^2*x^2+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 = 0.085515, 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((a*x+1)**2/(-a**2*x**2+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 = 1.15739, size = 49, normalized size = 1.4 \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((a*x+1)^2/(-a^2*x^2+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