3.1230 \(\int e^{-2 \tanh ^{-1}(a x)} (c-a^2 c x^2)^2 \, dx\)

Optimal. Leaf size=37 \[ \frac{c^2 (1-a x)^5}{5 a}-\frac{c^2 (1-a x)^4}{2 a} \]

[Out]

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

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Rubi [A]  time = 0.0367321, antiderivative size = 37, 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} \[ \frac{c^2 (1-a x)^5}{5 a}-\frac{c^2 (1-a x)^4}{2 a} \]

Antiderivative was successfully verified.

[In]

Int[(c - a^2*c*x^2)^2/E^(2*ArcTanh[a*x]),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)^3 (1+a x) \, 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 = 0.0160266, size = 32, normalized size = 0.86 \[ c^2 \left (-\frac{1}{5} a^4 x^5+\frac{a^3 x^4}{2}-a x^2+x\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.027, size = 29, 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^2*c*x^2+c)^2/(a*x+1)^2*(-a^2*x^2+1),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 = 0.967953, size = 50, normalized size = 1.35 \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^2*c*x^2+c)^2/(a*x+1)^2*(-a^2*x^2+1),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 = 2.19162, size = 76, normalized size = 2.05 \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^2*c*x^2+c)^2/(a*x+1)^2*(-a^2*x^2+1),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 = 0.097322, size = 36, normalized size = 0.97 \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**2*c*x**2+c)**2/(a*x+1)**2*(-a**2*x**2+1),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 = 1.1257, size = 73, normalized size = 1.97 \begin{align*} -\frac{{\left (2 \, c^{2} - \frac{15 \, c^{2}}{a x + 1} + \frac{40 \, c^{2}}{{\left (a x + 1\right )}^{2}} - \frac{40 \, c^{2}}{{\left (a x + 1\right )}^{3}}\right )}{\left (a x + 1\right )}^{5}}{10 \, a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/10*(2*c^2 - 15*c^2/(a*x + 1) + 40*c^2/(a*x + 1)^2 - 40*c^2/(a*x + 1)^3)*(a*x + 1)^5/a