3.1021 \(\int e^{2 \tanh ^{-1}(a x)} x^2 (c-a^2 c x^2) \, dx\)

Optimal. Leaf size=29 \[ \frac{1}{5} a^2 c x^5+\frac{1}{2} a c x^4+\frac{c x^3}{3} \]

[Out]

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

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Rubi [A]  time = 0.0534224, antiderivative size = 29, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.087, Rules used = {6150, 43} \[ \frac{1}{5} a^2 c x^5+\frac{1}{2} a c x^4+\frac{c x^3}{3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 6150

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[c^p, Int[x^m*(1 -
a*x)^(p - n/2)*(1 + a*x)^(p + n/2), x], x] /; FreeQ[{a, c, d, m, 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)} x^2 \left (c-a^2 c x^2\right ) \, dx &=c \int x^2 (1+a x)^2 \, dx\\ &=c \int \left (x^2+2 a x^3+a^2 x^4\right ) \, dx\\ &=\frac{c x^3}{3}+\frac{1}{2} a c x^4+\frac{1}{5} a^2 c x^5\\ \end{align*}

Mathematica [A]  time = 0.014817, size = 22, normalized size = 0.76 \[ \frac{1}{30} c x^3 \left (6 a^2 x^2+15 a x+10\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c*x^3*(10 + 15*a*x + 6*a^2*x^2))/30

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Maple [A]  time = 0.024, size = 23, normalized size = 0.8 \begin{align*} c \left ({\frac{{x}^{5}{a}^{2}}{5}}+{\frac{{x}^{4}a}{2}}+{\frac{{x}^{3}}{3}} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.939426, size = 31, normalized size = 1.07 \begin{align*} \frac{1}{5} \, a^{2} c x^{5} + \frac{1}{2} \, a c x^{4} + \frac{1}{3} \, c x^{3} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.94123, size = 55, normalized size = 1.9 \begin{align*} \frac{1}{5} \, a^{2} c x^{5} + \frac{1}{2} \, a c x^{4} + \frac{1}{3} \, c x^{3} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.077207, size = 24, normalized size = 0.83 \begin{align*} \frac{a^{2} c x^{5}}{5} + \frac{a c x^{4}}{2} + \frac{c x^{3}}{3} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**2*c*x**5/5 + a*c*x**4/2 + c*x**3/3

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Giac [A]  time = 1.12715, size = 31, normalized size = 1.07 \begin{align*} \frac{1}{5} \, a^{2} c x^{5} + \frac{1}{2} \, a c x^{4} + \frac{1}{3} \, c x^{3} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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