3.1022 \(\int e^{2 \tanh ^{-1}(a x)} x (c-a^2 c x^2) \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.040211, antiderivative size = 29, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {6150, 43} \[ \frac{1}{4} a^2 c x^4+\frac{2}{3} a c x^3+\frac{c x^2}{2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.014375, size = 22, normalized size = 0.76 \[ \frac{1}{12} c x^2 \left (3 a^2 x^2+8 a x+6\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c*x^2*(6 + 8*a*x + 3*a^2*x^2))/12

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Maple [A]  time = 0.025, size = 23, normalized size = 0.8 \begin{align*} c \left ({\frac{{a}^{2}{x}^{4}}{4}}+{\frac{2\,{x}^{3}a}{3}}+{\frac{{x}^{2}}{2}} \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*(-a^2*c*x^2+c),x)

[Out]

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

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Maxima [A]  time = 0.954782, size = 31, normalized size = 1.07 \begin{align*} \frac{1}{4} \, a^{2} c x^{4} + \frac{2}{3} \, a c x^{3} + \frac{1}{2} \, c x^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.91731, size = 55, normalized size = 1.9 \begin{align*} \frac{1}{4} \, a^{2} c x^{4} + \frac{2}{3} \, a c x^{3} + \frac{1}{2} \, c x^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.076318, size = 26, normalized size = 0.9 \begin{align*} \frac{a^{2} c x^{4}}{4} + \frac{2 a c x^{3}}{3} + \frac{c x^{2}}{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.13828, size = 31, normalized size = 1.07 \begin{align*} \frac{1}{4} \, a^{2} c x^{4} + \frac{2}{3} \, a c x^{3} + \frac{1}{2} \, c x^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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