3.1020 \(\int e^{2 \tanh ^{-1}(a x)} x^3 (c-a^2 c x^2) \, dx\)

Optimal. Leaf size=29 \[ \frac {1}{6} a^2 c x^6+\frac {2}{5} a c x^5+\frac {c x^4}{4} \]

[Out]

1/4*c*x^4+2/5*a*c*x^5+1/6*a^2*c*x^6

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Rubi [A]  time = 0.05, antiderivative size = 29, normalized size of antiderivative = 1.00, 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} \[ \frac {1}{6} a^2 c x^6+\frac {2}{5} a c x^5+\frac {c x^4}{4} \]

Antiderivative was successfully verified.

[In]

Int[E^(2*ArcTanh[a*x])*x^3*(c - a^2*c*x^2),x]

[Out]

(c*x^4)/4 + (2*a*c*x^5)/5 + (a^2*c*x^6)/6

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])

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])

Rubi steps

\begin {align*} \int e^{2 \tanh ^{-1}(a x)} x^3 \left (c-a^2 c x^2\right ) \, dx &=c \int x^3 (1+a x)^2 \, dx\\ &=c \int \left (x^3+2 a x^4+a^2 x^5\right ) \, dx\\ &=\frac {c x^4}{4}+\frac {2}{5} a c x^5+\frac {1}{6} a^2 c x^6\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 22, normalized size = 0.76 \[ \frac {1}{60} c x^4 \left (10 a^2 x^2+24 a x+15\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[E^(2*ArcTanh[a*x])*x^3*(c - a^2*c*x^2),x]

[Out]

(c*x^4*(15 + 24*a*x + 10*a^2*x^2))/60

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fricas [A]  time = 0.63, size = 23, normalized size = 0.79 \[ \frac {1}{6} \, a^{2} c x^{6} + \frac {2}{5} \, a c x^{5} + \frac {1}{4} \, c x^{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^2/(-a^2*x^2+1)*x^3*(-a^2*c*x^2+c),x, algorithm="fricas")

[Out]

1/6*a^2*c*x^6 + 2/5*a*c*x^5 + 1/4*c*x^4

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giac [A]  time = 0.19, size = 23, normalized size = 0.79 \[ \frac {1}{6} \, a^{2} c x^{6} + \frac {2}{5} \, a c x^{5} + \frac {1}{4} \, c x^{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^2/(-a^2*x^2+1)*x^3*(-a^2*c*x^2+c),x, algorithm="giac")

[Out]

1/6*a^2*c*x^6 + 2/5*a*c*x^5 + 1/4*c*x^4

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maple [A]  time = 0.02, size = 23, normalized size = 0.79 \[ c \left (\frac {1}{6} a^{2} x^{6}+\frac {2}{5} a \,x^{5}+\frac {1}{4} x^{4}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x+1)^2/(-a^2*x^2+1)*x^3*(-a^2*c*x^2+c),x)

[Out]

c*(1/6*a^2*x^6+2/5*a*x^5+1/4*x^4)

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maxima [A]  time = 0.33, size = 23, normalized size = 0.79 \[ \frac {1}{6} \, a^{2} c x^{6} + \frac {2}{5} \, a c x^{5} + \frac {1}{4} \, c x^{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^2/(-a^2*x^2+1)*x^3*(-a^2*c*x^2+c),x, algorithm="maxima")

[Out]

1/6*a^2*c*x^6 + 2/5*a*c*x^5 + 1/4*c*x^4

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mupad [B]  time = 0.04, size = 20, normalized size = 0.69 \[ \frac {c\,x^4\,\left (10\,a^2\,x^2+24\,a\,x+15\right )}{60} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(x^3*(c - a^2*c*x^2)*(a*x + 1)^2)/(a^2*x^2 - 1),x)

[Out]

(c*x^4*(24*a*x + 10*a^2*x^2 + 15))/60

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sympy [A]  time = 0.07, size = 26, normalized size = 0.90 \[ \frac {a^{2} c x^{6}}{6} + \frac {2 a c x^{5}}{5} + \frac {c x^{4}}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)**2/(-a**2*x**2+1)*x**3*(-a**2*c*x**2+c),x)

[Out]

a**2*c*x**6/6 + 2*a*c*x**5/5 + c*x**4/4

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