3.44.77 \(\int \frac {1}{6} (2 x-216 x^2+24 x^3+e^x (9 x^2+3 x^3)) \, dx\)

Optimal. Leaf size=22 \[ \frac {1}{2} x^3 \left (-24+e^x+\frac {1}{3 x}+2 x\right ) \]

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Rubi [A]  time = 0.08, antiderivative size = 26, normalized size of antiderivative = 1.18, number of steps used = 12, number of rules used = 5, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {12, 1593, 2196, 2176, 2194} \begin {gather*} x^4+\frac {e^x x^3}{2}-12 x^3+\frac {x^2}{6} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(2*x - 216*x^2 + 24*x^3 + E^x*(9*x^2 + 3*x^3))/6,x]

[Out]

x^2/6 - 12*x^3 + (E^x*x^3)/2 + x^4

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 2176

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^m
*(b*F^(g*(e + f*x)))^n)/(f*g*n*Log[F]), x] - Dist[(d*m)/(f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*(b*F^(g*(e + f*x
)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && IntegerQ[2*m] &&  !$UseGamma === True

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 2196

Int[(F_)^((c_.)*(v_))*(u_), x_Symbol] :> Int[ExpandIntegrand[F^(c*ExpandToSum[v, x]), u, x], x] /; FreeQ[{F, c
}, x] && PolynomialQ[u, x] && LinearQ[v, x] &&  !$UseGamma === True

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{6} \int \left (2 x-216 x^2+24 x^3+e^x \left (9 x^2+3 x^3\right )\right ) \, dx\\ &=\frac {x^2}{6}-12 x^3+x^4+\frac {1}{6} \int e^x \left (9 x^2+3 x^3\right ) \, dx\\ &=\frac {x^2}{6}-12 x^3+x^4+\frac {1}{6} \int e^x x^2 (9+3 x) \, dx\\ &=\frac {x^2}{6}-12 x^3+x^4+\frac {1}{6} \int \left (9 e^x x^2+3 e^x x^3\right ) \, dx\\ &=\frac {x^2}{6}-12 x^3+x^4+\frac {1}{2} \int e^x x^3 \, dx+\frac {3}{2} \int e^x x^2 \, dx\\ &=\frac {x^2}{6}+\frac {3 e^x x^2}{2}-12 x^3+\frac {e^x x^3}{2}+x^4-\frac {3}{2} \int e^x x^2 \, dx-3 \int e^x x \, dx\\ &=-3 e^x x+\frac {x^2}{6}-12 x^3+\frac {e^x x^3}{2}+x^4+3 \int e^x \, dx+3 \int e^x x \, dx\\ &=3 e^x+\frac {x^2}{6}-12 x^3+\frac {e^x x^3}{2}+x^4-3 \int e^x \, dx\\ &=\frac {x^2}{6}-12 x^3+\frac {e^x x^3}{2}+x^4\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 22, normalized size = 1.00 \begin {gather*} \frac {1}{6} x^2 \left (1+3 \left (-24+e^x\right ) x+6 x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2*x - 216*x^2 + 24*x^3 + E^x*(9*x^2 + 3*x^3))/6,x]

[Out]

(x^2*(1 + 3*(-24 + E^x)*x + 6*x^2))/6

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fricas [A]  time = 0.56, size = 21, normalized size = 0.95 \begin {gather*} x^{4} + \frac {1}{2} \, x^{3} e^{x} - 12 \, x^{3} + \frac {1}{6} \, x^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/6*(3*x^3+9*x^2)*exp(x)+4*x^3-36*x^2+1/3*x,x, algorithm="fricas")

[Out]

x^4 + 1/2*x^3*e^x - 12*x^3 + 1/6*x^2

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giac [A]  time = 0.19, size = 21, normalized size = 0.95 \begin {gather*} x^{4} + \frac {1}{2} \, x^{3} e^{x} - 12 \, x^{3} + \frac {1}{6} \, x^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/6*(3*x^3+9*x^2)*exp(x)+4*x^3-36*x^2+1/3*x,x, algorithm="giac")

[Out]

x^4 + 1/2*x^3*e^x - 12*x^3 + 1/6*x^2

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maple [A]  time = 0.02, size = 22, normalized size = 1.00

method	result	size
default	\(\frac {x^{2}}{6}-12 x^{3}+x^{4}+\frac {{\mathrm e}^{x} x^{3}}{2}\)	\(22\)
norman	\(\frac {x^{2}}{6}-12 x^{3}+x^{4}+\frac {{\mathrm e}^{x} x^{3}}{2}\)	\(22\)
risch	\(\frac {x^{2}}{6}-12 x^{3}+x^{4}+\frac {{\mathrm e}^{x} x^{3}}{2}\)	\(22\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/6*(3*x^3+9*x^2)*exp(x)+4*x^3-36*x^2+1/3*x,x,method=_RETURNVERBOSE)

[Out]

1/6*x^2-12*x^3+x^4+1/2*exp(x)*x^3

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maxima [A]  time = 0.42, size = 21, normalized size = 0.95 \begin {gather*} x^{4} + \frac {1}{2} \, x^{3} e^{x} - 12 \, x^{3} + \frac {1}{6} \, x^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/6*(3*x^3+9*x^2)*exp(x)+4*x^3-36*x^2+1/3*x,x, algorithm="maxima")

[Out]

x^4 + 1/2*x^3*e^x - 12*x^3 + 1/6*x^2

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mupad [B]  time = 0.04, size = 17, normalized size = 0.77 \begin {gather*} x^2\,\left (\frac {x\,{\mathrm {e}}^x}{2}-12\,x+x^2+\frac {1}{6}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/3 + (exp(x)*(9*x^2 + 3*x^3))/6 - 36*x^2 + 4*x^3,x)

[Out]

x^2*((x*exp(x))/2 - 12*x + x^2 + 1/6)

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sympy [A]  time = 0.09, size = 20, normalized size = 0.91 \begin {gather*} x^{4} + \frac {x^{3} e^{x}}{2} - 12 x^{3} + \frac {x^{2}}{6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/6*(3*x**3+9*x**2)*exp(x)+4*x**3-36*x**2+1/3*x,x)

[Out]

x**4 + x**3*exp(x)/2 - 12*x**3 + x**2/6

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