3.50.40 \(\int \frac {1}{6} (24 x+36 x^2+e^{3+2 x} (3 x^2+2 x^3)) \, dx\)

Optimal. Leaf size=22 \[ 2 x \left (x+x^2+\frac {1}{12} e^{3+2 x} x^2\right ) \]

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Rubi [A]  time = 0.12, antiderivative size = 25, normalized size of antiderivative = 1.14, number of steps used = 12, number of rules used = 5, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.156, Rules used = {12, 1593, 2196, 2176, 2194} \begin {gather*} \frac {1}{6} e^{2 x+3} x^3+2 x^3+2 x^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(24*x + 36*x^2 + E^(3 + 2*x)*(3*x^2 + 2*x^3))/6,x]

[Out]

2*x^2 + 2*x^3 + (E^(3 + 2*x)*x^3)/6

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 (24 x+36 x^2+e^{3+2 x} \left (3 x^2+2 x^3\right )\right ) \, dx\\ &=2 x^2+2 x^3+\frac {1}{6} \int e^{3+2 x} \left (3 x^2+2 x^3\right ) \, dx\\ &=2 x^2+2 x^3+\frac {1}{6} \int e^{3+2 x} x^2 (3+2 x) \, dx\\ &=2 x^2+2 x^3+\frac {1}{6} \int \left (3 e^{3+2 x} x^2+2 e^{3+2 x} x^3\right ) \, dx\\ &=2 x^2+2 x^3+\frac {1}{3} \int e^{3+2 x} x^3 \, dx+\frac {1}{2} \int e^{3+2 x} x^2 \, dx\\ &=2 x^2+\frac {1}{4} e^{3+2 x} x^2+2 x^3+\frac {1}{6} e^{3+2 x} x^3-\frac {1}{2} \int e^{3+2 x} x \, dx-\frac {1}{2} \int e^{3+2 x} x^2 \, dx\\ &=-\frac {1}{4} e^{3+2 x} x+2 x^2+2 x^3+\frac {1}{6} e^{3+2 x} x^3+\frac {1}{4} \int e^{3+2 x} \, dx+\frac {1}{2} \int e^{3+2 x} x \, dx\\ &=\frac {1}{8} e^{3+2 x}+2 x^2+2 x^3+\frac {1}{6} e^{3+2 x} x^3-\frac {1}{4} \int e^{3+2 x} \, dx\\ &=2 x^2+2 x^3+\frac {1}{6} e^{3+2 x} x^3\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 20, normalized size = 0.91 \begin {gather*} \frac {1}{6} x^2 \left (12+\left (12+e^{3+2 x}\right ) x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(24*x + 36*x^2 + E^(3 + 2*x)*(3*x^2 + 2*x^3))/6,x]

[Out]

(x^2*(12 + (12 + E^(3 + 2*x))*x))/6

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fricas [A]  time = 0.66, size = 22, normalized size = 1.00 \begin {gather*} \frac {1}{6} \, x^{3} e^{\left (2 \, x + 3\right )} + 2 \, x^{3} + 2 \, x^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*x^3*e^(2*x + 3) + 2*x^3 + 2*x^2

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giac [A]  time = 0.13, size = 22, normalized size = 1.00 \begin {gather*} \frac {1}{6} \, x^{3} e^{\left (2 \, x + 3\right )} + 2 \, x^{3} + 2 \, x^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*x^3*e^(2*x + 3) + 2*x^3 + 2*x^2

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maple [A]  time = 0.04, size = 23, normalized size = 1.05

method	result	size
norman	\(2 x^{2}+2 x^{3}+\frac {x^{3} {\mathrm e}^{3} {\mathrm e}^{2 x}}{6}\)	\(23\)
risch	\(2 x^{2}+2 x^{3}+\frac {x^{3} {\mathrm e}^{2 x +3}}{6}\)	\(23\)
default	\(2 x^{3}+2 x^{2}+\frac {{\mathrm e}^{3} \left (\frac {{\mathrm e}^{2 x} x^{2}}{2}-\frac {x \,{\mathrm e}^{2 x}}{2}+\frac {{\mathrm e}^{2 x}}{4}\right )}{2}+\frac {{\mathrm e}^{3} \left (\frac {{\mathrm e}^{2 x} x^{3}}{2}-\frac {3 \,{\mathrm e}^{2 x} x^{2}}{4}+\frac {3 x \,{\mathrm e}^{2 x}}{4}-\frac {3 \,{\mathrm e}^{2 x}}{8}\right )}{3}\)	\(75\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.35, size = 22, normalized size = 1.00 \begin {gather*} \frac {1}{6} \, x^{3} e^{\left (2 \, x + 3\right )} + 2 \, x^{3} + 2 \, x^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*x^3*e^(2*x + 3) + 2*x^3 + 2*x^2

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mupad [B]  time = 3.60, size = 18, normalized size = 0.82 \begin {gather*} \frac {x^2\,\left (12\,x+x\,{\mathrm {e}}^{2\,x+3}+12\right )}{6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(x^2*(12*x + x*exp(2*x + 3) + 12))/6

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sympy [A]  time = 0.14, size = 22, normalized size = 1.00 \begin {gather*} \frac {x^{3} e^{3} e^{2 x}}{6} + 2 x^{3} + 2 x^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x**3*exp(3)*exp(2*x)/6 + 2*x**3 + 2*x**2

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