3.18.86 \(\int \frac {1}{9} (e^{2+\frac {x}{2}} (12+6 x)+e^{4+x} (2 x+x^2)) \, dx\)

Optimal. Leaf size=20 \[ -5+\left (-2-\frac {1}{3} e^{2+\frac {x}{2}} x\right )^2 \]

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Rubi [B]  time = 0.08, antiderivative size = 42, normalized size of antiderivative = 2.10, 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, 2176, 2194, 1593, 2196} \begin {gather*} \frac {1}{9} e^{x+4} x^2-\frac {8}{3} e^{\frac {x}{2}+2}+\frac {4}{3} e^{\frac {x}{2}+2} (x+2) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(2 + x/2)*(12 + 6*x) + E^(4 + x)*(2*x + x^2))/9,x]

[Out]

(-8*E^(2 + x/2))/3 + (E^(4 + x)*x^2)/9 + (4*E^(2 + x/2)*(2 + x))/3

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}{9} \int \left (e^{2+\frac {x}{2}} (12+6 x)+e^{4+x} \left (2 x+x^2\right )\right ) \, dx\\ &=\frac {1}{9} \int e^{2+\frac {x}{2}} (12+6 x) \, dx+\frac {1}{9} \int e^{4+x} \left (2 x+x^2\right ) \, dx\\ &=\frac {4}{3} e^{2+\frac {x}{2}} (2+x)+\frac {1}{9} \int e^{4+x} x (2+x) \, dx-\frac {4}{3} \int e^{2+\frac {x}{2}} \, dx\\ &=-\frac {8}{3} e^{2+\frac {x}{2}}+\frac {4}{3} e^{2+\frac {x}{2}} (2+x)+\frac {1}{9} \int \left (2 e^{4+x} x+e^{4+x} x^2\right ) \, dx\\ &=-\frac {8}{3} e^{2+\frac {x}{2}}+\frac {4}{3} e^{2+\frac {x}{2}} (2+x)+\frac {1}{9} \int e^{4+x} x^2 \, dx+\frac {2}{9} \int e^{4+x} x \, dx\\ &=-\frac {8}{3} e^{2+\frac {x}{2}}+\frac {2}{9} e^{4+x} x+\frac {1}{9} e^{4+x} x^2+\frac {4}{3} e^{2+\frac {x}{2}} (2+x)-\frac {2}{9} \int e^{4+x} \, dx-\frac {2}{9} \int e^{4+x} x \, dx\\ &=-\frac {8}{3} e^{2+\frac {x}{2}}-\frac {2 e^{4+x}}{9}+\frac {1}{9} e^{4+x} x^2+\frac {4}{3} e^{2+\frac {x}{2}} (2+x)+\frac {2}{9} \int e^{4+x} \, dx\\ &=-\frac {8}{3} e^{2+\frac {x}{2}}+\frac {1}{9} e^{4+x} x^2+\frac {4}{3} e^{2+\frac {x}{2}} (2+x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 19, normalized size = 0.95 \begin {gather*} \frac {1}{9} \left (6+e^{2+\frac {x}{2}} x\right )^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(2 + x/2)*(12 + 6*x) + E^(4 + x)*(2*x + x^2))/9,x]

[Out]

(6 + E^(2 + x/2)*x)^2/9

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fricas [A]  time = 0.75, size = 19, normalized size = 0.95 \begin {gather*} \frac {1}{9} \, x^{2} e^{\left (x + 4\right )} + \frac {4}{3} \, x e^{\left (\frac {1}{2} \, x + 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/9*(x^2+2*x)*exp(2)^2*exp(1/2*x)^2+1/9*(6*x+12)*exp(2)*exp(1/2*x),x, algorithm="fricas")

[Out]

1/9*x^2*e^(x + 4) + 4/3*x*e^(1/2*x + 2)

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giac [A]  time = 0.27, size = 19, normalized size = 0.95 \begin {gather*} \frac {1}{9} \, x^{2} e^{\left (x + 4\right )} + \frac {4}{3} \, x e^{\left (\frac {1}{2} \, x + 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/9*(x^2+2*x)*exp(2)^2*exp(1/2*x)^2+1/9*(6*x+12)*exp(2)*exp(1/2*x),x, algorithm="giac")

[Out]

1/9*x^2*e^(x + 4) + 4/3*x*e^(1/2*x + 2)

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maple [A]  time = 0.05, size = 20, normalized size = 1.00

method	result	size
risch	\(\frac {4 x \,{\mathrm e}^{2+\frac {x}{2}}}{3}+\frac {x^{2} {\mathrm e}^{4+x}}{9}\)	\(20\)
derivativedivides	\(\frac {4 x \,{\mathrm e}^{2} {\mathrm e}^{\frac {x}{2}}}{3}+\frac {x^{2} {\mathrm e}^{4} {\mathrm e}^{x}}{9}\)	\(26\)
default	\(\frac {4 x \,{\mathrm e}^{2} {\mathrm e}^{\frac {x}{2}}}{3}+\frac {x^{2} {\mathrm e}^{4} {\mathrm e}^{x}}{9}\)	\(26\)
norman	\(\frac {4 x \,{\mathrm e}^{2} {\mathrm e}^{\frac {x}{2}}}{3}+\frac {x^{2} {\mathrm e}^{4} {\mathrm e}^{x}}{9}\)	\(26\)
meijerg	\(-\frac {{\mathrm e}^{4} \left (2-\frac {\left (3 x^{2}-6 x +6\right ) {\mathrm e}^{x}}{3}\right )}{9}+\frac {2 \,{\mathrm e}^{4} \left (1-\frac {\left (-2 x +2\right ) {\mathrm e}^{x}}{2}\right )}{9}-\frac {8 \,{\mathrm e}^{2} \left (1-{\mathrm e}^{\frac {x}{2}}\right )}{3}+\frac {8 \,{\mathrm e}^{2} \left (1-\frac {\left (2-x \right ) {\mathrm e}^{\frac {x}{2}}}{2}\right )}{3}\)	\(66\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/9*(x^2+2*x)*exp(2)^2*exp(1/2*x)^2+1/9*(6*x+12)*exp(2)*exp(1/2*x),x,method=_RETURNVERBOSE)

[Out]

4/3*x*exp(2+1/2*x)+1/9*x^2*exp(4+x)

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maxima [B]  time = 0.46, size = 33, normalized size = 1.65 \begin {gather*} \frac {1}{9} \, x^{2} e^{\left (x + 4\right )} + \frac {4}{3} \, {\left (x e^{2} - 2 \, e^{2}\right )} e^{\left (\frac {1}{2} \, x\right )} + \frac {8}{3} \, e^{\left (\frac {1}{2} \, x + 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/9*(x^2+2*x)*exp(2)^2*exp(1/2*x)^2+1/9*(6*x+12)*exp(2)*exp(1/2*x),x, algorithm="maxima")

[Out]

1/9*x^2*e^(x + 4) + 4/3*(x*e^2 - 2*e^2)*e^(1/2*x) + 8/3*e^(1/2*x + 2)

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mupad [B]  time = 0.07, size = 19, normalized size = 0.95 \begin {gather*} \frac {x\,{\mathrm {e}}^{\frac {x}{2}+2}\,\left (x\,{\mathrm {e}}^{\frac {x}{2}+2}+12\right )}{9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x/2)*exp(2)*(6*x + 12))/9 + (exp(4)*exp(x)*(2*x + x^2))/9,x)

[Out]

(x*exp(x/2 + 2)*(x*exp(x/2 + 2) + 12))/9

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sympy [A]  time = 0.14, size = 24, normalized size = 1.20 \begin {gather*} \frac {x^{2} e^{4} e^{x}}{9} + \frac {4 x e^{2} e^{\frac {x}{2}}}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/9*(x**2+2*x)*exp(2)**2*exp(1/2*x)**2+1/9*(6*x+12)*exp(2)*exp(1/2*x),x)

[Out]

x**2*exp(4)*exp(x)/9 + 4*x*exp(2)*exp(x/2)/3

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