3.76.95 \(\int \frac {1}{2} e^{70+\frac {20-x}{2}} (4 x-x^2) \, dx\)

Optimal. Leaf size=13 \[ e^{80-\frac {x}{2}} x^2 \]

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Rubi [A]  time = 0.07, antiderivative size = 13, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 5, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {12, 1593, 2196, 2176, 2194} \begin {gather*} e^{80-\frac {x}{2}} x^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(70 + (20 - x)/2)*(4*x - x^2))/2,x]

[Out]

E^(80 - x/2)*x^2

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}{2} \int e^{70+\frac {20-x}{2}} \left (4 x-x^2\right ) \, dx\\ &=\frac {1}{2} \int e^{70+\frac {20-x}{2}} (4-x) x \, dx\\ &=\frac {1}{2} \int \left (4 e^{80-\frac {x}{2}} x-e^{80-\frac {x}{2}} x^2\right ) \, dx\\ &=-\left (\frac {1}{2} \int e^{80-\frac {x}{2}} x^2 \, dx\right )+2 \int e^{80-\frac {x}{2}} x \, dx\\ &=-4 e^{80-\frac {x}{2}} x+e^{80-\frac {x}{2}} x^2-2 \int e^{80-\frac {x}{2}} x \, dx+4 \int e^{80-\frac {x}{2}} \, dx\\ &=-8 e^{80-\frac {x}{2}}+e^{80-\frac {x}{2}} x^2-4 \int e^{80-\frac {x}{2}} \, dx\\ &=e^{80-\frac {x}{2}} x^2\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 13, normalized size = 1.00 \begin {gather*} e^{80-\frac {x}{2}} x^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(70 + (20 - x)/2)*(4*x - x^2))/2,x]

[Out]

E^(80 - x/2)*x^2

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fricas [A]  time = 0.49, size = 10, normalized size = 0.77 \begin {gather*} x^{2} e^{\left (-\frac {1}{2} \, x + 80\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(-x^2+4*x)*exp(36)^2*exp(-1/4*x+5)^2/exp(1)^2,x, algorithm="fricas")

[Out]

x^2*e^(-1/2*x + 80)

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giac [A]  time = 0.20, size = 10, normalized size = 0.77 \begin {gather*} x^{2} e^{\left (-\frac {1}{2} \, x + 80\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(-x^2+4*x)*exp(36)^2*exp(-1/4*x+5)^2/exp(1)^2,x, algorithm="giac")

[Out]

x^2*e^(-1/2*x + 80)

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maple [A]  time = 0.07, size = 11, normalized size = 0.85

method	result	size
risch	\(x^{2} {\mathrm e}^{80-\frac {x}{2}}\)	\(11\)
gosper	\({\mathrm e}^{-\frac {x}{2}+10} x^{2} {\mathrm e}^{72} {\mathrm e}^{-2}\)	\(21\)
norman	\({\mathrm e}^{-\frac {x}{2}+10} x^{2} {\mathrm e}^{72} {\mathrm e}^{-2}\)	\(21\)
derivativedivides	\(-2 \,{\mathrm e}^{72} {\mathrm e}^{-2} \left (-200 \,{\mathrm e}^{-\frac {x}{2}+10}+80 \,{\mathrm e}^{-\frac {x}{2}+10} \left (-\frac {x}{4}+5\right )-8 \,{\mathrm e}^{-\frac {x}{2}+10} \left (-\frac {x}{4}+5\right )^{2}\right )\)	\(54\)
default	\(\frac {{\mathrm e}^{72} {\mathrm e}^{-2} \left (800 \,{\mathrm e}^{-\frac {x}{2}+10}-320 \,{\mathrm e}^{-\frac {x}{2}+10} \left (-\frac {x}{4}+5\right )+32 \,{\mathrm e}^{-\frac {x}{2}+10} \left (-\frac {x}{4}+5\right )^{2}\right )}{2}\)	\(54\)
meijerg	\(-4 \,{\mathrm e}^{50+\frac {x \,{\mathrm e}^{10}}{2}-\frac {x}{2}} \left (2-\frac {\left (\frac {3 x^{2} {\mathrm e}^{20}}{4}+3 x \,{\mathrm e}^{10}+6\right ) {\mathrm e}^{-\frac {x \,{\mathrm e}^{10}}{2}}}{3}\right )+8 \,{\mathrm e}^{60+\frac {x \,{\mathrm e}^{10}}{2}-\frac {x}{2}} \left (1-\frac {\left (x \,{\mathrm e}^{10}+2\right ) {\mathrm e}^{-\frac {x \,{\mathrm e}^{10}}{2}}}{2}\right )\)	\(68\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/2*(-x^2+4*x)*exp(36)^2*exp(-1/4*x+5)^2/exp(1)^2,x,method=_RETURNVERBOSE)

[Out]

x^2*exp(80-1/2*x)

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maxima [B]  time = 0.36, size = 37, normalized size = 2.85 \begin {gather*} {\left (x^{2} e^{80} + 4 \, x e^{80} + 8 \, e^{80}\right )} e^{\left (-\frac {1}{2} \, x\right )} - 4 \, {\left (x e^{80} + 2 \, e^{80}\right )} e^{\left (-\frac {1}{2} \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(-x^2+4*x)*exp(36)^2*exp(-1/4*x+5)^2/exp(1)^2,x, algorithm="maxima")

[Out]

(x^2*e^80 + 4*x*e^80 + 8*e^80)*e^(-1/2*x) - 4*(x*e^80 + 2*e^80)*e^(-1/2*x)

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mupad [B]  time = 0.04, size = 10, normalized size = 0.77 \begin {gather*} x^2\,{\mathrm {e}}^{-\frac {x}{2}}\,{\mathrm {e}}^{80} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(70)*exp(10 - x/2)*(4*x - x^2))/2,x)

[Out]

x^2*exp(-x/2)*exp(80)

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sympy [A]  time = 0.13, size = 12, normalized size = 0.92 \begin {gather*} x^{2} e^{70} e^{10 - \frac {x}{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(-x**2+4*x)*exp(36)**2*exp(-1/4*x+5)**2/exp(1)**2,x)

[Out]

x**2*exp(70)*exp(10 - x/2)

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