3.24.76 \(\int 2 e^{-2-x} (e^x (-1+\frac {e^{2+x}}{2})-5 x^4+x^5) \, dx\)

Optimal. Leaf size=22 \[ 5+e^x-2 e^{-2-x} x \left (e^x+x^4\right ) \]

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Rubi [A]  time = 0.22, antiderivative size = 22, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 4, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {12, 6742, 2194, 2176} \begin {gather*} -2 e^{-x-2} x^5-\frac {2 x}{e^2}+e^x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[2*E^(-2 - x)*(E^x*(-1 + E^(2 + x)/2) - 5*x^4 + x^5),x]

[Out]

E^x - (2*x)/E^2 - 2*E^(-2 - x)*x^5

Rule 12

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

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 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=2 \int e^{-2-x} \left (e^x \left (-1+\frac {e^{2+x}}{2}\right )-5 x^4+x^5\right ) \, dx\\ &=2 \int \left (\frac {-2+e^{2+x}}{2 e^2}-5 e^{-2-x} x^4+e^{-2-x} x^5\right ) \, dx\\ &=2 \int e^{-2-x} x^5 \, dx-10 \int e^{-2-x} x^4 \, dx+\frac {\int \left (-2+e^{2+x}\right ) \, dx}{e^2}\\ &=-\frac {2 x}{e^2}+10 e^{-2-x} x^4-2 e^{-2-x} x^5+10 \int e^{-2-x} x^4 \, dx-40 \int e^{-2-x} x^3 \, dx+\frac {\int e^{2+x} \, dx}{e^2}\\ &=e^x-\frac {2 x}{e^2}+40 e^{-2-x} x^3-2 e^{-2-x} x^5+40 \int e^{-2-x} x^3 \, dx-120 \int e^{-2-x} x^2 \, dx\\ &=e^x-\frac {2 x}{e^2}+120 e^{-2-x} x^2-2 e^{-2-x} x^5+120 \int e^{-2-x} x^2 \, dx-240 \int e^{-2-x} x \, dx\\ &=e^x-\frac {2 x}{e^2}+240 e^{-2-x} x-2 e^{-2-x} x^5-240 \int e^{-2-x} \, dx+240 \int e^{-2-x} x \, dx\\ &=240 e^{-2-x}+e^x-\frac {2 x}{e^2}-2 e^{-2-x} x^5+240 \int e^{-2-x} \, dx\\ &=e^x-\frac {2 x}{e^2}-2 e^{-2-x} x^5\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 22, normalized size = 1.00 \begin {gather*} e^x-\frac {2 x}{e^2}-2 e^{-2-x} x^5 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[2*E^(-2 - x)*(E^x*(-1 + E^(2 + x)/2) - 5*x^4 + x^5),x]

[Out]

E^x - (2*x)/E^2 - 2*E^(-2 - x)*x^5

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fricas [B]  time = 0.61, size = 43, normalized size = 1.95 \begin {gather*} -{\left (x^{5} e^{\left (-2 \, x + 2 \, \log \relax (2) - 4\right )} + {\left (x e^{\left (-x + \log \relax (2) - 2\right )} - 1\right )} e^{\left (\log \relax (2) - 2\right )}\right )} e^{\left (x - \log \relax (2) + 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(x^5*e^(-2*x + 2*log(2) - 4) + (x*e^(-x + log(2) - 2) - 1)*e^(log(2) - 2))*e^(x - log(2) + 2)

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giac [A]  time = 0.29, size = 23, normalized size = 1.05 \begin {gather*} -{\left (2 \, x^{5} e^{\left (-x\right )} + 2 \, x - e^{\left (x + 2\right )}\right )} e^{\left (-2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(2*x^5*e^(-x) + 2*x - e^(x + 2))*e^(-2)

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maple [A]  time = 0.07, size = 20, normalized size = 0.91

method	result	size
risch	\(-2 x \,{\mathrm e}^{-2}+{\mathrm e}^{x}-2 x^{5} {\mathrm e}^{-x -2}\)	\(20\)
norman	\(\left ({\mathrm e}^{2 x}-2 \,{\mathrm e}^{-2} x^{5}-2 x \,{\mathrm e}^{-2} {\mathrm e}^{x}\right ) {\mathrm e}^{-x}\)	\(29\)
default	\(-2 x \,{\mathrm e}^{-2}-10 \,{\mathrm e}^{-2} \left (-x^{4} {\mathrm e}^{-x}-4 x^{3} {\mathrm e}^{-x}-12 x^{2} {\mathrm e}^{-x}-24 x \,{\mathrm e}^{-x}-24 \,{\mathrm e}^{-x}\right )+2 \,{\mathrm e}^{-2} \left (-{\mathrm e}^{-x} x^{5}-5 x^{4} {\mathrm e}^{-x}-20 x^{3} {\mathrm e}^{-x}-60 x^{2} {\mathrm e}^{-x}-120 x \,{\mathrm e}^{-x}-120 \,{\mathrm e}^{-x}\right )+{\mathrm e}^{x}\)	\(114\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*x*exp(-2)+exp(x)-2*x^5*exp(-x-2)

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maxima [B]  time = 0.46, size = 65, normalized size = 2.95 \begin {gather*} -2 \, x e^{\left (-2\right )} - 2 \, {\left (x^{5} + 5 \, x^{4} + 20 \, x^{3} + 60 \, x^{2} + 120 \, x + 120\right )} e^{\left (-x - 2\right )} + 10 \, {\left (x^{4} + 4 \, x^{3} + 12 \, x^{2} + 24 \, x + 24\right )} e^{\left (-x - 2\right )} + e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*x*e^(-2) - 2*(x^5 + 5*x^4 + 20*x^3 + 60*x^2 + 120*x + 120)*e^(-x - 2) + 10*(x^4 + 4*x^3 + 12*x^2 + 24*x + 2
4)*e^(-x - 2) + e^x

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mupad [B]  time = 0.09, size = 19, normalized size = 0.86 \begin {gather*} {\mathrm {e}}^x-2\,x\,{\mathrm {e}}^{-2}-2\,x^5\,{\mathrm {e}}^{-x-2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(log(2) - x - 2)*(exp(x)*(exp(x - log(2) + 2) - 1) - 5*x^4 + x^5),x)

[Out]

exp(x) - 2*x*exp(-2) - 2*x^5*exp(- x - 2)

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sympy [A]  time = 0.14, size = 24, normalized size = 1.09 \begin {gather*} - \frac {2 x}{e^{2}} + \frac {- 2 x^{5} e^{- x} + e^{2} e^{x}}{e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*x*exp(-2) + (-2*x**5*exp(-x) + exp(2)*exp(x))*exp(-2)

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