3.43.75 \(\int e^{5-9 e^{10}+x} (5+7 x+x^2) \, dx\)

Optimal. Leaf size=15 \[ e^{5-9 e^{10}+x} x (5+x) \]

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Rubi [A]  time = 0.07, antiderivative size = 28, normalized size of antiderivative = 1.87, number of steps used = 8, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {2196, 2194, 2176} \begin {gather*} e^{x-9 e^{10}+5} x^2+5 e^{x-9 e^{10}+5} x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^(5 - 9*E^10 + x)*(5 + 7*x + x^2),x]

[Out]

5*E^(5 - 9*E^10 + x)*x + E^(5 - 9*E^10 + x)*x^2

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} &=\int \left (5 e^{5-9 e^{10}+x}+7 e^{5-9 e^{10}+x} x+e^{5-9 e^{10}+x} x^2\right ) \, dx\\ &=5 \int e^{5-9 e^{10}+x} \, dx+7 \int e^{5-9 e^{10}+x} x \, dx+\int e^{5-9 e^{10}+x} x^2 \, dx\\ &=5 e^{5-9 e^{10}+x}+7 e^{5-9 e^{10}+x} x+e^{5-9 e^{10}+x} x^2-2 \int e^{5-9 e^{10}+x} x \, dx-7 \int e^{5-9 e^{10}+x} \, dx\\ &=-2 e^{5-9 e^{10}+x}+5 e^{5-9 e^{10}+x} x+e^{5-9 e^{10}+x} x^2+2 \int e^{5-9 e^{10}+x} \, dx\\ &=5 e^{5-9 e^{10}+x} x+e^{5-9 e^{10}+x} x^2\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 15, normalized size = 1.00 \begin {gather*} e^{5-9 e^{10}+x} x (5+x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[E^(5 - 9*E^10 + x)*(5 + 7*x + x^2),x]

[Out]

E^(5 - 9*E^10 + x)*x*(5 + x)

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fricas [A]  time = 0.51, size = 16, normalized size = 1.07 \begin {gather*} {\left (x^{2} + 5 \, x\right )} e^{\left (x - 9 \, e^{10} + 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2+7*x+5)*exp(-9*exp(5)^2+5+x),x, algorithm="fricas")

[Out]

(x^2 + 5*x)*e^(x - 9*e^10 + 5)

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giac [A]  time = 0.12, size = 16, normalized size = 1.07 \begin {gather*} {\left (x^{2} + 5 \, x\right )} e^{\left (x - 9 \, e^{10} + 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2+7*x+5)*exp(-9*exp(5)^2+5+x),x, algorithm="giac")

[Out]

(x^2 + 5*x)*e^(x - 9*e^10 + 5)

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maple [A]  time = 0.03, size = 16, normalized size = 1.07

method	result	size
gosper	\(\left (5+x \right ) x \,{\mathrm e}^{-9 \,{\mathrm e}^{10}+5+x}\)	\(16\)
risch	\(\left (x^{2}+5 x \right ) {\mathrm e}^{-9 \,{\mathrm e}^{10}+5+x}\)	\(17\)
norman	\(x^{2} {\mathrm e}^{-9 \,{\mathrm e}^{10}+5+x}+5 x \,{\mathrm e}^{-9 \,{\mathrm e}^{10}+5+x}\)	\(29\)
meijerg	\(-{\mathrm e}^{-9 \,{\mathrm e}^{10}+5} \left (2-\frac {\left (3 x^{2}-6 x +6\right ) {\mathrm e}^{x}}{3}\right )+7 \,{\mathrm e}^{-9 \,{\mathrm e}^{10}+5} \left (1-\frac {\left (-2 x +2\right ) {\mathrm e}^{x}}{2}\right )-5 \,{\mathrm e}^{-9 \,{\mathrm e}^{10}+5} \left (1-{\mathrm e}^{x}\right )\)	\(62\)
derivativedivides	\({\mathrm e}^{-9 \,{\mathrm e}^{10}+5+x} \left (-9 \,{\mathrm e}^{10}+5+x \right )^{2}-5 \,{\mathrm e}^{-9 \,{\mathrm e}^{10}+5+x} \left (-9 \,{\mathrm e}^{10}+5+x \right )-27 \,{\mathrm e}^{-9 \,{\mathrm e}^{10}+5+x} {\mathrm e}^{10}+81 \,{\mathrm e}^{-9 \,{\mathrm e}^{10}+5+x} {\mathrm e}^{20}+18 \,{\mathrm e}^{10} \left ({\mathrm e}^{-9 \,{\mathrm e}^{10}+5+x} \left (-9 \,{\mathrm e}^{10}+5+x \right )-{\mathrm e}^{-9 \,{\mathrm e}^{10}+5+x}\right )\)	\(116\)
default	\({\mathrm e}^{-9 \,{\mathrm e}^{10}+5+x} \left (-9 \,{\mathrm e}^{10}+5+x \right )^{2}-5 \,{\mathrm e}^{-9 \,{\mathrm e}^{10}+5+x} \left (-9 \,{\mathrm e}^{10}+5+x \right )-27 \,{\mathrm e}^{-9 \,{\mathrm e}^{10}+5+x} {\mathrm e}^{10}+81 \,{\mathrm e}^{-9 \,{\mathrm e}^{10}+5+x} {\mathrm e}^{20}+18 \,{\mathrm e}^{10} \left ({\mathrm e}^{-9 \,{\mathrm e}^{10}+5+x} \left (-9 \,{\mathrm e}^{10}+5+x \right )-{\mathrm e}^{-9 \,{\mathrm e}^{10}+5+x}\right )\)	\(116\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2+7*x+5)*exp(-9*exp(5)^2+5+x),x,method=_RETURNVERBOSE)

[Out]

(5+x)*x*exp(-9*exp(5)^2+5+x)

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maxima [B]  time = 0.34, size = 53, normalized size = 3.53 \begin {gather*} {\left (x^{2} e^{5} - 2 \, x e^{5} + 2 \, e^{5}\right )} e^{\left (x - 9 \, e^{10}\right )} + 7 \, {\left (x e^{5} - e^{5}\right )} e^{\left (x - 9 \, e^{10}\right )} + 5 \, e^{\left (x - 9 \, e^{10} + 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2+7*x+5)*exp(-9*exp(5)^2+5+x),x, algorithm="maxima")

[Out]

(x^2*e^5 - 2*x*e^5 + 2*e^5)*e^(x - 9*e^10) + 7*(x*e^5 - e^5)*e^(x - 9*e^10) + 5*e^(x - 9*e^10 + 5)

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mupad [B]  time = 3.01, size = 14, normalized size = 0.93 \begin {gather*} x\,{\mathrm {e}}^{-9\,{\mathrm {e}}^{10}}\,{\mathrm {e}}^5\,{\mathrm {e}}^x\,\left (x+5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(x - 9*exp(10) + 5)*(7*x + x^2 + 5),x)

[Out]

x*exp(-9*exp(10))*exp(5)*exp(x)*(x + 5)

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sympy [A]  time = 0.09, size = 15, normalized size = 1.00 \begin {gather*} \left (x^{2} + 5 x\right ) e^{x - 9 e^{10} + 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**2+7*x+5)*exp(-9*exp(5)**2+5+x),x)

[Out]

(x**2 + 5*x)*exp(x - 9*exp(10) + 5)

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