3.83.67 \(\int e^{25+5 x+5 x^2-\frac {e^x x^2}{2}} (5+10 x+\frac {1}{2} e^x (-2 x-x^2)) \, dx\)

Optimal. Leaf size=19 \[ e^{25+x \left (5+5 x-\frac {e^x x}{2}\right )} \]

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Rubi [A]  time = 0.15, antiderivative size = 22, normalized size of antiderivative = 1.16, number of steps used = 1, number of rules used = 1, integrand size = 44, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.023, Rules used = {6706} \begin {gather*} e^{-\frac {1}{2} e^x x^2+5 x^2+5 x+25} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^(25 + 5*x + 5*x^2 - (E^x*x^2)/2)*(5 + 10*x + (E^x*(-2*x - x^2))/2),x]

[Out]

E^(25 + 5*x + 5*x^2 - (E^x*x^2)/2)

Rule 6706

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[(q*F^v)/Log[F], x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=e^{25+5 x+5 x^2-\frac {e^x x^2}{2}}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.31, size = 22, normalized size = 1.16 \begin {gather*} e^{25+5 x+5 x^2-\frac {e^x x^2}{2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[E^(25 + 5*x + 5*x^2 - (E^x*x^2)/2)*(5 + 10*x + (E^x*(-2*x - x^2))/2),x]

[Out]

E^(25 + 5*x + 5*x^2 - (E^x*x^2)/2)

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fricas [A]  time = 1.41, size = 23, normalized size = 1.21 \begin {gather*} e^{\left (-x^{2} e^{\left (x - \log \relax (2)\right )} + 5 \, x^{2} + 5 \, x + 25\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^2-2*x)*exp(x-log(2))+10*x+5)*exp(-x^2*exp(x-log(2))+5*x^2+5*x+25),x, algorithm="fricas")

[Out]

e^(-x^2*e^(x - log(2)) + 5*x^2 + 5*x + 25)

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giac [A]  time = 0.22, size = 23, normalized size = 1.21 \begin {gather*} e^{\left (-x^{2} e^{\left (x - \log \relax (2)\right )} + 5 \, x^{2} + 5 \, x + 25\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^2-2*x)*exp(x-log(2))+10*x+5)*exp(-x^2*exp(x-log(2))+5*x^2+5*x+25),x, algorithm="giac")

[Out]

e^(-x^2*e^(x - log(2)) + 5*x^2 + 5*x + 25)

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maple [A]  time = 0.06, size = 19, normalized size = 1.00

method	result	size
risch	\({\mathrm e}^{-\frac {{\mathrm e}^{x} x^{2}}{2}+5 x^{2}+5 x +25}\)	\(19\)
norman	\({\mathrm e}^{-x^{2} {\mathrm e}^{x -\ln \relax (2)}+5 x^{2}+5 x +25}\)	\(24\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-x^2-2*x)*exp(x-ln(2))+10*x+5)*exp(-x^2*exp(x-ln(2))+5*x^2+5*x+25),x,method=_RETURNVERBOSE)

[Out]

exp(-1/2*exp(x)*x^2+5*x^2+5*x+25)

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maxima [A]  time = 0.46, size = 18, normalized size = 0.95 \begin {gather*} e^{\left (-\frac {1}{2} \, x^{2} e^{x} + 5 \, x^{2} + 5 \, x + 25\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^2-2*x)*exp(x-log(2))+10*x+5)*exp(-x^2*exp(x-log(2))+5*x^2+5*x+25),x, algorithm="maxima")

[Out]

e^(-1/2*x^2*e^x + 5*x^2 + 5*x + 25)

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mupad [B]  time = 0.11, size = 21, normalized size = 1.11 \begin {gather*} {\mathrm {e}}^{5\,x}\,{\mathrm {e}}^{25}\,{\mathrm {e}}^{-\frac {x^2\,{\mathrm {e}}^x}{2}}\,{\mathrm {e}}^{5\,x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(5*x - x^2*exp(x - log(2)) + 5*x^2 + 25)*(10*x - exp(x - log(2))*(2*x + x^2) + 5),x)

[Out]

exp(5*x)*exp(25)*exp(-(x^2*exp(x))/2)*exp(5*x^2)

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sympy [A]  time = 0.20, size = 19, normalized size = 1.00 \begin {gather*} e^{- \frac {x^{2} e^{x}}{2} + 5 x^{2} + 5 x + 25} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x**2-2*x)*exp(x-ln(2))+10*x+5)*exp(-x**2*exp(x-ln(2))+5*x**2+5*x+25),x)

[Out]

exp(-x**2*exp(x)/2 + 5*x**2 + 5*x + 25)

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