\(\int e^{25+5 x+5 x^2-\frac {e^x x^2}{2}} (5+10 x+\frac {1}{2} e^x (-2 x-x^2)) \, dx\) [8267]

Optimal result

Integrand size = 44, antiderivative size = 19 \[ \int e^{25+5 x+5 x^2-\frac {e^x x^2}{2}} \left (5+10 x+\frac {1}{2} e^x \left (-2 x-x^2\right )\right ) \, dx=e^{25+x \left (5+5 x-\frac {e^x x}{2}\right )} \]

[Out]

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

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.16, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.023, Rules used = {6838} \[ \int e^{25+5 x+5 x^2-\frac {e^x x^2}{2}} \left (5+10 x+\frac {1}{2} e^x \left (-2 x-x^2\right )\right ) \, dx=e^{-\frac {1}{2} e^x x^2+5 x^2+5 x+25} \]

[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 6838

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

Mathematica [A] (verified)

Time = 0.46 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.16 \[ \int e^{25+5 x+5 x^2-\frac {e^x x^2}{2}} \left (5+10 x+\frac {1}{2} e^x \left (-2 x-x^2\right )\right ) \, dx=e^{25+5 x+5 x^2-\frac {e^x x^2}{2}} \]

[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)

Maple [A] (verified)

Time = 0.15 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00

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

[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)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.21 \[ \int e^{25+5 x+5 x^2-\frac {e^x x^2}{2}} \left (5+10 x+\frac {1}{2} e^x \left (-2 x-x^2\right )\right ) \, dx=e^{\left (-x^{2} e^{\left (x - \log \left (2\right )\right )} + 5 \, x^{2} + 5 \, x + 25\right )} \]

[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)

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int e^{25+5 x+5 x^2-\frac {e^x x^2}{2}} \left (5+10 x+\frac {1}{2} e^x \left (-2 x-x^2\right )\right ) \, dx=e^{- \frac {x^{2} e^{x}}{2} + 5 x^{2} + 5 x + 25} \]

[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)

Maxima [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.95 \[ \int e^{25+5 x+5 x^2-\frac {e^x x^2}{2}} \left (5+10 x+\frac {1}{2} e^x \left (-2 x-x^2\right )\right ) \, dx=e^{\left (-\frac {1}{2} \, x^{2} e^{x} + 5 \, x^{2} + 5 \, x + 25\right )} \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.21 \[ \int e^{25+5 x+5 x^2-\frac {e^x x^2}{2}} \left (5+10 x+\frac {1}{2} e^x \left (-2 x-x^2\right )\right ) \, dx=e^{\left (-x^{2} e^{\left (x - \log \left (2\right )\right )} + 5 \, x^{2} + 5 \, x + 25\right )} \]

[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)

Mupad [B] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.11 \[ \int e^{25+5 x+5 x^2-\frac {e^x x^2}{2}} \left (5+10 x+\frac {1}{2} e^x \left (-2 x-x^2\right )\right ) \, dx={\mathrm {e}}^{5\,x}\,{\mathrm {e}}^{25}\,{\mathrm {e}}^{-\frac {x^2\,{\mathrm {e}}^x}{2}}\,{\mathrm {e}}^{5\,x^2} \]

[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)