\(\int e^{-e-e^x+x+x^2} (1-e^x+2 x) \log ^2(25) \, dx\) [1178]

Optimal result

Integrand size = 30, antiderivative size = 20 \[ \int e^{-e-e^x+x+x^2} \left (1-e^x+2 x\right ) \log ^2(25) \, dx=e^{-e-e^x+x+x^2} \log ^2(25) \]

[Out]

exp(ln(4*ln(5)^2)-exp(x)-exp(1)+x^2+x)

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {12, 6838} \[ \int e^{-e-e^x+x+x^2} \left (1-e^x+2 x\right ) \log ^2(25) \, dx=e^{x^2+x-e^x-e} \log ^2(25) \]

[In]

Int[E^(-E - E^x + x + x^2)*(1 - E^x + 2*x)*Log[25]^2,x]

[Out]

E^(-E - E^x + x + x^2)*Log[25]^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 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}& = \log ^2(25) \int e^{-e-e^x+x+x^2} \left (1-e^x+2 x\right ) \, dx \\ & = e^{-e-e^x+x+x^2} \log ^2(25) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int e^{-e-e^x+x+x^2} \left (1-e^x+2 x\right ) \log ^2(25) \, dx=e^{-e-e^x+x+x^2} \log ^2(25) \]

[In]

Integrate[E^(-E - E^x + x + x^2)*(1 - E^x + 2*x)*Log[25]^2,x]

[Out]

E^(-E - E^x + x + x^2)*Log[25]^2

Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.05

method	result	size
risch	\(4 \ln \left (5\right )^{2} {\mathrm e}^{-{\mathrm e}^{x}-{\mathrm e}+x^{2}+x}\)	\(21\)
derivativedivides	\({\mathrm e}^{\ln \left (4 \ln \left (5\right )^{2}\right )-{\mathrm e}^{x}-{\mathrm e}+x^{2}+x}\)	\(22\)
default	\({\mathrm e}^{\ln \left (4 \ln \left (5\right )^{2}\right )-{\mathrm e}^{x}-{\mathrm e}+x^{2}+x}\)	\(22\)
norman	\({\mathrm e}^{\ln \left (4 \ln \left (5\right )^{2}\right )-{\mathrm e}^{x}-{\mathrm e}+x^{2}+x}\)	\(22\)
parallelrisch	\({\mathrm e}^{\ln \left (4 \ln \left (5\right )^{2}\right )-{\mathrm e}^{x}-{\mathrm e}+x^{2}+x}\)	\(22\)

[In]

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

[Out]

4*ln(5)^2*exp(-exp(x)-exp(1)+x^2+x)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.05 \[ \int e^{-e-e^x+x+x^2} \left (1-e^x+2 x\right ) \log ^2(25) \, dx=e^{\left (x^{2} + x - e - e^{x} + \log \left (4 \, \log \left (5\right )^{2}\right )\right )} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95 \[ \int e^{-e-e^x+x+x^2} \left (1-e^x+2 x\right ) \log ^2(25) \, dx=4 e^{x^{2} + x - e^{x} - e} \log {\left (5 \right )}^{2} \]

[In]

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

[Out]

4*exp(x**2 + x - exp(x) - E)*log(5)**2

Maxima [A] (verification not implemented)

none

Time = 0.17 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int e^{-e-e^x+x+x^2} \left (1-e^x+2 x\right ) \log ^2(25) \, dx=4 \, e^{\left (x^{2} + x - e - e^{x}\right )} \log \left (5\right )^{2} \]

[In]

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

[Out]

4*e^(x^2 + x - e - e^x)*log(5)^2

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.05 \[ \int e^{-e-e^x+x+x^2} \left (1-e^x+2 x\right ) \log ^2(25) \, dx=e^{\left (x^{2} + x - e - e^{x} + \log \left (4 \, \log \left (5\right )^{2}\right )\right )} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.10 \[ \int e^{-e-e^x+x+x^2} \left (1-e^x+2 x\right ) \log ^2(25) \, dx=4\,{\mathrm {e}}^{-\mathrm {e}}\,{\mathrm {e}}^{x^2}\,{\mathrm {e}}^{-{\mathrm {e}}^x}\,{\mathrm {e}}^x\,{\ln \left (5\right )}^2 \]

[In]

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

[Out]

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