\(\int (4+4 e^{10}+4 \log (5)+e^{2 e^e} (1+e^{10}+\log (5))) \, dx\) [5853]

Optimal result

Integrand size = 26, antiderivative size = 18 \[ \int \left (4+4 e^{10}+4 \log (5)+e^{2 e^e} \left (1+e^{10}+\log (5)\right )\right ) \, dx=\left (4+e^{2 e^e}\right ) x \left (1+e^{10}+\log (5)\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.038, Rules used = {8} \[ \int \left (4+4 e^{10}+4 \log (5)+e^{2 e^e} \left (1+e^{10}+\log (5)\right )\right ) \, dx=\left (4+e^{2 e^e}\right ) x \left (1+e^{10}+\log (5)\right ) \]

[In]

Int[4 + 4*E^10 + 4*Log[5] + E^(2*E^E)*(1 + E^10 + Log[5]),x]

[Out]

(4 + E^(2*E^E))*x*(1 + E^10 + Log[5])

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps \begin{align*} \text {integral}& = \left (4+e^{2 e^e}\right ) x \left (1+e^{10}+\log (5)\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.72 \[ \int \left (4+4 e^{10}+4 \log (5)+e^{2 e^e} \left (1+e^{10}+\log (5)\right )\right ) \, dx=4 x+4 e^{10} x+4 x \log (5)+e^{2 e^e} x \left (1+e^{10}+\log (5)\right ) \]

[In]

Integrate[4 + 4*E^10 + 4*Log[5] + E^(2*E^E)*(1 + E^10 + Log[5]),x]

[Out]

4*x + 4*E^10*x + 4*x*Log[5] + E^(2*E^E)*x*(1 + E^10 + Log[5])

Maple [A] (verified)

Time = 0.02 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.67

method	result	size
default	\(x \left (\left (\ln \left (5\right )+{\mathrm e}^{10}+1\right ) {\mathrm e}^{2 \,{\mathrm e}^{{\mathrm e}}}+4 \ln \left (5\right )+4 \,{\mathrm e}^{10}+4\right )\)	\(30\)
parallelrisch	\(x \left (\left (\ln \left (5\right )+{\mathrm e}^{10}+1\right ) {\mathrm e}^{2 \,{\mathrm e}^{{\mathrm e}}}+4 \ln \left (5\right )+4 \,{\mathrm e}^{10}+4\right )\)	\(30\)
norman	\(\left ({\mathrm e}^{2 \,{\mathrm e}^{{\mathrm e}}} {\mathrm e}^{10}+{\mathrm e}^{2 \,{\mathrm e}^{{\mathrm e}}} \ln \left (5\right )+{\mathrm e}^{2 \,{\mathrm e}^{{\mathrm e}}}+4 \,{\mathrm e}^{10}+4 \ln \left (5\right )+4\right ) x\)	\(41\)
risch	\({\mathrm e}^{2 \,{\mathrm e}^{{\mathrm e}}} x \,{\mathrm e}^{10}+{\mathrm e}^{2 \,{\mathrm e}^{{\mathrm e}}} x \ln \left (5\right )+4 x \,{\mathrm e}^{10}+{\mathrm e}^{2 \,{\mathrm e}^{{\mathrm e}}} x +4 x \ln \left (5\right )+4 x\)	\(43\)
parts	\({\mathrm e}^{2 \,{\mathrm e}^{{\mathrm e}}} x \,{\mathrm e}^{10}+{\mathrm e}^{2 \,{\mathrm e}^{{\mathrm e}}} x \ln \left (5\right )+4 x \,{\mathrm e}^{10}+{\mathrm e}^{2 \,{\mathrm e}^{{\mathrm e}}} x +4 x \ln \left (5\right )+4 x\)	\(47\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.72 \[ \int \left (4+4 e^{10}+4 \log (5)+e^{2 e^e} \left (1+e^{10}+\log (5)\right )\right ) \, dx=4 \, x e^{10} + {\left (x e^{10} + x \log \left (5\right ) + x\right )} e^{\left (2 \, e^{e}\right )} + 4 \, x \log \left (5\right ) + 4 \, x \]

[In]

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

[Out]

4*x*e^10 + (x*e^10 + x*log(5) + x)*e^(2*e^e) + 4*x*log(5) + 4*x

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.61 \[ \int \left (4+4 e^{10}+4 \log (5)+e^{2 e^e} \left (1+e^{10}+\log (5)\right )\right ) \, dx=x \left (4 + 4 \log {\left (5 \right )} + 4 e^{10} + \left (1 + \log {\left (5 \right )} + e^{10}\right ) e^{2 e^{e}}\right ) \]

[In]

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

[Out]

x*(4 + 4*log(5) + 4*exp(10) + (1 + log(5) + exp(10))*exp(2*exp(E)))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.39 \[ \int \left (4+4 e^{10}+4 \log (5)+e^{2 e^e} \left (1+e^{10}+\log (5)\right )\right ) \, dx={\left ({\left (e^{10} + \log \left (5\right ) + 1\right )} e^{\left (2 \, e^{e}\right )} + 4 \, e^{10} + 4 \, \log \left (5\right ) + 4\right )} x \]

[In]

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

[Out]

((e^10 + log(5) + 1)*e^(2*e^e) + 4*e^10 + 4*log(5) + 4)*x

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.39 \[ \int \left (4+4 e^{10}+4 \log (5)+e^{2 e^e} \left (1+e^{10}+\log (5)\right )\right ) \, dx={\left ({\left (e^{10} + \log \left (5\right ) + 1\right )} e^{\left (2 \, e^{e}\right )} + 4 \, e^{10} + 4 \, \log \left (5\right ) + 4\right )} x \]

[In]

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

[Out]

((e^10 + log(5) + 1)*e^(2*e^e) + 4*e^10 + 4*log(5) + 4)*x

Mupad [B] (verification not implemented)

Time = 0.00 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.39 \[ \int \left (4+4 e^{10}+4 \log (5)+e^{2 e^e} \left (1+e^{10}+\log (5)\right )\right ) \, dx=x\,\left (4\,{\mathrm {e}}^{10}+4\,\ln \left (5\right )+{\mathrm {e}}^{2\,{\mathrm {e}}^{\mathrm {e}}}\,\left ({\mathrm {e}}^{10}+\ln \left (5\right )+1\right )+4\right ) \]

[In]

int(4*exp(10) + 4*log(5) + exp(2*exp(exp(1)))*(exp(10) + log(5) + 1) + 4,x)

[Out]

x*(4*exp(10) + 4*log(5) + exp(2*exp(exp(1)))*(exp(10) + log(5) + 1) + 4)