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

Optimal. Leaf size=18 \[ \left (4+e^{2 e^e}\right ) x \left (1+e^{10}+\log (5)\right ) \]

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Rubi [A]  time = 0.01, antiderivative size = 18, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.038, Rules used = {8} \begin {gather*} \left (4+e^{2 e^e}\right ) x \left (1+e^{10}+\log (5)\right ) \end {gather*}

Antiderivative was successfully verified.

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

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Mathematica [A]  time = 0.00, size = 31, normalized size = 1.72 \begin {gather*} 4 x+4 e^{10} x+4 x \log (5)+e^{2 e^e} x \left (1+e^{10}+\log (5)\right ) \end {gather*}

Antiderivative was successfully verified.

[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])

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fricas [A]  time = 0.84, size = 31, normalized size = 1.72 \begin {gather*} 4 \, x e^{10} + {\left (x e^{10} + x \log \relax (5) + x\right )} e^{\left (2 \, e^{e}\right )} + 4 \, x \log \relax (5) + 4 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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giac [A]  time = 0.14, size = 25, normalized size = 1.39 \begin {gather*} {\left ({\left (e^{10} + \log \relax (5) + 1\right )} e^{\left (2 \, e^{e}\right )} + 4 \, e^{10} + 4 \, \log \relax (5) + 4\right )} x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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maple [A]  time = 0.03, size = 30, normalized size = 1.67

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

Verification of antiderivative is not currently implemented for this CAS.

[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]

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

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maxima [A]  time = 0.36, size = 25, normalized size = 1.39 \begin {gather*} {\left ({\left (e^{10} + \log \relax (5) + 1\right )} e^{\left (2 \, e^{e}\right )} + 4 \, e^{10} + 4 \, \log \relax (5) + 4\right )} x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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mupad [B]  time = 0.00, size = 25, normalized size = 1.39 \begin {gather*} x\,\left (4\,{\mathrm {e}}^{10}+4\,\ln \relax (5)+{\mathrm {e}}^{2\,{\mathrm {e}}^{\mathrm {e}}}\,\left ({\mathrm {e}}^{10}+\ln \relax (5)+1\right )+4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 0.05, size = 29, normalized size = 1.61 \begin {gather*} x \left (4 + 4 \log {\relax (5 )} + 4 e^{10} + \left (1 + \log {\relax (5 )} + e^{10}\right ) e^{2 e^{e}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)))

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