3.78.44 \(\int e^{1-e^x+e^{1-e^x+x+x \log (2 e^{2 x} x^2)}+x+x \log (2 e^{2 x} x^2)} (3-e^x+2 x+\log (2 e^{2 x} x^2)) \, dx\)

Optimal. Leaf size=25 \[ e^{e^{1-e^x+x+x \log \left (2 e^{2 x} x^2\right )}} \]

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Rubi [F]  time = 2.78, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \exp \left (1-e^x+e^{1-e^x+x+x \log \left (2 e^{2 x} x^2\right )}+x+x \log \left (2 e^{2 x} x^2\right )\right ) \left (3-e^x+2 x+\log \left (2 e^{2 x} x^2\right )\right ) \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[E^(1 - E^x + E^(1 - E^x + x + x*Log[2*E^(2*x)*x^2]) + x + x*Log[2*E^(2*x)*x^2])*(3 - E^x + 2*x + Log[2*E^(
2*x)*x^2]),x]

[Out]

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (3 \exp \left (1-e^x+e^{1-e^x+x+x \log \left (2 e^{2 x} x^2\right )}+x+x \log \left (2 e^{2 x} x^2\right )\right )-\exp \left (1-e^x+e^{1-e^x+x+x \log \left (2 e^{2 x} x^2\right )}+2 x+x \log \left (2 e^{2 x} x^2\right )\right )+2 \exp \left (1-e^x+e^{1-e^x+x+x \log \left (2 e^{2 x} x^2\right )}+x+x \log \left (2 e^{2 x} x^2\right )\right ) x+\exp \left (1-e^x+e^{1-e^x+x+x \log \left (2 e^{2 x} x^2\right )}+x+x \log \left (2 e^{2 x} x^2\right )\right ) \log \left (2 e^{2 x} x^2\right )\right ) \, dx\\ &=2 \int \exp \left (1-e^x+e^{1-e^x+x+x \log \left (2 e^{2 x} x^2\right )}+x+x \log \left (2 e^{2 x} x^2\right )\right ) x \, dx+3 \int \exp \left (1-e^x+e^{1-e^x+x+x \log \left (2 e^{2 x} x^2\right )}+x+x \log \left (2 e^{2 x} x^2\right )\right ) \, dx-\int \exp \left (1-e^x+e^{1-e^x+x+x \log \left (2 e^{2 x} x^2\right )}+2 x+x \log \left (2 e^{2 x} x^2\right )\right ) \, dx+\int \exp \left (1-e^x+e^{1-e^x+x+x \log \left (2 e^{2 x} x^2\right )}+x+x \log \left (2 e^{2 x} x^2\right )\right ) \log \left (2 e^{2 x} x^2\right ) \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 2.38, size = 27, normalized size = 1.08 \begin {gather*} e^{2^x e^{1-e^x+x} \left (e^{2 x} x^2\right )^x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[E^(1 - E^x + E^(1 - E^x + x + x*Log[2*E^(2*x)*x^2]) + x + x*Log[2*E^(2*x)*x^2])*(3 - E^x + 2*x + Log
[2*E^(2*x)*x^2]),x]

[Out]

E^(2^x*E^(1 - E^x + x)*(E^(2*x)*x^2)^x)

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fricas [A]  time = 1.04, size = 21, normalized size = 0.84 \begin {gather*} e^{\left (e^{\left (x \log \left (2 \, x^{2} e^{\left (2 \, x\right )}\right ) + x - e^{x} + 1\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((log(2*exp(2*x)*x^2)-exp(x)+2*x+3)*exp(x*log(2*exp(2*x)*x^2)+1-exp(x)+x)*exp(exp(x*log(2*exp(2*x)*x^
2)+1-exp(x)+x)),x, algorithm="fricas")

[Out]

e^(e^(x*log(2*x^2*e^(2*x)) + x - e^x + 1))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int {\left (2 \, x - e^{x} + \log \left (2 \, x^{2} e^{\left (2 \, x\right )}\right ) + 3\right )} e^{\left (x \log \left (2 \, x^{2} e^{\left (2 \, x\right )}\right ) + x + e^{\left (x \log \left (2 \, x^{2} e^{\left (2 \, x\right )}\right ) + x - e^{x} + 1\right )} - e^{x} + 1\right )}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((log(2*exp(2*x)*x^2)-exp(x)+2*x+3)*exp(x*log(2*exp(2*x)*x^2)+1-exp(x)+x)*exp(exp(x*log(2*exp(2*x)*x^
2)+1-exp(x)+x)),x, algorithm="giac")

[Out]

integrate((2*x - e^x + log(2*x^2*e^(2*x)) + 3)*e^(x*log(2*x^2*e^(2*x)) + x + e^(x*log(2*x^2*e^(2*x)) + x - e^x
+ 1) - e^x + 1), x)

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maple [C]  time = 0.47, size = 234, normalized size = 9.36

method	result	size
risch	\({\mathrm e}^{2^{x} x^{2 x} \left ({\mathrm e}^{x}\right )^{2 x} {\mathrm e}^{1-\frac {i \pi x \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )}{2}-\frac {i x \pi \,\mathrm {csgn}\left (i {\mathrm e}^{2 x}\right ) \mathrm {csgn}\left (i {\mathrm e}^{x}\right )^{2}}{2}-\frac {i \pi x \mathrm {csgn}\left (i x^{2}\right )^{3}}{2}+i x \pi \mathrm {csgn}\left (i x^{2}\right )^{2} \mathrm {csgn}\left (i x \right )-\frac {i x \pi \,\mathrm {csgn}\left (i {\mathrm e}^{2 x}\right ) \mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i x^{2} {\mathrm e}^{2 x}\right )}{2}+i x \pi \mathrm {csgn}\left (i {\mathrm e}^{2 x}\right )^{2} \mathrm {csgn}\left (i {\mathrm e}^{x}\right )-\frac {i x \pi \mathrm {csgn}\left (i {\mathrm e}^{2 x}\right )^{3}}{2}+\frac {i x \pi \,\mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i x^{2} {\mathrm e}^{2 x}\right )^{2}}{2}-\frac {i x \pi \mathrm {csgn}\left (i x^{2} {\mathrm e}^{2 x}\right )^{3}}{2}+\frac {i x \pi \,\mathrm {csgn}\left (i {\mathrm e}^{2 x}\right ) \mathrm {csgn}\left (i x^{2} {\mathrm e}^{2 x}\right )^{2}}{2}-{\mathrm e}^{x}+x}}\)	\(234\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((ln(2*exp(2*x)*x^2)-exp(x)+2*x+3)*exp(x*ln(2*exp(2*x)*x^2)+1-exp(x)+x)*exp(exp(x*ln(2*exp(2*x)*x^2)+1-exp(
x)+x)),x,method=_RETURNVERBOSE)

[Out]

exp(2^x*x^(2*x)*exp(x)^(2*x)*exp(1-1/2*I*Pi*x*csgn(I*x)^2*csgn(I*x^2)-1/2*I*x*Pi*csgn(I*exp(2*x))*csgn(I*exp(x
))^2-1/2*I*Pi*x*csgn(I*x^2)^3+I*x*Pi*csgn(I*x^2)^2*csgn(I*x)-1/2*I*x*Pi*csgn(I*exp(2*x))*csgn(I*x^2)*csgn(I*x^
2*exp(2*x))+I*x*Pi*csgn(I*exp(2*x))^2*csgn(I*exp(x))-1/2*I*x*Pi*csgn(I*exp(2*x))^3+1/2*I*x*Pi*csgn(I*x^2)*csgn
(I*x^2*exp(2*x))^2-1/2*I*x*Pi*csgn(I*x^2*exp(2*x))^3+1/2*I*x*Pi*csgn(I*exp(2*x))*csgn(I*x^2*exp(2*x))^2-exp(x)
+x))

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maxima [A]  time = 0.69, size = 23, normalized size = 0.92 \begin {gather*} e^{\left (e^{\left (2 \, x^{2} + x \log \relax (2) + 2 \, x \log \relax (x) + x - e^{x} + 1\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((log(2*exp(2*x)*x^2)-exp(x)+2*x+3)*exp(x*log(2*exp(2*x)*x^2)+1-exp(x)+x)*exp(exp(x*log(2*exp(2*x)*x^
2)+1-exp(x)+x)),x, algorithm="maxima")

[Out]

e^(e^(2*x^2 + x*log(2) + 2*x*log(x) + x - e^x + 1))

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mupad [B]  time = 5.18, size = 24, normalized size = 0.96 \begin {gather*} {\mathrm {e}}^{\mathrm {e}\,{\mathrm {e}}^{2\,x^2}\,{\mathrm {e}}^{-{\mathrm {e}}^x}\,{\mathrm {e}}^x\,{\left (2\,x^2\right )}^x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(x - exp(x) + x*log(2*x^2*exp(2*x)) + 1)*exp(exp(x - exp(x) + x*log(2*x^2*exp(2*x)) + 1))*(2*x - exp(x)
+ log(2*x^2*exp(2*x)) + 3),x)

[Out]

exp(exp(1)*exp(2*x^2)*exp(-exp(x))*exp(x)*(2*x^2)^x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((ln(2*exp(2*x)*x**2)-exp(x)+2*x+3)*exp(x*ln(2*exp(2*x)*x**2)+1-exp(x)+x)*exp(exp(x*ln(2*exp(2*x)*x**
2)+1-exp(x)+x)),x)

[Out]

Timed out

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