3.90.69 \(\int \frac {1}{2} (2+e^{2+2 x+e^{2 x} (2+2 x)+(-2-2 x) \log (x)} (-1+e^{2 x} (6 x+4 x^2)-2 x \log (x))) \, dx\)

Optimal. Leaf size=29 \[ \frac {1}{2} \left (2+e^{\left (2+\frac {2}{x}\right ) x \left (1+e^{2 x}-\log (x)\right )}\right ) x \]

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Rubi [F]  time = 1.59, 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 \frac {1}{2} \left (2+\exp \left (2+2 x+e^{2 x} (2+2 x)+(-2-2 x) \log (x)\right ) \left (-1+e^{2 x} \left (6 x+4 x^2\right )-2 x \log (x)\right )\right ) \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(2 + E^(2 + 2*x + E^(2*x)*(2 + 2*x) + (-2 - 2*x)*Log[x])*(-1 + E^(2*x)*(6*x + 4*x^2) - 2*x*Log[x]))/2,x]

[Out]

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{2} \int \left (2+\exp \left (2+2 x+e^{2 x} (2+2 x)+(-2-2 x) \log (x)\right ) \left (-1+e^{2 x} \left (6 x+4 x^2\right )-2 x \log (x)\right )\right ) \, dx\\ &=x+\frac {1}{2} \int \exp \left (2+2 x+e^{2 x} (2+2 x)+(-2-2 x) \log (x)\right ) \left (-1+e^{2 x} \left (6 x+4 x^2\right )-2 x \log (x)\right ) \, dx\\ &=x+\frac {1}{2} \int e^{2 (1+x) \left (1+e^{2 x}-\log (x)\right )} \left (-1+e^{2 x} \left (6 x+4 x^2\right )-2 x \log (x)\right ) \, dx\\ &=x+\frac {1}{2} \int \left (-e^{2 (1+x) \left (1+e^{2 x}-\log (x)\right )}+2 e^{2 x+2 (1+x) \left (1+e^{2 x}-\log (x)\right )} x (3+2 x)-2 e^{2 (1+x) \left (1+e^{2 x}-\log (x)\right )} x \log (x)\right ) \, dx\\ &=x-\frac {1}{2} \int e^{2 (1+x) \left (1+e^{2 x}-\log (x)\right )} \, dx+\int e^{2 x+2 (1+x) \left (1+e^{2 x}-\log (x)\right )} x (3+2 x) \, dx-\int e^{2 (1+x) \left (1+e^{2 x}-\log (x)\right )} x \log (x) \, dx\\ &=x-\frac {1}{2} \int e^{2 (1+x) \left (1+e^{2 x}-\log (x)\right )} \, dx+\int \left (3 e^{2 x+2 (1+x) \left (1+e^{2 x}-\log (x)\right )} x+2 e^{2 x+2 (1+x) \left (1+e^{2 x}-\log (x)\right )} x^2\right ) \, dx-\int e^{2 (1+x) \left (1+e^{2 x}-\log (x)\right )} x \log (x) \, dx\\ &=x-\frac {1}{2} \int e^{2 (1+x) \left (1+e^{2 x}-\log (x)\right )} \, dx+2 \int e^{2 x+2 (1+x) \left (1+e^{2 x}-\log (x)\right )} x^2 \, dx+3 \int e^{2 x+2 (1+x) \left (1+e^{2 x}-\log (x)\right )} x \, dx-\int e^{2 (1+x) \left (1+e^{2 x}-\log (x)\right )} x \log (x) \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.56, size = 32, normalized size = 1.10 \begin {gather*} x+\frac {1}{2} e^{2+2 x+2 e^{2 x} (1+x)} x^{1-2 (1+x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2 + E^(2 + 2*x + E^(2*x)*(2 + 2*x) + (-2 - 2*x)*Log[x])*(-1 + E^(2*x)*(6*x + 4*x^2) - 2*x*Log[x]))/
2,x]

[Out]

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

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fricas [A]  time = 0.80, size = 27, normalized size = 0.93 \begin {gather*} \frac {1}{2} \, x e^{\left (2 \, {\left (x + 1\right )} e^{\left (2 \, x\right )} - 2 \, {\left (x + 1\right )} \log \relax (x) + 2 \, x + 2\right )} + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(-2*x*log(x)+(4*x^2+6*x)*exp(2*x)-1)*exp((-2*x-2)*log(x)+(2*x+2)*exp(2*x)+2*x+2)+1,x, algorithm=
"fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(-2*x*log(x)+(4*x^2+6*x)*exp(2*x)-1)*exp((-2*x-2)*log(x)+(2*x+2)*exp(2*x)+2*x+2)+1,x, algorithm=
"giac")

[Out]

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

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maple [A]  time = 0.10, size = 25, normalized size = 0.86

method	result	size
risch	\(\frac {x \,x^{-2 x -2} {\mathrm e}^{2 \left (x +1\right ) \left ({\mathrm e}^{2 x}+1\right )}}{2}+x\)	\(25\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/2*(-2*x*ln(x)+(4*x^2+6*x)*exp(2*x)-1)*exp((-2*x-2)*ln(x)+(2*x+2)*exp(2*x)+2*x+2)+1,x,method=_RETURNVERBO
SE)

[Out]

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

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maxima [A]  time = 0.42, size = 31, normalized size = 1.07 \begin {gather*} x + \frac {e^{\left (2 \, x e^{\left (2 \, x\right )} - 2 \, x \log \relax (x) + 2 \, x + 2 \, e^{\left (2 \, x\right )} + 2\right )}}{2 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(-2*x*log(x)+(4*x^2+6*x)*exp(2*x)-1)*exp((-2*x-2)*log(x)+(2*x+2)*exp(2*x)+2*x+2)+1,x, algorithm=
"maxima")

[Out]

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

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mupad [B]  time = 7.44, size = 35, normalized size = 1.21 \begin {gather*} x+\frac {{\mathrm {e}}^{2\,{\mathrm {e}}^{2\,x}}\,{\mathrm {e}}^{2\,x}\,{\mathrm {e}}^2\,{\mathrm {e}}^{2\,x\,{\mathrm {e}}^{2\,x}}}{2\,x^{2\,x}\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1 - (exp(2*x - log(x)*(2*x + 2) + exp(2*x)*(2*x + 2) + 2)*(2*x*log(x) - exp(2*x)*(6*x + 4*x^2) + 1))/2,x)

[Out]

x + (exp(2*exp(2*x))*exp(2*x)*exp(2)*exp(2*x*exp(2*x)))/(2*x^(2*x)*x)

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sympy [A]  time = 0.47, size = 31, normalized size = 1.07 \begin {gather*} \frac {x e^{2 x + \left (- 2 x - 2\right ) \log {\relax (x )} + \left (2 x + 2\right ) e^{2 x} + 2}}{2} + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(-2*x*ln(x)+(4*x**2+6*x)*exp(2*x)-1)*exp((-2*x-2)*ln(x)+(2*x+2)*exp(2*x)+2*x+2)+1,x)

[Out]

x*exp(2*x + (-2*x - 2)*log(x) + (2*x + 2)*exp(2*x) + 2)/2 + x

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