∫ elog2(12(4+x2+log(2+ex(−2+x))))(8x+ex(−2−6x+4x2)) log(1 2(4+x2+log(2+ex(−2+x)))) ___________________________________________________________________________________________________________________8+2x2 +ex (−8+4x−2x2 +x3 )+(2+ex (−2+x)) log (2+ex (−2+x)) dx

3.46.79 \(\int \frac {e^{\log ^2(\frac {1}{2} (4+x^2+\log (2+e^x (-2+x))))} (8 x+e^x (-2-6 x+4 x^2)) \log (\frac {1}{2} (4+x^2+\log (2+e^x (-2+x))))}{8+2 x^2+e^x (-8+4 x-2 x^2+x^3)+(2+e^x (-2+x)) \log (2+e^x (-2+x))} \, dx\)

Optimal. Leaf size=28 \[ e^{\log ^2\left (2+\frac {1}{2} \left (x^2+\log \left (2-e^x (2-x)\right )\right )\right )} \]

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Rubi [F] time = 13.20, 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 {\exp \left (\log ^2\left (\frac {1}{2} \left (4+x^2+\log \left (2+e^x (-2+x)\right )\right )\right )\right ) \left (8 x+e^x \left (-2-6 x+4 x^2\right )\right ) \log \left (\frac {1}{2} \left (4+x^2+\log \left (2+e^x (-2+x)\right )\right )\right )}{8+2 x^2+e^x \left (-8+4 x-2 x^2+x^3\right )+\left (2+e^x (-2+x)\right ) \log \left (2+e^x (-2+x)\right )} \, dx \end {gather*}

Veriﬁcation is not applicable to the result.

[In]

Int[(E^Log[(4 + x^2 + Log[2 + E^x*(-2 + x)])/2]^2*(8*x + E^x*(-2 - 6*x + 4*x^2))*Log[(4 + x^2 + Log[2 + E^x*(-

2 + x)])/2])/(8 + 2*x^2 + E^x*(-8 + 4*x - 2*x^2 + x^3) + (2 + E^x*(-2 + x))*Log[2 + E^x*(-2 + x)]),x]

[Out]

2*Defer[Int][(E^Log[(4 + x^2 + Log[2 + E^x*(-2 + x)])/2]^2*Log[(4 + x^2 + Log[2 + E^x*(-2 + x)])/2])/(4 + x^2

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

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

+ E^x*(-2 + x)])/2]^2*x*Log[(4 + x^2 + Log[2 + E^x*(-2 + x)])/2])/(4 + x^2 + Log[2 + E^x*(-2 + x)]), x] - 4*De

fer[Int][(E^Log[(4 + x^2 + Log[2 + E^x*(-2 + x)])/2]^2*Log[(4 + x^2 + Log[2 + E^x*(-2 + x)])/2])/((2 - 2*E^x +

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

g[(4 + x^2 + Log[2 + E^x*(-2 + x)])/2])/((-2 + x)*(2 - 2*E^x + E^x*x)*(4 + x^2 + Log[2 + E^x*(-2 + x)])), x]

Rubi steps

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

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Mathematica [A] time = 0.05, size = 24, normalized size = 0.86 \begin {gather*} e^{\log ^2\left (\frac {1}{2} \left (4+x^2+\log \left (2+e^x (-2+x)\right )\right )\right )} \end {gather*}

Antiderivative was successfully veriﬁed.

[In]

Integrate[(E^Log[(4 + x^2 + Log[2 + E^x*(-2 + x)])/2]^2*(8*x + E^x*(-2 - 6*x + 4*x^2))*Log[(4 + x^2 + Log[2 +

E^x*(-2 + x)])/2])/(8 + 2*x^2 + E^x*(-8 + 4*x - 2*x^2 + x^3) + (2 + E^x*(-2 + x))*Log[2 + E^x*(-2 + x)]),x]

[Out]

E^Log[(4 + x^2 + Log[2 + E^x*(-2 + x)])/2]^2

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fricas [A] time = 0.47, size = 22, normalized size = 0.79 \begin {gather*} e^{\left (\log \left (\frac {1}{2} \, x^{2} + \frac {1}{2} \, \log \left ({\left (x - 2\right )} e^{x} + 2\right ) + 2\right )^{2}\right )} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x^2-6*x-2)*exp(x)+8*x)*log(1/2*log(exp(x)*(x-2)+2)+1/2*x^2+2)*exp(log(1/2*log(exp(x)*(x-2)+2)+1/

2*x^2+2)^2)/((exp(x)*(x-2)+2)*log(exp(x)*(x-2)+2)+(x^3-2*x^2+4*x-8)*exp(x)+2*x^2+8),x, algorithm="fricas")

[Out]

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

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giac [A] time = 0.28, size = 24, normalized size = 0.86 \begin {gather*} e^{\left (\log \left (\frac {1}{2} \, x^{2} + \frac {1}{2} \, \log \left (x e^{x} - 2 \, e^{x} + 2\right ) + 2\right )^{2}\right )} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x^2-6*x-2)*exp(x)+8*x)*log(1/2*log(exp(x)*(x-2)+2)+1/2*x^2+2)*exp(log(1/2*log(exp(x)*(x-2)+2)+1/

2*x^2+2)^2)/((exp(x)*(x-2)+2)*log(exp(x)*(x-2)+2)+(x^3-2*x^2+4*x-8)*exp(x)+2*x^2+8),x, algorithm="giac")

[Out]

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

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maple [A] time = 0.04, size = 23, normalized size = 0.82


method	result	size

risch	\({\mathrm e}^{\ln \left (\frac {\ln \left ({\mathrm e}^{x} \left (x -2\right )+2\right )}{2}+\frac {x^{2}}{2}+2\right )^{2}}\)	\(23\)

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(((4*x^2-6*x-2)*exp(x)+8*x)*ln(1/2*ln(exp(x)*(x-2)+2)+1/2*x^2+2)*exp(ln(1/2*ln(exp(x)*(x-2)+2)+1/2*x^2+2)^2

)/((exp(x)*(x-2)+2)*ln(exp(x)*(x-2)+2)+(x^3-2*x^2+4*x-8)*exp(x)+2*x^2+8),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A] time = 0.67, size = 42, normalized size = 1.50 \begin {gather*} e^{\left (\log \relax (2)^{2} - 2 \, \log \relax (2) \log \left (x^{2} + \log \left ({\left (x - 2\right )} e^{x} + 2\right ) + 4\right ) + \log \left (x^{2} + \log \left ({\left (x - 2\right )} e^{x} + 2\right ) + 4\right )^{2}\right )} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x^2-6*x-2)*exp(x)+8*x)*log(1/2*log(exp(x)*(x-2)+2)+1/2*x^2+2)*exp(log(1/2*log(exp(x)*(x-2)+2)+1/

2*x^2+2)^2)/((exp(x)*(x-2)+2)*log(exp(x)*(x-2)+2)+(x^3-2*x^2+4*x-8)*exp(x)+2*x^2+8),x, algorithm="maxima")

[Out]

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

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mupad [B] time = 3.59, size = 24, normalized size = 0.86 \begin {gather*} {\mathrm {e}}^{{\ln \left (\frac {\ln \left (x\,{\mathrm {e}}^x-2\,{\mathrm {e}}^x+2\right )}{2}+\frac {x^2}{2}+2\right )}^2} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((log(log(exp(x)*(x - 2) + 2)/2 + x^2/2 + 2)*exp(log(log(exp(x)*(x - 2) + 2)/2 + x^2/2 + 2)^2)*(8*x - exp(x

)*(6*x - 4*x^2 + 2)))/(exp(x)*(4*x - 2*x^2 + x^3 - 8) + log(exp(x)*(x - 2) + 2)*(exp(x)*(x - 2) + 2) + 2*x^2 +

8),x)

[Out]

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

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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(((4*x**2-6*x-2)*exp(x)+8*x)*ln(1/2*ln(exp(x)*(x-2)+2)+1/2*x**2+2)*exp(ln(1/2*ln(exp(x)*(x-2)+2)+1/2*

x**2+2)**2)/((exp(x)*(x-2)+2)*ln(exp(x)*(x-2)+2)+(x**3-2*x**2+4*x-8)*exp(x)+2*x**2+8),x)

[Out]

Timed out

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