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

Optimal. Leaf size=27 \[ \frac {1}{2} \log \left (x^2\right ) \log \left (x+\log \left (\left (1-\frac {e^2}{2}+e^x+x\right )^2\right )\right ) \]

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Rubi [F]  time = 2.71, 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 {\left (6 x-e^2 x+6 e^x x+2 x^2\right ) \log \left (x^2\right )+\left (4 x-2 e^2 x+4 e^x x+4 x^2+\left (4-2 e^2+4 e^x+4 x\right ) \log \left (\frac {1}{4} \left (4+e^4+4 e^{2 x}+e^2 (-4-4 x)+8 x+4 x^2+e^x \left (8-4 e^2+8 x\right )\right )\right )\right ) \log \left (x+\log \left (\frac {1}{4} \left (4+e^4+4 e^{2 x}+e^2 (-4-4 x)+8 x+4 x^2+e^x \left (8-4 e^2+8 x\right )\right )\right )\right )}{4 x^2-2 e^2 x^2+4 e^x x^2+4 x^3+\left (4 x-2 e^2 x+4 e^x x+4 x^2\right ) \log \left (\frac {1}{4} \left (4+e^4+4 e^{2 x}+e^2 (-4-4 x)+8 x+4 x^2+e^x \left (8-4 e^2+8 x\right )\right )\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[((6*x - E^2*x + 6*E^x*x + 2*x^2)*Log[x^2] + (4*x - 2*E^2*x + 4*E^x*x + 4*x^2 + (4 - 2*E^2 + 4*E^x + 4*x)*L
og[(4 + E^4 + 4*E^(2*x) + E^2*(-4 - 4*x) + 8*x + 4*x^2 + E^x*(8 - 4*E^2 + 8*x))/4])*Log[x + Log[(4 + E^4 + 4*E
^(2*x) + E^2*(-4 - 4*x) + 8*x + 4*x^2 + E^x*(8 - 4*E^2 + 8*x))/4]])/(4*x^2 - 2*E^2*x^2 + 4*E^x*x^2 + 4*x^3 + (
4*x - 2*E^2*x + 4*E^x*x + 4*x^2)*Log[(4 + E^4 + 4*E^(2*x) + E^2*(-4 - 4*x) + 8*x + 4*x^2 + E^x*(8 - 4*E^2 + 8*
x))/4]),x]

[Out]

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\left (6 x-e^2 x+6 e^x x+2 x^2\right ) \log \left (x^2\right )+\left (4 x-2 e^2 x+4 e^x x+4 x^2+\left (4-2 e^2+4 e^x+4 x\right ) \log \left (\frac {1}{4} \left (4+e^4+4 e^{2 x}+e^2 (-4-4 x)+8 x+4 x^2+e^x \left (8-4 e^2+8 x\right )\right )\right )\right ) \log \left (x+\log \left (\frac {1}{4} \left (4+e^4+4 e^{2 x}+e^2 (-4-4 x)+8 x+4 x^2+e^x \left (8-4 e^2+8 x\right )\right )\right )\right )}{4 e^x x^2+\left (4-2 e^2\right ) x^2+4 x^3+\left (4 x-2 e^2 x+4 e^x x+4 x^2\right ) \log \left (\frac {1}{4} \left (4+e^4+4 e^{2 x}+e^2 (-4-4 x)+8 x+4 x^2+e^x \left (8-4 e^2+8 x\right )\right )\right )} \, dx\\ &=\int \left (\frac {\left (6 e^x+6 \left (1-\frac {e^2}{6}\right )+2 x\right ) \log \left (x^2\right )}{2 \left (2 e^x+2 \left (1-\frac {e^2}{2}\right )+2 x\right ) \left (x+\log \left (\frac {1}{4} \left (2-e^2+2 e^x+2 x\right )^2\right )\right )}+\frac {\log \left (x+\log \left (\frac {1}{4} \left (2-e^2+2 e^x+2 x\right )^2\right )\right )}{x}\right ) \, dx\\ &=\frac {1}{2} \int \frac {\left (6 e^x+6 \left (1-\frac {e^2}{6}\right )+2 x\right ) \log \left (x^2\right )}{\left (2 e^x+2 \left (1-\frac {e^2}{2}\right )+2 x\right ) \left (x+\log \left (\frac {1}{4} \left (2-e^2+2 e^x+2 x\right )^2\right )\right )} \, dx+\int \frac {\log \left (x+\log \left (\frac {1}{4} \left (2-e^2+2 e^x+2 x\right )^2\right )\right )}{x} \, dx\\ &=\frac {1}{2} \int \left (\frac {3 \log \left (x^2\right )}{x+\log \left (\frac {1}{4} \left (2-e^2+2 e^x+2 x\right )^2\right )}+\frac {2 \left (e^2-2 x\right ) \log \left (x^2\right )}{\left (2 e^x+2 \left (1-\frac {e^2}{2}\right )+2 x\right ) \left (x+\log \left (\frac {1}{4} \left (2-e^2+2 e^x+2 x\right )^2\right )\right )}\right ) \, dx+\int \frac {\log \left (x+\log \left (\frac {1}{4} \left (2-e^2+2 e^x+2 x\right )^2\right )\right )}{x} \, dx\\ &=\frac {3}{2} \int \frac {\log \left (x^2\right )}{x+\log \left (\frac {1}{4} \left (2-e^2+2 e^x+2 x\right )^2\right )} \, dx+\int \frac {\left (e^2-2 x\right ) \log \left (x^2\right )}{\left (2 e^x+2 \left (1-\frac {e^2}{2}\right )+2 x\right ) \left (x+\log \left (\frac {1}{4} \left (2-e^2+2 e^x+2 x\right )^2\right )\right )} \, dx+\int \frac {\log \left (x+\log \left (\frac {1}{4} \left (2-e^2+2 e^x+2 x\right )^2\right )\right )}{x} \, dx\\ &=\frac {3}{2} \int \frac {\log \left (x^2\right )}{x+\log \left (\frac {1}{4} \left (2-e^2+2 e^x+2 x\right )^2\right )} \, dx+\int \left (\frac {2 x \log \left (x^2\right )}{\left (-2 e^x-2 \left (1-\frac {e^2}{2}\right )-2 x\right ) \left (x+\log \left (\frac {1}{4} \left (2-e^2+2 e^x+2 x\right )^2\right )\right )}+\frac {e^2 \log \left (x^2\right )}{\left (2 e^x+2 \left (1-\frac {e^2}{2}\right )+2 x\right ) \left (x+\log \left (\frac {1}{4} \left (2-e^2+2 e^x+2 x\right )^2\right )\right )}\right ) \, dx+\int \frac {\log \left (x+\log \left (\frac {1}{4} \left (2-e^2+2 e^x+2 x\right )^2\right )\right )}{x} \, dx\\ &=\frac {3}{2} \int \frac {\log \left (x^2\right )}{x+\log \left (\frac {1}{4} \left (2-e^2+2 e^x+2 x\right )^2\right )} \, dx+2 \int \frac {x \log \left (x^2\right )}{\left (-2 e^x-2 \left (1-\frac {e^2}{2}\right )-2 x\right ) \left (x+\log \left (\frac {1}{4} \left (2-e^2+2 e^x+2 x\right )^2\right )\right )} \, dx+e^2 \int \frac {\log \left (x^2\right )}{\left (2 e^x+2 \left (1-\frac {e^2}{2}\right )+2 x\right ) \left (x+\log \left (\frac {1}{4} \left (2-e^2+2 e^x+2 x\right )^2\right )\right )} \, dx+\int \frac {\log \left (x+\log \left (\frac {1}{4} \left (2-e^2+2 e^x+2 x\right )^2\right )\right )}{x} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 1.09, size = 33, normalized size = 1.22 \begin {gather*} \frac {1}{2} \log \left (x^2\right ) \log \left (x+\log \left (\frac {1}{4} \left (2-e^2+2 e^x+2 x\right )^2\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((6*x - E^2*x + 6*E^x*x + 2*x^2)*Log[x^2] + (4*x - 2*E^2*x + 4*E^x*x + 4*x^2 + (4 - 2*E^2 + 4*E^x +
4*x)*Log[(4 + E^4 + 4*E^(2*x) + E^2*(-4 - 4*x) + 8*x + 4*x^2 + E^x*(8 - 4*E^2 + 8*x))/4])*Log[x + Log[(4 + E^4
+ 4*E^(2*x) + E^2*(-4 - 4*x) + 8*x + 4*x^2 + E^x*(8 - 4*E^2 + 8*x))/4]])/(4*x^2 - 2*E^2*x^2 + 4*E^x*x^2 + 4*x
^3 + (4*x - 2*E^2*x + 4*E^x*x + 4*x^2)*Log[(4 + E^4 + 4*E^(2*x) + E^2*(-4 - 4*x) + 8*x + 4*x^2 + E^x*(8 - 4*E^
2 + 8*x))/4]),x]

[Out]

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

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fricas [A]  time = 1.23, size = 45, normalized size = 1.67 \begin {gather*} \frac {1}{2} \, \log \left (x^{2}\right ) \log \left (x + \log \left (x^{2} - {\left (x + 1\right )} e^{2} + {\left (2 \, x - e^{2} + 2\right )} e^{x} + 2 \, x + \frac {1}{4} \, e^{4} + e^{\left (2 \, x\right )} + 1\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((4*exp(x)-2*exp(2)+4*x+4)*log(exp(x)^2+1/4*(-4*exp(2)+8*x+8)*exp(x)+1/4*exp(2)^2+1/4*(-4*x-4)*exp(
2)+x^2+2*x+1)+4*exp(x)*x-2*exp(2)*x+4*x^2+4*x)*log(log(exp(x)^2+1/4*(-4*exp(2)+8*x+8)*exp(x)+1/4*exp(2)^2+1/4*
(-4*x-4)*exp(2)+x^2+2*x+1)+x)+(6*exp(x)*x-exp(2)*x+2*x^2+6*x)*log(x^2))/((4*exp(x)*x-2*exp(2)*x+4*x^2+4*x)*log
(exp(x)^2+1/4*(-4*exp(2)+8*x+8)*exp(x)+1/4*exp(2)^2+1/4*(-4*x-4)*exp(2)+x^2+2*x+1)+4*exp(x)*x^2-2*x^2*exp(2)+4
*x^3+4*x^2),x, algorithm="fricas")

[Out]

1/2*log(x^2)*log(x + log(x^2 - (x + 1)*e^2 + (2*x - e^2 + 2)*e^x + 2*x + 1/4*e^4 + e^(2*x) + 1))

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giac [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*exp(x)-2*exp(2)+4*x+4)*log(exp(x)^2+1/4*(-4*exp(2)+8*x+8)*exp(x)+1/4*exp(2)^2+1/4*(-4*x-4)*exp(
2)+x^2+2*x+1)+4*exp(x)*x-2*exp(2)*x+4*x^2+4*x)*log(log(exp(x)^2+1/4*(-4*exp(2)+8*x+8)*exp(x)+1/4*exp(2)^2+1/4*
(-4*x-4)*exp(2)+x^2+2*x+1)+x)+(6*exp(x)*x-exp(2)*x+2*x^2+6*x)*log(x^2))/((4*exp(x)*x-2*exp(2)*x+4*x^2+4*x)*log
(exp(x)^2+1/4*(-4*exp(2)+8*x+8)*exp(x)+1/4*exp(2)^2+1/4*(-4*x-4)*exp(2)+x^2+2*x+1)+4*exp(x)*x^2-2*x^2*exp(2)+4
*x^3+4*x^2),x, algorithm="giac")

[Out]

Timed out

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maple [C]  time = 2.05, size = 247, normalized size = 9.15

method	result	size
risch	\(\ln \relax (x ) \ln \left (-2 \ln \relax (2)+2 \ln \left (-2 \,{\mathrm e}^{x}+{\mathrm e}^{2}-2 x -2\right )-\frac {i \pi \,\mathrm {csgn}\left (i \left (-2 \,{\mathrm e}^{x}+{\mathrm e}^{2}-2 x -2\right )^{2}\right ) \left (-\mathrm {csgn}\left (i \left (-2 \,{\mathrm e}^{x}+{\mathrm e}^{2}-2 x -2\right )^{2}\right )+\mathrm {csgn}\left (i \left (-2 \,{\mathrm e}^{x}+{\mathrm e}^{2}-2 x -2\right )\right )\right )^{2}}{2}+x \right )-\frac {i \pi \,\mathrm {csgn}\left (i x^{2}\right ) \left (\mathrm {csgn}\left (i x \right )^{2}-2 \,\mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i x \right )+\mathrm {csgn}\left (i x^{2}\right )^{2}\right ) \ln \left (\ln \left (-2 \,{\mathrm e}^{x}+{\mathrm e}^{2}-2 x -2\right )-\frac {i \left (\pi \mathrm {csgn}\left (i \left (-2 \,{\mathrm e}^{x}+{\mathrm e}^{2}-2 x -2\right )\right )^{2} \mathrm {csgn}\left (i \left (-2 \,{\mathrm e}^{x}+{\mathrm e}^{2}-2 x -2\right )^{2}\right )-2 \pi \,\mathrm {csgn}\left (i \left (-2 \,{\mathrm e}^{x}+{\mathrm e}^{2}-2 x -2\right )\right ) \mathrm {csgn}\left (i \left (-2 \,{\mathrm e}^{x}+{\mathrm e}^{2}-2 x -2\right )^{2}\right )^{2}+\pi \mathrm {csgn}\left (i \left (-2 \,{\mathrm e}^{x}+{\mathrm e}^{2}-2 x -2\right )^{2}\right )^{3}-4 i \ln \relax (2)+2 i x \right )}{4}\right )}{4}\)	\(247\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((4*exp(x)-2*exp(2)+4*x+4)*ln(exp(x)^2+1/4*(-4*exp(2)+8*x+8)*exp(x)+1/4*exp(2)^2+1/4*(-4*x-4)*exp(2)+x^2+
2*x+1)+4*exp(x)*x-2*exp(2)*x+4*x^2+4*x)*ln(ln(exp(x)^2+1/4*(-4*exp(2)+8*x+8)*exp(x)+1/4*exp(2)^2+1/4*(-4*x-4)*
exp(2)+x^2+2*x+1)+x)+(6*exp(x)*x-exp(2)*x+2*x^2+6*x)*ln(x^2))/((4*exp(x)*x-2*exp(2)*x+4*x^2+4*x)*ln(exp(x)^2+1
/4*(-4*exp(2)+8*x+8)*exp(x)+1/4*exp(2)^2+1/4*(-4*x-4)*exp(2)+x^2+2*x+1)+4*exp(x)*x^2-2*x^2*exp(2)+4*x^3+4*x^2)
,x,method=_RETURNVERBOSE)

[Out]

ln(x)*ln(-2*ln(2)+2*ln(-2*exp(x)+exp(2)-2*x-2)-1/2*I*Pi*csgn(I*(-2*exp(x)+exp(2)-2*x-2)^2)*(-csgn(I*(-2*exp(x)
+exp(2)-2*x-2)^2)+csgn(I*(-2*exp(x)+exp(2)-2*x-2)))^2+x)-1/4*I*Pi*csgn(I*x^2)*(csgn(I*x)^2-2*csgn(I*x^2)*csgn(
I*x)+csgn(I*x^2)^2)*ln(ln(-2*exp(x)+exp(2)-2*x-2)-1/4*I*(Pi*csgn(I*(-2*exp(x)+exp(2)-2*x-2))^2*csgn(I*(-2*exp(
x)+exp(2)-2*x-2)^2)-2*Pi*csgn(I*(-2*exp(x)+exp(2)-2*x-2))*csgn(I*(-2*exp(x)+exp(2)-2*x-2)^2)^2+Pi*csgn(I*(-2*e
xp(x)+exp(2)-2*x-2)^2)^3-4*I*ln(2)+2*I*x))

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maxima [A]  time = 0.68, size = 26, normalized size = 0.96 \begin {gather*} \log \left (x - 2 \, \log \relax (2) + 2 \, \log \left (2 \, x - e^{2} + 2 \, e^{x} + 2\right )\right ) \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((4*exp(x)-2*exp(2)+4*x+4)*log(exp(x)^2+1/4*(-4*exp(2)+8*x+8)*exp(x)+1/4*exp(2)^2+1/4*(-4*x-4)*exp(
2)+x^2+2*x+1)+4*exp(x)*x-2*exp(2)*x+4*x^2+4*x)*log(log(exp(x)^2+1/4*(-4*exp(2)+8*x+8)*exp(x)+1/4*exp(2)^2+1/4*
(-4*x-4)*exp(2)+x^2+2*x+1)+x)+(6*exp(x)*x-exp(2)*x+2*x^2+6*x)*log(x^2))/((4*exp(x)*x-2*exp(2)*x+4*x^2+4*x)*log
(exp(x)^2+1/4*(-4*exp(2)+8*x+8)*exp(x)+1/4*exp(2)^2+1/4*(-4*x-4)*exp(2)+x^2+2*x+1)+4*exp(x)*x^2-2*x^2*exp(2)+4
*x^3+4*x^2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 4.81, size = 48, normalized size = 1.78 \begin {gather*} \frac {\ln \left (x^2\right )\,\ln \left (x+\ln \left (2\,x+{\mathrm {e}}^{2\,x}+\frac {{\mathrm {e}}^4}{4}+\frac {{\mathrm {e}}^x\,\left (8\,x-4\,{\mathrm {e}}^2+8\right )}{4}+x^2-\frac {{\mathrm {e}}^2\,\left (4\,x+4\right )}{4}+1\right )\right )}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(x^2)*(6*x - x*exp(2) + 6*x*exp(x) + 2*x^2) + log(x + log(2*x + exp(2*x) + exp(4)/4 + (exp(x)*(8*x - 4
*exp(2) + 8))/4 + x^2 - (exp(2)*(4*x + 4))/4 + 1))*(4*x - 2*x*exp(2) + log(2*x + exp(2*x) + exp(4)/4 + (exp(x)
*(8*x - 4*exp(2) + 8))/4 + x^2 - (exp(2)*(4*x + 4))/4 + 1)*(4*x - 2*exp(2) + 4*exp(x) + 4) + 4*x*exp(x) + 4*x^
2))/(4*x^2*exp(x) + log(2*x + exp(2*x) + exp(4)/4 + (exp(x)*(8*x - 4*exp(2) + 8))/4 + x^2 - (exp(2)*(4*x + 4))
/4 + 1)*(4*x - 2*x*exp(2) + 4*x*exp(x) + 4*x^2) - 2*x^2*exp(2) + 4*x^2 + 4*x^3),x)

[Out]

(log(x^2)*log(x + log(2*x + exp(2*x) + exp(4)/4 + (exp(x)*(8*x - 4*exp(2) + 8))/4 + x^2 - (exp(2)*(4*x + 4))/4
+ 1)))/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*exp(x)-2*exp(2)+4*x+4)*ln(exp(x)**2+1/4*(-4*exp(2)+8*x+8)*exp(x)+1/4*exp(2)**2+1/4*(-4*x-4)*exp
(2)+x**2+2*x+1)+4*exp(x)*x-2*exp(2)*x+4*x**2+4*x)*ln(ln(exp(x)**2+1/4*(-4*exp(2)+8*x+8)*exp(x)+1/4*exp(2)**2+1
/4*(-4*x-4)*exp(2)+x**2+2*x+1)+x)+(6*exp(x)*x-exp(2)*x+2*x**2+6*x)*ln(x**2))/((4*exp(x)*x-2*exp(2)*x+4*x**2+4*
x)*ln(exp(x)**2+1/4*(-4*exp(2)+8*x+8)*exp(x)+1/4*exp(2)**2+1/4*(-4*x-4)*exp(2)+x**2+2*x+1)+4*exp(x)*x**2-2*x**
2*exp(2)+4*x**3+4*x**2),x)

[Out]

Timed out

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