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

Optimal. Leaf size=30 \[ \frac {1}{2} e^{-x} \log (2) \left (x (3+\log (4))+\log \left (-x+x^2+x \log (x)\right )\right ) \]

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Rubi [F]  time = 3.34, 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 (-x+6 x^2-3 x^3\right ) \log (2)+\left (-x+2 x^2-x^3\right ) \log (2) \log (4)+\left (\left (1+3 x-3 x^2\right ) \log (2)+\left (x-x^2\right ) \log (2) \log (4)\right ) \log (x)+\left (\left (x-x^2\right ) \log (2)-x \log (2) \log (x)\right ) \log \left (-x+x^2+x \log (x)\right )}{e^x \left (-2 x+2 x^2\right )+2 e^x x \log (x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[((-x + 6*x^2 - 3*x^3)*Log[2] + (-x + 2*x^2 - x^3)*Log[2]*Log[4] + ((1 + 3*x - 3*x^2)*Log[2] + (x - x^2)*Lo
g[2]*Log[4])*Log[x] + ((x - x^2)*Log[2] - x*Log[2]*Log[x])*Log[-x + x^2 + x*Log[x]])/(E^x*(-2*x + 2*x^2) + 2*E
^x*x*Log[x]),x]

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 0.09, size = 26, normalized size = 0.87 \begin {gather*} \frac {1}{2} e^{-x} \log (2) (x (3+\log (4))+\log (x (-1+x+\log (x)))) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((-x + 6*x^2 - 3*x^3)*Log[2] + (-x + 2*x^2 - x^3)*Log[2]*Log[4] + ((1 + 3*x - 3*x^2)*Log[2] + (x - x
^2)*Log[2]*Log[4])*Log[x] + ((x - x^2)*Log[2] - x*Log[2]*Log[x])*Log[-x + x^2 + x*Log[x]])/(E^x*(-2*x + 2*x^2)
+ 2*E^x*x*Log[x]),x]

[Out]

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

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fricas [A]  time = 1.14, size = 34, normalized size = 1.13 \begin {gather*} \frac {1}{2} \, {\left (2 \, x \log \relax (2)^{2} + 3 \, x \log \relax (2) + \log \relax (2) \log \left (x^{2} + x \log \relax (x) - x\right )\right )} e^{\left (-x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.20, size = 44, normalized size = 1.47 \begin {gather*} x e^{\left (-x\right )} \log \relax (2)^{2} + \frac {3}{2} \, x e^{\left (-x\right )} \log \relax (2) + \frac {1}{2} \, e^{\left (-x\right )} \log \relax (2) \log \left (x + \log \relax (x) - 1\right ) + \frac {1}{2} \, e^{\left (-x\right )} \log \relax (2) \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.09, size = 127, normalized size = 4.23

method	result	size
risch	\(\frac {\ln \relax (2) {\mathrm e}^{-x} \ln \left (-1+\ln \relax (x )+x \right )}{2}+\frac {\ln \relax (2) \left (-i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i \left (-1+\ln \relax (x )+x \right )\right ) \mathrm {csgn}\left (i x \left (-1+\ln \relax (x )+x \right )\right )+i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x \left (-1+\ln \relax (x )+x \right )\right )^{2}+i \pi \,\mathrm {csgn}\left (i \left (-1+\ln \relax (x )+x \right )\right ) \mathrm {csgn}\left (i x \left (-1+\ln \relax (x )+x \right )\right )^{2}-i \pi \mathrm {csgn}\left (i x \left (-1+\ln \relax (x )+x \right )\right )^{3}+4 x \ln \relax (2)+6 x +2 \ln \relax (x )\right ) {\mathrm e}^{-x}}{4}\)	\(127\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*ln(2)*exp(-x)*ln(-1+ln(x)+x)+1/4*ln(2)*(-I*Pi*csgn(I*x)*csgn(I*(-1+ln(x)+x))*csgn(I*x*(-1+ln(x)+x))+I*Pi*c
sgn(I*x)*csgn(I*x*(-1+ln(x)+x))^2+I*Pi*csgn(I*(-1+ln(x)+x))*csgn(I*x*(-1+ln(x)+x))^2-I*Pi*csgn(I*x*(-1+ln(x)+x
))^3+4*x*ln(2)+6*x+2*ln(x))*exp(-x)

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maxima [A]  time = 0.52, size = 34, normalized size = 1.13 \begin {gather*} \frac {1}{2} \, {\left ({\left (2 \, \log \relax (2)^{2} + 3 \, \log \relax (2)\right )} x + \log \relax (2) \log \left (x + \log \relax (x) - 1\right ) + \log \relax (2) \log \relax (x)\right )} e^{\left (-x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int \frac {\ln \relax (2)\,\left (3\,x^3-6\,x^2+x\right )+2\,{\ln \relax (2)}^2\,\left (x^3-2\,x^2+x\right )-\ln \relax (x)\,\left (\ln \relax (2)\,\left (-3\,x^2+3\,x+1\right )+2\,{\ln \relax (2)}^2\,\left (x-x^2\right )\right )-\ln \left (x\,\ln \relax (x)-x+x^2\right )\,\left (\ln \relax (2)\,\left (x-x^2\right )-x\,\ln \relax (2)\,\ln \relax (x)\right )}{{\mathrm {e}}^x\,\left (2\,x-2\,x^2\right )-2\,x\,{\mathrm {e}}^x\,\ln \relax (x)} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 20.63, size = 34, normalized size = 1.13 \begin {gather*} \frac {\left (2 x \log {\relax (2 )}^{2} + 3 x \log {\relax (2 )} + \log {\relax (2 )} \log {\left (x^{2} + x \log {\relax (x )} - x \right )}\right ) e^{- x}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x*ln(2)*ln(x)+(-x**2+x)*ln(2))*ln(x*ln(x)+x**2-x)+(2*(-x**2+x)*ln(2)**2+(-3*x**2+3*x+1)*ln(2))*ln
(x)+2*(-x**3+2*x**2-x)*ln(2)**2+(-3*x**3+6*x**2-x)*ln(2))/(2*x*exp(x)*ln(x)+(2*x**2-2*x)*exp(x)),x)

[Out]

(2*x*log(2)**2 + 3*x*log(2) + log(2)*log(x**2 + x*log(x) - x))*exp(-x)/2

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