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

Optimal. Leaf size=33 \[ \left (1-e^3\right ) \left (-x+\log \left (x+x \log (x)-\log \left (\frac {x}{2}+\log (2+\log (2))\right )\right )\right ) \]

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Rubi [A]  time = 0.57, antiderivative size = 40, normalized size of antiderivative = 1.21, number of steps used = 5, number of rules used = 4, integrand size = 172, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.023, Rules used = {6688, 12, 6742, 6684} \begin {gather*} \left (1-e^3\right ) \log \left (x+x \log (x)-\log \left (\frac {x}{2}+\log (2+\log (2))\right )\right )-\left (1-e^3\right ) x \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 6684

Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*Log[RemoveContent[y, x]], x] /;  !Fa
lseQ[q]]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

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

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Mathematica [A]  time = 0.05, size = 31, normalized size = 0.94 \begin {gather*} \left (-1+e^3\right ) \left (x-\log \left (x+x \log (x)-\log \left (\frac {x}{2}+\log (2+\log (2))\right )\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.51, size = 34, normalized size = 1.03 \begin {gather*} x e^{3} - {\left (e^{3} - 1\right )} \log \left (-x \log \relax (x) - x + \log \left (\frac {1}{2} \, x + \log \left (\log \relax (2) + 2\right )\right )\right ) - x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.23, size = 54, normalized size = 1.64 \begin {gather*} x e^{3} - e^{3} \log \left (x \log \relax (x) + x + \log \relax (2) - \log \left (x + 2 \, \log \left (\log \relax (2) + 2\right )\right )\right ) - x + \log \left (x \log \relax (x) + x + \log \relax (2) - \log \left (x + 2 \, \log \left (\log \relax (2) + 2\right )\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.28, size = 34, normalized size = 1.03

method	result	size
norman	\(\left ({\mathrm e}^{3}-1\right ) x +\left (-{\mathrm e}^{3}+1\right ) \ln \left (x +x \ln \relax (x )-\ln \left (\ln \left (\ln \relax (2)+2\right )+\frac {x}{2}\right )\right )\)	\(34\)
risch	\(x \,{\mathrm e}^{3}-x -\ln \left (\ln \left (\ln \left (\ln \relax (2)+2\right )+\frac {x}{2}\right )-\left (\ln \relax (x )+1\right ) x \right ) {\mathrm e}^{3}+\ln \left (\ln \left (\ln \left (\ln \relax (2)+2\right )+\frac {x}{2}\right )-\left (\ln \relax (x )+1\right ) x \right )\)	\(51\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.50, size = 37, normalized size = 1.12 \begin {gather*} x {\left (e^{3} - 1\right )} - {\left (e^{3} - 1\right )} \log \left (-x \log \relax (x) - x - \log \relax (2) + \log \left (x + 2 \, \log \left (\log \relax (2) + 2\right )\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int -\frac {\ln \relax (x)\,\left ({\mathrm {e}}^3\,\left (x-x^2\right )-x+x^2\right )-2\,x-\ln \left (\ln \relax (2)+2\right )\,\left (\ln \relax (x)\,\left ({\mathrm {e}}^3\,\left (2\,x-2\right )-2\,x+2\right )-2\,x+{\mathrm {e}}^3\,\left (2\,x-4\right )+4\right )+\ln \left (\frac {x}{2}+\ln \left (\ln \relax (2)+2\right )\right )\,\left (x\,{\mathrm {e}}^3-x+\ln \left (\ln \relax (2)+2\right )\,\left (2\,{\mathrm {e}}^3-2\right )\right )-{\mathrm {e}}^3\,\left (x^2-2\,x+1\right )+x^2+1}{x^2\,\ln \relax (x)+\ln \left (\ln \relax (2)+2\right )\,\left (2\,x+2\,x\,\ln \relax (x)\right )-\ln \left (\frac {x}{2}+\ln \left (\ln \relax (2)+2\right )\right )\,\left (x+2\,\ln \left (\ln \relax (2)+2\right )\right )+x^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.53, size = 39, normalized size = 1.18 \begin {gather*} x \left (-1 + e^{3}\right ) - \left (-1 + e\right ) \left (1 + e + e^{2}\right ) \log {\left (- x \log {\relax (x )} - x + \log {\left (\frac {x}{2} + \log {\left (\log {\relax (2 )} + 2 \right )} \right )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*exp(3)-2)*ln(ln(2)+2)+x*exp(3)-x)*ln(ln(ln(2)+2)+1/2*x)+(((-2*x+2)*exp(3)+2*x-2)*ln(x)+(4-2*x)*
exp(3)+2*x-4)*ln(ln(2)+2)+((-x**2+x)*exp(3)+x**2-x)*ln(x)+(-x**2+2*x-1)*exp(3)+x**2-2*x+1)/((2*ln(ln(2)+2)+x)*
ln(ln(ln(2)+2)+1/2*x)+(-2*x*ln(x)-2*x)*ln(ln(2)+2)-x**2*ln(x)-x**2),x)

[Out]

x*(-1 + exp(3)) - (-1 + E)*(1 + E + exp(2))*log(-x*log(x) - x + log(x/2 + log(log(2) + 2)))

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