3.75.96 \(\int \frac {(8-2 e^x-2 x) \log (-4+e^x+x)+(12 x+21 x^2-6 x^3+e^x (-3 x-6 x^2)) \log (x) \log ^2(-4+e^x+x)+(x+e^x x) \log (x) \log (\log ^2(x))}{(-12 x+3 e^x x+3 x^2) \log (x) \log ^2(-4+e^x+x)} \, dx\)

Optimal. Leaf size=27 \[ -x-x^2-\frac {\log \left (\log ^2(x)\right )}{3 \log \left (-4+e^x+x\right )} \]

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Rubi [F]  time = 1.74, 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 (8-2 e^x-2 x\right ) \log \left (-4+e^x+x\right )+\left (12 x+21 x^2-6 x^3+e^x \left (-3 x-6 x^2\right )\right ) \log (x) \log ^2\left (-4+e^x+x\right )+\left (x+e^x x\right ) \log (x) \log \left (\log ^2(x)\right )}{\left (-12 x+3 e^x x+3 x^2\right ) \log (x) \log ^2\left (-4+e^x+x\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[((8 - 2*E^x - 2*x)*Log[-4 + E^x + x] + (12*x + 21*x^2 - 6*x^3 + E^x*(-3*x - 6*x^2))*Log[x]*Log[-4 + E^x +
x]^2 + (x + E^x*x)*Log[x]*Log[Log[x]^2])/((-12*x + 3*E^x*x + 3*x^2)*Log[x]*Log[-4 + E^x + x]^2),x]

[Out]

-x - x^2 - (2*Defer[Int][1/(x*Log[x]*Log[-4 + E^x + x]), x])/3 + Defer[Int][Log[Log[x]^2]/Log[-4 + E^x + x]^2,
x]/3 + (5*Defer[Int][Log[Log[x]^2]/((-4 + E^x + x)*Log[-4 + E^x + x]^2), x])/3 - Defer[Int][(x*Log[Log[x]^2])
/((-4 + E^x + x)*Log[-4 + E^x + x]^2), x]/3

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {1}{3} \left (-3-6 x-\frac {2}{x \log (x) \log \left (-4+e^x+x\right )}+\frac {\left (1+e^x\right ) \log \left (\log ^2(x)\right )}{\left (-4+e^x+x\right ) \log ^2\left (-4+e^x+x\right )}\right ) \, dx\\ &=\frac {1}{3} \int \left (-3-6 x-\frac {2}{x \log (x) \log \left (-4+e^x+x\right )}+\frac {\left (1+e^x\right ) \log \left (\log ^2(x)\right )}{\left (-4+e^x+x\right ) \log ^2\left (-4+e^x+x\right )}\right ) \, dx\\ &=-x-x^2+\frac {1}{3} \int \frac {\left (1+e^x\right ) \log \left (\log ^2(x)\right )}{\left (-4+e^x+x\right ) \log ^2\left (-4+e^x+x\right )} \, dx-\frac {2}{3} \int \frac {1}{x \log (x) \log \left (-4+e^x+x\right )} \, dx\\ &=-x-x^2+\frac {1}{3} \int \left (\frac {\log \left (\log ^2(x)\right )}{\log ^2\left (-4+e^x+x\right )}-\frac {(-5+x) \log \left (\log ^2(x)\right )}{\left (-4+e^x+x\right ) \log ^2\left (-4+e^x+x\right )}\right ) \, dx-\frac {2}{3} \int \frac {1}{x \log (x) \log \left (-4+e^x+x\right )} \, dx\\ &=-x-x^2+\frac {1}{3} \int \frac {\log \left (\log ^2(x)\right )}{\log ^2\left (-4+e^x+x\right )} \, dx-\frac {1}{3} \int \frac {(-5+x) \log \left (\log ^2(x)\right )}{\left (-4+e^x+x\right ) \log ^2\left (-4+e^x+x\right )} \, dx-\frac {2}{3} \int \frac {1}{x \log (x) \log \left (-4+e^x+x\right )} \, dx\\ &=-x-x^2+\frac {1}{3} \int \frac {\log \left (\log ^2(x)\right )}{\log ^2\left (-4+e^x+x\right )} \, dx-\frac {1}{3} \int \left (-\frac {5 \log \left (\log ^2(x)\right )}{\left (-4+e^x+x\right ) \log ^2\left (-4+e^x+x\right )}+\frac {x \log \left (\log ^2(x)\right )}{\left (-4+e^x+x\right ) \log ^2\left (-4+e^x+x\right )}\right ) \, dx-\frac {2}{3} \int \frac {1}{x \log (x) \log \left (-4+e^x+x\right )} \, dx\\ &=-x-x^2+\frac {1}{3} \int \frac {\log \left (\log ^2(x)\right )}{\log ^2\left (-4+e^x+x\right )} \, dx-\frac {1}{3} \int \frac {x \log \left (\log ^2(x)\right )}{\left (-4+e^x+x\right ) \log ^2\left (-4+e^x+x\right )} \, dx-\frac {2}{3} \int \frac {1}{x \log (x) \log \left (-4+e^x+x\right )} \, dx+\frac {5}{3} \int \frac {\log \left (\log ^2(x)\right )}{\left (-4+e^x+x\right ) \log ^2\left (-4+e^x+x\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.15, size = 29, normalized size = 1.07 \begin {gather*} \frac {1}{3} \left (-3 x-3 x^2-\frac {\log \left (\log ^2(x)\right )}{\log \left (-4+e^x+x\right )}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((8 - 2*E^x - 2*x)*Log[-4 + E^x + x] + (12*x + 21*x^2 - 6*x^3 + E^x*(-3*x - 6*x^2))*Log[x]*Log[-4 +
E^x + x]^2 + (x + E^x*x)*Log[x]*Log[Log[x]^2])/((-12*x + 3*E^x*x + 3*x^2)*Log[x]*Log[-4 + E^x + x]^2),x]

[Out]

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

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fricas [A]  time = 0.83, size = 29, normalized size = 1.07 \begin {gather*} -\frac {3 \, {\left (x^{2} + x\right )} \log \left (x + e^{x} - 4\right ) + \log \left (\log \relax (x)^{2}\right )}{3 \, \log \left (x + e^{x} - 4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(x)*x+x)*log(x)*log(log(x)^2)+((-6*x^2-3*x)*exp(x)-6*x^3+21*x^2+12*x)*log(x)*log(exp(x)+x-4)^2+
(-2*exp(x)-2*x+8)*log(exp(x)+x-4))/(3*exp(x)*x+3*x^2-12*x)/log(x)/log(exp(x)+x-4)^2,x, algorithm="fricas")

[Out]

-1/3*(3*(x^2 + x)*log(x + e^x - 4) + log(log(x)^2))/log(x + e^x - 4)

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giac [A]  time = 0.43, size = 36, normalized size = 1.33 \begin {gather*} -\frac {3 \, x^{2} \log \left (x + e^{x} - 4\right ) + 3 \, x \log \left (x + e^{x} - 4\right ) + \log \left (\log \relax (x)^{2}\right )}{3 \, \log \left (x + e^{x} - 4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(x)*x+x)*log(x)*log(log(x)^2)+((-6*x^2-3*x)*exp(x)-6*x^3+21*x^2+12*x)*log(x)*log(exp(x)+x-4)^2+
(-2*exp(x)-2*x+8)*log(exp(x)+x-4))/(3*exp(x)*x+3*x^2-12*x)/log(x)/log(exp(x)+x-4)^2,x, algorithm="giac")

[Out]

-1/3*(3*x^2*log(x + e^x - 4) + 3*x*log(x + e^x - 4) + log(log(x)^2))/log(x + e^x - 4)

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maple [C]  time = 0.17, size = 80, normalized size = 2.96

method	result	size
risch	\(-x^{2}-x -\frac {-i \pi \mathrm {csgn}\left (i \ln \relax (x )\right )^{2} \mathrm {csgn}\left (i \ln \relax (x )^{2}\right )+2 i \pi \,\mathrm {csgn}\left (i \ln \relax (x )\right ) \mathrm {csgn}\left (i \ln \relax (x )^{2}\right )^{2}-i \pi \mathrm {csgn}\left (i \ln \relax (x )^{2}\right )^{3}+4 \ln \left (\ln \relax (x )\right )}{6 \ln \left ({\mathrm e}^{x}+x -4\right )}\)	\(80\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((exp(x)*x+x)*ln(x)*ln(ln(x)^2)+((-6*x^2-3*x)*exp(x)-6*x^3+21*x^2+12*x)*ln(x)*ln(exp(x)+x-4)^2+(-2*exp(x)-
2*x+8)*ln(exp(x)+x-4))/(3*exp(x)*x+3*x^2-12*x)/ln(x)/ln(exp(x)+x-4)^2,x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.45, size = 29, normalized size = 1.07 \begin {gather*} -\frac {3 \, {\left (x^{2} + x\right )} \log \left (x + e^{x} - 4\right ) + 2 \, \log \left (\log \relax (x)\right )}{3 \, \log \left (x + e^{x} - 4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(x)*x+x)*log(x)*log(log(x)^2)+((-6*x^2-3*x)*exp(x)-6*x^3+21*x^2+12*x)*log(x)*log(exp(x)+x-4)^2+
(-2*exp(x)-2*x+8)*log(exp(x)+x-4))/(3*exp(x)*x+3*x^2-12*x)/log(x)/log(exp(x)+x-4)^2,x, algorithm="maxima")

[Out]

-1/3*(3*(x^2 + x)*log(x + e^x - 4) + 2*log(log(x)))/log(x + e^x - 4)

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mupad [B]  time = 4.86, size = 171, normalized size = 6.33 \begin {gather*} \frac {2}{3\,x}-\frac {\frac {\ln \left ({\ln \relax (x)}^2\right )}{3}-\frac {2\,\ln \left (x+{\mathrm {e}}^x-4\right )\,\left (x+{\mathrm {e}}^x-4\right )}{3\,x\,\ln \relax (x)\,\left ({\mathrm {e}}^x+1\right )}}{\ln \left (x+{\mathrm {e}}^x-4\right )}-\frac {\frac {2\,\left (x+{\mathrm {e}}^x-4\right )}{3\,x\,\left ({\mathrm {e}}^x+1\right )}-\frac {2\,\ln \relax (x)\,\left (3\,{\mathrm {e}}^x-{\mathrm {e}}^{2\,x}-x^2\,{\mathrm {e}}^x+5\,x\,{\mathrm {e}}^x+4\right )}{3\,x\,{\left ({\mathrm {e}}^x+1\right )}^2}}{\ln \relax (x)}-x-x^2+\frac {2\,\left (5\,x^2-x^3\right )}{3\,x^2\,\left ({\mathrm {e}}^{2\,x}+2\,{\mathrm {e}}^x+1\right )}-\frac {2\,\left (-x^3+5\,x^2+5\,x\right )}{3\,x^2\,\left ({\mathrm {e}}^x+1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(x + exp(x) - 4)^2*log(x)*(12*x - exp(x)*(3*x + 6*x^2) + 21*x^2 - 6*x^3) - log(x + exp(x) - 4)*(2*x +
2*exp(x) - 8) + log(log(x)^2)*log(x)*(x + x*exp(x)))/(log(x + exp(x) - 4)^2*log(x)*(3*x*exp(x) - 12*x + 3*x^2)
),x)

[Out]

2/(3*x) - (log(log(x)^2)/3 - (2*log(x + exp(x) - 4)*(x + exp(x) - 4))/(3*x*log(x)*(exp(x) + 1)))/log(x + exp(x
) - 4) - ((2*(x + exp(x) - 4))/(3*x*(exp(x) + 1)) - (2*log(x)*(3*exp(x) - exp(2*x) - x^2*exp(x) + 5*x*exp(x) +
4))/(3*x*(exp(x) + 1)^2))/log(x) - x - x^2 + (2*(5*x^2 - x^3))/(3*x^2*(exp(2*x) + 2*exp(x) + 1)) - (2*(5*x +
5*x^2 - x^3))/(3*x^2*(exp(x) + 1))

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sympy [A]  time = 2.06, size = 22, normalized size = 0.81 \begin {gather*} - x^{2} - x - \frac {\log {\left (\log {\relax (x )}^{2} \right )}}{3 \log {\left (x + e^{x} - 4 \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(x)*x+x)*ln(x)*ln(ln(x)**2)+((-6*x**2-3*x)*exp(x)-6*x**3+21*x**2+12*x)*ln(x)*ln(exp(x)+x-4)**2+
(-2*exp(x)-2*x+8)*ln(exp(x)+x-4))/(3*exp(x)*x+3*x**2-12*x)/ln(x)/ln(exp(x)+x-4)**2,x)

[Out]

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

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