3.25.23 \(\int \frac {(16-12 x^2+4 x^3+e^3 (-16 x^2-16 x^3+28 x^4-8 x^5)+e^3 (-16 x-16 x^2+28 x^3-8 x^4) \log (x)) \log (\log ^2(e^{-e^3 (2+x+x^2)} (x+\log (x))))+(-4 x^2+2 x^3+(-4 x+2 x^2) \log (x)) \log (e^{-e^3 (2+x+x^2)} (x+\log (x))) \log ^2(\log ^2(e^{-e^3 (2+x+x^2)} (x+\log (x))))}{(x^2+x \log (x)) \log (e^{-e^3 (2+x+x^2)} (x+\log (x)))} \, dx\)

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

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Rubi [F]  time = 8.08, 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 (16-12 x^2+4 x^3+e^3 \left (-16 x^2-16 x^3+28 x^4-8 x^5\right )+e^3 \left (-16 x-16 x^2+28 x^3-8 x^4\right ) \log (x)\right ) \log \left (\log ^2\left (e^{-e^3 \left (2+x+x^2\right )} (x+\log (x))\right )\right )+\left (-4 x^2+2 x^3+\left (-4 x+2 x^2\right ) \log (x)\right ) \log \left (e^{-e^3 \left (2+x+x^2\right )} (x+\log (x))\right ) \log ^2\left (\log ^2\left (e^{-e^3 \left (2+x+x^2\right )} (x+\log (x))\right )\right )}{\left (x^2+x \log (x)\right ) \log \left (e^{-e^3 \left (2+x+x^2\right )} (x+\log (x))\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[((16 - 12*x^2 + 4*x^3 + E^3*(-16*x^2 - 16*x^3 + 28*x^4 - 8*x^5) + E^3*(-16*x - 16*x^2 + 28*x^3 - 8*x^4)*Lo
g[x])*Log[Log[(x + Log[x])/E^(E^3*(2 + x + x^2))]^2] + (-4*x^2 + 2*x^3 + (-4*x + 2*x^2)*Log[x])*Log[(x + Log[x
])/E^(E^3*(2 + x + x^2))]*Log[Log[(x + Log[x])/E^(E^3*(2 + x + x^2))]^2]^2)/((x^2 + x*Log[x])*Log[(x + Log[x])
/E^(E^3*(2 + x + x^2))]),x]

[Out]

16*Defer[Int][Log[Log[(x + Log[x])/E^(E^3*(2 + x + x^2))]^2]/(x*(x + Log[x])*Log[(x + Log[x])/E^(E^3*(2 + x +
x^2))]), x] - 12*Defer[Int][(x*Log[Log[(x + Log[x])/E^(E^3*(2 + x + x^2))]^2])/((x + Log[x])*Log[(x + Log[x])/
E^(E^3*(2 + x + x^2))]), x] - 16*E^3*Defer[Int][(x*Log[Log[(x + Log[x])/E^(E^3*(2 + x + x^2))]^2])/((x + Log[x
])*Log[(x + Log[x])/E^(E^3*(2 + x + x^2))]), x] + 4*Defer[Int][(x^2*Log[Log[(x + Log[x])/E^(E^3*(2 + x + x^2))
]^2])/((x + Log[x])*Log[(x + Log[x])/E^(E^3*(2 + x + x^2))]), x] - 16*E^3*Defer[Int][(x^2*Log[Log[(x + Log[x])
/E^(E^3*(2 + x + x^2))]^2])/((x + Log[x])*Log[(x + Log[x])/E^(E^3*(2 + x + x^2))]), x] + 28*E^3*Defer[Int][(x^
3*Log[Log[(x + Log[x])/E^(E^3*(2 + x + x^2))]^2])/((x + Log[x])*Log[(x + Log[x])/E^(E^3*(2 + x + x^2))]), x] -
8*E^3*Defer[Int][(x^4*Log[Log[(x + Log[x])/E^(E^3*(2 + x + x^2))]^2])/((x + Log[x])*Log[(x + Log[x])/E^(E^3*(
2 + x + x^2))]), x] - 16*E^3*Defer[Int][(Log[x]*Log[Log[(x + Log[x])/E^(E^3*(2 + x + x^2))]^2])/((x + Log[x])*
Log[(x + Log[x])/E^(E^3*(2 + x + x^2))]), x] - 16*E^3*Defer[Int][(x*Log[x]*Log[Log[(x + Log[x])/E^(E^3*(2 + x
+ x^2))]^2])/((x + Log[x])*Log[(x + Log[x])/E^(E^3*(2 + x + x^2))]), x] + 28*E^3*Defer[Int][(x^2*Log[x]*Log[Lo
g[(x + Log[x])/E^(E^3*(2 + x + x^2))]^2])/((x + Log[x])*Log[(x + Log[x])/E^(E^3*(2 + x + x^2))]), x] - 8*E^3*D
efer[Int][(x^3*Log[x]*Log[Log[(x + Log[x])/E^(E^3*(2 + x + x^2))]^2])/((x + Log[x])*Log[(x + Log[x])/E^(E^3*(2
+ x + x^2))]), x] - 4*Defer[Int][Log[Log[(x + Log[x])/E^(E^3*(2 + x + x^2))]^2]^2, x] + 2*Defer[Int][x*Log[Lo
g[(x + Log[x])/E^(E^3*(2 + x + x^2))]^2]^2, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\left (16-12 x^2+4 x^3+e^3 \left (-16 x^2-16 x^3+28 x^4-8 x^5\right )+e^3 \left (-16 x-16 x^2+28 x^3-8 x^4\right ) \log (x)\right ) \log \left (\log ^2\left (e^{-e^3 \left (2+x+x^2\right )} (x+\log (x))\right )\right )+\left (-4 x^2+2 x^3+\left (-4 x+2 x^2\right ) \log (x)\right ) \log \left (e^{-e^3 \left (2+x+x^2\right )} (x+\log (x))\right ) \log ^2\left (\log ^2\left (e^{-e^3 \left (2+x+x^2\right )} (x+\log (x))\right )\right )}{x (x+\log (x)) \log \left (e^{-e^3 \left (2+x+x^2\right )} (x+\log (x))\right )} \, dx\\ &=\int \frac {2 (2-x) \log \left (\log ^2\left (e^{-e^3 \left (2+x+x^2\right )} (x+\log (x))\right )\right ) \left (\frac {2 (-2+x) \left (-1-x+e^3 x^2+2 e^3 x^3+e^3 x (1+2 x) \log (x)\right )}{\log \left (e^{-e^3 \left (2+x+x^2\right )} (x+\log (x))\right )}-x (x+\log (x)) \log \left (\log ^2\left (e^{-e^3 \left (2+x+x^2\right )} (x+\log (x))\right )\right )\right )}{x (x+\log (x))} \, dx\\ &=2 \int \frac {(2-x) \log \left (\log ^2\left (e^{-e^3 \left (2+x+x^2\right )} (x+\log (x))\right )\right ) \left (\frac {2 (-2+x) \left (-1-x+e^3 x^2+2 e^3 x^3+e^3 x (1+2 x) \log (x)\right )}{\log \left (e^{-e^3 \left (2+x+x^2\right )} (x+\log (x))\right )}-x (x+\log (x)) \log \left (\log ^2\left (e^{-e^3 \left (2+x+x^2\right )} (x+\log (x))\right )\right )\right )}{x (x+\log (x))} \, dx\\ &=2 \int \left (-\frac {2 (-2+x)^2 \left (-1-x+e^3 x^2+2 e^3 x^3+e^3 x \log (x)+2 e^3 x^2 \log (x)\right ) \log \left (\log ^2\left (e^{-e^3 \left (2+x+x^2\right )} (x+\log (x))\right )\right )}{x (x+\log (x)) \log \left (e^{-e^3 \left (2+x+x^2\right )} (x+\log (x))\right )}+(-2+x) \log ^2\left (\log ^2\left (e^{-e^3 \left (2+x+x^2\right )} (x+\log (x))\right )\right )\right ) \, dx\\ &=2 \int (-2+x) \log ^2\left (\log ^2\left (e^{-e^3 \left (2+x+x^2\right )} (x+\log (x))\right )\right ) \, dx-4 \int \frac {(-2+x)^2 \left (-1-x+e^3 x^2+2 e^3 x^3+e^3 x \log (x)+2 e^3 x^2 \log (x)\right ) \log \left (\log ^2\left (e^{-e^3 \left (2+x+x^2\right )} (x+\log (x))\right )\right )}{x (x+\log (x)) \log \left (e^{-e^3 \left (2+x+x^2\right )} (x+\log (x))\right )} \, dx\\ &=2 \int \left (-2 \log ^2\left (\log ^2\left (e^{-e^3 \left (2+x+x^2\right )} (x+\log (x))\right )\right )+x \log ^2\left (\log ^2\left (e^{-e^3 \left (2+x+x^2\right )} (x+\log (x))\right )\right )\right ) \, dx-4 \int \left (-\frac {4 \left (-1-x+e^3 x^2+2 e^3 x^3+e^3 x \log (x)+2 e^3 x^2 \log (x)\right ) \log \left (\log ^2\left (e^{-e^3 \left (2+x+x^2\right )} (x+\log (x))\right )\right )}{(x+\log (x)) \log \left (e^{-e^3 \left (2+x+x^2\right )} (x+\log (x))\right )}+\frac {4 \left (-1-x+e^3 x^2+2 e^3 x^3+e^3 x \log (x)+2 e^3 x^2 \log (x)\right ) \log \left (\log ^2\left (e^{-e^3 \left (2+x+x^2\right )} (x+\log (x))\right )\right )}{x (x+\log (x)) \log \left (e^{-e^3 \left (2+x+x^2\right )} (x+\log (x))\right )}+\frac {x \left (-1-x+e^3 x^2+2 e^3 x^3+e^3 x \log (x)+2 e^3 x^2 \log (x)\right ) \log \left (\log ^2\left (e^{-e^3 \left (2+x+x^2\right )} (x+\log (x))\right )\right )}{(x+\log (x)) \log \left (e^{-e^3 \left (2+x+x^2\right )} (x+\log (x))\right )}\right ) \, dx\\ &=2 \int x \log ^2\left (\log ^2\left (e^{-e^3 \left (2+x+x^2\right )} (x+\log (x))\right )\right ) \, dx-4 \int \frac {x \left (-1-x+e^3 x^2+2 e^3 x^3+e^3 x \log (x)+2 e^3 x^2 \log (x)\right ) \log \left (\log ^2\left (e^{-e^3 \left (2+x+x^2\right )} (x+\log (x))\right )\right )}{(x+\log (x)) \log \left (e^{-e^3 \left (2+x+x^2\right )} (x+\log (x))\right )} \, dx-4 \int \log ^2\left (\log ^2\left (e^{-e^3 \left (2+x+x^2\right )} (x+\log (x))\right )\right ) \, dx+16 \int \frac {\left (-1-x+e^3 x^2+2 e^3 x^3+e^3 x \log (x)+2 e^3 x^2 \log (x)\right ) \log \left (\log ^2\left (e^{-e^3 \left (2+x+x^2\right )} (x+\log (x))\right )\right )}{(x+\log (x)) \log \left (e^{-e^3 \left (2+x+x^2\right )} (x+\log (x))\right )} \, dx-16 \int \frac {\left (-1-x+e^3 x^2+2 e^3 x^3+e^3 x \log (x)+2 e^3 x^2 \log (x)\right ) \log \left (\log ^2\left (e^{-e^3 \left (2+x+x^2\right )} (x+\log (x))\right )\right )}{x (x+\log (x)) \log \left (e^{-e^3 \left (2+x+x^2\right )} (x+\log (x))\right )} \, dx\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.18, size = 30, normalized size = 1.00 \begin {gather*} (-2+x)^2 \log ^2\left (\log ^2\left (e^{-e^3 \left (2+x+x^2\right )} (x+\log (x))\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((16 - 12*x^2 + 4*x^3 + E^3*(-16*x^2 - 16*x^3 + 28*x^4 - 8*x^5) + E^3*(-16*x - 16*x^2 + 28*x^3 - 8*x
^4)*Log[x])*Log[Log[(x + Log[x])/E^(E^3*(2 + x + x^2))]^2] + (-4*x^2 + 2*x^3 + (-4*x + 2*x^2)*Log[x])*Log[(x +
Log[x])/E^(E^3*(2 + x + x^2))]*Log[Log[(x + Log[x])/E^(E^3*(2 + x + x^2))]^2]^2)/((x^2 + x*Log[x])*Log[(x + L
og[x])/E^(E^3*(2 + x + x^2))]),x]

[Out]

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

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fricas [A]  time = 0.62, size = 43, normalized size = 1.43 \begin {gather*} {\left (x^{2} - 4 \, x + 4\right )} \log \left (\log \left (x e^{\left (-{\left (x^{2} + x + 2\right )} e^{3}\right )} + e^{\left (-{\left (x^{2} + x + 2\right )} e^{3}\right )} \log \relax (x)\right )^{2}\right )^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x^2-4*x)*log(x)+2*x^3-4*x^2)*log((x+log(x))/exp((x^2+x+2)*exp(3)))*log(log((x+log(x))/exp((x^2+
x+2)*exp(3)))^2)^2+((-8*x^4+28*x^3-16*x^2-16*x)*exp(3)*log(x)+(-8*x^5+28*x^4-16*x^3-16*x^2)*exp(3)+4*x^3-12*x^
2+16)*log(log((x+log(x))/exp((x^2+x+2)*exp(3)))^2))/(x*log(x)+x^2)/log((x+log(x))/exp((x^2+x+2)*exp(3))),x, al
gorithm="fricas")

[Out]

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

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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((((2*x^2-4*x)*log(x)+2*x^3-4*x^2)*log((x+log(x))/exp((x^2+x+2)*exp(3)))*log(log((x+log(x))/exp((x^2+
x+2)*exp(3)))^2)^2+((-8*x^4+28*x^3-16*x^2-16*x)*exp(3)*log(x)+(-8*x^5+28*x^4-16*x^3-16*x^2)*exp(3)+4*x^3-12*x^
2+16)*log(log((x+log(x))/exp((x^2+x+2)*exp(3)))^2))/(x*log(x)+x^2)/log((x+log(x))/exp((x^2+x+2)*exp(3))),x, al
gorithm="giac")

[Out]

Timed out

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maple [F]  time = 0.14, size = 0, normalized size = 0.00 \[\int \frac {\left (\left (2 x^{2}-4 x \right ) \ln \relax (x )+2 x^{3}-4 x^{2}\right ) \ln \left (\left (x +\ln \relax (x )\right ) {\mathrm e}^{-\left (x^{2}+x +2\right ) {\mathrm e}^{3}}\right ) \ln \left (\ln \left (\left (x +\ln \relax (x )\right ) {\mathrm e}^{-\left (x^{2}+x +2\right ) {\mathrm e}^{3}}\right )^{2}\right )^{2}+\left (\left (-8 x^{4}+28 x^{3}-16 x^{2}-16 x \right ) {\mathrm e}^{3} \ln \relax (x )+\left (-8 x^{5}+28 x^{4}-16 x^{3}-16 x^{2}\right ) {\mathrm e}^{3}+4 x^{3}-12 x^{2}+16\right ) \ln \left (\ln \left (\left (x +\ln \relax (x )\right ) {\mathrm e}^{-\left (x^{2}+x +2\right ) {\mathrm e}^{3}}\right )^{2}\right )}{\left (x \ln \relax (x )+x^{2}\right ) \ln \left (\left (x +\ln \relax (x )\right ) {\mathrm e}^{-\left (x^{2}+x +2\right ) {\mathrm e}^{3}}\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((2*x^2-4*x)*ln(x)+2*x^3-4*x^2)*ln((x+ln(x))/exp((x^2+x+2)*exp(3)))*ln(ln((x+ln(x))/exp((x^2+x+2)*exp(3))
)^2)^2+((-8*x^4+28*x^3-16*x^2-16*x)*exp(3)*ln(x)+(-8*x^5+28*x^4-16*x^3-16*x^2)*exp(3)+4*x^3-12*x^2+16)*ln(ln((
x+ln(x))/exp((x^2+x+2)*exp(3)))^2))/(x*ln(x)+x^2)/ln((x+ln(x))/exp((x^2+x+2)*exp(3))),x)

[Out]

int((((2*x^2-4*x)*ln(x)+2*x^3-4*x^2)*ln((x+ln(x))/exp((x^2+x+2)*exp(3)))*ln(ln((x+ln(x))/exp((x^2+x+2)*exp(3))
)^2)^2+((-8*x^4+28*x^3-16*x^2-16*x)*exp(3)*ln(x)+(-8*x^5+28*x^4-16*x^3-16*x^2)*exp(3)+4*x^3-12*x^2+16)*ln(ln((
x+ln(x))/exp((x^2+x+2)*exp(3)))^2))/(x*ln(x)+x^2)/ln((x+ln(x))/exp((x^2+x+2)*exp(3))),x)

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maxima [A]  time = 0.69, size = 35, normalized size = 1.17 \begin {gather*} 4 \, {\left (x^{2} - 4 \, x + 4\right )} \log \left (x^{2} e^{3} + x e^{3} + 2 \, e^{3} - \log \left (x + \log \relax (x)\right )\right )^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x^2-4*x)*log(x)+2*x^3-4*x^2)*log((x+log(x))/exp((x^2+x+2)*exp(3)))*log(log((x+log(x))/exp((x^2+
x+2)*exp(3)))^2)^2+((-8*x^4+28*x^3-16*x^2-16*x)*exp(3)*log(x)+(-8*x^5+28*x^4-16*x^3-16*x^2)*exp(3)+4*x^3-12*x^
2+16)*log(log((x+log(x))/exp((x^2+x+2)*exp(3)))^2))/(x*log(x)+x^2)/log((x+log(x))/exp((x^2+x+2)*exp(3))),x, al
gorithm="maxima")

[Out]

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

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mupad [B]  time = 3.26, size = 36, normalized size = 1.20 \begin {gather*} {\ln \left ({\ln \left ({\mathrm {e}}^{-x^2\,{\mathrm {e}}^3}\,{\mathrm {e}}^{-2\,{\mathrm {e}}^3}\,{\mathrm {e}}^{-x\,{\mathrm {e}}^3}\,\left (x+\ln \relax (x)\right )\right )}^2\right )}^2\,{\left (x-2\right )}^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(log(exp(-exp(3)*(x + x^2 + 2))*(x + log(x)))^2)*(12*x^2 - 4*x^3 + exp(3)*(16*x^2 + 16*x^3 - 28*x^4 +
8*x^5) + exp(3)*log(x)*(16*x + 16*x^2 - 28*x^3 + 8*x^4) - 16) + log(log(exp(-exp(3)*(x + x^2 + 2))*(x + log(x
)))^2)^2*log(exp(-exp(3)*(x + x^2 + 2))*(x + log(x)))*(log(x)*(4*x - 2*x^2) + 4*x^2 - 2*x^3))/(log(exp(-exp(3)
*(x + x^2 + 2))*(x + log(x)))*(x*log(x) + x^2)),x)

[Out]

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

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sympy [A]  time = 19.14, size = 31, normalized size = 1.03 \begin {gather*} \left (x^{2} - 4 x + 4\right ) \log {\left (\log {\left (\left (x + \log {\relax (x )}\right ) e^{- \left (x^{2} + x + 2\right ) e^{3}} \right )}^{2} \right )}^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x**2-4*x)*ln(x)+2*x**3-4*x**2)*ln((x+ln(x))/exp((x**2+x+2)*exp(3)))*ln(ln((x+ln(x))/exp((x**2+x
+2)*exp(3)))**2)**2+((-8*x**4+28*x**3-16*x**2-16*x)*exp(3)*ln(x)+(-8*x**5+28*x**4-16*x**3-16*x**2)*exp(3)+4*x*
*3-12*x**2+16)*ln(ln((x+ln(x))/exp((x**2+x+2)*exp(3)))**2))/(x*ln(x)+x**2)/ln((x+ln(x))/exp((x**2+x+2)*exp(3))
),x)

[Out]

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

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