∫ ee3∕xx2+e3∕x log(log(eex x)) ______________________________________ 8+e3∕xx2 (8e3∕xx+e6∕xx3+ex(8e3∕xx2+e6∕xx4)+e3∕x(−24x2+16x3) log(eex x)+(−24e3∕x−2e6∕xx3) log(eex x) log(log(eex x)))_____________________________________________________________________________________________________ (64x2+16e3∕xx4+e6∕xx6) log(eexx) dx [2189]

3.22.89 \(\int \frac {e^{\frac {e^{3/x} x^2+e^{3/x} \log (\log (e^{e^x} x))}{8+e^{3/x} x^2}} (8 e^{3/x} x+e^{6/x} x^3+e^x (8 e^{3/x} x^2+e^{6/x} x^4)+e^{3/x} (-24 x^2+16 x^3) \log (e^{e^x} x)+(-24 e^{3/x}-2 e^{6/x} x^3) \log (e^{e^x} x) \log (\log (e^{e^x} x)))}{(64 x^2+16 e^{3/x} x^4+e^{6/x} x^6) \log (e^{e^x} x)} \, dx\) [2189]

Optimal. Leaf size=31 \[ e^{\frac {x^2+\log \left (\log \left (e^{e^x} x\right )\right )}{8 e^{-3/x}+x^2}} \]

[Out]

exp((x^2+ln(ln(x*exp(exp(x)))))/(8/exp(3/x)+x^2))

________________________________________________________________________________________

Rubi [F]
time = 21.90, 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 {\exp \left (\frac {e^{3/x} x^2+e^{3/x} \log \left (\log \left (e^{e^x} x\right )\right )}{8+e^{3/x} x^2}\right ) \left (8 e^{3/x} x+e^{6/x} x^3+e^x \left (8 e^{3/x} x^2+e^{6/x} x^4\right )+e^{3/x} \left (-24 x^2+16 x^3\right ) \log \left (e^{e^x} x\right )+\left (-24 e^{3/x}-2 e^{6/x} x^3\right ) \log \left (e^{e^x} x\right ) \log \left (\log \left (e^{e^x} x\right )\right )\right )}{\left (64 x^2+16 e^{3/x} x^4+e^{6/x} x^6\right ) \log \left (e^{e^x} x\right )} \, dx \end {gather*}

Veriﬁcation is not applicable to the result.

[In]

Int[(E^((E^(3/x)*x^2 + E^(3/x)*Log[Log[E^E^x*x]])/(8 + E^(3/x)*x^2))*(8*E^(3/x)*x + E^(6/x)*x^3 + E^x*(8*E^(3/

x)*x^2 + E^(6/x)*x^4) + E^(3/x)*(-24*x^2 + 16*x^3)*Log[E^E^x*x] + (-24*E^(3/x) - 2*E^(6/x)*x^3)*Log[E^E^x*x]*L

og[Log[E^E^x*x]]))/((64*x^2 + 16*E^(3/x)*x^4 + E^(6/x)*x^6)*Log[E^E^x*x]),x]

[Out]

-24*Defer[Int][(E^(3/x + (E^(3/x)*x^2)/(8 + E^(3/x)*x^2))*Log[E^E^x*x]^(E^(3/x)/(8 + E^(3/x)*x^2)))/(8 + E^(3/

x)*x^2)^2, x] + 16*Defer[Int][(E^(3/x + (E^(3/x)*x^2)/(8 + E^(3/x)*x^2))*x*Log[E^E^x*x]^(E^(3/x)/(8 + E^(3/x)*

x^2)))/(8 + E^(3/x)*x^2)^2, x] + Defer[Int][(E^(3/x + x + (E^(3/x)*x^2)/(8 + E^(3/x)*x^2))*Log[E^E^x*x]^(-1 +

E^(3/x)/(8 + E^(3/x)*x^2)))/(8 + E^(3/x)*x^2), x] + Defer[Int][(E^(3/x + (E^(3/x)*x^2)/(8 + E^(3/x)*x^2))*Log[

E^E^x*x]^(-1 + E^(3/x)/(8 + E^(3/x)*x^2)))/(x*(8 + E^(3/x)*x^2)), x] - 24*Defer[Int][(E^(3/x + (E^(3/x)*x^2)/(

8 + E^(3/x)*x^2))*Log[E^E^x*x]^(E^(3/x)/(8 + E^(3/x)*x^2))*Log[Log[E^E^x*x]])/(x^2*(8 + E^(3/x)*x^2)^2), x] +

16*Defer[Int][(E^(3/x + (E^(3/x)*x^2)/(8 + E^(3/x)*x^2))*Log[E^E^x*x]^(E^(3/x)/(8 + E^(3/x)*x^2))*Log[Log[E^E^

x*x]])/(x*(8 + E^(3/x)*x^2)^2), x] - 2*Defer[Int][(E^(3/x + (E^(3/x)*x^2)/(8 + E^(3/x)*x^2))*Log[E^E^x*x]^(E^(

3/x)/(8 + E^(3/x)*x^2))*Log[Log[E^E^x*x]])/(x*(8 + E^(3/x)*x^2)), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{\frac {3}{x}+\frac {e^{3/x} x^2}{8+e^{3/x} x^2}} \log ^{-1+\frac {e^{3/x}}{8+e^{3/x} x^2}}\left (e^{e^x} x\right ) \left (x \left (1+e^x x\right ) \left (8+e^{3/x} x^2\right )-2 \log \left (e^{e^x} x\right ) \left (4 (3-2 x) x^2+\left (12+e^{3/x} x^3\right ) \log \left (\log \left (e^{e^x} x\right )\right )\right )\right )}{x^2 \left (8+e^{3/x} x^2\right )^2} \, dx\\ &=\int \left (\frac {e^{\frac {3}{x}+x+\frac {e^{3/x} x^2}{8+e^{3/x} x^2}} \log ^{-1+\frac {e^{3/x}}{8+e^{3/x} x^2}}\left (e^{e^x} x\right )}{8+e^{3/x} x^2}-\frac {e^{\frac {3}{x}+\frac {e^{3/x} x^2}{8+e^{3/x} x^2}} \log ^{-1+\frac {e^{3/x}}{8+e^{3/x} x^2}}\left (e^{e^x} x\right ) \left (-8 x-e^{3/x} x^3+24 x^2 \log \left (e^{e^x} x\right )-16 x^3 \log \left (e^{e^x} x\right )+24 \log \left (e^{e^x} x\right ) \log \left (\log \left (e^{e^x} x\right )\right )+2 e^{3/x} x^3 \log \left (e^{e^x} x\right ) \log \left (\log \left (e^{e^x} x\right )\right )\right )}{x^2 \left (8+e^{3/x} x^2\right )^2}\right ) \, dx\\ &=\int \frac {e^{\frac {3}{x}+x+\frac {e^{3/x} x^2}{8+e^{3/x} x^2}} \log ^{-1+\frac {e^{3/x}}{8+e^{3/x} x^2}}\left (e^{e^x} x\right )}{8+e^{3/x} x^2} \, dx-\int \frac {e^{\frac {3}{x}+\frac {e^{3/x} x^2}{8+e^{3/x} x^2}} \log ^{-1+\frac {e^{3/x}}{8+e^{3/x} x^2}}\left (e^{e^x} x\right ) \left (-8 x-e^{3/x} x^3+24 x^2 \log \left (e^{e^x} x\right )-16 x^3 \log \left (e^{e^x} x\right )+24 \log \left (e^{e^x} x\right ) \log \left (\log \left (e^{e^x} x\right )\right )+2 e^{3/x} x^3 \log \left (e^{e^x} x\right ) \log \left (\log \left (e^{e^x} x\right )\right )\right )}{x^2 \left (8+e^{3/x} x^2\right )^2} \, dx\\ &=\int \frac {e^{\frac {3}{x}+x+\frac {e^{3/x} x^2}{8+e^{3/x} x^2}} \log ^{-1+\frac {e^{3/x}}{8+e^{3/x} x^2}}\left (e^{e^x} x\right )}{8+e^{3/x} x^2} \, dx-\int \frac {e^{\frac {3}{x}+\frac {e^{3/x} x^2}{8+e^{3/x} x^2}} \log ^{-1+\frac {e^{3/x}}{8+e^{3/x} x^2}}\left (e^{e^x} x\right ) \left (-x \left (8+e^{3/x} x^2\right )+2 \log \left (e^{e^x} x\right ) \left (4 (3-2 x) x^2+\left (12+e^{3/x} x^3\right ) \log \left (\log \left (e^{e^x} x\right )\right )\right )\right )}{x^2 \left (8+e^{3/x} x^2\right )^2} \, dx\\ &=\int \frac {e^{\frac {3}{x}+x+\frac {e^{3/x} x^2}{8+e^{3/x} x^2}} \log ^{-1+\frac {e^{3/x}}{8+e^{3/x} x^2}}\left (e^{e^x} x\right )}{8+e^{3/x} x^2} \, dx-\int \left (-\frac {8 e^{\frac {3}{x}+\frac {e^{3/x} x^2}{8+e^{3/x} x^2}} (-3+2 x) \log ^{\frac {e^{3/x}}{8+e^{3/x} x^2}}\left (e^{e^x} x\right ) \left (x^2+\log \left (\log \left (e^{e^x} x\right )\right )\right )}{x^2 \left (8+e^{3/x} x^2\right )^2}+\frac {e^{\frac {3}{x}+\frac {e^{3/x} x^2}{8+e^{3/x} x^2}} \log ^{-1+\frac {e^{3/x}}{8+e^{3/x} x^2}}\left (e^{e^x} x\right ) \left (-1+2 \log \left (e^{e^x} x\right ) \log \left (\log \left (e^{e^x} x\right )\right )\right )}{x \left (8+e^{3/x} x^2\right )}\right ) \, dx\\ &=8 \int \frac {e^{\frac {3}{x}+\frac {e^{3/x} x^2}{8+e^{3/x} x^2}} (-3+2 x) \log ^{\frac {e^{3/x}}{8+e^{3/x} x^2}}\left (e^{e^x} x\right ) \left (x^2+\log \left (\log \left (e^{e^x} x\right )\right )\right )}{x^2 \left (8+e^{3/x} x^2\right )^2} \, dx+\int \frac {e^{\frac {3}{x}+x+\frac {e^{3/x} x^2}{8+e^{3/x} x^2}} \log ^{-1+\frac {e^{3/x}}{8+e^{3/x} x^2}}\left (e^{e^x} x\right )}{8+e^{3/x} x^2} \, dx-\int \frac {e^{\frac {3}{x}+\frac {e^{3/x} x^2}{8+e^{3/x} x^2}} \log ^{-1+\frac {e^{3/x}}{8+e^{3/x} x^2}}\left (e^{e^x} x\right ) \left (-1+2 \log \left (e^{e^x} x\right ) \log \left (\log \left (e^{e^x} x\right )\right )\right )}{x \left (8+e^{3/x} x^2\right )} \, dx\\ &=8 \int \left (-\frac {3 e^{\frac {3}{x}+\frac {e^{3/x} x^2}{8+e^{3/x} x^2}} \log ^{\frac {e^{3/x}}{8+e^{3/x} x^2}}\left (e^{e^x} x\right ) \left (x^2+\log \left (\log \left (e^{e^x} x\right )\right )\right )}{x^2 \left (8+e^{3/x} x^2\right )^2}+\frac {2 e^{\frac {3}{x}+\frac {e^{3/x} x^2}{8+e^{3/x} x^2}} \log ^{\frac {e^{3/x}}{8+e^{3/x} x^2}}\left (e^{e^x} x\right ) \left (x^2+\log \left (\log \left (e^{e^x} x\right )\right )\right )}{x \left (8+e^{3/x} x^2\right )^2}\right ) \, dx+\int \frac {e^{\frac {3}{x}+x+\frac {e^{3/x} x^2}{8+e^{3/x} x^2}} \log ^{-1+\frac {e^{3/x}}{8+e^{3/x} x^2}}\left (e^{e^x} x\right )}{8+e^{3/x} x^2} \, dx-\int \left (-\frac {e^{\frac {3}{x}+\frac {e^{3/x} x^2}{8+e^{3/x} x^2}} \log ^{-1+\frac {e^{3/x}}{8+e^{3/x} x^2}}\left (e^{e^x} x\right )}{x \left (8+e^{3/x} x^2\right )}+\frac {2 e^{\frac {3}{x}+\frac {e^{3/x} x^2}{8+e^{3/x} x^2}} \log ^{\frac {e^{3/x}}{8+e^{3/x} x^2}}\left (e^{e^x} x\right ) \log \left (\log \left (e^{e^x} x\right )\right )}{x \left (8+e^{3/x} x^2\right )}\right ) \, dx\\ &=-\left (2 \int \frac {e^{\frac {3}{x}+\frac {e^{3/x} x^2}{8+e^{3/x} x^2}} \log ^{\frac {e^{3/x}}{8+e^{3/x} x^2}}\left (e^{e^x} x\right ) \log \left (\log \left (e^{e^x} x\right )\right )}{x \left (8+e^{3/x} x^2\right )} \, dx\right )+16 \int \frac {e^{\frac {3}{x}+\frac {e^{3/x} x^2}{8+e^{3/x} x^2}} \log ^{\frac {e^{3/x}}{8+e^{3/x} x^2}}\left (e^{e^x} x\right ) \left (x^2+\log \left (\log \left (e^{e^x} x\right )\right )\right )}{x \left (8+e^{3/x} x^2\right )^2} \, dx-24 \int \frac {e^{\frac {3}{x}+\frac {e^{3/x} x^2}{8+e^{3/x} x^2}} \log ^{\frac {e^{3/x}}{8+e^{3/x} x^2}}\left (e^{e^x} x\right ) \left (x^2+\log \left (\log \left (e^{e^x} x\right )\right )\right )}{x^2 \left (8+e^{3/x} x^2\right )^2} \, dx+\int \frac {e^{\frac {3}{x}+x+\frac {e^{3/x} x^2}{8+e^{3/x} x^2}} \log ^{-1+\frac {e^{3/x}}{8+e^{3/x} x^2}}\left (e^{e^x} x\right )}{8+e^{3/x} x^2} \, dx+\int \frac {e^{\frac {3}{x}+\frac {e^{3/x} x^2}{8+e^{3/x} x^2}} \log ^{-1+\frac {e^{3/x}}{8+e^{3/x} x^2}}\left (e^{e^x} x\right )}{x \left (8+e^{3/x} x^2\right )} \, dx\\ &=-\left (2 \int \frac {e^{\frac {3}{x}+\frac {e^{3/x} x^2}{8+e^{3/x} x^2}} \log ^{\frac {e^{3/x}}{8+e^{3/x} x^2}}\left (e^{e^x} x\right ) \log \left (\log \left (e^{e^x} x\right )\right )}{x \left (8+e^{3/x} x^2\right )} \, dx\right )+16 \int \left (\frac {e^{\frac {3}{x}+\frac {e^{3/x} x^2}{8+e^{3/x} x^2}} x \log ^{\frac {e^{3/x}}{8+e^{3/x} x^2}}\left (e^{e^x} x\right )}{\left (8+e^{3/x} x^2\right )^2}+\frac {e^{\frac {3}{x}+\frac {e^{3/x} x^2}{8+e^{3/x} x^2}} \log ^{\frac {e^{3/x}}{8+e^{3/x} x^2}}\left (e^{e^x} x\right ) \log \left (\log \left (e^{e^x} x\right )\right )}{x \left (8+e^{3/x} x^2\right )^2}\right ) \, dx-24 \int \left (\frac {e^{\frac {3}{x}+\frac {e^{3/x} x^2}{8+e^{3/x} x^2}} \log ^{\frac {e^{3/x}}{8+e^{3/x} x^2}}\left (e^{e^x} x\right )}{\left (8+e^{3/x} x^2\right )^2}+\frac {e^{\frac {3}{x}+\frac {e^{3/x} x^2}{8+e^{3/x} x^2}} \log ^{\frac {e^{3/x}}{8+e^{3/x} x^2}}\left (e^{e^x} x\right ) \log \left (\log \left (e^{e^x} x\right )\right )}{x^2 \left (8+e^{3/x} x^2\right )^2}\right ) \, dx+\int \frac {e^{\frac {3}{x}+x+\frac {e^{3/x} x^2}{8+e^{3/x} x^2}} \log ^{-1+\frac {e^{3/x}}{8+e^{3/x} x^2}}\left (e^{e^x} x\right )}{8+e^{3/x} x^2} \, dx+\int \frac {e^{\frac {3}{x}+\frac {e^{3/x} x^2}{8+e^{3/x} x^2}} \log ^{-1+\frac {e^{3/x}}{8+e^{3/x} x^2}}\left (e^{e^x} x\right )}{x \left (8+e^{3/x} x^2\right )} \, dx\\ &=-\left (2 \int \frac {e^{\frac {3}{x}+\frac {e^{3/x} x^2}{8+e^{3/x} x^2}} \log ^{\frac {e^{3/x}}{8+e^{3/x} x^2}}\left (e^{e^x} x\right ) \log \left (\log \left (e^{e^x} x\right )\right )}{x \left (8+e^{3/x} x^2\right )} \, dx\right )+16 \int \frac {e^{\frac {3}{x}+\frac {e^{3/x} x^2}{8+e^{3/x} x^2}} x \log ^{\frac {e^{3/x}}{8+e^{3/x} x^2}}\left (e^{e^x} x\right )}{\left (8+e^{3/x} x^2\right )^2} \, dx+16 \int \frac {e^{\frac {3}{x}+\frac {e^{3/x} x^2}{8+e^{3/x} x^2}} \log ^{\frac {e^{3/x}}{8+e^{3/x} x^2}}\left (e^{e^x} x\right ) \log \left (\log \left (e^{e^x} x\right )\right )}{x \left (8+e^{3/x} x^2\right )^2} \, dx-24 \int \frac {e^{\frac {3}{x}+\frac {e^{3/x} x^2}{8+e^{3/x} x^2}} \log ^{\frac {e^{3/x}}{8+e^{3/x} x^2}}\left (e^{e^x} x\right )}{\left (8+e^{3/x} x^2\right )^2} \, dx-24 \int \frac {e^{\frac {3}{x}+\frac {e^{3/x} x^2}{8+e^{3/x} x^2}} \log ^{\frac {e^{3/x}}{8+e^{3/x} x^2}}\left (e^{e^x} x\right ) \log \left (\log \left (e^{e^x} x\right )\right )}{x^2 \left (8+e^{3/x} x^2\right )^2} \, dx+\int \frac {e^{\frac {3}{x}+x+\frac {e^{3/x} x^2}{8+e^{3/x} x^2}} \log ^{-1+\frac {e^{3/x}}{8+e^{3/x} x^2}}\left (e^{e^x} x\right )}{8+e^{3/x} x^2} \, dx+\int \frac {e^{\frac {3}{x}+\frac {e^{3/x} x^2}{8+e^{3/x} x^2}} \log ^{-1+\frac {e^{3/x}}{8+e^{3/x} x^2}}\left (e^{e^x} x\right )}{x \left (8+e^{3/x} x^2\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]
time = 0.27, size = 54, normalized size = 1.74 \begin {gather*} e^{1-\frac {8}{8+e^{3/x} x^2}} \log ^{\frac {e^{3/x}}{8+e^{3/x} x^2}}\left (e^{e^x} x\right ) \end {gather*}

Antiderivative was successfully veriﬁed.

[In]

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

*E^(3/x)*x^2 + E^(6/x)*x^4) + E^(3/x)*(-24*x^2 + 16*x^3)*Log[E^E^x*x] + (-24*E^(3/x) - 2*E^(6/x)*x^3)*Log[E^E^

x*x]*Log[Log[E^E^x*x]]))/((64*x^2 + 16*E^(3/x)*x^4 + E^(6/x)*x^6)*Log[E^E^x*x]),x]

[Out]

E^(1 - 8/(8 + E^(3/x)*x^2))*Log[E^E^x*x]^(E^(3/x)/(8 + E^(3/x)*x^2))

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 0.29, size = 81, normalized size = 2.61


method	result	size

risch	\({\mathrm e}^{\frac {{\mathrm e}^{\frac {3}{x}} \left (x^{2}+\ln \left (\ln \left (x \right )+\ln \left ({\mathrm e}^{{\mathrm e}^{x}}\right )-\frac {i \pi \,\mathrm {csgn}\left (i x \,{\mathrm e}^{{\mathrm e}^{x}}\right ) \left (-\mathrm {csgn}\left (i x \,{\mathrm e}^{{\mathrm e}^{x}}\right )+\mathrm {csgn}\left (i x \right )\right ) \left (-\mathrm {csgn}\left (i x \,{\mathrm e}^{{\mathrm e}^{x}}\right )+\mathrm {csgn}\left (i {\mathrm e}^{{\mathrm e}^{x}}\right )\right )}{2}\right )\right )}{x^{2} {\mathrm e}^{\frac {3}{x}}+8}}\)	\(81\)

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(((-2*x^3*exp(3/x)^2-24*exp(3/x))*ln(x*exp(exp(x)))*ln(ln(x*exp(exp(x))))+(16*x^3-24*x^2)*exp(3/x)*ln(x*exp

(exp(x)))+(x^4*exp(3/x)^2+8*x^2*exp(3/x))*exp(x)+x^3*exp(3/x)^2+8*x*exp(3/x))*exp((exp(3/x)*ln(ln(x*exp(exp(x)

)))+x^2*exp(3/x))/(x^2*exp(3/x)+8))/(x^6*exp(3/x)^2+16*x^4*exp(3/x)+64*x^2)/ln(x*exp(exp(x))),x,method=_RETURN

VERBOSE)

[Out]

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

-csgn(I*x*exp(exp(x)))+csgn(I*exp(exp(x))))))/(x^2*exp(3/x)+8))

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Maxima [A]
time = 0.46, size = 46, normalized size = 1.48 \begin {gather*} e^{\left (\frac {e^{\frac {3}{x}} \log \left (e^{x} + \log \left (x\right )\right )}{x^{2} e^{\frac {3}{x}} + 8} - \frac {8}{x^{2} e^{\frac {3}{x}} + 8} + 1\right )} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x^3*exp(3/x)^2-24*exp(3/x))*log(x*exp(exp(x)))*log(log(x*exp(exp(x))))+(16*x^3-24*x^2)*exp(3/x)

*log(x*exp(exp(x)))+(x^4*exp(3/x)^2+8*x^2*exp(3/x))*exp(x)+x^3*exp(3/x)^2+8*x*exp(3/x))*exp((exp(3/x)*log(log(

x*exp(exp(x))))+x^2*exp(3/x))/(x^2*exp(3/x)+8))/(x^6*exp(3/x)^2+16*x^4*exp(3/x)+64*x^2)/log(x*exp(exp(x))),x,

algorithm="maxima")

[Out]

e^(e^(3/x)*log(e^x + log(x))/(x^2*e^(3/x) + 8) - 8/(x^2*e^(3/x) + 8) + 1)

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Fricas [A]
time = 0.41, size = 41, normalized size = 1.32 \begin {gather*} e^{\left (\frac {x^{2} e^{\frac {3}{x}} + e^{\frac {3}{x}} \log \left (\log \left (x e^{\left (e^{x}\right )}\right )\right )}{x^{2} e^{\frac {3}{x}} + 8}\right )} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x^3*exp(3/x)^2-24*exp(3/x))*log(x*exp(exp(x)))*log(log(x*exp(exp(x))))+(16*x^3-24*x^2)*exp(3/x)

*log(x*exp(exp(x)))+(x^4*exp(3/x)^2+8*x^2*exp(3/x))*exp(x)+x^3*exp(3/x)^2+8*x*exp(3/x))*exp((exp(3/x)*log(log(

x*exp(exp(x))))+x^2*exp(3/x))/(x^2*exp(3/x)+8))/(x^6*exp(3/x)^2+16*x^4*exp(3/x)+64*x^2)/log(x*exp(exp(x))),x,

algorithm="fricas")

[Out]

e^((x^2*e^(3/x) + e^(3/x)*log(log(x*e^(e^x))))/(x^2*e^(3/x) + 8))

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Sympy [F(-1)] Timed out
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**3*exp(3/x)**2-24*exp(3/x))*ln(x*exp(exp(x)))*ln(ln(x*exp(exp(x))))+(16*x**3-24*x**2)*exp(3/x

)*ln(x*exp(exp(x)))+(x**4*exp(3/x)**2+8*x**2*exp(3/x))*exp(x)+x**3*exp(3/x)**2+8*x*exp(3/x))*exp((exp(3/x)*ln(

ln(x*exp(exp(x))))+x**2*exp(3/x))/(x**2*exp(3/x)+8))/(x**6*exp(3/x)**2+16*x**4*exp(3/x)+64*x**2)/ln(x*exp(exp(

x))),x)

[Out]

Timed out

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x^3*exp(3/x)^2-24*exp(3/x))*log(x*exp(exp(x)))*log(log(x*exp(exp(x))))+(16*x^3-24*x^2)*exp(3/x)

*log(x*exp(exp(x)))+(x^4*exp(3/x)^2+8*x^2*exp(3/x))*exp(x)+x^3*exp(3/x)^2+8*x*exp(3/x))*exp((exp(3/x)*log(log(

x*exp(exp(x))))+x^2*exp(3/x))/(x^2*exp(3/x)+8))/(x^6*exp(3/x)^2+16*x^4*exp(3/x)+64*x^2)/log(x*exp(exp(x))),x,

algorithm="giac")

[Out]

integrate((x^3*e^(6/x) + 8*(2*x^3 - 3*x^2)*e^(3/x)*log(x*e^(e^x)) - 2*(x^3*e^(6/x) + 12*e^(3/x))*log(x*e^(e^x)

)*log(log(x*e^(e^x))) + (x^4*e^(6/x) + 8*x^2*e^(3/x))*e^x + 8*x*e^(3/x))*e^((x^2*e^(3/x) + e^(3/x)*log(log(x*e

^(e^x))))/(x^2*e^(3/x) + 8))/((x^6*e^(6/x) + 16*x^4*e^(3/x) + 64*x^2)*log(x*e^(e^x))), x)

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Mupad [B]
time = 1.67, size = 53, normalized size = 1.71 \begin {gather*} {\mathrm {e}}^{\frac {x^2\,{\mathrm {e}}^{3/x}}{x^2\,{\mathrm {e}}^{3/x}+8}}\,{\left ({\mathrm {e}}^x+\ln \left (x\right )\right )}^{\frac {{\mathrm {e}}^{3/x}}{x^2\,{\mathrm {e}}^{3/x}+8}} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((exp((x^2*exp(3/x) + exp(3/x)*log(log(x*exp(exp(x)))))/(x^2*exp(3/x) + 8))*(exp(x)*(8*x^2*exp(3/x) + x^4*e

xp(6/x)) + 8*x*exp(3/x) + x^3*exp(6/x) - log(x*exp(exp(x)))*log(log(x*exp(exp(x))))*(24*exp(3/x) + 2*x^3*exp(6

/x)) - exp(3/x)*log(x*exp(exp(x)))*(24*x^2 - 16*x^3)))/(log(x*exp(exp(x)))*(16*x^4*exp(3/x) + x^6*exp(6/x) + 6

4*x^2)),x)

[Out]

exp((x^2*exp(3/x))/(x^2*exp(3/x) + 8))*(exp(x) + log(x))^(exp(3/x)/(x^2*exp(3/x) + 8))

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