3.25.6 \(\int \frac {e^{\frac {1}{5} x^2 \log (x) \log (\log (-\frac {x}{20+5 x}))} (4 x \log (x)+((4 x+x^2) \log (-\frac {x}{20+5 x})+(8 x+2 x^2) \log (x) \log (-\frac {x}{20+5 x})) \log (\log (-\frac {x}{20+5 x})))}{(20+5 x) \log (-\frac {x}{20+5 x})} \, dx\)

Optimal. Leaf size=25 \[ e^{\frac {1}{5} x^2 \log (x) \log \left (\log \left (\frac {x}{5 (-4-x)}\right )\right )} \]

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Rubi [F]  time = 2.14, 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 {e^{\frac {1}{5} x^2 \log (x) \log \left (\log \left (-\frac {x}{20+5 x}\right )\right )} \left (4 x \log (x)+\left (\left (4 x+x^2\right ) \log \left (-\frac {x}{20+5 x}\right )+\left (8 x+2 x^2\right ) \log (x) \log \left (-\frac {x}{20+5 x}\right )\right ) \log \left (\log \left (-\frac {x}{20+5 x}\right )\right )\right )}{(20+5 x) \log \left (-\frac {x}{20+5 x}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^((x^2*Log[x]*Log[Log[-(x/(20 + 5*x))]])/5)*(4*x*Log[x] + ((4*x + x^2)*Log[-(x/(20 + 5*x))] + (8*x + 2*x
^2)*Log[x]*Log[-(x/(20 + 5*x))])*Log[Log[-(x/(20 + 5*x))]]))/((20 + 5*x)*Log[-(x/(20 + 5*x))]),x]

[Out]

(4*Defer[Int][(E^((x^2*Log[x]*Log[Log[-(x/(20 + 5*x))]])/5)*Log[x])/Log[-1/5*x/(4 + x)], x])/5 - (16*Defer[Int
][(E^((x^2*Log[x]*Log[Log[-(x/(20 + 5*x))]])/5)*Log[x])/((4 + x)*Log[-1/5*x/(4 + x)]), x])/5 + Defer[Int][E^((
x^2*Log[x]*Log[Log[-(x/(20 + 5*x))]])/5)*x*Log[Log[-1/5*x/(4 + x)]], x]/5 + (2*Defer[Int][E^((x^2*Log[x]*Log[L
og[-(x/(20 + 5*x))]])/5)*x*Log[x]*Log[Log[-1/5*x/(4 + x)]], x])/5

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {4 e^{\frac {1}{5} x^2 \log (x) \log \left (\log \left (-\frac {x}{20+5 x}\right )\right )} x \log (x)}{5 (4+x) \log \left (-\frac {x}{5 (4+x)}\right )}+\frac {1}{5} e^{\frac {1}{5} x^2 \log (x) \log \left (\log \left (-\frac {x}{20+5 x}\right )\right )} x (1+2 \log (x)) \log \left (\log \left (-\frac {x}{5 (4+x)}\right )\right )\right ) \, dx\\ &=\frac {1}{5} \int e^{\frac {1}{5} x^2 \log (x) \log \left (\log \left (-\frac {x}{20+5 x}\right )\right )} x (1+2 \log (x)) \log \left (\log \left (-\frac {x}{5 (4+x)}\right )\right ) \, dx+\frac {4}{5} \int \frac {e^{\frac {1}{5} x^2 \log (x) \log \left (\log \left (-\frac {x}{20+5 x}\right )\right )} x \log (x)}{(4+x) \log \left (-\frac {x}{5 (4+x)}\right )} \, dx\\ &=\frac {1}{5} \int \left (e^{\frac {1}{5} x^2 \log (x) \log \left (\log \left (-\frac {x}{20+5 x}\right )\right )} x \log \left (\log \left (-\frac {x}{5 (4+x)}\right )\right )+2 e^{\frac {1}{5} x^2 \log (x) \log \left (\log \left (-\frac {x}{20+5 x}\right )\right )} x \log (x) \log \left (\log \left (-\frac {x}{5 (4+x)}\right )\right )\right ) \, dx+\frac {4}{5} \int \left (\frac {e^{\frac {1}{5} x^2 \log (x) \log \left (\log \left (-\frac {x}{20+5 x}\right )\right )} \log (x)}{\log \left (-\frac {x}{5 (4+x)}\right )}-\frac {4 e^{\frac {1}{5} x^2 \log (x) \log \left (\log \left (-\frac {x}{20+5 x}\right )\right )} \log (x)}{(4+x) \log \left (-\frac {x}{5 (4+x)}\right )}\right ) \, dx\\ &=\frac {1}{5} \int e^{\frac {1}{5} x^2 \log (x) \log \left (\log \left (-\frac {x}{20+5 x}\right )\right )} x \log \left (\log \left (-\frac {x}{5 (4+x)}\right )\right ) \, dx+\frac {2}{5} \int e^{\frac {1}{5} x^2 \log (x) \log \left (\log \left (-\frac {x}{20+5 x}\right )\right )} x \log (x) \log \left (\log \left (-\frac {x}{5 (4+x)}\right )\right ) \, dx+\frac {4}{5} \int \frac {e^{\frac {1}{5} x^2 \log (x) \log \left (\log \left (-\frac {x}{20+5 x}\right )\right )} \log (x)}{\log \left (-\frac {x}{5 (4+x)}\right )} \, dx-\frac {16}{5} \int \frac {e^{\frac {1}{5} x^2 \log (x) \log \left (\log \left (-\frac {x}{20+5 x}\right )\right )} \log (x)}{(4+x) \log \left (-\frac {x}{5 (4+x)}\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.12, size = 21, normalized size = 0.84 \begin {gather*} x^{\frac {1}{5} x^2 \log \left (\log \left (-\frac {x}{5 (4+x)}\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((x^2*Log[x]*Log[Log[-(x/(20 + 5*x))]])/5)*(4*x*Log[x] + ((4*x + x^2)*Log[-(x/(20 + 5*x))] + (8*x
+ 2*x^2)*Log[x]*Log[-(x/(20 + 5*x))])*Log[Log[-(x/(20 + 5*x))]]))/((20 + 5*x)*Log[-(x/(20 + 5*x))]),x]

[Out]

x^((x^2*Log[Log[-1/5*x/(4 + x)]])/5)

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fricas [A]  time = 0.74, size = 18, normalized size = 0.72 \begin {gather*} e^{\left (\frac {1}{5} \, x^{2} \log \relax (x) \log \left (\log \left (-\frac {x}{5 \, {\left (x + 4\right )}}\right )\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x^2+8*x)*log(-x/(20+5*x))*log(x)+(x^2+4*x)*log(-x/(20+5*x)))*log(log(-x/(20+5*x)))+4*x*log(x))*
exp(1/5*x^2*log(x)*log(log(-x/(20+5*x))))/(20+5*x)/log(-x/(20+5*x)),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.52, size = 18, normalized size = 0.72 \begin {gather*} e^{\left (\frac {1}{5} \, x^{2} \log \relax (x) \log \left (\log \left (-\frac {x}{5 \, {\left (x + 4\right )}}\right )\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x^2+8*x)*log(-x/(20+5*x))*log(x)+(x^2+4*x)*log(-x/(20+5*x)))*log(log(-x/(20+5*x)))+4*x*log(x))*
exp(1/5*x^2*log(x)*log(log(-x/(20+5*x))))/(20+5*x)/log(-x/(20+5*x)),x, algorithm="giac")

[Out]

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

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maple [C]  time = 0.80, size = 108, normalized size = 4.32

method	result	size
risch	\(\left (-\ln \relax (5)+i \pi +\ln \relax (x )-\ln \left (4+x \right )-\frac {i \pi \,\mathrm {csgn}\left (\frac {i x}{4+x}\right ) \left (-\mathrm {csgn}\left (\frac {i x}{4+x}\right )+\mathrm {csgn}\left (i x \right )\right ) \left (-\mathrm {csgn}\left (\frac {i x}{4+x}\right )+\mathrm {csgn}\left (\frac {i}{4+x}\right )\right )}{2}+i \pi \mathrm {csgn}\left (\frac {i x}{4+x}\right )^{2} \left (\mathrm {csgn}\left (\frac {i x}{4+x}\right )-1\right )\right )^{\frac {x^{2} \ln \relax (x )}{5}}\)	\(108\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((2*x^2+8*x)*ln(-x/(20+5*x))*ln(x)+(x^2+4*x)*ln(-x/(20+5*x)))*ln(ln(-x/(20+5*x)))+4*x*ln(x))*exp(1/5*x^2*
ln(x)*ln(ln(-x/(20+5*x))))/(20+5*x)/ln(-x/(20+5*x)),x,method=_RETURNVERBOSE)

[Out]

(-ln(5)+I*Pi+ln(x)-ln(4+x)-1/2*I*Pi*csgn(I*x/(4+x))*(-csgn(I*x/(4+x))+csgn(I*x))*(-csgn(I*x/(4+x))+csgn(I/(4+x
)))+I*Pi*csgn(I*x/(4+x))^2*(csgn(I*x/(4+x))-1))^(1/5*x^2*ln(x))

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maxima [A]  time = 0.69, size = 24, normalized size = 0.96 \begin {gather*} e^{\left (\frac {1}{5} \, x^{2} \log \relax (x) \log \left (-\log \relax (5) + \log \relax (x) - \log \left (-x - 4\right )\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x^2+8*x)*log(-x/(20+5*x))*log(x)+(x^2+4*x)*log(-x/(20+5*x)))*log(log(-x/(20+5*x)))+4*x*log(x))*
exp(1/5*x^2*log(x)*log(log(-x/(20+5*x))))/(20+5*x)/log(-x/(20+5*x)),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.89, size = 20, normalized size = 0.80 \begin {gather*} {\mathrm {e}}^{\frac {x^2\,\ln \left (\ln \left (-\frac {x}{5\,x+20}\right )\right )\,\ln \relax (x)}{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp((x^2*log(log(-x/(5*x + 20)))*log(x))/5)*(4*x*log(x) + log(log(-x/(5*x + 20)))*(log(-x/(5*x + 20))*(4*
x + x^2) + log(-x/(5*x + 20))*log(x)*(8*x + 2*x^2))))/(log(-x/(5*x + 20))*(5*x + 20)),x)

[Out]

exp((x^2*log(log(-x/(5*x + 20)))*log(x))/5)

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sympy [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+8*x)*ln(-x/(20+5*x))*ln(x)+(x**2+4*x)*ln(-x/(20+5*x)))*ln(ln(-x/(20+5*x)))+4*x*ln(x))*exp(
1/5*x**2*ln(x)*ln(ln(-x/(20+5*x))))/(20+5*x)/ln(-x/(20+5*x)),x)

[Out]

Timed out

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