3.46.50 \(\int \frac {e^{e^x} (250-200 x^2) \log (\log (\frac {x}{20+16 x^2}))+e^{e^x+x} (125 x+100 x^3) \log (\frac {x}{20+16 x^2}) \log ^2(\log (\frac {x}{20+16 x^2}))}{(5 x+4 x^3) \log (\frac {x}{20+16 x^2})} \, dx\)

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

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Rubi [F]  time = 2.96, 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^{e^x} \left (250-200 x^2\right ) \log \left (\log \left (\frac {x}{20+16 x^2}\right )\right )+e^{e^x+x} \left (125 x+100 x^3\right ) \log \left (\frac {x}{20+16 x^2}\right ) \log ^2\left (\log \left (\frac {x}{20+16 x^2}\right )\right )}{\left (5 x+4 x^3\right ) \log \left (\frac {x}{20+16 x^2}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^E^x*(250 - 200*x^2)*Log[Log[x/(20 + 16*x^2)]] + E^(E^x + x)*(125*x + 100*x^3)*Log[x/(20 + 16*x^2)]*Log[
Log[x/(20 + 16*x^2)]]^2)/((5*x + 4*x^3)*Log[x/(20 + 16*x^2)]),x]

[Out]

100*Defer[Int][(E^E^x*Log[Log[x/(20 + 16*x^2)]])/((I*Sqrt[5] - 2*x)*Log[x/(20 + 16*x^2)]), x] + 50*Defer[Int][
(E^E^x*Log[Log[x/(20 + 16*x^2)]])/(x*Log[x/(20 + 16*x^2)]), x] - 100*Defer[Int][(E^E^x*Log[Log[x/(20 + 16*x^2)
]])/((I*Sqrt[5] + 2*x)*Log[x/(20 + 16*x^2)]), x] + 25*Defer[Int][E^(E^x + x)*Log[Log[x/(20 + 16*x^2)]]^2, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{e^x} \left (250-200 x^2\right ) \log \left (\log \left (\frac {x}{20+16 x^2}\right )\right )+e^{e^x+x} \left (125 x+100 x^3\right ) \log \left (\frac {x}{20+16 x^2}\right ) \log ^2\left (\log \left (\frac {x}{20+16 x^2}\right )\right )}{x \left (5+4 x^2\right ) \log \left (\frac {x}{20+16 x^2}\right )} \, dx\\ &=\int \left (-\frac {50 e^{e^x} \left (-5+4 x^2\right ) \log \left (\log \left (\frac {x}{20+16 x^2}\right )\right )}{x \left (5+4 x^2\right ) \log \left (\frac {x}{20+16 x^2}\right )}+25 e^{e^x+x} \log ^2\left (\log \left (\frac {x}{20+16 x^2}\right )\right )\right ) \, dx\\ &=25 \int e^{e^x+x} \log ^2\left (\log \left (\frac {x}{20+16 x^2}\right )\right ) \, dx-50 \int \frac {e^{e^x} \left (-5+4 x^2\right ) \log \left (\log \left (\frac {x}{20+16 x^2}\right )\right )}{x \left (5+4 x^2\right ) \log \left (\frac {x}{20+16 x^2}\right )} \, dx\\ &=25 \int e^{e^x+x} \log ^2\left (\log \left (\frac {x}{20+16 x^2}\right )\right ) \, dx-50 \int \left (-\frac {e^{e^x} \log \left (\log \left (\frac {x}{20+16 x^2}\right )\right )}{x \log \left (\frac {x}{20+16 x^2}\right )}+\frac {8 e^{e^x} x \log \left (\log \left (\frac {x}{20+16 x^2}\right )\right )}{\left (5+4 x^2\right ) \log \left (\frac {x}{20+16 x^2}\right )}\right ) \, dx\\ &=25 \int e^{e^x+x} \log ^2\left (\log \left (\frac {x}{20+16 x^2}\right )\right ) \, dx+50 \int \frac {e^{e^x} \log \left (\log \left (\frac {x}{20+16 x^2}\right )\right )}{x \log \left (\frac {x}{20+16 x^2}\right )} \, dx-400 \int \frac {e^{e^x} x \log \left (\log \left (\frac {x}{20+16 x^2}\right )\right )}{\left (5+4 x^2\right ) \log \left (\frac {x}{20+16 x^2}\right )} \, dx\\ &=25 \int e^{e^x+x} \log ^2\left (\log \left (\frac {x}{20+16 x^2}\right )\right ) \, dx+50 \int \frac {e^{e^x} \log \left (\log \left (\frac {x}{20+16 x^2}\right )\right )}{x \log \left (\frac {x}{20+16 x^2}\right )} \, dx-400 \int \left (-\frac {e^{e^x} \log \left (\log \left (\frac {x}{20+16 x^2}\right )\right )}{4 \left (i \sqrt {5}-2 x\right ) \log \left (\frac {x}{20+16 x^2}\right )}+\frac {e^{e^x} \log \left (\log \left (\frac {x}{20+16 x^2}\right )\right )}{4 \left (i \sqrt {5}+2 x\right ) \log \left (\frac {x}{20+16 x^2}\right )}\right ) \, dx\\ &=25 \int e^{e^x+x} \log ^2\left (\log \left (\frac {x}{20+16 x^2}\right )\right ) \, dx+50 \int \frac {e^{e^x} \log \left (\log \left (\frac {x}{20+16 x^2}\right )\right )}{x \log \left (\frac {x}{20+16 x^2}\right )} \, dx+100 \int \frac {e^{e^x} \log \left (\log \left (\frac {x}{20+16 x^2}\right )\right )}{\left (i \sqrt {5}-2 x\right ) \log \left (\frac {x}{20+16 x^2}\right )} \, dx-100 \int \frac {e^{e^x} \log \left (\log \left (\frac {x}{20+16 x^2}\right )\right )}{\left (i \sqrt {5}+2 x\right ) \log \left (\frac {x}{20+16 x^2}\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [F]  time = 0.17, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {e^{e^x} \left (250-200 x^2\right ) \log \left (\log \left (\frac {x}{20+16 x^2}\right )\right )+e^{e^x+x} \left (125 x+100 x^3\right ) \log \left (\frac {x}{20+16 x^2}\right ) \log ^2\left (\log \left (\frac {x}{20+16 x^2}\right )\right )}{\left (5 x+4 x^3\right ) \log \left (\frac {x}{20+16 x^2}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[(E^E^x*(250 - 200*x^2)*Log[Log[x/(20 + 16*x^2)]] + E^(E^x + x)*(125*x + 100*x^3)*Log[x/(20 + 16*x^2)
]*Log[Log[x/(20 + 16*x^2)]]^2)/((5*x + 4*x^3)*Log[x/(20 + 16*x^2)]),x]

[Out]

Integrate[(E^E^x*(250 - 200*x^2)*Log[Log[x/(20 + 16*x^2)]] + E^(E^x + x)*(125*x + 100*x^3)*Log[x/(20 + 16*x^2)
]*Log[Log[x/(20 + 16*x^2)]]^2)/((5*x + 4*x^3)*Log[x/(20 + 16*x^2)]), x]

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fricas [A]  time = 0.61, size = 21, normalized size = 0.84 \begin {gather*} 25 \, e^{\left (e^{x}\right )} \log \left (\log \left (\frac {x}{4 \, {\left (4 \, x^{2} + 5\right )}}\right )\right )^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((100*x^3+125*x)*exp(x)*log(x/(16*x^2+20))*exp(exp(x))*log(log(x/(16*x^2+20)))^2+(-200*x^2+250)*exp(
exp(x))*log(log(x/(16*x^2+20))))/(4*x^3+5*x)/log(x/(16*x^2+20)),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {25 \, {\left ({\left (4 \, x^{3} + 5 \, x\right )} e^{\left (x + e^{x}\right )} \log \left (\frac {x}{4 \, {\left (4 \, x^{2} + 5\right )}}\right ) \log \left (\log \left (\frac {x}{4 \, {\left (4 \, x^{2} + 5\right )}}\right )\right )^{2} - 2 \, {\left (4 \, x^{2} - 5\right )} e^{\left (e^{x}\right )} \log \left (\log \left (\frac {x}{4 \, {\left (4 \, x^{2} + 5\right )}}\right )\right )\right )}}{{\left (4 \, x^{3} + 5 \, x\right )} \log \left (\frac {x}{4 \, {\left (4 \, x^{2} + 5\right )}}\right )}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((100*x^3+125*x)*exp(x)*log(x/(16*x^2+20))*exp(exp(x))*log(log(x/(16*x^2+20)))^2+(-200*x^2+250)*exp(
exp(x))*log(log(x/(16*x^2+20))))/(4*x^3+5*x)/log(x/(16*x^2+20)),x, algorithm="giac")

[Out]

integrate(25*((4*x^3 + 5*x)*e^(x + e^x)*log(1/4*x/(4*x^2 + 5))*log(log(1/4*x/(4*x^2 + 5)))^2 - 2*(4*x^2 - 5)*e
^(e^x)*log(log(1/4*x/(4*x^2 + 5))))/((4*x^3 + 5*x)*log(1/4*x/(4*x^2 + 5))), x)

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maple [C]  time = 0.24, size = 86, normalized size = 3.44

method	result	size
risch	\(25 \,{\mathrm e}^{{\mathrm e}^{x}} \ln \left (-4 \ln \relax (2)+\ln \relax (x )-\ln \left (x^{2}+\frac {5}{4}\right )-\frac {i \pi \,\mathrm {csgn}\left (\frac {i x}{x^{2}+\frac {5}{4}}\right ) \left (-\mathrm {csgn}\left (\frac {i x}{x^{2}+\frac {5}{4}}\right )+\mathrm {csgn}\left (i x \right )\right ) \left (-\mathrm {csgn}\left (\frac {i x}{x^{2}+\frac {5}{4}}\right )+\mathrm {csgn}\left (\frac {i}{x^{2}+\frac {5}{4}}\right )\right )}{2}\right )^{2}\)	\(86\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((100*x^3+125*x)*exp(x)*ln(x/(16*x^2+20))*exp(exp(x))*ln(ln(x/(16*x^2+20)))^2+(-200*x^2+250)*exp(exp(x))*l
n(ln(x/(16*x^2+20))))/(4*x^3+5*x)/ln(x/(16*x^2+20)),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.62, size = 25, normalized size = 1.00 \begin {gather*} 25 \, e^{\left (e^{x}\right )} \log \left (-2 \, \log \relax (2) - \log \left (4 \, x^{2} + 5\right ) + \log \relax (x)\right )^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((100*x^3+125*x)*exp(x)*log(x/(16*x^2+20))*exp(exp(x))*log(log(x/(16*x^2+20)))^2+(-200*x^2+250)*exp(
exp(x))*log(log(x/(16*x^2+20))))/(4*x^3+5*x)/log(x/(16*x^2+20)),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 3.40, size = 21, normalized size = 0.84 \begin {gather*} 25\,{\ln \left (\ln \relax (x)-\ln \left (16\,x^2+20\right )\right )}^2\,{\mathrm {e}}^{{\mathrm {e}}^x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(exp(x))*log(log(x/(16*x^2 + 20)))*(200*x^2 - 250) - log(x/(16*x^2 + 20))*exp(exp(x))*exp(x)*log(log(
x/(16*x^2 + 20)))^2*(125*x + 100*x^3))/(log(x/(16*x^2 + 20))*(5*x + 4*x^3)),x)

[Out]

25*log(log(x) - log(16*x^2 + 20))^2*exp(exp(x))

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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(((100*x**3+125*x)*exp(x)*ln(x/(16*x**2+20))*exp(exp(x))*ln(ln(x/(16*x**2+20)))**2+(-200*x**2+250)*ex
p(exp(x))*ln(ln(x/(16*x**2+20))))/(4*x**3+5*x)/ln(x/(16*x**2+20)),x)

[Out]

Timed out

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