Integrand size = 97, antiderivative size = 25 \[ \int \frac {e^{e^x} \left (250-200 x^2\right ) \log \left (\log \left (\frac {x}{20+16 x^2}\right )\right )+e^{e^x+x} \left (125 x+100 x^3\right ) \log \left (\frac {x}{20+16 x^2}\right ) \log ^2\left (\log \left (\frac {x}{20+16 x^2}\right )\right )}{\left (5 x+4 x^3\right ) \log \left (\frac {x}{20+16 x^2}\right )} \, dx=25 e^{e^x} \log ^2\left (\log \left (\frac {x}{4 \left (5+4 x^2\right )}\right )\right ) \]
\[ \int \frac {e^{e^x} \left (250-200 x^2\right ) \log \left (\log \left (\frac {x}{20+16 x^2}\right )\right )+e^{e^x+x} \left (125 x+100 x^3\right ) \log \left (\frac {x}{20+16 x^2}\right ) \log ^2\left (\log \left (\frac {x}{20+16 x^2}\right )\right )}{\left (5 x+4 x^3\right ) \log \left (\frac {x}{20+16 x^2}\right )} \, dx=\int \frac {e^{e^x} \left (250-200 x^2\right ) \log \left (\log \left (\frac {x}{20+16 x^2}\right )\right )+e^{e^x+x} \left (125 x+100 x^3\right ) \log \left (\frac {x}{20+16 x^2}\right ) \log ^2\left (\log \left (\frac {x}{20+16 x^2}\right )\right )}{\left (5 x+4 x^3\right ) \log \left (\frac {x}{20+16 x^2}\right )} \, dx \]
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]
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]
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {e^{e^x} \left (250-200 x^2\right ) \log \left (\log \left (\frac {x}{16 x^2+20}\right )\right )+e^{x+e^x} \left (100 x^3+125 x\right ) \log \left (\frac {x}{16 x^2+20}\right ) \log ^2\left (\log \left (\frac {x}{16 x^2+20}\right )\right )}{\left (4 x^3+5 x\right ) \log \left (\frac {x}{16 x^2+20}\right )} \, dx\) |
| \(\Big \downarrow \) 2026 |
| \(\displaystyle \int \frac {e^{e^x} \left (250-200 x^2\right ) \log \left (\log \left (\frac {x}{16 x^2+20}\right )\right )+e^{x+e^x} \left (100 x^3+125 x\right ) \log \left (\frac {x}{16 x^2+20}\right ) \log ^2\left (\log \left (\frac {x}{16 x^2+20}\right )\right )}{x \left (4 x^2+5\right ) \log \left (\frac {x}{16 x^2+20}\right )}dx\) |
| \(\Big \downarrow \) 7276 |
| \(\displaystyle \int \left (25 e^{x+e^x} \log ^2\left (\log \left (\frac {x}{16 x^2+20}\right )\right )-\frac {50 e^{e^x} \left (4 x^2-5\right ) \log \left (\log \left (\frac {x}{16 x^2+20}\right )\right )}{x \left (4 x^2+5\right ) \log \left (\frac {x}{16 x^2+20}\right )}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle 25 \int e^{x+e^x} \log ^2\left (\log \left (\frac {x}{16 x^2+20}\right )\right )dx+100 \int \frac {e^{e^x} \log \left (\log \left (\frac {x}{16 x^2+20}\right )\right )}{\left (i \sqrt {5}-2 x\right ) \log \left (\frac {x}{16 x^2+20}\right )}dx+50 \int \frac {e^{e^x} \log \left (\log \left (\frac {x}{16 x^2+20}\right )\right )}{x \log \left (\frac {x}{16 x^2+20}\right )}dx-100 \int \frac {e^{e^x} \log \left (\log \left (\frac {x}{16 x^2+20}\right )\right )}{\left (2 x+i \sqrt {5}\right ) \log \left (\frac {x}{16 x^2+20}\right )}dx\) |
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]
3.16.50.3.1 Defintions of rubi rules used
Int[(Fx_.)*(Px_)^(p_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Int[x^(p *r)*ExpandToSum[Px/x^r, x]^p*Fx, x] /; IGtQ[r, 0]] /; PolyQ[Px, x] && Integ erQ[p] && !MonomialQ[Px, x] && (ILtQ[p, 0] || !PolyQ[u, x])
Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionE xpand[u/(a + b*x^n), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ [n, 0]
Time = 66.83 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88
| method | result | size |
| parallelrisch | \(25 {\ln \left (\ln \left (\frac {x}{16 x^{2}+20}\right )\right )}^{2} {\mathrm e}^{{\mathrm e}^{x}}\) | \(22\) |
| risch | \(25 \,{\mathrm e}^{{\mathrm e}^{x}} {\ln \left (-4 \ln \left (2\right )+\ln \left (x \right )-\ln \left (x^{2}+\frac {5}{4}\right )-\frac {i \pi \,\operatorname {csgn}\left (\frac {i x}{x^{2}+\frac {5}{4}}\right ) \left (-\operatorname {csgn}\left (\frac {i x}{x^{2}+\frac {5}{4}}\right )+\operatorname {csgn}\left (i x \right )\right ) \left (-\operatorname {csgn}\left (\frac {i x}{x^{2}+\frac {5}{4}}\right )+\operatorname {csgn}\left (\frac {i}{x^{2}+\frac {5}{4}}\right )\right )}{2}\right )}^{2}\) | \(86\) |
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))*ln(ln(x/(16*x^2+20))))/(4*x^3+5*x)/ln(x /(16*x^2+20)),x,method=_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84 \[ \int \frac {e^{e^x} \left (250-200 x^2\right ) \log \left (\log \left (\frac {x}{20+16 x^2}\right )\right )+e^{e^x+x} \left (125 x+100 x^3\right ) \log \left (\frac {x}{20+16 x^2}\right ) \log ^2\left (\log \left (\frac {x}{20+16 x^2}\right )\right )}{\left (5 x+4 x^3\right ) \log \left (\frac {x}{20+16 x^2}\right )} \, dx=25 \, e^{\left (e^{x}\right )} \log \left (\log \left (\frac {x}{4 \, {\left (4 \, x^{2} + 5\right )}}\right )\right )^{2} \]
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=\
Timed out. \[ \int \frac {e^{e^x} \left (250-200 x^2\right ) \log \left (\log \left (\frac {x}{20+16 x^2}\right )\right )+e^{e^x+x} \left (125 x+100 x^3\right ) \log \left (\frac {x}{20+16 x^2}\right ) \log ^2\left (\log \left (\frac {x}{20+16 x^2}\right )\right )}{\left (5 x+4 x^3\right ) \log \left (\frac {x}{20+16 x^2}\right )} \, dx=\text {Timed out} \]
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)*exp(exp(x))*ln(ln(x/(16*x**2+20))))/(4*x **3+5*x)/ln(x/(16*x**2+20)),x)
Time = 0.36 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {e^{e^x} \left (250-200 x^2\right ) \log \left (\log \left (\frac {x}{20+16 x^2}\right )\right )+e^{e^x+x} \left (125 x+100 x^3\right ) \log \left (\frac {x}{20+16 x^2}\right ) \log ^2\left (\log \left (\frac {x}{20+16 x^2}\right )\right )}{\left (5 x+4 x^3\right ) \log \left (\frac {x}{20+16 x^2}\right )} \, dx=25 \, e^{\left (e^{x}\right )} \log \left (-2 \, \log \left (2\right ) - \log \left (4 \, x^{2} + 5\right ) + \log \left (x\right )\right )^{2} \]
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=\
\[ \int \frac {e^{e^x} \left (250-200 x^2\right ) \log \left (\log \left (\frac {x}{20+16 x^2}\right )\right )+e^{e^x+x} \left (125 x+100 x^3\right ) \log \left (\frac {x}{20+16 x^2}\right ) \log ^2\left (\log \left (\frac {x}{20+16 x^2}\right )\right )}{\left (5 x+4 x^3\right ) \log \left (\frac {x}{20+16 x^2}\right )} \, dx=\int { \frac {25 \, {\left ({\left (4 \, x^{3} + 5 \, x\right )} e^{\left (x + e^{x}\right )} \log \left (\frac {x}{4 \, {\left (4 \, x^{2} + 5\right )}}\right ) \log \left (\log \left (\frac {x}{4 \, {\left (4 \, x^{2} + 5\right )}}\right )\right )^{2} - 2 \, {\left (4 \, x^{2} - 5\right )} e^{\left (e^{x}\right )} \log \left (\log \left (\frac {x}{4 \, {\left (4 \, x^{2} + 5\right )}}\right )\right )\right )}}{{\left (4 \, x^{3} + 5 \, x\right )} \log \left (\frac {x}{4 \, {\left (4 \, x^{2} + 5\right )}}\right )} \,d x } \]
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=\
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)
Time = 13.06 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84 \[ \int \frac {e^{e^x} \left (250-200 x^2\right ) \log \left (\log \left (\frac {x}{20+16 x^2}\right )\right )+e^{e^x+x} \left (125 x+100 x^3\right ) \log \left (\frac {x}{20+16 x^2}\right ) \log ^2\left (\log \left (\frac {x}{20+16 x^2}\right )\right )}{\left (5 x+4 x^3\right ) \log \left (\frac {x}{20+16 x^2}\right )} \, dx=25\,{\ln \left (\ln \left (x\right )-\ln \left (16\,x^2+20\right )\right )}^2\,{\mathrm {e}}^{{\mathrm {e}}^x} \]