Integrand size = 136, antiderivative size = 29 \[ \int \frac {e^{\frac {1}{9} \left (225 e \log ^2\left (x^2\right )-150 e \log \left (x^2\right ) \log \left (\log \left (\frac {4-x}{4}\right )\right )+25 e \log ^2\left (\log \left (\frac {4-x}{4}\right )\right )\right )} \left (\left (-150 e x+e (-3600+900 x) \log \left (\frac {4-x}{4}\right )\right ) \log \left (x^2\right )+\left (50 e x+e (1200-300 x) \log \left (\frac {4-x}{4}\right )\right ) \log \left (\log \left (\frac {4-x}{4}\right )\right )\right )}{\left (-36 x+9 x^2\right ) \log \left (\frac {4-x}{4}\right )} \, dx=e^{25 e \left (-\log \left (x^2\right )+\frac {1}{3} \log \left (\log \left (\frac {4-x}{4}\right )\right )\right )^2} \] Output:
exp(5*(1/3*ln(ln(-1/4*x+1))-ln(x^2))*(5/3*ln(ln(-1/4*x+1))-5*ln(x^2))*exp( 1))
Time = 0.11 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.86 \[ \int \frac {e^{\frac {1}{9} \left (225 e \log ^2\left (x^2\right )-150 e \log \left (x^2\right ) \log \left (\log \left (\frac {4-x}{4}\right )\right )+25 e \log ^2\left (\log \left (\frac {4-x}{4}\right )\right )\right )} \left (\left (-150 e x+e (-3600+900 x) \log \left (\frac {4-x}{4}\right )\right ) \log \left (x^2\right )+\left (50 e x+e (1200-300 x) \log \left (\frac {4-x}{4}\right )\right ) \log \left (\log \left (\frac {4-x}{4}\right )\right )\right )}{\left (-36 x+9 x^2\right ) \log \left (\frac {4-x}{4}\right )} \, dx=e^{\frac {25}{9} e \left (-3 \log \left (x^2\right )+\log \left (\log \left (1-\frac {x}{4}\right )\right )\right )^2} \] Input:
Integrate[(E^((225*E*Log[x^2]^2 - 150*E*Log[x^2]*Log[Log[(4 - x)/4]] + 25* E*Log[Log[(4 - x)/4]]^2)/9)*((-150*E*x + E*(-3600 + 900*x)*Log[(4 - x)/4]) *Log[x^2] + (50*E*x + E*(1200 - 300*x)*Log[(4 - x)/4])*Log[Log[(4 - x)/4]] ))/((-36*x + 9*x^2)*Log[(4 - x)/4]),x]
Output:
E^((25*E*(-3*Log[x^2] + Log[Log[1 - x/4]])^2)/9)
Time = 3.13 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.93, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.037, Rules used = {2026, 7292, 27, 27, 7257}
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.
| \(\displaystyle \int \frac {\left (\left (e (900 x-3600) \log \left (\frac {4-x}{4}\right )-150 e x\right ) \log \left (x^2\right )+\left (50 e x+e (1200-300 x) \log \left (\frac {4-x}{4}\right )\right ) \log \left (\log \left (\frac {4-x}{4}\right )\right )\right ) \exp \left (\frac {1}{9} \left (225 e \log ^2\left (x^2\right )-150 e \log \left (\log \left (\frac {4-x}{4}\right )\right ) \log \left (x^2\right )+25 e \log ^2\left (\log \left (\frac {4-x}{4}\right )\right )\right )\right )}{\left (9 x^2-36 x\right ) \log \left (\frac {4-x}{4}\right )} \, dx\) |
| \(\Big \downarrow \) 2026 |
| \(\displaystyle \int \frac {\left (\left (e (900 x-3600) \log \left (\frac {4-x}{4}\right )-150 e x\right ) \log \left (x^2\right )+\left (50 e x+e (1200-300 x) \log \left (\frac {4-x}{4}\right )\right ) \log \left (\log \left (\frac {4-x}{4}\right )\right )\right ) \exp \left (\frac {1}{9} \left (225 e \log ^2\left (x^2\right )-150 e \log \left (\log \left (\frac {4-x}{4}\right )\right ) \log \left (x^2\right )+25 e \log ^2\left (\log \left (\frac {4-x}{4}\right )\right )\right )\right )}{x (9 x-36) \log \left (\frac {4-x}{4}\right )}dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle \int \frac {50 e^{\frac {25}{9} e \left (3 \log \left (x^2\right )-\log \left (\log \left (1-\frac {x}{4}\right )\right )\right )^2+1} \left (x-6 x \log \left (1-\frac {x}{4}\right )+24 \log \left (1-\frac {x}{4}\right )\right ) \left (3 \log \left (x^2\right )-\log \left (\log \left (1-\frac {x}{4}\right )\right )\right )}{(36-9 x) x \log \left (1-\frac {x}{4}\right )}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle 50 \int \frac {e^{\frac {25}{9} e \left (3 \log \left (x^2\right )-\log \left (\log \left (1-\frac {x}{4}\right )\right )\right )^2+1} \left (-6 \log \left (1-\frac {x}{4}\right ) x+x+24 \log \left (1-\frac {x}{4}\right )\right ) \left (3 \log \left (x^2\right )-\log \left (\log \left (1-\frac {x}{4}\right )\right )\right )}{9 (4-x) x \log \left (1-\frac {x}{4}\right )}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {50}{9} \int \frac {e^{\frac {25}{9} e \left (3 \log \left (x^2\right )-\log \left (\log \left (1-\frac {x}{4}\right )\right )\right )^2+1} \left (-6 \log \left (1-\frac {x}{4}\right ) x+x+24 \log \left (1-\frac {x}{4}\right )\right ) \left (3 \log \left (x^2\right )-\log \left (\log \left (1-\frac {x}{4}\right )\right )\right )}{(4-x) x \log \left (1-\frac {x}{4}\right )}dx\) |
| \(\Big \downarrow \) 7257 |
| \(\displaystyle e^{\frac {25}{9} e \left (3 \log \left (x^2\right )-\log \left (\log \left (1-\frac {x}{4}\right )\right )\right )^2}\) |
Input:
Int[(E^((225*E*Log[x^2]^2 - 150*E*Log[x^2]*Log[Log[(4 - x)/4]] + 25*E*Log[ Log[(4 - x)/4]]^2)/9)*((-150*E*x + E*(-3600 + 900*x)*Log[(4 - x)/4])*Log[x ^2] + (50*E*x + E*(1200 - 300*x)*Log[(4 - x)/4])*Log[Log[(4 - x)/4]]))/((- 36*x + 9*x^2)*Log[(4 - x)/4]),x]
Output:
E^((25*E*(3*Log[x^2] - Log[Log[1 - x/4]])^2)/9)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Sim p[q*(F^v/Log[F]), x] /; !FalseQ[q]] /; FreeQ[F, x]
Time = 39.65 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.28
| method | result | size |
| parallelrisch | \({\mathrm e}^{\frac {25 \,{\mathrm e} \left (\ln \left (\ln \left (-\frac {x}{4}+1\right )\right )^{2}-6 \ln \left (\ln \left (-\frac {x}{4}+1\right )\right ) \ln \left (x^{2}\right )+9 \ln \left (x^{2}\right )^{2}\right )}{9}}\) | \(37\) |
| risch | \(\ln \left (-\frac {x}{4}+1\right )^{-\frac {50 \,{\mathrm e} \left (-i \pi \,\operatorname {csgn}\left (i x^{2}\right )+i \operatorname {csgn}\left (i x \right ) \pi +2 \ln \left (x \right )\right )}{3}} x^{-100 i {\mathrm e} \pi \,\operatorname {csgn}\left (i x^{2}\right )} x^{100 i {\mathrm e} \pi \,\operatorname {csgn}\left (i x \right )} {\mathrm e}^{\frac {25 \,{\mathrm e} \left (-9 \pi ^{2} \operatorname {csgn}\left (i x^{2}\right )^{6}+36 \pi ^{2} \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{5}-54 \pi ^{2} \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )^{4}+36 \pi ^{2} \operatorname {csgn}\left (i x \right )^{3} \operatorname {csgn}\left (i x^{2}\right )^{3}-9 \pi ^{2} \operatorname {csgn}\left (i x \right )^{4} \operatorname {csgn}\left (i x^{2}\right )^{2}+4 \ln \left (\ln \left (-\frac {x}{4}+1\right )\right )^{2}+144 \ln \left (x \right )^{2}\right )}{36}}\) | \(187\) |
Input:
int((((-300*x+1200)*exp(1)*ln(-1/4*x+1)+50*x*exp(1))*ln(ln(-1/4*x+1))+((90 0*x-3600)*exp(1)*ln(-1/4*x+1)-150*x*exp(1))*ln(x^2))*exp(25/9*exp(1)*ln(ln (-1/4*x+1))^2-50/3*exp(1)*ln(x^2)*ln(ln(-1/4*x+1))+25*exp(1)*ln(x^2)^2)/(9 *x^2-36*x)/ln(-1/4*x+1),x,method=_RETURNVERBOSE)
Output:
exp(25/9*exp(1)*(ln(ln(-1/4*x+1))^2-6*ln(ln(-1/4*x+1))*ln(x^2)+9*ln(x^2)^2 ))
Time = 0.10 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.38 \[ \int \frac {e^{\frac {1}{9} \left (225 e \log ^2\left (x^2\right )-150 e \log \left (x^2\right ) \log \left (\log \left (\frac {4-x}{4}\right )\right )+25 e \log ^2\left (\log \left (\frac {4-x}{4}\right )\right )\right )} \left (\left (-150 e x+e (-3600+900 x) \log \left (\frac {4-x}{4}\right )\right ) \log \left (x^2\right )+\left (50 e x+e (1200-300 x) \log \left (\frac {4-x}{4}\right )\right ) \log \left (\log \left (\frac {4-x}{4}\right )\right )\right )}{\left (-36 x+9 x^2\right ) \log \left (\frac {4-x}{4}\right )} \, dx=e^{\left (25 \, e \log \left (x^{2}\right )^{2} - \frac {50}{3} \, e \log \left (x^{2}\right ) \log \left (\log \left (-\frac {1}{4} \, x + 1\right )\right ) + \frac {25}{9} \, e \log \left (\log \left (-\frac {1}{4} \, x + 1\right )\right )^{2}\right )} \] Input:
integrate((((-300*x+1200)*exp(1)*log(-1/4*x+1)+50*exp(1)*x)*log(log(-1/4*x +1))+((900*x-3600)*exp(1)*log(-1/4*x+1)-150*exp(1)*x)*log(x^2))*exp(25/9*e xp(1)*log(log(-1/4*x+1))^2-50/3*exp(1)*log(x^2)*log(log(-1/4*x+1))+25*exp( 1)*log(x^2)^2)/(9*x^2-36*x)/log(-1/4*x+1),x, algorithm="fricas")
Output:
e^(25*e*log(x^2)^2 - 50/3*e*log(x^2)*log(log(-1/4*x + 1)) + 25/9*e*log(log (-1/4*x + 1))^2)
Leaf count of result is larger than twice the leaf count of optimal. 49 vs. \(2 (24) = 48\).
Time = 1.31 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.69 \[ \int \frac {e^{\frac {1}{9} \left (225 e \log ^2\left (x^2\right )-150 e \log \left (x^2\right ) \log \left (\log \left (\frac {4-x}{4}\right )\right )+25 e \log ^2\left (\log \left (\frac {4-x}{4}\right )\right )\right )} \left (\left (-150 e x+e (-3600+900 x) \log \left (\frac {4-x}{4}\right )\right ) \log \left (x^2\right )+\left (50 e x+e (1200-300 x) \log \left (\frac {4-x}{4}\right )\right ) \log \left (\log \left (\frac {4-x}{4}\right )\right )\right )}{\left (-36 x+9 x^2\right ) \log \left (\frac {4-x}{4}\right )} \, dx=e^{25 e \log {\left (x^{2} \right )}^{2} - \frac {50 e \log {\left (x^{2} \right )} \log {\left (\log {\left (1 - \frac {x}{4} \right )} \right )}}{3} + \frac {25 e \log {\left (\log {\left (1 - \frac {x}{4} \right )} \right )}^{2}}{9}} \] Input:
integrate((((-300*x+1200)*exp(1)*ln(-1/4*x+1)+50*exp(1)*x)*ln(ln(-1/4*x+1) )+((900*x-3600)*exp(1)*ln(-1/4*x+1)-150*exp(1)*x)*ln(x**2))*exp(25/9*exp(1 )*ln(ln(-1/4*x+1))**2-50/3*exp(1)*ln(x**2)*ln(ln(-1/4*x+1))+25*exp(1)*ln(x **2)**2)/(9*x**2-36*x)/ln(-1/4*x+1),x)
Output:
exp(25*E*log(x**2)**2 - 50*E*log(x**2)*log(log(1 - x/4))/3 + 25*E*log(log( 1 - x/4))**2/9)
Time = 0.57 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.59 \[ \int \frac {e^{\frac {1}{9} \left (225 e \log ^2\left (x^2\right )-150 e \log \left (x^2\right ) \log \left (\log \left (\frac {4-x}{4}\right )\right )+25 e \log ^2\left (\log \left (\frac {4-x}{4}\right )\right )\right )} \left (\left (-150 e x+e (-3600+900 x) \log \left (\frac {4-x}{4}\right )\right ) \log \left (x^2\right )+\left (50 e x+e (1200-300 x) \log \left (\frac {4-x}{4}\right )\right ) \log \left (\log \left (\frac {4-x}{4}\right )\right )\right )}{\left (-36 x+9 x^2\right ) \log \left (\frac {4-x}{4}\right )} \, dx=e^{\left (100 \, e \log \left (x\right )^{2} - \frac {100}{3} \, e \log \left (x\right ) \log \left (-2 \, \log \left (2\right ) + \log \left (-x + 4\right )\right ) + \frac {25}{9} \, e \log \left (-2 \, \log \left (2\right ) + \log \left (-x + 4\right )\right )^{2}\right )} \] Input:
integrate((((-300*x+1200)*exp(1)*log(-1/4*x+1)+50*exp(1)*x)*log(log(-1/4*x +1))+((900*x-3600)*exp(1)*log(-1/4*x+1)-150*exp(1)*x)*log(x^2))*exp(25/9*e xp(1)*log(log(-1/4*x+1))^2-50/3*exp(1)*log(x^2)*log(log(-1/4*x+1))+25*exp( 1)*log(x^2)^2)/(9*x^2-36*x)/log(-1/4*x+1),x, algorithm="maxima")
Output:
e^(100*e*log(x)^2 - 100/3*e*log(x)*log(-2*log(2) + log(-x + 4)) + 25/9*e*l og(-2*log(2) + log(-x + 4))^2)
Time = 0.47 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.38 \[ \int \frac {e^{\frac {1}{9} \left (225 e \log ^2\left (x^2\right )-150 e \log \left (x^2\right ) \log \left (\log \left (\frac {4-x}{4}\right )\right )+25 e \log ^2\left (\log \left (\frac {4-x}{4}\right )\right )\right )} \left (\left (-150 e x+e (-3600+900 x) \log \left (\frac {4-x}{4}\right )\right ) \log \left (x^2\right )+\left (50 e x+e (1200-300 x) \log \left (\frac {4-x}{4}\right )\right ) \log \left (\log \left (\frac {4-x}{4}\right )\right )\right )}{\left (-36 x+9 x^2\right ) \log \left (\frac {4-x}{4}\right )} \, dx=e^{\left (25 \, e \log \left (x^{2}\right )^{2} - \frac {50}{3} \, e \log \left (x^{2}\right ) \log \left (\log \left (-\frac {1}{4} \, x + 1\right )\right ) + \frac {25}{9} \, e \log \left (\log \left (-\frac {1}{4} \, x + 1\right )\right )^{2}\right )} \] Input:
integrate((((-300*x+1200)*exp(1)*log(-1/4*x+1)+50*exp(1)*x)*log(log(-1/4*x +1))+((900*x-3600)*exp(1)*log(-1/4*x+1)-150*exp(1)*x)*log(x^2))*exp(25/9*e xp(1)*log(log(-1/4*x+1))^2-50/3*exp(1)*log(x^2)*log(log(-1/4*x+1))+25*exp( 1)*log(x^2)^2)/(9*x^2-36*x)/log(-1/4*x+1),x, algorithm="giac")
Output:
e^(25*e*log(x^2)^2 - 50/3*e*log(x^2)*log(log(-1/4*x + 1)) + 25/9*e*log(log (-1/4*x + 1))^2)
Time = 7.82 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.45 \[ \int \frac {e^{\frac {1}{9} \left (225 e \log ^2\left (x^2\right )-150 e \log \left (x^2\right ) \log \left (\log \left (\frac {4-x}{4}\right )\right )+25 e \log ^2\left (\log \left (\frac {4-x}{4}\right )\right )\right )} \left (\left (-150 e x+e (-3600+900 x) \log \left (\frac {4-x}{4}\right )\right ) \log \left (x^2\right )+\left (50 e x+e (1200-300 x) \log \left (\frac {4-x}{4}\right )\right ) \log \left (\log \left (\frac {4-x}{4}\right )\right )\right )}{\left (-36 x+9 x^2\right ) \log \left (\frac {4-x}{4}\right )} \, dx={\mathrm {e}}^{-\frac {50\,\ln \left (\ln \left (1-\frac {x}{4}\right )\right )\,\ln \left (x^2\right )\,\mathrm {e}}{3}}\,{\mathrm {e}}^{\frac {25\,{\ln \left (\ln \left (1-\frac {x}{4}\right )\right )}^2\,\mathrm {e}}{9}}\,{\mathrm {e}}^{25\,{\ln \left (x^2\right )}^2\,\mathrm {e}} \] Input:
int(-(exp((25*log(log(1 - x/4))^2*exp(1))/9 + 25*log(x^2)^2*exp(1) - (50*l og(log(1 - x/4))*log(x^2)*exp(1))/3)*(log(log(1 - x/4))*(50*x*exp(1) - exp (1)*log(1 - x/4)*(300*x - 1200)) - log(x^2)*(150*x*exp(1) - exp(1)*log(1 - x/4)*(900*x - 3600))))/(log(1 - x/4)*(36*x - 9*x^2)),x)
Output:
exp(-(50*log(log(1 - x/4))*log(x^2)*exp(1))/3)*exp((25*log(log(1 - x/4))^2 *exp(1))/9)*exp(25*log(x^2)^2*exp(1))
\[ \int \frac {e^{\frac {1}{9} \left (225 e \log ^2\left (x^2\right )-150 e \log \left (x^2\right ) \log \left (\log \left (\frac {4-x}{4}\right )\right )+25 e \log ^2\left (\log \left (\frac {4-x}{4}\right )\right )\right )} \left (\left (-150 e x+e (-3600+900 x) \log \left (\frac {4-x}{4}\right )\right ) \log \left (x^2\right )+\left (50 e x+e (1200-300 x) \log \left (\frac {4-x}{4}\right )\right ) \log \left (\log \left (\frac {4-x}{4}\right )\right )\right )}{\left (-36 x+9 x^2\right ) \log \left (\frac {4-x}{4}\right )} \, dx=\int \frac {\left (\left (\left (-300 x +1200\right ) {\mathrm e} \,\mathrm {log}\left (-\frac {x}{4}+1\right )+50 \,{\mathrm e} x \right ) \mathrm {log}\left (\mathrm {log}\left (-\frac {x}{4}+1\right )\right )+\left (\left (900 x -3600\right ) {\mathrm e} \,\mathrm {log}\left (-\frac {x}{4}+1\right )-150 \,{\mathrm e} x \right ) \mathrm {log}\left (x^{2}\right )\right ) {\mathrm e}^{\frac {25 \,{\mathrm e} \mathrm {log}\left (\mathrm {log}\left (-\frac {x}{4}+1\right )\right )^{2}}{9}-\frac {50 \,{\mathrm e} \,\mathrm {log}\left (x^{2}\right ) \mathrm {log}\left (\mathrm {log}\left (-\frac {x}{4}+1\right )\right )}{3}+25 \,{\mathrm e} \mathrm {log}\left (x^{2}\right )^{2}}}{\left (9 x^{2}-36 x \right ) \mathrm {log}\left (-\frac {x}{4}+1\right )}d x \] Input:
int((((-300*x+1200)*exp(1)*log(-1/4*x+1)+50*exp(1)*x)*log(log(-1/4*x+1))+( (900*x-3600)*exp(1)*log(-1/4*x+1)-150*exp(1)*x)*log(x^2))*exp(25/9*exp(1)* log(log(-1/4*x+1))^2-50/3*exp(1)*log(x^2)*log(log(-1/4*x+1))+25*exp(1)*log (x^2)^2)/(9*x^2-36*x)/log(-1/4*x+1),x)
Output:
int((((-300*x+1200)*exp(1)*log(-1/4*x+1)+50*exp(1)*x)*log(log(-1/4*x+1))+( (900*x-3600)*exp(1)*log(-1/4*x+1)-150*exp(1)*x)*log(x^2))*exp(25/9*exp(1)* log(log(-1/4*x+1))^2-50/3*exp(1)*log(x^2)*log(log(-1/4*x+1))+25*exp(1)*log (x^2)^2)/(9*x^2-36*x)/log(-1/4*x+1),x)