Integrand size = 103, antiderivative size = 22 \[ \int \frac {e^{\frac {4+(1+x) \log \left (\frac {x^2}{7}\right )+\log \left (\frac {x^2}{7}\right ) \log (\log (4+x))}{\log \left (\frac {x^2}{7}\right )}} \left ((-32-8 x) \log (4+x)+\log ^2\left (\frac {x^2}{7}\right ) \left (x+\left (4 x+x^2\right ) \log (4+x)\right )\right )}{\left (4 x+x^2\right ) \log ^2\left (\frac {x^2}{7}\right ) \log (4+x)} \, dx=e^{1+x+\frac {4}{\log \left (\frac {x^2}{7}\right )}} \log (4+x) \]
Time = 0.42 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {4+(1+x) \log \left (\frac {x^2}{7}\right )+\log \left (\frac {x^2}{7}\right ) \log (\log (4+x))}{\log \left (\frac {x^2}{7}\right )}} \left ((-32-8 x) \log (4+x)+\log ^2\left (\frac {x^2}{7}\right ) \left (x+\left (4 x+x^2\right ) \log (4+x)\right )\right )}{\left (4 x+x^2\right ) \log ^2\left (\frac {x^2}{7}\right ) \log (4+x)} \, dx=e^{1+x+\frac {4}{\log \left (\frac {x^2}{7}\right )}} \log (4+x) \]
Integrate[(E^((4 + (1 + x)*Log[x^2/7] + Log[x^2/7]*Log[Log[4 + x]])/Log[x^ 2/7])*((-32 - 8*x)*Log[4 + x] + Log[x^2/7]^2*(x + (4*x + x^2)*Log[4 + x])) )/((4*x + x^2)*Log[x^2/7]^2*Log[4 + x]),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.
\(\displaystyle \int \frac {\left (\left (\left (x^2+4 x\right ) \log (x+4)+x\right ) \log ^2\left (\frac {x^2}{7}\right )+(-8 x-32) \log (x+4)\right ) \exp \left (\frac {(x+1) \log \left (\frac {x^2}{7}\right )+\log (\log (x+4)) \log \left (\frac {x^2}{7}\right )+4}{\log \left (\frac {x^2}{7}\right )}\right )}{\left (x^2+4 x\right ) \log ^2\left (\frac {x^2}{7}\right ) \log (x+4)} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {\left (\left (\left (x^2+4 x\right ) \log (x+4)+x\right ) \log ^2\left (\frac {x^2}{7}\right )+(-8 x-32) \log (x+4)\right ) \exp \left (\frac {(x+1) \log \left (\frac {x^2}{7}\right )+\log (\log (x+4)) \log \left (\frac {x^2}{7}\right )+4}{\log \left (\frac {x^2}{7}\right )}\right )}{x (x+4) \log ^2\left (\frac {x^2}{7}\right ) \log (x+4)}dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {e^{\frac {4}{\log \left (\frac {x^2}{7}\right )}+x+1} \left (\log ^2\left (\frac {x^2}{7}\right ) (x+(x+4) x \log (x+4))-8 (x+4) \log (x+4)\right )}{x (x+4) \log ^2\left (\frac {x^2}{7}\right )}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {e^{\frac {4}{\log \left (\frac {x^2}{7}\right )}+x+1} \left (x \log ^2\left (\frac {x^2}{7}\right )-8\right ) \log (x+4)}{x \log ^2\left (\frac {x^2}{7}\right )}+\frac {e^{\frac {4}{\log \left (\frac {x^2}{7}\right )}+x+1}}{x+4}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -8 \int \frac {e^{x+\frac {4}{\log \left (\frac {x^2}{7}\right )}+1} \log (x+4)}{x \log ^2\left (\frac {x^2}{7}\right )}dx+\int \frac {e^{x+\frac {4}{\log \left (\frac {x^2}{7}\right )}+1}}{x+4}dx+\int e^{x+\frac {4}{\log \left (\frac {x^2}{7}\right )}+1} \log (x+4)dx\) |
Int[(E^((4 + (1 + x)*Log[x^2/7] + Log[x^2/7]*Log[Log[4 + x]])/Log[x^2/7])* ((-32 - 8*x)*Log[4 + x] + Log[x^2/7]^2*(x + (4*x + x^2)*Log[4 + x])))/((4* x + x^2)*Log[x^2/7]^2*Log[4 + x]),x]
3.1.31.3.1 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_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 55.74 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.59
method | result | size |
parallelrisch | \({\mathrm e}^{\frac {\ln \left (\frac {x^{2}}{7}\right ) \ln \left (\ln \left (4+x \right )\right )+\left (1+x \right ) \ln \left (\frac {x^{2}}{7}\right )+4}{\ln \left (\frac {x^{2}}{7}\right )}}\) | \(35\) |
risch | \({\mathrm e}^{\frac {-2 i \ln \left (\ln \left (4+x \right )\right ) \pi \operatorname {csgn}\left (i x^{2}\right )^{2} \operatorname {csgn}\left (i x \right )+i \ln \left (\ln \left (4+x \right )\right ) \pi \,\operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i x \right )^{2}-2 i x \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2}+i x \pi \operatorname {csgn}\left (i x^{2}\right )^{3}+i \pi \operatorname {csgn}\left (i x^{2}\right )^{3}+i x \pi \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )+i \pi \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )+i \ln \left (\ln \left (4+x \right )\right ) \pi \operatorname {csgn}\left (i x^{2}\right )^{3}-2 i \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2}-4 \ln \left (\ln \left (4+x \right )\right ) \ln \left (x \right )-4 x \ln \left (x \right )+2 \ln \left (\ln \left (4+x \right )\right ) \ln \left (7\right )+2 x \ln \left (7\right )-4 \ln \left (x \right )+2 \ln \left (7\right )-8}{i \pi \operatorname {csgn}\left (i x^{2}\right )^{3}-2 i \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2}+i \pi \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )-4 \ln \left (x \right )+2 \ln \left (7\right )}}\) | \(266\) |
int((((x^2+4*x)*ln(4+x)+x)*ln(1/7*x^2)^2+(-8*x-32)*ln(4+x))*exp((ln(1/7*x^ 2)*ln(ln(4+x))+(1+x)*ln(1/7*x^2)+4)/ln(1/7*x^2))/(x^2+4*x)/ln(4+x)/ln(1/7* x^2)^2,x,method=_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.55 \[ \int \frac {e^{\frac {4+(1+x) \log \left (\frac {x^2}{7}\right )+\log \left (\frac {x^2}{7}\right ) \log (\log (4+x))}{\log \left (\frac {x^2}{7}\right )}} \left ((-32-8 x) \log (4+x)+\log ^2\left (\frac {x^2}{7}\right ) \left (x+\left (4 x+x^2\right ) \log (4+x)\right )\right )}{\left (4 x+x^2\right ) \log ^2\left (\frac {x^2}{7}\right ) \log (4+x)} \, dx=e^{\left (\frac {{\left (x + 1\right )} \log \left (\frac {1}{7} \, x^{2}\right ) + \log \left (\frac {1}{7} \, x^{2}\right ) \log \left (\log \left (x + 4\right )\right ) + 4}{\log \left (\frac {1}{7} \, x^{2}\right )}\right )} \]
integrate((((x^2+4*x)*log(4+x)+x)*log(1/7*x^2)^2+(-8*x-32)*log(4+x))*exp(( log(1/7*x^2)*log(log(4+x))+(1+x)*log(1/7*x^2)+4)/log(1/7*x^2))/(x^2+4*x)/l og(4+x)/log(1/7*x^2)^2,x, algorithm=\
Time = 0.87 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.45 \[ \int \frac {e^{\frac {4+(1+x) \log \left (\frac {x^2}{7}\right )+\log \left (\frac {x^2}{7}\right ) \log (\log (4+x))}{\log \left (\frac {x^2}{7}\right )}} \left ((-32-8 x) \log (4+x)+\log ^2\left (\frac {x^2}{7}\right ) \left (x+\left (4 x+x^2\right ) \log (4+x)\right )\right )}{\left (4 x+x^2\right ) \log ^2\left (\frac {x^2}{7}\right ) \log (4+x)} \, dx=e^{\frac {\left (x + 1\right ) \log {\left (\frac {x^{2}}{7} \right )} + \log {\left (\frac {x^{2}}{7} \right )} \log {\left (\log {\left (x + 4 \right )} \right )} + 4}{\log {\left (\frac {x^{2}}{7} \right )}}} \]
integrate((((x**2+4*x)*ln(4+x)+x)*ln(1/7*x**2)**2+(-8*x-32)*ln(4+x))*exp(( ln(1/7*x**2)*ln(ln(4+x))+(1+x)*ln(1/7*x**2)+4)/ln(1/7*x**2))/(x**2+4*x)/ln (4+x)/ln(1/7*x**2)**2,x)
Time = 0.43 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {e^{\frac {4+(1+x) \log \left (\frac {x^2}{7}\right )+\log \left (\frac {x^2}{7}\right ) \log (\log (4+x))}{\log \left (\frac {x^2}{7}\right )}} \left ((-32-8 x) \log (4+x)+\log ^2\left (\frac {x^2}{7}\right ) \left (x+\left (4 x+x^2\right ) \log (4+x)\right )\right )}{\left (4 x+x^2\right ) \log ^2\left (\frac {x^2}{7}\right ) \log (4+x)} \, dx=e^{\left (x - \frac {4}{\log \left (7\right ) - 2 \, \log \left (x\right )} + 1\right )} \log \left (x + 4\right ) \]
integrate((((x^2+4*x)*log(4+x)+x)*log(1/7*x^2)^2+(-8*x-32)*log(4+x))*exp(( log(1/7*x^2)*log(log(4+x))+(1+x)*log(1/7*x^2)+4)/log(1/7*x^2))/(x^2+4*x)/l og(4+x)/log(1/7*x^2)^2,x, algorithm=\
Time = 1.64 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.86 \[ \int \frac {e^{\frac {4+(1+x) \log \left (\frac {x^2}{7}\right )+\log \left (\frac {x^2}{7}\right ) \log (\log (4+x))}{\log \left (\frac {x^2}{7}\right )}} \left ((-32-8 x) \log (4+x)+\log ^2\left (\frac {x^2}{7}\right ) \left (x+\left (4 x+x^2\right ) \log (4+x)\right )\right )}{\left (4 x+x^2\right ) \log ^2\left (\frac {x^2}{7}\right ) \log (4+x)} \, dx=e^{\left (x + \frac {4}{\log \left (\frac {1}{7} \, x^{2}\right )} + \log \left (\log \left (x + 4\right )\right ) + 1\right )} \]
integrate((((x^2+4*x)*log(4+x)+x)*log(1/7*x^2)^2+(-8*x-32)*log(4+x))*exp(( log(1/7*x^2)*log(log(4+x))+(1+x)*log(1/7*x^2)+4)/log(1/7*x^2))/(x^2+4*x)/l og(4+x)/log(1/7*x^2)^2,x, algorithm=\
Time = 12.03 (sec) , antiderivative size = 121, normalized size of antiderivative = 5.50 \[ \int \frac {e^{\frac {4+(1+x) \log \left (\frac {x^2}{7}\right )+\log \left (\frac {x^2}{7}\right ) \log (\log (4+x))}{\log \left (\frac {x^2}{7}\right )}} \left ((-32-8 x) \log (4+x)+\log ^2\left (\frac {x^2}{7}\right ) \left (x+\left (4 x+x^2\right ) \log (4+x)\right )\right )}{\left (4 x+x^2\right ) \log ^2\left (\frac {x^2}{7}\right ) \log (4+x)} \, dx=\frac {{\ln \left (x+4\right )}^{\frac {\ln \left (x^2\right )}{\ln \left (x^2\right )-\ln \left (7\right )}}\,{\mathrm {e}}^{\frac {4}{\ln \left (x^2\right )-\ln \left (7\right )}}\,{\left (x^2\right )}^{\frac {1}{\ln \left (x^2\right )-\ln \left (7\right )}}\,{\left (x^2\right )}^{\frac {x}{\ln \left (x^2\right )-\ln \left (7\right )}}}{7^{\frac {x}{\ln \left (x^2\right )-\ln \left (7\right )}}\,7^{\frac {1}{\ln \left (x^2\right )-\ln \left (7\right )}}\,{\ln \left (x+4\right )}^{\frac {\ln \left (7\right )}{\ln \left (x^2\right )-\ln \left (7\right )}}} \]
int(-(exp((log(x^2/7)*log(log(x + 4)) + log(x^2/7)*(x + 1) + 4)/log(x^2/7) )*(log(x + 4)*(8*x + 32) - log(x^2/7)^2*(x + log(x + 4)*(4*x + x^2))))/(lo g(x + 4)*log(x^2/7)^2*(4*x + x^2)),x)