Integrand size = 132, antiderivative size = 25 \[ \int \frac {-72+e^{e^x} \left (e^{2+x} \left (468+444 x+144 x^2+16 x^3\right )+e^x \left (468+444 x+144 x^2+16 x^3\right ) \log \left (\frac {39+24 x+4 x^2}{9+6 x+x^2}\right )\right )}{e^2 \left (117+111 x+36 x^2+4 x^3\right )+\left (117+111 x+36 x^2+4 x^3\right ) \log \left (\frac {39+24 x+4 x^2}{9+6 x+x^2}\right )} \, dx=4 \left (e^{e^x}+3 \log \left (e^2+\log \left (4+\frac {3}{(3+x)^2}\right )\right )\right ) \]
Time = 0.12 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.28 \[ \int \frac {-72+e^{e^x} \left (e^{2+x} \left (468+444 x+144 x^2+16 x^3\right )+e^x \left (468+444 x+144 x^2+16 x^3\right ) \log \left (\frac {39+24 x+4 x^2}{9+6 x+x^2}\right )\right )}{e^2 \left (117+111 x+36 x^2+4 x^3\right )+\left (117+111 x+36 x^2+4 x^3\right ) \log \left (\frac {39+24 x+4 x^2}{9+6 x+x^2}\right )} \, dx=4 \left (e^{e^x}+3 \log \left (e^2+\log \left (\frac {39+24 x+4 x^2}{(3+x)^2}\right )\right )\right ) \]
Integrate[(-72 + E^E^x*(E^(2 + x)*(468 + 444*x + 144*x^2 + 16*x^3) + E^x*( 468 + 444*x + 144*x^2 + 16*x^3)*Log[(39 + 24*x + 4*x^2)/(9 + 6*x + x^2)])) /(E^2*(117 + 111*x + 36*x^2 + 4*x^3) + (117 + 111*x + 36*x^2 + 4*x^3)*Log[ (39 + 24*x + 4*x^2)/(9 + 6*x + 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.
\(\displaystyle \int \frac {e^{e^x} \left (e^{x+2} \left (16 x^3+144 x^2+444 x+468\right )+e^x \left (16 x^3+144 x^2+444 x+468\right ) \log \left (\frac {4 x^2+24 x+39}{x^2+6 x+9}\right )\right )-72}{e^2 \left (4 x^3+36 x^2+111 x+117\right )+\left (4 x^3+36 x^2+111 x+117\right ) \log \left (\frac {4 x^2+24 x+39}{x^2+6 x+9}\right )} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {e^{e^x} \left (e^{x+2} \left (16 x^3+144 x^2+444 x+468\right )+e^x \left (16 x^3+144 x^2+444 x+468\right ) \log \left (\frac {4 x^2+24 x+39}{x^2+6 x+9}\right )\right )-72}{\left (4 x^3+36 x^2+111 x+117\right ) \left (\log \left (\frac {4 x^2+24 x+39}{(x+3)^2}\right )+e^2\right )}dx\) |
\(\Big \downarrow \) 2463 |
\(\displaystyle \int \left (\frac {e^{e^x} \left (e^{x+2} \left (16 x^3+144 x^2+444 x+468\right )+e^x \left (16 x^3+144 x^2+444 x+468\right ) \log \left (\frac {4 x^2+24 x+39}{x^2+6 x+9}\right )\right )-72}{3 (x+3) \left (\log \left (\frac {4 x^2+24 x+39}{(x+3)^2}\right )+e^2\right )}-\frac {4 (x+3) \left (e^{e^x} \left (e^{x+2} \left (16 x^3+144 x^2+444 x+468\right )+e^x \left (16 x^3+144 x^2+444 x+468\right ) \log \left (\frac {4 x^2+24 x+39}{x^2+6 x+9}\right )\right )-72\right )}{3 \left (4 x^2+24 x+39\right ) \left (\log \left (\frac {4 x^2+24 x+39}{(x+3)^2}\right )+e^2\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 192 i \sqrt {3} \int \frac {1}{\left (-8 x+4 i \sqrt {3}-24\right ) \left (\log \left (\frac {4 x^2+24 x+39}{(x+3)^2}\right )+e^2\right )}dx-24 \int \frac {1}{(x+3) \left (\log \left (\frac {4 x^2+24 x+39}{(x+3)^2}\right )+e^2\right )}dx+96 \left (1+2 i \sqrt {3}\right ) \int \frac {1}{\left (8 x-4 i \sqrt {3}+24\right ) \left (\log \left (\frac {4 x^2+24 x+39}{(x+3)^2}\right )+e^2\right )}dx+96 \left (1-2 i \sqrt {3}\right ) \int \frac {1}{\left (8 x+4 i \sqrt {3}+24\right ) \left (\log \left (\frac {4 x^2+24 x+39}{(x+3)^2}\right )+e^2\right )}dx+192 i \sqrt {3} \int \frac {1}{\left (8 x+4 i \sqrt {3}+24\right ) \left (\log \left (\frac {4 x^2+24 x+39}{(x+3)^2}\right )+e^2\right )}dx+4 e^{e^x}\) |
Int[(-72 + E^E^x*(E^(2 + x)*(468 + 444*x + 144*x^2 + 16*x^3) + E^x*(468 + 444*x + 144*x^2 + 16*x^3)*Log[(39 + 24*x + 4*x^2)/(9 + 6*x + x^2)]))/(E^2* (117 + 111*x + 36*x^2 + 4*x^3) + (117 + 111*x + 36*x^2 + 4*x^3)*Log[(39 + 24*x + 4*x^2)/(9 + 6*x + x^2)]),x]
3.10.20.3.1 Defintions of rubi rules used
Int[(u_.)*(Px_)^(p_), x_Symbol] :> With[{Qx = Factor[Px]}, Int[ExpandIntegr and[u, Qx^p, x], x] /; !SumQ[NonfreeFactors[Qx, x]]] /; PolyQ[Px, x] && Gt Q[Expon[Px, x], 2] && !BinomialQ[Px, x] && !TrinomialQ[Px, x] && ILtQ[p, 0]
Time = 36.17 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.40
method | result | size |
parallelrisch | \(12 \ln \left (\ln \left (\frac {4 x^{2}+24 x +39}{x^{2}+6 x +9}\right )+{\mathrm e}^{2}\right )+4 \,{\mathrm e}^{{\mathrm e}^{x}}\) | \(35\) |
risch | \(12 \ln \left (\ln \left (x^{2}+6 x +\frac {39}{4}\right )-\frac {i \left (-\pi \,\operatorname {csgn}\left (i \left (x^{2}+6 x +\frac {39}{4}\right )\right ) {\operatorname {csgn}\left (\frac {i \left (x^{2}+6 x +\frac {39}{4}\right )}{\left (3+x \right )^{2}}\right )}^{2}+\pi \,\operatorname {csgn}\left (i \left (x^{2}+6 x +\frac {39}{4}\right )\right ) \operatorname {csgn}\left (\frac {i \left (x^{2}+6 x +\frac {39}{4}\right )}{\left (3+x \right )^{2}}\right ) \operatorname {csgn}\left (\frac {i}{\left (3+x \right )^{2}}\right )+\pi {\operatorname {csgn}\left (\frac {i \left (x^{2}+6 x +\frac {39}{4}\right )}{\left (3+x \right )^{2}}\right )}^{3}-\pi {\operatorname {csgn}\left (\frac {i \left (x^{2}+6 x +\frac {39}{4}\right )}{\left (3+x \right )^{2}}\right )}^{2} \operatorname {csgn}\left (\frac {i}{\left (3+x \right )^{2}}\right )-\pi \operatorname {csgn}\left (i \left (3+x \right )\right )^{2} \operatorname {csgn}\left (i \left (3+x \right )^{2}\right )+2 \pi \,\operatorname {csgn}\left (i \left (3+x \right )\right ) \operatorname {csgn}\left (i \left (3+x \right )^{2}\right )^{2}-\pi \operatorname {csgn}\left (i \left (3+x \right )^{2}\right )^{3}+2 i {\mathrm e}^{2}+4 i \ln \left (2\right )-4 i \ln \left (3+x \right )\right )}{2}\right )+4 \,{\mathrm e}^{{\mathrm e}^{x}}\) | \(223\) |
int((((16*x^3+144*x^2+444*x+468)*exp(x)*ln((4*x^2+24*x+39)/(x^2+6*x+9))+(1 6*x^3+144*x^2+444*x+468)*exp(2)*exp(x))*exp(exp(x))-72)/((4*x^3+36*x^2+111 *x+117)*ln((4*x^2+24*x+39)/(x^2+6*x+9))+(4*x^3+36*x^2+111*x+117)*exp(2)),x ,method=_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.36 \[ \int \frac {-72+e^{e^x} \left (e^{2+x} \left (468+444 x+144 x^2+16 x^3\right )+e^x \left (468+444 x+144 x^2+16 x^3\right ) \log \left (\frac {39+24 x+4 x^2}{9+6 x+x^2}\right )\right )}{e^2 \left (117+111 x+36 x^2+4 x^3\right )+\left (117+111 x+36 x^2+4 x^3\right ) \log \left (\frac {39+24 x+4 x^2}{9+6 x+x^2}\right )} \, dx=4 \, e^{\left (e^{x}\right )} + 12 \, \log \left (e^{2} + \log \left (\frac {4 \, x^{2} + 24 \, x + 39}{x^{2} + 6 \, x + 9}\right )\right ) \]
integrate((((16*x^3+144*x^2+444*x+468)*exp(x)*log((4*x^2+24*x+39)/(x^2+6*x +9))+(16*x^3+144*x^2+444*x+468)*exp(2)*exp(x))*exp(exp(x))-72)/((4*x^3+36* x^2+111*x+117)*log((4*x^2+24*x+39)/(x^2+6*x+9))+(4*x^3+36*x^2+111*x+117)*e xp(2)),x, algorithm=\
Time = 0.34 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.28 \[ \int \frac {-72+e^{e^x} \left (e^{2+x} \left (468+444 x+144 x^2+16 x^3\right )+e^x \left (468+444 x+144 x^2+16 x^3\right ) \log \left (\frac {39+24 x+4 x^2}{9+6 x+x^2}\right )\right )}{e^2 \left (117+111 x+36 x^2+4 x^3\right )+\left (117+111 x+36 x^2+4 x^3\right ) \log \left (\frac {39+24 x+4 x^2}{9+6 x+x^2}\right )} \, dx=4 e^{e^{x}} + 12 \log {\left (\log {\left (\frac {4 x^{2} + 24 x + 39}{x^{2} + 6 x + 9} \right )} + e^{2} \right )} \]
integrate((((16*x**3+144*x**2+444*x+468)*exp(x)*ln((4*x**2+24*x+39)/(x**2+ 6*x+9))+(16*x**3+144*x**2+444*x+468)*exp(2)*exp(x))*exp(exp(x))-72)/((4*x* *3+36*x**2+111*x+117)*ln((4*x**2+24*x+39)/(x**2+6*x+9))+(4*x**3+36*x**2+11 1*x+117)*exp(2)),x)
Time = 0.35 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.16 \[ \int \frac {-72+e^{e^x} \left (e^{2+x} \left (468+444 x+144 x^2+16 x^3\right )+e^x \left (468+444 x+144 x^2+16 x^3\right ) \log \left (\frac {39+24 x+4 x^2}{9+6 x+x^2}\right )\right )}{e^2 \left (117+111 x+36 x^2+4 x^3\right )+\left (117+111 x+36 x^2+4 x^3\right ) \log \left (\frac {39+24 x+4 x^2}{9+6 x+x^2}\right )} \, dx=4 \, e^{\left (e^{x}\right )} + 12 \, \log \left (e^{2} + \log \left (4 \, x^{2} + 24 \, x + 39\right ) - 2 \, \log \left (x + 3\right )\right ) \]
integrate((((16*x^3+144*x^2+444*x+468)*exp(x)*log((4*x^2+24*x+39)/(x^2+6*x +9))+(16*x^3+144*x^2+444*x+468)*exp(2)*exp(x))*exp(exp(x))-72)/((4*x^3+36* x^2+111*x+117)*log((4*x^2+24*x+39)/(x^2+6*x+9))+(4*x^3+36*x^2+111*x+117)*e xp(2)),x, algorithm=\
Time = 0.47 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.68 \[ \int \frac {-72+e^{e^x} \left (e^{2+x} \left (468+444 x+144 x^2+16 x^3\right )+e^x \left (468+444 x+144 x^2+16 x^3\right ) \log \left (\frac {39+24 x+4 x^2}{9+6 x+x^2}\right )\right )}{e^2 \left (117+111 x+36 x^2+4 x^3\right )+\left (117+111 x+36 x^2+4 x^3\right ) \log \left (\frac {39+24 x+4 x^2}{9+6 x+x^2}\right )} \, dx=4 \, {\left (3 \, e^{x} \log \left (e^{2} + \log \left (\frac {4 \, x^{2} + 24 \, x + 39}{x^{2} + 6 \, x + 9}\right )\right ) + e^{\left (x + e^{x}\right )}\right )} e^{\left (-x\right )} \]
integrate((((16*x^3+144*x^2+444*x+468)*exp(x)*log((4*x^2+24*x+39)/(x^2+6*x +9))+(16*x^3+144*x^2+444*x+468)*exp(2)*exp(x))*exp(exp(x))-72)/((4*x^3+36* x^2+111*x+117)*log((4*x^2+24*x+39)/(x^2+6*x+9))+(4*x^3+36*x^2+111*x+117)*e xp(2)),x, algorithm=\
Time = 12.19 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.36 \[ \int \frac {-72+e^{e^x} \left (e^{2+x} \left (468+444 x+144 x^2+16 x^3\right )+e^x \left (468+444 x+144 x^2+16 x^3\right ) \log \left (\frac {39+24 x+4 x^2}{9+6 x+x^2}\right )\right )}{e^2 \left (117+111 x+36 x^2+4 x^3\right )+\left (117+111 x+36 x^2+4 x^3\right ) \log \left (\frac {39+24 x+4 x^2}{9+6 x+x^2}\right )} \, dx=4\,{\mathrm {e}}^{{\mathrm {e}}^x}+12\,\ln \left ({\mathrm {e}}^2+\ln \left (\frac {4\,x^2+24\,x+39}{x^2+6\,x+9}\right )\right ) \]