Integrand size = 101, antiderivative size = 24 \[ \int \frac {120+360 x^3+90 x^4+e^{x^{x/2}} \left (30+120 x^3+x^{x/2} \left (15-15 x-15 x^4+\left (15-15 x-15 x^4\right ) \log (x)\right )\right )}{18+2 e^{2 x^{x/2}}+12 x+2 x^2+e^{x^{x/2}} (12+4 x)} \, dx=5+\frac {15 \left (-1+x+x^4\right )}{3+e^{x^{x/2}}+x} \]
Time = 0.10 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int \frac {120+360 x^3+90 x^4+e^{x^{x/2}} \left (30+120 x^3+x^{x/2} \left (15-15 x-15 x^4+\left (15-15 x-15 x^4\right ) \log (x)\right )\right )}{18+2 e^{2 x^{x/2}}+12 x+2 x^2+e^{x^{x/2}} (12+4 x)} \, dx=\frac {15 \left (-1+x+x^4\right )}{3+e^{x^{x/2}}+x} \]
Integrate[(120 + 360*x^3 + 90*x^4 + E^x^(x/2)*(30 + 120*x^3 + x^(x/2)*(15 - 15*x - 15*x^4 + (15 - 15*x - 15*x^4)*Log[x])))/(18 + 2*E^(2*x^(x/2)) + 1 2*x + 2*x^2 + E^x^(x/2)*(12 + 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.
\(\displaystyle \int \frac {90 x^4+360 x^3+e^{x^{x/2}} \left (120 x^3+x^{x/2} \left (-15 x^4+\left (-15 x^4-15 x+15\right ) \log (x)-15 x+15\right )+30\right )+120}{2 e^{2 x^{x/2}}+e^{x^{x/2}} (4 x+12)+2 x^2+12 x+18} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {90 x^4+360 x^3+e^{x^{x/2}} \left (120 x^3+x^{x/2} \left (-15 x^4+\left (-15 x^4-15 x+15\right ) \log (x)-15 x+15\right )+30\right )+120}{2 \left (e^{x^{x/2}}+x+3\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{2} \int \frac {15 \left (6 x^4+24 x^3+e^{x^{x/2}} \left (\left (-x^4-x+\left (-x^4-x+1\right ) \log (x)+1\right ) x^{x/2}+8 x^3+2\right )+8\right )}{\left (x+e^{x^{x/2}}+3\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {15}{2} \int \frac {6 x^4+24 x^3+e^{x^{x/2}} \left (\left (-x^4-x+\left (-x^4-x+1\right ) \log (x)+1\right ) x^{x/2}+8 x^3+2\right )+8}{\left (x+e^{x^{x/2}}+3\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {15}{2} \int \left (-\frac {e^{x^{x/2}} \left (x^4+x-1\right ) (\log (x)+1) x^{x/2}}{\left (x+e^{x^{x/2}}+3\right )^2}+\frac {6 x^4}{\left (x+e^{x^{x/2}}+3\right )^2}+\frac {8 e^{x^{x/2}} x^3}{\left (x+e^{x^{x/2}}+3\right )^2}+\frac {24 x^3}{\left (x+e^{x^{x/2}}+3\right )^2}+\frac {2 e^{x^{x/2}}}{\left (x+e^{x^{x/2}}+3\right )^2}+\frac {8}{\left (x+e^{x^{x/2}}+3\right )^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {15}{2} \left (16 \text {Subst}\left (\int \frac {1}{\left (2 x+e^{2^x x^x}+3\right )^2}dx,x,\frac {x}{2}\right )+4 \text {Subst}\left (\int \frac {e^{2^x x^x}}{\left (2 x+e^{2^x x^x}+3\right )^2}dx,x,\frac {x}{2}\right )+2 \text {Subst}\left (\int \frac {2^x e^{2^x x^x} x^x}{\left (2 x+e^{2^x x^x}+3\right )^2}dx,x,\frac {x}{2}\right )-2 \text {Subst}\left (\int \frac {\int \frac {2^x e^{2^x x^x} x^x}{\left (2 x+e^{2^x x^x}+3\right )^2}dx}{x}dx,x,\frac {x}{2}\right )+2 \log (x) \text {Subst}\left (\int \frac {2^x e^{2^x x^x} x^x}{\left (2 x+e^{2^x x^x}+3\right )^2}dx,x,\frac {x}{2}\right )-2 \text {Subst}\left (\int \frac {2^{x+1} e^{2^x x^x} x^{x+1}}{\left (2 x+e^{2^x x^x}+3\right )^2}dx,x,\frac {x}{2}\right )-2 \text {Subst}\left (\int \frac {2^{x+4} e^{2^x x^x} x^{x+4}}{\left (2 x+e^{2^x x^x}+3\right )^2}dx,x,\frac {x}{2}\right )+2 \text {Subst}\left (\int \frac {\int \frac {2^{x+1} e^{2^x x^x} x^{x+1}}{\left (2 x+e^{2^x x^x}+3\right )^2}dx}{x}dx,x,\frac {x}{2}\right )+2 \text {Subst}\left (\int \frac {\int \frac {2^{x+4} e^{2^x x^x} x^{x+4}}{\left (2 x+e^{2^x x^x}+3\right )^2}dx}{x}dx,x,\frac {x}{2}\right )-2 \log (x) \text {Subst}\left (\int \frac {2^{x+1} e^{2^x x^x} x^{x+1}}{\left (2 x+e^{2^x x^x}+3\right )^2}dx,x,\frac {x}{2}\right )-2 \log (x) \text {Subst}\left (\int \frac {2^{x+4} e^{2^x x^x} x^{x+4}}{\left (2 x+e^{2^x x^x}+3\right )^2}dx,x,\frac {x}{2}\right )+6 \int \frac {x^4}{\left (x+e^{x^{x/2}}+3\right )^2}dx+24 \int \frac {x^3}{\left (x+e^{x^{x/2}}+3\right )^2}dx+8 \int \frac {e^{x^{x/2}} x^3}{\left (x+e^{x^{x/2}}+3\right )^2}dx\right )\) |
Int[(120 + 360*x^3 + 90*x^4 + E^x^(x/2)*(30 + 120*x^3 + x^(x/2)*(15 - 15*x - 15*x^4 + (15 - 15*x - 15*x^4)*Log[x])))/(18 + 2*E^(2*x^(x/2)) + 12*x + 2*x^2 + E^x^(x/2)*(12 + 4*x)),x]
3.26.92.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Time = 3.10 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83
method | result | size |
risch | \(\frac {15 x^{4}+15 x -15}{x +3+{\mathrm e}^{x^{\frac {x}{2}}}}\) | \(20\) |
parallelrisch | \(-\frac {-30 x^{4}-30 x +30}{2 \left (x +3+{\mathrm e}^{{\mathrm e}^{\frac {x \ln \left (x \right )}{2}}}\right )}\) | \(25\) |
int(((((-15*x^4-15*x+15)*ln(x)-15*x^4-15*x+15)*exp(1/2*x*ln(x))+120*x^3+30 )*exp(exp(1/2*x*ln(x)))+90*x^4+360*x^3+120)/(2*exp(exp(1/2*x*ln(x)))^2+(4* x+12)*exp(exp(1/2*x*ln(x)))+2*x^2+12*x+18),x,method=_RETURNVERBOSE)
Time = 0.26 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.79 \[ \int \frac {120+360 x^3+90 x^4+e^{x^{x/2}} \left (30+120 x^3+x^{x/2} \left (15-15 x-15 x^4+\left (15-15 x-15 x^4\right ) \log (x)\right )\right )}{18+2 e^{2 x^{x/2}}+12 x+2 x^2+e^{x^{x/2}} (12+4 x)} \, dx=\frac {15 \, {\left (x^{4} + x - 1\right )}}{x + e^{\left (x^{\frac {1}{2} \, x}\right )} + 3} \]
integrate(((((-15*x^4-15*x+15)*log(x)-15*x^4-15*x+15)*exp(1/2*x*log(x))+12 0*x^3+30)*exp(exp(1/2*x*log(x)))+90*x^4+360*x^3+120)/(2*exp(exp(1/2*x*log( x)))^2+(4*x+12)*exp(exp(1/2*x*log(x)))+2*x^2+12*x+18),x, algorithm=\
Time = 0.26 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int \frac {120+360 x^3+90 x^4+e^{x^{x/2}} \left (30+120 x^3+x^{x/2} \left (15-15 x-15 x^4+\left (15-15 x-15 x^4\right ) \log (x)\right )\right )}{18+2 e^{2 x^{x/2}}+12 x+2 x^2+e^{x^{x/2}} (12+4 x)} \, dx=\frac {15 x^{4} + 15 x - 15}{x + e^{e^{\frac {x \log {\left (x \right )}}{2}}} + 3} \]
integrate(((((-15*x**4-15*x+15)*ln(x)-15*x**4-15*x+15)*exp(1/2*x*ln(x))+12 0*x**3+30)*exp(exp(1/2*x*ln(x)))+90*x**4+360*x**3+120)/(2*exp(exp(1/2*x*ln (x)))**2+(4*x+12)*exp(exp(1/2*x*ln(x)))+2*x**2+12*x+18),x)
Time = 0.25 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.79 \[ \int \frac {120+360 x^3+90 x^4+e^{x^{x/2}} \left (30+120 x^3+x^{x/2} \left (15-15 x-15 x^4+\left (15-15 x-15 x^4\right ) \log (x)\right )\right )}{18+2 e^{2 x^{x/2}}+12 x+2 x^2+e^{x^{x/2}} (12+4 x)} \, dx=\frac {15 \, {\left (x^{4} + x - 1\right )}}{x + e^{\left (x^{\frac {1}{2} \, x}\right )} + 3} \]
integrate(((((-15*x^4-15*x+15)*log(x)-15*x^4-15*x+15)*exp(1/2*x*log(x))+12 0*x^3+30)*exp(exp(1/2*x*log(x)))+90*x^4+360*x^3+120)/(2*exp(exp(1/2*x*log( x)))^2+(4*x+12)*exp(exp(1/2*x*log(x)))+2*x^2+12*x+18),x, algorithm=\
Timed out. \[ \int \frac {120+360 x^3+90 x^4+e^{x^{x/2}} \left (30+120 x^3+x^{x/2} \left (15-15 x-15 x^4+\left (15-15 x-15 x^4\right ) \log (x)\right )\right )}{18+2 e^{2 x^{x/2}}+12 x+2 x^2+e^{x^{x/2}} (12+4 x)} \, dx=\text {Timed out} \]
integrate(((((-15*x^4-15*x+15)*log(x)-15*x^4-15*x+15)*exp(1/2*x*log(x))+12 0*x^3+30)*exp(exp(1/2*x*log(x)))+90*x^4+360*x^3+120)/(2*exp(exp(1/2*x*log( x)))^2+(4*x+12)*exp(exp(1/2*x*log(x)))+2*x^2+12*x+18),x, algorithm=\
Time = 8.42 (sec) , antiderivative size = 170, normalized size of antiderivative = 7.08 \[ \int \frac {120+360 x^3+90 x^4+e^{x^{x/2}} \left (30+120 x^3+x^{x/2} \left (15-15 x-15 x^4+\left (15-15 x-15 x^4\right ) \log (x)\right )\right )}{18+2 e^{2 x^{x/2}}+12 x+2 x^2+e^{x^{x/2}} (12+4 x)} \, dx=\frac {15\,\left (x^2\,{\mathrm {e}}^{\frac {x\,\ln \left (x\right )}{2}}-3\,{\mathrm {e}}^{\frac {x\,\ln \left (x\right )}{2}}-2\,x+3\,x^4\,{\mathrm {e}}^{\frac {x\,\ln \left (x\right )}{2}}+x^5\,{\mathrm {e}}^{\frac {x\,\ln \left (x\right )}{2}}-3\,{\mathrm {e}}^{\frac {x\,\ln \left (x\right )}{2}}\,\ln \left (x\right )-2\,x^4+2\,x\,{\mathrm {e}}^{\frac {x\,\ln \left (x\right )}{2}}+x^2\,{\mathrm {e}}^{\frac {x\,\ln \left (x\right )}{2}}\,\ln \left (x\right )+3\,x^4\,{\mathrm {e}}^{\frac {x\,\ln \left (x\right )}{2}}\,\ln \left (x\right )+x^5\,{\mathrm {e}}^{\frac {x\,\ln \left (x\right )}{2}}\,\ln \left (x\right )+2\,x\,{\mathrm {e}}^{\frac {x\,\ln \left (x\right )}{2}}\,\ln \left (x\right )+2\right )}{\left (x+{\mathrm {e}}^{{\mathrm {e}}^{\frac {x\,\ln \left (x\right )}{2}}}+3\right )\,\left (3\,{\mathrm {e}}^{\frac {x\,\ln \left (x\right )}{2}}+3\,{\mathrm {e}}^{\frac {x\,\ln \left (x\right )}{2}}\,\ln \left (x\right )+x\,{\mathrm {e}}^{\frac {x\,\ln \left (x\right )}{2}}+x\,{\mathrm {e}}^{\frac {x\,\ln \left (x\right )}{2}}\,\ln \left (x\right )-2\right )} \]
int((exp(exp((x*log(x))/2))*(120*x^3 - exp((x*log(x))/2)*(15*x + log(x)*(1 5*x + 15*x^4 - 15) + 15*x^4 - 15) + 30) + 360*x^3 + 90*x^4 + 120)/(12*x + 2*exp(2*exp((x*log(x))/2)) + exp(exp((x*log(x))/2))*(4*x + 12) + 2*x^2 + 1 8),x)
(15*(x^2*exp((x*log(x))/2) - 3*exp((x*log(x))/2) - 2*x + 3*x^4*exp((x*log( x))/2) + x^5*exp((x*log(x))/2) - 3*exp((x*log(x))/2)*log(x) - 2*x^4 + 2*x* exp((x*log(x))/2) + x^2*exp((x*log(x))/2)*log(x) + 3*x^4*exp((x*log(x))/2) *log(x) + x^5*exp((x*log(x))/2)*log(x) + 2*x*exp((x*log(x))/2)*log(x) + 2) )/((x + exp(exp((x*log(x))/2)) + 3)*(3*exp((x*log(x))/2) + 3*exp((x*log(x) )/2)*log(x) + x*exp((x*log(x))/2) + x*exp((x*log(x))/2)*log(x) - 2))