Integrand size = 98, antiderivative size = 27 \[ \int \frac {10+20 x+20 x^2+10 x^3+(5+10 x) \log \left (x^2\right )+e^{e^x} \left (2+4 x+4 x^2+2 x^3+\left (1+2 x+e^x \left (x+2 x^2+2 x^3+x^4\right )\right ) \log \left (x^2\right )\right )}{3+6 x+9 x^2+6 x^3+3 x^4} \, dx=\frac {\left (5+e^{e^x}\right ) x \log \left (x^2\right )}{3+\frac {3 x^2}{1+x}} \] Output:
(exp(exp(x))+5)*ln(x^2)/(3+3*x^2/(1+x))*x
Time = 0.46 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.26 \[ \int \frac {10+20 x+20 x^2+10 x^3+(5+10 x) \log \left (x^2\right )+e^{e^x} \left (2+4 x+4 x^2+2 x^3+\left (1+2 x+e^x \left (x+2 x^2+2 x^3+x^4\right )\right ) \log \left (x^2\right )\right )}{3+6 x+9 x^2+6 x^3+3 x^4} \, dx=\frac {1}{3} \left (10 \log (x)+\frac {\left (-5+e^{e^x} x (1+x)\right ) \log \left (x^2\right )}{1+x+x^2}\right ) \] Input:
Integrate[(10 + 20*x + 20*x^2 + 10*x^3 + (5 + 10*x)*Log[x^2] + E^E^x*(2 + 4*x + 4*x^2 + 2*x^3 + (1 + 2*x + E^x*(x + 2*x^2 + 2*x^3 + x^4))*Log[x^2])) /(3 + 6*x + 9*x^2 + 6*x^3 + 3*x^4),x]
Output:
(10*Log[x] + ((-5 + E^E^x*x*(1 + x))*Log[x^2])/(1 + x + x^2))/3
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.
| \(\displaystyle \int \frac {10 x^3+20 x^2+(10 x+5) \log \left (x^2\right )+e^{e^x} \left (2 x^3+4 x^2+\left (e^x \left (x^4+2 x^3+2 x^2+x\right )+2 x+1\right ) \log \left (x^2\right )+4 x+2\right )+20 x+10}{3 x^4+6 x^3+9 x^2+6 x+3} \, dx\) |
| \(\Big \downarrow \) 2463 |
| \(\displaystyle \int \frac {10 x^3+20 x^2+(10 x+5) \log \left (x^2\right )+e^{e^x} \left (2 x^3+4 x^2+\left (e^x \left (x^4+2 x^3+2 x^2+x\right )+2 x+1\right ) \log \left (x^2\right )+4 x+2\right )+20 x+10}{3 \left (x^2+x+1\right )^2}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{3} \int \frac {10 x^3+20 x^2+20 x+5 (2 x+1) \log \left (x^2\right )+e^{e^x} \left (2 x^3+4 x^2+4 x+\left (2 x+e^x \left (x^4+2 x^3+2 x^2+x\right )+1\right ) \log \left (x^2\right )+2\right )+10}{\left (x^2+x+1\right )^2}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \frac {1}{3} \int \left (\frac {2 e^{e^x} x^3}{\left (x^2+x+1\right )^2}+\frac {10 x^3}{\left (x^2+x+1\right )^2}+\frac {4 e^{e^x} x^2}{\left (x^2+x+1\right )^2}+\frac {20 x^2}{\left (x^2+x+1\right )^2}+\frac {e^{x+e^x} (x+1) \log \left (x^2\right ) x}{x^2+x+1}+\frac {2 e^{e^x} \log \left (x^2\right ) x}{\left (x^2+x+1\right )^2}+\frac {4 e^{e^x} x}{\left (x^2+x+1\right )^2}+\frac {20 x}{\left (x^2+x+1\right )^2}+\frac {e^{e^x} \log \left (x^2\right )}{\left (x^2+x+1\right )^2}+\frac {5 (2 x+1) \log \left (x^2\right )}{\left (x^2+x+1\right )^2}+\frac {2 e^{e^x}}{\left (x^2+x+1\right )^2}+\frac {10}{\left (x^2+x+1\right )^2}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{3} \left (-\frac {10 (x+2) x^2}{3 \left (x^2+x+1\right )}-\frac {20 \log \left (x^2\right ) x}{3 \left (1-i \sqrt {3}\right ) \left (2 x-i \sqrt {3}+1\right )}+\frac {20 \log \left (x^2\right ) x}{3 \left (2 x-i \sqrt {3}+1\right )}-\frac {20 \log \left (x^2\right ) x}{3 \left (1+i \sqrt {3}\right ) \left (2 x+i \sqrt {3}+1\right )}+\frac {20 \log \left (x^2\right ) x}{3 \left (2 x+i \sqrt {3}+1\right )}-\frac {20 (x+2) x}{3 \left (x^2+x+1\right )}+\frac {10 x}{3}+\frac {10 \arctan \left (\frac {2 x+1}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {20 \log \left (2 i x+\sqrt {3}+i\right )}{3 \left (1-i \sqrt {3}\right )}-\frac {20}{3} \log \left (2 i x+\sqrt {3}+i\right )+e^{e^x} \log \left (x^2\right )+\frac {20 \log \left (2 x+i \sqrt {3}+1\right )}{3 \left (1+i \sqrt {3}\right )}-\frac {20}{3} \log \left (2 x+i \sqrt {3}+1\right )+5 \log \left (x^2+x+1\right )+\frac {4}{3} \left (1-i \sqrt {3}\right ) \log \left (x^2\right ) \int \frac {e^{e^x}}{\left (-2 x+i \sqrt {3}-1\right )^2}dx-\frac {4}{3} \log \left (x^2\right ) \int \frac {e^{e^x}}{\left (-2 x+i \sqrt {3}-1\right )^2}dx+\frac {8 i \int \frac {e^{e^x}}{-2 x+i \sqrt {3}-1}dx}{\sqrt {3}}-\frac {2 i \log \left (x^2\right ) \int \frac {e^{x+e^x}}{-2 x+i \sqrt {3}-1}dx}{\sqrt {3}}-2 \int \frac {e^{e^x}}{x}dx+2 \left (1+i \sqrt {3}\right ) \int \frac {e^{e^x}}{2 x-i \sqrt {3}+1}dx+\frac {4}{3} \left (1+i \sqrt {3}\right ) \log \left (x^2\right ) \int \frac {e^{e^x}}{\left (2 x+i \sqrt {3}+1\right )^2}dx-\frac {4}{3} \log \left (x^2\right ) \int \frac {e^{e^x}}{\left (2 x+i \sqrt {3}+1\right )^2}dx+2 \left (1-i \sqrt {3}\right ) \int \frac {e^{e^x}}{2 x+i \sqrt {3}+1}dx+\frac {8 i \int \frac {e^{e^x}}{2 x+i \sqrt {3}+1}dx}{\sqrt {3}}-\frac {2 i \log \left (x^2\right ) \int \frac {e^{x+e^x}}{2 x+i \sqrt {3}+1}dx}{\sqrt {3}}+\frac {4 i \int \frac {\int \frac {e^{x+e^x}}{-2 x+i \sqrt {3}-1}dx}{x}dx}{\sqrt {3}}-\frac {8}{3} \left (1-i \sqrt {3}\right ) \int \frac {\int -\frac {e^{e^x}}{\left (2 i x+\sqrt {3}+i\right )^2}dx}{x}dx+\frac {8}{3} \int \frac {\int -\frac {e^{e^x}}{\left (2 i x+\sqrt {3}+i\right )^2}dx}{x}dx-\frac {8}{3} \left (1+i \sqrt {3}\right ) \int \frac {\int \frac {e^{e^x}}{\left (2 x+i \sqrt {3}+1\right )^2}dx}{x}dx+\frac {8}{3} \int \frac {\int \frac {e^{e^x}}{\left (2 x+i \sqrt {3}+1\right )^2}dx}{x}dx+\frac {4 i \int \frac {\int \frac {e^{x+e^x}}{2 x+i \sqrt {3}+1}dx}{x}dx}{\sqrt {3}}-\frac {20 (x+2)}{3 \left (x^2+x+1\right )}+\frac {10 (2 x+1)}{3 \left (x^2+x+1\right )}\right )\) |
Input:
Int[(10 + 20*x + 20*x^2 + 10*x^3 + (5 + 10*x)*Log[x^2] + E^E^x*(2 + 4*x + 4*x^2 + 2*x^3 + (1 + 2*x + E^x*(x + 2*x^2 + 2*x^3 + x^4))*Log[x^2]))/(3 + 6*x + 9*x^2 + 6*x^3 + 3*x^4),x]
Output:
$Aborted
Leaf count of result is larger than twice the leaf count of optimal. \(55\) vs. \(2(25)=50\).
Time = 4.44 (sec) , antiderivative size = 56, normalized size of antiderivative = 2.07
| method | result | size |
| parallelrisch | \(-\frac {-\ln \left (x^{2}\right ) {\mathrm e}^{{\mathrm e}^{x}} x^{2}-10 x^{2} \ln \left (x \right )-\ln \left (x^{2}\right ) {\mathrm e}^{{\mathrm e}^{x}} x -10 x \ln \left (x \right )-10 \ln \left (x \right )+5 \ln \left (x^{2}\right )}{3 \left (x^{2}+x +1\right )}\) | \(56\) |
| risch | \(-\frac {10 \ln \left (x \right )}{3 \left (x^{2}+x +1\right )}+\frac {\frac {5 i \pi \,\operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i x \right )^{2}}{6}-\frac {5 i \pi \operatorname {csgn}\left (i x^{2}\right )^{2} \operatorname {csgn}\left (i x \right )}{3}+\frac {5 i \pi \operatorname {csgn}\left (i x^{2}\right )^{3}}{6}+\frac {10 x^{2} \ln \left (x \right )}{3}+\frac {10 x \ln \left (x \right )}{3}+\frac {10 \ln \left (x \right )}{3}}{x^{2}+x +1}+\frac {x \left (-i \pi x \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )+2 i \pi x \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2}-i \pi x \operatorname {csgn}\left (i x^{2}\right )^{3}-i \pi \,\operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i x \right )^{2}+2 i \pi \operatorname {csgn}\left (i x^{2}\right )^{2} \operatorname {csgn}\left (i x \right )-i \pi \operatorname {csgn}\left (i x^{2}\right )^{3}+4 x \ln \left (x \right )+4 \ln \left (x \right )\right ) {\mathrm e}^{{\mathrm e}^{x}}}{6 x^{2}+6 x +6}\) | \(215\) |
Input:
int(((((x^4+2*x^3+2*x^2+x)*exp(x)+2*x+1)*ln(x^2)+2*x^3+4*x^2+4*x+2)*exp(ex p(x))+(10*x+5)*ln(x^2)+10*x^3+20*x^2+20*x+10)/(3*x^4+6*x^3+9*x^2+6*x+3),x, method=_RETURNVERBOSE)
Output:
-1/3*(-ln(x^2)*exp(exp(x))*x^2-10*x^2*ln(x)-ln(x^2)*exp(exp(x))*x-10*x*ln( x)-10*ln(x)+5*ln(x^2))/(x^2+x+1)
Time = 0.11 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.30 \[ \int \frac {10+20 x+20 x^2+10 x^3+(5+10 x) \log \left (x^2\right )+e^{e^x} \left (2+4 x+4 x^2+2 x^3+\left (1+2 x+e^x \left (x+2 x^2+2 x^3+x^4\right )\right ) \log \left (x^2\right )\right )}{3+6 x+9 x^2+6 x^3+3 x^4} \, dx=\frac {{\left (x^{2} + x\right )} e^{\left (e^{x}\right )} \log \left (x^{2}\right ) + 5 \, {\left (x^{2} + x\right )} \log \left (x^{2}\right )}{3 \, {\left (x^{2} + x + 1\right )}} \] Input:
integrate(((((x^4+2*x^3+2*x^2+x)*exp(x)+2*x+1)*log(x^2)+2*x^3+4*x^2+4*x+2) *exp(exp(x))+(10*x+5)*log(x^2)+10*x^3+20*x^2+20*x+10)/(3*x^4+6*x^3+9*x^2+6 *x+3),x, algorithm="fricas")
Output:
1/3*((x^2 + x)*e^(e^x)*log(x^2) + 5*(x^2 + x)*log(x^2))/(x^2 + x + 1)
Leaf count of result is larger than twice the leaf count of optimal. 53 vs. \(2 (22) = 44\).
Time = 0.18 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.96 \[ \int \frac {10+20 x+20 x^2+10 x^3+(5+10 x) \log \left (x^2\right )+e^{e^x} \left (2+4 x+4 x^2+2 x^3+\left (1+2 x+e^x \left (x+2 x^2+2 x^3+x^4\right )\right ) \log \left (x^2\right )\right )}{3+6 x+9 x^2+6 x^3+3 x^4} \, dx=\frac {\left (x^{2} \log {\left (x^{2} \right )} + x \log {\left (x^{2} \right )}\right ) e^{e^{x}}}{3 x^{2} + 3 x + 3} + \frac {10 \log {\left (x \right )}}{3} - \frac {5 \log {\left (x^{2} \right )}}{3 x^{2} + 3 x + 3} \] Input:
integrate(((((x**4+2*x**3+2*x**2+x)*exp(x)+2*x+1)*ln(x**2)+2*x**3+4*x**2+4 *x+2)*exp(exp(x))+(10*x+5)*ln(x**2)+10*x**3+20*x**2+20*x+10)/(3*x**4+6*x** 3+9*x**2+6*x+3),x)
Output:
(x**2*log(x**2) + x*log(x**2))*exp(exp(x))/(3*x**2 + 3*x + 3) + 10*log(x)/ 3 - 5*log(x**2)/(3*x**2 + 3*x + 3)
Leaf count of result is larger than twice the leaf count of optimal. 73 vs. \(2 (25) = 50\).
Time = 0.15 (sec) , antiderivative size = 73, normalized size of antiderivative = 2.70 \[ \int \frac {10+20 x+20 x^2+10 x^3+(5+10 x) \log \left (x^2\right )+e^{e^x} \left (2+4 x+4 x^2+2 x^3+\left (1+2 x+e^x \left (x+2 x^2+2 x^3+x^4\right )\right ) \log \left (x^2\right )\right )}{3+6 x+9 x^2+6 x^3+3 x^4} \, dx=\frac {2 \, {\left ({\left (x^{2} + x\right )} e^{\left (e^{x}\right )} \log \left (x\right ) + 5 \, {\left (x^{2} + x\right )} \log \left (x\right )\right )}}{3 \, {\left (x^{2} + x + 1\right )}} + \frac {20 \, {\left (2 \, x + 1\right )}}{9 \, {\left (x^{2} + x + 1\right )}} - \frac {20 \, {\left (x + 2\right )}}{9 \, {\left (x^{2} + x + 1\right )}} - \frac {20 \, {\left (x - 1\right )}}{9 \, {\left (x^{2} + x + 1\right )}} \] Input:
integrate(((((x^4+2*x^3+2*x^2+x)*exp(x)+2*x+1)*log(x^2)+2*x^3+4*x^2+4*x+2) *exp(exp(x))+(10*x+5)*log(x^2)+10*x^3+20*x^2+20*x+10)/(3*x^4+6*x^3+9*x^2+6 *x+3),x, algorithm="maxima")
Output:
2/3*((x^2 + x)*e^(e^x)*log(x) + 5*(x^2 + x)*log(x))/(x^2 + x + 1) + 20/9*( 2*x + 1)/(x^2 + x + 1) - 20/9*(x + 2)/(x^2 + x + 1) - 20/9*(x - 1)/(x^2 + x + 1)
\[ \int \frac {10+20 x+20 x^2+10 x^3+(5+10 x) \log \left (x^2\right )+e^{e^x} \left (2+4 x+4 x^2+2 x^3+\left (1+2 x+e^x \left (x+2 x^2+2 x^3+x^4\right )\right ) \log \left (x^2\right )\right )}{3+6 x+9 x^2+6 x^3+3 x^4} \, dx=\int { \frac {10 \, x^{3} + 20 \, x^{2} + {\left (2 \, x^{3} + 4 \, x^{2} + {\left ({\left (x^{4} + 2 \, x^{3} + 2 \, x^{2} + x\right )} e^{x} + 2 \, x + 1\right )} \log \left (x^{2}\right ) + 4 \, x + 2\right )} e^{\left (e^{x}\right )} + 5 \, {\left (2 \, x + 1\right )} \log \left (x^{2}\right ) + 20 \, x + 10}{3 \, {\left (x^{4} + 2 \, x^{3} + 3 \, x^{2} + 2 \, x + 1\right )}} \,d x } \] Input:
integrate(((((x^4+2*x^3+2*x^2+x)*exp(x)+2*x+1)*log(x^2)+2*x^3+4*x^2+4*x+2) *exp(exp(x))+(10*x+5)*log(x^2)+10*x^3+20*x^2+20*x+10)/(3*x^4+6*x^3+9*x^2+6 *x+3),x, algorithm="giac")
Output:
integrate(1/3*(10*x^3 + 20*x^2 + (2*x^3 + 4*x^2 + ((x^4 + 2*x^3 + 2*x^2 + x)*e^x + 2*x + 1)*log(x^2) + 4*x + 2)*e^(e^x) + 5*(2*x + 1)*log(x^2) + 20* x + 10)/(x^4 + 2*x^3 + 3*x^2 + 2*x + 1), x)
Timed out. \[ \int \frac {10+20 x+20 x^2+10 x^3+(5+10 x) \log \left (x^2\right )+e^{e^x} \left (2+4 x+4 x^2+2 x^3+\left (1+2 x+e^x \left (x+2 x^2+2 x^3+x^4\right )\right ) \log \left (x^2\right )\right )}{3+6 x+9 x^2+6 x^3+3 x^4} \, dx=\int \frac {20\,x+{\mathrm {e}}^{{\mathrm {e}}^x}\,\left (4\,x+\ln \left (x^2\right )\,\left (2\,x+{\mathrm {e}}^x\,\left (x^4+2\,x^3+2\,x^2+x\right )+1\right )+4\,x^2+2\,x^3+2\right )+20\,x^2+10\,x^3+\ln \left (x^2\right )\,\left (10\,x+5\right )+10}{3\,x^4+6\,x^3+9\,x^2+6\,x+3} \,d x \] Input:
int((20*x + exp(exp(x))*(4*x + log(x^2)*(2*x + exp(x)*(x + 2*x^2 + 2*x^3 + x^4) + 1) + 4*x^2 + 2*x^3 + 2) + 20*x^2 + 10*x^3 + log(x^2)*(10*x + 5) + 10)/(6*x + 9*x^2 + 6*x^3 + 3*x^4 + 3),x)
Output:
int((20*x + exp(exp(x))*(4*x + log(x^2)*(2*x + exp(x)*(x + 2*x^2 + 2*x^3 + x^4) + 1) + 4*x^2 + 2*x^3 + 2) + 20*x^2 + 10*x^3 + log(x^2)*(10*x + 5) + 10)/(6*x + 9*x^2 + 6*x^3 + 3*x^4 + 3), x)
Time = 0.19 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.30 \[ \int \frac {10+20 x+20 x^2+10 x^3+(5+10 x) \log \left (x^2\right )+e^{e^x} \left (2+4 x+4 x^2+2 x^3+\left (1+2 x+e^x \left (x+2 x^2+2 x^3+x^4\right )\right ) \log \left (x^2\right )\right )}{3+6 x+9 x^2+6 x^3+3 x^4} \, dx=\frac {\mathrm {log}\left (x^{2}\right ) x \left (e^{e^{x}} x +e^{e^{x}}+5 x +5\right )}{3 x^{2}+3 x +3} \] Input:
int(((((x^4+2*x^3+2*x^2+x)*exp(x)+2*x+1)*log(x^2)+2*x^3+4*x^2+4*x+2)*exp(e xp(x))+(10*x+5)*log(x^2)+10*x^3+20*x^2+20*x+10)/(3*x^4+6*x^3+9*x^2+6*x+3), x)
Output:
(log(x**2)*x*(e**(e**x)*x + e**(e**x) + 5*x + 5))/(3*(x**2 + x + 1))