Integrand size = 56, antiderivative size = 23 \[ \int \frac {\left (-20+100 x-180 x^4+100 x^5\right ) \log \left (x^2\right )+\left (40+40 x^4\right ) \log \left (\frac {e^{5 x}}{x+2 x^5+x^9}\right )}{x+x^5} \, dx=20 \log \left (x^2\right ) \log \left (\frac {e^{5 x}}{x \left (1+x^4\right )^2}\right ) \] Output:
20*ln(exp(5*x)/(x^4+1)^2/x)*ln(x^2)
Leaf count is larger than twice the leaf count of optimal. \(63\) vs. \(2(23)=46\).
Time = 0.18 (sec) , antiderivative size = 63, normalized size of antiderivative = 2.74 \[ \int \frac {\left (-20+100 x-180 x^4+100 x^5\right ) \log \left (x^2\right )+\left (40+40 x^4\right ) \log \left (\frac {e^{5 x}}{x+2 x^5+x^9}\right )}{x+x^5} \, dx=20 \left (\log \left (x^2\right ) \left (5 x+\log \left (\frac {1}{x \left (1+x^4\right )^2}\right )\right )-2 \log (x) \left (5 x+\log \left (\frac {1}{x \left (1+x^4\right )^2}\right )-\log \left (\frac {e^{5 x}}{x \left (1+x^4\right )^2}\right )\right )\right ) \] Input:
Integrate[((-20 + 100*x - 180*x^4 + 100*x^5)*Log[x^2] + (40 + 40*x^4)*Log[ E^(5*x)/(x + 2*x^5 + x^9)])/(x + x^5),x]
Output:
20*(Log[x^2]*(5*x + Log[1/(x*(1 + x^4)^2)]) - 2*Log[x]*(5*x + Log[1/(x*(1 + x^4)^2)] - Log[E^(5*x)/(x*(1 + x^4)^2)]))
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.
\(\displaystyle \int \frac {\left (40 x^4+40\right ) \log \left (\frac {e^{5 x}}{x^9+2 x^5+x}\right )+\left (100 x^5-180 x^4+100 x-20\right ) \log \left (x^2\right )}{x^5+x} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {\left (40 x^4+40\right ) \log \left (\frac {e^{5 x}}{x^9+2 x^5+x}\right )+\left (100 x^5-180 x^4+100 x-20\right ) \log \left (x^2\right )}{x \left (x^4+1\right )}dx\) |
\(\Big \downarrow \) 7276 |
\(\displaystyle \int \left (\frac {40 \log \left (\frac {e^{5 x}}{x \left (x^4+1\right )^2}\right )}{x}+\frac {20 (x-1) \left (5 x^4-4 x^3-4 x^2-4 x+1\right ) \log \left (x^2\right )}{x \left (x^4+1\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 40 \int \frac {\log \left (\frac {e^{5 x}}{x \left (x^4+1\right )^2}\right )}{x}dx-20 \operatorname {PolyLog}\left (2,-x^4\right )-5 \log ^2\left (x^2\right )+100 x \log \left (x^2\right )-40 \log \left (x^4+1\right ) \log \left (x^2\right )-200 x\) |
Input:
Int[((-20 + 100*x - 180*x^4 + 100*x^5)*Log[x^2] + (40 + 40*x^4)*Log[E^(5*x )/(x + 2*x^5 + x^9)])/(x + x^5),x]
Output:
$Aborted
Time = 24.26 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.22
method | result | size |
parallelrisch | \(20 \ln \left (x^{2}\right ) \ln \left (\frac {{\mathrm e}^{5 x}}{x \left (x^{8}+2 x^{4}+1\right )}\right )\) | \(28\) |
default | \(40 \ln \left (\frac {{\mathrm e}^{5 x}}{x^{9}+2 x^{5}+x}\right ) \ln \left (x \right )+40 \ln \left (x \right )^{2}-200 x \ln \left (x \right )+80 \ln \left (x \right ) \ln \left (\frac {-x +\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}}{\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}}\right )+80 \operatorname {dilog}\left (\frac {-x +\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}}{\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}}\right )+80 \ln \left (x \right ) \ln \left (\frac {-x -\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}}{-\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}}\right )+80 \operatorname {dilog}\left (\frac {-x -\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}}{-\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}}\right )+80 \ln \left (x \right ) \ln \left (\frac {-x -\frac {\sqrt {2}}{2}-\frac {i \sqrt {2}}{2}}{-\frac {\sqrt {2}}{2}-\frac {i \sqrt {2}}{2}}\right )+80 \operatorname {dilog}\left (\frac {-x -\frac {\sqrt {2}}{2}-\frac {i \sqrt {2}}{2}}{-\frac {\sqrt {2}}{2}-\frac {i \sqrt {2}}{2}}\right )+80 \ln \left (x \right ) \ln \left (\frac {-x +\frac {\sqrt {2}}{2}-\frac {i \sqrt {2}}{2}}{\frac {\sqrt {2}}{2}-\frac {i \sqrt {2}}{2}}\right )+80 \operatorname {dilog}\left (\frac {-x +\frac {\sqrt {2}}{2}-\frac {i \sqrt {2}}{2}}{\frac {\sqrt {2}}{2}-\frac {i \sqrt {2}}{2}}\right )+100 x \ln \left (x^{2}\right )-20 \ln \left (x \right ) \ln \left (x^{2}\right )+20 \left (\munderset {\underline {\hspace {1.25 ex}}\alpha =\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )}{\sum }\left (-2 \ln \left (x -\underline {\hspace {1.25 ex}}\alpha \right ) \ln \left (x^{2}\right )+4 \operatorname {dilog}\left (\frac {x}{\underline {\hspace {1.25 ex}}\alpha }\right )+4 \ln \left (x -\underline {\hspace {1.25 ex}}\alpha \right ) \ln \left (\frac {x}{\underline {\hspace {1.25 ex}}\alpha }\right )\right )\right )\) | \(368\) |
parts | \(40 \ln \left (\frac {{\mathrm e}^{5 x}}{x^{9}+2 x^{5}+x}\right ) \ln \left (x \right )+40 \ln \left (x \right )^{2}-200 x \ln \left (x \right )+80 \ln \left (x \right ) \ln \left (\frac {-x +\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}}{\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}}\right )+80 \operatorname {dilog}\left (\frac {-x +\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}}{\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}}\right )+80 \ln \left (x \right ) \ln \left (\frac {-x -\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}}{-\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}}\right )+80 \operatorname {dilog}\left (\frac {-x -\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}}{-\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}}\right )+80 \ln \left (x \right ) \ln \left (\frac {-x -\frac {\sqrt {2}}{2}-\frac {i \sqrt {2}}{2}}{-\frac {\sqrt {2}}{2}-\frac {i \sqrt {2}}{2}}\right )+80 \operatorname {dilog}\left (\frac {-x -\frac {\sqrt {2}}{2}-\frac {i \sqrt {2}}{2}}{-\frac {\sqrt {2}}{2}-\frac {i \sqrt {2}}{2}}\right )+80 \ln \left (x \right ) \ln \left (\frac {-x +\frac {\sqrt {2}}{2}-\frac {i \sqrt {2}}{2}}{\frac {\sqrt {2}}{2}-\frac {i \sqrt {2}}{2}}\right )+80 \operatorname {dilog}\left (\frac {-x +\frac {\sqrt {2}}{2}-\frac {i \sqrt {2}}{2}}{\frac {\sqrt {2}}{2}-\frac {i \sqrt {2}}{2}}\right )+100 x \ln \left (x^{2}\right )-20 \ln \left (x \right ) \ln \left (x^{2}\right )+20 \left (\munderset {\underline {\hspace {1.25 ex}}\alpha =\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )}{\sum }\left (-2 \ln \left (x -\underline {\hspace {1.25 ex}}\alpha \right ) \ln \left (x^{2}\right )+4 \operatorname {dilog}\left (\frac {x}{\underline {\hspace {1.25 ex}}\alpha }\right )+4 \ln \left (x -\underline {\hspace {1.25 ex}}\alpha \right ) \ln \left (\frac {x}{\underline {\hspace {1.25 ex}}\alpha }\right )\right )\right )\) | \(368\) |
risch | \(\text {Expression too large to display}\) | \(2623\) |
Input:
int(((40*x^4+40)*ln(exp(5*x)/(x^9+2*x^5+x))+(100*x^5-180*x^4+100*x-20)*ln( x^2))/(x^5+x),x,method=_RETURNVERBOSE)
Output:
20*ln(x^2)*ln(exp(5*x)/x/(x^8+2*x^4+1))
Time = 0.10 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int \frac {\left (-20+100 x-180 x^4+100 x^5\right ) \log \left (x^2\right )+\left (40+40 x^4\right ) \log \left (\frac {e^{5 x}}{x+2 x^5+x^9}\right )}{x+x^5} \, dx=20 \, \log \left (x^{2}\right ) \log \left (\frac {e^{\left (5 \, x\right )}}{x^{9} + 2 \, x^{5} + x}\right ) \] Input:
integrate(((40*x^4+40)*log(exp(5*x)/(x^9+2*x^5+x))+(100*x^5-180*x^4+100*x- 20)*log(x^2))/(x^5+x),x, algorithm="fricas")
Output:
20*log(x^2)*log(e^(5*x)/(x^9 + 2*x^5 + x))
Time = 0.24 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {\left (-20+100 x-180 x^4+100 x^5\right ) \log \left (x^2\right )+\left (40+40 x^4\right ) \log \left (\frac {e^{5 x}}{x+2 x^5+x^9}\right )}{x+x^5} \, dx=20 \log {\left (x^{2} \right )} \log {\left (\frac {e^{5 x}}{x^{9} + 2 x^{5} + x} \right )} \] Input:
integrate(((40*x**4+40)*ln(exp(5*x)/(x**9+2*x**5+x))+(100*x**5-180*x**4+10 0*x-20)*ln(x**2))/(x**5+x),x)
Output:
20*log(x**2)*log(exp(5*x)/(x**9 + 2*x**5 + x))
Time = 0.15 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {\left (-20+100 x-180 x^4+100 x^5\right ) \log \left (x^2\right )+\left (40+40 x^4\right ) \log \left (\frac {e^{5 x}}{x+2 x^5+x^9}\right )}{x+x^5} \, dx=200 \, x \log \left (x\right ) - 80 \, \log \left (x^{4} + 1\right ) \log \left (x\right ) - 40 \, \log \left (x\right )^{2} \] Input:
integrate(((40*x^4+40)*log(exp(5*x)/(x^9+2*x^5+x))+(100*x^5-180*x^4+100*x- 20)*log(x^2))/(x^5+x),x, algorithm="maxima")
Output:
200*x*log(x) - 80*log(x^4 + 1)*log(x) - 40*log(x)^2
Time = 0.13 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.17 \[ \int \frac {\left (-20+100 x-180 x^4+100 x^5\right ) \log \left (x^2\right )+\left (40+40 x^4\right ) \log \left (\frac {e^{5 x}}{x+2 x^5+x^9}\right )}{x+x^5} \, dx=200 \, x \log \left (x\right ) - 40 \, \log \left (x^{8} + 2 \, x^{4} + 1\right ) \log \left (x\right ) - 40 \, \log \left (x\right )^{2} \] Input:
integrate(((40*x^4+40)*log(exp(5*x)/(x^9+2*x^5+x))+(100*x^5-180*x^4+100*x- 20)*log(x^2))/(x^5+x),x, algorithm="giac")
Output:
200*x*log(x) - 40*log(x^8 + 2*x^4 + 1)*log(x) - 40*log(x)^2
Time = 2.89 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {\left (-20+100 x-180 x^4+100 x^5\right ) \log \left (x^2\right )+\left (40+40 x^4\right ) \log \left (\frac {e^{5 x}}{x+2 x^5+x^9}\right )}{x+x^5} \, dx=20\,\ln \left (x^2\right )\,\left (5\,x+\ln \left (\frac {1}{x^9+2\,x^5+x}\right )\right ) \] Input:
int((log(exp(5*x)/(x + 2*x^5 + x^9))*(40*x^4 + 40) + log(x^2)*(100*x - 180 *x^4 + 100*x^5 - 20))/(x + x^5),x)
Output:
20*log(x^2)*(5*x + log(1/(x + 2*x^5 + x^9)))
Time = 0.20 (sec) , antiderivative size = 81, normalized size of antiderivative = 3.52 \[ \int \frac {\left (-20+100 x-180 x^4+100 x^5\right ) \log \left (x^2\right )+\left (40+40 x^4\right ) \log \left (\frac {e^{5 x}}{x+2 x^5+x^9}\right )}{x+x^5} \, dx=20 \,\mathrm {log}\left (x^{2}\right ) \mathrm {log}\left (\frac {e^{5 x}}{x^{9}+2 x^{5}+x}\right )+20 \,\mathrm {log}\left (x^{2}\right )+80 \,\mathrm {log}\left (-\sqrt {2}\, x +x^{2}+1\right )+80 \,\mathrm {log}\left (\sqrt {2}\, x +x^{2}+1\right )+40 \,\mathrm {log}\left (\frac {e^{5 x}}{x^{9}+2 x^{5}+x}\right )-200 x \] Input:
int(((40*x^4+40)*log(exp(5*x)/(x^9+2*x^5+x))+(100*x^5-180*x^4+100*x-20)*lo g(x^2))/(x^5+x),x)
Output:
20*(log(x**2)*log(e**(5*x)/(x**9 + 2*x**5 + x)) + log(x**2) + 4*log( - sqr t(2)*x + x**2 + 1) + 4*log(sqrt(2)*x + x**2 + 1) + 2*log(e**(5*x)/(x**9 + 2*x**5 + x)) - 10*x)