Integrand size = 91, antiderivative size = 23 \[ \int \frac {4 x-5 x^2-8 x^3+10 x^4+\left (-x+2 x^3\right ) \log (5)+\left (4 x-10 x^2+e^{x^2} (40-100 x-10 \log (5))-x \log (5)\right ) \log \left (e^{-x^2} \left (10 e^{x^2}+x\right )\right )}{10 e^{x^2}+x} \, dx=x (4-5 x-\log (5)) \log \left (10+e^{-x^2} x\right ) \]
\[ \int \frac {4 x-5 x^2-8 x^3+10 x^4+\left (-x+2 x^3\right ) \log (5)+\left (4 x-10 x^2+e^{x^2} (40-100 x-10 \log (5))-x \log (5)\right ) \log \left (e^{-x^2} \left (10 e^{x^2}+x\right )\right )}{10 e^{x^2}+x} \, dx=\int \frac {4 x-5 x^2-8 x^3+10 x^4+\left (-x+2 x^3\right ) \log (5)+\left (4 x-10 x^2+e^{x^2} (40-100 x-10 \log (5))-x \log (5)\right ) \log \left (e^{-x^2} \left (10 e^{x^2}+x\right )\right )}{10 e^{x^2}+x} \, dx \]
Integrate[(4*x - 5*x^2 - 8*x^3 + 10*x^4 + (-x + 2*x^3)*Log[5] + (4*x - 10* x^2 + E^x^2*(40 - 100*x - 10*Log[5]) - x*Log[5])*Log[(10*E^x^2 + x)/E^x^2] )/(10*E^x^2 + x),x]
Integrate[(4*x - 5*x^2 - 8*x^3 + 10*x^4 + (-x + 2*x^3)*Log[5] + (4*x - 10* x^2 + E^x^2*(40 - 100*x - 10*Log[5]) - x*Log[5])*Log[(10*E^x^2 + x)/E^x^2] )/(10*E^x^2 + 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.
\(\displaystyle \int \frac {10 x^4-8 x^3+\left (2 x^3-x\right ) \log (5)-5 x^2+\left (-10 x^2+e^{x^2} (-100 x+40-10 \log (5))+4 x-x \log (5)\right ) \log \left (e^{-x^2} \left (10 e^{x^2}+x\right )\right )+4 x}{10 e^{x^2}+x} \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {x \left (2 x^2-1\right ) (5 x-4+\log (5))}{10 e^{x^2}+x}-(10 x-4+\log (5)) \log \left (e^{-x^2} x+10\right )\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -5 \int \frac {x^2}{x+10 e^{x^2}}dx+\frac {1}{10} \left (34-\log ^2(5)+8 \log (5)\right ) \int \frac {x^2}{x+10 e^{x^2}}dx+\frac {1}{20} (4-\log (5))^2 \int \frac {1}{x+10 e^{x^2}}dx-\frac {1}{20} \log \left (e^{-x^2} x+10\right ) (-10 x+4-\log (5))^2\) |
Int[(4*x - 5*x^2 - 8*x^3 + 10*x^4 + (-x + 2*x^3)*Log[5] + (4*x - 10*x^2 + E^x^2*(40 - 100*x - 10*Log[5]) - x*Log[5])*Log[(10*E^x^2 + x)/E^x^2])/(10* E^x^2 + x),x]
3.9.38.3.1 Defintions of rubi rules used
Leaf count of result is larger than twice the leaf count of optimal. \(46\) vs. \(2(22)=44\).
Time = 0.53 (sec) , antiderivative size = 47, normalized size of antiderivative = 2.04
method | result | size |
norman | \(\left (-\ln \left (5\right )+4\right ) x \ln \left (\left (10 \,{\mathrm e}^{x^{2}}+x \right ) {\mathrm e}^{-x^{2}}\right )-5 \ln \left (\left (10 \,{\mathrm e}^{x^{2}}+x \right ) {\mathrm e}^{-x^{2}}\right ) x^{2}\) | \(47\) |
parallelrisch | \(-\ln \left (5\right ) \ln \left (\left (10 \,{\mathrm e}^{x^{2}}+x \right ) {\mathrm e}^{-x^{2}}\right ) x -5 \ln \left (\left (10 \,{\mathrm e}^{x^{2}}+x \right ) {\mathrm e}^{-x^{2}}\right ) x^{2}+4 \ln \left (\left (10 \,{\mathrm e}^{x^{2}}+x \right ) {\mathrm e}^{-x^{2}}\right ) x\) | \(63\) |
risch | \(\left (x \ln \left (5\right )+5 x^{2}-4 x \right ) \ln \left ({\mathrm e}^{x^{2}}\right )-\ln \left (5\right ) x \ln \left (10 \,{\mathrm e}^{x^{2}}+x \right )-5 x^{2} \ln \left (10 \,{\mathrm e}^{x^{2}}+x \right )+4 x \ln \left (10 \,{\mathrm e}^{x^{2}}+x \right )+\frac {i \ln \left (5\right ) \pi x \,\operatorname {csgn}\left (i \left (10 \,{\mathrm e}^{x^{2}}+x \right )\right ) \operatorname {csgn}\left (i {\mathrm e}^{-x^{2}} \left (10 \,{\mathrm e}^{x^{2}}+x \right )\right ) \operatorname {csgn}\left (i {\mathrm e}^{-x^{2}}\right )}{2}+\frac {5 i \pi \,x^{2} \operatorname {csgn}\left (i \left (10 \,{\mathrm e}^{x^{2}}+x \right )\right ) \operatorname {csgn}\left (i {\mathrm e}^{-x^{2}} \left (10 \,{\mathrm e}^{x^{2}}+x \right )\right ) \operatorname {csgn}\left (i {\mathrm e}^{-x^{2}}\right )}{2}+2 i \pi x \,\operatorname {csgn}\left (i \left (10 \,{\mathrm e}^{x^{2}}+x \right )\right ) \operatorname {csgn}\left (i {\mathrm e}^{-x^{2}} \left (10 \,{\mathrm e}^{x^{2}}+x \right )\right )^{2}-\frac {i \ln \left (5\right ) \pi x \operatorname {csgn}\left (i {\mathrm e}^{-x^{2}} \left (10 \,{\mathrm e}^{x^{2}}+x \right )\right )^{2} \operatorname {csgn}\left (i {\mathrm e}^{-x^{2}}\right )}{2}-2 i \pi x \,\operatorname {csgn}\left (i \left (10 \,{\mathrm e}^{x^{2}}+x \right )\right ) \operatorname {csgn}\left (i {\mathrm e}^{-x^{2}} \left (10 \,{\mathrm e}^{x^{2}}+x \right )\right ) \operatorname {csgn}\left (i {\mathrm e}^{-x^{2}}\right )+2 i \pi x \operatorname {csgn}\left (i {\mathrm e}^{-x^{2}} \left (10 \,{\mathrm e}^{x^{2}}+x \right )\right )^{2} \operatorname {csgn}\left (i {\mathrm e}^{-x^{2}}\right )+\frac {5 i \pi \,x^{2} \operatorname {csgn}\left (i {\mathrm e}^{-x^{2}} \left (10 \,{\mathrm e}^{x^{2}}+x \right )\right )^{3}}{2}-\frac {i \ln \left (5\right ) \pi x \,\operatorname {csgn}\left (i \left (10 \,{\mathrm e}^{x^{2}}+x \right )\right ) \operatorname {csgn}\left (i {\mathrm e}^{-x^{2}} \left (10 \,{\mathrm e}^{x^{2}}+x \right )\right )^{2}}{2}-\frac {5 i \pi \,x^{2} \operatorname {csgn}\left (i \left (10 \,{\mathrm e}^{x^{2}}+x \right )\right ) \operatorname {csgn}\left (i {\mathrm e}^{-x^{2}} \left (10 \,{\mathrm e}^{x^{2}}+x \right )\right )^{2}}{2}-2 i \pi x \operatorname {csgn}\left (i {\mathrm e}^{-x^{2}} \left (10 \,{\mathrm e}^{x^{2}}+x \right )\right )^{3}+\frac {i \ln \left (5\right ) \pi x \operatorname {csgn}\left (i {\mathrm e}^{-x^{2}} \left (10 \,{\mathrm e}^{x^{2}}+x \right )\right )^{3}}{2}-\frac {5 i \pi \,x^{2} \operatorname {csgn}\left (i {\mathrm e}^{-x^{2}} \left (10 \,{\mathrm e}^{x^{2}}+x \right )\right )^{2} \operatorname {csgn}\left (i {\mathrm e}^{-x^{2}}\right )}{2}\) | \(503\) |
int((((-10*ln(5)-100*x+40)*exp(x^2)-x*ln(5)-10*x^2+4*x)*ln((10*exp(x^2)+x) /exp(x^2))+(2*x^3-x)*ln(5)+10*x^4-8*x^3-5*x^2+4*x)/(10*exp(x^2)+x),x,metho d=_RETURNVERBOSE)
Time = 0.24 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.35 \[ \int \frac {4 x-5 x^2-8 x^3+10 x^4+\left (-x+2 x^3\right ) \log (5)+\left (4 x-10 x^2+e^{x^2} (40-100 x-10 \log (5))-x \log (5)\right ) \log \left (e^{-x^2} \left (10 e^{x^2}+x\right )\right )}{10 e^{x^2}+x} \, dx=-{\left (5 \, x^{2} + x \log \left (5\right ) - 4 \, x\right )} \log \left ({\left (x + 10 \, e^{\left (x^{2}\right )}\right )} e^{\left (-x^{2}\right )}\right ) \]
integrate((((-10*log(5)-100*x+40)*exp(x^2)-x*log(5)-10*x^2+4*x)*log((10*ex p(x^2)+x)/exp(x^2))+(2*x^3-x)*log(5)+10*x^4-8*x^3-5*x^2+4*x)/(10*exp(x^2)+ x),x, algorithm=\
Time = 0.17 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.17 \[ \int \frac {4 x-5 x^2-8 x^3+10 x^4+\left (-x+2 x^3\right ) \log (5)+\left (4 x-10 x^2+e^{x^2} (40-100 x-10 \log (5))-x \log (5)\right ) \log \left (e^{-x^2} \left (10 e^{x^2}+x\right )\right )}{10 e^{x^2}+x} \, dx=\left (- 5 x^{2} - x \log {\left (5 \right )} + 4 x\right ) \log {\left (\left (x + 10 e^{x^{2}}\right ) e^{- x^{2}} \right )} \]
integrate((((-10*ln(5)-100*x+40)*exp(x**2)-x*ln(5)-10*x**2+4*x)*ln((10*exp (x**2)+x)/exp(x**2))+(2*x**3-x)*ln(5)+10*x**4-8*x**3-5*x**2+4*x)/(10*exp(x **2)+x),x)
Time = 0.30 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.61 \[ \int \frac {4 x-5 x^2-8 x^3+10 x^4+\left (-x+2 x^3\right ) \log (5)+\left (4 x-10 x^2+e^{x^2} (40-100 x-10 \log (5))-x \log (5)\right ) \log \left (e^{-x^2} \left (10 e^{x^2}+x\right )\right )}{10 e^{x^2}+x} \, dx=5 \, x^{4} + x^{3} {\left (\log \left (5\right ) - 4\right )} - {\left (5 \, x^{2} + x {\left (\log \left (5\right ) - 4\right )}\right )} \log \left (x + 10 \, e^{\left (x^{2}\right )}\right ) \]
integrate((((-10*log(5)-100*x+40)*exp(x^2)-x*log(5)-10*x^2+4*x)*log((10*ex p(x^2)+x)/exp(x^2))+(2*x^3-x)*log(5)+10*x^4-8*x^3-5*x^2+4*x)/(10*exp(x^2)+ x),x, algorithm=\
Leaf count of result is larger than twice the leaf count of optimal. 62 vs. \(2 (21) = 42\).
Time = 0.29 (sec) , antiderivative size = 62, normalized size of antiderivative = 2.70 \[ \int \frac {4 x-5 x^2-8 x^3+10 x^4+\left (-x+2 x^3\right ) \log (5)+\left (4 x-10 x^2+e^{x^2} (40-100 x-10 \log (5))-x \log (5)\right ) \log \left (e^{-x^2} \left (10 e^{x^2}+x\right )\right )}{10 e^{x^2}+x} \, dx=-5 \, x^{2} \log \left ({\left (x + 10 \, e^{\left (x^{2}\right )}\right )} e^{\left (-x^{2}\right )}\right ) - x \log \left (5\right ) \log \left ({\left (x + 10 \, e^{\left (x^{2}\right )}\right )} e^{\left (-x^{2}\right )}\right ) + 4 \, x \log \left ({\left (x + 10 \, e^{\left (x^{2}\right )}\right )} e^{\left (-x^{2}\right )}\right ) \]
integrate((((-10*log(5)-100*x+40)*exp(x^2)-x*log(5)-10*x^2+4*x)*log((10*ex p(x^2)+x)/exp(x^2))+(2*x^3-x)*log(5)+10*x^4-8*x^3-5*x^2+4*x)/(10*exp(x^2)+ x),x, algorithm=\
-5*x^2*log((x + 10*e^(x^2))*e^(-x^2)) - x*log(5)*log((x + 10*e^(x^2))*e^(- x^2)) + 4*x*log((x + 10*e^(x^2))*e^(-x^2))
Time = 13.00 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.26 \[ \int \frac {4 x-5 x^2-8 x^3+10 x^4+\left (-x+2 x^3\right ) \log (5)+\left (4 x-10 x^2+e^{x^2} (40-100 x-10 \log (5))-x \log (5)\right ) \log \left (e^{-x^2} \left (10 e^{x^2}+x\right )\right )}{10 e^{x^2}+x} \, dx=-\left (5\,x^2+\left (\ln \left (5\right )-4\right )\,x\right )\,\left (\ln \left (x+10\,{\mathrm {e}}^{x^2}\right )-x^2\right ) \]