Integrand size = 139, antiderivative size = 33 \begin {dmath*} \int \frac {-e^8-20 e^4 x-75 x^2+e^5 \left (-e^8-10 e^4 x-25 x^2\right )+e^{x^2} \left (1+e^5+2 x^2\right )+\left (e^8-e^{x^2}+10 e^4 x+25 x^2\right ) \log \left (e^8 x-e^{x^2} x+10 e^4 x^2+25 x^3\right )}{-e^8 x^2+e^{x^2} x^2-10 e^4 x^3-25 x^4} \, dx=3+\frac {-e^5+x+\log \left (x \left (-e^{x^2}+\left (e^4+5 x\right )^2\right )\right )}{x} \end {dmath*}
Time = 0.09 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.15 \begin {dmath*} \int \frac {-e^8-20 e^4 x-75 x^2+e^5 \left (-e^8-10 e^4 x-25 x^2\right )+e^{x^2} \left (1+e^5+2 x^2\right )+\left (e^8-e^{x^2}+10 e^4 x+25 x^2\right ) \log \left (e^8 x-e^{x^2} x+10 e^4 x^2+25 x^3\right )}{-e^8 x^2+e^{x^2} x^2-10 e^4 x^3-25 x^4} \, dx=-\frac {e^5}{x}+\frac {\log \left (x \left (e^8-e^{x^2}+10 e^4 x+25 x^2\right )\right )}{x} \end {dmath*}
Integrate[(-E^8 - 20*E^4*x - 75*x^2 + E^5*(-E^8 - 10*E^4*x - 25*x^2) + E^x ^2*(1 + E^5 + 2*x^2) + (E^8 - E^x^2 + 10*E^4*x + 25*x^2)*Log[E^8*x - E^x^2 *x + 10*E^4*x^2 + 25*x^3])/(-(E^8*x^2) + E^x^2*x^2 - 10*E^4*x^3 - 25*x^4), x]
Time = 1.80 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.39, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.014, Rules used = {7293, 2009}
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.
\(\displaystyle \int \frac {-75 x^2+e^5 \left (-25 x^2-10 e^4 x-e^8\right )+e^{x^2} \left (2 x^2+e^5+1\right )+\left (25 x^2-e^{x^2}+10 e^4 x+e^8\right ) \log \left (25 x^3+10 e^4 x^2-e^{x^2} x+e^8 x\right )-20 e^4 x-e^8}{-25 x^4-10 e^4 x^3+e^{x^2} x^2-e^8 x^2} \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {2 x^2-\log \left (x \left (25 x^2-e^{x^2}+10 e^4 x+e^8\right )\right )+e^5+1}{x^2}+\frac {2 \left (-25 x^3-10 e^4 x^2+\left (25-e^8\right ) x+5 e^4\right )}{x \left (25 x^2-e^{x^2}+10 e^4 x+e^8\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\log \left (25 x^3+10 e^4 x^2-e^{x^2} x+e^8 x\right )}{x}-\frac {1+e^5}{x}+\frac {1}{x}\) |
Int[(-E^8 - 20*E^4*x - 75*x^2 + E^5*(-E^8 - 10*E^4*x - 25*x^2) + E^x^2*(1 + E^5 + 2*x^2) + (E^8 - E^x^2 + 10*E^4*x + 25*x^2)*Log[E^8*x - E^x^2*x + 1 0*E^4*x^2 + 25*x^3])/(-(E^8*x^2) + E^x^2*x^2 - 10*E^4*x^3 - 25*x^4),x]
3.3.70.3.1 Defintions of rubi rules used
Time = 1.18 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.06
method | result | size |
parallelrisch | \(-\frac {{\mathrm e}^{5}-\ln \left (x \left (-{\mathrm e}^{x^{2}}+{\mathrm e}^{8}+10 x \,{\mathrm e}^{4}+25 x^{2}\right )\right )}{x}\) | \(35\) |
norman | \(\frac {-{\mathrm e}^{5}+\ln \left (-{\mathrm e}^{x^{2}} x +x \,{\mathrm e}^{8}+10 x^{2} {\mathrm e}^{4}+25 x^{3}\right )}{x}\) | \(37\) |
risch | \(\frac {\ln \left (-{\mathrm e}^{x^{2}}+{\mathrm e}^{8}+10 x \,{\mathrm e}^{4}+25 x^{2}\right )}{x}-\frac {-i \pi \,\operatorname {csgn}\left (i \left (-{\mathrm e}^{x^{2}}+{\mathrm e}^{8}+10 x \,{\mathrm e}^{4}+25 x^{2}\right )\right ) {\operatorname {csgn}\left (i x \left (-{\mathrm e}^{x^{2}}+{\mathrm e}^{8}+10 x \,{\mathrm e}^{4}+25 x^{2}\right )\right )}^{2}+i \pi \,\operatorname {csgn}\left (i \left (-{\mathrm e}^{x^{2}}+{\mathrm e}^{8}+10 x \,{\mathrm e}^{4}+25 x^{2}\right )\right ) \operatorname {csgn}\left (i x \left (-{\mathrm e}^{x^{2}}+{\mathrm e}^{8}+10 x \,{\mathrm e}^{4}+25 x^{2}\right )\right ) \operatorname {csgn}\left (i x \right )+i \pi {\operatorname {csgn}\left (i x \left (-{\mathrm e}^{x^{2}}+{\mathrm e}^{8}+10 x \,{\mathrm e}^{4}+25 x^{2}\right )\right )}^{3}-i \pi {\operatorname {csgn}\left (i x \left (-{\mathrm e}^{x^{2}}+{\mathrm e}^{8}+10 x \,{\mathrm e}^{4}+25 x^{2}\right )\right )}^{2} \operatorname {csgn}\left (i x \right )+2 \,{\mathrm e}^{5}-2 \ln \left (x \right )}{2 x}\) | \(214\) |
int(((-exp(x^2)+exp(4)^2+10*x*exp(4)+25*x^2)*ln(-exp(x^2)*x+x*exp(4)^2+10* x^2*exp(4)+25*x^3)+(exp(5)+2*x^2+1)*exp(x^2)+(-exp(4)^2-10*x*exp(4)-25*x^2 )*exp(5)-exp(4)^2-20*x*exp(4)-75*x^2)/(x^2*exp(x^2)-x^2*exp(4)^2-10*x^3*ex p(4)-25*x^4),x,method=_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.06 \begin {dmath*} \int \frac {-e^8-20 e^4 x-75 x^2+e^5 \left (-e^8-10 e^4 x-25 x^2\right )+e^{x^2} \left (1+e^5+2 x^2\right )+\left (e^8-e^{x^2}+10 e^4 x+25 x^2\right ) \log \left (e^8 x-e^{x^2} x+10 e^4 x^2+25 x^3\right )}{-e^8 x^2+e^{x^2} x^2-10 e^4 x^3-25 x^4} \, dx=-\frac {e^{5} - \log \left (25 \, x^{3} + 10 \, x^{2} e^{4} + x e^{8} - x e^{\left (x^{2}\right )}\right )}{x} \end {dmath*}
integrate(((-exp(x^2)+exp(4)^2+10*x*exp(4)+25*x^2)*log(-exp(x^2)*x+x*exp(4 )^2+10*x^2*exp(4)+25*x^3)+(exp(5)+2*x^2+1)*exp(x^2)+(-exp(4)^2-10*x*exp(4) -25*x^2)*exp(5)-exp(4)^2-20*x*exp(4)-75*x^2)/(x^2*exp(x^2)-x^2*exp(4)^2-10 *x^3*exp(4)-25*x^4),x, algorithm=\
Time = 0.19 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.97 \begin {dmath*} \int \frac {-e^8-20 e^4 x-75 x^2+e^5 \left (-e^8-10 e^4 x-25 x^2\right )+e^{x^2} \left (1+e^5+2 x^2\right )+\left (e^8-e^{x^2}+10 e^4 x+25 x^2\right ) \log \left (e^8 x-e^{x^2} x+10 e^4 x^2+25 x^3\right )}{-e^8 x^2+e^{x^2} x^2-10 e^4 x^3-25 x^4} \, dx=\frac {\log {\left (25 x^{3} + 10 x^{2} e^{4} - x e^{x^{2}} + x e^{8} \right )}}{x} - \frac {e^{5}}{x} \end {dmath*}
integrate(((-exp(x**2)+exp(4)**2+10*x*exp(4)+25*x**2)*ln(-exp(x**2)*x+x*ex p(4)**2+10*x**2*exp(4)+25*x**3)+(exp(5)+2*x**2+1)*exp(x**2)+(-exp(4)**2-10 *x*exp(4)-25*x**2)*exp(5)-exp(4)**2-20*x*exp(4)-75*x**2)/(x**2*exp(x**2)-x **2*exp(4)**2-10*x**3*exp(4)-25*x**4),x)
Time = 0.26 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.03 \begin {dmath*} \int \frac {-e^8-20 e^4 x-75 x^2+e^5 \left (-e^8-10 e^4 x-25 x^2\right )+e^{x^2} \left (1+e^5+2 x^2\right )+\left (e^8-e^{x^2}+10 e^4 x+25 x^2\right ) \log \left (e^8 x-e^{x^2} x+10 e^4 x^2+25 x^3\right )}{-e^8 x^2+e^{x^2} x^2-10 e^4 x^3-25 x^4} \, dx=-\frac {e^{5} - \log \left (25 \, x^{2} + 10 \, x e^{4} + e^{8} - e^{\left (x^{2}\right )}\right ) - \log \left (x\right )}{x} \end {dmath*}
integrate(((-exp(x^2)+exp(4)^2+10*x*exp(4)+25*x^2)*log(-exp(x^2)*x+x*exp(4 )^2+10*x^2*exp(4)+25*x^3)+(exp(5)+2*x^2+1)*exp(x^2)+(-exp(4)^2-10*x*exp(4) -25*x^2)*exp(5)-exp(4)^2-20*x*exp(4)-75*x^2)/(x^2*exp(x^2)-x^2*exp(4)^2-10 *x^3*exp(4)-25*x^4),x, algorithm=\
Time = 0.33 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.06 \begin {dmath*} \int \frac {-e^8-20 e^4 x-75 x^2+e^5 \left (-e^8-10 e^4 x-25 x^2\right )+e^{x^2} \left (1+e^5+2 x^2\right )+\left (e^8-e^{x^2}+10 e^4 x+25 x^2\right ) \log \left (e^8 x-e^{x^2} x+10 e^4 x^2+25 x^3\right )}{-e^8 x^2+e^{x^2} x^2-10 e^4 x^3-25 x^4} \, dx=-\frac {e^{5} - \log \left (25 \, x^{3} + 10 \, x^{2} e^{4} + x e^{8} - x e^{\left (x^{2}\right )}\right )}{x} \end {dmath*}
integrate(((-exp(x^2)+exp(4)^2+10*x*exp(4)+25*x^2)*log(-exp(x^2)*x+x*exp(4 )^2+10*x^2*exp(4)+25*x^3)+(exp(5)+2*x^2+1)*exp(x^2)+(-exp(4)^2-10*x*exp(4) -25*x^2)*exp(5)-exp(4)^2-20*x*exp(4)-75*x^2)/(x^2*exp(x^2)-x^2*exp(4)^2-10 *x^3*exp(4)-25*x^4),x, algorithm=\
Time = 13.22 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.06 \begin {dmath*} \int \frac {-e^8-20 e^4 x-75 x^2+e^5 \left (-e^8-10 e^4 x-25 x^2\right )+e^{x^2} \left (1+e^5+2 x^2\right )+\left (e^8-e^{x^2}+10 e^4 x+25 x^2\right ) \log \left (e^8 x-e^{x^2} x+10 e^4 x^2+25 x^3\right )}{-e^8 x^2+e^{x^2} x^2-10 e^4 x^3-25 x^4} \, dx=-\frac {{\mathrm {e}}^5-\ln \left (x\,{\mathrm {e}}^8-x\,{\mathrm {e}}^{x^2}+10\,x^2\,{\mathrm {e}}^4+25\,x^3\right )}{x} \end {dmath*}