Integrand size = 166, antiderivative size = 35 \[ \int \frac {\left (-7 x+\frac {1}{9} e^{5+2 x} x+e^5 \left (24 x-10 x^2+x^3\right )+\frac {1}{3} e^x \left (x+x^2+e^5 \left (11 x-2 x^2\right )\right )\right ) \log ^2(x)+e^{\frac {4}{\log (x)}} \left (-\frac {4 e^{5+x}}{3}+e^5 (-20+4 x)+\left (e^5 x-\frac {1}{3} e^{5+x} x\right ) \log ^2(x)\right )}{\left (\frac {1}{9} e^{5+2 x} x+\frac {1}{3} e^{5+x} \left (10 x-2 x^2\right )+e^5 \left (25 x-10 x^2+x^3\right )\right ) \log ^2(x)} \, dx=x+\frac {-1+e^{\frac {4}{\log (x)}}-\frac {2+x}{e^5}}{5+\frac {e^x}{3}-x} \]
Time = 0.21 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.23 \[ \int \frac {\left (-7 x+\frac {1}{9} e^{5+2 x} x+e^5 \left (24 x-10 x^2+x^3\right )+\frac {1}{3} e^x \left (x+x^2+e^5 \left (11 x-2 x^2\right )\right )\right ) \log ^2(x)+e^{\frac {4}{\log (x)}} \left (-\frac {4 e^{5+x}}{3}+e^5 (-20+4 x)+\left (e^5 x-\frac {1}{3} e^{5+x} x\right ) \log ^2(x)\right )}{\left (\frac {1}{9} e^{5+2 x} x+\frac {1}{3} e^{5+x} \left (10 x-2 x^2\right )+e^5 \left (25 x-10 x^2+x^3\right )\right ) \log ^2(x)} \, dx=\frac {3 e^{\frac {4}{\log (x)}}}{15+e^x-3 x}+x-\frac {3 \left (2+e^5+x\right )}{e^5 \left (15+e^x-3 x\right )} \]
Integrate[((-7*x + (E^(5 + 2*x)*x)/9 + E^5*(24*x - 10*x^2 + x^3) + (E^x*(x + x^2 + E^5*(11*x - 2*x^2)))/3)*Log[x]^2 + E^(4/Log[x])*((-4*E^(5 + x))/3 + E^5*(-20 + 4*x) + (E^5*x - (E^(5 + x)*x)/3)*Log[x]^2))/(((E^(5 + 2*x)*x )/9 + (E^(5 + x)*(10*x - 2*x^2))/3 + E^5*(25*x - 10*x^2 + x^3))*Log[x]^2), 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.
| \(\displaystyle \int \frac {\left (\frac {1}{3} e^x \left (x^2+e^5 \left (11 x-2 x^2\right )+x\right )+e^5 \left (x^3-10 x^2+24 x\right )+\frac {1}{9} e^{2 x+5} x-7 x\right ) \log ^2(x)+e^{\frac {4}{\log (x)}} \left (-\frac {4 e^{x+5}}{3}+e^5 (4 x-20)+\left (e^5 x-\frac {1}{3} e^{x+5} x\right ) \log ^2(x)\right )}{\left (\frac {1}{3} e^{x+5} \left (10 x-2 x^2\right )+e^5 \left (x^3-10 x^2+25 x\right )+\frac {1}{9} e^{2 x+5} x\right ) \log ^2(x)} \, dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle \int \frac {9 \left (x^2-10 x+24\right )+e^x (33-6 x)+e^{2 x}+3 e^{x-5} (x+1)-\frac {12 \left (-3 x+e^x+15\right ) e^{\frac {4}{\log (x)}}}{x \log ^2(x)}-3 e^{x+\frac {4}{\log (x)}}+9 e^{\frac {4}{\log (x)}}-\frac {63}{e^5}}{\left (-3 x+e^x+15\right )^2}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (\frac {9 (x-6) (x-4)}{\left (-3 x+e^x+15\right )^2}+\frac {3 e^{x-5} (x+1)}{\left (-3 x+e^x+15\right )^2}-\frac {3 e^x (2 x-11)}{\left (-3 x+e^x+15\right )^2}+\frac {e^{2 x}}{\left (-3 x+e^x+15\right )^2}-\frac {63}{e^5 \left (-3 x+e^x+15\right )^2}-\frac {3 e^{\frac {4}{\log (x)}} \left (4 e^x-12 x+e^x x \log ^2(x)-3 x \log ^2(x)+60\right )}{\left (-3 x+e^x+15\right )^2 x \log ^2(x)}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle 9 \int \frac {x^2}{\left (-3 x+e^x+15\right )^2}dx-\frac {63 \int \frac {1}{\left (-3 x+e^x+15\right )^2}dx}{e^5}+216 \int \frac {1}{\left (-3 x+e^x+15\right )^2}dx+3 \int \frac {e^{x-5}}{\left (-3 x+e^x+15\right )^2}dx+33 \int \frac {e^x}{\left (-3 x+e^x+15\right )^2}dx+\int \frac {e^{2 x}}{\left (-3 x+e^x+15\right )^2}dx-90 \int \frac {x}{\left (-3 x+e^x+15\right )^2}dx+3 \int \frac {e^{x-5} x}{\left (-3 x+e^x+15\right )^2}dx-6 \int \frac {e^x x}{\left (-3 x+e^x+15\right )^2}dx+\frac {3 e^{\frac {4}{\log (x)}}}{-3 x+e^x+15}\) |
Int[((-7*x + (E^(5 + 2*x)*x)/9 + E^5*(24*x - 10*x^2 + x^3) + (E^x*(x + x^2 + E^5*(11*x - 2*x^2)))/3)*Log[x]^2 + E^(4/Log[x])*((-4*E^(5 + x))/3 + E^5 *(-20 + 4*x) + (E^5*x - (E^(5 + x)*x)/3)*Log[x]^2))/(((E^(5 + 2*x)*x)/9 + (E^(5 + x)*(10*x - 2*x^2))/3 + E^5*(25*x - 10*x^2 + x^3))*Log[x]^2),x]
3.4.15.3.1 Defintions of rubi rules used
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 10.57 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.29
| method | result | size |
| risch | \(-\frac {3 x^{2}-15 x -{\mathrm e}^{x} x +3+3 \,{\mathrm e}^{-5} x -3 \,{\mathrm e}^{\frac {4}{\ln \left (x \right )}}+6 \,{\mathrm e}^{-5}}{-3 x +{\mathrm e}^{x}+15}\) | \(45\) |
| parallelrisch | \(\frac {\left (x^{2} {\mathrm e}^{5}-x \,{\mathrm e}^{5} {\mathrm e}^{-\ln \left (3\right )+x}-5 x \,{\mathrm e}^{5}-{\mathrm e}^{5} {\mathrm e}^{\frac {4}{\ln \left (x \right )}}+2+{\mathrm e}^{5}+x \right ) {\mathrm e}^{-5}}{x -{\mathrm e}^{-\ln \left (3\right )+x}-5}\) | \(59\) |
int((((-x*exp(5)*exp(-ln(3)+x)+x*exp(5))*ln(x)^2-4*exp(5)*exp(-ln(3)+x)+(4 *x-20)*exp(5))*exp(4/ln(x))+(x*exp(5)*exp(-ln(3)+x)^2+((-2*x^2+11*x)*exp(5 )+x^2+x)*exp(-ln(3)+x)+(x^3-10*x^2+24*x)*exp(5)-7*x)*ln(x)^2)/(x*exp(5)*ex p(-ln(3)+x)^2+(-2*x^2+10*x)*exp(5)*exp(-ln(3)+x)+(x^3-10*x^2+25*x)*exp(5)) /ln(x)^2,x,method=_RETURNVERBOSE)
Time = 0.24 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.60 \[ \int \frac {\left (-7 x+\frac {1}{9} e^{5+2 x} x+e^5 \left (24 x-10 x^2+x^3\right )+\frac {1}{3} e^x \left (x+x^2+e^5 \left (11 x-2 x^2\right )\right )\right ) \log ^2(x)+e^{\frac {4}{\log (x)}} \left (-\frac {4 e^{5+x}}{3}+e^5 (-20+4 x)+\left (e^5 x-\frac {1}{3} e^{5+x} x\right ) \log ^2(x)\right )}{\left (\frac {1}{9} e^{5+2 x} x+\frac {1}{3} e^{5+x} \left (10 x-2 x^2\right )+e^5 \left (25 x-10 x^2+x^3\right )\right ) \log ^2(x)} \, dx=\frac {{\left (x^{2} - 5 \, x + 1\right )} e^{5} - x e^{\left (x - \log \left (3\right ) + 5\right )} + x - e^{\left (\frac {4}{\log \left (x\right )} + 5\right )} + 2}{{\left (x - 5\right )} e^{5} - e^{\left (x - \log \left (3\right ) + 5\right )}} \]
integrate((((-x*exp(5)*exp(-log(3)+x)+x*exp(5))*log(x)^2-4*exp(5)*exp(-log (3)+x)+(4*x-20)*exp(5))*exp(4/log(x))+(x*exp(5)*exp(-log(3)+x)^2+((-2*x^2+ 11*x)*exp(5)+x^2+x)*exp(-log(3)+x)+(x^3-10*x^2+24*x)*exp(5)-7*x)*log(x)^2) /(x*exp(5)*exp(-log(3)+x)^2+(-2*x^2+10*x)*exp(5)*exp(-log(3)+x)+(x^3-10*x^ 2+25*x)*exp(5))/log(x)^2,x, algorithm=\
((x^2 - 5*x + 1)*e^5 - x*e^(x - log(3) + 5) + x - e^(4/log(x) + 5) + 2)/(( x - 5)*e^5 - e^(x - log(3) + 5))
Time = 0.14 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.17 \[ \int \frac {\left (-7 x+\frac {1}{9} e^{5+2 x} x+e^5 \left (24 x-10 x^2+x^3\right )+\frac {1}{3} e^x \left (x+x^2+e^5 \left (11 x-2 x^2\right )\right )\right ) \log ^2(x)+e^{\frac {4}{\log (x)}} \left (-\frac {4 e^{5+x}}{3}+e^5 (-20+4 x)+\left (e^5 x-\frac {1}{3} e^{5+x} x\right ) \log ^2(x)\right )}{\left (\frac {1}{9} e^{5+2 x} x+\frac {1}{3} e^{5+x} \left (10 x-2 x^2\right )+e^5 \left (25 x-10 x^2+x^3\right )\right ) \log ^2(x)} \, dx=x + \frac {- 3 x + 3 e^{5} e^{\frac {4}{\log {\left (x \right )}}} - 3 e^{5} - 6}{- 3 x e^{5} + e^{5} e^{x} + 15 e^{5}} \]
integrate((((-x*exp(5)*exp(-ln(3)+x)+x*exp(5))*ln(x)**2-4*exp(5)*exp(-ln(3 )+x)+(4*x-20)*exp(5))*exp(4/ln(x))+(x*exp(5)*exp(-ln(3)+x)**2+((-2*x**2+11 *x)*exp(5)+x**2+x)*exp(-ln(3)+x)+(x**3-10*x**2+24*x)*exp(5)-7*x)*ln(x)**2) /(x*exp(5)*exp(-ln(3)+x)**2+(-2*x**2+10*x)*exp(5)*exp(-ln(3)+x)+(x**3-10*x **2+25*x)*exp(5))/ln(x)**2,x)
x + (-3*x + 3*exp(5)*exp(4/log(x)) - 3*exp(5) - 6)/(-3*x*exp(5) + exp(5)*e xp(x) + 15*exp(5))
Time = 0.25 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.69 \[ \int \frac {\left (-7 x+\frac {1}{9} e^{5+2 x} x+e^5 \left (24 x-10 x^2+x^3\right )+\frac {1}{3} e^x \left (x+x^2+e^5 \left (11 x-2 x^2\right )\right )\right ) \log ^2(x)+e^{\frac {4}{\log (x)}} \left (-\frac {4 e^{5+x}}{3}+e^5 (-20+4 x)+\left (e^5 x-\frac {1}{3} e^{5+x} x\right ) \log ^2(x)\right )}{\left (\frac {1}{9} e^{5+2 x} x+\frac {1}{3} e^{5+x} \left (10 x-2 x^2\right )+e^5 \left (25 x-10 x^2+x^3\right )\right ) \log ^2(x)} \, dx=\frac {3 \, x^{2} e^{5} - 3 \, x {\left (5 \, e^{5} - 1\right )} - x e^{\left (x + 5\right )} + 3 \, e^{5} - 3 \, e^{\left (\frac {4}{\log \left (x\right )} + 5\right )} + 6}{3 \, x e^{5} - 15 \, e^{5} - e^{\left (x + 5\right )}} \]
integrate((((-x*exp(5)*exp(-log(3)+x)+x*exp(5))*log(x)^2-4*exp(5)*exp(-log (3)+x)+(4*x-20)*exp(5))*exp(4/log(x))+(x*exp(5)*exp(-log(3)+x)^2+((-2*x^2+ 11*x)*exp(5)+x^2+x)*exp(-log(3)+x)+(x^3-10*x^2+24*x)*exp(5)-7*x)*log(x)^2) /(x*exp(5)*exp(-log(3)+x)^2+(-2*x^2+10*x)*exp(5)*exp(-log(3)+x)+(x^3-10*x^ 2+25*x)*exp(5))/log(x)^2,x, algorithm=\
(3*x^2*e^5 - 3*x*(5*e^5 - 1) - x*e^(x + 5) + 3*e^5 - 3*e^(4/log(x) + 5) + 6)/(3*x*e^5 - 15*e^5 - e^(x + 5))
Exception generated. \[ \int \frac {\left (-7 x+\frac {1}{9} e^{5+2 x} x+e^5 \left (24 x-10 x^2+x^3\right )+\frac {1}{3} e^x \left (x+x^2+e^5 \left (11 x-2 x^2\right )\right )\right ) \log ^2(x)+e^{\frac {4}{\log (x)}} \left (-\frac {4 e^{5+x}}{3}+e^5 (-20+4 x)+\left (e^5 x-\frac {1}{3} e^{5+x} x\right ) \log ^2(x)\right )}{\left (\frac {1}{9} e^{5+2 x} x+\frac {1}{3} e^{5+x} \left (10 x-2 x^2\right )+e^5 \left (25 x-10 x^2+x^3\right )\right ) \log ^2(x)} \, dx=\text {Exception raised: TypeError} \]
integrate((((-x*exp(5)*exp(-log(3)+x)+x*exp(5))*log(x)^2-4*exp(5)*exp(-log (3)+x)+(4*x-20)*exp(5))*exp(4/log(x))+(x*exp(5)*exp(-log(3)+x)^2+((-2*x^2+ 11*x)*exp(5)+x^2+x)*exp(-log(3)+x)+(x^3-10*x^2+24*x)*exp(5)-7*x)*log(x)^2) /(x*exp(5)*exp(-log(3)+x)^2+(-2*x^2+10*x)*exp(5)*exp(-log(3)+x)+(x^3-10*x^ 2+25*x)*exp(5))/log(x)^2,x, algorithm=\
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Unable to divide, perhaps due to ro unding error%%%{-128,[0,6,2,0,0]%%%}+%%%{-256,[0,6,1,1,0]%%%}+%%%{768,[0,6 ,1,0,0]%%
Time = 8.07 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.49 \[ \int \frac {\left (-7 x+\frac {1}{9} e^{5+2 x} x+e^5 \left (24 x-10 x^2+x^3\right )+\frac {1}{3} e^x \left (x+x^2+e^5 \left (11 x-2 x^2\right )\right )\right ) \log ^2(x)+e^{\frac {4}{\log (x)}} \left (-\frac {4 e^{5+x}}{3}+e^5 (-20+4 x)+\left (e^5 x-\frac {1}{3} e^{5+x} x\right ) \log ^2(x)\right )}{\left (\frac {1}{9} e^{5+2 x} x+\frac {1}{3} e^{5+x} \left (10 x-2 x^2\right )+e^5 \left (25 x-10 x^2+x^3\right )\right ) \log ^2(x)} \, dx=-\frac {{\mathrm {e}}^{-5}\,\left (3\,x+3\,{\mathrm {e}}^5-15\,x\,{\mathrm {e}}^5+3\,x^2\,{\mathrm {e}}^5-3\,{\mathrm {e}}^{\frac {4}{\ln \left (x\right )}}\,{\mathrm {e}}^5-x\,{\mathrm {e}}^5\,{\mathrm {e}}^x+6\right )}{{\mathrm {e}}^x-3\,x+15} \]
int((exp(4/log(x))*(log(x)^2*(x*exp(5) - x*exp(x - log(3))*exp(5)) - 4*exp (x - log(3))*exp(5) + exp(5)*(4*x - 20)) + log(x)^2*(exp(5)*(24*x - 10*x^2 + x^3) - 7*x + exp(x - log(3))*(x + exp(5)*(11*x - 2*x^2) + x^2) + x*exp( 5)*exp(2*x - 2*log(3))))/(log(x)^2*(exp(5)*(25*x - 10*x^2 + x^3) + exp(x - log(3))*exp(5)*(10*x - 2*x^2) + x*exp(5)*exp(2*x - 2*log(3)))),x)