Integrand size = 257, antiderivative size = 37 \[ \int \frac {75 x-90 x^2-30 x^4-24 x^5-3 x^8+e^x \left (125 x-25 x^2+50 x^4-10 x^5+5 x^7-x^8\right )+\left (-125+50 x-50 x^3+20 x^4-5 x^6+2 x^7\right ) \log \left (\frac {1}{3} e^{\frac {3 x+e^x \left (5+x^3\right )}{5+x^3}}\right )}{100 x^2-100 x^3+25 x^4+40 x^5-40 x^6+10 x^7+4 x^8-4 x^9+x^{10}+\left (100 x-50 x^2+40 x^4-20 x^5+4 x^7-2 x^8\right ) \log \left (\frac {1}{3} e^{\frac {3 x+e^x \left (5+x^3\right )}{5+x^3}}\right )+\left (25+10 x^3+x^6\right ) \log ^2\left (\frac {1}{3} e^{\frac {3 x+e^x \left (5+x^3\right )}{5+x^3}}\right )} \, dx=\frac {5-x}{-2+x-\frac {\log \left (\frac {1}{3} e^{e^x+\frac {3 x}{5+x^3}}\right )}{x}} \]
Time = 0.16 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.00 \[ \int \frac {75 x-90 x^2-30 x^4-24 x^5-3 x^8+e^x \left (125 x-25 x^2+50 x^4-10 x^5+5 x^7-x^8\right )+\left (-125+50 x-50 x^3+20 x^4-5 x^6+2 x^7\right ) \log \left (\frac {1}{3} e^{\frac {3 x+e^x \left (5+x^3\right )}{5+x^3}}\right )}{100 x^2-100 x^3+25 x^4+40 x^5-40 x^6+10 x^7+4 x^8-4 x^9+x^{10}+\left (100 x-50 x^2+40 x^4-20 x^5+4 x^7-2 x^8\right ) \log \left (\frac {1}{3} e^{\frac {3 x+e^x \left (5+x^3\right )}{5+x^3}}\right )+\left (25+10 x^3+x^6\right ) \log ^2\left (\frac {1}{3} e^{\frac {3 x+e^x \left (5+x^3\right )}{5+x^3}}\right )} \, dx=-\frac {(-5+x) x}{(-2+x) x-\log \left (\frac {1}{3} e^{e^x+\frac {3 x}{5+x^3}}\right )} \]
Integrate[(75*x - 90*x^2 - 30*x^4 - 24*x^5 - 3*x^8 + E^x*(125*x - 25*x^2 + 50*x^4 - 10*x^5 + 5*x^7 - x^8) + (-125 + 50*x - 50*x^3 + 20*x^4 - 5*x^6 + 2*x^7)*Log[E^((3*x + E^x*(5 + x^3))/(5 + x^3))/3])/(100*x^2 - 100*x^3 + 2 5*x^4 + 40*x^5 - 40*x^6 + 10*x^7 + 4*x^8 - 4*x^9 + x^10 + (100*x - 50*x^2 + 40*x^4 - 20*x^5 + 4*x^7 - 2*x^8)*Log[E^((3*x + E^x*(5 + x^3))/(5 + x^3)) /3] + (25 + 10*x^3 + x^6)*Log[E^((3*x + E^x*(5 + x^3))/(5 + x^3))/3]^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.
| \(\displaystyle \int \frac {-3 x^8-24 x^5-30 x^4-90 x^2+\left (2 x^7-5 x^6+20 x^4-50 x^3+50 x-125\right ) \log \left (\frac {1}{3} e^{\frac {e^x \left (x^3+5\right )+3 x}{x^3+5}}\right )+e^x \left (-x^8+5 x^7-10 x^5+50 x^4-25 x^2+125 x\right )+75 x}{x^{10}-4 x^9+4 x^8+10 x^7-40 x^6+40 x^5+25 x^4-100 x^3+100 x^2+\left (x^6+10 x^3+25\right ) \log ^2\left (\frac {1}{3} e^{\frac {e^x \left (x^3+5\right )+3 x}{x^3+5}}\right )+\left (-2 x^8+4 x^7-20 x^5+40 x^4-50 x^2+100 x\right ) \log \left (\frac {1}{3} e^{\frac {e^x \left (x^3+5\right )+3 x}{x^3+5}}\right )} \, dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle \int \frac {(2 x-5) \left (x^3+5\right )^2 \log \left (\frac {1}{3} e^{\frac {3 x}{x^3+5}+e^x}\right )-x \left (e^x (x-5) \left (x^3+5\right )^2+3 \left (x^7+8 x^4+10 x^3+30 x-25\right )\right )}{\left (x^3+5\right )^2 \left ((x-2) x-\log \left (\frac {1}{3} e^{\frac {3 x}{x^3+5}+e^x}\right )\right )^2}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (\frac {-3 x^8-24 x^5-30 x^4-50 x^3 \log \left (\frac {1}{3} e^{\frac {3 x}{x^3+5}+e^x}\right )+50 x \log \left (\frac {1}{3} e^{\frac {3 x}{x^3+5}+e^x}\right )-125 \log \left (\frac {1}{3} e^{\frac {3 x}{x^3+5}+e^x}\right )-90 x^2+2 x^7 \log \left (\frac {1}{3} e^{\frac {3 x}{x^3+5}+e^x}\right )-5 x^6 \log \left (\frac {1}{3} e^{\frac {3 x}{x^3+5}+e^x}\right )+20 x^4 \log \left (\frac {1}{3} e^{\frac {3 x}{x^3+5}+e^x}\right )+75 x}{\left (x^3+5\right )^2 \left (-\log \left (\frac {1}{3} e^{\frac {3 x}{x^3+5}+e^x}\right )+x^2-2 x\right )^2}-\frac {e^x (x-5) x}{\left (-\log \left (\frac {1}{3} e^{\frac {3 x}{x^3+5}+e^x}\right )+x^2-2 x\right )^2}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle 10 \int \frac {x}{\left (x^2-2 x-\log \left (\frac {1}{3} e^{\frac {3 x}{x^3+5}+e^x}\right )\right )^2}dx+5 \int \frac {e^x x}{\left (x^2-2 x-\log \left (\frac {1}{3} e^{\frac {3 x}{x^3+5}+e^x}\right )\right )^2}dx-12 \int \frac {x^2}{\left (x^2-2 x-\log \left (\frac {1}{3} e^{\frac {3 x}{x^3+5}+e^x}\right )\right )^2}dx-\int \frac {e^x x^2}{\left (x^2-2 x-\log \left (\frac {1}{3} e^{\frac {3 x}{x^3+5}+e^x}\right )\right )^2}dx+2 \int \frac {x^3}{\left (x^2-2 x-\log \left (\frac {1}{3} e^{\frac {3 x}{x^3+5}+e^x}\right )\right )^2}dx+2 \int \frac {1}{\left (x-\sqrt [3]{-5}\right ) \left (x^2-2 x-\log \left (\frac {1}{3} e^{\frac {3 x}{x^3+5}+e^x}\right )\right )^2}dx+2\ 5^{2/3} \int \frac {1}{\left (x+\sqrt [3]{5}\right ) \left (x^2-2 x-\log \left (\frac {1}{3} e^{\frac {3 x}{x^3+5}+e^x}\right )\right )^2}dx+2 \int \frac {1}{\left (x+\sqrt [3]{5}\right ) \left (x^2-2 x-\log \left (\frac {1}{3} e^{\frac {3 x}{x^3+5}+e^x}\right )\right )^2}dx+2 \int \frac {1}{\left (x+(-1)^{2/3} \sqrt [3]{5}\right ) \left (x^2-2 x-\log \left (\frac {1}{3} e^{\frac {3 x}{x^3+5}+e^x}\right )\right )^2}dx+2 (-5)^{2/3} \int \frac {1}{\left (\sqrt [3]{5}-\sqrt [3]{-1} x\right ) \left (x^2-2 x-\log \left (\frac {1}{3} e^{\frac {3 x}{x^3+5}+e^x}\right )\right )^2}dx-2 \sqrt [3]{-1} 5^{2/3} \int \frac {1}{\left ((-1)^{2/3} x+\sqrt [3]{5}\right ) \left (x^2-2 x-\log \left (\frac {1}{3} e^{\frac {3 x}{x^3+5}+e^x}\right )\right )^2}dx+225 \int \frac {x}{\left (x^3+5\right )^2 \left (x^2-2 x-\log \left (\frac {1}{3} e^{\frac {3 x}{x^3+5}+e^x}\right )\right )^2}dx-45 \int \frac {x^2}{\left (x^3+5\right )^2 \left (x^2-2 x-\log \left (\frac {1}{3} e^{\frac {3 x}{x^3+5}+e^x}\right )\right )^2}dx+5 \int \frac {1}{x^2-2 x-\log \left (\frac {1}{3} e^{\frac {3 x}{x^3+5}+e^x}\right )}dx-2 \int \frac {x}{x^2-2 x-\log \left (\frac {1}{3} e^{\frac {3 x}{x^3+5}+e^x}\right )}dx\) |
Int[(75*x - 90*x^2 - 30*x^4 - 24*x^5 - 3*x^8 + E^x*(125*x - 25*x^2 + 50*x^ 4 - 10*x^5 + 5*x^7 - x^8) + (-125 + 50*x - 50*x^3 + 20*x^4 - 5*x^6 + 2*x^7 )*Log[E^((3*x + E^x*(5 + x^3))/(5 + x^3))/3])/(100*x^2 - 100*x^3 + 25*x^4 + 40*x^5 - 40*x^6 + 10*x^7 + 4*x^8 - 4*x^9 + x^10 + (100*x - 50*x^2 + 40*x ^4 - 20*x^5 + 4*x^7 - 2*x^8)*Log[E^((3*x + E^x*(5 + x^3))/(5 + x^3))/3] + (25 + 10*x^3 + x^6)*Log[E^((3*x + E^x*(5 + x^3))/(5 + x^3))/3]^2),x]
3.19.39.3.1 Defintions of rubi rules used
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 0.29 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.30
\[-\frac {2 \left (-5+x \right ) x}{2 x^{2}+2 \ln \left (3\right )-4 x -2 \ln \left ({\mathrm e}^{\frac {{\mathrm e}^{x} x^{3}+5 \,{\mathrm e}^{x}+3 x}{x^{3}+5}}\right )}\]
int(((2*x^7-5*x^6+20*x^4-50*x^3+50*x-125)*ln(1/3*exp(((x^3+5)*exp(x)+3*x)/ (x^3+5)))+(-x^8+5*x^7-10*x^5+50*x^4-25*x^2+125*x)*exp(x)-3*x^8-24*x^5-30*x ^4-90*x^2+75*x)/((x^6+10*x^3+25)*ln(1/3*exp(((x^3+5)*exp(x)+3*x)/(x^3+5))) ^2+(-2*x^8+4*x^7-20*x^5+40*x^4-50*x^2+100*x)*ln(1/3*exp(((x^3+5)*exp(x)+3* x)/(x^3+5)))+x^10-4*x^9+4*x^8+10*x^7-40*x^6+40*x^5+25*x^4-100*x^3+100*x^2) ,x)
Time = 0.26 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.49 \[ \int \frac {75 x-90 x^2-30 x^4-24 x^5-3 x^8+e^x \left (125 x-25 x^2+50 x^4-10 x^5+5 x^7-x^8\right )+\left (-125+50 x-50 x^3+20 x^4-5 x^6+2 x^7\right ) \log \left (\frac {1}{3} e^{\frac {3 x+e^x \left (5+x^3\right )}{5+x^3}}\right )}{100 x^2-100 x^3+25 x^4+40 x^5-40 x^6+10 x^7+4 x^8-4 x^9+x^{10}+\left (100 x-50 x^2+40 x^4-20 x^5+4 x^7-2 x^8\right ) \log \left (\frac {1}{3} e^{\frac {3 x+e^x \left (5+x^3\right )}{5+x^3}}\right )+\left (25+10 x^3+x^6\right ) \log ^2\left (\frac {1}{3} e^{\frac {3 x+e^x \left (5+x^3\right )}{5+x^3}}\right )} \, dx=-\frac {x^{5} - 5 \, x^{4} + 5 \, x^{2} - 25 \, x}{x^{5} - 2 \, x^{4} + 5 \, x^{2} - {\left (x^{3} + 5\right )} e^{x} + {\left (x^{3} + 5\right )} \log \left (3\right ) - 13 \, x} \]
integrate(((2*x^7-5*x^6+20*x^4-50*x^3+50*x-125)*log(1/3*exp(((x^3+5)*exp(x )+3*x)/(x^3+5)))+(-x^8+5*x^7-10*x^5+50*x^4-25*x^2+125*x)*exp(x)-3*x^8-24*x ^5-30*x^4-90*x^2+75*x)/((x^6+10*x^3+25)*log(1/3*exp(((x^3+5)*exp(x)+3*x)/( x^3+5)))^2+(-2*x^8+4*x^7-20*x^5+40*x^4-50*x^2+100*x)*log(1/3*exp(((x^3+5)* exp(x)+3*x)/(x^3+5)))+x^10-4*x^9+4*x^8+10*x^7-40*x^6+40*x^5+25*x^4-100*x^3 +100*x^2),x, algorithm=\
-(x^5 - 5*x^4 + 5*x^2 - 25*x)/(x^5 - 2*x^4 + 5*x^2 - (x^3 + 5)*e^x + (x^3 + 5)*log(3) - 13*x)
Leaf count of result is larger than twice the leaf count of optimal. 53 vs. \(2 (24) = 48\).
Time = 0.53 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.43 \[ \int \frac {75 x-90 x^2-30 x^4-24 x^5-3 x^8+e^x \left (125 x-25 x^2+50 x^4-10 x^5+5 x^7-x^8\right )+\left (-125+50 x-50 x^3+20 x^4-5 x^6+2 x^7\right ) \log \left (\frac {1}{3} e^{\frac {3 x+e^x \left (5+x^3\right )}{5+x^3}}\right )}{100 x^2-100 x^3+25 x^4+40 x^5-40 x^6+10 x^7+4 x^8-4 x^9+x^{10}+\left (100 x-50 x^2+40 x^4-20 x^5+4 x^7-2 x^8\right ) \log \left (\frac {1}{3} e^{\frac {3 x+e^x \left (5+x^3\right )}{5+x^3}}\right )+\left (25+10 x^3+x^6\right ) \log ^2\left (\frac {1}{3} e^{\frac {3 x+e^x \left (5+x^3\right )}{5+x^3}}\right )} \, dx=\frac {x^{5} - 5 x^{4} + 5 x^{2} - 25 x}{- x^{5} + 2 x^{4} - x^{3} \log {\left (3 \right )} - 5 x^{2} + 13 x + \left (x^{3} + 5\right ) e^{x} - 5 \log {\left (3 \right )}} \]
integrate(((2*x**7-5*x**6+20*x**4-50*x**3+50*x-125)*ln(1/3*exp(((x**3+5)*e xp(x)+3*x)/(x**3+5)))+(-x**8+5*x**7-10*x**5+50*x**4-25*x**2+125*x)*exp(x)- 3*x**8-24*x**5-30*x**4-90*x**2+75*x)/((x**6+10*x**3+25)*ln(1/3*exp(((x**3+ 5)*exp(x)+3*x)/(x**3+5)))**2+(-2*x**8+4*x**7-20*x**5+40*x**4-50*x**2+100*x )*ln(1/3*exp(((x**3+5)*exp(x)+3*x)/(x**3+5)))+x**10-4*x**9+4*x**8+10*x**7- 40*x**6+40*x**5+25*x**4-100*x**3+100*x**2),x)
(x**5 - 5*x**4 + 5*x**2 - 25*x)/(-x**5 + 2*x**4 - x**3*log(3) - 5*x**2 + 1 3*x + (x**3 + 5)*exp(x) - 5*log(3))
Time = 0.48 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.54 \[ \int \frac {75 x-90 x^2-30 x^4-24 x^5-3 x^8+e^x \left (125 x-25 x^2+50 x^4-10 x^5+5 x^7-x^8\right )+\left (-125+50 x-50 x^3+20 x^4-5 x^6+2 x^7\right ) \log \left (\frac {1}{3} e^{\frac {3 x+e^x \left (5+x^3\right )}{5+x^3}}\right )}{100 x^2-100 x^3+25 x^4+40 x^5-40 x^6+10 x^7+4 x^8-4 x^9+x^{10}+\left (100 x-50 x^2+40 x^4-20 x^5+4 x^7-2 x^8\right ) \log \left (\frac {1}{3} e^{\frac {3 x+e^x \left (5+x^3\right )}{5+x^3}}\right )+\left (25+10 x^3+x^6\right ) \log ^2\left (\frac {1}{3} e^{\frac {3 x+e^x \left (5+x^3\right )}{5+x^3}}\right )} \, dx=-\frac {x^{5} - 5 \, x^{4} + 5 \, x^{2} - 25 \, x}{x^{5} - 2 \, x^{4} + x^{3} \log \left (3\right ) + 5 \, x^{2} - {\left (x^{3} + 5\right )} e^{x} - 13 \, x + 5 \, \log \left (3\right )} \]
integrate(((2*x^7-5*x^6+20*x^4-50*x^3+50*x-125)*log(1/3*exp(((x^3+5)*exp(x )+3*x)/(x^3+5)))+(-x^8+5*x^7-10*x^5+50*x^4-25*x^2+125*x)*exp(x)-3*x^8-24*x ^5-30*x^4-90*x^2+75*x)/((x^6+10*x^3+25)*log(1/3*exp(((x^3+5)*exp(x)+3*x)/( x^3+5)))^2+(-2*x^8+4*x^7-20*x^5+40*x^4-50*x^2+100*x)*log(1/3*exp(((x^3+5)* exp(x)+3*x)/(x^3+5)))+x^10-4*x^9+4*x^8+10*x^7-40*x^6+40*x^5+25*x^4-100*x^3 +100*x^2),x, algorithm=\
-(x^5 - 5*x^4 + 5*x^2 - 25*x)/(x^5 - 2*x^4 + x^3*log(3) + 5*x^2 - (x^3 + 5 )*e^x - 13*x + 5*log(3))
Time = 0.57 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.59 \[ \int \frac {75 x-90 x^2-30 x^4-24 x^5-3 x^8+e^x \left (125 x-25 x^2+50 x^4-10 x^5+5 x^7-x^8\right )+\left (-125+50 x-50 x^3+20 x^4-5 x^6+2 x^7\right ) \log \left (\frac {1}{3} e^{\frac {3 x+e^x \left (5+x^3\right )}{5+x^3}}\right )}{100 x^2-100 x^3+25 x^4+40 x^5-40 x^6+10 x^7+4 x^8-4 x^9+x^{10}+\left (100 x-50 x^2+40 x^4-20 x^5+4 x^7-2 x^8\right ) \log \left (\frac {1}{3} e^{\frac {3 x+e^x \left (5+x^3\right )}{5+x^3}}\right )+\left (25+10 x^3+x^6\right ) \log ^2\left (\frac {1}{3} e^{\frac {3 x+e^x \left (5+x^3\right )}{5+x^3}}\right )} \, dx=-\frac {x^{5} - 5 \, x^{4} + 5 \, x^{2} - 25 \, x}{x^{5} - 2 \, x^{4} - x^{3} e^{x} + x^{3} \log \left (3\right ) + 5 \, x^{2} - 13 \, x - 5 \, e^{x} + 5 \, \log \left (3\right )} \]
integrate(((2*x^7-5*x^6+20*x^4-50*x^3+50*x-125)*log(1/3*exp(((x^3+5)*exp(x )+3*x)/(x^3+5)))+(-x^8+5*x^7-10*x^5+50*x^4-25*x^2+125*x)*exp(x)-3*x^8-24*x ^5-30*x^4-90*x^2+75*x)/((x^6+10*x^3+25)*log(1/3*exp(((x^3+5)*exp(x)+3*x)/( x^3+5)))^2+(-2*x^8+4*x^7-20*x^5+40*x^4-50*x^2+100*x)*log(1/3*exp(((x^3+5)* exp(x)+3*x)/(x^3+5)))+x^10-4*x^9+4*x^8+10*x^7-40*x^6+40*x^5+25*x^4-100*x^3 +100*x^2),x, algorithm=\
-(x^5 - 5*x^4 + 5*x^2 - 25*x)/(x^5 - 2*x^4 - x^3*e^x + x^3*log(3) + 5*x^2 - 13*x - 5*e^x + 5*log(3))
Time = 11.76 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.70 \[ \int \frac {75 x-90 x^2-30 x^4-24 x^5-3 x^8+e^x \left (125 x-25 x^2+50 x^4-10 x^5+5 x^7-x^8\right )+\left (-125+50 x-50 x^3+20 x^4-5 x^6+2 x^7\right ) \log \left (\frac {1}{3} e^{\frac {3 x+e^x \left (5+x^3\right )}{5+x^3}}\right )}{100 x^2-100 x^3+25 x^4+40 x^5-40 x^6+10 x^7+4 x^8-4 x^9+x^{10}+\left (100 x-50 x^2+40 x^4-20 x^5+4 x^7-2 x^8\right ) \log \left (\frac {1}{3} e^{\frac {3 x+e^x \left (5+x^3\right )}{5+x^3}}\right )+\left (25+10 x^3+x^6\right ) \log ^2\left (\frac {1}{3} e^{\frac {3 x+e^x \left (5+x^3\right )}{5+x^3}}\right )} \, dx=-\frac {-x^5+5\,x^4-5\,x^2+25\,x}{13\,x-5\,\ln \left (3\right )+5\,{\mathrm {e}}^x+x^3\,{\mathrm {e}}^x-x^3\,\ln \left (3\right )-5\,x^2+2\,x^4-x^5} \]
int(-(90*x^2 - log(exp((3*x + exp(x)*(x^3 + 5))/(x^3 + 5))/3)*(50*x - 50*x ^3 + 20*x^4 - 5*x^6 + 2*x^7 - 125) - exp(x)*(125*x - 25*x^2 + 50*x^4 - 10* x^5 + 5*x^7 - x^8) - 75*x + 30*x^4 + 24*x^5 + 3*x^8)/(log(exp((3*x + exp(x )*(x^3 + 5))/(x^3 + 5))/3)*(100*x - 50*x^2 + 40*x^4 - 20*x^5 + 4*x^7 - 2*x ^8) + 100*x^2 - 100*x^3 + 25*x^4 + 40*x^5 - 40*x^6 + 10*x^7 + 4*x^8 - 4*x^ 9 + x^10 + log(exp((3*x + exp(x)*(x^3 + 5))/(x^3 + 5))/3)^2*(10*x^3 + x^6 + 25)),x)