Integrand size = 135, antiderivative size = 34 \[ \int \frac {100-160 x-20 x^2+6 x^3+e^x \left (150 x+30 x^2-10 x^3\right )}{\left (-25 x+30 x^2-x^4+e^x \left (-25 x^2+5 x^3\right )\right ) \log ^2\left (\frac {25 x^4-50 x^5+15 x^6+25 e^{2 x} x^6+10 x^7+x^8+e^x \left (50 x^5-50 x^6-10 x^7\right )}{625-250 x+25 x^2}\right )} \, dx=\frac {1}{\log \left (\frac {x^2 \left (-x+\left (1-e^x+\frac {x}{5}\right ) x^2\right )^2}{(-5+x)^2}\right )} \] Output:
1/ln((x^2*(1/5*x+1-exp(x))-x)^2*x^2/(-5+x)^2)
Time = 0.06 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.88 \[ \int \frac {100-160 x-20 x^2+6 x^3+e^x \left (150 x+30 x^2-10 x^3\right )}{\left (-25 x+30 x^2-x^4+e^x \left (-25 x^2+5 x^3\right )\right ) \log ^2\left (\frac {25 x^4-50 x^5+15 x^6+25 e^{2 x} x^6+10 x^7+x^8+e^x \left (50 x^5-50 x^6-10 x^7\right )}{625-250 x+25 x^2}\right )} \, dx=\frac {1}{\log \left (\frac {x^4 \left (-5-5 \left (-1+e^x\right ) x+x^2\right )^2}{25 (-5+x)^2}\right )} \] Input:
Integrate[(100 - 160*x - 20*x^2 + 6*x^3 + E^x*(150*x + 30*x^2 - 10*x^3))/( (-25*x + 30*x^2 - x^4 + E^x*(-25*x^2 + 5*x^3))*Log[(25*x^4 - 50*x^5 + 15*x ^6 + 25*E^(2*x)*x^6 + 10*x^7 + x^8 + E^x*(50*x^5 - 50*x^6 - 10*x^7))/(625 - 250*x + 25*x^2)]^2),x]
Output:
Log[(x^4*(-5 - 5*(-1 + E^x)*x + x^2)^2)/(25*(-5 + x)^2)]^(-1)
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.
\(\displaystyle \int \frac {6 x^3-20 x^2+e^x \left (-10 x^3+30 x^2+150 x\right )-160 x+100}{\left (-x^4+30 x^2+e^x \left (5 x^3-25 x^2\right )-25 x\right ) \log ^2\left (\frac {x^8+10 x^7+25 e^{2 x} x^6+15 x^6-50 x^5+25 x^4+e^x \left (-10 x^7-50 x^6+50 x^5\right )}{25 x^2-250 x+625}\right )} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {-6 x^3+20 x^2-e^x \left (-10 x^3+30 x^2+150 x\right )+160 x-100}{(5-x) x \left (-x^2+5 e^x x-5 x+5\right ) \log ^2\left (\frac {x^4 \left (x^2-5 e^x x+5 x-5\right )^2}{25 x^2-250 x+625}\right )}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {2 \left (x^3+4 x^2-5 x-5\right )}{x \left (x^2-5 e^x x+5 x-5\right ) \log ^2\left (\frac {x^4 \left (x^2-5 \left (e^x-1\right ) x-5\right )^2}{25 (x-5)^2}\right )}-\frac {2 \left (x^2-3 x-15\right )}{(x-5) x \log ^2\left (\frac {x^4 \left (x^2-5 \left (e^x-1\right ) x-5\right )^2}{25 (x-5)^2}\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -2 \int \frac {1}{\log ^2\left (\frac {x^4 \left (x^2-5 \left (-1+e^x\right ) x-5\right )^2}{25 (x-5)^2}\right )}dx+2 \int \frac {1}{(x-5) \log ^2\left (\frac {x^4 \left (x^2-5 \left (-1+e^x\right ) x-5\right )^2}{25 (x-5)^2}\right )}dx-6 \int \frac {1}{x \log ^2\left (\frac {x^4 \left (x^2-5 \left (-1+e^x\right ) x-5\right )^2}{25 (x-5)^2}\right )}dx-10 \int \frac {1}{\left (x^2-5 e^x x+5 x-5\right ) \log ^2\left (\frac {x^4 \left (x^2-5 \left (-1+e^x\right ) x-5\right )^2}{25 (x-5)^2}\right )}dx-10 \int \frac {1}{x \left (x^2-5 e^x x+5 x-5\right ) \log ^2\left (\frac {x^4 \left (x^2-5 \left (-1+e^x\right ) x-5\right )^2}{25 (x-5)^2}\right )}dx+8 \int \frac {x}{\left (x^2-5 e^x x+5 x-5\right ) \log ^2\left (\frac {x^4 \left (x^2-5 \left (-1+e^x\right ) x-5\right )^2}{25 (x-5)^2}\right )}dx+2 \int \frac {x^2}{\left (x^2-5 e^x x+5 x-5\right ) \log ^2\left (\frac {x^4 \left (x^2-5 \left (-1+e^x\right ) x-5\right )^2}{25 (x-5)^2}\right )}dx\) |
Input:
Int[(100 - 160*x - 20*x^2 + 6*x^3 + E^x*(150*x + 30*x^2 - 10*x^3))/((-25*x + 30*x^2 - x^4 + E^x*(-25*x^2 + 5*x^3))*Log[(25*x^4 - 50*x^5 + 15*x^6 + 2 5*E^(2*x)*x^6 + 10*x^7 + x^8 + E^x*(50*x^5 - 50*x^6 - 10*x^7))/(625 - 250* x + 25*x^2)]^2),x]
Output:
$Aborted
Leaf count of result is larger than twice the leaf count of optimal. \(67\) vs. \(2(31)=62\).
Time = 1.96 (sec) , antiderivative size = 68, normalized size of antiderivative = 2.00
method | result | size |
parallelrisch | \(\frac {1}{\ln \left (\frac {25 \,{\mathrm e}^{2 x} x^{6}+\left (-10 x^{7}-50 x^{6}+50 x^{5}\right ) {\mathrm e}^{x}+x^{8}+10 x^{7}+15 x^{6}-50 x^{5}+25 x^{4}}{25 x^{2}-250 x +625}\right )}\) | \(68\) |
risch | \(\text {Expression too large to display}\) | \(786\) |
Input:
int(((-10*x^3+30*x^2+150*x)*exp(x)+6*x^3-20*x^2-160*x+100)/((5*x^3-25*x^2) *exp(x)-x^4+30*x^2-25*x)/ln((25*x^6*exp(x)^2+(-10*x^7-50*x^6+50*x^5)*exp(x )+x^8+10*x^7+15*x^6-50*x^5+25*x^4)/(25*x^2-250*x+625))^2,x,method=_RETURNV ERBOSE)
Output:
1/ln(1/25/(x^2-10*x+25)*(25*x^6*exp(x)^2+(-10*x^7-50*x^6+50*x^5)*exp(x)+x^ 8+10*x^7+15*x^6-50*x^5+25*x^4))
Leaf count of result is larger than twice the leaf count of optimal. 66 vs. \(2 (30) = 60\).
Time = 0.08 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.94 \[ \int \frac {100-160 x-20 x^2+6 x^3+e^x \left (150 x+30 x^2-10 x^3\right )}{\left (-25 x+30 x^2-x^4+e^x \left (-25 x^2+5 x^3\right )\right ) \log ^2\left (\frac {25 x^4-50 x^5+15 x^6+25 e^{2 x} x^6+10 x^7+x^8+e^x \left (50 x^5-50 x^6-10 x^7\right )}{625-250 x+25 x^2}\right )} \, dx=\frac {1}{\log \left (\frac {x^{8} + 10 \, x^{7} + 25 \, x^{6} e^{\left (2 \, x\right )} + 15 \, x^{6} - 50 \, x^{5} + 25 \, x^{4} - 10 \, {\left (x^{7} + 5 \, x^{6} - 5 \, x^{5}\right )} e^{x}}{25 \, {\left (x^{2} - 10 \, x + 25\right )}}\right )} \] Input:
integrate(((-10*x^3+30*x^2+150*x)*exp(x)+6*x^3-20*x^2-160*x+100)/((5*x^3-2 5*x^2)*exp(x)-x^4+30*x^2-25*x)/log((25*x^6*exp(x)^2+(-10*x^7-50*x^6+50*x^5 )*exp(x)+x^8+10*x^7+15*x^6-50*x^5+25*x^4)/(25*x^2-250*x+625))^2,x, algorit hm="fricas")
Output:
1/log(1/25*(x^8 + 10*x^7 + 25*x^6*e^(2*x) + 15*x^6 - 50*x^5 + 25*x^4 - 10* (x^7 + 5*x^6 - 5*x^5)*e^x)/(x^2 - 10*x + 25))
Leaf count of result is larger than twice the leaf count of optimal. 65 vs. \(2 (26) = 52\).
Time = 0.29 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.91 \[ \int \frac {100-160 x-20 x^2+6 x^3+e^x \left (150 x+30 x^2-10 x^3\right )}{\left (-25 x+30 x^2-x^4+e^x \left (-25 x^2+5 x^3\right )\right ) \log ^2\left (\frac {25 x^4-50 x^5+15 x^6+25 e^{2 x} x^6+10 x^7+x^8+e^x \left (50 x^5-50 x^6-10 x^7\right )}{625-250 x+25 x^2}\right )} \, dx=\frac {1}{\log {\left (\frac {x^{8} + 10 x^{7} + 25 x^{6} e^{2 x} + 15 x^{6} - 50 x^{5} + 25 x^{4} + \left (- 10 x^{7} - 50 x^{6} + 50 x^{5}\right ) e^{x}}{25 x^{2} - 250 x + 625} \right )}} \] Input:
integrate(((-10*x**3+30*x**2+150*x)*exp(x)+6*x**3-20*x**2-160*x+100)/((5*x **3-25*x**2)*exp(x)-x**4+30*x**2-25*x)/ln((25*x**6*exp(x)**2+(-10*x**7-50* x**6+50*x**5)*exp(x)+x**8+10*x**7+15*x**6-50*x**5+25*x**4)/(25*x**2-250*x+ 625))**2,x)
Output:
1/log((x**8 + 10*x**7 + 25*x**6*exp(2*x) + 15*x**6 - 50*x**5 + 25*x**4 + ( -10*x**7 - 50*x**6 + 50*x**5)*exp(x))/(25*x**2 - 250*x + 625))
Time = 0.44 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.97 \[ \int \frac {100-160 x-20 x^2+6 x^3+e^x \left (150 x+30 x^2-10 x^3\right )}{\left (-25 x+30 x^2-x^4+e^x \left (-25 x^2+5 x^3\right )\right ) \log ^2\left (\frac {25 x^4-50 x^5+15 x^6+25 e^{2 x} x^6+10 x^7+x^8+e^x \left (50 x^5-50 x^6-10 x^7\right )}{625-250 x+25 x^2}\right )} \, dx=-\frac {1}{2 \, {\left (\log \left (5\right ) - \log \left (-x^{2} + 5 \, x e^{x} - 5 \, x + 5\right ) + \log \left (x - 5\right ) - 2 \, \log \left (x\right )\right )}} \] Input:
integrate(((-10*x^3+30*x^2+150*x)*exp(x)+6*x^3-20*x^2-160*x+100)/((5*x^3-2 5*x^2)*exp(x)-x^4+30*x^2-25*x)/log((25*x^6*exp(x)^2+(-10*x^7-50*x^6+50*x^5 )*exp(x)+x^8+10*x^7+15*x^6-50*x^5+25*x^4)/(25*x^2-250*x+625))^2,x, algorit hm="maxima")
Output:
-1/2/(log(5) - log(-x^2 + 5*x*e^x - 5*x + 5) + log(x - 5) - 2*log(x))
Leaf count of result is larger than twice the leaf count of optimal. 69 vs. \(2 (30) = 60\).
Time = 1.32 (sec) , antiderivative size = 69, normalized size of antiderivative = 2.03 \[ \int \frac {100-160 x-20 x^2+6 x^3+e^x \left (150 x+30 x^2-10 x^3\right )}{\left (-25 x+30 x^2-x^4+e^x \left (-25 x^2+5 x^3\right )\right ) \log ^2\left (\frac {25 x^4-50 x^5+15 x^6+25 e^{2 x} x^6+10 x^7+x^8+e^x \left (50 x^5-50 x^6-10 x^7\right )}{625-250 x+25 x^2}\right )} \, dx=\frac {1}{\log \left (\frac {x^{8} - 10 \, x^{7} e^{x} + 10 \, x^{7} + 25 \, x^{6} e^{\left (2 \, x\right )} - 50 \, x^{6} e^{x} + 15 \, x^{6} + 50 \, x^{5} e^{x} - 50 \, x^{5} + 25 \, x^{4}}{25 \, {\left (x^{2} - 10 \, x + 25\right )}}\right )} \] Input:
integrate(((-10*x^3+30*x^2+150*x)*exp(x)+6*x^3-20*x^2-160*x+100)/((5*x^3-2 5*x^2)*exp(x)-x^4+30*x^2-25*x)/log((25*x^6*exp(x)^2+(-10*x^7-50*x^6+50*x^5 )*exp(x)+x^8+10*x^7+15*x^6-50*x^5+25*x^4)/(25*x^2-250*x+625))^2,x, algorit hm="giac")
Output:
1/log(1/25*(x^8 - 10*x^7*e^x + 10*x^7 + 25*x^6*e^(2*x) - 50*x^6*e^x + 15*x ^6 + 50*x^5*e^x - 50*x^5 + 25*x^4)/(x^2 - 10*x + 25))
Time = 3.82 (sec) , antiderivative size = 69, normalized size of antiderivative = 2.03 \[ \int \frac {100-160 x-20 x^2+6 x^3+e^x \left (150 x+30 x^2-10 x^3\right )}{\left (-25 x+30 x^2-x^4+e^x \left (-25 x^2+5 x^3\right )\right ) \log ^2\left (\frac {25 x^4-50 x^5+15 x^6+25 e^{2 x} x^6+10 x^7+x^8+e^x \left (50 x^5-50 x^6-10 x^7\right )}{625-250 x+25 x^2}\right )} \, dx=\frac {1}{\ln \left (\frac {25\,x^6\,{\mathrm {e}}^{2\,x}-{\mathrm {e}}^x\,\left (10\,x^7+50\,x^6-50\,x^5\right )+25\,x^4-50\,x^5+15\,x^6+10\,x^7+x^8}{25\,x^2-250\,x+625}\right )} \] Input:
int(-(6*x^3 - 20*x^2 - 160*x + exp(x)*(150*x + 30*x^2 - 10*x^3) + 100)/(lo g((25*x^6*exp(2*x) - exp(x)*(50*x^6 - 50*x^5 + 10*x^7) + 25*x^4 - 50*x^5 + 15*x^6 + 10*x^7 + x^8)/(25*x^2 - 250*x + 625))^2*(25*x + exp(x)*(25*x^2 - 5*x^3) - 30*x^2 + x^4)),x)
Output:
1/log((25*x^6*exp(2*x) - exp(x)*(50*x^6 - 50*x^5 + 10*x^7) + 25*x^4 - 50*x ^5 + 15*x^6 + 10*x^7 + x^8)/(25*x^2 - 250*x + 625))
\[ \int \frac {100-160 x-20 x^2+6 x^3+e^x \left (150 x+30 x^2-10 x^3\right )}{\left (-25 x+30 x^2-x^4+e^x \left (-25 x^2+5 x^3\right )\right ) \log ^2\left (\frac {25 x^4-50 x^5+15 x^6+25 e^{2 x} x^6+10 x^7+x^8+e^x \left (50 x^5-50 x^6-10 x^7\right )}{625-250 x+25 x^2}\right )} \, dx=\text {too large to display} \] Input:
int(((-10*x^3+30*x^2+150*x)*exp(x)+6*x^3-20*x^2-160*x+100)/((5*x^3-25*x^2) *exp(x)-x^4+30*x^2-25*x)/log((25*x^6*exp(x)^2+(-10*x^7-50*x^6+50*x^5)*exp( x)+x^8+10*x^7+15*x^6-50*x^5+25*x^4)/(25*x^2-250*x+625))^2,x)
Output:
2*(75*int(e**x/(5*e**x*log((25*e**(2*x)*x**6 - 10*e**x*x**7 - 50*e**x*x**6 + 50*e**x*x**5 + x**8 + 10*x**7 + 15*x**6 - 50*x**5 + 25*x**4)/(25*x**2 - 250*x + 625))**2*x**2 - 25*e**x*log((25*e**(2*x)*x**6 - 10*e**x*x**7 - 50 *e**x*x**6 + 50*e**x*x**5 + x**8 + 10*x**7 + 15*x**6 - 50*x**5 + 25*x**4)/ (25*x**2 - 250*x + 625))**2*x - log((25*e**(2*x)*x**6 - 10*e**x*x**7 - 50* e**x*x**6 + 50*e**x*x**5 + x**8 + 10*x**7 + 15*x**6 - 50*x**5 + 25*x**4)/( 25*x**2 - 250*x + 625))**2*x**3 + 30*log((25*e**(2*x)*x**6 - 10*e**x*x**7 - 50*e**x*x**6 + 50*e**x*x**5 + x**8 + 10*x**7 + 15*x**6 - 50*x**5 + 25*x* *4)/(25*x**2 - 250*x + 625))**2*x - 25*log((25*e**(2*x)*x**6 - 10*e**x*x** 7 - 50*e**x*x**6 + 50*e**x*x**5 + x**8 + 10*x**7 + 15*x**6 - 50*x**5 + 25* x**4)/(25*x**2 - 250*x + 625))**2),x) + 3*int(x**2/(5*e**x*log((25*e**(2*x )*x**6 - 10*e**x*x**7 - 50*e**x*x**6 + 50*e**x*x**5 + x**8 + 10*x**7 + 15* x**6 - 50*x**5 + 25*x**4)/(25*x**2 - 250*x + 625))**2*x**2 - 25*e**x*log(( 25*e**(2*x)*x**6 - 10*e**x*x**7 - 50*e**x*x**6 + 50*e**x*x**5 + x**8 + 10* x**7 + 15*x**6 - 50*x**5 + 25*x**4)/(25*x**2 - 250*x + 625))**2*x - log((2 5*e**(2*x)*x**6 - 10*e**x*x**7 - 50*e**x*x**6 + 50*e**x*x**5 + x**8 + 10*x **7 + 15*x**6 - 50*x**5 + 25*x**4)/(25*x**2 - 250*x + 625))**2*x**3 + 30*l og((25*e**(2*x)*x**6 - 10*e**x*x**7 - 50*e**x*x**6 + 50*e**x*x**5 + x**8 + 10*x**7 + 15*x**6 - 50*x**5 + 25*x**4)/(25*x**2 - 250*x + 625))**2*x - 25 *log((25*e**(2*x)*x**6 - 10*e**x*x**7 - 50*e**x*x**6 + 50*e**x*x**5 + x...