Integrand size = 147, antiderivative size = 26 \[ \int \frac {4 x^2+e^x \left (1-4 x+4 x^2\right )+\left (-6 x^2+12 x^3\right ) \log \left (\frac {-5+10 x}{2 x}\right )}{e^{3 x} (-1+2 x)+e^{2 x} \left (-6 x^2+12 x^3\right ) \log \left (\frac {-5+10 x}{2 x}\right )+e^x \left (-12 x^4+24 x^5\right ) \log ^2\left (\frac {-5+10 x}{2 x}\right )+\left (-8 x^6+16 x^7\right ) \log ^3\left (\frac {-5+10 x}{2 x}\right )} \, dx=2-\frac {x}{\left (e^x+2 x^2 \log \left (5-\frac {5}{2 x}\right )\right )^2} \] Output:
2-x/(2*x^2*ln(5-5/2/x)+exp(x))^2
Time = 0.52 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \[ \int \frac {4 x^2+e^x \left (1-4 x+4 x^2\right )+\left (-6 x^2+12 x^3\right ) \log \left (\frac {-5+10 x}{2 x}\right )}{e^{3 x} (-1+2 x)+e^{2 x} \left (-6 x^2+12 x^3\right ) \log \left (\frac {-5+10 x}{2 x}\right )+e^x \left (-12 x^4+24 x^5\right ) \log ^2\left (\frac {-5+10 x}{2 x}\right )+\left (-8 x^6+16 x^7\right ) \log ^3\left (\frac {-5+10 x}{2 x}\right )} \, dx=-\frac {x}{\left (e^x+2 x^2 \log \left (5-\frac {5}{2 x}\right )\right )^2} \] Input:
Integrate[(4*x^2 + E^x*(1 - 4*x + 4*x^2) + (-6*x^2 + 12*x^3)*Log[(-5 + 10* x)/(2*x)])/(E^(3*x)*(-1 + 2*x) + E^(2*x)*(-6*x^2 + 12*x^3)*Log[(-5 + 10*x) /(2*x)] + E^x*(-12*x^4 + 24*x^5)*Log[(-5 + 10*x)/(2*x)]^2 + (-8*x^6 + 16*x ^7)*Log[(-5 + 10*x)/(2*x)]^3),x]
Output:
-(x/(E^x + 2*x^2*Log[5 - 5/(2*x)])^2)
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.
| \(\displaystyle \int \frac {4 x^2+e^x \left (4 x^2-4 x+1\right )+\left (12 x^3-6 x^2\right ) \log \left (\frac {10 x-5}{2 x}\right )}{\left (16 x^7-8 x^6\right ) \log ^3\left (\frac {10 x-5}{2 x}\right )+e^x \left (24 x^5-12 x^4\right ) \log ^2\left (\frac {10 x-5}{2 x}\right )+e^{2 x} \left (12 x^3-6 x^2\right ) \log \left (\frac {10 x-5}{2 x}\right )+e^{3 x} (2 x-1)} \, dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle \int \frac {-4 x^2-6 x^2 (2 x-1) \log \left (5-\frac {5}{2 x}\right )-e^x (1-2 x)^2}{(1-2 x) \left (2 x^2 \log \left (5-\frac {5}{2 x}\right )+e^x\right )^3}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (\frac {2 x-1}{\left (2 x^2 \log \left (5-\frac {5}{2 x}\right )+e^x\right )^2}-\frac {4 x^2 \left (2 x^2 \log \left (5-\frac {5}{2 x}\right )-5 x \log \left (5-\frac {5}{2 x}\right )+2 \log \left (5-\frac {5}{2 x}\right )-1\right )}{(2 x-1) \left (2 x^2 \log \left (5-\frac {5}{2 x}\right )+e^x\right )^3}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \int \frac {1}{\left (2 \log \left (5-\frac {5}{2 x}\right ) x^2+e^x\right )^3}dx+2 \int \frac {x}{\left (2 \log \left (5-\frac {5}{2 x}\right ) x^2+e^x\right )^3}dx+\int \frac {1}{(2 x-1) \left (2 \log \left (5-\frac {5}{2 x}\right ) x^2+e^x\right )^3}dx+8 \int \frac {x^2 \log \left (5-\frac {5}{2 x}\right )}{\left (2 \log \left (5-\frac {5}{2 x}\right ) x^2+e^x\right )^3}dx-\int \frac {1}{\left (2 \log \left (5-\frac {5}{2 x}\right ) x^2+e^x\right )^2}dx+2 \int \frac {x}{\left (2 \log \left (5-\frac {5}{2 x}\right ) x^2+e^x\right )^2}dx-4 \int \frac {x^3 \log \left (5-\frac {5}{2 x}\right )}{\left (2 \log \left (5-\frac {5}{2 x}\right ) x^2+e^x\right )^3}dx\) |
Input:
Int[(4*x^2 + E^x*(1 - 4*x + 4*x^2) + (-6*x^2 + 12*x^3)*Log[(-5 + 10*x)/(2* x)])/(E^(3*x)*(-1 + 2*x) + E^(2*x)*(-6*x^2 + 12*x^3)*Log[(-5 + 10*x)/(2*x) ] + E^x*(-12*x^4 + 24*x^5)*Log[(-5 + 10*x)/(2*x)]^2 + (-8*x^6 + 16*x^7)*Lo g[(-5 + 10*x)/(2*x)]^3),x]
Output:
$Aborted
Time = 4.81 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.81
| method | result | size |
| parallelrisch | \(-\frac {x}{4 x^{4} \ln \left (\frac {-\frac {5}{2}+5 x}{x}\right )^{2}+4 \,{\mathrm e}^{x} x^{2} \ln \left (\frac {-\frac {5}{2}+5 x}{x}\right )+{\mathrm e}^{2 x}}\) | \(47\) |
| risch | \(\frac {x}{\left (\pi \,x^{2} \operatorname {csgn}\left (\frac {i}{x}\right ) \operatorname {csgn}\left (i \left (x -\frac {1}{2}\right )\right ) \operatorname {csgn}\left (\frac {i \left (x -\frac {1}{2}\right )}{x}\right )-\pi \,x^{2} \operatorname {csgn}\left (\frac {i}{x}\right ) \operatorname {csgn}\left (\frac {i \left (x -\frac {1}{2}\right )}{x}\right )^{2}-\pi \,x^{2} \operatorname {csgn}\left (i \left (x -\frac {1}{2}\right )\right ) \operatorname {csgn}\left (\frac {i \left (x -\frac {1}{2}\right )}{x}\right )^{2}+\pi \,x^{2} \operatorname {csgn}\left (\frac {i \left (x -\frac {1}{2}\right )}{x}\right )^{3}+2 i \ln \left (5\right ) x^{2}-2 i x^{2} \ln \left (x \right )+2 i x^{2} \ln \left (x -\frac {1}{2}\right )+i {\mathrm e}^{x}\right )^{2}}\) | \(133\) |
Input:
int(((12*x^3-6*x^2)*ln(1/2*(10*x-5)/x)+(4*x^2-4*x+1)*exp(x)+4*x^2)/((16*x^ 7-8*x^6)*ln(1/2*(10*x-5)/x)^3+(24*x^5-12*x^4)*exp(x)*ln(1/2*(10*x-5)/x)^2+ (12*x^3-6*x^2)*exp(x)^2*ln(1/2*(10*x-5)/x)+(-1+2*x)*exp(x)^3),x,method=_RE TURNVERBOSE)
Output:
-x/(4*x^4*ln(5/2*(-1+2*x)/x)^2+4*exp(x)*x^2*ln(5/2*(-1+2*x)/x)+exp(x)^2)
Time = 0.07 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.77 \[ \int \frac {4 x^2+e^x \left (1-4 x+4 x^2\right )+\left (-6 x^2+12 x^3\right ) \log \left (\frac {-5+10 x}{2 x}\right )}{e^{3 x} (-1+2 x)+e^{2 x} \left (-6 x^2+12 x^3\right ) \log \left (\frac {-5+10 x}{2 x}\right )+e^x \left (-12 x^4+24 x^5\right ) \log ^2\left (\frac {-5+10 x}{2 x}\right )+\left (-8 x^6+16 x^7\right ) \log ^3\left (\frac {-5+10 x}{2 x}\right )} \, dx=-\frac {x}{4 \, x^{4} \log \left (\frac {5 \, {\left (2 \, x - 1\right )}}{2 \, x}\right )^{2} + 4 \, x^{2} e^{x} \log \left (\frac {5 \, {\left (2 \, x - 1\right )}}{2 \, x}\right ) + e^{\left (2 \, x\right )}} \] Input:
integrate(((12*x^3-6*x^2)*log(1/2*(10*x-5)/x)+(4*x^2-4*x+1)*exp(x)+4*x^2)/ ((16*x^7-8*x^6)*log(1/2*(10*x-5)/x)^3+(24*x^5-12*x^4)*exp(x)*log(1/2*(10*x -5)/x)^2+(12*x^3-6*x^2)*exp(x)^2*log(1/2*(10*x-5)/x)+(-1+2*x)*exp(x)^3),x, algorithm="fricas")
Output:
-x/(4*x^4*log(5/2*(2*x - 1)/x)^2 + 4*x^2*e^x*log(5/2*(2*x - 1)/x) + e^(2*x ))
Leaf count of result is larger than twice the leaf count of optimal. 42 vs. \(2 (20) = 40\).
Time = 0.14 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.62 \[ \int \frac {4 x^2+e^x \left (1-4 x+4 x^2\right )+\left (-6 x^2+12 x^3\right ) \log \left (\frac {-5+10 x}{2 x}\right )}{e^{3 x} (-1+2 x)+e^{2 x} \left (-6 x^2+12 x^3\right ) \log \left (\frac {-5+10 x}{2 x}\right )+e^x \left (-12 x^4+24 x^5\right ) \log ^2\left (\frac {-5+10 x}{2 x}\right )+\left (-8 x^6+16 x^7\right ) \log ^3\left (\frac {-5+10 x}{2 x}\right )} \, dx=- \frac {x}{4 x^{4} \log {\left (\frac {5 x - \frac {5}{2}}{x} \right )}^{2} + 4 x^{2} e^{x} \log {\left (\frac {5 x - \frac {5}{2}}{x} \right )} + e^{2 x}} \] Input:
integrate(((12*x**3-6*x**2)*ln(1/2*(10*x-5)/x)+(4*x**2-4*x+1)*exp(x)+4*x** 2)/((16*x**7-8*x**6)*ln(1/2*(10*x-5)/x)**3+(24*x**5-12*x**4)*exp(x)*ln(1/2 *(10*x-5)/x)**2+(12*x**3-6*x**2)*exp(x)**2*ln(1/2*(10*x-5)/x)+(-1+2*x)*exp (x)**3),x)
Output:
-x/(4*x**4*log((5*x - 5/2)/x)**2 + 4*x**2*exp(x)*log((5*x - 5/2)/x) + exp( 2*x))
Leaf count of result is larger than twice the leaf count of optimal. 123 vs. \(2 (23) = 46\).
Time = 0.23 (sec) , antiderivative size = 123, normalized size of antiderivative = 4.73 \[ \int \frac {4 x^2+e^x \left (1-4 x+4 x^2\right )+\left (-6 x^2+12 x^3\right ) \log \left (\frac {-5+10 x}{2 x}\right )}{e^{3 x} (-1+2 x)+e^{2 x} \left (-6 x^2+12 x^3\right ) \log \left (\frac {-5+10 x}{2 x}\right )+e^x \left (-12 x^4+24 x^5\right ) \log ^2\left (\frac {-5+10 x}{2 x}\right )+\left (-8 x^6+16 x^7\right ) \log ^3\left (\frac {-5+10 x}{2 x}\right )} \, dx=-\frac {x}{4 \, x^{4} \log \left (2 \, x - 1\right )^{2} - 8 \, x^{4} {\left (\log \left (5\right ) - \log \left (2\right )\right )} \log \left (x\right ) + 4 \, x^{4} \log \left (x\right )^{2} + 4 \, {\left (\log \left (5\right )^{2} - 2 \, \log \left (5\right ) \log \left (2\right ) + \log \left (2\right )^{2}\right )} x^{4} + 4 \, {\left (x^{2} {\left (\log \left (5\right ) - \log \left (2\right )\right )} - x^{2} \log \left (x\right )\right )} e^{x} + 4 \, {\left (2 \, x^{4} {\left (\log \left (5\right ) - \log \left (2\right )\right )} - 2 \, x^{4} \log \left (x\right ) + x^{2} e^{x}\right )} \log \left (2 \, x - 1\right ) + e^{\left (2 \, x\right )}} \] Input:
integrate(((12*x^3-6*x^2)*log(1/2*(10*x-5)/x)+(4*x^2-4*x+1)*exp(x)+4*x^2)/ ((16*x^7-8*x^6)*log(1/2*(10*x-5)/x)^3+(24*x^5-12*x^4)*exp(x)*log(1/2*(10*x -5)/x)^2+(12*x^3-6*x^2)*exp(x)^2*log(1/2*(10*x-5)/x)+(-1+2*x)*exp(x)^3),x, algorithm="maxima")
Output:
-x/(4*x^4*log(2*x - 1)^2 - 8*x^4*(log(5) - log(2))*log(x) + 4*x^4*log(x)^2 + 4*(log(5)^2 - 2*log(5)*log(2) + log(2)^2)*x^4 + 4*(x^2*(log(5) - log(2) ) - x^2*log(x))*e^x + 4*(2*x^4*(log(5) - log(2)) - 2*x^4*log(x) + x^2*e^x) *log(2*x - 1) + e^(2*x))
Time = 0.28 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.77 \[ \int \frac {4 x^2+e^x \left (1-4 x+4 x^2\right )+\left (-6 x^2+12 x^3\right ) \log \left (\frac {-5+10 x}{2 x}\right )}{e^{3 x} (-1+2 x)+e^{2 x} \left (-6 x^2+12 x^3\right ) \log \left (\frac {-5+10 x}{2 x}\right )+e^x \left (-12 x^4+24 x^5\right ) \log ^2\left (\frac {-5+10 x}{2 x}\right )+\left (-8 x^6+16 x^7\right ) \log ^3\left (\frac {-5+10 x}{2 x}\right )} \, dx=-\frac {x}{4 \, x^{4} \log \left (\frac {5 \, {\left (2 \, x - 1\right )}}{2 \, x}\right )^{2} + 4 \, x^{2} e^{x} \log \left (\frac {5 \, {\left (2 \, x - 1\right )}}{2 \, x}\right ) + e^{\left (2 \, x\right )}} \] Input:
integrate(((12*x^3-6*x^2)*log(1/2*(10*x-5)/x)+(4*x^2-4*x+1)*exp(x)+4*x^2)/ ((16*x^7-8*x^6)*log(1/2*(10*x-5)/x)^3+(24*x^5-12*x^4)*exp(x)*log(1/2*(10*x -5)/x)^2+(12*x^3-6*x^2)*exp(x)^2*log(1/2*(10*x-5)/x)+(-1+2*x)*exp(x)^3),x, algorithm="giac")
Output:
-x/(4*x^4*log(5/2*(2*x - 1)/x)^2 + 4*x^2*e^x*log(5/2*(2*x - 1)/x) + e^(2*x ))
Time = 2.76 (sec) , antiderivative size = 116, normalized size of antiderivative = 4.46 \[ \int \frac {4 x^2+e^x \left (1-4 x+4 x^2\right )+\left (-6 x^2+12 x^3\right ) \log \left (\frac {-5+10 x}{2 x}\right )}{e^{3 x} (-1+2 x)+e^{2 x} \left (-6 x^2+12 x^3\right ) \log \left (\frac {-5+10 x}{2 x}\right )+e^x \left (-12 x^4+24 x^5\right ) \log ^2\left (\frac {-5+10 x}{2 x}\right )+\left (-8 x^6+16 x^7\right ) \log ^3\left (\frac {-5+10 x}{2 x}\right )} \, dx=\frac {{\mathrm {e}}^x\,\left (-4\,x^5+12\,x^4-9\,x^3+2\,x^2\right )+2\,x^4-4\,x^5}{\left ({\mathrm {e}}^{2\,x}+4\,x^4\,{\ln \left (\frac {5\,x-\frac {5}{2}}{x}\right )}^2+4\,x^2\,{\mathrm {e}}^x\,\ln \left (\frac {5\,x-\frac {5}{2}}{x}\right )\right )\,\left (9\,x^2\,{\mathrm {e}}^x-12\,x^3\,{\mathrm {e}}^x+4\,x^4\,{\mathrm {e}}^x-2\,x\,{\mathrm {e}}^x-2\,x^3+4\,x^4\right )} \] Input:
int(-(exp(x)*(4*x^2 - 4*x + 1) - log((5*x - 5/2)/x)*(6*x^2 - 12*x^3) + 4*x ^2)/(log((5*x - 5/2)/x)^3*(8*x^6 - 16*x^7) - exp(3*x)*(2*x - 1) + exp(2*x) *log((5*x - 5/2)/x)*(6*x^2 - 12*x^3) + exp(x)*log((5*x - 5/2)/x)^2*(12*x^4 - 24*x^5)),x)
Output:
(exp(x)*(2*x^2 - 9*x^3 + 12*x^4 - 4*x^5) + 2*x^4 - 4*x^5)/((exp(2*x) + 4*x ^4*log((5*x - 5/2)/x)^2 + 4*x^2*exp(x)*log((5*x - 5/2)/x))*(9*x^2*exp(x) - 12*x^3*exp(x) + 4*x^4*exp(x) - 2*x*exp(x) - 2*x^3 + 4*x^4))
Time = 0.20 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.85 \[ \int \frac {4 x^2+e^x \left (1-4 x+4 x^2\right )+\left (-6 x^2+12 x^3\right ) \log \left (\frac {-5+10 x}{2 x}\right )}{e^{3 x} (-1+2 x)+e^{2 x} \left (-6 x^2+12 x^3\right ) \log \left (\frac {-5+10 x}{2 x}\right )+e^x \left (-12 x^4+24 x^5\right ) \log ^2\left (\frac {-5+10 x}{2 x}\right )+\left (-8 x^6+16 x^7\right ) \log ^3\left (\frac {-5+10 x}{2 x}\right )} \, dx=-\frac {x}{e^{2 x}+4 e^{x} \mathrm {log}\left (\frac {10 x -5}{2 x}\right ) x^{2}+4 \mathrm {log}\left (\frac {10 x -5}{2 x}\right )^{2} x^{4}} \] Input:
int(((12*x^3-6*x^2)*log(1/2*(10*x-5)/x)+(4*x^2-4*x+1)*exp(x)+4*x^2)/((16*x ^7-8*x^6)*log(1/2*(10*x-5)/x)^3+(24*x^5-12*x^4)*exp(x)*log(1/2*(10*x-5)/x) ^2+(12*x^3-6*x^2)*exp(x)^2*log(1/2*(10*x-5)/x)+(-1+2*x)*exp(x)^3),x)
Output:
( - x)/(e**(2*x) + 4*e**x*log((10*x - 5)/(2*x))*x**2 + 4*log((10*x - 5)/(2 *x))**2*x**4)