Integrand size = 176, antiderivative size = 29 \[ \int \frac {e^x \left (-1600 x^2-64 x^5\right )+e^{5 e^{-x}} \left (e^x \left (3200-256 x^3\right )+\left (16000 x+640 x^4\right ) \log \left (\frac {25+x^3}{x}\right )\right )}{e^x \left (25 x^7+x^{10}\right )+e^{5 e^{-x}+x} \left (300 x^5+12 x^8\right ) \log \left (\frac {25+x^3}{x}\right )+e^{10 e^{-x}+x} \left (1200 x^3+48 x^6\right ) \log ^2\left (\frac {25+x^3}{x}\right )+e^{15 e^{-x}+x} \left (1600 x+64 x^4\right ) \log ^3\left (\frac {25+x^3}{x}\right )} \, dx=\frac {16}{\left (x^2+4 e^{5 e^{-x}} \log \left (\frac {25}{x}+x^2\right )\right )^2} \] Output:
16/(4*exp(5/exp(x))*ln(x^2+25/x)+x^2)^2
Time = 0.08 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00 \[ \int \frac {e^x \left (-1600 x^2-64 x^5\right )+e^{5 e^{-x}} \left (e^x \left (3200-256 x^3\right )+\left (16000 x+640 x^4\right ) \log \left (\frac {25+x^3}{x}\right )\right )}{e^x \left (25 x^7+x^{10}\right )+e^{5 e^{-x}+x} \left (300 x^5+12 x^8\right ) \log \left (\frac {25+x^3}{x}\right )+e^{10 e^{-x}+x} \left (1200 x^3+48 x^6\right ) \log ^2\left (\frac {25+x^3}{x}\right )+e^{15 e^{-x}+x} \left (1600 x+64 x^4\right ) \log ^3\left (\frac {25+x^3}{x}\right )} \, dx=\frac {16}{\left (x^2+4 e^{5 e^{-x}} \log \left (\frac {25+x^3}{x}\right )\right )^2} \] Input:
Integrate[(E^x*(-1600*x^2 - 64*x^5) + E^(5/E^x)*(E^x*(3200 - 256*x^3) + (1 6000*x + 640*x^4)*Log[(25 + x^3)/x]))/(E^x*(25*x^7 + x^10) + E^(5/E^x + x) *(300*x^5 + 12*x^8)*Log[(25 + x^3)/x] + E^(10/E^x + x)*(1200*x^3 + 48*x^6) *Log[(25 + x^3)/x]^2 + E^(15/E^x + x)*(1600*x + 64*x^4)*Log[(25 + x^3)/x]^ 3),x]
Output:
16/(x^2 + 4*E^(5/E^x)*Log[(25 + x^3)/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.
\(\displaystyle \int \frac {e^x \left (-64 x^5-1600 x^2\right )+e^{5 e^{-x}} \left (e^x \left (3200-256 x^3\right )+\left (640 x^4+16000 x\right ) \log \left (\frac {x^3+25}{x}\right )\right )}{e^x \left (x^{10}+25 x^7\right )+e^{x+10 e^{-x}} \left (48 x^6+1200 x^3\right ) \log ^2\left (\frac {x^3+25}{x}\right )+e^{x+15 e^{-x}} \left (64 x^4+1600 x\right ) \log ^3\left (\frac {x^3+25}{x}\right )+e^{x+5 e^{-x}} \left (12 x^8+300 x^5\right ) \log \left (\frac {x^3+25}{x}\right )} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {e^{-x} \left (640 e^{5 e^{-x}} x \left (x^3+25\right ) \log \left (\frac {x^3+25}{x}\right )-64 e^x \left (e^{5 e^{-x}} \left (4 x^3-50\right )+\left (x^3+25\right ) x^2\right )\right )}{x \left (x^3+25\right ) \left (4 e^{5 e^{-x}} \log \left (\frac {x^3+25}{x}\right )+x^2\right )^3}dx\) |
\(\Big \downarrow \) 7276 |
\(\displaystyle \int \left (\frac {640 e^{5 e^{-x}-x} \log \left (\frac {x^3+25}{x}\right )}{\left (4 e^{5 e^{-x}} \log \left (\frac {x^3+25}{x}\right )+x^2\right )^3}-\frac {64 \left (x^5+4 e^{5 e^{-x}} x^3+25 x^2-50 e^{5 e^{-x}}\right )}{x \left (x^3+25\right ) \left (4 e^{5 e^{-x}} \log \left (\frac {x^3+25}{x}\right )+x^2\right )^3}\right )dx\) |
\(\Big \downarrow \) 7299 |
\(\displaystyle \int \left (\frac {640 e^{5 e^{-x}-x} \log \left (\frac {x^3+25}{x}\right )}{\left (4 e^{5 e^{-x}} \log \left (\frac {x^3+25}{x}\right )+x^2\right )^3}-\frac {64 \left (x^5+4 e^{5 e^{-x}} x^3+25 x^2-50 e^{5 e^{-x}}\right )}{x \left (x^3+25\right ) \left (4 e^{5 e^{-x}} \log \left (\frac {x^3+25}{x}\right )+x^2\right )^3}\right )dx\) |
Input:
Int[(E^x*(-1600*x^2 - 64*x^5) + E^(5/E^x)*(E^x*(3200 - 256*x^3) + (16000*x + 640*x^4)*Log[(25 + x^3)/x]))/(E^x*(25*x^7 + x^10) + E^(5/E^x + x)*(300* x^5 + 12*x^8)*Log[(25 + x^3)/x] + E^(10/E^x + x)*(1200*x^3 + 48*x^6)*Log[( 25 + x^3)/x]^2 + E^(15/E^x + x)*(1600*x + 64*x^4)*Log[(25 + x^3)/x]^3),x]
Output:
$Aborted
Timed out.
\[\int \frac {\left (\left (640 x^{4}+16000 x \right ) \ln \left (\frac {x^{3}+25}{x}\right )+\left (-256 x^{3}+3200\right ) {\mathrm e}^{x}\right ) {\mathrm e}^{5 \,{\mathrm e}^{-x}}+\left (-64 x^{5}-1600 x^{2}\right ) {\mathrm e}^{x}}{\left (64 x^{4}+1600 x \right ) {\mathrm e}^{x} \ln \left (\frac {x^{3}+25}{x}\right )^{3} {\mathrm e}^{15 \,{\mathrm e}^{-x}}+\left (48 x^{6}+1200 x^{3}\right ) {\mathrm e}^{x} \ln \left (\frac {x^{3}+25}{x}\right )^{2} {\mathrm e}^{10 \,{\mathrm e}^{-x}}+\left (12 x^{8}+300 x^{5}\right ) {\mathrm e}^{x} \ln \left (\frac {x^{3}+25}{x}\right ) {\mathrm e}^{5 \,{\mathrm e}^{-x}}+\left (x^{10}+25 x^{7}\right ) {\mathrm e}^{x}}d x\]
Input:
int((((640*x^4+16000*x)*ln((x^3+25)/x)+(-256*x^3+3200)*exp(x))*exp(5/exp(x ))+(-64*x^5-1600*x^2)*exp(x))/((64*x^4+1600*x)*exp(x)*ln((x^3+25)/x)^3*exp (5/exp(x))^3+(48*x^6+1200*x^3)*exp(x)*ln((x^3+25)/x)^2*exp(5/exp(x))^2+(12 *x^8+300*x^5)*exp(x)*ln((x^3+25)/x)*exp(5/exp(x))+(x^10+25*x^7)*exp(x)),x)
Output:
int((((640*x^4+16000*x)*ln((x^3+25)/x)+(-256*x^3+3200)*exp(x))*exp(5/exp(x ))+(-64*x^5-1600*x^2)*exp(x))/((64*x^4+1600*x)*exp(x)*ln((x^3+25)/x)^3*exp (5/exp(x))^3+(48*x^6+1200*x^3)*exp(x)*ln((x^3+25)/x)^2*exp(5/exp(x))^2+(12 *x^8+300*x^5)*exp(x)*ln((x^3+25)/x)*exp(5/exp(x))+(x^10+25*x^7)*exp(x)),x)
Leaf count of result is larger than twice the leaf count of optimal. 73 vs. \(2 (27) = 54\).
Time = 0.09 (sec) , antiderivative size = 73, normalized size of antiderivative = 2.52 \[ \int \frac {e^x \left (-1600 x^2-64 x^5\right )+e^{5 e^{-x}} \left (e^x \left (3200-256 x^3\right )+\left (16000 x+640 x^4\right ) \log \left (\frac {25+x^3}{x}\right )\right )}{e^x \left (25 x^7+x^{10}\right )+e^{5 e^{-x}+x} \left (300 x^5+12 x^8\right ) \log \left (\frac {25+x^3}{x}\right )+e^{10 e^{-x}+x} \left (1200 x^3+48 x^6\right ) \log ^2\left (\frac {25+x^3}{x}\right )+e^{15 e^{-x}+x} \left (1600 x+64 x^4\right ) \log ^3\left (\frac {25+x^3}{x}\right )} \, dx=\frac {16 \, e^{\left (2 \, x\right )}}{x^{4} e^{\left (2 \, x\right )} + 8 \, x^{2} e^{\left ({\left (x e^{x} + 5\right )} e^{\left (-x\right )} + x\right )} \log \left (\frac {x^{3} + 25}{x}\right ) + 16 \, e^{\left (2 \, {\left (x e^{x} + 5\right )} e^{\left (-x\right )}\right )} \log \left (\frac {x^{3} + 25}{x}\right )^{2}} \] Input:
integrate((((640*x^4+16000*x)*log((x^3+25)/x)+(-256*x^3+3200)*exp(x))*exp( 5/exp(x))+(-64*x^5-1600*x^2)*exp(x))/((64*x^4+1600*x)*exp(x)*log((x^3+25)/ x)^3*exp(5/exp(x))^3+(48*x^6+1200*x^3)*exp(x)*log((x^3+25)/x)^2*exp(5/exp( x))^2+(12*x^8+300*x^5)*exp(x)*log((x^3+25)/x)*exp(5/exp(x))+(x^10+25*x^7)* exp(x)),x, algorithm="fricas")
Output:
16*e^(2*x)/(x^4*e^(2*x) + 8*x^2*e^((x*e^x + 5)*e^(-x) + x)*log((x^3 + 25)/ x) + 16*e^(2*(x*e^x + 5)*e^(-x))*log((x^3 + 25)/x)^2)
Time = 0.33 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.45 \[ \int \frac {e^x \left (-1600 x^2-64 x^5\right )+e^{5 e^{-x}} \left (e^x \left (3200-256 x^3\right )+\left (16000 x+640 x^4\right ) \log \left (\frac {25+x^3}{x}\right )\right )}{e^x \left (25 x^7+x^{10}\right )+e^{5 e^{-x}+x} \left (300 x^5+12 x^8\right ) \log \left (\frac {25+x^3}{x}\right )+e^{10 e^{-x}+x} \left (1200 x^3+48 x^6\right ) \log ^2\left (\frac {25+x^3}{x}\right )+e^{15 e^{-x}+x} \left (1600 x+64 x^4\right ) \log ^3\left (\frac {25+x^3}{x}\right )} \, dx=\frac {16}{x^{4} + 8 x^{2} e^{5 e^{- x}} \log {\left (\frac {x^{3} + 25}{x} \right )} + 16 e^{10 e^{- x}} \log {\left (\frac {x^{3} + 25}{x} \right )}^{2}} \] Input:
integrate((((640*x**4+16000*x)*ln((x**3+25)/x)+(-256*x**3+3200)*exp(x))*ex p(5/exp(x))+(-64*x**5-1600*x**2)*exp(x))/((64*x**4+1600*x)*exp(x)*ln((x**3 +25)/x)**3*exp(5/exp(x))**3+(48*x**6+1200*x**3)*exp(x)*ln((x**3+25)/x)**2* exp(5/exp(x))**2+(12*x**8+300*x**5)*exp(x)*ln((x**3+25)/x)*exp(5/exp(x))+( x**10+25*x**7)*exp(x)),x)
Output:
16/(x**4 + 8*x**2*exp(5*exp(-x))*log((x**3 + 25)/x) + 16*exp(10*exp(-x))*l og((x**3 + 25)/x)**2)
Leaf count of result is larger than twice the leaf count of optimal. 67 vs. \(2 (27) = 54\).
Time = 1.23 (sec) , antiderivative size = 67, normalized size of antiderivative = 2.31 \[ \int \frac {e^x \left (-1600 x^2-64 x^5\right )+e^{5 e^{-x}} \left (e^x \left (3200-256 x^3\right )+\left (16000 x+640 x^4\right ) \log \left (\frac {25+x^3}{x}\right )\right )}{e^x \left (25 x^7+x^{10}\right )+e^{5 e^{-x}+x} \left (300 x^5+12 x^8\right ) \log \left (\frac {25+x^3}{x}\right )+e^{10 e^{-x}+x} \left (1200 x^3+48 x^6\right ) \log ^2\left (\frac {25+x^3}{x}\right )+e^{15 e^{-x}+x} \left (1600 x+64 x^4\right ) \log ^3\left (\frac {25+x^3}{x}\right )} \, dx=\frac {16}{x^{4} + 16 \, {\left (\log \left (x^{3} + 25\right )^{2} - 2 \, \log \left (x^{3} + 25\right ) \log \left (x\right ) + \log \left (x\right )^{2}\right )} e^{\left (10 \, e^{\left (-x\right )}\right )} + 8 \, {\left (x^{2} \log \left (x^{3} + 25\right ) - x^{2} \log \left (x\right )\right )} e^{\left (5 \, e^{\left (-x\right )}\right )}} \] Input:
integrate((((640*x^4+16000*x)*log((x^3+25)/x)+(-256*x^3+3200)*exp(x))*exp( 5/exp(x))+(-64*x^5-1600*x^2)*exp(x))/((64*x^4+1600*x)*exp(x)*log((x^3+25)/ x)^3*exp(5/exp(x))^3+(48*x^6+1200*x^3)*exp(x)*log((x^3+25)/x)^2*exp(5/exp( x))^2+(12*x^8+300*x^5)*exp(x)*log((x^3+25)/x)*exp(5/exp(x))+(x^10+25*x^7)* exp(x)),x, algorithm="maxima")
Output:
16/(x^4 + 16*(log(x^3 + 25)^2 - 2*log(x^3 + 25)*log(x) + log(x)^2)*e^(10*e ^(-x)) + 8*(x^2*log(x^3 + 25) - x^2*log(x))*e^(5*e^(-x)))
Timed out. \[ \int \frac {e^x \left (-1600 x^2-64 x^5\right )+e^{5 e^{-x}} \left (e^x \left (3200-256 x^3\right )+\left (16000 x+640 x^4\right ) \log \left (\frac {25+x^3}{x}\right )\right )}{e^x \left (25 x^7+x^{10}\right )+e^{5 e^{-x}+x} \left (300 x^5+12 x^8\right ) \log \left (\frac {25+x^3}{x}\right )+e^{10 e^{-x}+x} \left (1200 x^3+48 x^6\right ) \log ^2\left (\frac {25+x^3}{x}\right )+e^{15 e^{-x}+x} \left (1600 x+64 x^4\right ) \log ^3\left (\frac {25+x^3}{x}\right )} \, dx=\text {Timed out} \] Input:
integrate((((640*x^4+16000*x)*log((x^3+25)/x)+(-256*x^3+3200)*exp(x))*exp( 5/exp(x))+(-64*x^5-1600*x^2)*exp(x))/((64*x^4+1600*x)*exp(x)*log((x^3+25)/ x)^3*exp(5/exp(x))^3+(48*x^6+1200*x^3)*exp(x)*log((x^3+25)/x)^2*exp(5/exp( x))^2+(12*x^8+300*x^5)*exp(x)*log((x^3+25)/x)*exp(5/exp(x))+(x^10+25*x^7)* exp(x)),x, algorithm="giac")
Output:
Timed out
Time = 3.01 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.76 \[ \int \frac {e^x \left (-1600 x^2-64 x^5\right )+e^{5 e^{-x}} \left (e^x \left (3200-256 x^3\right )+\left (16000 x+640 x^4\right ) \log \left (\frac {25+x^3}{x}\right )\right )}{e^x \left (25 x^7+x^{10}\right )+e^{5 e^{-x}+x} \left (300 x^5+12 x^8\right ) \log \left (\frac {25+x^3}{x}\right )+e^{10 e^{-x}+x} \left (1200 x^3+48 x^6\right ) \log ^2\left (\frac {25+x^3}{x}\right )+e^{15 e^{-x}+x} \left (1600 x+64 x^4\right ) \log ^3\left (\frac {25+x^3}{x}\right )} \, dx=\frac {16}{16\,{\mathrm {e}}^{10\,{\mathrm {e}}^{-x}}\,{\ln \left (\frac {x^3+25}{x}\right )}^2+x^4+8\,x^2\,{\mathrm {e}}^{5\,{\mathrm {e}}^{-x}}\,\ln \left (\frac {x^3+25}{x}\right )} \] Input:
int(-(exp(x)*(1600*x^2 + 64*x^5) - exp(5*exp(-x))*(log((x^3 + 25)/x)*(1600 0*x + 640*x^4) - exp(x)*(256*x^3 - 3200)))/(exp(x)*(25*x^7 + x^10) + exp(1 0*exp(-x))*exp(x)*log((x^3 + 25)/x)^2*(1200*x^3 + 48*x^6) + exp(5*exp(-x)) *exp(x)*log((x^3 + 25)/x)*(300*x^5 + 12*x^8) + exp(15*exp(-x))*exp(x)*log( (x^3 + 25)/x)^3*(1600*x + 64*x^4)),x)
Output:
16/(16*exp(10*exp(-x))*log((x^3 + 25)/x)^2 + x^4 + 8*x^2*exp(5*exp(-x))*lo g((x^3 + 25)/x))
Time = 0.25 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.90 \[ \int \frac {e^x \left (-1600 x^2-64 x^5\right )+e^{5 e^{-x}} \left (e^x \left (3200-256 x^3\right )+\left (16000 x+640 x^4\right ) \log \left (\frac {25+x^3}{x}\right )\right )}{e^x \left (25 x^7+x^{10}\right )+e^{5 e^{-x}+x} \left (300 x^5+12 x^8\right ) \log \left (\frac {25+x^3}{x}\right )+e^{10 e^{-x}+x} \left (1200 x^3+48 x^6\right ) \log ^2\left (\frac {25+x^3}{x}\right )+e^{15 e^{-x}+x} \left (1600 x+64 x^4\right ) \log ^3\left (\frac {25+x^3}{x}\right )} \, dx=\frac {16}{16 e^{\frac {10}{e^{x}}} \mathrm {log}\left (\frac {x^{3}+25}{x}\right )^{2}+8 e^{\frac {5}{e^{x}}} \mathrm {log}\left (\frac {x^{3}+25}{x}\right ) x^{2}+x^{4}} \] Input:
int((((640*x^4+16000*x)*log((x^3+25)/x)+(-256*x^3+3200)*exp(x))*exp(5/exp( x))+(-64*x^5-1600*x^2)*exp(x))/((64*x^4+1600*x)*exp(x)*log((x^3+25)/x)^3*e xp(5/exp(x))^3+(48*x^6+1200*x^3)*exp(x)*log((x^3+25)/x)^2*exp(5/exp(x))^2+ (12*x^8+300*x^5)*exp(x)*log((x^3+25)/x)*exp(5/exp(x))+(x^10+25*x^7)*exp(x) ),x)
Output:
16/(16*e**(10/e**x)*log((x**3 + 25)/x)**2 + 8*e**(5/e**x)*log((x**3 + 25)/ x)*x**2 + x**4)