Integrand size = 89, antiderivative size = 26 \[ \int \frac {e^{\frac {e^{-\frac {5 x}{3 x+15 \log (4)}} (4+8 x)}{x}-\frac {5 x}{3 x+15 \log (4)}} \left (-12 x^2+\left (-220 x-200 x^2\right ) \log (4)-300 \log ^2(4)\right )}{3 x^4+30 x^3 \log (4)+75 x^2 \log ^2(4)} \, dx=e^{2 e^{-\frac {5 x}{3 (x+5 \log (4))}} \left (4+\frac {2}{x}\right )} \] Output:
exp(2/exp(x/(3/5*x+6*ln(2)))*(2/x+4))
Time = 1.16 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.23 \[ \int \frac {e^{\frac {e^{-\frac {5 x}{3 x+15 \log (4)}} (4+8 x)}{x}-\frac {5 x}{3 x+15 \log (4)}} \left (-12 x^2+\left (-220 x-200 x^2\right ) \log (4)-300 \log ^2(4)\right )}{3 x^4+30 x^3 \log (4)+75 x^2 \log ^2(4)} \, dx=e^{\frac {2^{2+\frac {50}{3 (x+5 \log (4))}} (1+2 x)}{e^{5/3} x}} \] Input:
Integrate[(E^((4 + 8*x)/(E^((5*x)/(3*x + 15*Log[4]))*x) - (5*x)/(3*x + 15* Log[4]))*(-12*x^2 + (-220*x - 200*x^2)*Log[4] - 300*Log[4]^2))/(3*x^4 + 30 *x^3*Log[4] + 75*x^2*Log[4]^2),x]
Output:
E^((2^(2 + 50/(3*(x + 5*Log[4])))*(1 + 2*x))/(E^(5/3)*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.
| \(\displaystyle \int \frac {\left (-12 x^2+\left (-200 x^2-220 x\right ) \log (4)-300 \log ^2(4)\right ) \exp \left (\frac {(8 x+4) e^{-\frac {5 x}{3 x+15 \log (4)}}}{x}-\frac {5 x}{3 x+15 \log (4)}\right )}{3 x^4+30 x^3 \log (4)+75 x^2 \log ^2(4)} \, dx\) |
| \(\Big \downarrow \) 2026 |
| \(\displaystyle \int \frac {\left (-12 x^2+\left (-200 x^2-220 x\right ) \log (4)-300 \log ^2(4)\right ) \exp \left (\frac {(8 x+4) e^{-\frac {5 x}{3 x+15 \log (4)}}}{x}-\frac {5 x}{3 x+15 \log (4)}\right )}{x^2 \left (3 x^2+30 x \log (4)+75 \log ^2(4)\right )}dx\) |
| \(\Big \downarrow \) 2007 |
| \(\displaystyle \int \frac {\left (-12 x^2+\left (-200 x^2-220 x\right ) \log (4)-300 \log ^2(4)\right ) \exp \left (\frac {(8 x+4) e^{-\frac {5 x}{3 x+15 \log (4)}}}{x}-\frac {5 x}{3 x+15 \log (4)}\right )}{x^2 \left (\sqrt {3} x+5 \sqrt {3} \log (4)\right )^2}dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle \int \frac {\left (-4 x^2 (3+50 \log (4))-220 x \log (4)-300 \log ^2(4)\right ) \exp \left (\frac {(8 x+4) e^{-\frac {5 x}{3 x+15 \log (4)}}}{x}-\frac {5 x}{3 x+15 \log (4)}\right )}{x^2 \left (\sqrt {3} x+5 \sqrt {3} \log (4)\right )^2}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (-\frac {4 \exp \left (\frac {(8 x+4) e^{-\frac {5 x}{3 x+15 \log (4)}}}{x}-\frac {5 x}{3 x+15 \log (4)}\right )}{x^2}+\frac {4 \exp \left (\frac {(8 x+4) e^{-\frac {5 x}{3 x+15 \log (4)}}}{x}-\frac {5 x}{3 x+15 \log (4)}\right )}{3 \log (4) (x+5 \log (4))}-\frac {20 (10 \log (4)-1) \exp \left (\frac {(8 x+4) e^{-\frac {5 x}{3 x+15 \log (4)}}}{x}-\frac {5 x}{3 x+15 \log (4)}\right )}{3 (x+5 \log (4))^2}-\frac {4 \exp \left (\frac {(8 x+4) e^{-\frac {5 x}{3 x+15 \log (4)}}}{x}-\frac {5 x}{3 x+15 \log (4)}\right )}{x \log (64)}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -4 \int \frac {\exp \left (\frac {e^{-\frac {5 x}{3 x+15 \log (4)}} (8 x+4)}{x}-\frac {5 x}{3 x+15 \log (4)}\right )}{x^2}dx-\frac {4 \int \frac {\exp \left (\frac {e^{-\frac {5 x}{3 x+15 \log (4)}} (8 x+4)}{x}-\frac {5 x}{3 x+15 \log (4)}\right )}{x}dx}{\log (64)}+\frac {20}{3} (1-10 \log (4)) \int \frac {\exp \left (\frac {e^{-\frac {5 x}{3 x+15 \log (4)}} (8 x+4)}{x}-\frac {5 x}{3 x+15 \log (4)}\right )}{(x+5 \log (4))^2}dx+\frac {4 \int \frac {\exp \left (\frac {e^{-\frac {5 x}{3 x+15 \log (4)}} (8 x+4)}{x}-\frac {5 x}{3 x+15 \log (4)}\right )}{x+5 \log (4)}dx}{3 \log (4)}\) |
Input:
Int[(E^((4 + 8*x)/(E^((5*x)/(3*x + 15*Log[4]))*x) - (5*x)/(3*x + 15*Log[4] ))*(-12*x^2 + (-220*x - 200*x^2)*Log[4] - 300*Log[4]^2))/(3*x^4 + 30*x^3*L og[4] + 75*x^2*Log[4]^2),x]
Output:
$Aborted
Time = 4.82 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92
| method | result | size |
| risch | \({\mathrm e}^{\frac {4 \left (1+2 x \right ) {\mathrm e}^{-\frac {5 x}{3 \left (10 \ln \left (2\right )+x \right )}}}{x}}\) | \(24\) |
| parallelrisch | \(\frac {30 \ln \left (2\right ) {\mathrm e}^{\frac {4 \left (1+2 x \right ) {\mathrm e}^{-\frac {5 x}{3 \left (10 \ln \left (2\right )+x \right )}}}{x}}+3 \,{\mathrm e}^{\frac {4 \left (1+2 x \right ) {\mathrm e}^{-\frac {5 x}{3 \left (10 \ln \left (2\right )+x \right )}}}{x}} x}{30 \ln \left (2\right )+3 x}\) | \(69\) |
| norman | \(\frac {\left (x^{2} {\mathrm e}^{\frac {5 x}{30 \ln \left (2\right )+3 x}} {\mathrm e}^{\frac {\left (8 x +4\right ) {\mathrm e}^{-\frac {5 x}{30 \ln \left (2\right )+3 x}}}{x}}+10 x \ln \left (2\right ) {\mathrm e}^{\frac {5 x}{30 \ln \left (2\right )+3 x}} {\mathrm e}^{\frac {\left (8 x +4\right ) {\mathrm e}^{-\frac {5 x}{30 \ln \left (2\right )+3 x}}}{x}}\right ) {\mathrm e}^{-\frac {5 x}{30 \ln \left (2\right )+3 x}}}{x \left (10 \ln \left (2\right )+x \right )}\) | \(119\) |
Input:
int((-1200*ln(2)^2+2*(-200*x^2-220*x)*ln(2)-12*x^2)*exp((8*x+4)/x/exp(5*x/ (30*ln(2)+3*x)))/(300*x^2*ln(2)^2+60*x^3*ln(2)+3*x^4)/exp(5*x/(30*ln(2)+3* x)),x,method=_RETURNVERBOSE)
Output:
exp(4*(1+2*x)*exp(-5/3*x/(10*ln(2)+x))/x)
Leaf count of result is larger than twice the leaf count of optimal. 62 vs. \(2 (20) = 40\).
Time = 0.09 (sec) , antiderivative size = 62, normalized size of antiderivative = 2.38 \[ \int \frac {e^{\frac {e^{-\frac {5 x}{3 x+15 \log (4)}} (4+8 x)}{x}-\frac {5 x}{3 x+15 \log (4)}} \left (-12 x^2+\left (-220 x-200 x^2\right ) \log (4)-300 \log ^2(4)\right )}{3 x^4+30 x^3 \log (4)+75 x^2 \log ^2(4)} \, dx=e^{\left (-\frac {5 \, x^{2} - 12 \, {\left (2 \, x^{2} + 10 \, {\left (2 \, x + 1\right )} \log \left (2\right ) + x\right )} e^{\left (-\frac {5 \, x}{3 \, {\left (x + 10 \, \log \left (2\right )\right )}}\right )}}{3 \, {\left (x^{2} + 10 \, x \log \left (2\right )\right )}} + \frac {5 \, x}{3 \, {\left (x + 10 \, \log \left (2\right )\right )}}\right )} \] Input:
integrate((-1200*log(2)^2+2*(-200*x^2-220*x)*log(2)-12*x^2)*exp((8*x+4)/x/ exp(5*x/(30*log(2)+3*x)))/(300*x^2*log(2)^2+60*x^3*log(2)+3*x^4)/exp(5*x/( 30*log(2)+3*x)),x, algorithm="fricas")
Output:
e^(-1/3*(5*x^2 - 12*(2*x^2 + 10*(2*x + 1)*log(2) + x)*e^(-5/3*x/(x + 10*lo g(2))))/(x^2 + 10*x*log(2)) + 5/3*x/(x + 10*log(2)))
Time = 0.21 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.77 \[ \int \frac {e^{\frac {e^{-\frac {5 x}{3 x+15 \log (4)}} (4+8 x)}{x}-\frac {5 x}{3 x+15 \log (4)}} \left (-12 x^2+\left (-220 x-200 x^2\right ) \log (4)-300 \log ^2(4)\right )}{3 x^4+30 x^3 \log (4)+75 x^2 \log ^2(4)} \, dx=e^{\frac {\left (8 x + 4\right ) e^{- \frac {5 x}{3 x + 30 \log {\left (2 \right )}}}}{x}} \] Input:
integrate((-1200*ln(2)**2+2*(-200*x**2-220*x)*ln(2)-12*x**2)*exp((8*x+4)/x /exp(5*x/(30*ln(2)+3*x)))/(300*x**2*ln(2)**2+60*x**3*ln(2)+3*x**4)/exp(5*x /(30*ln(2)+3*x)),x)
Output:
exp((8*x + 4)*exp(-5*x/(3*x + 30*log(2)))/x)
Time = 0.33 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.50 \[ \int \frac {e^{\frac {e^{-\frac {5 x}{3 x+15 \log (4)}} (4+8 x)}{x}-\frac {5 x}{3 x+15 \log (4)}} \left (-12 x^2+\left (-220 x-200 x^2\right ) \log (4)-300 \log ^2(4)\right )}{3 x^4+30 x^3 \log (4)+75 x^2 \log ^2(4)} \, dx=e^{\left (\frac {4 \, e^{\left (\frac {50 \, \log \left (2\right )}{3 \, {\left (x + 10 \, \log \left (2\right )\right )}} - \frac {5}{3}\right )}}{x} + 8 \, e^{\left (\frac {50 \, \log \left (2\right )}{3 \, {\left (x + 10 \, \log \left (2\right )\right )}} - \frac {5}{3}\right )}\right )} \] Input:
integrate((-1200*log(2)^2+2*(-200*x^2-220*x)*log(2)-12*x^2)*exp((8*x+4)/x/ exp(5*x/(30*log(2)+3*x)))/(300*x^2*log(2)^2+60*x^3*log(2)+3*x^4)/exp(5*x/( 30*log(2)+3*x)),x, algorithm="maxima")
Output:
e^(4*e^(50/3*log(2)/(x + 10*log(2)) - 5/3)/x + 8*e^(50/3*log(2)/(x + 10*lo g(2)) - 5/3))
\[ \int \frac {e^{\frac {e^{-\frac {5 x}{3 x+15 \log (4)}} (4+8 x)}{x}-\frac {5 x}{3 x+15 \log (4)}} \left (-12 x^2+\left (-220 x-200 x^2\right ) \log (4)-300 \log ^2(4)\right )}{3 x^4+30 x^3 \log (4)+75 x^2 \log ^2(4)} \, dx=\int { -\frac {4 \, {\left (3 \, x^{2} + 10 \, {\left (10 \, x^{2} + 11 \, x\right )} \log \left (2\right ) + 300 \, \log \left (2\right )^{2}\right )} e^{\left (\frac {4 \, {\left (2 \, x + 1\right )} e^{\left (-\frac {5 \, x}{3 \, {\left (x + 10 \, \log \left (2\right )\right )}}\right )}}{x} - \frac {5 \, x}{3 \, {\left (x + 10 \, \log \left (2\right )\right )}}\right )}}{3 \, {\left (x^{4} + 20 \, x^{3} \log \left (2\right ) + 100 \, x^{2} \log \left (2\right )^{2}\right )}} \,d x } \] Input:
integrate((-1200*log(2)^2+2*(-200*x^2-220*x)*log(2)-12*x^2)*exp((8*x+4)/x/ exp(5*x/(30*log(2)+3*x)))/(300*x^2*log(2)^2+60*x^3*log(2)+3*x^4)/exp(5*x/( 30*log(2)+3*x)),x, algorithm="giac")
Output:
integrate(-4/3*(3*x^2 + 10*(10*x^2 + 11*x)*log(2) + 300*log(2)^2)*e^(4*(2* x + 1)*e^(-5/3*x/(x + 10*log(2)))/x - 5/3*x/(x + 10*log(2)))/(x^4 + 20*x^3 *log(2) + 100*x^2*log(2)^2), x)
Time = 3.20 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.46 \[ \int \frac {e^{\frac {e^{-\frac {5 x}{3 x+15 \log (4)}} (4+8 x)}{x}-\frac {5 x}{3 x+15 \log (4)}} \left (-12 x^2+\left (-220 x-200 x^2\right ) \log (4)-300 \log ^2(4)\right )}{3 x^4+30 x^3 \log (4)+75 x^2 \log ^2(4)} \, dx={\mathrm {e}}^{\frac {4\,{\mathrm {e}}^{-\frac {5\,x}{3\,x+30\,\ln \left (2\right )}}}{x}}\,{\mathrm {e}}^{8\,{\mathrm {e}}^{-\frac {5\,x}{3\,x+30\,\ln \left (2\right )}}} \] Input:
int(-(exp((exp(-(5*x)/(3*x + 30*log(2)))*(8*x + 4))/x)*exp(-(5*x)/(3*x + 3 0*log(2)))*(2*log(2)*(220*x + 200*x^2) + 1200*log(2)^2 + 12*x^2))/(300*x^2 *log(2)^2 + 60*x^3*log(2) + 3*x^4),x)
Output:
exp((4*exp(-(5*x)/(3*x + 30*log(2))))/x)*exp(8*exp(-(5*x)/(3*x + 30*log(2) )))
\[ \int \frac {e^{\frac {e^{-\frac {5 x}{3 x+15 \log (4)}} (4+8 x)}{x}-\frac {5 x}{3 x+15 \log (4)}} \left (-12 x^2+\left (-220 x-200 x^2\right ) \log (4)-300 \log ^2(4)\right )}{3 x^4+30 x^3 \log (4)+75 x^2 \log ^2(4)} \, dx=-400 \left (\int \frac {e^{\frac {8 x +4}{e^{\frac {5 x}{30 \,\mathrm {log}\left (2\right )+3 x}} x}}}{100 e^{\frac {5 x}{30 \,\mathrm {log}\left (2\right )+3 x}} \mathrm {log}\left (2\right )^{2} x^{2}+20 e^{\frac {5 x}{30 \,\mathrm {log}\left (2\right )+3 x}} \mathrm {log}\left (2\right ) x^{3}+e^{\frac {5 x}{30 \,\mathrm {log}\left (2\right )+3 x}} x^{4}}d x \right ) \mathrm {log}\left (2\right )^{2}-\frac {440 \left (\int \frac {e^{\frac {8 x +4}{e^{\frac {5 x}{30 \,\mathrm {log}\left (2\right )+3 x}} x}}}{100 e^{\frac {5 x}{30 \,\mathrm {log}\left (2\right )+3 x}} \mathrm {log}\left (2\right )^{2} x +20 e^{\frac {5 x}{30 \,\mathrm {log}\left (2\right )+3 x}} \mathrm {log}\left (2\right ) x^{2}+e^{\frac {5 x}{30 \,\mathrm {log}\left (2\right )+3 x}} x^{3}}d x \right ) \mathrm {log}\left (2\right )}{3}-\frac {400 \left (\int \frac {e^{\frac {8 x +4}{e^{\frac {5 x}{30 \,\mathrm {log}\left (2\right )+3 x}} x}}}{100 e^{\frac {5 x}{30 \,\mathrm {log}\left (2\right )+3 x}} \mathrm {log}\left (2\right )^{2}+20 e^{\frac {5 x}{30 \,\mathrm {log}\left (2\right )+3 x}} \mathrm {log}\left (2\right ) x +e^{\frac {5 x}{30 \,\mathrm {log}\left (2\right )+3 x}} x^{2}}d x \right ) \mathrm {log}\left (2\right )}{3}-4 \left (\int \frac {e^{\frac {8 x +4}{e^{\frac {5 x}{30 \,\mathrm {log}\left (2\right )+3 x}} x}}}{100 e^{\frac {5 x}{30 \,\mathrm {log}\left (2\right )+3 x}} \mathrm {log}\left (2\right )^{2}+20 e^{\frac {5 x}{30 \,\mathrm {log}\left (2\right )+3 x}} \mathrm {log}\left (2\right ) x +e^{\frac {5 x}{30 \,\mathrm {log}\left (2\right )+3 x}} x^{2}}d x \right ) \] Input:
int((-1200*log(2)^2+2*(-200*x^2-220*x)*log(2)-12*x^2)*exp((8*x+4)/x/exp(5* x/(30*log(2)+3*x)))/(300*x^2*log(2)^2+60*x^3*log(2)+3*x^4)/exp(5*x/(30*log (2)+3*x)),x)
Output:
(4*( - 300*int(e**((8*x + 4)/(e**((5*x)/(30*log(2) + 3*x))*x))/(100*e**((5 *x)/(30*log(2) + 3*x))*log(2)**2*x**2 + 20*e**((5*x)/(30*log(2) + 3*x))*lo g(2)*x**3 + e**((5*x)/(30*log(2) + 3*x))*x**4),x)*log(2)**2 - 110*int(e**( (8*x + 4)/(e**((5*x)/(30*log(2) + 3*x))*x))/(100*e**((5*x)/(30*log(2) + 3* x))*log(2)**2*x + 20*e**((5*x)/(30*log(2) + 3*x))*log(2)*x**2 + e**((5*x)/ (30*log(2) + 3*x))*x**3),x)*log(2) - 100*int(e**((8*x + 4)/(e**((5*x)/(30* log(2) + 3*x))*x))/(100*e**((5*x)/(30*log(2) + 3*x))*log(2)**2 + 20*e**((5 *x)/(30*log(2) + 3*x))*log(2)*x + e**((5*x)/(30*log(2) + 3*x))*x**2),x)*lo g(2) - 3*int(e**((8*x + 4)/(e**((5*x)/(30*log(2) + 3*x))*x))/(100*e**((5*x )/(30*log(2) + 3*x))*log(2)**2 + 20*e**((5*x)/(30*log(2) + 3*x))*log(2)*x + e**((5*x)/(30*log(2) + 3*x))*x**2),x)))/3