Integrand size = 104, antiderivative size = 25 \[ \int \frac {-3 e^4 x^2+2 x^4-x^4 \log \left (\frac {64}{25 x}\right )}{e^8+9 x^4+6 x^5+x^6+e^4 \left (-6 x^2-2 x^3\right )+\left (2 e^4 x^2-6 x^4-2 x^5\right ) \log \left (\frac {64}{25 x}\right )+x^4 \log ^2\left (\frac {64}{25 x}\right )} \, dx=\frac {x}{3-\frac {e^4}{x^2}+x-\log \left (\frac {64}{25 x}\right )} \] Output:
x/(3+x-exp(2)^2/x^2-ln(64/25/x))
Time = 0.57 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.32 \[ \int \frac {-3 e^4 x^2+2 x^4-x^4 \log \left (\frac {64}{25 x}\right )}{e^8+9 x^4+6 x^5+x^6+e^4 \left (-6 x^2-2 x^3\right )+\left (2 e^4 x^2-6 x^4-2 x^5\right ) \log \left (\frac {64}{25 x}\right )+x^4 \log ^2\left (\frac {64}{25 x}\right )} \, dx=-\frac {x^3}{e^4-3 x^2-x^3+x^2 \log \left (\frac {64}{25 x}\right )} \] Input:
Integrate[(-3*E^4*x^2 + 2*x^4 - x^4*Log[64/(25*x)])/(E^8 + 9*x^4 + 6*x^5 + x^6 + E^4*(-6*x^2 - 2*x^3) + (2*E^4*x^2 - 6*x^4 - 2*x^5)*Log[64/(25*x)] + x^4*Log[64/(25*x)]^2),x]
Output:
-(x^3/(E^4 - 3*x^2 - x^3 + x^2*Log[64/(25*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.
\(\displaystyle \int \frac {2 x^4+x^4 \left (-\log \left (\frac {64}{25 x}\right )\right )-3 e^4 x^2}{x^6+6 x^5+9 x^4+x^4 \log ^2\left (\frac {64}{25 x}\right )+e^4 \left (-2 x^3-6 x^2\right )+\left (-2 x^5-6 x^4+2 e^4 x^2\right ) \log \left (\frac {64}{25 x}\right )+e^8} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {2 x^4+x^4 \left (-\log \left (\frac {64}{25 x}\right )\right )-3 e^4 x^2}{\left (-\left ((x+3) x^2\right )+x^2 \log \left (\frac {64}{25 x}\right )+e^4\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {x^2}{x^3+3 x^2-x^2 \log \left (\frac {64}{25 x}\right )-e^4}-\frac {x^2 \left (x^3+x^2+2 e^4\right )}{\left (-x^3-3 x^2+x^2 \log \left (\frac {64}{25 x}\right )+e^4\right )^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \int \frac {x^2}{x^3-\log \left (\frac {64}{25 x}\right ) x^2+3 x^2-e^4}dx-2 e^4 \int \frac {x^2}{\left (-x^3+\log \left (\frac {64}{25 x}\right ) x^2-3 x^2+e^4\right )^2}dx-\int \frac {x^5}{\left (-x^3+\log \left (\frac {64}{25 x}\right ) x^2-3 x^2+e^4\right )^2}dx-\int \frac {x^4}{\left (-x^3+\log \left (\frac {64}{25 x}\right ) x^2-3 x^2+e^4\right )^2}dx\) |
Input:
Int[(-3*E^4*x^2 + 2*x^4 - x^4*Log[64/(25*x)])/(E^8 + 9*x^4 + 6*x^5 + x^6 + E^4*(-6*x^2 - 2*x^3) + (2*E^4*x^2 - 6*x^4 - 2*x^5)*Log[64/(25*x)] + x^4*L og[64/(25*x)]^2),x]
Output:
$Aborted
Time = 1.04 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.24
method | result | size |
risch | \(-\frac {x^{3}}{x^{2} \ln \left (\frac {64}{25 x}\right )-x^{3}+{\mathrm e}^{4}-3 x^{2}}\) | \(31\) |
derivativedivides | \(-\frac {262144}{\frac {262144 \,{\mathrm e}^{4}}{x^{3}}+\frac {262144 \ln \left (\frac {64}{25 x}\right )}{x}-\frac {786432}{x}-262144}\) | \(32\) |
default | \(-\frac {262144}{\frac {262144 \,{\mathrm e}^{4}}{x^{3}}+\frac {262144 \ln \left (\frac {64}{25 x}\right )}{x}-\frac {786432}{x}-262144}\) | \(32\) |
parallelrisch | \(-\frac {x^{3}}{x^{2} \ln \left (\frac {64}{25 x}\right )-x^{3}+{\mathrm e}^{4}-3 x^{2}}\) | \(33\) |
norman | \(\frac {3 x^{2}-x^{2} \ln \left (\frac {64}{25 x}\right )-{\mathrm e}^{4}}{x^{2} \ln \left (\frac {64}{25 x}\right )-x^{3}+{\mathrm e}^{4}-3 x^{2}}\) | \(52\) |
Input:
int((-x^4*ln(64/25/x)-3*x^2*exp(2)^2+2*x^4)/(x^4*ln(64/25/x)^2+(2*x^2*exp( 2)^2-2*x^5-6*x^4)*ln(64/25/x)+exp(2)^4+(-2*x^3-6*x^2)*exp(2)^2+x^6+6*x^5+9 *x^4),x,method=_RETURNVERBOSE)
Output:
-x^3/(x^2*ln(64/25/x)-x^3+exp(4)-3*x^2)
Time = 0.06 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.20 \[ \int \frac {-3 e^4 x^2+2 x^4-x^4 \log \left (\frac {64}{25 x}\right )}{e^8+9 x^4+6 x^5+x^6+e^4 \left (-6 x^2-2 x^3\right )+\left (2 e^4 x^2-6 x^4-2 x^5\right ) \log \left (\frac {64}{25 x}\right )+x^4 \log ^2\left (\frac {64}{25 x}\right )} \, dx=\frac {x^{3}}{x^{3} - x^{2} \log \left (\frac {64}{25 \, x}\right ) + 3 \, x^{2} - e^{4}} \] Input:
integrate((-x^4*log(64/25/x)-3*x^2*exp(2)^2+2*x^4)/(x^4*log(64/25/x)^2+(2* x^2*exp(2)^2-2*x^5-6*x^4)*log(64/25/x)+exp(2)^4+(-2*x^3-6*x^2)*exp(2)^2+x^ 6+6*x^5+9*x^4),x, algorithm="fricas")
Output:
x^3/(x^3 - x^2*log(64/25/x) + 3*x^2 - e^4)
Time = 0.13 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.04 \[ \int \frac {-3 e^4 x^2+2 x^4-x^4 \log \left (\frac {64}{25 x}\right )}{e^8+9 x^4+6 x^5+x^6+e^4 \left (-6 x^2-2 x^3\right )+\left (2 e^4 x^2-6 x^4-2 x^5\right ) \log \left (\frac {64}{25 x}\right )+x^4 \log ^2\left (\frac {64}{25 x}\right )} \, dx=- \frac {x^{3}}{- x^{3} + x^{2} \log {\left (\frac {64}{25 x} \right )} - 3 x^{2} + e^{4}} \] Input:
integrate((-x**4*ln(64/25/x)-3*x**2*exp(2)**2+2*x**4)/(x**4*ln(64/25/x)**2 +(2*x**2*exp(2)**2-2*x**5-6*x**4)*ln(64/25/x)+exp(2)**4+(-2*x**3-6*x**2)*e xp(2)**2+x**6+6*x**5+9*x**4),x)
Output:
-x**3/(-x**3 + x**2*log(64/(25*x)) - 3*x**2 + exp(4))
Time = 0.15 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.36 \[ \int \frac {-3 e^4 x^2+2 x^4-x^4 \log \left (\frac {64}{25 x}\right )}{e^8+9 x^4+6 x^5+x^6+e^4 \left (-6 x^2-2 x^3\right )+\left (2 e^4 x^2-6 x^4-2 x^5\right ) \log \left (\frac {64}{25 x}\right )+x^4 \log ^2\left (\frac {64}{25 x}\right )} \, dx=\frac {x^{3}}{x^{3} + x^{2} {\left (2 \, \log \left (5\right ) - 6 \, \log \left (2\right ) + 3\right )} + x^{2} \log \left (x\right ) - e^{4}} \] Input:
integrate((-x^4*log(64/25/x)-3*x^2*exp(2)^2+2*x^4)/(x^4*log(64/25/x)^2+(2* x^2*exp(2)^2-2*x^5-6*x^4)*log(64/25/x)+exp(2)^4+(-2*x^3-6*x^2)*exp(2)^2+x^ 6+6*x^5+9*x^4),x, algorithm="maxima")
Output:
x^3/(x^3 + x^2*(2*log(5) - 6*log(2) + 3) + x^2*log(x) - e^4)
Time = 0.15 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08 \[ \int \frac {-3 e^4 x^2+2 x^4-x^4 \log \left (\frac {64}{25 x}\right )}{e^8+9 x^4+6 x^5+x^6+e^4 \left (-6 x^2-2 x^3\right )+\left (2 e^4 x^2-6 x^4-2 x^5\right ) \log \left (\frac {64}{25 x}\right )+x^4 \log ^2\left (\frac {64}{25 x}\right )} \, dx=-\frac {1}{\frac {\log \left (\frac {64}{25 \, x}\right )}{x} - \frac {3}{x} + \frac {e^{4}}{x^{3}} - 1} \] Input:
integrate((-x^4*log(64/25/x)-3*x^2*exp(2)^2+2*x^4)/(x^4*log(64/25/x)^2+(2* x^2*exp(2)^2-2*x^5-6*x^4)*log(64/25/x)+exp(2)^4+(-2*x^3-6*x^2)*exp(2)^2+x^ 6+6*x^5+9*x^4),x, algorithm="giac")
Output:
-1/(log(64/25/x)/x - 3/x + e^4/x^3 - 1)
Time = 2.23 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.20 \[ \int \frac {-3 e^4 x^2+2 x^4-x^4 \log \left (\frac {64}{25 x}\right )}{e^8+9 x^4+6 x^5+x^6+e^4 \left (-6 x^2-2 x^3\right )+\left (2 e^4 x^2-6 x^4-2 x^5\right ) \log \left (\frac {64}{25 x}\right )+x^4 \log ^2\left (\frac {64}{25 x}\right )} \, dx=-\frac {x^3}{{\mathrm {e}}^4-3\,x^2-x^3+x^2\,\ln \left (\frac {64}{25\,x}\right )} \] Input:
int(-(3*x^2*exp(4) - 2*x^4 + x^4*log(64/(25*x)))/(exp(8) + x^4*log(64/(25* x))^2 - log(64/(25*x))*(6*x^4 - 2*x^2*exp(4) + 2*x^5) - exp(4)*(6*x^2 + 2* x^3) + 9*x^4 + 6*x^5 + x^6),x)
Output:
-x^3/(exp(4) - 3*x^2 - x^3 + x^2*log(64/(25*x)))
Time = 0.16 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.24 \[ \int \frac {-3 e^4 x^2+2 x^4-x^4 \log \left (\frac {64}{25 x}\right )}{e^8+9 x^4+6 x^5+x^6+e^4 \left (-6 x^2-2 x^3\right )+\left (2 e^4 x^2-6 x^4-2 x^5\right ) \log \left (\frac {64}{25 x}\right )+x^4 \log ^2\left (\frac {64}{25 x}\right )} \, dx=-\frac {x^{3}}{\mathrm {log}\left (\frac {64}{25 x}\right ) x^{2}+e^{4}-x^{3}-3 x^{2}} \] Input:
int((-x^4*log(64/25/x)-3*x^2*exp(2)^2+2*x^4)/(x^4*log(64/25/x)^2+(2*x^2*ex p(2)^2-2*x^5-6*x^4)*log(64/25/x)+exp(2)^4+(-2*x^3-6*x^2)*exp(2)^2+x^6+6*x^ 5+9*x^4),x)
Output:
( - x**3)/(log(64/(25*x))*x**2 + e**4 - x**3 - 3*x**2)