Integrand size = 77, antiderivative size = 24 \[ \int \frac {25 x^3+e^{\frac {2-50 x^2-25 x^3}{25 x^2}} \left (-4-25 x^3\right )}{25 e^{\frac {2-50 x^2-25 x^3}{25 x^2}} x^3+25 x^4+225 x^3 \log (2)} \, dx=\log \left (9+\frac {e^{-2+\frac {2}{25 x^2}-x}+x}{\log (2)}\right ) \] Output:
ln(9+(x+exp(2/25/x^2-2-x))/ln(2))
Time = 2.79 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.50 \[ \int \frac {25 x^3+e^{\frac {2-50 x^2-25 x^3}{25 x^2}} \left (-4-25 x^3\right )}{25 e^{\frac {2-50 x^2-25 x^3}{25 x^2}} x^3+25 x^4+225 x^3 \log (2)} \, dx=\frac {1}{25} \left (-25 x+25 \log \left (e^{\frac {2}{25 x^2}}+e^{2+x} x+e^{2+x} \log (512)\right )\right ) \] Input:
Integrate[(25*x^3 + E^((2 - 50*x^2 - 25*x^3)/(25*x^2))*(-4 - 25*x^3))/(25* E^((2 - 50*x^2 - 25*x^3)/(25*x^2))*x^3 + 25*x^4 + 225*x^3*Log[2]),x]
Output:
(-25*x + 25*Log[E^(2/(25*x^2)) + E^(2 + x)*x + E^(2 + x)*Log[512]])/25
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.
\(\displaystyle \int \frac {25 x^3+e^{\frac {-25 x^3-50 x^2+2}{25 x^2}} \left (-25 x^3-4\right )}{25 x^4+225 x^3 \log (2)+25 e^{\frac {-25 x^3-50 x^2+2}{25 x^2}} x^3} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {e^{x+2} \left (25 x^3+e^{\frac {-25 x^3-50 x^2+2}{25 x^2}} \left (-25 x^3-4\right )\right )}{25 x^3 \left (e^{\frac {2}{25 x^2}}+e^{x+2} x+e^{x+2} \log (512)\right )}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{25} \int \frac {e^{x+2} \left (25 x^3-e^{\frac {-25 x^3-50 x^2+2}{25 x^2}} \left (25 x^3+4\right )\right )}{x^3 \left (e^{x+2} x+e^{\frac {2}{25 x^2}}+e^{x+2} \log (512)\right )}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {1}{25} \int \left (\frac {e^{x+2} \left (25 x^4+25 (1+\log (512)) x^3+4 x+4 \log (512)\right )}{x^3 \left (e^{x+2} x+e^{\frac {2}{25 x^2}}+e^{x+2} \log (512)\right )}-\frac {25 x^3+4}{x^3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{25} \left (25 (1+\log (512)) \int \frac {e^{x+2}}{e^{x+2} x+e^{\frac {2}{25 x^2}}+e^{x+2} \log (512)}dx+4 \int \frac {e^{x+2}}{x^2 \left (e^{x+2} x+e^{\frac {2}{25 x^2}}+e^{x+2} \log (512)\right )}dx+25 \int \frac {e^{x+2} x}{e^{x+2} x+e^{\frac {2}{25 x^2}}+e^{x+2} \log (512)}dx+4 \log (512) \int \frac {e^{x+2}}{x^3 \left (e^{x+2} x+e^{\frac {2}{25 x^2}}+e^{x+2} \log (512)\right )}dx+\frac {2}{x^2}-25 x\right )\) |
Input:
Int[(25*x^3 + E^((2 - 50*x^2 - 25*x^3)/(25*x^2))*(-4 - 25*x^3))/(25*E^((2 - 50*x^2 - 25*x^3)/(25*x^2))*x^3 + 25*x^4 + 225*x^3*Log[2]),x]
Output:
$Aborted
Time = 0.48 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08
method | result | size |
norman | \(\ln \left (9 \ln \left (2\right )+x +{\mathrm e}^{\frac {-25 x^{3}-50 x^{2}+2}{25 x^{2}}}\right )\) | \(26\) |
parallelrisch | \(\ln \left ({\mathrm e}^{-\frac {25 x^{3}+50 x^{2}-2}{25 x^{2}}}+9 \ln \left (2\right )+x \right )\) | \(26\) |
risch | \(-x +\frac {2}{25 x^{2}}-\frac {-25 x^{3}-50 x^{2}+2}{25 x^{2}}+\ln \left ({\mathrm e}^{-\frac {25 x^{3}+50 x^{2}-2}{25 x^{2}}}+9 \ln \left (2\right )+x \right )\) | \(52\) |
Input:
int(((-25*x^3-4)*exp(1/25*(-25*x^3-50*x^2+2)/x^2)+25*x^3)/(25*x^3*exp(1/25 *(-25*x^3-50*x^2+2)/x^2)+225*x^3*ln(2)+25*x^4),x,method=_RETURNVERBOSE)
Output:
ln(9*ln(2)+x+exp(1/25*(-25*x^3-50*x^2+2)/x^2))
Time = 0.15 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.04 \[ \int \frac {25 x^3+e^{\frac {2-50 x^2-25 x^3}{25 x^2}} \left (-4-25 x^3\right )}{25 e^{\frac {2-50 x^2-25 x^3}{25 x^2}} x^3+25 x^4+225 x^3 \log (2)} \, dx=\log \left (x + e^{\left (-\frac {25 \, x^{3} + 50 \, x^{2} - 2}{25 \, x^{2}}\right )} + 9 \, \log \left (2\right )\right ) \] Input:
integrate(((-25*x^3-4)*exp(1/25*(-25*x^3-50*x^2+2)/x^2)+25*x^3)/(25*x^3*ex p(1/25*(-25*x^3-50*x^2+2)/x^2)+225*x^3*log(2)+25*x^4),x, algorithm="fricas ")
Output:
log(x + e^(-1/25*(25*x^3 + 50*x^2 - 2)/x^2) + 9*log(2))
Time = 0.12 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {25 x^3+e^{\frac {2-50 x^2-25 x^3}{25 x^2}} \left (-4-25 x^3\right )}{25 e^{\frac {2-50 x^2-25 x^3}{25 x^2}} x^3+25 x^4+225 x^3 \log (2)} \, dx=\log {\left (x + e^{\frac {- x^{3} - 2 x^{2} + \frac {2}{25}}{x^{2}}} + 9 \log {\left (2 \right )} \right )} \] Input:
integrate(((-25*x**3-4)*exp(1/25*(-25*x**3-50*x**2+2)/x**2)+25*x**3)/(25*x **3*exp(1/25*(-25*x**3-50*x**2+2)/x**2)+225*x**3*ln(2)+25*x**4),x)
Output:
log(x + exp((-x**3 - 2*x**2 + 2/25)/x**2) + 9*log(2))
Time = 0.15 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08 \[ \int \frac {25 x^3+e^{\frac {2-50 x^2-25 x^3}{25 x^2}} \left (-4-25 x^3\right )}{25 e^{\frac {2-50 x^2-25 x^3}{25 x^2}} x^3+25 x^4+225 x^3 \log (2)} \, dx=-x + \log \left ({\left (x e^{2} + 9 \, e^{2} \log \left (2\right )\right )} e^{x} + e^{\left (\frac {2}{25 \, x^{2}}\right )}\right ) \] Input:
integrate(((-25*x^3-4)*exp(1/25*(-25*x^3-50*x^2+2)/x^2)+25*x^3)/(25*x^3*ex p(1/25*(-25*x^3-50*x^2+2)/x^2)+225*x^3*log(2)+25*x^4),x, algorithm="maxima ")
Output:
-x + log((x*e^2 + 9*e^2*log(2))*e^x + e^(2/25/x^2))
Time = 0.15 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.04 \[ \int \frac {25 x^3+e^{\frac {2-50 x^2-25 x^3}{25 x^2}} \left (-4-25 x^3\right )}{25 e^{\frac {2-50 x^2-25 x^3}{25 x^2}} x^3+25 x^4+225 x^3 \log (2)} \, dx=\log \left (x + e^{\left (-\frac {25 \, x^{3} + 50 \, x^{2} - 2}{25 \, x^{2}}\right )} + 9 \, \log \left (2\right )\right ) \] Input:
integrate(((-25*x^3-4)*exp(1/25*(-25*x^3-50*x^2+2)/x^2)+25*x^3)/(25*x^3*ex p(1/25*(-25*x^3-50*x^2+2)/x^2)+225*x^3*log(2)+25*x^4),x, algorithm="giac")
Output:
log(x + e^(-1/25*(25*x^3 + 50*x^2 - 2)/x^2) + 9*log(2))
Time = 2.79 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.67 \[ \int \frac {25 x^3+e^{\frac {2-50 x^2-25 x^3}{25 x^2}} \left (-4-25 x^3\right )}{25 e^{\frac {2-50 x^2-25 x^3}{25 x^2}} x^3+25 x^4+225 x^3 \log (2)} \, dx=\ln \left (x+\ln \left (512\right )+{\mathrm {e}}^{\frac {2}{25\,x^2}-x-2}\right ) \] Input:
int((25*x^3 - exp(-(2*x^2 + x^3 - 2/25)/x^2)*(25*x^3 + 4))/(25*x^3*exp(-(2 *x^2 + x^3 - 2/25)/x^2) + 225*x^3*log(2) + 25*x^4),x)
Output:
log(x + log(512) + exp(2/(25*x^2) - x - 2))
Time = 0.28 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.29 \[ \int \frac {25 x^3+e^{\frac {2-50 x^2-25 x^3}{25 x^2}} \left (-4-25 x^3\right )}{25 e^{\frac {2-50 x^2-25 x^3}{25 x^2}} x^3+25 x^4+225 x^3 \log (2)} \, dx=\mathrm {log}\left (e^{\frac {2}{25 x^{2}}}+9 e^{x} \mathrm {log}\left (2\right ) e^{2}+e^{x} e^{2} x \right )-x \] Input:
int(((-25*x^3-4)*exp(1/25*(-25*x^3-50*x^2+2)/x^2)+25*x^3)/(25*x^3*exp(1/25 *(-25*x^3-50*x^2+2)/x^2)+225*x^3*log(2)+25*x^4),x)
Output:
log(e**(2/(25*x**2)) + 9*e**x*log(2)*e**2 + e**x*e**2*x) - x