Integrand size = 111, antiderivative size = 26 \[ \int \frac {20-18 \log (9)+4 \log ^2(9)+e^{4+50 x^2} \log ^2(9)+e^{25 x^2} \left (e^2 \left (-9-50 x^2\right ) \log (9)+4 e^2 \log ^2(9)\right )}{25-20 \log (9)+4 \log ^2(9)+e^{4+50 x^2} \log ^2(9)+e^{25 x^2} \left (-10 e^2 \log (9)+4 e^2 \log ^2(9)\right )} \, dx=4+x-\frac {x}{5+\left (-2-e^{2+25 x^2}\right ) \log (9)} \] Output:
x+4-x/(5+2*(-2-exp(2)*exp(25*x^2))*ln(3))
Time = 0.75 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.38 \[ \int \frac {20-18 \log (9)+4 \log ^2(9)+e^{4+50 x^2} \log ^2(9)+e^{25 x^2} \left (e^2 \left (-9-50 x^2\right ) \log (9)+4 e^2 \log ^2(9)\right )}{25-20 \log (9)+4 \log ^2(9)+e^{4+50 x^2} \log ^2(9)+e^{25 x^2} \left (-10 e^2 \log (9)+4 e^2 \log ^2(9)\right )} \, dx=\frac {x \left (-4+e^{2+25 x^2} \log (9)+\log (81)\right )}{-5+e^{2+25 x^2} \log (9)+\log (81)} \] Input:
Integrate[(20 - 18*Log[9] + 4*Log[9]^2 + E^(4 + 50*x^2)*Log[9]^2 + E^(25*x ^2)*(E^2*(-9 - 50*x^2)*Log[9] + 4*E^2*Log[9]^2))/(25 - 20*Log[9] + 4*Log[9 ]^2 + E^(4 + 50*x^2)*Log[9]^2 + E^(25*x^2)*(-10*E^2*Log[9] + 4*E^2*Log[9]^ 2)),x]
Output:
(x*(-4 + E^(2 + 25*x^2)*Log[9] + Log[81]))/(-5 + E^(2 + 25*x^2)*Log[9] + L og[81])
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.
\(\displaystyle \int \frac {e^{25 x^2} \left (e^2 \left (-50 x^2-9\right ) \log (9)+4 e^2 \log ^2(9)\right )+e^{50 x^2+4} \log ^2(9)+20+4 \log ^2(9)-18 \log (9)}{e^{50 x^2+4} \log ^2(9)+e^{25 x^2} \left (4 e^2 \log ^2(9)-10 e^2 \log (9)\right )+25+4 \log ^2(9)-20 \log (9)} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {e^{25 x^2} \left (e^2 \left (-50 x^2-9\right ) \log (9)+4 e^2 \log ^2(9)\right )+e^{50 x^2+4} \log ^2(9)+20 \left (1+\frac {1}{10} \log (9) (\log (81)-9)\right )}{\left (5 \left (1-\frac {4 \log (3)}{5}\right )-e^{25 x^2+2} \log (9)\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {2 \left (-25 x^2 (5-\log (81))+8 \log ^2(3)+\log (9) \log (81)-8 \log (3) \log (9)\right )}{\left (5 \left (1-\frac {4 \log (3)}{5}\right )-e^{25 x^2+2} \log (9)\right )^2}+\frac {50 x^2-1}{5 \left (1-\frac {4 \log (3)}{5}\right )-e^{25 x^2+2} \log (9)}+1\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -50 (5-\log (81)) \int \frac {x^2}{\left (5 \left (1-\frac {4 \log (3)}{5}\right )-e^{25 x^2+2} \log (9)\right )^2}dx+50 \int \frac {x^2}{5 \left (1-\frac {4 \log (3)}{5}\right )-e^{25 x^2+2} \log (9)}dx+\int \frac {1}{-5 \left (1-\frac {4 \log (3)}{5}\right )+e^{25 x^2+2} \log (9)}dx+x\) |
Input:
Int[(20 - 18*Log[9] + 4*Log[9]^2 + E^(4 + 50*x^2)*Log[9]^2 + E^(25*x^2)*(E ^2*(-9 - 50*x^2)*Log[9] + 4*E^2*Log[9]^2))/(25 - 20*Log[9] + 4*Log[9]^2 + E^(4 + 50*x^2)*Log[9]^2 + E^(25*x^2)*(-10*E^2*Log[9] + 4*E^2*Log[9]^2)),x]
Output:
$Aborted
Time = 0.37 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96
method | result | size |
risch | \(x +\frac {x}{2 \ln \left (3\right ) {\mathrm e}^{25 x^{2}+2}+4 \ln \left (3\right )-5}\) | \(25\) |
norman | \(\frac {\left (4 \ln \left (3\right )-4\right ) x +2 x \,{\mathrm e}^{2} \ln \left (3\right ) {\mathrm e}^{25 x^{2}}}{2 \,{\mathrm e}^{2} \ln \left (3\right ) {\mathrm e}^{25 x^{2}}+4 \ln \left (3\right )-5}\) | \(44\) |
parallelrisch | \(\frac {8 \,{\mathrm e}^{2} {\mathrm e}^{25 x^{2}} \ln \left (3\right )^{2} x -10 x \,{\mathrm e}^{2} \ln \left (3\right ) {\mathrm e}^{25 x^{2}}+16 x \ln \left (3\right )^{2}-36 x \ln \left (3\right )+20 x}{\left (4 \ln \left (3\right )-5\right ) \left (2 \,{\mathrm e}^{2} \ln \left (3\right ) {\mathrm e}^{25 x^{2}}+4 \ln \left (3\right )-5\right )}\) | \(74\) |
Input:
int((4*exp(2)^2*ln(3)^2*exp(25*x^2)^2+(16*exp(2)*ln(3)^2+2*(-50*x^2-9)*exp (2)*ln(3))*exp(25*x^2)+16*ln(3)^2-36*ln(3)+20)/(4*exp(2)^2*ln(3)^2*exp(25* x^2)^2+(16*exp(2)*ln(3)^2-20*exp(2)*ln(3))*exp(25*x^2)+16*ln(3)^2-40*ln(3) +25),x,method=_RETURNVERBOSE)
Output:
x+x/(2*ln(3)*exp(25*x^2+2)+4*ln(3)-5)
Time = 0.09 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.65 \[ \int \frac {20-18 \log (9)+4 \log ^2(9)+e^{4+50 x^2} \log ^2(9)+e^{25 x^2} \left (e^2 \left (-9-50 x^2\right ) \log (9)+4 e^2 \log ^2(9)\right )}{25-20 \log (9)+4 \log ^2(9)+e^{4+50 x^2} \log ^2(9)+e^{25 x^2} \left (-10 e^2 \log (9)+4 e^2 \log ^2(9)\right )} \, dx=\frac {2 \, {\left (x e^{\left (25 \, x^{2} + 2\right )} \log \left (3\right ) + 2 \, x \log \left (3\right ) - 2 \, x\right )}}{2 \, e^{\left (25 \, x^{2} + 2\right )} \log \left (3\right ) + 4 \, \log \left (3\right ) - 5} \] Input:
integrate((4*exp(2)^2*log(3)^2*exp(25*x^2)^2+(16*exp(2)*log(3)^2+2*(-50*x^ 2-9)*exp(2)*log(3))*exp(25*x^2)+16*log(3)^2-36*log(3)+20)/(4*exp(2)^2*log( 3)^2*exp(25*x^2)^2+(16*exp(2)*log(3)^2-20*exp(2)*log(3))*exp(25*x^2)+16*lo g(3)^2-40*log(3)+25),x, algorithm="fricas")
Output:
2*(x*e^(25*x^2 + 2)*log(3) + 2*x*log(3) - 2*x)/(2*e^(25*x^2 + 2)*log(3) + 4*log(3) - 5)
Time = 0.08 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \[ \int \frac {20-18 \log (9)+4 \log ^2(9)+e^{4+50 x^2} \log ^2(9)+e^{25 x^2} \left (e^2 \left (-9-50 x^2\right ) \log (9)+4 e^2 \log ^2(9)\right )}{25-20 \log (9)+4 \log ^2(9)+e^{4+50 x^2} \log ^2(9)+e^{25 x^2} \left (-10 e^2 \log (9)+4 e^2 \log ^2(9)\right )} \, dx=x + \frac {x}{2 e^{2} e^{25 x^{2}} \log {\left (3 \right )} - 5 + 4 \log {\left (3 \right )}} \] Input:
integrate((4*exp(2)**2*ln(3)**2*exp(25*x**2)**2+(16*exp(2)*ln(3)**2+2*(-50 *x**2-9)*exp(2)*ln(3))*exp(25*x**2)+16*ln(3)**2-36*ln(3)+20)/(4*exp(2)**2* ln(3)**2*exp(25*x**2)**2+(16*exp(2)*ln(3)**2-20*exp(2)*ln(3))*exp(25*x**2) +16*ln(3)**2-40*ln(3)+25),x)
Output:
x + x/(2*exp(2)*exp(25*x**2)*log(3) - 5 + 4*log(3))
Time = 0.17 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.62 \[ \int \frac {20-18 \log (9)+4 \log ^2(9)+e^{4+50 x^2} \log ^2(9)+e^{25 x^2} \left (e^2 \left (-9-50 x^2\right ) \log (9)+4 e^2 \log ^2(9)\right )}{25-20 \log (9)+4 \log ^2(9)+e^{4+50 x^2} \log ^2(9)+e^{25 x^2} \left (-10 e^2 \log (9)+4 e^2 \log ^2(9)\right )} \, dx=\frac {2 \, {\left (x e^{\left (25 \, x^{2} + 2\right )} \log \left (3\right ) + 2 \, x {\left (\log \left (3\right ) - 1\right )}\right )}}{2 \, e^{\left (25 \, x^{2} + 2\right )} \log \left (3\right ) + 4 \, \log \left (3\right ) - 5} \] Input:
integrate((4*exp(2)^2*log(3)^2*exp(25*x^2)^2+(16*exp(2)*log(3)^2+2*(-50*x^ 2-9)*exp(2)*log(3))*exp(25*x^2)+16*log(3)^2-36*log(3)+20)/(4*exp(2)^2*log( 3)^2*exp(25*x^2)^2+(16*exp(2)*log(3)^2-20*exp(2)*log(3))*exp(25*x^2)+16*lo g(3)^2-40*log(3)+25),x, algorithm="maxima")
Output:
2*(x*e^(25*x^2 + 2)*log(3) + 2*x*(log(3) - 1))/(2*e^(25*x^2 + 2)*log(3) + 4*log(3) - 5)
Time = 0.12 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.65 \[ \int \frac {20-18 \log (9)+4 \log ^2(9)+e^{4+50 x^2} \log ^2(9)+e^{25 x^2} \left (e^2 \left (-9-50 x^2\right ) \log (9)+4 e^2 \log ^2(9)\right )}{25-20 \log (9)+4 \log ^2(9)+e^{4+50 x^2} \log ^2(9)+e^{25 x^2} \left (-10 e^2 \log (9)+4 e^2 \log ^2(9)\right )} \, dx=\frac {2 \, {\left (x e^{\left (25 \, x^{2} + 2\right )} \log \left (3\right ) + 2 \, x \log \left (3\right ) - 2 \, x\right )}}{2 \, e^{\left (25 \, x^{2} + 2\right )} \log \left (3\right ) + 4 \, \log \left (3\right ) - 5} \] Input:
integrate((4*exp(2)^2*log(3)^2*exp(25*x^2)^2+(16*exp(2)*log(3)^2+2*(-50*x^ 2-9)*exp(2)*log(3))*exp(25*x^2)+16*log(3)^2-36*log(3)+20)/(4*exp(2)^2*log( 3)^2*exp(25*x^2)^2+(16*exp(2)*log(3)^2-20*exp(2)*log(3))*exp(25*x^2)+16*lo g(3)^2-40*log(3)+25),x, algorithm="giac")
Output:
2*(x*e^(25*x^2 + 2)*log(3) + 2*x*log(3) - 2*x)/(2*e^(25*x^2 + 2)*log(3) + 4*log(3) - 5)
Time = 0.37 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \[ \int \frac {20-18 \log (9)+4 \log ^2(9)+e^{4+50 x^2} \log ^2(9)+e^{25 x^2} \left (e^2 \left (-9-50 x^2\right ) \log (9)+4 e^2 \log ^2(9)\right )}{25-20 \log (9)+4 \log ^2(9)+e^{4+50 x^2} \log ^2(9)+e^{25 x^2} \left (-10 e^2 \log (9)+4 e^2 \log ^2(9)\right )} \, dx=x+\frac {x}{4\,\ln \left (3\right )+2\,{\mathrm {e}}^{25\,x^2+2}\,\ln \left (3\right )-5} \] Input:
int((exp(25*x^2)*(16*exp(2)*log(3)^2 - 2*exp(2)*log(3)*(50*x^2 + 9)) - 36* log(3) + 16*log(3)^2 + 4*exp(4)*exp(50*x^2)*log(3)^2 + 20)/(16*log(3)^2 - exp(25*x^2)*(20*exp(2)*log(3) - 16*exp(2)*log(3)^2) - 40*log(3) + 4*exp(4) *exp(50*x^2)*log(3)^2 + 25),x)
Output:
x + x/(4*log(3) + 2*exp(25*x^2 + 2)*log(3) - 5)
Time = 0.18 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.69 \[ \int \frac {20-18 \log (9)+4 \log ^2(9)+e^{4+50 x^2} \log ^2(9)+e^{25 x^2} \left (e^2 \left (-9-50 x^2\right ) \log (9)+4 e^2 \log ^2(9)\right )}{25-20 \log (9)+4 \log ^2(9)+e^{4+50 x^2} \log ^2(9)+e^{25 x^2} \left (-10 e^2 \log (9)+4 e^2 \log ^2(9)\right )} \, dx=\frac {2 x \left (e^{25 x^{2}} \mathrm {log}\left (3\right ) e^{2}+2 \,\mathrm {log}\left (3\right )-2\right )}{2 e^{25 x^{2}} \mathrm {log}\left (3\right ) e^{2}+4 \,\mathrm {log}\left (3\right )-5} \] Input:
int((4*exp(2)^2*log(3)^2*exp(25*x^2)^2+(16*exp(2)*log(3)^2+2*(-50*x^2-9)*e xp(2)*log(3))*exp(25*x^2)+16*log(3)^2-36*log(3)+20)/(4*exp(2)^2*log(3)^2*e xp(25*x^2)^2+(16*exp(2)*log(3)^2-20*exp(2)*log(3))*exp(25*x^2)+16*log(3)^2 -40*log(3)+25),x)
Output:
(2*x*(e**(25*x**2)*log(3)*e**2 + 2*log(3) - 2))/(2*e**(25*x**2)*log(3)*e** 2 + 4*log(3) - 5)