Integrand size = 111, antiderivative size = 27 \[ \int \frac {21+e^{-8+50 x^2} \left (3-300 x^2\right )-3 \log (5)-3 \log \left (x^2\right )}{25+e^{-16+100 x^2}+10 x+x^2+e^{-8+50 x^2} (10+2 x-2 \log (5))+(-10-2 x) \log (5)+\log ^2(5)+\left (-10-2 e^{-8+50 x^2}-2 x+2 \log (5)\right ) \log \left (x^2\right )+\log ^2\left (x^2\right )} \, dx=\frac {3 x}{5+e^{-8+50 x^2}+x-\log (5)-\log \left (x^2\right )} \] Output:
3*x/(5-ln(5)-ln(x^2)+exp(25*x^2-4)^2+x)
Time = 0.33 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.33 \[ \int \frac {21+e^{-8+50 x^2} \left (3-300 x^2\right )-3 \log (5)-3 \log \left (x^2\right )}{25+e^{-16+100 x^2}+10 x+x^2+e^{-8+50 x^2} (10+2 x-2 \log (5))+(-10-2 x) \log (5)+\log ^2(5)+\left (-10-2 e^{-8+50 x^2}-2 x+2 \log (5)\right ) \log \left (x^2\right )+\log ^2\left (x^2\right )} \, dx=\frac {3 e^8 x}{e^{50 x^2}+e^8 (5+x-\log (5))-e^8 \log \left (x^2\right )} \] Input:
Integrate[(21 + E^(-8 + 50*x^2)*(3 - 300*x^2) - 3*Log[5] - 3*Log[x^2])/(25 + E^(-16 + 100*x^2) + 10*x + x^2 + E^(-8 + 50*x^2)*(10 + 2*x - 2*Log[5]) + (-10 - 2*x)*Log[5] + Log[5]^2 + (-10 - 2*E^(-8 + 50*x^2) - 2*x + 2*Log[5 ])*Log[x^2] + Log[x^2]^2),x]
Output:
(3*E^8*x)/(E^(50*x^2) + E^8*(5 + x - Log[5]) - E^8*Log[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.
| \(\displaystyle \int \frac {e^{50 x^2-8} \left (3-300 x^2\right )-3 \log \left (x^2\right )+21-3 \log (5)}{x^2+e^{100 x^2-16}+\log ^2\left (x^2\right )+e^{50 x^2-8} (2 x+10-2 \log (5))+\left (-2 e^{50 x^2-8}-2 x-10+2 \log (5)\right ) \log \left (x^2\right )+10 x+(-2 x-10) \log (5)+25+\log ^2(5)} \, dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle \int \frac {e^{16} \left (e^{50 x^2-8} \left (3-300 x^2\right )-3 \log \left (x^2\right )+21 \left (1-\frac {\log (5)}{7}\right )\right )}{\left (e^{50 x^2}-e^8 \log \left (x^2\right )+e^8 x+5 e^8 \left (1-\frac {\log (5)}{5}\right )\right )^2}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle e^{16} \int \frac {3 \left (e^{50 x^2-8} \left (1-100 x^2\right )-\log \left (x^2\right )-\log (5)+7\right )}{\left (e^8 x+e^{50 x^2}-e^8 \log \left (x^2\right )+e^8 (5-\log (5))\right )^2}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle 3 e^{16} \int \frac {e^{50 x^2-8} \left (1-100 x^2\right )-\log \left (x^2\right )-\log (5)+7}{\left (e^8 x+e^{50 x^2}-e^8 \log \left (x^2\right )+e^8 (5-\log (5))\right )^2}dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle 3 e^{16} \int \frac {e^{50 x^2-8} \left (1-100 x^2\right )-\log \left (x^2\right )+7 \left (1-\frac {\log (5)}{7}\right )}{\left (e^8 x+e^{50 x^2}-e^8 \log \left (x^2\right )+e^8 (5-\log (5))\right )^2}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle 3 e^{16} \int \left (\frac {1-100 x^2}{e^8 \left (e^8 x+e^{50 x^2}-e^8 \log \left (x^2\right )+5 e^8 \left (1-\frac {\log (5)}{5}\right )\right )}+\frac {100 x^3-100 \log \left (x^2\right ) x^2+500 \left (1-\frac {\log (5)}{5}\right ) x^2-x+2}{\left (e^8 x+e^{50 x^2}-e^8 \log \left (x^2\right )+5 e^8 \left (1-\frac {\log (5)}{5}\right )\right )^2}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle 3 e^{16} \left (2 \int \frac {1}{\left (e^8 x+e^{50 x^2}-e^8 \log \left (x^2\right )+5 e^8 \left (1-\frac {\log (5)}{5}\right )\right )^2}dx-\int \frac {x}{\left (e^8 x+e^{50 x^2}-e^8 \log \left (x^2\right )+5 e^8 \left (1-\frac {\log (5)}{5}\right )\right )^2}dx+100 (5-\log (5)) \int \frac {x^2}{\left (e^8 x+e^{50 x^2}-e^8 \log \left (x^2\right )+5 e^8 \left (1-\frac {\log (5)}{5}\right )\right )^2}dx-100 \int \frac {x^2 \log \left (x^2\right )}{\left (e^8 x+e^{50 x^2}-e^8 \log \left (x^2\right )+5 e^8 \left (1-\frac {\log (5)}{5}\right )\right )^2}dx+\frac {\int \frac {1}{e^8 x+e^{50 x^2}-e^8 \log \left (x^2\right )+5 e^8 \left (1-\frac {\log (5)}{5}\right )}dx}{e^8}+\frac {100 \int \frac {x^2}{-e^8 x-e^{50 x^2}+e^8 \log \left (x^2\right )-5 e^8 \left (1-\frac {\log (5)}{5}\right )}dx}{e^8}+100 \int \frac {x^3}{\left (e^8 x+e^{50 x^2}-e^8 \log \left (x^2\right )+5 e^8 \left (1-\frac {\log (5)}{5}\right )\right )^2}dx\right )\) |
Input:
Int[(21 + E^(-8 + 50*x^2)*(3 - 300*x^2) - 3*Log[5] - 3*Log[x^2])/(25 + E^( -16 + 100*x^2) + 10*x + x^2 + E^(-8 + 50*x^2)*(10 + 2*x - 2*Log[5]) + (-10 - 2*x)*Log[5] + Log[5]^2 + (-10 - 2*E^(-8 + 50*x^2) - 2*x + 2*Log[5])*Log [x^2] + Log[x^2]^2),x]
Output:
$Aborted
Time = 0.61 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.07
| method | result | size |
| parallelrisch | \(-\frac {3 x}{-x -5+\ln \left (5\right )+\ln \left (x^{2}\right )-{\mathrm e}^{50 x^{2}-8}}\) | \(29\) |
| risch | \(\frac {6 x}{i \pi \,\operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i x \right )^{2}-2 i \operatorname {csgn}\left (i x^{2}\right )^{2} \operatorname {csgn}\left (i x \right ) \pi +i \pi \operatorname {csgn}\left (i x^{2}\right )^{3}+2 \,{\mathrm e}^{2 \left (5 x -2\right ) \left (5 x +2\right )}-2 \ln \left (5\right )+2 x -4 \ln \left (x \right )+10}\) | \(83\) |
Input:
int((-3*ln(x^2)+(-300*x^2+3)*exp(25*x^2-4)^2-3*ln(5)+21)/(ln(x^2)^2+(-2*ex p(25*x^2-4)^2+2*ln(5)-2*x-10)*ln(x^2)+exp(25*x^2-4)^4+(-2*ln(5)+2*x+10)*ex p(25*x^2-4)^2+ln(5)^2+(-2*x-10)*ln(5)+x^2+10*x+25),x,method=_RETURNVERBOSE )
Output:
-3*x/(-exp(25*x^2-4)^2+ln(5)-x+ln(x^2)-5)
Time = 0.09 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \[ \int \frac {21+e^{-8+50 x^2} \left (3-300 x^2\right )-3 \log (5)-3 \log \left (x^2\right )}{25+e^{-16+100 x^2}+10 x+x^2+e^{-8+50 x^2} (10+2 x-2 \log (5))+(-10-2 x) \log (5)+\log ^2(5)+\left (-10-2 e^{-8+50 x^2}-2 x+2 \log (5)\right ) \log \left (x^2\right )+\log ^2\left (x^2\right )} \, dx=\frac {3 \, x}{x + e^{\left (50 \, x^{2} - 8\right )} - \log \left (5\right ) - \log \left (x^{2}\right ) + 5} \] Input:
integrate((-3*log(x^2)+(-300*x^2+3)*exp(25*x^2-4)^2-3*log(5)+21)/(log(x^2) ^2+(-2*exp(25*x^2-4)^2+2*log(5)-2*x-10)*log(x^2)+exp(25*x^2-4)^4+(-2*log(5 )+2*x+10)*exp(25*x^2-4)^2+log(5)^2+(-2*x-10)*log(5)+x^2+10*x+25),x, algori thm="fricas")
Output:
3*x/(x + e^(50*x^2 - 8) - log(5) - log(x^2) + 5)
Time = 0.11 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.81 \[ \int \frac {21+e^{-8+50 x^2} \left (3-300 x^2\right )-3 \log (5)-3 \log \left (x^2\right )}{25+e^{-16+100 x^2}+10 x+x^2+e^{-8+50 x^2} (10+2 x-2 \log (5))+(-10-2 x) \log (5)+\log ^2(5)+\left (-10-2 e^{-8+50 x^2}-2 x+2 \log (5)\right ) \log \left (x^2\right )+\log ^2\left (x^2\right )} \, dx=\frac {3 x}{x + e^{50 x^{2} - 8} - \log {\left (x^{2} \right )} - \log {\left (5 \right )} + 5} \] Input:
integrate((-3*ln(x**2)+(-300*x**2+3)*exp(25*x**2-4)**2-3*ln(5)+21)/(ln(x** 2)**2+(-2*exp(25*x**2-4)**2+2*ln(5)-2*x-10)*ln(x**2)+exp(25*x**2-4)**4+(-2 *ln(5)+2*x+10)*exp(25*x**2-4)**2+ln(5)**2+(-2*x-10)*ln(5)+x**2+10*x+25),x)
Output:
3*x/(x + exp(50*x**2 - 8) - log(x**2) - log(5) + 5)
Time = 0.16 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.19 \[ \int \frac {21+e^{-8+50 x^2} \left (3-300 x^2\right )-3 \log (5)-3 \log \left (x^2\right )}{25+e^{-16+100 x^2}+10 x+x^2+e^{-8+50 x^2} (10+2 x-2 \log (5))+(-10-2 x) \log (5)+\log ^2(5)+\left (-10-2 e^{-8+50 x^2}-2 x+2 \log (5)\right ) \log \left (x^2\right )+\log ^2\left (x^2\right )} \, dx=\frac {3 \, x e^{8}}{x e^{8} - {\left (\log \left (5\right ) - 5\right )} e^{8} - 2 \, e^{8} \log \left (x\right ) + e^{\left (50 \, x^{2}\right )}} \] Input:
integrate((-3*log(x^2)+(-300*x^2+3)*exp(25*x^2-4)^2-3*log(5)+21)/(log(x^2) ^2+(-2*exp(25*x^2-4)^2+2*log(5)-2*x-10)*log(x^2)+exp(25*x^2-4)^4+(-2*log(5 )+2*x+10)*exp(25*x^2-4)^2+log(5)^2+(-2*x-10)*log(5)+x^2+10*x+25),x, algori thm="maxima")
Output:
3*x*e^8/(x*e^8 - (log(5) - 5)*e^8 - 2*e^8*log(x) + e^(50*x^2))
Time = 0.20 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.33 \[ \int \frac {21+e^{-8+50 x^2} \left (3-300 x^2\right )-3 \log (5)-3 \log \left (x^2\right )}{25+e^{-16+100 x^2}+10 x+x^2+e^{-8+50 x^2} (10+2 x-2 \log (5))+(-10-2 x) \log (5)+\log ^2(5)+\left (-10-2 e^{-8+50 x^2}-2 x+2 \log (5)\right ) \log \left (x^2\right )+\log ^2\left (x^2\right )} \, dx=\frac {3 \, x e^{8}}{x e^{8} - e^{8} \log \left (5\right ) - e^{8} \log \left (x^{2}\right ) + 5 \, e^{8} + e^{\left (50 \, x^{2}\right )}} \] Input:
integrate((-3*log(x^2)+(-300*x^2+3)*exp(25*x^2-4)^2-3*log(5)+21)/(log(x^2) ^2+(-2*exp(25*x^2-4)^2+2*log(5)-2*x-10)*log(x^2)+exp(25*x^2-4)^4+(-2*log(5 )+2*x+10)*exp(25*x^2-4)^2+log(5)^2+(-2*x-10)*log(5)+x^2+10*x+25),x, algori thm="giac")
Output:
3*x*e^8/(x*e^8 - e^8*log(5) - e^8*log(x^2) + 5*e^8 + e^(50*x^2))
Timed out. \[ \int \frac {21+e^{-8+50 x^2} \left (3-300 x^2\right )-3 \log (5)-3 \log \left (x^2\right )}{25+e^{-16+100 x^2}+10 x+x^2+e^{-8+50 x^2} (10+2 x-2 \log (5))+(-10-2 x) \log (5)+\log ^2(5)+\left (-10-2 e^{-8+50 x^2}-2 x+2 \log (5)\right ) \log \left (x^2\right )+\log ^2\left (x^2\right )} \, dx=\int -\frac {3\,\ln \left (x^2\right )+3\,\ln \left (5\right )+{\mathrm {e}}^{50\,x^2-8}\,\left (300\,x^2-3\right )-21}{10\,x+{\mathrm {e}}^{100\,x^2-16}-\ln \left (x^2\right )\,\left (2\,x-2\,\ln \left (5\right )+2\,{\mathrm {e}}^{50\,x^2-8}+10\right )-\ln \left (5\right )\,\left (2\,x+10\right )+{\mathrm {e}}^{50\,x^2-8}\,\left (2\,x-2\,\ln \left (5\right )+10\right )+{\ln \left (x^2\right )}^2+{\ln \left (5\right )}^2+x^2+25} \,d x \] Input:
int(-(3*log(x^2) + 3*log(5) + exp(50*x^2 - 8)*(300*x^2 - 3) - 21)/(10*x + exp(100*x^2 - 16) - log(x^2)*(2*x - 2*log(5) + 2*exp(50*x^2 - 8) + 10) - l og(5)*(2*x + 10) + exp(50*x^2 - 8)*(2*x - 2*log(5) + 10) + log(x^2)^2 + lo g(5)^2 + x^2 + 25),x)
Output:
int(-(3*log(x^2) + 3*log(5) + exp(50*x^2 - 8)*(300*x^2 - 3) - 21)/(10*x + exp(100*x^2 - 16) - log(x^2)*(2*x - 2*log(5) + 2*exp(50*x^2 - 8) + 10) - l og(5)*(2*x + 10) + exp(50*x^2 - 8)*(2*x - 2*log(5) + 10) + log(x^2)^2 + lo g(5)^2 + x^2 + 25), x)
Time = 0.28 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.56 \[ \int \frac {21+e^{-8+50 x^2} \left (3-300 x^2\right )-3 \log (5)-3 \log \left (x^2\right )}{25+e^{-16+100 x^2}+10 x+x^2+e^{-8+50 x^2} (10+2 x-2 \log (5))+(-10-2 x) \log (5)+\log ^2(5)+\left (-10-2 e^{-8+50 x^2}-2 x+2 \log (5)\right ) \log \left (x^2\right )+\log ^2\left (x^2\right )} \, dx=\frac {3 e^{8} x}{e^{50 x^{2}}-\mathrm {log}\left (x^{2}\right ) e^{8}-\mathrm {log}\left (5\right ) e^{8}+e^{8} x +5 e^{8}} \] Input:
int((-3*log(x^2)+(-300*x^2+3)*exp(25*x^2-4)^2-3*log(5)+21)/(log(x^2)^2+(-2 *exp(25*x^2-4)^2+2*log(5)-2*x-10)*log(x^2)+exp(25*x^2-4)^4+(-2*log(5)+2*x+ 10)*exp(25*x^2-4)^2+log(5)^2+(-2*x-10)*log(5)+x^2+10*x+25),x)
Output:
(3*e**8*x)/(e**(50*x**2) - log(x**2)*e**8 - log(5)*e**8 + e**8*x + 5*e**8)