Integrand size = 182, antiderivative size = 29 \[ \int \frac {75+225 x+\left (150 x-30 e x+30 x^2-150 x \log (5)\right ) \log \left (\frac {3}{x}\right )+\left (30 x+e^2 x+10 x^2+x^3+e \left (-10 x-2 x^2\right )+\left (-50 x+10 e x-10 x^2\right ) \log (5)+25 x \log ^2(5)\right ) \log ^2\left (\frac {3}{x}\right )}{225 x+\left (150 x-30 e x+30 x^2-150 x \log (5)\right ) \log \left (\frac {3}{x}\right )+\left (25 x+e^2 x+10 x^2+x^3+e \left (-10 x-2 x^2\right )+\left (-50 x+10 e x-10 x^2\right ) \log (5)+25 x \log ^2(5)\right ) \log ^2\left (\frac {3}{x}\right )} \, dx=x+\frac {5}{-5+e-x-5 \left (-\log (5)+\frac {3}{\log \left (\frac {3}{x}\right )}\right )} \] Output:
x+5/(exp(1)-5-x-15/ln(3/x)+5*ln(5))
Time = 0.06 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.66 \[ \int \frac {75+225 x+\left (150 x-30 e x+30 x^2-150 x \log (5)\right ) \log \left (\frac {3}{x}\right )+\left (30 x+e^2 x+10 x^2+x^3+e \left (-10 x-2 x^2\right )+\left (-50 x+10 e x-10 x^2\right ) \log (5)+25 x \log ^2(5)\right ) \log ^2\left (\frac {3}{x}\right )}{225 x+\left (150 x-30 e x+30 x^2-150 x \log (5)\right ) \log \left (\frac {3}{x}\right )+\left (25 x+e^2 x+10 x^2+x^3+e \left (-10 x-2 x^2\right )+\left (-50 x+10 e x-10 x^2\right ) \log (5)+25 x \log ^2(5)\right ) \log ^2\left (\frac {3}{x}\right )} \, dx=\frac {15 x+\left (-5+x^2-x (-5+e+5 \log (5))\right ) \log \left (\frac {3}{x}\right )}{15+(5-e+x-5 \log (5)) \log \left (\frac {3}{x}\right )} \] Input:
Integrate[(75 + 225*x + (150*x - 30*E*x + 30*x^2 - 150*x*Log[5])*Log[3/x] + (30*x + E^2*x + 10*x^2 + x^3 + E*(-10*x - 2*x^2) + (-50*x + 10*E*x - 10* x^2)*Log[5] + 25*x*Log[5]^2)*Log[3/x]^2)/(225*x + (150*x - 30*E*x + 30*x^2 - 150*x*Log[5])*Log[3/x] + (25*x + E^2*x + 10*x^2 + x^3 + E*(-10*x - 2*x^ 2) + (-50*x + 10*E*x - 10*x^2)*Log[5] + 25*x*Log[5]^2)*Log[3/x]^2),x]
Output:
(15*x + (-5 + x^2 - x*(-5 + E + 5*Log[5]))*Log[3/x])/(15 + (5 - E + x - 5* Log[5])*Log[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.
| \(\displaystyle \int \frac {\left (30 x^2-30 e x+150 x-150 x \log (5)\right ) \log \left (\frac {3}{x}\right )+\left (x^3+10 x^2+e \left (-2 x^2-10 x\right )+\left (-10 x^2+10 e x-50 x\right ) \log (5)+e^2 x+30 x+25 x \log ^2(5)\right ) \log ^2\left (\frac {3}{x}\right )+225 x+75}{\left (30 x^2-30 e x+150 x-150 x \log (5)\right ) \log \left (\frac {3}{x}\right )+\left (x^3+10 x^2+e \left (-2 x^2-10 x\right )+\left (-10 x^2+10 e x-50 x\right ) \log (5)+e^2 x+25 x+25 x \log ^2(5)\right ) \log ^2\left (\frac {3}{x}\right )+225 x} \, dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle \int \frac {x \left (x^2-10 x (\log (5)-1)-2 e (x+5-5 \log (5))+e^2+5 \left (6+5 \log ^2(5)-10 \log (5)\right )\right ) \log ^2\left (\frac {3}{x}\right )+75 (3 x+1)+30 x (x-e+5-5 \log (5)) \log \left (\frac {3}{x}\right )}{x \left ((x-e+5-5 \log (5)) \log \left (\frac {3}{x}\right )+15\right )^2}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (\frac {x^2+2 x (5-e-5 \log (5))+e^2-10 e+30+25 \log ^2(5)+10 e \log (5)-50 \log (5)}{(x-e+5-5 \log (5))^2}+\frac {75 \left (x^2+x (25-2 e-5 \log (25))+(-5+e+5 \log (5))^2\right )}{x (x-e+5-5 \log (5))^2 \left (x \log \left (\frac {3}{x}\right )+5 \left (1-\frac {e}{5}-\log (5)\right ) \log \left (\frac {3}{x}\right )+15\right )^2}+\frac {150}{(x-e+5-5 \log (5))^2 \left (-x \log \left (\frac {3}{x}\right )-5 \left (1-\frac {e}{5}-\log (5)\right ) \log \left (\frac {3}{x}\right )-15\right )}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle 150 \int \frac {1}{(x-5 \log (5)-e+5)^2 \left (-x \log \left (\frac {3}{x}\right )-5 \left (1-\frac {e}{5}-\log (5)\right ) \log \left (\frac {3}{x}\right )-15\right )}dx+75 \int \frac {1}{x \left (x \log \left (\frac {3}{x}\right )+5 \left (1-\frac {e}{5}-\log (5)\right ) \log \left (\frac {3}{x}\right )+15\right )^2}dx+1125 \int \frac {1}{(x-5 \log (5)-e+5)^2 \left (x \log \left (\frac {3}{x}\right )+5 \left (1-\frac {e}{5}-\log (5)\right ) \log \left (\frac {3}{x}\right )+15\right )^2}dx+x-\frac {5}{x-e+5-5 \log (5)}\) |
Input:
Int[(75 + 225*x + (150*x - 30*E*x + 30*x^2 - 150*x*Log[5])*Log[3/x] + (30* x + E^2*x + 10*x^2 + x^3 + E*(-10*x - 2*x^2) + (-50*x + 10*E*x - 10*x^2)*L og[5] + 25*x*Log[5]^2)*Log[3/x]^2)/(225*x + (150*x - 30*E*x + 30*x^2 - 150 *x*Log[5])*Log[3/x] + (25*x + E^2*x + 10*x^2 + x^3 + E*(-10*x - 2*x^2) + ( -50*x + 10*E*x - 10*x^2)*Log[5] + 25*x*Log[5]^2)*Log[3/x]^2),x]
Output:
$Aborted
Leaf count of result is larger than twice the leaf count of optimal. \(89\) vs. \(2(27)=54\).
Time = 9.15 (sec) , antiderivative size = 90, normalized size of antiderivative = 3.10
| method | result | size |
| risch | \(\frac {5 x \ln \left (5\right )+x \,{\mathrm e}-x^{2}-5 x +5}{5 \ln \left (5\right )+{\mathrm e}-x -5}+\frac {75}{\left (5 \ln \left (5\right )+{\mathrm e}-x -5\right ) \left (5 \ln \left (\frac {3}{x}\right ) \ln \left (5\right )+\ln \left (\frac {3}{x}\right ) {\mathrm e}-x \ln \left (\frac {3}{x}\right )-5 \ln \left (\frac {3}{x}\right )-15\right )}\) | \(90\) |
| norman | \(\frac {\left (25 \ln \left (5\right )^{2}+10 \,{\mathrm e} \ln \left (5\right )+{\mathrm e}^{2}-50 \ln \left (5\right )-10 \,{\mathrm e}+30\right ) \ln \left (\frac {3}{x}\right )+75-15 x -x^{2} \ln \left (\frac {3}{x}\right )-15 \,{\mathrm e}-75 \ln \left (5\right )}{5 \ln \left (\frac {3}{x}\right ) \ln \left (5\right )+\ln \left (\frac {3}{x}\right ) {\mathrm e}-x \ln \left (\frac {3}{x}\right )-5 \ln \left (\frac {3}{x}\right )-15}\) | \(99\) |
| derivativedivides | \(-\frac {\left (\frac {225 \ln \left (5\right )+45 \,{\mathrm e}-225}{x^{2}}+\frac {9 \left (-10+\frac {10 \,{\mathrm e}}{3}-\frac {10 \,{\mathrm e} \ln \left (5\right )}{3}-\frac {25 \ln \left (5\right )^{2}}{3}+\frac {50 \ln \left (5\right )}{3}-\frac {{\mathrm e}^{2}}{3}\right ) \ln \left (\frac {3}{x}\right )}{x^{2}}+3 \ln \left (\frac {3}{x}\right )+\frac {45}{x}\right ) x}{\frac {15 \ln \left (5\right ) \ln \left (\frac {3}{x}\right )}{x}+\frac {3 \,{\mathrm e} \ln \left (\frac {3}{x}\right )}{x}-\frac {15 \ln \left (\frac {3}{x}\right )}{x}-\frac {45}{x}-3 \ln \left (\frac {3}{x}\right )}\) | \(125\) |
| default | \(-\frac {\left (\frac {225 \ln \left (5\right )+45 \,{\mathrm e}-225}{x^{2}}+\frac {9 \left (-10+\frac {10 \,{\mathrm e}}{3}-\frac {10 \,{\mathrm e} \ln \left (5\right )}{3}-\frac {25 \ln \left (5\right )^{2}}{3}+\frac {50 \ln \left (5\right )}{3}-\frac {{\mathrm e}^{2}}{3}\right ) \ln \left (\frac {3}{x}\right )}{x^{2}}+3 \ln \left (\frac {3}{x}\right )+\frac {45}{x}\right ) x}{\frac {15 \ln \left (5\right ) \ln \left (\frac {3}{x}\right )}{x}+\frac {3 \,{\mathrm e} \ln \left (\frac {3}{x}\right )}{x}-\frac {15 \ln \left (\frac {3}{x}\right )}{x}-\frac {45}{x}-3 \ln \left (\frac {3}{x}\right )}\) | \(125\) |
| parallelrisch | \(\frac {75-15 x -15 \,{\mathrm e}-x^{2} \ln \left (\frac {3}{x}\right )+30 \ln \left (\frac {3}{x}\right )-75 \ln \left (5\right )+10 \ln \left (5\right ) \ln \left (\frac {3}{x}\right ) {\mathrm e}+{\mathrm e}^{2} \ln \left (\frac {3}{x}\right )+25 \ln \left (5\right )^{2} \ln \left (\frac {3}{x}\right )-50 \ln \left (\frac {3}{x}\right ) \ln \left (5\right )-10 \ln \left (\frac {3}{x}\right ) {\mathrm e}}{5 \ln \left (\frac {3}{x}\right ) \ln \left (5\right )+\ln \left (\frac {3}{x}\right ) {\mathrm e}-x \ln \left (\frac {3}{x}\right )-5 \ln \left (\frac {3}{x}\right )-15}\) | \(129\) |
Input:
int(((25*x*ln(5)^2+(10*x*exp(1)-10*x^2-50*x)*ln(5)+x*exp(1)^2+(-2*x^2-10*x )*exp(1)+x^3+10*x^2+30*x)*ln(3/x)^2+(-150*x*ln(5)-30*x*exp(1)+30*x^2+150*x )*ln(3/x)+225*x+75)/((25*x*ln(5)^2+(10*x*exp(1)-10*x^2-50*x)*ln(5)+x*exp(1 )^2+(-2*x^2-10*x)*exp(1)+x^3+10*x^2+25*x)*ln(3/x)^2+(-150*x*ln(5)-30*x*exp (1)+30*x^2+150*x)*ln(3/x)+225*x),x,method=_RETURNVERBOSE)
Output:
(5*x*ln(5)+x*exp(1)-x^2-5*x+5)/(5*ln(5)+exp(1)-x-5)+75/(5*ln(5)+exp(1)-x-5 )/(5*ln(3/x)*ln(5)+ln(3/x)*exp(1)-x*ln(3/x)-5*ln(3/x)-15)
Time = 0.10 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.79 \[ \int \frac {75+225 x+\left (150 x-30 e x+30 x^2-150 x \log (5)\right ) \log \left (\frac {3}{x}\right )+\left (30 x+e^2 x+10 x^2+x^3+e \left (-10 x-2 x^2\right )+\left (-50 x+10 e x-10 x^2\right ) \log (5)+25 x \log ^2(5)\right ) \log ^2\left (\frac {3}{x}\right )}{225 x+\left (150 x-30 e x+30 x^2-150 x \log (5)\right ) \log \left (\frac {3}{x}\right )+\left (25 x+e^2 x+10 x^2+x^3+e \left (-10 x-2 x^2\right )+\left (-50 x+10 e x-10 x^2\right ) \log (5)+25 x \log ^2(5)\right ) \log ^2\left (\frac {3}{x}\right )} \, dx=\frac {{\left (x^{2} - x e - 5 \, x \log \left (5\right ) + 5 \, x - 5\right )} \log \left (\frac {3}{x}\right ) + 15 \, x}{{\left (x - e - 5 \, \log \left (5\right ) + 5\right )} \log \left (\frac {3}{x}\right ) + 15} \] Input:
integrate(((25*x*log(5)^2+(10*exp(1)*x-10*x^2-50*x)*log(5)+x*exp(1)^2+(-2* x^2-10*x)*exp(1)+x^3+10*x^2+30*x)*log(3/x)^2+(-150*x*log(5)-30*exp(1)*x+30 *x^2+150*x)*log(3/x)+225*x+75)/((25*x*log(5)^2+(10*exp(1)*x-10*x^2-50*x)*l og(5)+x*exp(1)^2+(-2*x^2-10*x)*exp(1)+x^3+10*x^2+25*x)*log(3/x)^2+(-150*x* log(5)-30*exp(1)*x+30*x^2+150*x)*log(3/x)+225*x),x, algorithm="fricas")
Output:
((x^2 - x*e - 5*x*log(5) + 5*x - 5)*log(3/x) + 15*x)/((x - e - 5*log(5) + 5)*log(3/x) + 15)
Leaf count of result is larger than twice the leaf count of optimal. 87 vs. \(2 (20) = 40\).
Time = 0.32 (sec) , antiderivative size = 87, normalized size of antiderivative = 3.00 \[ \int \frac {75+225 x+\left (150 x-30 e x+30 x^2-150 x \log (5)\right ) \log \left (\frac {3}{x}\right )+\left (30 x+e^2 x+10 x^2+x^3+e \left (-10 x-2 x^2\right )+\left (-50 x+10 e x-10 x^2\right ) \log (5)+25 x \log ^2(5)\right ) \log ^2\left (\frac {3}{x}\right )}{225 x+\left (150 x-30 e x+30 x^2-150 x \log (5)\right ) \log \left (\frac {3}{x}\right )+\left (25 x+e^2 x+10 x^2+x^3+e \left (-10 x-2 x^2\right )+\left (-50 x+10 e x-10 x^2\right ) \log (5)+25 x \log ^2(5)\right ) \log ^2\left (\frac {3}{x}\right )} \, dx=x + \frac {75}{15 x + \left (x^{2} - 10 x \log {\left (5 \right )} - 2 e x + 10 x - 50 \log {\left (5 \right )} - 10 e + e^{2} + 25 + 10 e \log {\left (5 \right )} + 25 \log {\left (5 \right )}^{2}\right ) \log {\left (\frac {3}{x} \right )} - 75 \log {\left (5 \right )} - 15 e + 75} - \frac {5}{x - 5 \log {\left (5 \right )} - e + 5} \] Input:
integrate(((25*x*ln(5)**2+(10*exp(1)*x-10*x**2-50*x)*ln(5)+x*exp(1)**2+(-2 *x**2-10*x)*exp(1)+x**3+10*x**2+30*x)*ln(3/x)**2+(-150*x*ln(5)-30*exp(1)*x +30*x**2+150*x)*ln(3/x)+225*x+75)/((25*x*ln(5)**2+(10*exp(1)*x-10*x**2-50* x)*ln(5)+x*exp(1)**2+(-2*x**2-10*x)*exp(1)+x**3+10*x**2+25*x)*ln(3/x)**2+( -150*x*ln(5)-30*exp(1)*x+30*x**2+150*x)*ln(3/x)+225*x),x)
Output:
x + 75/(15*x + (x**2 - 10*x*log(5) - 2*E*x + 10*x - 50*log(5) - 10*E + exp (2) + 25 + 10*E*log(5) + 25*log(5)**2)*log(3/x) - 75*log(5) - 15*E + 75) - 5/(x - 5*log(5) - E + 5)
Leaf count of result is larger than twice the leaf count of optimal. 87 vs. \(2 (27) = 54\).
Time = 0.19 (sec) , antiderivative size = 87, normalized size of antiderivative = 3.00 \[ \int \frac {75+225 x+\left (150 x-30 e x+30 x^2-150 x \log (5)\right ) \log \left (\frac {3}{x}\right )+\left (30 x+e^2 x+10 x^2+x^3+e \left (-10 x-2 x^2\right )+\left (-50 x+10 e x-10 x^2\right ) \log (5)+25 x \log ^2(5)\right ) \log ^2\left (\frac {3}{x}\right )}{225 x+\left (150 x-30 e x+30 x^2-150 x \log (5)\right ) \log \left (\frac {3}{x}\right )+\left (25 x+e^2 x+10 x^2+x^3+e \left (-10 x-2 x^2\right )+\left (-50 x+10 e x-10 x^2\right ) \log (5)+25 x \log ^2(5)\right ) \log ^2\left (\frac {3}{x}\right )} \, dx=\frac {x^{2} \log \left (3\right ) - {\left (5 \, {\left (\log \left (5\right ) - 1\right )} \log \left (3\right ) + e \log \left (3\right ) - 15\right )} x - {\left (x^{2} - x {\left (e + 5 \, \log \left (5\right ) - 5\right )} - 5\right )} \log \left (x\right ) - 5 \, \log \left (3\right )}{x \log \left (3\right ) - 5 \, {\left (\log \left (5\right ) - 1\right )} \log \left (3\right ) - e \log \left (3\right ) - {\left (x - e - 5 \, \log \left (5\right ) + 5\right )} \log \left (x\right ) + 15} \] Input:
integrate(((25*x*log(5)^2+(10*exp(1)*x-10*x^2-50*x)*log(5)+x*exp(1)^2+(-2* x^2-10*x)*exp(1)+x^3+10*x^2+30*x)*log(3/x)^2+(-150*x*log(5)-30*exp(1)*x+30 *x^2+150*x)*log(3/x)+225*x+75)/((25*x*log(5)^2+(10*exp(1)*x-10*x^2-50*x)*l og(5)+x*exp(1)^2+(-2*x^2-10*x)*exp(1)+x^3+10*x^2+25*x)*log(3/x)^2+(-150*x* log(5)-30*exp(1)*x+30*x^2+150*x)*log(3/x)+225*x),x, algorithm="maxima")
Output:
(x^2*log(3) - (5*(log(5) - 1)*log(3) + e*log(3) - 15)*x - (x^2 - x*(e + 5* log(5) - 5) - 5)*log(x) - 5*log(3))/(x*log(3) - 5*(log(5) - 1)*log(3) - e* log(3) - (x - e - 5*log(5) + 5)*log(x) + 15)
Leaf count of result is larger than twice the leaf count of optimal. 271 vs. \(2 (27) = 54\).
Time = 0.44 (sec) , antiderivative size = 271, normalized size of antiderivative = 9.34 \[ \int \frac {75+225 x+\left (150 x-30 e x+30 x^2-150 x \log (5)\right ) \log \left (\frac {3}{x}\right )+\left (30 x+e^2 x+10 x^2+x^3+e \left (-10 x-2 x^2\right )+\left (-50 x+10 e x-10 x^2\right ) \log (5)+25 x \log ^2(5)\right ) \log ^2\left (\frac {3}{x}\right )}{225 x+\left (150 x-30 e x+30 x^2-150 x \log (5)\right ) \log \left (\frac {3}{x}\right )+\left (25 x+e^2 x+10 x^2+x^3+e \left (-10 x-2 x^2\right )+\left (-50 x+10 e x-10 x^2\right ) \log (5)+25 x \log ^2(5)\right ) \log ^2\left (\frac {3}{x}\right )} \, dx=\frac {e \log \left (\frac {3}{x}\right ) - \frac {10 \, e \log \left (5\right ) \log \left (\frac {3}{x}\right )}{x} + 5 \, \log \left (5\right ) \log \left (\frac {3}{x}\right ) - \frac {25 \, \log \left (5\right )^{2} \log \left (\frac {3}{x}\right )}{x} - \frac {e^{2} \log \left (\frac {3}{x}\right )}{x} + \frac {10 \, e \log \left (\frac {3}{x}\right )}{x} + \frac {50 \, \log \left (5\right ) \log \left (\frac {3}{x}\right )}{x} + \frac {15 \, e}{x} + \frac {75 \, \log \left (5\right )}{x} - \frac {30 \, \log \left (\frac {3}{x}\right )}{x} - \frac {75}{x} - \frac {75}{x^{2}} - 5 \, \log \left (\frac {3}{x}\right )}{\frac {e \log \left (\frac {3}{x}\right )}{x} + \frac {5 \, \log \left (5\right ) \log \left (\frac {3}{x}\right )}{x} - \frac {10 \, e \log \left (5\right ) \log \left (\frac {3}{x}\right )}{x^{2}} - \frac {25 \, \log \left (5\right )^{2} \log \left (\frac {3}{x}\right )}{x^{2}} - \frac {5 \, \log \left (\frac {3}{x}\right )}{x} - \frac {e^{2} \log \left (\frac {3}{x}\right )}{x^{2}} + \frac {10 \, e \log \left (\frac {3}{x}\right )}{x^{2}} + \frac {50 \, \log \left (5\right ) \log \left (\frac {3}{x}\right )}{x^{2}} + \frac {15 \, e}{x^{2}} + \frac {75 \, \log \left (5\right )}{x^{2}} - \frac {25 \, \log \left (\frac {3}{x}\right )}{x^{2}} - \frac {75}{x^{2}}} \] Input:
integrate(((25*x*log(5)^2+(10*exp(1)*x-10*x^2-50*x)*log(5)+x*exp(1)^2+(-2* x^2-10*x)*exp(1)+x^3+10*x^2+30*x)*log(3/x)^2+(-150*x*log(5)-30*exp(1)*x+30 *x^2+150*x)*log(3/x)+225*x+75)/((25*x*log(5)^2+(10*exp(1)*x-10*x^2-50*x)*l og(5)+x*exp(1)^2+(-2*x^2-10*x)*exp(1)+x^3+10*x^2+25*x)*log(3/x)^2+(-150*x* log(5)-30*exp(1)*x+30*x^2+150*x)*log(3/x)+225*x),x, algorithm="giac")
Output:
(e*log(3/x) - 10*e*log(5)*log(3/x)/x + 5*log(5)*log(3/x) - 25*log(5)^2*log (3/x)/x - e^2*log(3/x)/x + 10*e*log(3/x)/x + 50*log(5)*log(3/x)/x + 15*e/x + 75*log(5)/x - 30*log(3/x)/x - 75/x - 75/x^2 - 5*log(3/x))/(e*log(3/x)/x + 5*log(5)*log(3/x)/x - 10*e*log(5)*log(3/x)/x^2 - 25*log(5)^2*log(3/x)/x ^2 - 5*log(3/x)/x - e^2*log(3/x)/x^2 + 10*e*log(3/x)/x^2 + 50*log(5)*log(3 /x)/x^2 + 15*e/x^2 + 75*log(5)/x^2 - 25*log(3/x)/x^2 - 75/x^2)
Time = 7.28 (sec) , antiderivative size = 94, normalized size of antiderivative = 3.24 \[ \int \frac {75+225 x+\left (150 x-30 e x+30 x^2-150 x \log (5)\right ) \log \left (\frac {3}{x}\right )+\left (30 x+e^2 x+10 x^2+x^3+e \left (-10 x-2 x^2\right )+\left (-50 x+10 e x-10 x^2\right ) \log (5)+25 x \log ^2(5)\right ) \log ^2\left (\frac {3}{x}\right )}{225 x+\left (150 x-30 e x+30 x^2-150 x \log (5)\right ) \log \left (\frac {3}{x}\right )+\left (25 x+e^2 x+10 x^2+x^3+e \left (-10 x-2 x^2\right )+\left (-50 x+10 e x-10 x^2\right ) \log (5)+25 x \log ^2(5)\right ) \log ^2\left (\frac {3}{x}\right )} \, dx=\frac {15\,x-5\,\ln \left (\frac {3}{x}\right )+5\,x\,\ln \left (\frac {3}{x}\right )+x^2\,\ln \left (\frac {3}{x}\right )-x\,\mathrm {e}\,\ln \left (\frac {3}{x}\right )-5\,x\,\ln \left (5\right )\,\ln \left (\frac {3}{x}\right )}{5\,\ln \left (\frac {3}{x}\right )-\mathrm {e}\,\ln \left (\frac {3}{x}\right )-5\,\ln \left (5\right )\,\ln \left (\frac {3}{x}\right )+x\,\ln \left (\frac {3}{x}\right )+15} \] Input:
int((225*x + log(3/x)^2*(30*x - exp(1)*(10*x + 2*x^2) + x*exp(2) - log(5)* (50*x - 10*x*exp(1) + 10*x^2) + 25*x*log(5)^2 + 10*x^2 + x^3) + log(3/x)*( 150*x - 30*x*exp(1) - 150*x*log(5) + 30*x^2) + 75)/(225*x + log(3/x)^2*(25 *x - exp(1)*(10*x + 2*x^2) + x*exp(2) - log(5)*(50*x - 10*x*exp(1) + 10*x^ 2) + 25*x*log(5)^2 + 10*x^2 + x^3) + log(3/x)*(150*x - 30*x*exp(1) - 150*x *log(5) + 30*x^2)),x)
Output:
(15*x - 5*log(3/x) + 5*x*log(3/x) + x^2*log(3/x) - x*exp(1)*log(3/x) - 5*x *log(5)*log(3/x))/(5*log(3/x) - exp(1)*log(3/x) - 5*log(5)*log(3/x) + x*lo g(3/x) + 15)
Time = 0.18 (sec) , antiderivative size = 176, normalized size of antiderivative = 6.07 \[ \int \frac {75+225 x+\left (150 x-30 e x+30 x^2-150 x \log (5)\right ) \log \left (\frac {3}{x}\right )+\left (30 x+e^2 x+10 x^2+x^3+e \left (-10 x-2 x^2\right )+\left (-50 x+10 e x-10 x^2\right ) \log (5)+25 x \log ^2(5)\right ) \log ^2\left (\frac {3}{x}\right )}{225 x+\left (150 x-30 e x+30 x^2-150 x \log (5)\right ) \log \left (\frac {3}{x}\right )+\left (25 x+e^2 x+10 x^2+x^3+e \left (-10 x-2 x^2\right )+\left (-50 x+10 e x-10 x^2\right ) \log (5)+25 x \log ^2(5)\right ) \log ^2\left (\frac {3}{x}\right )} \, dx=\frac {5 \mathrm {log}\left (\frac {3}{x}\right )^{2} \mathrm {log}\left (5\right )+\mathrm {log}\left (\frac {3}{x}\right )^{2} e -\mathrm {log}\left (\frac {3}{x}\right )^{2} x -5 \mathrm {log}\left (\frac {3}{x}\right )^{2}+5 \,\mathrm {log}\left (\frac {3}{x}\right ) \mathrm {log}\left (x \right ) \mathrm {log}\left (5\right )+\mathrm {log}\left (\frac {3}{x}\right ) \mathrm {log}\left (x \right ) e -\mathrm {log}\left (\frac {3}{x}\right ) \mathrm {log}\left (x \right ) x -5 \,\mathrm {log}\left (\frac {3}{x}\right ) \mathrm {log}\left (x \right )+15 \,\mathrm {log}\left (\frac {3}{x}\right ) \mathrm {log}\left (5\right ) x +3 \,\mathrm {log}\left (\frac {3}{x}\right ) e x -3 \,\mathrm {log}\left (\frac {3}{x}\right ) x^{2}-15 \,\mathrm {log}\left (\frac {3}{x}\right ) x -15 \,\mathrm {log}\left (x \right )-45 x}{15 \,\mathrm {log}\left (\frac {3}{x}\right ) \mathrm {log}\left (5\right )+3 \,\mathrm {log}\left (\frac {3}{x}\right ) e -3 \,\mathrm {log}\left (\frac {3}{x}\right ) x -15 \,\mathrm {log}\left (\frac {3}{x}\right )-45} \] Input:
int(((25*x*log(5)^2+(10*exp(1)*x-10*x^2-50*x)*log(5)+x*exp(1)^2+(-2*x^2-10 *x)*exp(1)+x^3+10*x^2+30*x)*log(3/x)^2+(-150*x*log(5)-30*exp(1)*x+30*x^2+1 50*x)*log(3/x)+225*x+75)/((25*x*log(5)^2+(10*exp(1)*x-10*x^2-50*x)*log(5)+ x*exp(1)^2+(-2*x^2-10*x)*exp(1)+x^3+10*x^2+25*x)*log(3/x)^2+(-150*x*log(5) -30*exp(1)*x+30*x^2+150*x)*log(3/x)+225*x),x)
Output:
(5*log(3/x)**2*log(5) + log(3/x)**2*e - log(3/x)**2*x - 5*log(3/x)**2 + 5* log(3/x)*log(x)*log(5) + log(3/x)*log(x)*e - log(3/x)*log(x)*x - 5*log(3/x )*log(x) + 15*log(3/x)*log(5)*x + 3*log(3/x)*e*x - 3*log(3/x)*x**2 - 15*lo g(3/x)*x - 15*log(x) - 45*x)/(3*(5*log(3/x)*log(5) + log(3/x)*e - log(3/x) *x - 5*log(3/x) - 15))