Integrand size = 152, antiderivative size = 27 \[ \int \frac {-100 e^{48} x \log (x)+200 e^{48} x \log ^2(x)+e^x (-100-100 x) \log ^3(x)}{-8 e^{144} x^6+\left (24 e^{96+x} x^5+12 e^{96} x^4 \log (3)\right ) \log (x)+\left (-24 e^{48+2 x} x^4-24 e^{48+x} x^3 \log (3)-6 e^{48} x^2 \log ^2(3)\right ) \log ^2(x)+\left (8 e^{3 x} x^3+12 e^{2 x} x^2 \log (3)+6 e^x x \log ^2(3)+\log ^3(3)\right ) \log ^3(x)} \, dx=\frac {25}{\left (-\log (3)+2 x \left (-e^x+\frac {e^{48} x}{\log (x)}\right )\right )^2} \] Output:
25/(2*x*(exp(48)/ln(x)*x-exp(x))-ln(3))^2
Time = 0.52 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.07 \[ \int \frac {-100 e^{48} x \log (x)+200 e^{48} x \log ^2(x)+e^x (-100-100 x) \log ^3(x)}{-8 e^{144} x^6+\left (24 e^{96+x} x^5+12 e^{96} x^4 \log (3)\right ) \log (x)+\left (-24 e^{48+2 x} x^4-24 e^{48+x} x^3 \log (3)-6 e^{48} x^2 \log ^2(3)\right ) \log ^2(x)+\left (8 e^{3 x} x^3+12 e^{2 x} x^2 \log (3)+6 e^x x \log ^2(3)+\log ^3(3)\right ) \log ^3(x)} \, dx=\frac {25 \log ^2(x)}{\left (-2 e^{48} x^2+\left (2 e^x x+\log (3)\right ) \log (x)\right )^2} \] Input:
Integrate[(-100*E^48*x*Log[x] + 200*E^48*x*Log[x]^2 + E^x*(-100 - 100*x)*L og[x]^3)/(-8*E^144*x^6 + (24*E^(96 + x)*x^5 + 12*E^96*x^4*Log[3])*Log[x] + (-24*E^(48 + 2*x)*x^4 - 24*E^(48 + x)*x^3*Log[3] - 6*E^48*x^2*Log[3]^2)*L og[x]^2 + (8*E^(3*x)*x^3 + 12*E^(2*x)*x^2*Log[3] + 6*E^x*x*Log[3]^2 + Log[ 3]^3)*Log[x]^3),x]
Output:
(25*Log[x]^2)/(-2*E^48*x^2 + (2*E^x*x + Log[3])*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.
| \(\displaystyle \int \frac {e^x (-100 x-100) \log ^3(x)+200 e^{48} x \log ^2(x)-100 e^{48} x \log (x)}{-8 e^{144} x^6+\left (24 e^{x+96} x^5+12 e^{96} x^4 \log (3)\right ) \log (x)+\left (8 e^{3 x} x^3+12 e^{2 x} x^2 \log (3)+6 e^x x \log ^2(3)+\log ^3(3)\right ) \log ^3(x)+\left (-24 e^{2 x+48} x^4-24 e^{x+48} x^3 \log (3)-6 e^{48} x^2 \log ^2(3)\right ) \log ^2(x)} \, dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle \int \frac {100 \log (x) \left (e^{48} x+e^x (x+1) \log ^2(x)-2 e^{48} x \log (x)\right )}{\left (2 e^{48} x^2-\left (2 e^x x+\log (3)\right ) \log (x)\right )^3}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle 100 \int \frac {\log (x) \left (e^x (x+1) \log ^2(x)-2 e^{48} x \log (x)+e^{48} x\right )}{\left (2 e^{48} x^2-\left (2 e^x x+\log (3)\right ) \log (x)\right )^3}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle 100 \int \left (\frac {\log (x) \left (2 e^{48} \log (x) x^3-2 e^{48} \log (x) x^2+2 e^{48} x^2-\log (3) \log ^2(x) x-\log (3) \log ^2(x)\right )}{2 x \left (2 e^{48} x^2-2 e^x \log (x) x-\log (3) \log (x)\right )^3}-\frac {(x+1) \log ^2(x)}{2 x \left (2 e^{48} x^2-2 e^x \log (x) x-\log (3) \log (x)\right )^2}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle 100 \left (-\frac {1}{2} \log (3) \int \frac {\log ^3(x)}{x \left (2 e^{48} x^2-2 e^x \log (x) x-\log (3) \log (x)\right )^3}dx+\frac {1}{2} \log (3) \int \frac {\log ^3(x)}{\left (-2 e^{48} x^2+2 e^x \log (x) x+\log (3) \log (x)\right )^3}dx-e^{48} \int \frac {x \log ^2(x)}{\left (2 e^{48} x^2-2 e^x \log (x) x-\log (3) \log (x)\right )^3}dx+e^{48} \int \frac {x^2 \log ^2(x)}{\left (2 e^{48} x^2-2 e^x \log (x) x-\log (3) \log (x)\right )^3}dx-\frac {1}{2} \int \frac {\log ^2(x)}{x \left (2 e^{48} x^2-2 e^x \log (x) x-\log (3) \log (x)\right )^2}dx-\frac {1}{2} \int \frac {\log ^2(x)}{\left (-2 e^{48} x^2+2 e^x \log (x) x+\log (3) \log (x)\right )^2}dx+e^{48} \int \frac {x \log (x)}{\left (2 e^{48} x^2-2 e^x \log (x) x-\log (3) \log (x)\right )^3}dx\right )\) |
Input:
Int[(-100*E^48*x*Log[x] + 200*E^48*x*Log[x]^2 + E^x*(-100 - 100*x)*Log[x]^ 3)/(-8*E^144*x^6 + (24*E^(96 + x)*x^5 + 12*E^96*x^4*Log[3])*Log[x] + (-24* E^(48 + 2*x)*x^4 - 24*E^(48 + x)*x^3*Log[3] - 6*E^48*x^2*Log[3]^2)*Log[x]^ 2 + (8*E^(3*x)*x^3 + 12*E^(2*x)*x^2*Log[3] + 6*E^x*x*Log[3]^2 + Log[3]^3)* Log[x]^3),x]
Output:
$Aborted
Leaf count of result is larger than twice the leaf count of optimal. \(73\) vs. \(2(25)=50\).
Time = 4.63 (sec) , antiderivative size = 74, normalized size of antiderivative = 2.74
| method | result | size |
| parallelrisch | \(\frac {25 \ln \left (x \right )^{2}}{4 x^{4} {\mathrm e}^{96}-8 x^{3} {\mathrm e}^{48} \ln \left (x \right ) {\mathrm e}^{x}+4 \,{\mathrm e}^{2 x} \ln \left (x \right )^{2} x^{2}-4 x^{2} {\mathrm e}^{48} \ln \left (3\right ) \ln \left (x \right )+4 x \ln \left (3\right ) {\mathrm e}^{x} \ln \left (x \right )^{2}+\ln \left (x \right )^{2} \ln \left (3\right )^{2}}\) | \(74\) |
| risch | \(\frac {25}{\left (2 \,{\mathrm e}^{x} x +\ln \left (3\right )\right )^{2}}-\frac {100 \left ({\mathrm e}^{48} x^{2}-2 x \,{\mathrm e}^{x} \ln \left (x \right )-\ln \left (3\right ) \ln \left (x \right )\right ) {\mathrm e}^{48} x^{2}}{\left (4 \,{\mathrm e}^{2 x} x^{2}+4 x \ln \left (3\right ) {\mathrm e}^{x}+\ln \left (3\right )^{2}\right ) \left (2 \,{\mathrm e}^{48} x^{2}-2 x \,{\mathrm e}^{x} \ln \left (x \right )-\ln \left (3\right ) \ln \left (x \right )\right )^{2}}\) | \(87\) |
Input:
int(((-100*x-100)*exp(x)*ln(x)^3+200*x*exp(48)*ln(x)^2-100*x*exp(48)*ln(x) )/((8*x^3*exp(x)^3+12*x^2*ln(3)*exp(x)^2+6*x*ln(3)^2*exp(x)+ln(3)^3)*ln(x) ^3+(-24*x^4*exp(48)*exp(x)^2-24*x^3*exp(48)*ln(3)*exp(x)-6*x^2*exp(48)*ln( 3)^2)*ln(x)^2+(24*x^5*exp(48)^2*exp(x)+12*x^4*exp(48)^2*ln(3))*ln(x)-8*x^6 *exp(48)^3),x,method=_RETURNVERBOSE)
Output:
25*ln(x)^2/(4*x^4*exp(48)^2-8*x^3*exp(48)*ln(x)*exp(x)+4*x^2*ln(x)^2*exp(x )^2-4*x^2*exp(48)*ln(3)*ln(x)+4*x*ln(3)*exp(x)*ln(x)^2+ln(x)^2*ln(3)^2)
Leaf count of result is larger than twice the leaf count of optimal. 73 vs. \(2 (25) = 50\).
Time = 0.11 (sec) , antiderivative size = 73, normalized size of antiderivative = 2.70 \[ \int \frac {-100 e^{48} x \log (x)+200 e^{48} x \log ^2(x)+e^x (-100-100 x) \log ^3(x)}{-8 e^{144} x^6+\left (24 e^{96+x} x^5+12 e^{96} x^4 \log (3)\right ) \log (x)+\left (-24 e^{48+2 x} x^4-24 e^{48+x} x^3 \log (3)-6 e^{48} x^2 \log ^2(3)\right ) \log ^2(x)+\left (8 e^{3 x} x^3+12 e^{2 x} x^2 \log (3)+6 e^x x \log ^2(3)+\log ^3(3)\right ) \log ^3(x)} \, dx=\frac {25 \, e^{192} \log \left (x\right )^{2}}{4 \, x^{4} e^{288} + {\left (4 \, x^{2} e^{\left (2 \, x + 192\right )} + 4 \, x e^{\left (x + 192\right )} \log \left (3\right ) + e^{192} \log \left (3\right )^{2}\right )} \log \left (x\right )^{2} - 4 \, {\left (2 \, x^{3} e^{\left (x + 240\right )} + x^{2} e^{240} \log \left (3\right )\right )} \log \left (x\right )} \] Input:
integrate(((-100*x-100)*exp(x)*log(x)^3+200*x*exp(48)*log(x)^2-100*x*exp(4 8)*log(x))/((8*x^3*exp(x)^3+12*x^2*log(3)*exp(x)^2+6*x*log(3)^2*exp(x)+log (3)^3)*log(x)^3+(-24*x^4*exp(48)*exp(x)^2-24*x^3*exp(48)*log(3)*exp(x)-6*x ^2*exp(48)*log(3)^2)*log(x)^2+(24*x^5*exp(48)^2*exp(x)+12*x^4*exp(48)^2*lo g(3))*log(x)-8*x^6*exp(48)^3),x, algorithm="fricas")
Output:
25*e^192*log(x)^2/(4*x^4*e^288 + (4*x^2*e^(2*x + 192) + 4*x*e^(x + 192)*lo g(3) + e^192*log(3)^2)*log(x)^2 - 4*(2*x^3*e^(x + 240) + x^2*e^240*log(3)) *log(x))
Leaf count of result is larger than twice the leaf count of optimal. 82 vs. \(2 (20) = 40\).
Time = 0.18 (sec) , antiderivative size = 82, normalized size of antiderivative = 3.04 \[ \int \frac {-100 e^{48} x \log (x)+200 e^{48} x \log ^2(x)+e^x (-100-100 x) \log ^3(x)}{-8 e^{144} x^6+\left (24 e^{96+x} x^5+12 e^{96} x^4 \log (3)\right ) \log (x)+\left (-24 e^{48+2 x} x^4-24 e^{48+x} x^3 \log (3)-6 e^{48} x^2 \log ^2(3)\right ) \log ^2(x)+\left (8 e^{3 x} x^3+12 e^{2 x} x^2 \log (3)+6 e^x x \log ^2(3)+\log ^3(3)\right ) \log ^3(x)} \, dx=\frac {25 \log {\left (x \right )}^{2}}{4 x^{4} e^{96} + 4 x^{2} e^{2 x} \log {\left (x \right )}^{2} - 4 x^{2} e^{48} \log {\left (3 \right )} \log {\left (x \right )} + \left (- 8 x^{3} e^{48} \log {\left (x \right )} + 4 x \log {\left (3 \right )} \log {\left (x \right )}^{2}\right ) e^{x} + \log {\left (3 \right )}^{2} \log {\left (x \right )}^{2}} \] Input:
integrate(((-100*x-100)*exp(x)*ln(x)**3+200*x*exp(48)*ln(x)**2-100*x*exp(4 8)*ln(x))/((8*x**3*exp(x)**3+12*x**2*ln(3)*exp(x)**2+6*x*ln(3)**2*exp(x)+l n(3)**3)*ln(x)**3+(-24*x**4*exp(48)*exp(x)**2-24*x**3*exp(48)*ln(3)*exp(x) -6*x**2*exp(48)*ln(3)**2)*ln(x)**2+(24*x**5*exp(48)**2*exp(x)+12*x**4*exp( 48)**2*ln(3))*ln(x)-8*x**6*exp(48)**3),x)
Output:
25*log(x)**2/(4*x**4*exp(96) + 4*x**2*exp(2*x)*log(x)**2 - 4*x**2*exp(48)* log(3)*log(x) + (-8*x**3*exp(48)*log(x) + 4*x*log(3)*log(x)**2)*exp(x) + l og(3)**2*log(x)**2)
Leaf count of result is larger than twice the leaf count of optimal. 72 vs. \(2 (25) = 50\).
Time = 0.47 (sec) , antiderivative size = 72, normalized size of antiderivative = 2.67 \[ \int \frac {-100 e^{48} x \log (x)+200 e^{48} x \log ^2(x)+e^x (-100-100 x) \log ^3(x)}{-8 e^{144} x^6+\left (24 e^{96+x} x^5+12 e^{96} x^4 \log (3)\right ) \log (x)+\left (-24 e^{48+2 x} x^4-24 e^{48+x} x^3 \log (3)-6 e^{48} x^2 \log ^2(3)\right ) \log ^2(x)+\left (8 e^{3 x} x^3+12 e^{2 x} x^2 \log (3)+6 e^x x \log ^2(3)+\log ^3(3)\right ) \log ^3(x)} \, dx=\frac {25 \, \log \left (x\right )^{2}}{4 \, x^{4} e^{96} - 4 \, x^{2} e^{48} \log \left (3\right ) \log \left (x\right ) + 4 \, x^{2} e^{\left (2 \, x\right )} \log \left (x\right )^{2} + \log \left (3\right )^{2} \log \left (x\right )^{2} - 4 \, {\left (2 \, x^{3} e^{48} \log \left (x\right ) - x \log \left (3\right ) \log \left (x\right )^{2}\right )} e^{x}} \] Input:
integrate(((-100*x-100)*exp(x)*log(x)^3+200*x*exp(48)*log(x)^2-100*x*exp(4 8)*log(x))/((8*x^3*exp(x)^3+12*x^2*log(3)*exp(x)^2+6*x*log(3)^2*exp(x)+log (3)^3)*log(x)^3+(-24*x^4*exp(48)*exp(x)^2-24*x^3*exp(48)*log(3)*exp(x)-6*x ^2*exp(48)*log(3)^2)*log(x)^2+(24*x^5*exp(48)^2*exp(x)+12*x^4*exp(48)^2*lo g(3))*log(x)-8*x^6*exp(48)^3),x, algorithm="maxima")
Output:
25*log(x)^2/(4*x^4*e^96 - 4*x^2*e^48*log(3)*log(x) + 4*x^2*e^(2*x)*log(x)^ 2 + log(3)^2*log(x)^2 - 4*(2*x^3*e^48*log(x) - x*log(3)*log(x)^2)*e^x)
Leaf count of result is larger than twice the leaf count of optimal. 71 vs. \(2 (25) = 50\).
Time = 22.38 (sec) , antiderivative size = 71, normalized size of antiderivative = 2.63 \[ \int \frac {-100 e^{48} x \log (x)+200 e^{48} x \log ^2(x)+e^x (-100-100 x) \log ^3(x)}{-8 e^{144} x^6+\left (24 e^{96+x} x^5+12 e^{96} x^4 \log (3)\right ) \log (x)+\left (-24 e^{48+2 x} x^4-24 e^{48+x} x^3 \log (3)-6 e^{48} x^2 \log ^2(3)\right ) \log ^2(x)+\left (8 e^{3 x} x^3+12 e^{2 x} x^2 \log (3)+6 e^x x \log ^2(3)+\log ^3(3)\right ) \log ^3(x)} \, dx=\frac {25 \, \log \left (x\right )^{2}}{4 \, x^{4} e^{96} - 8 \, x^{3} e^{\left (x + 48\right )} \log \left (x\right ) - 4 \, x^{2} e^{48} \log \left (3\right ) \log \left (x\right ) + 4 \, x^{2} e^{\left (2 \, x\right )} \log \left (x\right )^{2} + 4 \, x e^{x} \log \left (3\right ) \log \left (x\right )^{2} + \log \left (3\right )^{2} \log \left (x\right )^{2}} \] Input:
integrate(((-100*x-100)*exp(x)*log(x)^3+200*x*exp(48)*log(x)^2-100*x*exp(4 8)*log(x))/((8*x^3*exp(x)^3+12*x^2*log(3)*exp(x)^2+6*x*log(3)^2*exp(x)+log (3)^3)*log(x)^3+(-24*x^4*exp(48)*exp(x)^2-24*x^3*exp(48)*log(3)*exp(x)-6*x ^2*exp(48)*log(3)^2)*log(x)^2+(24*x^5*exp(48)^2*exp(x)+12*x^4*exp(48)^2*lo g(3))*log(x)-8*x^6*exp(48)^3),x, algorithm="giac")
Output:
25*log(x)^2/(4*x^4*e^96 - 8*x^3*e^(x + 48)*log(x) - 4*x^2*e^48*log(3)*log( x) + 4*x^2*e^(2*x)*log(x)^2 + 4*x*e^x*log(3)*log(x)^2 + log(3)^2*log(x)^2)
Timed out. \[ \int \frac {-100 e^{48} x \log (x)+200 e^{48} x \log ^2(x)+e^x (-100-100 x) \log ^3(x)}{-8 e^{144} x^6+\left (24 e^{96+x} x^5+12 e^{96} x^4 \log (3)\right ) \log (x)+\left (-24 e^{48+2 x} x^4-24 e^{48+x} x^3 \log (3)-6 e^{48} x^2 \log ^2(3)\right ) \log ^2(x)+\left (8 e^{3 x} x^3+12 e^{2 x} x^2 \log (3)+6 e^x x \log ^2(3)+\log ^3(3)\right ) \log ^3(x)} \, dx=\int -\frac {{\mathrm {e}}^x\,\left (100\,x+100\right )\,{\ln \left (x\right )}^3-200\,x\,{\mathrm {e}}^{48}\,{\ln \left (x\right )}^2+100\,x\,{\mathrm {e}}^{48}\,\ln \left (x\right )}{\ln \left (x\right )\,\left (12\,x^4\,{\mathrm {e}}^{96}\,\ln \left (3\right )+24\,x^5\,{\mathrm {e}}^{96}\,{\mathrm {e}}^x\right )-{\ln \left (x\right )}^2\,\left (6\,x^2\,{\mathrm {e}}^{48}\,{\ln \left (3\right )}^2+24\,x^4\,{\mathrm {e}}^{2\,x}\,{\mathrm {e}}^{48}+24\,x^3\,{\mathrm {e}}^{48}\,{\mathrm {e}}^x\,\ln \left (3\right )\right )-8\,x^6\,{\mathrm {e}}^{144}+{\ln \left (x\right )}^3\,\left (8\,x^3\,{\mathrm {e}}^{3\,x}+{\ln \left (3\right )}^3+6\,x\,{\mathrm {e}}^x\,{\ln \left (3\right )}^2+12\,x^2\,{\mathrm {e}}^{2\,x}\,\ln \left (3\right )\right )} \,d x \] Input:
int(-(100*x*exp(48)*log(x) - 200*x*exp(48)*log(x)^2 + exp(x)*log(x)^3*(100 *x + 100))/(log(x)*(12*x^4*exp(96)*log(3) + 24*x^5*exp(96)*exp(x)) - log(x )^2*(6*x^2*exp(48)*log(3)^2 + 24*x^4*exp(2*x)*exp(48) + 24*x^3*exp(48)*exp (x)*log(3)) - 8*x^6*exp(144) + log(x)^3*(8*x^3*exp(3*x) + log(3)^3 + 6*x*e xp(x)*log(3)^2 + 12*x^2*exp(2*x)*log(3))),x)
Output:
int(-(100*x*exp(48)*log(x) - 200*x*exp(48)*log(x)^2 + exp(x)*log(x)^3*(100 *x + 100))/(log(x)*(12*x^4*exp(96)*log(3) + 24*x^5*exp(96)*exp(x)) - log(x )^2*(6*x^2*exp(48)*log(3)^2 + 24*x^4*exp(2*x)*exp(48) + 24*x^3*exp(48)*exp (x)*log(3)) - 8*x^6*exp(144) + log(x)^3*(8*x^3*exp(3*x) + log(3)^3 + 6*x*e xp(x)*log(3)^2 + 12*x^2*exp(2*x)*log(3))), x)
Time = 0.17 (sec) , antiderivative size = 77, normalized size of antiderivative = 2.85 \[ \int \frac {-100 e^{48} x \log (x)+200 e^{48} x \log ^2(x)+e^x (-100-100 x) \log ^3(x)}{-8 e^{144} x^6+\left (24 e^{96+x} x^5+12 e^{96} x^4 \log (3)\right ) \log (x)+\left (-24 e^{48+2 x} x^4-24 e^{48+x} x^3 \log (3)-6 e^{48} x^2 \log ^2(3)\right ) \log ^2(x)+\left (8 e^{3 x} x^3+12 e^{2 x} x^2 \log (3)+6 e^x x \log ^2(3)+\log ^3(3)\right ) \log ^3(x)} \, dx=\frac {25 \mathrm {log}\left (x \right )^{2}}{4 e^{2 x} \mathrm {log}\left (x \right )^{2} x^{2}+4 e^{x} \mathrm {log}\left (x \right )^{2} \mathrm {log}\left (3\right ) x -8 e^{x} \mathrm {log}\left (x \right ) e^{48} x^{3}+\mathrm {log}\left (x \right )^{2} \mathrm {log}\left (3\right )^{2}-4 \,\mathrm {log}\left (x \right ) \mathrm {log}\left (3\right ) e^{48} x^{2}+4 e^{96} x^{4}} \] Input:
int(((-100*x-100)*exp(x)*log(x)^3+200*x*exp(48)*log(x)^2-100*x*exp(48)*log (x))/((8*x^3*exp(x)^3+12*x^2*log(3)*exp(x)^2+6*x*log(3)^2*exp(x)+log(3)^3) *log(x)^3+(-24*x^4*exp(48)*exp(x)^2-24*x^3*exp(48)*log(3)*exp(x)-6*x^2*exp (48)*log(3)^2)*log(x)^2+(24*x^5*exp(48)^2*exp(x)+12*x^4*exp(48)^2*log(3))* log(x)-8*x^6*exp(48)^3),x)
Output:
(25*log(x)**2)/(4*e**(2*x)*log(x)**2*x**2 + 4*e**x*log(x)**2*log(3)*x - 8* e**x*log(x)*e**48*x**3 + log(x)**2*log(3)**2 - 4*log(x)*log(3)*e**48*x**2 + 4*e**96*x**4)