Integrand size = 130, antiderivative size = 24 \[ \int \frac {10 e^4+e^8 \left (-15 x+18 x^4\right )+12 e^8 x^3 \log (4)+2 e^8 x^2 \log ^2(4)+\left (-24 e^4 x^3-8 e^4 x^2 \log (4)\right ) \log (x)+8 x^2 \log ^2(x)}{9 e^8 x^3+6 e^8 x^2 \log (4)+e^8 x \log ^2(4)+\left (-12 e^4 x^2-4 e^4 x \log (4)\right ) \log (x)+4 x \log ^2(x)} \, dx=30+x^2+\frac {5}{x+\log (4)+2 \left (x-\frac {\log (x)}{e^4}\right )} \]
Time = 1.14 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08 \[ \int \frac {10 e^4+e^8 \left (-15 x+18 x^4\right )+12 e^8 x^3 \log (4)+2 e^8 x^2 \log ^2(4)+\left (-24 e^4 x^3-8 e^4 x^2 \log (4)\right ) \log (x)+8 x^2 \log ^2(x)}{9 e^8 x^3+6 e^8 x^2 \log (4)+e^8 x \log ^2(4)+\left (-12 e^4 x^2-4 e^4 x \log (4)\right ) \log (x)+4 x \log ^2(x)} \, dx=x^2+\frac {5 e^4}{e^4 (3 x+\log (4))-2 \log (x)} \]
Integrate[(10*E^4 + E^8*(-15*x + 18*x^4) + 12*E^8*x^3*Log[4] + 2*E^8*x^2*L og[4]^2 + (-24*E^4*x^3 - 8*E^4*x^2*Log[4])*Log[x] + 8*x^2*Log[x]^2)/(9*E^8 *x^3 + 6*E^8*x^2*Log[4] + E^8*x*Log[4]^2 + (-12*E^4*x^2 - 4*E^4*x*Log[4])* Log[x] + 4*x*Log[x]^2),x]
Time = 0.89 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.17, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.023, Rules used = {7292, 7293, 2009}
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.
\(\displaystyle \int \frac {e^8 \left (18 x^4-15 x\right )+12 e^8 x^3 \log (4)+8 x^2 \log ^2(x)+2 e^8 x^2 \log ^2(4)+\left (-24 e^4 x^3-8 e^4 x^2 \log (4)\right ) \log (x)+10 e^4}{9 e^8 x^3+6 e^8 x^2 \log (4)+\left (-12 e^4 x^2-4 e^4 x \log (4)\right ) \log (x)+4 x \log ^2(x)+e^8 x \log ^2(4)} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {e^8 \left (18 x^4-15 x\right )+12 e^8 x^3 \log (4)+8 x^2 \log ^2(x)+2 e^8 x^2 \log ^2(4)+\left (-24 e^4 x^3-8 e^4 x^2 \log (4)\right ) \log (x)+10 e^4}{x \left (3 e^4 x-2 \log (x)+e^4 \log (4)\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (2 x-\frac {5 e^4 \left (3 e^4 x-2\right )}{x \left (3 e^4 x-2 \log (x)+e^4 \log (4)\right )^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle x^2+\frac {5 e^4}{3 e^4 x-2 \log (x)+e^4 \log (4)}\) |
Int[(10*E^4 + E^8*(-15*x + 18*x^4) + 12*E^8*x^3*Log[4] + 2*E^8*x^2*Log[4]^ 2 + (-24*E^4*x^3 - 8*E^4*x^2*Log[4])*Log[x] + 8*x^2*Log[x]^2)/(9*E^8*x^3 + 6*E^8*x^2*Log[4] + E^8*x*Log[4]^2 + (-12*E^4*x^2 - 4*E^4*x*Log[4])*Log[x] + 4*x*Log[x]^2),x]
3.27.39.3.1 Defintions of rubi rules used
Time = 1.04 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.12
method | result | size |
risch | \(x^{2}+\frac {5 \,{\mathrm e}^{4}}{2 \,{\mathrm e}^{4} \ln \left (2\right )+3 x \,{\mathrm e}^{4}-2 \ln \left (x \right )}\) | \(27\) |
default | \(\frac {3 x^{3} {\mathrm e}^{4}-2 x^{2} \ln \left (x \right )+2 x^{2} {\mathrm e}^{4} \ln \left (2\right )+5 \,{\mathrm e}^{4}}{2 \,{\mathrm e}^{4} \ln \left (2\right )+3 x \,{\mathrm e}^{4}-2 \ln \left (x \right )}\) | \(48\) |
norman | \(\frac {3 x^{3} {\mathrm e}^{4}-2 x^{2} \ln \left (x \right )+2 x^{2} {\mathrm e}^{4} \ln \left (2\right )+5 \,{\mathrm e}^{4}}{2 \,{\mathrm e}^{4} \ln \left (2\right )+3 x \,{\mathrm e}^{4}-2 \ln \left (x \right )}\) | \(48\) |
parallelrisch | \(-\frac {-4 x^{2} {\mathrm e}^{4} \ln \left (2\right )-6 x^{3} {\mathrm e}^{4}+4 x^{2} \ln \left (x \right )-10 \,{\mathrm e}^{4}}{2 \left (2 \,{\mathrm e}^{4} \ln \left (2\right )+3 x \,{\mathrm e}^{4}-2 \ln \left (x \right )\right )}\) | \(49\) |
int((8*x^2*ln(x)^2+(-16*x^2*exp(4)*ln(2)-24*x^3*exp(4))*ln(x)+8*x^2*exp(4) ^2*ln(2)^2+24*x^3*exp(4)^2*ln(2)+(18*x^4-15*x)*exp(4)^2+10*exp(4))/(4*x*ln (x)^2+(-8*x*exp(4)*ln(2)-12*x^2*exp(4))*ln(x)+4*x*exp(4)^2*ln(2)^2+12*x^2* exp(4)^2*ln(2)+9*x^3*exp(4)^2),x,method=_RETURNVERBOSE)
Time = 0.26 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.92 \[ \int \frac {10 e^4+e^8 \left (-15 x+18 x^4\right )+12 e^8 x^3 \log (4)+2 e^8 x^2 \log ^2(4)+\left (-24 e^4 x^3-8 e^4 x^2 \log (4)\right ) \log (x)+8 x^2 \log ^2(x)}{9 e^8 x^3+6 e^8 x^2 \log (4)+e^8 x \log ^2(4)+\left (-12 e^4 x^2-4 e^4 x \log (4)\right ) \log (x)+4 x \log ^2(x)} \, dx=\frac {2 \, x^{2} e^{4} \log \left (2\right ) - 2 \, x^{2} \log \left (x\right ) + {\left (3 \, x^{3} + 5\right )} e^{4}}{3 \, x e^{4} + 2 \, e^{4} \log \left (2\right ) - 2 \, \log \left (x\right )} \]
integrate((8*x^2*log(x)^2+(-16*x^2*exp(4)*log(2)-24*x^3*exp(4))*log(x)+8*x ^2*exp(4)^2*log(2)^2+24*x^3*exp(4)^2*log(2)+(18*x^4-15*x)*exp(4)^2+10*exp( 4))/(4*x*log(x)^2+(-8*x*exp(4)*log(2)-12*x^2*exp(4))*log(x)+4*x*exp(4)^2*l og(2)^2+12*x^2*exp(4)^2*log(2)+9*x^3*exp(4)^2),x, algorithm=\
Time = 0.07 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.12 \[ \int \frac {10 e^4+e^8 \left (-15 x+18 x^4\right )+12 e^8 x^3 \log (4)+2 e^8 x^2 \log ^2(4)+\left (-24 e^4 x^3-8 e^4 x^2 \log (4)\right ) \log (x)+8 x^2 \log ^2(x)}{9 e^8 x^3+6 e^8 x^2 \log (4)+e^8 x \log ^2(4)+\left (-12 e^4 x^2-4 e^4 x \log (4)\right ) \log (x)+4 x \log ^2(x)} \, dx=x^{2} - \frac {5 e^{4}}{- 3 x e^{4} + 2 \log {\left (x \right )} - 2 e^{4} \log {\left (2 \right )}} \]
integrate((8*x**2*ln(x)**2+(-16*x**2*exp(4)*ln(2)-24*x**3*exp(4))*ln(x)+8* x**2*exp(4)**2*ln(2)**2+24*x**3*exp(4)**2*ln(2)+(18*x**4-15*x)*exp(4)**2+1 0*exp(4))/(4*x*ln(x)**2+(-8*x*exp(4)*ln(2)-12*x**2*exp(4))*ln(x)+4*x*exp(4 )**2*ln(2)**2+12*x**2*exp(4)**2*ln(2)+9*x**3*exp(4)**2),x)
Leaf count of result is larger than twice the leaf count of optimal. 47 vs. \(2 (23) = 46\).
Time = 0.31 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.96 \[ \int \frac {10 e^4+e^8 \left (-15 x+18 x^4\right )+12 e^8 x^3 \log (4)+2 e^8 x^2 \log ^2(4)+\left (-24 e^4 x^3-8 e^4 x^2 \log (4)\right ) \log (x)+8 x^2 \log ^2(x)}{9 e^8 x^3+6 e^8 x^2 \log (4)+e^8 x \log ^2(4)+\left (-12 e^4 x^2-4 e^4 x \log (4)\right ) \log (x)+4 x \log ^2(x)} \, dx=\frac {3 \, x^{3} e^{4} + 2 \, x^{2} e^{4} \log \left (2\right ) - 2 \, x^{2} \log \left (x\right ) + 5 \, e^{4}}{3 \, x e^{4} + 2 \, e^{4} \log \left (2\right ) - 2 \, \log \left (x\right )} \]
integrate((8*x^2*log(x)^2+(-16*x^2*exp(4)*log(2)-24*x^3*exp(4))*log(x)+8*x ^2*exp(4)^2*log(2)^2+24*x^3*exp(4)^2*log(2)+(18*x^4-15*x)*exp(4)^2+10*exp( 4))/(4*x*log(x)^2+(-8*x*exp(4)*log(2)-12*x^2*exp(4))*log(x)+4*x*exp(4)^2*l og(2)^2+12*x^2*exp(4)^2*log(2)+9*x^3*exp(4)^2),x, algorithm=\
Leaf count of result is larger than twice the leaf count of optimal. 47 vs. \(2 (23) = 46\).
Time = 0.26 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.96 \[ \int \frac {10 e^4+e^8 \left (-15 x+18 x^4\right )+12 e^8 x^3 \log (4)+2 e^8 x^2 \log ^2(4)+\left (-24 e^4 x^3-8 e^4 x^2 \log (4)\right ) \log (x)+8 x^2 \log ^2(x)}{9 e^8 x^3+6 e^8 x^2 \log (4)+e^8 x \log ^2(4)+\left (-12 e^4 x^2-4 e^4 x \log (4)\right ) \log (x)+4 x \log ^2(x)} \, dx=\frac {3 \, x^{3} e^{4} + 2 \, x^{2} e^{4} \log \left (2\right ) - 2 \, x^{2} \log \left (x\right ) + 5 \, e^{4}}{3 \, x e^{4} + 2 \, e^{4} \log \left (2\right ) - 2 \, \log \left (x\right )} \]
integrate((8*x^2*log(x)^2+(-16*x^2*exp(4)*log(2)-24*x^3*exp(4))*log(x)+8*x ^2*exp(4)^2*log(2)^2+24*x^3*exp(4)^2*log(2)+(18*x^4-15*x)*exp(4)^2+10*exp( 4))/(4*x*log(x)^2+(-8*x*exp(4)*log(2)-12*x^2*exp(4))*log(x)+4*x*exp(4)^2*l og(2)^2+12*x^2*exp(4)^2*log(2)+9*x^3*exp(4)^2),x, algorithm=\
Timed out. \[ \int \frac {10 e^4+e^8 \left (-15 x+18 x^4\right )+12 e^8 x^3 \log (4)+2 e^8 x^2 \log ^2(4)+\left (-24 e^4 x^3-8 e^4 x^2 \log (4)\right ) \log (x)+8 x^2 \log ^2(x)}{9 e^8 x^3+6 e^8 x^2 \log (4)+e^8 x \log ^2(4)+\left (-12 e^4 x^2-4 e^4 x \log (4)\right ) \log (x)+4 x \log ^2(x)} \, dx=\int \frac {10\,{\mathrm {e}}^4-{\mathrm {e}}^8\,\left (15\,x-18\,x^4\right )-\ln \left (x\right )\,\left (24\,{\mathrm {e}}^4\,x^3+16\,{\mathrm {e}}^4\,\ln \left (2\right )\,x^2\right )+8\,x^2\,{\ln \left (x\right )}^2+24\,x^3\,{\mathrm {e}}^8\,\ln \left (2\right )+8\,x^2\,{\mathrm {e}}^8\,{\ln \left (2\right )}^2}{4\,x\,{\ln \left (x\right )}^2+9\,x^3\,{\mathrm {e}}^8-\ln \left (x\right )\,\left (12\,{\mathrm {e}}^4\,x^2+8\,{\mathrm {e}}^4\,\ln \left (2\right )\,x\right )+4\,x\,{\mathrm {e}}^8\,{\ln \left (2\right )}^2+12\,x^2\,{\mathrm {e}}^8\,\ln \left (2\right )} \,d x \]
int((10*exp(4) - exp(8)*(15*x - 18*x^4) - log(x)*(24*x^3*exp(4) + 16*x^2*e xp(4)*log(2)) + 8*x^2*log(x)^2 + 24*x^3*exp(8)*log(2) + 8*x^2*exp(8)*log(2 )^2)/(4*x*log(x)^2 + 9*x^3*exp(8) - log(x)*(12*x^2*exp(4) + 8*x*exp(4)*log (2)) + 4*x*exp(8)*log(2)^2 + 12*x^2*exp(8)*log(2)),x)