Integrand size = 137, antiderivative size = 30 \[ \int \frac {e^6 \left (64 x^7-64 x^8+16 x^9\right )+e^x \left (4 e^3 x^2+e^6 \left (-24 x^6+8 x^7\right )\right )+\left (e^{3+x} \left (-12 x^2-4 x^3\right )+e^3 \left (32 x^4-16 x^5\right )\right ) \log (x)+4 x \log ^2(x)}{e^6 \left (16 x^6-16 x^7+4 x^8\right )+e^3 \left (8 x^3-4 x^4\right ) \log (x)+\log ^2(x)} \, dx=2 x \left (x+\frac {e^x}{-2 x+x^2-\frac {\log (x)}{2 e^3 x^2}}\right ) \]
Time = 0.12 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.10 \[ \int \frac {e^6 \left (64 x^7-64 x^8+16 x^9\right )+e^x \left (4 e^3 x^2+e^6 \left (-24 x^6+8 x^7\right )\right )+\left (e^{3+x} \left (-12 x^2-4 x^3\right )+e^3 \left (32 x^4-16 x^5\right )\right ) \log (x)+4 x \log ^2(x)}{e^6 \left (16 x^6-16 x^7+4 x^8\right )+e^3 \left (8 x^3-4 x^4\right ) \log (x)+\log ^2(x)} \, dx=2 x^2 \left (1+\frac {2 e^{3+x} x}{2 e^3 (-2+x) x^3-\log (x)}\right ) \]
Integrate[(E^6*(64*x^7 - 64*x^8 + 16*x^9) + E^x*(4*E^3*x^2 + E^6*(-24*x^6 + 8*x^7)) + (E^(3 + x)*(-12*x^2 - 4*x^3) + E^3*(32*x^4 - 16*x^5))*Log[x] + 4*x*Log[x]^2)/(E^6*(16*x^6 - 16*x^7 + 4*x^8) + E^3*(8*x^3 - 4*x^4)*Log[x] + Log[x]^2),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.
\(\displaystyle \int \frac {e^6 \left (16 x^9-64 x^8+64 x^7\right )+e^x \left (4 e^3 x^2+e^6 \left (8 x^7-24 x^6\right )\right )+\left (e^3 \left (32 x^4-16 x^5\right )+e^{x+3} \left (-4 x^3-12 x^2\right )\right ) \log (x)+4 x \log ^2(x)}{e^3 \left (8 x^3-4 x^4\right ) \log (x)+e^6 \left (4 x^8-16 x^7+16 x^6\right )+\log ^2(x)} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {e^6 \left (16 x^9-64 x^8+64 x^7\right )+e^x \left (4 e^3 x^2+e^6 \left (8 x^7-24 x^6\right )\right )+\left (e^3 \left (32 x^4-16 x^5\right )+e^{x+3} \left (-4 x^3-12 x^2\right )\right ) \log (x)+4 x \log ^2(x)}{\left (-2 e^3 x^4+4 e^3 x^3+\log (x)\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {4 e^{x+3} x^2 \left (2 e^3 x^5-6 e^3 x^4-x \log (x)-3 \log (x)+1\right )}{\left (2 e^3 x^4-4 e^3 x^3-\log (x)\right )^2}+4 x\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 4 \int \frac {e^{x+3} x^3}{2 e^3 x^4-4 e^3 x^3-\log (x)}dx-32 \int \frac {e^{x+6} x^6}{\left (2 e^3 x^4-4 e^3 x^3-\log (x)\right )^2}dx+48 \int \frac {e^{x+6} x^5}{\left (2 e^3 x^4-4 e^3 x^3-\log (x)\right )^2}dx+4 \int \frac {e^{x+3} x^2}{\left (2 e^3 x^4-4 e^3 x^3-\log (x)\right )^2}dx+12 \int \frac {e^{x+3} x^2}{2 e^3 x^4-4 e^3 x^3-\log (x)}dx+2 x^2\) |
Int[(E^6*(64*x^7 - 64*x^8 + 16*x^9) + E^x*(4*E^3*x^2 + E^6*(-24*x^6 + 8*x^ 7)) + (E^(3 + x)*(-12*x^2 - 4*x^3) + E^3*(32*x^4 - 16*x^5))*Log[x] + 4*x*L og[x]^2)/(E^6*(16*x^6 - 16*x^7 + 4*x^8) + E^3*(8*x^3 - 4*x^4)*Log[x] + Log [x]^2),x]
3.27.22.3.1 Defintions of rubi rules used
Time = 5.99 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.23
method | result | size |
risch | \(2 x^{2}+\frac {4 x^{3} {\mathrm e}^{3+x}}{2 x^{4} {\mathrm e}^{3}-4 x^{3} {\mathrm e}^{3}-\ln \left (x \right )}\) | \(37\) |
parallelrisch | \(\frac {4 x^{6} {\mathrm e}^{3}-8 x^{5} {\mathrm e}^{3}+4 \,{\mathrm e}^{3} x^{3} {\mathrm e}^{x}-2 x^{2} \ln \left (x \right )}{2 x^{4} {\mathrm e}^{3}-4 x^{3} {\mathrm e}^{3}-\ln \left (x \right )}\) | \(54\) |
int((4*x*ln(x)^2+((-4*x^3-12*x^2)*exp(3)*exp(x)+(-16*x^5+32*x^4)*exp(3))*l n(x)+((8*x^7-24*x^6)*exp(3)^2+4*x^2*exp(3))*exp(x)+(16*x^9-64*x^8+64*x^7)* exp(3)^2)/(ln(x)^2+(-4*x^4+8*x^3)*exp(3)*ln(x)+(4*x^8-16*x^7+16*x^6)*exp(3 )^2),x,method=_RETURNVERBOSE)
Time = 0.24 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.73 \[ \int \frac {e^6 \left (64 x^7-64 x^8+16 x^9\right )+e^x \left (4 e^3 x^2+e^6 \left (-24 x^6+8 x^7\right )\right )+\left (e^{3+x} \left (-12 x^2-4 x^3\right )+e^3 \left (32 x^4-16 x^5\right )\right ) \log (x)+4 x \log ^2(x)}{e^6 \left (16 x^6-16 x^7+4 x^8\right )+e^3 \left (8 x^3-4 x^4\right ) \log (x)+\log ^2(x)} \, dx=\frac {2 \, {\left (2 \, x^{3} e^{\left (x + 3\right )} - x^{2} \log \left (x\right ) + 2 \, {\left (x^{6} - 2 \, x^{5}\right )} e^{3}\right )}}{2 \, {\left (x^{4} - 2 \, x^{3}\right )} e^{3} - \log \left (x\right )} \]
integrate((4*x*log(x)^2+((-4*x^3-12*x^2)*exp(3)*exp(x)+(-16*x^5+32*x^4)*ex p(3))*log(x)+((8*x^7-24*x^6)*exp(3)^2+4*x^2*exp(3))*exp(x)+(16*x^9-64*x^8+ 64*x^7)*exp(3)^2)/(log(x)^2+(-4*x^4+8*x^3)*exp(3)*log(x)+(4*x^8-16*x^7+16* x^6)*exp(3)^2),x, algorithm=\
Time = 0.12 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.20 \[ \int \frac {e^6 \left (64 x^7-64 x^8+16 x^9\right )+e^x \left (4 e^3 x^2+e^6 \left (-24 x^6+8 x^7\right )\right )+\left (e^{3+x} \left (-12 x^2-4 x^3\right )+e^3 \left (32 x^4-16 x^5\right )\right ) \log (x)+4 x \log ^2(x)}{e^6 \left (16 x^6-16 x^7+4 x^8\right )+e^3 \left (8 x^3-4 x^4\right ) \log (x)+\log ^2(x)} \, dx=\frac {4 x^{3} e^{3} e^{x}}{2 x^{4} e^{3} - 4 x^{3} e^{3} - \log {\left (x \right )}} + 2 x^{2} \]
integrate((4*x*ln(x)**2+((-4*x**3-12*x**2)*exp(3)*exp(x)+(-16*x**5+32*x**4 )*exp(3))*ln(x)+((8*x**7-24*x**6)*exp(3)**2+4*x**2*exp(3))*exp(x)+(16*x**9 -64*x**8+64*x**7)*exp(3)**2)/(ln(x)**2+(-4*x**4+8*x**3)*exp(3)*ln(x)+(4*x* *8-16*x**7+16*x**6)*exp(3)**2),x)
Time = 0.24 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.80 \[ \int \frac {e^6 \left (64 x^7-64 x^8+16 x^9\right )+e^x \left (4 e^3 x^2+e^6 \left (-24 x^6+8 x^7\right )\right )+\left (e^{3+x} \left (-12 x^2-4 x^3\right )+e^3 \left (32 x^4-16 x^5\right )\right ) \log (x)+4 x \log ^2(x)}{e^6 \left (16 x^6-16 x^7+4 x^8\right )+e^3 \left (8 x^3-4 x^4\right ) \log (x)+\log ^2(x)} \, dx=\frac {2 \, {\left (2 \, x^{6} e^{3} - 4 \, x^{5} e^{3} + 2 \, x^{3} e^{\left (x + 3\right )} - x^{2} \log \left (x\right )\right )}}{2 \, x^{4} e^{3} - 4 \, x^{3} e^{3} - \log \left (x\right )} \]
integrate((4*x*log(x)^2+((-4*x^3-12*x^2)*exp(3)*exp(x)+(-16*x^5+32*x^4)*ex p(3))*log(x)+((8*x^7-24*x^6)*exp(3)^2+4*x^2*exp(3))*exp(x)+(16*x^9-64*x^8+ 64*x^7)*exp(3)^2)/(log(x)^2+(-4*x^4+8*x^3)*exp(3)*log(x)+(4*x^8-16*x^7+16* x^6)*exp(3)^2),x, algorithm=\
Time = 0.29 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.80 \[ \int \frac {e^6 \left (64 x^7-64 x^8+16 x^9\right )+e^x \left (4 e^3 x^2+e^6 \left (-24 x^6+8 x^7\right )\right )+\left (e^{3+x} \left (-12 x^2-4 x^3\right )+e^3 \left (32 x^4-16 x^5\right )\right ) \log (x)+4 x \log ^2(x)}{e^6 \left (16 x^6-16 x^7+4 x^8\right )+e^3 \left (8 x^3-4 x^4\right ) \log (x)+\log ^2(x)} \, dx=\frac {2 \, {\left (2 \, x^{6} e^{3} - 4 \, x^{5} e^{3} + 2 \, x^{3} e^{\left (x + 3\right )} - x^{2} \log \left (x\right )\right )}}{2 \, x^{4} e^{3} - 4 \, x^{3} e^{3} - \log \left (x\right )} \]
integrate((4*x*log(x)^2+((-4*x^3-12*x^2)*exp(3)*exp(x)+(-16*x^5+32*x^4)*ex p(3))*log(x)+((8*x^7-24*x^6)*exp(3)^2+4*x^2*exp(3))*exp(x)+(16*x^9-64*x^8+ 64*x^7)*exp(3)^2)/(log(x)^2+(-4*x^4+8*x^3)*exp(3)*log(x)+(4*x^8-16*x^7+16* x^6)*exp(3)^2),x, algorithm=\
Timed out. \[ \int \frac {e^6 \left (64 x^7-64 x^8+16 x^9\right )+e^x \left (4 e^3 x^2+e^6 \left (-24 x^6+8 x^7\right )\right )+\left (e^{3+x} \left (-12 x^2-4 x^3\right )+e^3 \left (32 x^4-16 x^5\right )\right ) \log (x)+4 x \log ^2(x)}{e^6 \left (16 x^6-16 x^7+4 x^8\right )+e^3 \left (8 x^3-4 x^4\right ) \log (x)+\log ^2(x)} \, dx=\int \frac {4\,x\,{\ln \left (x\right )}^2+\left ({\mathrm {e}}^3\,\left (32\,x^4-16\,x^5\right )-{\mathrm {e}}^{x+3}\,\left (4\,x^3+12\,x^2\right )\right )\,\ln \left (x\right )+{\mathrm {e}}^6\,\left (16\,x^9-64\,x^8+64\,x^7\right )-{\mathrm {e}}^x\,\left ({\mathrm {e}}^6\,\left (24\,x^6-8\,x^7\right )-4\,x^2\,{\mathrm {e}}^3\right )}{{\ln \left (x\right )}^2+{\mathrm {e}}^3\,\left (8\,x^3-4\,x^4\right )\,\ln \left (x\right )+{\mathrm {e}}^6\,\left (4\,x^8-16\,x^7+16\,x^6\right )} \,d x \]
int((4*x*log(x)^2 + log(x)*(exp(3)*(32*x^4 - 16*x^5) - exp(3)*exp(x)*(12*x ^2 + 4*x^3)) + exp(6)*(64*x^7 - 64*x^8 + 16*x^9) - exp(x)*(exp(6)*(24*x^6 - 8*x^7) - 4*x^2*exp(3)))/(log(x)^2 + exp(6)*(16*x^6 - 16*x^7 + 4*x^8) + e xp(3)*log(x)*(8*x^3 - 4*x^4)),x)