Integrand size = 113, antiderivative size = 26 \[ \int \frac {-4 e x^2+16 x^3+4 x^4+e^{1+x} \left (-48 x^2+16 x^3\right )+12 e x^2 \log (x)}{16 e^{2+2 x}+4 x^2+4 x^3+x^4+e^{1+x} \left (-16 x-8 x^2\right )+\left (-8 e^{2+x}+e \left (4 x+2 x^2\right )\right ) \log (x)+e^2 \log ^2(x)} \, dx=\frac {4 x^2}{2+x \left (1+\frac {e \left (-4 e^x+\log (x)\right )}{x^2}\right )} \]
Time = 0.47 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96 \[ \int \frac {-4 e x^2+16 x^3+4 x^4+e^{1+x} \left (-48 x^2+16 x^3\right )+12 e x^2 \log (x)}{16 e^{2+2 x}+4 x^2+4 x^3+x^4+e^{1+x} \left (-16 x-8 x^2\right )+\left (-8 e^{2+x}+e \left (4 x+2 x^2\right )\right ) \log (x)+e^2 \log ^2(x)} \, dx=\frac {4 x^3}{-4 e^{1+x}+2 x+x^2+e \log (x)} \]
Integrate[(-4*E*x^2 + 16*x^3 + 4*x^4 + E^(1 + x)*(-48*x^2 + 16*x^3) + 12*E *x^2*Log[x])/(16*E^(2 + 2*x) + 4*x^2 + 4*x^3 + x^4 + E^(1 + x)*(-16*x - 8* x^2) + (-8*E^(2 + x) + E*(4*x + 2*x^2))*Log[x] + E^2*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.
\(\displaystyle \int \frac {4 x^4+16 x^3-4 e x^2+12 e x^2 \log (x)+e^{x+1} \left (16 x^3-48 x^2\right )}{x^4+4 x^3+4 x^2+e^{x+1} \left (-8 x^2-16 x\right )+\left (e \left (2 x^2+4 x\right )-8 e^{x+2}\right ) \log (x)+16 e^{2 x+2}+e^2 \log ^2(x)} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {4 x^2 \left (4 e^{x+1} (x-3)+x (x+4)+3 e \log (x)-e\right )}{\left (-x (x+2)+4 e^{x+1}-e \log (x)\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 4 \int -\frac {x^2 \left (4 e^{x+1} (3-x)-x (x+4)-3 e \log (x)+e\right )}{\left (-x (x+2)+4 e^{x+1}-e \log (x)\right )^2}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -4 \int \frac {x^2 \left (4 e^{x+1} (3-x)-x (x+4)-3 e \log (x)+e\right )}{\left (-x (x+2)+4 e^{x+1}-e \log (x)\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -4 \int \left (\frac {(x-3) x^2}{x^2+2 x-4 e^{x+1}+e \log (x)}-\frac {x^2 \left (x^3+e \log (x) x-2 x-e\right )}{\left (x^2+2 x-4 e^{x+1}+e \log (x)\right )^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -4 \left (e \int \frac {x^2}{\left (-x^2-2 x+4 e^{x+1}-e \log (x)\right )^2}dx-3 \int \frac {x^2}{x^2+2 x-4 e^{x+1}+e \log (x)}dx-\int \frac {x^5}{\left (x^2+2 x-4 e^{x+1}+e \log (x)\right )^2}dx+2 \int \frac {x^3}{\left (x^2+2 x-4 e^{x+1}+e \log (x)\right )^2}dx-e \int \frac {x^3 \log (x)}{\left (x^2+2 x-4 e^{x+1}+e \log (x)\right )^2}dx+\int \frac {x^3}{x^2+2 x-4 e^{x+1}+e \log (x)}dx\right )\) |
Int[(-4*E*x^2 + 16*x^3 + 4*x^4 + E^(1 + x)*(-48*x^2 + 16*x^3) + 12*E*x^2*L og[x])/(16*E^(2 + 2*x) + 4*x^2 + 4*x^3 + x^4 + E^(1 + x)*(-16*x - 8*x^2) + (-8*E^(2 + x) + E*(4*x + 2*x^2))*Log[x] + E^2*Log[x]^2),x]
3.4.43.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 0.39 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00
method | result | size |
parallelrisch | \(\frac {4 x^{3}}{{\mathrm e} \ln \left (x \right )-4 \,{\mathrm e} \,{\mathrm e}^{x}+x^{2}+2 x}\) | \(26\) |
risch | \(-\frac {4 x^{3}}{4 \,{\mathrm e}^{1+x}-{\mathrm e} \ln \left (x \right )-x^{2}-2 x}\) | \(29\) |
int((12*x^2*exp(1)*ln(x)+(16*x^3-48*x^2)*exp(1)*exp(x)-4*x^2*exp(1)+4*x^4+ 16*x^3)/(exp(1)^2*ln(x)^2+(-8*exp(1)^2*exp(x)+(2*x^2+4*x)*exp(1))*ln(x)+16 *exp(1)^2*exp(x)^2+(-8*x^2-16*x)*exp(1)*exp(x)+x^4+4*x^3+4*x^2),x,method=_ RETURNVERBOSE)
Time = 0.27 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.19 \[ \int \frac {-4 e x^2+16 x^3+4 x^4+e^{1+x} \left (-48 x^2+16 x^3\right )+12 e x^2 \log (x)}{16 e^{2+2 x}+4 x^2+4 x^3+x^4+e^{1+x} \left (-16 x-8 x^2\right )+\left (-8 e^{2+x}+e \left (4 x+2 x^2\right )\right ) \log (x)+e^2 \log ^2(x)} \, dx=\frac {4 \, x^{3} e}{{\left (x^{2} + 2 \, x\right )} e + e^{2} \log \left (x\right ) - 4 \, e^{\left (x + 2\right )}} \]
integrate((12*x^2*exp(1)*log(x)+(16*x^3-48*x^2)*exp(1)*exp(x)-4*x^2*exp(1) +4*x^4+16*x^3)/(exp(1)^2*log(x)^2+(-8*exp(1)^2*exp(x)+(2*x^2+4*x)*exp(1))* log(x)+16*exp(1)^2*exp(x)^2+(-8*x^2-16*x)*exp(1)*exp(x)+x^4+4*x^3+4*x^2),x , algorithm=\
Time = 0.11 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.04 \[ \int \frac {-4 e x^2+16 x^3+4 x^4+e^{1+x} \left (-48 x^2+16 x^3\right )+12 e x^2 \log (x)}{16 e^{2+2 x}+4 x^2+4 x^3+x^4+e^{1+x} \left (-16 x-8 x^2\right )+\left (-8 e^{2+x}+e \left (4 x+2 x^2\right )\right ) \log (x)+e^2 \log ^2(x)} \, dx=- \frac {4 x^{3}}{- x^{2} - 2 x + 4 e e^{x} - e \log {\left (x \right )}} \]
integrate((12*x**2*exp(1)*ln(x)+(16*x**3-48*x**2)*exp(1)*exp(x)-4*x**2*exp (1)+4*x**4+16*x**3)/(exp(1)**2*ln(x)**2+(-8*exp(1)**2*exp(x)+(2*x**2+4*x)* exp(1))*ln(x)+16*exp(1)**2*exp(x)**2+(-8*x**2-16*x)*exp(1)*exp(x)+x**4+4*x **3+4*x**2),x)
Time = 0.23 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96 \[ \int \frac {-4 e x^2+16 x^3+4 x^4+e^{1+x} \left (-48 x^2+16 x^3\right )+12 e x^2 \log (x)}{16 e^{2+2 x}+4 x^2+4 x^3+x^4+e^{1+x} \left (-16 x-8 x^2\right )+\left (-8 e^{2+x}+e \left (4 x+2 x^2\right )\right ) \log (x)+e^2 \log ^2(x)} \, dx=\frac {4 \, x^{3}}{x^{2} + e \log \left (x\right ) + 2 \, x - 4 \, e^{\left (x + 1\right )}} \]
integrate((12*x^2*exp(1)*log(x)+(16*x^3-48*x^2)*exp(1)*exp(x)-4*x^2*exp(1) +4*x^4+16*x^3)/(exp(1)^2*log(x)^2+(-8*exp(1)^2*exp(x)+(2*x^2+4*x)*exp(1))* log(x)+16*exp(1)^2*exp(x)^2+(-8*x^2-16*x)*exp(1)*exp(x)+x^4+4*x^3+4*x^2),x , algorithm=\
Time = 0.30 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96 \[ \int \frac {-4 e x^2+16 x^3+4 x^4+e^{1+x} \left (-48 x^2+16 x^3\right )+12 e x^2 \log (x)}{16 e^{2+2 x}+4 x^2+4 x^3+x^4+e^{1+x} \left (-16 x-8 x^2\right )+\left (-8 e^{2+x}+e \left (4 x+2 x^2\right )\right ) \log (x)+e^2 \log ^2(x)} \, dx=\frac {4 \, x^{3}}{x^{2} + e \log \left (x\right ) + 2 \, x - 4 \, e^{\left (x + 1\right )}} \]
integrate((12*x^2*exp(1)*log(x)+(16*x^3-48*x^2)*exp(1)*exp(x)-4*x^2*exp(1) +4*x^4+16*x^3)/(exp(1)^2*log(x)^2+(-8*exp(1)^2*exp(x)+(2*x^2+4*x)*exp(1))* log(x)+16*exp(1)^2*exp(x)^2+(-8*x^2-16*x)*exp(1)*exp(x)+x^4+4*x^3+4*x^2),x , algorithm=\
Timed out. \[ \int \frac {-4 e x^2+16 x^3+4 x^4+e^{1+x} \left (-48 x^2+16 x^3\right )+12 e x^2 \log (x)}{16 e^{2+2 x}+4 x^2+4 x^3+x^4+e^{1+x} \left (-16 x-8 x^2\right )+\left (-8 e^{2+x}+e \left (4 x+2 x^2\right )\right ) \log (x)+e^2 \log ^2(x)} \, dx=\int \frac {16\,x^3-4\,x^2\,\mathrm {e}-{\mathrm {e}}^{x+1}\,\left (48\,x^2-16\,x^3\right )+4\,x^4+12\,x^2\,\mathrm {e}\,\ln \left (x\right )}{16\,{\mathrm {e}}^{2\,x+2}+{\mathrm {e}}^2\,{\ln \left (x\right )}^2-{\mathrm {e}}^{x+1}\,\left (8\,x^2+16\,x\right )-\ln \left (x\right )\,\left (8\,{\mathrm {e}}^{x+2}-\mathrm {e}\,\left (2\,x^2+4\,x\right )\right )+4\,x^2+4\,x^3+x^4} \,d x \]
int((16*x^3 - 4*x^2*exp(1) + 4*x^4 - exp(1)*exp(x)*(48*x^2 - 16*x^3) + 12* x^2*exp(1)*log(x))/(16*exp(2*x)*exp(2) + exp(2)*log(x)^2 + log(x)*(exp(1)* (4*x + 2*x^2) - 8*exp(2)*exp(x)) + 4*x^2 + 4*x^3 + x^4 - exp(1)*exp(x)*(16 *x + 8*x^2)),x)