Integrand size = 61, antiderivative size = 19 \[ \int \frac {8 x^3+4 e x^3+4 x^4+e^2 x^4+e^2 \left (8 x^3+4 e x^3+4 x^4\right ) \log \left (e^3 (2+e+x)\right )}{2+e+x} \, dx=x^4 \left (1+e^2 \log \left (e^3 (2+e+x)\right )\right ) \] Output:
x^4*(1+ln((2+x+exp(1))*exp(3))/exp(-2))
Time = 0.07 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.26 \[ \int \frac {8 x^3+4 e x^3+4 x^4+e^2 x^4+e^2 \left (8 x^3+4 e x^3+4 x^4\right ) \log \left (e^3 (2+e+x)\right )}{2+e+x} \, dx=x^4+3 e^2 x^4+e^2 x^4 \log (2+e+x) \] Input:
Integrate[(8*x^3 + 4*E*x^3 + 4*x^4 + E^2*x^4 + E^2*(8*x^3 + 4*E*x^3 + 4*x^ 4)*Log[E^3*(2 + E + x)])/(2 + E + x),x]
Output:
x^4 + 3*E^2*x^4 + E^2*x^4*Log[2 + E + x]
Time = 0.37 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.95, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.066, Rules used = {6, 6, 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.
\(\displaystyle \int \frac {e^2 x^4+4 x^4+4 e x^3+8 x^3+e^2 \left (4 x^4+4 e x^3+8 x^3\right ) \log \left (e^3 (x+e+2)\right )}{x+e+2} \, dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {e^2 x^4+4 x^4+(8+4 e) x^3+e^2 \left (4 x^4+4 e x^3+8 x^3\right ) \log \left (e^3 (x+e+2)\right )}{x+e+2}dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {\left (4+e^2\right ) x^4+(8+4 e) x^3+e^2 \left (4 x^4+4 e x^3+8 x^3\right ) \log \left (e^3 (x+e+2)\right )}{x+e+2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {\left (\left (4+e^2\right ) x+4 (2+e)\right ) x^3}{x+e+2}+4 e^2 x^3 (\log (x+e+2)+3)\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{4} \left (4+e^2\right ) x^4-\frac {e^2 x^4}{4}+e^2 x^4 (\log (x+e+2)+3)\) |
Input:
Int[(8*x^3 + 4*E*x^3 + 4*x^4 + E^2*x^4 + E^2*(8*x^3 + 4*E*x^3 + 4*x^4)*Log [E^3*(2 + E + x)])/(2 + E + x),x]
Output:
-1/4*(E^2*x^4) + ((4 + E^2)*x^4)/4 + E^2*x^4*(3 + Log[2 + E + x])
Int[(u_.)*((v_.) + (a_.)*(Fx_) + (b_.)*(Fx_))^(p_.), x_Symbol] :> Int[u*(v + (a + b)*Fx)^p, x] /; FreeQ[{a, b}, x] && !FreeQ[Fx, x]
Time = 0.60 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.05
method | result | size |
norman | \(x^{4}+x^{4} {\mathrm e}^{2} \ln \left (\left (2+x +{\mathrm e}\right ) {\mathrm e}^{3}\right )\) | \(20\) |
risch | \(x^{4}+x^{4} {\mathrm e}^{2} \ln \left (\left (2+x +{\mathrm e}\right ) {\mathrm e}^{3}\right )\) | \(20\) |
parallelrisch | \(x^{4}+x^{4} {\mathrm e}^{2} \ln \left (\left (2+x +{\mathrm e}\right ) {\mathrm e}^{3}\right )\) | \(20\) |
parts | \(-16 \,{\mathrm e}^{2} \ln \left (x \,{\mathrm e}^{3}+\left ({\mathrm e}+2\right ) {\mathrm e}^{3}\right )+\frac {25 \,{\mathrm e}^{2} {\mathrm e}^{4}}{12}+\frac {200 \,{\mathrm e} \,{\mathrm e}^{2}}{3}+\frac {50 \,{\mathrm e}^{3} {\mathrm e}^{2}}{3}+50 \left ({\mathrm e}^{2}\right )^{2}+\frac {100 \,{\mathrm e}^{2}}{3}+x^{4}+{\mathrm e}^{2} \left ({\mathrm e}^{4}+8 \,{\mathrm e}^{3}+24 \,{\mathrm e}^{2}+32 \,{\mathrm e}+16\right ) \ln \left (2+x +{\mathrm e}\right )+{\mathrm e}^{2} \ln \left (x \,{\mathrm e}^{3}+\left ({\mathrm e}+2\right ) {\mathrm e}^{3}\right ) x^{4}-{\mathrm e}^{2} {\mathrm e}^{4} \ln \left (x \,{\mathrm e}^{3}+\left ({\mathrm e}+2\right ) {\mathrm e}^{3}\right )-8 \,{\mathrm e}^{3} {\mathrm e}^{2} \ln \left (x \,{\mathrm e}^{3}+\left ({\mathrm e}+2\right ) {\mathrm e}^{3}\right )-32 \,{\mathrm e}^{2} {\mathrm e} \ln \left (x \,{\mathrm e}^{3}+\left ({\mathrm e}+2\right ) {\mathrm e}^{3}\right )-24 \left ({\mathrm e}^{2}\right )^{2} \ln \left (x \,{\mathrm e}^{3}+\left ({\mathrm e}+2\right ) {\mathrm e}^{3}\right )\) | \(188\) |
derivativedivides | \(\text {Expression too large to display}\) | \(1039\) |
default | \(\text {Expression too large to display}\) | \(1039\) |
Input:
int(((4*x^3*exp(1)+4*x^4+8*x^3)*exp(2)*ln((2+x+exp(1))*exp(3))+x^4*exp(2)+ 4*x^3*exp(1)+4*x^4+8*x^3)/(2+x+exp(1)),x,method=_RETURNVERBOSE)
Output:
x^4+x^4*exp(2)*ln((2+x+exp(1))*exp(3))
Time = 0.08 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.05 \[ \int \frac {8 x^3+4 e x^3+4 x^4+e^2 x^4+e^2 \left (8 x^3+4 e x^3+4 x^4\right ) \log \left (e^3 (2+e+x)\right )}{2+e+x} \, dx=x^{4} e^{2} \log \left ({\left (x + 2\right )} e^{3} + e^{4}\right ) + x^{4} \] Input:
integrate(((4*x^3*exp(1)+4*x^4+8*x^3)*exp(2)*log((2+x+exp(1))*exp(3))+x^4* exp(2)+4*x^3*exp(1)+4*x^4+8*x^3)/(2+x+exp(1)),x, algorithm="fricas")
Output:
x^4*e^2*log((x + 2)*e^3 + e^4) + x^4
Time = 0.07 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.05 \[ \int \frac {8 x^3+4 e x^3+4 x^4+e^2 x^4+e^2 \left (8 x^3+4 e x^3+4 x^4\right ) \log \left (e^3 (2+e+x)\right )}{2+e+x} \, dx=x^{4} e^{2} \log {\left (\left (x + 2 + e\right ) e^{3} \right )} + x^{4} \] Input:
integrate(((4*x**3*exp(1)+4*x**4+8*x**3)*exp(2)*ln((2+x+exp(1))*exp(3))+x* *4*exp(2)+4*x**3*exp(1)+4*x**4+8*x**3)/(2+x+exp(1)),x)
Output:
x**4*exp(2)*log((x + 2 + E)*exp(3)) + x**4
Leaf count of result is larger than twice the leaf count of optimal. 679 vs. \(2 (18) = 36\).
Time = 0.05 (sec) , antiderivative size = 679, normalized size of antiderivative = 35.74 \[ \int \frac {8 x^3+4 e x^3+4 x^4+e^2 x^4+e^2 \left (8 x^3+4 e x^3+4 x^4\right ) \log \left (e^3 (2+e+x)\right )}{2+e+x} \, dx =\text {Too large to display} \] Input:
integrate(((4*x^3*exp(1)+4*x^4+8*x^3)*exp(2)*log((2+x+exp(1))*exp(3))+x^4* exp(2)+4*x^3*exp(1)+4*x^4+8*x^3)/(2+x+exp(1)),x, algorithm="maxima")
Output:
x^4 - 4/3*x^3*(e + 2) + 8/3*x^3 + 2*x^2*(e^2 + 4*e + 4) - 4*x^2*(e + 2) + 2/3*(2*x^3 - 3*x^2*(e + 2) + 6*x*(e^2 + 4*e + 4) - 6*(e^3 + 6*e^2 + 12*e + 8)*log(x + e + 2))*e^3*log(x*e^3 + e^4 + 2*e^3) + 1/3*(3*x^4 - 4*x^3*(e + 2) + 6*x^2*(e^2 + 4*e + 4) - 12*x*(e^3 + 6*e^2 + 12*e + 8) + 12*(e^4 + 8* e^3 + 24*e^2 + 32*e + 16)*log(x + e + 2))*e^2*log(x*e^3 + e^4 + 2*e^3) + 4 /3*(2*x^3 - 3*x^2*(e + 2) + 6*x*(e^2 + 4*e + 4) - 6*(e^3 + 6*e^2 + 12*e + 8)*log(x + e + 2))*e^2*log(x*e^3 + e^4 + 2*e^3) - 4*x*(e^3 + 6*e^2 + 12*e + 8) + 8*x*(e^2 + 4*e + 4) - 1/9*(4*x^3 - 15*x^2*(e + 2) - 18*(e^3 + 6*e^2 + 12*e + 8)*log(x + e + 2)^2 + 66*x*(e^2 + 4*e + 4) - 66*(e^3 + 6*e^2 + 1 2*e + 8)*log(x + e + 2))*e^3 - 1/36*(9*x^4 - 28*x^3*(e + 2) + 78*x^2*(e^2 + 4*e + 4) + 72*(e^4 + 8*e^3 + 24*e^2 + 32*e + 16)*log(x + e + 2)^2 - 300* x*(e^3 + 6*e^2 + 12*e + 8) + 300*(e^4 + 8*e^3 + 24*e^2 + 32*e + 16)*log(x + e + 2))*e^2 + 1/12*(3*x^4 - 4*x^3*(e + 2) + 6*x^2*(e^2 + 4*e + 4) - 12*x *(e^3 + 6*e^2 + 12*e + 8) + 12*(e^4 + 8*e^3 + 24*e^2 + 32*e + 16)*log(x + e + 2))*e^2 - 2/9*(4*x^3 - 15*x^2*(e + 2) - 18*(e^3 + 6*e^2 + 12*e + 8)*lo g(x + e + 2)^2 + 66*x*(e^2 + 4*e + 4) - 66*(e^3 + 6*e^2 + 12*e + 8)*log(x + e + 2))*e^2 + 2/3*(2*x^3 - 3*x^2*(e + 2) + 6*x*(e^2 + 4*e + 4) - 6*(e^3 + 6*e^2 + 12*e + 8)*log(x + e + 2))*e + 4*(e^4 + 8*e^3 + 24*e^2 + 32*e + 1 6)*log(x + e + 2) - 8*(e^3 + 6*e^2 + 12*e + 8)*log(x + e + 2)
Leaf count of result is larger than twice the leaf count of optimal. 399 vs. \(2 (18) = 36\).
Time = 0.12 (sec) , antiderivative size = 399, normalized size of antiderivative = 21.00 \[ \int \frac {8 x^3+4 e x^3+4 x^4+e^2 x^4+e^2 \left (8 x^3+4 e x^3+4 x^4\right ) \log \left (e^3 (2+e+x)\right )}{2+e+x} \, dx=x^{4} e^{2} \log \left (x + e + 2\right ) + 3 \, x^{4} e^{2} + \frac {4}{3} \, x^{3} e^{3} \log \left (x e^{3} + e^{4} + 2 \, e^{3}\right ) - \frac {4}{3} \, x^{3} e^{3} \log \left (x + e + 2\right ) + x^{4} - 4 \, x^{3} e^{3} - 2 \, x^{2} e^{4} \log \left (x e^{3} + e^{4} + 2 \, e^{3}\right ) - 4 \, x^{2} e^{3} \log \left (x e^{3} + e^{4} + 2 \, e^{3}\right ) + 2 \, x^{2} e^{4} \log \left (x + e + 2\right ) + 4 \, x^{2} e^{3} \log \left (x + e + 2\right ) + 6 \, x^{2} e^{4} + 12 \, x^{2} e^{3} + 4 \, x e^{5} \log \left (x e^{3} + e^{4} + 2 \, e^{3}\right ) + 16 \, x e^{4} \log \left (x e^{3} + e^{4} + 2 \, e^{3}\right ) + 16 \, x e^{3} \log \left (x e^{3} + e^{4} + 2 \, e^{3}\right ) - 2 \, e^{6} \log \left (x e^{3} + e^{4} + 2 \, e^{3}\right )^{2} - 12 \, e^{5} \log \left (x e^{3} + e^{4} + 2 \, e^{3}\right )^{2} - 24 \, e^{4} \log \left (x e^{3} + e^{4} + 2 \, e^{3}\right )^{2} - 16 \, e^{3} \log \left (x e^{3} + e^{4} + 2 \, e^{3}\right )^{2} - 4 \, x e^{5} \log \left (x + e + 2\right ) - 16 \, x e^{4} \log \left (x + e + 2\right ) - 16 \, x e^{3} \log \left (x + e + 2\right ) + 2 \, e^{6} \log \left (x + e + 2\right )^{2} + 12 \, e^{5} \log \left (x + e + 2\right )^{2} + 24 \, e^{4} \log \left (x + e + 2\right )^{2} + 16 \, e^{3} \log \left (x + e + 2\right )^{2} - 12 \, x e^{5} - 48 \, x e^{4} - 48 \, x e^{3} + 12 \, e^{6} \log \left (x + e + 2\right ) + 72 \, e^{5} \log \left (x + e + 2\right ) + 144 \, e^{4} \log \left (x + e + 2\right ) + 96 \, e^{3} \log \left (x + e + 2\right ) \] Input:
integrate(((4*x^3*exp(1)+4*x^4+8*x^3)*exp(2)*log((2+x+exp(1))*exp(3))+x^4* exp(2)+4*x^3*exp(1)+4*x^4+8*x^3)/(2+x+exp(1)),x, algorithm="giac")
Output:
x^4*e^2*log(x + e + 2) + 3*x^4*e^2 + 4/3*x^3*e^3*log(x*e^3 + e^4 + 2*e^3) - 4/3*x^3*e^3*log(x + e + 2) + x^4 - 4*x^3*e^3 - 2*x^2*e^4*log(x*e^3 + e^4 + 2*e^3) - 4*x^2*e^3*log(x*e^3 + e^4 + 2*e^3) + 2*x^2*e^4*log(x + e + 2) + 4*x^2*e^3*log(x + e + 2) + 6*x^2*e^4 + 12*x^2*e^3 + 4*x*e^5*log(x*e^3 + e^4 + 2*e^3) + 16*x*e^4*log(x*e^3 + e^4 + 2*e^3) + 16*x*e^3*log(x*e^3 + e^ 4 + 2*e^3) - 2*e^6*log(x*e^3 + e^4 + 2*e^3)^2 - 12*e^5*log(x*e^3 + e^4 + 2 *e^3)^2 - 24*e^4*log(x*e^3 + e^4 + 2*e^3)^2 - 16*e^3*log(x*e^3 + e^4 + 2*e ^3)^2 - 4*x*e^5*log(x + e + 2) - 16*x*e^4*log(x + e + 2) - 16*x*e^3*log(x + e + 2) + 2*e^6*log(x + e + 2)^2 + 12*e^5*log(x + e + 2)^2 + 24*e^4*log(x + e + 2)^2 + 16*e^3*log(x + e + 2)^2 - 12*x*e^5 - 48*x*e^4 - 48*x*e^3 + 1 2*e^6*log(x + e + 2) + 72*e^5*log(x + e + 2) + 144*e^4*log(x + e + 2) + 96 *e^3*log(x + e + 2)
Time = 4.21 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.95 \[ \int \frac {8 x^3+4 e x^3+4 x^4+e^2 x^4+e^2 \left (8 x^3+4 e x^3+4 x^4\right ) \log \left (e^3 (2+e+x)\right )}{2+e+x} \, dx=x^4\,\left ({\mathrm {e}}^2\,\ln \left ({\mathrm {e}}^3\,\left (x+\mathrm {e}+2\right )\right )+1\right ) \] Input:
int((4*x^3*exp(1) + x^4*exp(2) + 8*x^3 + 4*x^4 + exp(2)*log(exp(3)*(x + ex p(1) + 2))*(4*x^3*exp(1) + 8*x^3 + 4*x^4))/(x + exp(1) + 2),x)
Output:
x^4*(exp(2)*log(exp(3)*(x + exp(1) + 2)) + 1)
Time = 0.18 (sec) , antiderivative size = 175, normalized size of antiderivative = 9.21 \[ \int \frac {8 x^3+4 e x^3+4 x^4+e^2 x^4+e^2 \left (8 x^3+4 e x^3+4 x^4\right ) \log \left (e^3 (2+e+x)\right )}{2+e+x} \, dx=-\mathrm {log}\left (e^{4}+e^{3} x +2 e^{3}\right ) e^{6}-8 \,\mathrm {log}\left (e^{4}+e^{3} x +2 e^{3}\right ) e^{5}-24 \,\mathrm {log}\left (e^{4}+e^{3} x +2 e^{3}\right ) e^{4}-32 \,\mathrm {log}\left (e^{4}+e^{3} x +2 e^{3}\right ) e^{3}+\mathrm {log}\left (e^{4}+e^{3} x +2 e^{3}\right ) e^{2} x^{4}-16 \,\mathrm {log}\left (e^{4}+e^{3} x +2 e^{3}\right ) e^{2}+\mathrm {log}\left (e +x +2\right ) e^{6}+8 \,\mathrm {log}\left (e +x +2\right ) e^{5}+24 \,\mathrm {log}\left (e +x +2\right ) e^{4}+32 \,\mathrm {log}\left (e +x +2\right ) e^{3}+16 \,\mathrm {log}\left (e +x +2\right ) e^{2}+x^{4} \] Input:
int(((4*x^3*exp(1)+4*x^4+8*x^3)*exp(2)*log((2+x+exp(1))*exp(3))+x^4*exp(2) +4*x^3*exp(1)+4*x^4+8*x^3)/(2+x+exp(1)),x)
Output:
- log(e**4 + e**3*x + 2*e**3)*e**6 - 8*log(e**4 + e**3*x + 2*e**3)*e**5 - 24*log(e**4 + e**3*x + 2*e**3)*e**4 - 32*log(e**4 + e**3*x + 2*e**3)*e**3 + log(e**4 + e**3*x + 2*e**3)*e**2*x**4 - 16*log(e**4 + e**3*x + 2*e**3)* e**2 + log(e + x + 2)*e**6 + 8*log(e + x + 2)*e**5 + 24*log(e + x + 2)*e** 4 + 32*log(e + x + 2)*e**3 + 16*log(e + x + 2)*e**2 + x**4