Integrand size = 75, antiderivative size = 17 \begin {dmath*} \int \frac {10 x+2 x^2+e^3 \left (8 x+4 x^2+2 x^3\right )+\left (10+2 x+e^3 \left (8+4 x+2 x^2\right )\right ) \log \left (4+x+e^3 \left (4+x^2\right )\right )}{4+x+e^3 \left (4+x^2\right )} \, dx=\left (x+\log \left (4+x+e^3 \left (4+x^2\right )\right )\right )^2 \end {dmath*}
Time = 0.03 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \begin {dmath*} \int \frac {10 x+2 x^2+e^3 \left (8 x+4 x^2+2 x^3\right )+\left (10+2 x+e^3 \left (8+4 x+2 x^2\right )\right ) \log \left (4+x+e^3 \left (4+x^2\right )\right )}{4+x+e^3 \left (4+x^2\right )} \, dx=\left (x+\log \left (4+x+e^3 \left (4+x^2\right )\right )\right )^2 \end {dmath*}
Integrate[(10*x + 2*x^2 + E^3*(8*x + 4*x^2 + 2*x^3) + (10 + 2*x + E^3*(8 + 4*x + 2*x^2))*Log[4 + x + E^3*(4 + x^2)])/(4 + x + E^3*(4 + x^2)),x]
Time = 0.40 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.040, Rules used = {7292, 27, 7237}
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.
\(\displaystyle \int \frac {2 x^2+\left (e^3 \left (2 x^2+4 x+8\right )+2 x+10\right ) \log \left (e^3 \left (x^2+4\right )+x+4\right )+e^3 \left (2 x^3+4 x^2+8 x\right )+10 x}{e^3 \left (x^2+4\right )+x+4} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {2 \left (e^3 x^2+\left (1+2 e^3\right ) x+4 e^3+5\right ) \left (\log \left (e^3 \left (x^2+4\right )+x+4\right )+x\right )}{e^3 x^2+x+4 \left (1+e^3\right )}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 2 \int \frac {\left (e^3 x^2+\left (1+2 e^3\right ) x+4 e^3+5\right ) \left (x+\log \left (x+e^3 \left (x^2+4\right )+4\right )\right )}{e^3 x^2+x+4 \left (1+e^3\right )}dx\) |
\(\Big \downarrow \) 7237 |
\(\displaystyle \left (\log \left (e^3 \left (x^2+4\right )+x+4\right )+x\right )^2\) |
Int[(10*x + 2*x^2 + E^3*(8*x + 4*x^2 + 2*x^3) + (10 + 2*x + E^3*(8 + 4*x + 2*x^2))*Log[4 + x + E^3*(4 + x^2)])/(4 + x + E^3*(4 + x^2)),x]
3.13.67.3.1 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_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Si mp[q*(y^(m + 1)/(m + 1)), x] /; !FalseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]
Leaf count of result is larger than twice the leaf count of optimal. \(33\) vs. \(2(16)=32\).
Time = 0.51 (sec) , antiderivative size = 34, normalized size of antiderivative = 2.00
method | result | size |
norman | \(x^{2}+{\ln \left (\left (x^{2}+4\right ) {\mathrm e}^{3}+4+x \right )}^{2}+2 x \ln \left (\left (x^{2}+4\right ) {\mathrm e}^{3}+4+x \right )\) | \(34\) |
risch | \(x^{2}+{\ln \left (\left (x^{2}+4\right ) {\mathrm e}^{3}+4+x \right )}^{2}+2 x \ln \left (\left (x^{2}+4\right ) {\mathrm e}^{3}+4+x \right )\) | \(34\) |
parallelrisch | \(\frac {\left (-8+2 x^{2} {\mathrm e}^{3}+4 \,{\mathrm e}^{3} \ln \left (x^{2} {\mathrm e}^{3}+4 \,{\mathrm e}^{3}+x +4\right ) x +2 \ln \left (x^{2} {\mathrm e}^{3}+4 \,{\mathrm e}^{3}+x +4\right )^{2} {\mathrm e}^{3}-8 \,{\mathrm e}^{3}\right ) {\mathrm e}^{-3}}{2}\) | \(59\) |
default | \(x^{2}+2 \ln \left (x^{2} {\mathrm e}^{3}+4 \,{\mathrm e}^{3}+x +4\right ) x +2 \left (\munderset {\underline {\hspace {1.25 ex}}\alpha =\operatorname {RootOf}\left (\textit {\_Z}^{2} {\mathrm e}^{3}+4 \,{\mathrm e}^{3}+\textit {\_Z} +4\right )}{\sum }\left (\ln \left (x -\underline {\hspace {1.25 ex}}\alpha \right ) \ln \left (x^{2} {\mathrm e}^{3}+4 \,{\mathrm e}^{3}+x +4\right )-\frac {\ln \left (x -\underline {\hspace {1.25 ex}}\alpha \right )^{2}}{2 \left (2 \,{\mathrm e}^{3} \underline {\hspace {1.25 ex}}\alpha +1\right )}-\frac {\left (2 \,{\mathrm e}^{3} \underline {\hspace {1.25 ex}}\alpha +1\right ) \ln \left (x -\underline {\hspace {1.25 ex}}\alpha \right ) \ln \left (\frac {2 \,{\mathrm e}^{3} \underline {\hspace {1.25 ex}}\alpha +\left (x -\underline {\hspace {1.25 ex}}\alpha \right ) {\mathrm e}^{3}+1}{2 \,{\mathrm e}^{3} \underline {\hspace {1.25 ex}}\alpha +1}\right )}{16 \,{\mathrm e}^{6}+16 \,{\mathrm e}^{3}-1}-\frac {\left (2 \,{\mathrm e}^{3} \underline {\hspace {1.25 ex}}\alpha +1\right ) \operatorname {dilog}\left (\frac {2 \,{\mathrm e}^{3} \underline {\hspace {1.25 ex}}\alpha +\left (x -\underline {\hspace {1.25 ex}}\alpha \right ) {\mathrm e}^{3}+1}{2 \,{\mathrm e}^{3} \underline {\hspace {1.25 ex}}\alpha +1}\right )}{16 \,{\mathrm e}^{6}+16 \,{\mathrm e}^{3}-1}\right )\right )\) | \(186\) |
parts | \(x^{2}+2 \ln \left (x^{2} {\mathrm e}^{3}+4 \,{\mathrm e}^{3}+x +4\right ) x +2 \left (\munderset {\underline {\hspace {1.25 ex}}\alpha =\operatorname {RootOf}\left (\textit {\_Z}^{2} {\mathrm e}^{3}+4 \,{\mathrm e}^{3}+\textit {\_Z} +4\right )}{\sum }\left (\ln \left (x -\underline {\hspace {1.25 ex}}\alpha \right ) \ln \left (x^{2} {\mathrm e}^{3}+4 \,{\mathrm e}^{3}+x +4\right )-\frac {\ln \left (x -\underline {\hspace {1.25 ex}}\alpha \right )^{2}}{2 \left (2 \,{\mathrm e}^{3} \underline {\hspace {1.25 ex}}\alpha +1\right )}-\frac {\left (2 \,{\mathrm e}^{3} \underline {\hspace {1.25 ex}}\alpha +1\right ) \ln \left (x -\underline {\hspace {1.25 ex}}\alpha \right ) \ln \left (\frac {2 \,{\mathrm e}^{3} \underline {\hspace {1.25 ex}}\alpha +\left (x -\underline {\hspace {1.25 ex}}\alpha \right ) {\mathrm e}^{3}+1}{2 \,{\mathrm e}^{3} \underline {\hspace {1.25 ex}}\alpha +1}\right )}{16 \,{\mathrm e}^{6}+16 \,{\mathrm e}^{3}-1}-\frac {\left (2 \,{\mathrm e}^{3} \underline {\hspace {1.25 ex}}\alpha +1\right ) \operatorname {dilog}\left (\frac {2 \,{\mathrm e}^{3} \underline {\hspace {1.25 ex}}\alpha +\left (x -\underline {\hspace {1.25 ex}}\alpha \right ) {\mathrm e}^{3}+1}{2 \,{\mathrm e}^{3} \underline {\hspace {1.25 ex}}\alpha +1}\right )}{16 \,{\mathrm e}^{6}+16 \,{\mathrm e}^{3}-1}\right )\right )\) | \(186\) |
int((((2*x^2+4*x+8)*exp(3)+2*x+10)*ln((x^2+4)*exp(3)+4+x)+(2*x^3+4*x^2+8*x )*exp(3)+2*x^2+10*x)/((x^2+4)*exp(3)+4+x),x,method=_RETURNVERBOSE)
Leaf count of result is larger than twice the leaf count of optimal. 33 vs. \(2 (16) = 32\).
Time = 0.25 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.94 \begin {dmath*} \int \frac {10 x+2 x^2+e^3 \left (8 x+4 x^2+2 x^3\right )+\left (10+2 x+e^3 \left (8+4 x+2 x^2\right )\right ) \log \left (4+x+e^3 \left (4+x^2\right )\right )}{4+x+e^3 \left (4+x^2\right )} \, dx=x^{2} + 2 \, x \log \left ({\left (x^{2} + 4\right )} e^{3} + x + 4\right ) + \log \left ({\left (x^{2} + 4\right )} e^{3} + x + 4\right )^{2} \end {dmath*}
integrate((((2*x^2+4*x+8)*exp(3)+2*x+10)*log((x^2+4)*exp(3)+4+x)+(2*x^3+4* x^2+8*x)*exp(3)+2*x^2+10*x)/((x^2+4)*exp(3)+4+x),x, algorithm=\
Leaf count of result is larger than twice the leaf count of optimal. 34 vs. \(2 (15) = 30\).
Time = 0.12 (sec) , antiderivative size = 34, normalized size of antiderivative = 2.00 \begin {dmath*} \int \frac {10 x+2 x^2+e^3 \left (8 x+4 x^2+2 x^3\right )+\left (10+2 x+e^3 \left (8+4 x+2 x^2\right )\right ) \log \left (4+x+e^3 \left (4+x^2\right )\right )}{4+x+e^3 \left (4+x^2\right )} \, dx=x^{2} + 2 x \log {\left (x + \left (x^{2} + 4\right ) e^{3} + 4 \right )} + \log {\left (x + \left (x^{2} + 4\right ) e^{3} + 4 \right )}^{2} \end {dmath*}
integrate((((2*x**2+4*x+8)*exp(3)+2*x+10)*ln((x**2+4)*exp(3)+4+x)+(2*x**3+ 4*x**2+8*x)*exp(3)+2*x**2+10*x)/((x**2+4)*exp(3)+4+x),x)
Leaf count of result is larger than twice the leaf count of optimal. 439 vs. \(2 (16) = 32\).
Time = 0.48 (sec) , antiderivative size = 439, normalized size of antiderivative = 25.82 \begin {dmath*} \int \frac {10 x+2 x^2+e^3 \left (8 x+4 x^2+2 x^3\right )+\left (10+2 x+e^3 \left (8+4 x+2 x^2\right )\right ) \log \left (4+x+e^3 \left (4+x^2\right )\right )}{4+x+e^3 \left (4+x^2\right )} \, dx=2 \, \sqrt {16 \, e^{6} + 16 \, e^{3} - 1} \arctan \left (\frac {2 \, x e^{3} + 1}{\sqrt {16 \, e^{6} + 16 \, e^{3} - 1}}\right ) e^{\left (-3\right )} - \frac {2 \, {\left (8 \, e^{6} + 8 \, e^{3} - 1\right )} \arctan \left (\frac {2 \, x e^{3} + 1}{\sqrt {16 \, e^{6} + 16 \, e^{3} - 1}}\right ) e^{\left (-6\right )}}{\sqrt {16 \, e^{6} + 16 \, e^{3} - 1}} - {\left ({\left (4 \, e^{6} + 4 \, e^{3} - 1\right )} e^{\left (-9\right )} \log \left (x^{2} e^{3} + x + 4 \, e^{3} + 4\right ) - \frac {2 \, {\left (12 \, e^{6} + 12 \, e^{3} - 1\right )} \arctan \left (\frac {2 \, x e^{3} + 1}{\sqrt {16 \, e^{6} + 16 \, e^{3} - 1}}\right ) e^{\left (-9\right )}}{\sqrt {16 \, e^{6} + 16 \, e^{3} - 1}} - {\left (x^{2} e^{3} - 2 \, x\right )} e^{\left (-6\right )}\right )} e^{3} - 2 \, {\left (\frac {2 \, {\left (8 \, e^{6} + 8 \, e^{3} - 1\right )} \arctan \left (\frac {2 \, x e^{3} + 1}{\sqrt {16 \, e^{6} + 16 \, e^{3} - 1}}\right ) e^{\left (-6\right )}}{\sqrt {16 \, e^{6} + 16 \, e^{3} - 1}} - 2 \, x e^{\left (-3\right )} + e^{\left (-6\right )} \log \left (x^{2} e^{3} + x + 4 \, e^{3} + 4\right )\right )} e^{3} + 4 \, {\left (e^{\left (-3\right )} \log \left (x^{2} e^{3} + x + 4 \, e^{3} + 4\right ) - \frac {2 \, \arctan \left (\frac {2 \, x e^{3} + 1}{\sqrt {16 \, e^{6} + 16 \, e^{3} - 1}}\right ) e^{\left (-3\right )}}{\sqrt {16 \, e^{6} + 16 \, e^{3} - 1}}\right )} e^{3} + {\left (e^{3} \log \left (x^{2} e^{3} + x + 4 \, e^{3} + 4\right )^{2} - 4 \, x e^{3} + {\left (2 \, x e^{3} + 1\right )} \log \left (x^{2} e^{3} + x + 4 \, e^{3} + 4\right )\right )} e^{\left (-3\right )} + 2 \, x e^{\left (-3\right )} + 5 \, e^{\left (-3\right )} \log \left (x^{2} e^{3} + x + 4 \, e^{3} + 4\right ) - e^{\left (-6\right )} \log \left (x^{2} e^{3} + x + 4 \, e^{3} + 4\right ) - \frac {10 \, \arctan \left (\frac {2 \, x e^{3} + 1}{\sqrt {16 \, e^{6} + 16 \, e^{3} - 1}}\right ) e^{\left (-3\right )}}{\sqrt {16 \, e^{6} + 16 \, e^{3} - 1}} \end {dmath*}
integrate((((2*x^2+4*x+8)*exp(3)+2*x+10)*log((x^2+4)*exp(3)+4+x)+(2*x^3+4* x^2+8*x)*exp(3)+2*x^2+10*x)/((x^2+4)*exp(3)+4+x),x, algorithm=\
2*sqrt(16*e^6 + 16*e^3 - 1)*arctan((2*x*e^3 + 1)/sqrt(16*e^6 + 16*e^3 - 1) )*e^(-3) - 2*(8*e^6 + 8*e^3 - 1)*arctan((2*x*e^3 + 1)/sqrt(16*e^6 + 16*e^3 - 1))*e^(-6)/sqrt(16*e^6 + 16*e^3 - 1) - ((4*e^6 + 4*e^3 - 1)*e^(-9)*log( x^2*e^3 + x + 4*e^3 + 4) - 2*(12*e^6 + 12*e^3 - 1)*arctan((2*x*e^3 + 1)/sq rt(16*e^6 + 16*e^3 - 1))*e^(-9)/sqrt(16*e^6 + 16*e^3 - 1) - (x^2*e^3 - 2*x )*e^(-6))*e^3 - 2*(2*(8*e^6 + 8*e^3 - 1)*arctan((2*x*e^3 + 1)/sqrt(16*e^6 + 16*e^3 - 1))*e^(-6)/sqrt(16*e^6 + 16*e^3 - 1) - 2*x*e^(-3) + e^(-6)*log( x^2*e^3 + x + 4*e^3 + 4))*e^3 + 4*(e^(-3)*log(x^2*e^3 + x + 4*e^3 + 4) - 2 *arctan((2*x*e^3 + 1)/sqrt(16*e^6 + 16*e^3 - 1))*e^(-3)/sqrt(16*e^6 + 16*e ^3 - 1))*e^3 + (e^3*log(x^2*e^3 + x + 4*e^3 + 4)^2 - 4*x*e^3 + (2*x*e^3 + 1)*log(x^2*e^3 + x + 4*e^3 + 4))*e^(-3) + 2*x*e^(-3) + 5*e^(-3)*log(x^2*e^ 3 + x + 4*e^3 + 4) - e^(-6)*log(x^2*e^3 + x + 4*e^3 + 4) - 10*arctan((2*x* e^3 + 1)/sqrt(16*e^6 + 16*e^3 - 1))*e^(-3)/sqrt(16*e^6 + 16*e^3 - 1)
Leaf count of result is larger than twice the leaf count of optimal. 37 vs. \(2 (16) = 32\).
Time = 0.28 (sec) , antiderivative size = 37, normalized size of antiderivative = 2.18 \begin {dmath*} \int \frac {10 x+2 x^2+e^3 \left (8 x+4 x^2+2 x^3\right )+\left (10+2 x+e^3 \left (8+4 x+2 x^2\right )\right ) \log \left (4+x+e^3 \left (4+x^2\right )\right )}{4+x+e^3 \left (4+x^2\right )} \, dx=x^{2} + 2 \, x \log \left (x^{2} e^{3} + x + 4 \, e^{3} + 4\right ) + \log \left (x^{2} e^{3} + x + 4 \, e^{3} + 4\right )^{2} \end {dmath*}
integrate((((2*x^2+4*x+8)*exp(3)+2*x+10)*log((x^2+4)*exp(3)+4+x)+(2*x^3+4* x^2+8*x)*exp(3)+2*x^2+10*x)/((x^2+4)*exp(3)+4+x),x, algorithm=\
Time = 0.44 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.94 \begin {dmath*} \int \frac {10 x+2 x^2+e^3 \left (8 x+4 x^2+2 x^3\right )+\left (10+2 x+e^3 \left (8+4 x+2 x^2\right )\right ) \log \left (4+x+e^3 \left (4+x^2\right )\right )}{4+x+e^3 \left (4+x^2\right )} \, dx={\left (x+\ln \left (x+{\mathrm {e}}^3\,\left (x^2+4\right )+4\right )\right )}^2 \end {dmath*}