Optimal. Leaf size=24 \[ \log \left (\left (-2+2 x-(4+2 x-e (1+\log (x)))^2\right )^2\right ) \]
________________________________________________________________________________________
Rubi [F] time = 1.64, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {4 e^2+e (-16-16 x)+28 x+16 x^2+\left (4 e^2-8 e x\right ) \log (x)}{18 x+e^2 x+14 x^2+4 x^3+e \left (-8 x-4 x^2\right )+\left (2 e^2 x+e \left (-8 x-4 x^2\right )\right ) \log (x)+e^2 x \log ^2(x)} \, dx \end {gather*}
Verification is not applicable to the result.
[In]
[Out]
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {4 e^2+e (-16-16 x)+28 x+16 x^2+\left (4 e^2-8 e x\right ) \log (x)}{\left (18+e^2\right ) x+14 x^2+4 x^3+e \left (-8 x-4 x^2\right )+\left (2 e^2 x+e \left (-8 x-4 x^2\right )\right ) \log (x)+e^2 x \log ^2(x)} \, dx\\ &=\int \frac {4 \left (-4 \left (1-\frac {e}{4}\right ) e+7 \left (1-\frac {4 e}{7}\right ) x+4 x^2+e^2 \log (x)-2 e x \log (x)\right )}{\left (18+e^2\right ) x+14 x^2+4 x^3+e \left (-8 x-4 x^2\right )+\left (2 e^2 x+e \left (-8 x-4 x^2\right )\right ) \log (x)+e^2 x \log ^2(x)} \, dx\\ &=4 \int \frac {-4 \left (1-\frac {e}{4}\right ) e+7 \left (1-\frac {4 e}{7}\right ) x+4 x^2+e^2 \log (x)-2 e x \log (x)}{\left (18+e^2\right ) x+14 x^2+4 x^3+e \left (-8 x-4 x^2\right )+\left (2 e^2 x+e \left (-8 x-4 x^2\right )\right ) \log (x)+e^2 x \log ^2(x)} \, dx\\ &=4 \int \frac {e^2-4 e (1+x)+x (7+4 x)+e (e-2 x) \log (x)}{x \left (e^2-4 e (2+x)+2 \left (9+7 x+2 x^2\right )+2 e (e-2 (2+x)) \log (x)+e^2 \log ^2(x)\right )} \, dx\\ &=4 \int \left (\frac {4 \left (1-\frac {7}{4 e}\right ) e}{-18 \left (1+\frac {1}{18} (-8+e) e\right )-14 \left (1-\frac {2 e}{7}\right ) x-4 x^2+8 \left (1-\frac {e}{4}\right ) e \log (x)+4 e x \log (x)-e^2 \log ^2(x)}+\frac {4 \left (1-\frac {e}{4}\right ) e}{x \left (-18 \left (1+\frac {1}{18} (-8+e) e\right )-14 \left (1-\frac {2 e}{7}\right ) x-4 x^2+8 \left (1-\frac {e}{4}\right ) e \log (x)+4 e x \log (x)-e^2 \log ^2(x)\right )}+\frac {2 e \log (x)}{-18 \left (1+\frac {1}{18} (-8+e) e\right )-14 \left (1-\frac {2 e}{7}\right ) x-4 x^2+8 \left (1-\frac {e}{4}\right ) e \log (x)+4 e x \log (x)-e^2 \log ^2(x)}+\frac {4 x}{18 \left (1+\frac {1}{18} (-8+e) e\right )+14 \left (1-\frac {2 e}{7}\right ) x+4 x^2-8 \left (1-\frac {e}{4}\right ) e \log (x)-4 e x \log (x)+e^2 \log ^2(x)}+\frac {e^2 \log (x)}{x \left (18 \left (1+\frac {1}{18} (-8+e) e\right )+14 \left (1-\frac {2 e}{7}\right ) x+4 x^2-8 \left (1-\frac {e}{4}\right ) e \log (x)-4 e x \log (x)+e^2 \log ^2(x)\right )}\right ) \, dx\\ &=16 \int \frac {x}{18 \left (1+\frac {1}{18} (-8+e) e\right )+14 \left (1-\frac {2 e}{7}\right ) x+4 x^2-8 \left (1-\frac {e}{4}\right ) e \log (x)-4 e x \log (x)+e^2 \log ^2(x)} \, dx-(4 (7-4 e)) \int \frac {1}{-18 \left (1+\frac {1}{18} (-8+e) e\right )-14 \left (1-\frac {2 e}{7}\right ) x-4 x^2+8 \left (1-\frac {e}{4}\right ) e \log (x)+4 e x \log (x)-e^2 \log ^2(x)} \, dx+(8 e) \int \frac {\log (x)}{-18 \left (1+\frac {1}{18} (-8+e) e\right )-14 \left (1-\frac {2 e}{7}\right ) x-4 x^2+8 \left (1-\frac {e}{4}\right ) e \log (x)+4 e x \log (x)-e^2 \log ^2(x)} \, dx+(4 (4-e) e) \int \frac {1}{x \left (-18 \left (1+\frac {1}{18} (-8+e) e\right )-14 \left (1-\frac {2 e}{7}\right ) x-4 x^2+8 \left (1-\frac {e}{4}\right ) e \log (x)+4 e x \log (x)-e^2 \log ^2(x)\right )} \, dx+\left (4 e^2\right ) \int \frac {\log (x)}{x \left (18 \left (1+\frac {1}{18} (-8+e) e\right )+14 \left (1-\frac {2 e}{7}\right ) x+4 x^2-8 \left (1-\frac {e}{4}\right ) e \log (x)-4 e x \log (x)+e^2 \log ^2(x)\right )} \, dx\\ \end {aligned} \end {gather*}
________________________________________________________________________________________
Mathematica [B] time = 1.47, size = 49, normalized size = 2.04 \begin {gather*} 2 \log \left (18-8 e+e^2+14 x-4 e x+4 x^2-8 e \log (x)+2 e^2 \log (x)-4 e x \log (x)+e^2 \log ^2(x)\right ) \end {gather*}
Antiderivative was successfully verified.
[In]
[Out]
________________________________________________________________________________________
fricas [B] time = 0.65, size = 45, normalized size = 1.88 \begin {gather*} 2 \, \log \left (e^{2} \log \relax (x)^{2} + 4 \, x^{2} - 4 \, {\left (x + 2\right )} e - 2 \, {\left (2 \, {\left (x + 2\right )} e - e^{2}\right )} \log \relax (x) + 14 \, x + e^{2} + 18\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
giac [B] time = 0.23, size = 50, normalized size = 2.08 \begin {gather*} 2 \, \log \left (-4 \, x e \log \relax (x) + e^{2} \log \relax (x)^{2} + 4 \, x^{2} - 4 \, x e + 2 \, e^{2} \log \relax (x) - 8 \, e \log \relax (x) + 14 \, x + e^{2} - 8 \, e + 18\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
maple [A] time = 0.09, size = 46, normalized size = 1.92
method | result | size |
risch | \(2 \ln \left (\ln \relax (x )^{2}+2 \,{\mathrm e}^{-1} \left ({\mathrm e}-2 x -4\right ) \ln \relax (x )+\left ({\mathrm e}^{2}-4 x \,{\mathrm e}+4 x^{2}-8 \,{\mathrm e}+14 x +18\right ) {\mathrm e}^{-2}\right )\) | \(46\) |
norman | \(2 \ln \left ({\mathrm e}^{2} \ln \relax (x )^{2}+2 \,{\mathrm e}^{2} \ln \relax (x )-4 x \,{\mathrm e} \ln \relax (x )-8 \,{\mathrm e} \ln \relax (x )+{\mathrm e}^{2}-4 x \,{\mathrm e}+4 x^{2}-8 \,{\mathrm e}+14 x +18\right )\) | \(57\) |
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
maxima [B] time = 0.84, size = 53, normalized size = 2.21 \begin {gather*} 2 \, \log \left ({\left (e^{2} \log \relax (x)^{2} + 4 \, x^{2} - 2 \, x {\left (2 \, e - 7\right )} - 2 \, {\left (2 \, x e - e^{2} + 4 \, e\right )} \log \relax (x) + e^{2} - 8 \, e + 18\right )} e^{\left (-2\right )}\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
mupad [B] time = 5.67, size = 50, normalized size = 2.08 \begin {gather*} 2\,\ln \left (14\,x-8\,\mathrm {e}+{\mathrm {e}}^2+{\mathrm {e}}^2\,{\ln \relax (x)}^2-4\,x\,\mathrm {e}-8\,\mathrm {e}\,\ln \relax (x)+2\,{\mathrm {e}}^2\,\ln \relax (x)+4\,x^2-4\,x\,\mathrm {e}\,\ln \relax (x)+18\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
sympy [B] time = 0.31, size = 53, normalized size = 2.21 \begin {gather*} 2 \log {\left (\frac {\left (- 4 x - 8 + 2 e\right ) \log {\relax (x )}}{e} + \frac {4 x^{2} - 4 e x + 14 x - 8 e + e^{2} + 18}{e^{2}} + \log {\relax (x )}^{2} \right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________