Optimal. Leaf size=22 \[ (1-x) \left (-e+x^2+\log \left (x \left (5+\log ^2(5)\right )\right )\right ) \]
________________________________________________________________________________________
Rubi [A] time = 0.03, antiderivative size = 32, normalized size of antiderivative = 1.45, number of steps used = 6, number of rules used = 3, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {6, 14, 2295} \begin {gather*} -x^3+x^2-(1-e) x+x-x \log \left (x \left (5+\log ^2(5)\right )\right )+\log (x) \end {gather*}
Antiderivative was successfully verified.
[In]
[Out]
Rule 6
Rule 14
Rule 2295
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {1+(-1+e) x+2 x^2-3 x^3-x \log \left (5 x+x \log ^2(5)\right )}{x} \, dx\\ &=\int \left (\frac {1-(1-e) x+2 x^2-3 x^3}{x}-\log \left (x \left (5+\log ^2(5)\right )\right )\right ) \, dx\\ &=\int \frac {1-(1-e) x+2 x^2-3 x^3}{x} \, dx-\int \log \left (x \left (5+\log ^2(5)\right )\right ) \, dx\\ &=x-x \log \left (x \left (5+\log ^2(5)\right )\right )+\int \left (-1+e+\frac {1}{x}+2 x-3 x^2\right ) \, dx\\ &=x-(1-e) x+x^2-x^3+\log (x)-x \log \left (x \left (5+\log ^2(5)\right )\right )\\ \end {aligned} \end {gather*}
________________________________________________________________________________________
Mathematica [A] time = 0.01, size = 26, normalized size = 1.18 \begin {gather*} e x+x^2-x^3+\log (x)-x \log \left (x \left (5+\log ^2(5)\right )\right ) \end {gather*}
Antiderivative was successfully verified.
[In]
[Out]
________________________________________________________________________________________
fricas [A] time = 0.85, size = 29, normalized size = 1.32 \begin {gather*} -x^{3} + x^{2} + x e - {\left (x - 1\right )} \log \left (x \log \relax (5)^{2} + 5 \, x\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
giac [A] time = 0.13, size = 29, normalized size = 1.32 \begin {gather*} -x^{3} + x^{2} + x e - x \log \left (x \log \relax (5)^{2} + 5 \, x\right ) + \log \relax (x) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
maple [A] time = 0.09, size = 30, normalized size = 1.36
method | result | size |
risch | \(-x \ln \left (x \ln \relax (5)^{2}+5 x \right )-x^{3}+x \,{\mathrm e}+x^{2}+\ln \relax (x )\) | \(30\) |
norman | \(x^{2}+x \,{\mathrm e}+\ln \left (x \ln \relax (5)^{2}+5 x \right )-x^{3}-x \ln \left (x \ln \relax (5)^{2}+5 x \right )\) | \(39\) |
derivativedivides | \(\frac {\ln \relax (5)^{6} \ln \left (\left (\ln \relax (5)^{2}+5\right ) x \right )}{\left (\ln \relax (5)^{2}+5\right )^{3}}+\frac {\ln \relax (5)^{4} {\mathrm e} x}{\left (\ln \relax (5)^{2}+5\right )^{2}}-\frac {\ln \relax (5)^{4} \left (\left (\ln \relax (5)^{2}+5\right ) x \ln \left (\left (\ln \relax (5)^{2}+5\right ) x \right )-\left (\ln \relax (5)^{2}+5\right ) x \right )}{\left (\ln \relax (5)^{2}+5\right )^{3}}-\frac {\ln \relax (5)^{4} x}{\left (\ln \relax (5)^{2}+5\right )^{2}}+\frac {15 \ln \relax (5)^{4} \ln \left (\left (\ln \relax (5)^{2}+5\right ) x \right )}{\left (\ln \relax (5)^{2}+5\right )^{3}}+\frac {10 \ln \relax (5)^{2} {\mathrm e} x}{\left (\ln \relax (5)^{2}+5\right )^{2}}-\frac {10 \ln \relax (5)^{2} \left (\left (\ln \relax (5)^{2}+5\right ) x \ln \left (\left (\ln \relax (5)^{2}+5\right ) x \right )-\left (\ln \relax (5)^{2}+5\right ) x \right )}{\left (\ln \relax (5)^{2}+5\right )^{3}}+\frac {\ln \relax (5)^{2} x^{2}}{\ln \relax (5)^{2}+5}-\frac {10 x \ln \relax (5)^{2}}{\left (\ln \relax (5)^{2}+5\right )^{2}}-x^{3}+\frac {75 \ln \relax (5)^{2} \ln \left (\left (\ln \relax (5)^{2}+5\right ) x \right )}{\left (\ln \relax (5)^{2}+5\right )^{3}}+\frac {25 \,{\mathrm e} x}{\left (\ln \relax (5)^{2}+5\right )^{2}}-\frac {25 \left (\left (\ln \relax (5)^{2}+5\right ) x \ln \left (\left (\ln \relax (5)^{2}+5\right ) x \right )-\left (\ln \relax (5)^{2}+5\right ) x \right )}{\left (\ln \relax (5)^{2}+5\right )^{3}}+\frac {5 x^{2}}{\ln \relax (5)^{2}+5}-\frac {25 x}{\left (\ln \relax (5)^{2}+5\right )^{2}}+\frac {125 \ln \left (\left (\ln \relax (5)^{2}+5\right ) x \right )}{\left (\ln \relax (5)^{2}+5\right )^{3}}\) | \(329\) |
default | \(\frac {\ln \relax (5)^{6} \ln \left (\left (\ln \relax (5)^{2}+5\right ) x \right )}{\left (\ln \relax (5)^{2}+5\right )^{3}}+\frac {\ln \relax (5)^{4} {\mathrm e} x}{\left (\ln \relax (5)^{2}+5\right )^{2}}-\frac {\ln \relax (5)^{4} \left (\left (\ln \relax (5)^{2}+5\right ) x \ln \left (\left (\ln \relax (5)^{2}+5\right ) x \right )-\left (\ln \relax (5)^{2}+5\right ) x \right )}{\left (\ln \relax (5)^{2}+5\right )^{3}}-\frac {\ln \relax (5)^{4} x}{\left (\ln \relax (5)^{2}+5\right )^{2}}+\frac {15 \ln \relax (5)^{4} \ln \left (\left (\ln \relax (5)^{2}+5\right ) x \right )}{\left (\ln \relax (5)^{2}+5\right )^{3}}+\frac {10 \ln \relax (5)^{2} {\mathrm e} x}{\left (\ln \relax (5)^{2}+5\right )^{2}}-\frac {10 \ln \relax (5)^{2} \left (\left (\ln \relax (5)^{2}+5\right ) x \ln \left (\left (\ln \relax (5)^{2}+5\right ) x \right )-\left (\ln \relax (5)^{2}+5\right ) x \right )}{\left (\ln \relax (5)^{2}+5\right )^{3}}+\frac {\ln \relax (5)^{2} x^{2}}{\ln \relax (5)^{2}+5}-\frac {10 x \ln \relax (5)^{2}}{\left (\ln \relax (5)^{2}+5\right )^{2}}-x^{3}+\frac {75 \ln \relax (5)^{2} \ln \left (\left (\ln \relax (5)^{2}+5\right ) x \right )}{\left (\ln \relax (5)^{2}+5\right )^{3}}+\frac {25 \,{\mathrm e} x}{\left (\ln \relax (5)^{2}+5\right )^{2}}-\frac {25 \left (\left (\ln \relax (5)^{2}+5\right ) x \ln \left (\left (\ln \relax (5)^{2}+5\right ) x \right )-\left (\ln \relax (5)^{2}+5\right ) x \right )}{\left (\ln \relax (5)^{2}+5\right )^{3}}+\frac {5 x^{2}}{\ln \relax (5)^{2}+5}-\frac {25 x}{\left (\ln \relax (5)^{2}+5\right )^{2}}+\frac {125 \ln \left (\left (\ln \relax (5)^{2}+5\right ) x \right )}{\left (\ln \relax (5)^{2}+5\right )^{3}}\) | \(329\) |
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
maxima [B] time = 0.37, size = 60, normalized size = 2.73 \begin {gather*} -x^{3} + x^{2} + x e - x + \frac {x \log \relax (5)^{2} - {\left (x \log \relax (5)^{2} + 5 \, x\right )} \log \left (x \log \relax (5)^{2} + 5 \, x\right ) + 5 \, x}{\log \relax (5)^{2} + 5} + \log \relax (x) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
mupad [B] time = 3.20, size = 29, normalized size = 1.32 \begin {gather*} \ln \relax (x)-x\,\ln \left (5\,x+x\,{\ln \relax (5)}^2\right )+x\,\mathrm {e}+x^2-x^3 \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
sympy [A] time = 0.14, size = 27, normalized size = 1.23 \begin {gather*} - x^{3} + x^{2} - x \log {\left (x \log {\relax (5 )}^{2} + 5 x \right )} + e x + \log {\relax (x )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________