Optimal. Leaf size=20 \[ 5-\log \left (e^{2 x}\right ) \log \left (x+e^{5 x} x\right ) \]
________________________________________________________________________________________
Rubi [A] time = 1.69, antiderivative size = 13, normalized size of antiderivative = 0.65, number of steps used = 18, number of rules used = 6, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.140, Rules used = {6742, 2184, 2190, 2279, 2391, 2548} \begin {gather*} -2 x \log \left (\left (e^{5 x}+1\right ) x\right ) \end {gather*}
Antiderivative was successfully verified.
[In]
[Out]
Rule 2184
Rule 2190
Rule 2279
Rule 2391
Rule 2548
Rule 6742
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {2 x}{1+e^x}-\frac {2 \left (-4+3 e^x-2 e^{2 x}+e^{3 x}\right ) x}{1-e^x+e^{2 x}-e^{3 x}+e^{4 x}}-2 \left (1+5 x+\log \left (\left (1+e^{5 x}\right ) x\right )\right )\right ) \, dx\\ &=2 \int \frac {x}{1+e^x} \, dx-2 \int \frac {\left (-4+3 e^x-2 e^{2 x}+e^{3 x}\right ) x}{1-e^x+e^{2 x}-e^{3 x}+e^{4 x}} \, dx-2 \int \left (1+5 x+\log \left (\left (1+e^{5 x}\right ) x\right )\right ) \, dx\\ &=-2 x-4 x^2-2 \int \frac {e^x x}{1+e^x} \, dx-2 \int \left (-\frac {4 x}{1-e^x+e^{2 x}-e^{3 x}+e^{4 x}}+\frac {3 e^x x}{1-e^x+e^{2 x}-e^{3 x}+e^{4 x}}-\frac {2 e^{2 x} x}{1-e^x+e^{2 x}-e^{3 x}+e^{4 x}}+\frac {e^{3 x} x}{1-e^x+e^{2 x}-e^{3 x}+e^{4 x}}\right ) \, dx-2 \int \log \left (\left (1+e^{5 x}\right ) x\right ) \, dx\\ &=-2 x-4 x^2-2 x \log \left (1+e^x\right )-2 x \log \left (\left (1+e^{5 x}\right ) x\right )-2 \int \frac {e^{3 x} x}{1-e^x+e^{2 x}-e^{3 x}+e^{4 x}} \, dx+2 \int \frac {1+e^{5 x} (1+5 x)}{1+e^{5 x}} \, dx+2 \int \log \left (1+e^x\right ) \, dx+4 \int \frac {e^{2 x} x}{1-e^x+e^{2 x}-e^{3 x}+e^{4 x}} \, dx-6 \int \frac {e^x x}{1-e^x+e^{2 x}-e^{3 x}+e^{4 x}} \, dx+8 \int \frac {x}{1-e^x+e^{2 x}-e^{3 x}+e^{4 x}} \, dx\\ &=-2 x-4 x^2-2 x \log \left (1+e^x\right )-2 x \log \left (\left (1+e^{5 x}\right ) x\right )-2 \int \frac {e^{3 x} x}{1-e^x+e^{2 x}-e^{3 x}+e^{4 x}} \, dx+2 \int \left (1+5 x-\frac {x}{1+e^x}+\frac {\left (-4+3 e^x-2 e^{2 x}+e^{3 x}\right ) x}{1-e^x+e^{2 x}-e^{3 x}+e^{4 x}}\right ) \, dx+2 \operatorname {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^x\right )+4 \int \frac {e^{2 x} x}{1-e^x+e^{2 x}-e^{3 x}+e^{4 x}} \, dx-6 \int \frac {e^x x}{1-e^x+e^{2 x}-e^{3 x}+e^{4 x}} \, dx+8 \int \frac {x}{1-e^x+e^{2 x}-e^{3 x}+e^{4 x}} \, dx\\ &=x^2-2 x \log \left (1+e^x\right )-2 x \log \left (\left (1+e^{5 x}\right ) x\right )-2 \text {Li}_2\left (-e^x\right )-2 \int \frac {x}{1+e^x} \, dx-2 \int \frac {e^{3 x} x}{1-e^x+e^{2 x}-e^{3 x}+e^{4 x}} \, dx+2 \int \frac {\left (-4+3 e^x-2 e^{2 x}+e^{3 x}\right ) x}{1-e^x+e^{2 x}-e^{3 x}+e^{4 x}} \, dx+4 \int \frac {e^{2 x} x}{1-e^x+e^{2 x}-e^{3 x}+e^{4 x}} \, dx-6 \int \frac {e^x x}{1-e^x+e^{2 x}-e^{3 x}+e^{4 x}} \, dx+8 \int \frac {x}{1-e^x+e^{2 x}-e^{3 x}+e^{4 x}} \, dx\\ &=-2 x \log \left (1+e^x\right )-2 x \log \left (\left (1+e^{5 x}\right ) x\right )-2 \text {Li}_2\left (-e^x\right )+2 \int \frac {e^x x}{1+e^x} \, dx-2 \int \frac {e^{3 x} x}{1-e^x+e^{2 x}-e^{3 x}+e^{4 x}} \, dx+2 \int \left (-\frac {4 x}{1-e^x+e^{2 x}-e^{3 x}+e^{4 x}}+\frac {3 e^x x}{1-e^x+e^{2 x}-e^{3 x}+e^{4 x}}-\frac {2 e^{2 x} x}{1-e^x+e^{2 x}-e^{3 x}+e^{4 x}}+\frac {e^{3 x} x}{1-e^x+e^{2 x}-e^{3 x}+e^{4 x}}\right ) \, dx+4 \int \frac {e^{2 x} x}{1-e^x+e^{2 x}-e^{3 x}+e^{4 x}} \, dx-6 \int \frac {e^x x}{1-e^x+e^{2 x}-e^{3 x}+e^{4 x}} \, dx+8 \int \frac {x}{1-e^x+e^{2 x}-e^{3 x}+e^{4 x}} \, dx\\ &=-2 x \log \left (\left (1+e^{5 x}\right ) x\right )-2 \text {Li}_2\left (-e^x\right )-2 \int \log \left (1+e^x\right ) \, dx\\ &=-2 x \log \left (\left (1+e^{5 x}\right ) x\right )-2 \text {Li}_2\left (-e^x\right )-2 \operatorname {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^x\right )\\ &=-2 x \log \left (\left (1+e^{5 x}\right ) x\right )\\ \end {aligned} \end {gather*}
________________________________________________________________________________________
Mathematica [A] time = 0.25, size = 13, normalized size = 0.65 \begin {gather*} -2 x \log \left (\left (1+e^{5 x}\right ) x\right ) \end {gather*}
Antiderivative was successfully verified.
[In]
[Out]
________________________________________________________________________________________
fricas [A] time = 0.46, size = 12, normalized size = 0.60 \begin {gather*} -2 \, x \log \left (x e^{\left (5 \, x\right )} + x\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
giac [A] time = 0.19, size = 12, normalized size = 0.60 \begin {gather*} -2 \, x \log \left (x e^{\left (5 \, x\right )} + x\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
maple [A] time = 0.07, size = 13, normalized size = 0.65
method | result | size |
norman | \(-2 x \ln \left (x \,{\mathrm e}^{5 x}+x \right )\) | \(13\) |
risch | \(-2 x \ln \left ({\mathrm e}^{5 x}+1\right )+i \pi x \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i \left ({\mathrm e}^{5 x}+1\right )\right ) \mathrm {csgn}\left (i x \left ({\mathrm e}^{5 x}+1\right )\right )-i \pi x \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x \left ({\mathrm e}^{5 x}+1\right )\right )^{2}-i \pi x \,\mathrm {csgn}\left (i \left ({\mathrm e}^{5 x}+1\right )\right ) \mathrm {csgn}\left (i x \left ({\mathrm e}^{5 x}+1\right )\right )^{2}+i \pi x \mathrm {csgn}\left (i x \left ({\mathrm e}^{5 x}+1\right )\right )^{3}-2 x \ln \relax (x )\) | \(117\) |
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
maxima [B] time = 0.47, size = 38, normalized size = 1.90 \begin {gather*} -2 \, x \log \relax (x) - 2 \, x \log \left (e^{\left (4 \, x\right )} - e^{\left (3 \, x\right )} + e^{\left (2 \, x\right )} - e^{x} + 1\right ) - 2 \, x \log \left (e^{x} + 1\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
mupad [B] time = 3.16, size = 12, normalized size = 0.60 \begin {gather*} -2\,x\,\ln \left (x\,\left ({\mathrm {e}}^{5\,x}+1\right )\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
sympy [A] time = 0.34, size = 14, normalized size = 0.70 \begin {gather*} - 2 x \log {\left (x e^{5 x} + x \right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________