Optimal. Leaf size=19 \[ x \left (x+4 e^{-2 x} \log (x)\right ) (9+\log (16+x)) \]
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Rubi [A] time = 4.14, antiderivative size = 38, normalized size of antiderivative = 2.00, number of steps used = 45, number of rules used = 14, integrand size = 82, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {6742, 77, 2395, 43, 6688, 2194, 2199, 2176, 2178, 2554, 12, 2288, 2557, 14} \begin {gather*} 9 x^2+x^2 \log (x+16)+36 e^{-2 x} x \log (x)+4 e^{-2 x} x \log (x) \log (x+16) \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 14
Rule 43
Rule 77
Rule 2176
Rule 2178
Rule 2194
Rule 2199
Rule 2288
Rule 2395
Rule 2554
Rule 2557
Rule 6688
Rule 6742
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {x (288+19 x+32 \log (16+x)+2 x \log (16+x))}{16+x}-\frac {4 e^{-2 x} \left (-144-9 x-144 \log (x)+278 x \log (x)+18 x^2 \log (x)-16 \log (16+x)-x \log (16+x)-16 \log (x) \log (16+x)+31 x \log (x) \log (16+x)+2 x^2 \log (x) \log (16+x)\right )}{16+x}\right ) \, dx\\ &=-\left (4 \int \frac {e^{-2 x} \left (-144-9 x-144 \log (x)+278 x \log (x)+18 x^2 \log (x)-16 \log (16+x)-x \log (16+x)-16 \log (x) \log (16+x)+31 x \log (x) \log (16+x)+2 x^2 \log (x) \log (16+x)\right )}{16+x} \, dx\right )+\int \frac {x (288+19 x+32 \log (16+x)+2 x \log (16+x))}{16+x} \, dx\\ &=-\left (4 \int \frac {e^{-2 x} \left (-((16+x) (9+\log (16+x)))+\log (x) \left (2 \left (-72+139 x+9 x^2\right )+\left (-16+31 x+2 x^2\right ) \log (16+x)\right )\right )}{16+x} \, dx\right )+\int \left (\frac {x (288+19 x)}{16+x}+2 x \log (16+x)\right ) \, dx\\ &=2 \int x \log (16+x) \, dx-4 \int \left (\frac {e^{-2 x} \left (-144-9 x-144 \log (x)+278 x \log (x)+18 x^2 \log (x)\right )}{16+x}+e^{-2 x} (-1-\log (x)+2 x \log (x)) \log (16+x)\right ) \, dx+\int \frac {x (288+19 x)}{16+x} \, dx\\ &=x^2 \log (16+x)-4 \int \frac {e^{-2 x} \left (-144-9 x-144 \log (x)+278 x \log (x)+18 x^2 \log (x)\right )}{16+x} \, dx-4 \int e^{-2 x} (-1-\log (x)+2 x \log (x)) \log (16+x) \, dx-\int \frac {x^2}{16+x} \, dx+\int \left (-16+19 x+\frac {256}{16+x}\right ) \, dx\\ &=-16 x+\frac {19 x^2}{2}+256 \log (16+x)+x^2 \log (16+x)-4 \int \left (-9 e^{-2 x}+\frac {2 e^{-2 x} \left (-72+139 x+9 x^2\right ) \log (x)}{16+x}\right ) \, dx-4 \int \left (-e^{-2 x} \log (16+x)-e^{-2 x} \log (x) \log (16+x)+2 e^{-2 x} x \log (x) \log (16+x)\right ) \, dx-\int \left (-16+x+\frac {256}{16+x}\right ) \, dx\\ &=9 x^2+x^2 \log (16+x)+4 \int e^{-2 x} \log (16+x) \, dx+4 \int e^{-2 x} \log (x) \log (16+x) \, dx-8 \int \frac {e^{-2 x} \left (-72+139 x+9 x^2\right ) \log (x)}{16+x} \, dx-8 \int e^{-2 x} x \log (x) \log (16+x) \, dx+36 \int e^{-2 x} \, dx\\ &=-18 e^{-2 x}+9 x^2-2 e^{-2 x} \log (x)+36 e^{-2 x} x \log (x)-64 e^{32} \text {Ei}(-2 (16+x)) \log (x)-2 e^{-2 x} \log (16+x)+x^2 \log (16+x)+4 e^{-2 x} x \log (x) \log (16+x)-4 \int -\frac {e^{-2 x}}{2 (16+x)} \, dx-4 \int -\frac {e^{-2 x} \log (x)}{2 (16+x)} \, dx-4 \int -\frac {e^{-2 x} \log (16+x)}{2 x} \, dx+8 \int \frac {e^{-2 x} \left (1-18 x+32 e^{32+2 x} \text {Ei}(-2 (16+x))\right )}{4 x} \, dx+8 \int \frac {e^{-2 x} (-1-2 x) \log (x)}{4 (16+x)} \, dx+8 \int \frac {e^{-2 x} (-1-2 x) \log (16+x)}{4 x} \, dx\\ &=-18 e^{-2 x}+9 x^2-2 e^{-2 x} \log (x)+36 e^{-2 x} x \log (x)-64 e^{32} \text {Ei}(-2 (16+x)) \log (x)-2 e^{-2 x} \log (16+x)+x^2 \log (16+x)+4 e^{-2 x} x \log (x) \log (16+x)+2 \int \frac {e^{-2 x}}{16+x} \, dx+2 \int \frac {e^{-2 x} \left (1-18 x+32 e^{32+2 x} \text {Ei}(-2 (16+x))\right )}{x} \, dx+2 \int \frac {e^{-2 x} \log (x)}{16+x} \, dx+2 \int \frac {e^{-2 x} (-1-2 x) \log (x)}{16+x} \, dx+2 \int \frac {e^{-2 x} \log (16+x)}{x} \, dx+2 \int \frac {e^{-2 x} (-1-2 x) \log (16+x)}{x} \, dx\\ &=-18 e^{-2 x}+9 x^2+2 e^{32} \text {Ei}(-2 (16+x))+36 e^{-2 x} x \log (x)+x^2 \log (16+x)+4 e^{-2 x} x \log (x) \log (16+x)-2 \int \frac {e^{32} \text {Ei}(-32-2 x)}{x} \, dx+2 \int \left (\frac {e^{-2 x} (1-18 x)}{x}+\frac {32 e^{32} \text {Ei}(-32-2 x)}{x}\right ) \, dx-2 \int \frac {e^{-2 x}-\text {Ei}(-2 x)}{16+x} \, dx-2 \int \frac {\text {Ei}(-2 x)}{16+x} \, dx-2 \int \frac {e^{-2 x}+31 e^{32} \text {Ei}(-2 (16+x))}{x} \, dx\\ &=-18 e^{-2 x}+9 x^2+2 e^{32} \text {Ei}(-2 (16+x))+36 e^{-2 x} x \log (x)+x^2 \log (16+x)+4 e^{-2 x} x \log (x) \log (16+x)+2 \int \frac {e^{-2 x} (1-18 x)}{x} \, dx-2 \int \left (\frac {e^{-2 x}}{x}+\frac {31 e^{32} \text {Ei}(-32-2 x)}{x}\right ) \, dx-2 \int \frac {\text {Ei}(-2 x)}{16+x} \, dx-2 \int \left (\frac {e^{-2 x}}{16+x}-\frac {\text {Ei}(-2 x)}{16+x}\right ) \, dx-\left (2 e^{32}\right ) \int \frac {\text {Ei}(-32-2 x)}{x} \, dx+\left (64 e^{32}\right ) \int \frac {\text {Ei}(-32-2 x)}{x} \, dx\\ &=-18 e^{-2 x}+9 x^2+2 e^{32} \text {Ei}(-2 (16+x))+36 e^{-2 x} x \log (x)+x^2 \log (16+x)+4 e^{-2 x} x \log (x) \log (16+x)+2 \int \left (-18 e^{-2 x}+\frac {e^{-2 x}}{x}\right ) \, dx-2 \int \frac {e^{-2 x}}{x} \, dx-2 \int \frac {e^{-2 x}}{16+x} \, dx-\left (2 e^{32}\right ) \int \frac {\text {Ei}(-32-2 x)}{x} \, dx-\left (62 e^{32}\right ) \int \frac {\text {Ei}(-32-2 x)}{x} \, dx+\left (64 e^{32}\right ) \int \frac {\text {Ei}(-32-2 x)}{x} \, dx\\ &=-18 e^{-2 x}+9 x^2-2 \text {Ei}(-2 x)+36 e^{-2 x} x \log (x)+x^2 \log (16+x)+4 e^{-2 x} x \log (x) \log (16+x)+2 \int \frac {e^{-2 x}}{x} \, dx-36 \int e^{-2 x} \, dx-\left (2 e^{32}\right ) \int \frac {\text {Ei}(-32-2 x)}{x} \, dx-\left (62 e^{32}\right ) \int \frac {\text {Ei}(-32-2 x)}{x} \, dx+\left (64 e^{32}\right ) \int \frac {\text {Ei}(-32-2 x)}{x} \, dx\\ &=9 x^2+36 e^{-2 x} x \log (x)+x^2 \log (16+x)+4 e^{-2 x} x \log (x) \log (16+x)-\left (2 e^{32}\right ) \int \frac {\text {Ei}(-32-2 x)}{x} \, dx-\left (62 e^{32}\right ) \int \frac {\text {Ei}(-32-2 x)}{x} \, dx+\left (64 e^{32}\right ) \int \frac {\text {Ei}(-32-2 x)}{x} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.07, size = 25, normalized size = 1.32 \begin {gather*} e^{-2 x} x \left (e^{2 x} x+4 \log (x)\right ) (9+\log (16+x)) \end {gather*}
Antiderivative was successfully verified.
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fricas [B] time = 0.73, size = 39, normalized size = 2.05 \begin {gather*} {\left (9 \, x^{2} e^{\left (2 \, x\right )} + {\left (x^{2} e^{\left (2 \, x\right )} + 4 \, x \log \relax (x)\right )} \log \left (x + 16\right ) + 36 \, x \log \relax (x)\right )} e^{\left (-2 \, x\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.15, size = 36, normalized size = 1.89 \begin {gather*} 4 \, x e^{\left (-2 \, x\right )} \log \left (x + 16\right ) \log \relax (x) + x^{2} \log \left (x + 16\right ) + 40 \, x e^{\left (-2 \, x\right )} \log \relax (x) + 9 \, x^{2} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.11, size = 41, normalized size = 2.16
method | result | size |
risch | \(x \left (x \,{\mathrm e}^{2 x}+4 \ln \relax (x )\right ) {\mathrm e}^{-2 x} \ln \left (x +16\right )+9 x \left (x \,{\mathrm e}^{2 x}+4 \ln \relax (x )\right ) {\mathrm e}^{-2 x}\) | \(41\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} 36 \, x e^{\left (-2 \, x\right )} \log \relax (x) + 9 \, x^{2} - 576 \, e^{32} E_{1}\left (2 \, x + 32\right ) + {\left (4 \, x e^{\left (-2 \, x\right )} \log \relax (x) + x^{2}\right )} \log \left (x + 16\right ) - 576 \, \int \frac {e^{\left (-2 \, x\right )}}{x + 16}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.05 \begin {gather*} \int \frac {{\mathrm {e}}^{-2\,x}\,\left (36\,x+{\mathrm {e}}^{2\,x}\,\left (19\,x^2+288\,x\right )+\ln \left (x+16\right )\,\left (4\,x+{\mathrm {e}}^{2\,x}\,\left (2\,x^2+32\,x\right )-\ln \relax (x)\,\left (8\,x^2+124\,x-64\right )+64\right )-\ln \relax (x)\,\left (72\,x^2+1112\,x-576\right )+576\right )}{x+16} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [B] time = 19.64, size = 48, normalized size = 2.53 \begin {gather*} 9 x^{2} + \left (x^{2} - \frac {256}{3}\right ) \log {\left (x + 16 \right )} + \left (4 x \log {\relax (x )} \log {\left (x + 16 \right )} + 36 x \log {\relax (x )}\right ) e^{- 2 x} + \frac {256 \log {\left (x + 16 \right )}}{3} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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