Optimal. Leaf size=28 \[ 5 \left (-5 e^2+e^{4 x} x+\frac {25}{(2+x)^2 (15+\log (4))^2}\right ) \]
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Rubi [A] time = 0.19, antiderivative size = 37, normalized size of antiderivative = 1.32, number of steps used = 4, number of rules used = 3, integrand size = 130, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.023, Rules used = {6688, 2176, 2194} \begin {gather*} \frac {5}{4} e^{4 x} (4 x+1)-\frac {5 e^{4 x}}{4}+\frac {125}{(x+2)^2 (15+\log (4))^2} \end {gather*}
Antiderivative was successfully verified.
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Rule 2176
Rule 2194
Rule 6688
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \left (5 e^{4 x} (1+4 x)-\frac {250}{(2+x)^3 (15+\log (4))^2}\right ) \, dx\\ &=\frac {125}{(2+x)^2 (15+\log (4))^2}+5 \int e^{4 x} (1+4 x) \, dx\\ &=\frac {5}{4} e^{4 x} (1+4 x)+\frac {125}{(2+x)^2 (15+\log (4))^2}-5 \int e^{4 x} \, dx\\ &=-\frac {5 e^{4 x}}{4}+\frac {5}{4} e^{4 x} (1+4 x)+\frac {125}{(2+x)^2 (15+\log (4))^2}\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.05, size = 22, normalized size = 0.79 \begin {gather*} 5 e^{4 x} x+\frac {125}{(2+x)^2 (15+\log (4))^2} \end {gather*}
Antiderivative was successfully verified.
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fricas [B] time = 0.70, size = 95, normalized size = 3.39 \begin {gather*} \frac {5 \, {\left ({\left (225 \, x^{3} + 4 \, {\left (x^{3} + 4 \, x^{2} + 4 \, x\right )} \log \relax (2)^{2} + 900 \, x^{2} + 60 \, {\left (x^{3} + 4 \, x^{2} + 4 \, x\right )} \log \relax (2) + 900 \, x\right )} e^{\left (4 \, x\right )} + 25\right )}}{4 \, {\left (x^{2} + 4 \, x + 4\right )} \log \relax (2)^{2} + 225 \, x^{2} + 60 \, {\left (x^{2} + 4 \, x + 4\right )} \log \relax (2) + 900 \, x + 900} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.14, size = 147, normalized size = 5.25 \begin {gather*} \frac {5 \, {\left (4 \, x^{3} e^{\left (4 \, x\right )} \log \relax (2)^{2} + 60 \, x^{3} e^{\left (4 \, x\right )} \log \relax (2) + 16 \, x^{2} e^{\left (4 \, x\right )} \log \relax (2)^{2} + 225 \, x^{3} e^{\left (4 \, x\right )} + 240 \, x^{2} e^{\left (4 \, x\right )} \log \relax (2) + 16 \, x e^{\left (4 \, x\right )} \log \relax (2)^{2} + 900 \, x^{2} e^{\left (4 \, x\right )} + 240 \, x e^{\left (4 \, x\right )} \log \relax (2) + 900 \, x e^{\left (4 \, x\right )} + 25\right )}}{4 \, x^{2} \log \relax (2)^{2} + 60 \, x^{2} \log \relax (2) + 16 \, x \log \relax (2)^{2} + 225 \, x^{2} + 240 \, x \log \relax (2) + 16 \, \log \relax (2)^{2} + 900 \, x + 240 \, \log \relax (2) + 900} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.15, size = 60, normalized size = 2.14
method | result | size |
risch | \(\frac {125}{4 \left (x^{2} \ln \relax (2)^{2}+4 x \ln \relax (2)^{2}+15 x^{2} \ln \relax (2)+4 \ln \relax (2)^{2}+60 x \ln \relax (2)+\frac {225 x^{2}}{4}+60 \ln \relax (2)+225 x +225\right )}+5 x \,{\mathrm e}^{4 x}\) | \(60\) |
norman | \(\frac {\left (10 \ln \relax (2)+75\right ) x^{3} {\mathrm e}^{4 x}+\left (40 \ln \relax (2)+300\right ) x \,{\mathrm e}^{4 x}+\left (40 \ln \relax (2)+300\right ) x^{2} {\mathrm e}^{4 x}+\frac {125}{2 \ln \relax (2)+15}}{\left (2+x \right )^{2} \left (2 \ln \relax (2)+15\right )}\) | \(66\) |
derivativedivides | \(\frac {2000}{\left (4 \ln \relax (2)^{2}+60 \ln \relax (2)+225\right ) \left (4 x +8\right )^{2}}+\frac {1125 \,{\mathrm e}^{4 x} x}{4 \ln \relax (2)^{2}+60 \ln \relax (2)+225}+\frac {300 \ln \relax (2) {\mathrm e}^{4 x} x}{4 \ln \relax (2)^{2}+60 \ln \relax (2)+225}+\frac {20 \ln \relax (2)^{2} {\mathrm e}^{4 x} x}{4 \ln \relax (2)^{2}+60 \ln \relax (2)+225}\) | \(94\) |
default | \(\frac {2000}{\left (4 \ln \relax (2)^{2}+60 \ln \relax (2)+225\right ) \left (4 x +8\right )^{2}}+\frac {1125 \,{\mathrm e}^{4 x} x}{4 \ln \relax (2)^{2}+60 \ln \relax (2)+225}+\frac {300 \ln \relax (2) {\mathrm e}^{4 x} x}{4 \ln \relax (2)^{2}+60 \ln \relax (2)+225}+\frac {20 \ln \relax (2)^{2} {\mathrm e}^{4 x} x}{4 \ln \relax (2)^{2}+60 \ln \relax (2)+225}\) | \(94\) |
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*} 5 \, x e^{\left (4 \, x\right )} - \frac {160 \, e^{\left (-8\right )} E_{3}\left (-4 \, x - 8\right ) \log \relax (2)^{2}}{{\left (x + 2\right )}^{2} {\left (2 \, \log \relax (2) + 15\right )}^{2}} + \frac {125}{{\left (4 \, \log \relax (2)^{2} + 60 \, \log \relax (2) + 225\right )} x^{2} + 4 \, {\left (4 \, \log \relax (2)^{2} + 60 \, \log \relax (2) + 225\right )} x + 16 \, \log \relax (2)^{2} + 240 \, \log \relax (2) + 900} - \frac {2400 \, e^{\left (-8\right )} E_{3}\left (-4 \, x - 8\right ) \log \relax (2)}{{\left (x + 2\right )}^{2} {\left (2 \, \log \relax (2) + 15\right )}^{2}} - \frac {9000 \, e^{\left (-8\right )} E_{3}\left (-4 \, x - 8\right )}{{\left (x + 2\right )}^{2} {\left (2 \, \log \relax (2) + 15\right )}^{2}} - 40 \, \int \frac {e^{\left (4 \, x\right )}}{x^{3} + 6 \, x^{2} + 12 \, x + 8}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 3.48, size = 52, normalized size = 1.86 \begin {gather*} \frac {125}{\left (30\,\ln \relax (4)+{\ln \relax (4)}^2+225\right )\,x^2+\left (120\,\ln \relax (4)+4\,{\ln \relax (4)}^2+900\right )\,x+120\,\ln \relax (4)+4\,{\ln \relax (4)}^2+900}+5\,x\,{\mathrm {e}}^{4\,x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.43, size = 54, normalized size = 1.93 \begin {gather*} 5 x e^{4 x} + \frac {250}{x^{2} \left (8 \log {\relax (2 )}^{2} + 120 \log {\relax (2 )} + 450\right ) + x \left (32 \log {\relax (2 )}^{2} + 480 \log {\relax (2 )} + 1800\right ) + 32 \log {\relax (2 )}^{2} + 480 \log {\relax (2 )} + 1800} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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