Optimal. Leaf size=28 \[ 16 \left (-e^x+\frac {1}{4} (-2+x) x-\frac {1}{5} \log \left ((25+x)^2\right )\right )^2 \]
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Rubi [A] time = 0.30, antiderivative size = 28, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 86, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.035, Rules used = {6688, 12, 6686} \begin {gather*} \frac {1}{25} \left (5 (2-x) x+20 e^x+4 \log \left ((x+25)^2\right )\right )^2 \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 6686
Rule 6688
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {4 \left (129-120 x-5 x^2+10 e^x (25+x)\right ) \left (20 e^x-5 (-2+x) x+4 \log \left ((25+x)^2\right )\right )}{25 (25+x)} \, dx\\ &=\frac {4}{25} \int \frac {\left (129-120 x-5 x^2+10 e^x (25+x)\right ) \left (20 e^x-5 (-2+x) x+4 \log \left ((25+x)^2\right )\right )}{25+x} \, dx\\ &=\frac {1}{25} \left (20 e^x+5 (2-x) x+4 \log \left ((25+x)^2\right )\right )^2\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.06, size = 26, normalized size = 0.93 \begin {gather*} \frac {1}{25} \left (20 e^x-5 (-2+x) x+4 \log \left ((25+x)^2\right )\right )^2 \end {gather*}
Antiderivative was successfully verified.
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fricas [B] time = 0.71, size = 66, normalized size = 2.36 \begin {gather*} x^{4} - 4 \, x^{3} + 4 \, x^{2} - 8 \, {\left (x^{2} - 2 \, x\right )} e^{x} - \frac {8}{5} \, {\left (x^{2} - 2 \, x - 4 \, e^{x}\right )} \log \left (x^{2} + 50 \, x + 625\right ) + \frac {16}{25} \, \log \left (x^{2} + 50 \, x + 625\right )^{2} + 16 \, e^{\left (2 \, x\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.36, size = 84, normalized size = 3.00 \begin {gather*} x^{4} - 4 \, x^{3} - 8 \, x^{2} e^{x} - \frac {8}{5} \, x^{2} \log \left (x^{2} + 50 \, x + 625\right ) + 4 \, x^{2} + 16 \, x e^{x} + \frac {16}{5} \, x \log \left (x^{2} + 50 \, x + 625\right ) + \frac {32}{5} \, e^{x} \log \left (x^{2} + 50 \, x + 625\right ) + \frac {16}{25} \, \log \left (x^{2} + 50 \, x + 625\right )^{2} + 16 \, e^{\left (2 \, x\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.25, size = 107, normalized size = 3.82
method | result | size |
default | \(\frac {32 \left (\ln \left (\left (x +25\right )^{2}\right )-2 \ln \left (x +25\right )\right ) {\mathrm e}^{x}}{5}+16 \,{\mathrm e}^{x} x -8 \,{\mathrm e}^{x} x^{2}+\frac {64 \ln \left (x +25\right ) {\mathrm e}^{x}}{5}+16 \,{\mathrm e}^{2 x}+x^{4}-4 x^{3}+4 x^{2}-\frac {8 \ln \left (x^{2}+50 x +625\right ) x^{2}}{5}+\frac {16 \ln \left (x^{2}+50 x +625\right ) x}{5}+\frac {64 \ln \left (x +25\right ) \ln \left (x^{2}+50 x +625\right )}{25}-\frac {64 \ln \left (x +25\right )^{2}}{25}\) | \(107\) |
risch | \(\frac {64 \ln \left (x +25\right )^{2}}{25}+\left (-\frac {16 x^{2}}{5}+\frac {32 x}{5}+\frac {64 \,{\mathrm e}^{x}}{5}\right ) \ln \left (x +25\right )+\frac {64 i \ln \left (x +25\right ) \pi \,\mathrm {csgn}\left (i \left (x +25\right )\right ) \mathrm {csgn}\left (i \left (x +25\right )^{2}\right )^{2}}{25}-\frac {8 i \pi x \mathrm {csgn}\left (i \left (x +25\right )^{2}\right )^{3}}{5}-\frac {32 i \ln \left (x +25\right ) \pi \mathrm {csgn}\left (i \left (x +25\right )\right )^{2} \mathrm {csgn}\left (i \left (x +25\right )^{2}\right )}{25}-\frac {16 i \pi \mathrm {csgn}\left (i \left (x +25\right )^{2}\right )^{3} {\mathrm e}^{x}}{5}-\frac {16 i \pi \mathrm {csgn}\left (i \left (x +25\right )\right )^{2} \mathrm {csgn}\left (i \left (x +25\right )^{2}\right ) {\mathrm e}^{x}}{5}+\frac {4 i \pi \,x^{2} \mathrm {csgn}\left (i \left (x +25\right )^{2}\right )^{3}}{5}+x^{4}-4 x^{3}+4 x^{2}-\frac {8 i \pi x \mathrm {csgn}\left (i \left (x +25\right )\right )^{2} \mathrm {csgn}\left (i \left (x +25\right )^{2}\right )}{5}-\frac {32 i \ln \left (x +25\right ) \pi \mathrm {csgn}\left (i \left (x +25\right )^{2}\right )^{3}}{25}-\frac {8 i \pi \,x^{2} \mathrm {csgn}\left (i \left (x +25\right )\right ) \mathrm {csgn}\left (i \left (x +25\right )^{2}\right )^{2}}{5}+16 \,{\mathrm e}^{2 x}+\frac {32 i \pi \,\mathrm {csgn}\left (i \left (x +25\right )\right ) \mathrm {csgn}\left (i \left (x +25\right )^{2}\right )^{2} {\mathrm e}^{x}}{5}+\frac {4 i \pi \,x^{2} \mathrm {csgn}\left (i \left (x +25\right )\right )^{2} \mathrm {csgn}\left (i \left (x +25\right )^{2}\right )}{5}+\frac {16 i \pi x \,\mathrm {csgn}\left (i \left (x +25\right )\right ) \mathrm {csgn}\left (i \left (x +25\right )^{2}\right )^{2}}{5}+16 \,{\mathrm e}^{x} x -8 \,{\mathrm e}^{x} x^{2}\) | \(325\) |
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*} x^{4} - 4 \, x^{3} + 4 \, x^{2} - 8 \, {\left (x^{2} - 2 \, x\right )} e^{x} - \frac {2064}{5} \, e^{\left (-25\right )} E_{1}\left (-x - 25\right ) - \frac {16}{5} \, {\left (x^{2} - 2 \, x - 4 \, e^{x} - 675\right )} \log \left (x + 25\right ) + \frac {64}{25} \, \log \left (x + 25\right )^{2} + 16 \, e^{\left (2 \, x\right )} - \frac {2064}{5} \, \int \frac {e^{x}}{x + 25}\,{d x} - 2160 \, \log \left (x + 25\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.08, size = 68, normalized size = 2.43 \begin {gather*} 16\,{\mathrm {e}}^{2\,x}+\ln \left (x^2+50\,x+625\right )\,\left (\frac {16\,x}{5}+\frac {32\,{\mathrm {e}}^x}{5}-\frac {8\,x^2}{5}\right )+{\mathrm {e}}^x\,\left (16\,x-8\,x^2\right )+4\,x^2-4\,x^3+x^4+\frac {16\,{\ln \left (x^2+50\,x+625\right )}^2}{25} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.48, size = 82, normalized size = 2.93 \begin {gather*} x^{4} - 4 x^{3} + 4 x^{2} + \left (- \frac {8 x^{2}}{5} + \frac {16 x}{5}\right ) \log {\left (x^{2} + 50 x + 625 \right )} + \frac {\left (- 40 x^{2} + 80 x + 32 \log {\left (x^{2} + 50 x + 625 \right )}\right ) e^{x}}{5} + 16 e^{2 x} + \frac {16 \log {\left (x^{2} + 50 x + 625 \right )}^{2}}{25} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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