Optimal. Leaf size=28 \[ \frac {(5-x) x^2}{\left (\frac {3}{4}+\frac {5 x}{e^2}\right ) \left (4+x^2\right )^2} \]
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Rubi [C] time = 0.26, antiderivative size = 243, normalized size of antiderivative = 8.68, number of steps used = 8, number of rules used = 4, integrand size = 117, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.034, Rules used = {2074, 639, 199, 203} \begin {gather*} \frac {6 e^2 \left (100+3 e^2\right ) x}{\left (1600+9 e^4\right ) \left (x^2+4\right )}-\frac {10 e^2 \left (3 \left (100+3 e^2\right ) \left (320+3 e^4\right ) x+2 \left (25600+288 e^4-27 e^6\right )\right )}{\left (1600+9 e^4\right )^2 \left (x^2+4\right )}+\frac {16 e^2 \left (\left (100+3 e^2\right ) x+5 \left (16-3 e^2\right )\right )}{\left (1600+9 e^4\right ) \left (x^2+4\right )^2}+\frac {720 e^6 \left (100+3 e^2\right )}{\left (1600+9 e^4\right )^2 \left (20 x+3 e^2\right )}+\frac {3 e^2 \left (100+3 e^2\right ) \tan ^{-1}\left (\frac {x}{2}\right )}{1600+9 e^4}-\frac {15 e^2 \left (100+3 e^2\right ) \left (320+3 e^4\right ) \tan ^{-1}\left (\frac {x}{2}\right )}{\left (1600+9 e^4\right )^2}+\frac {18 e^6 \left (100+3 e^2\right ) \tan ^{-1}\left (\frac {x}{2}\right )}{\left (1600+9 e^4\right )^2} \end {gather*}
Antiderivative was successfully verified.
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Rule 199
Rule 203
Rule 639
Rule 2074
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-\frac {14400 e^6 \left (100+3 e^2\right )}{\left (1600+9 e^4\right )^2 \left (3 e^2+20 x\right )^2}+\frac {64 \left (4 e^2 \left (100+3 e^2\right )-5 e^2 \left (16-3 e^2\right ) x\right )}{\left (1600+9 e^4\right ) \left (4+x^2\right )^3}+\frac {40 \left (-6 e^2 \left (100+3 e^2\right ) \left (320+3 e^4\right )+e^2 \left (25600+288 e^4-27 e^6\right ) x\right )}{\left (1600+9 e^4\right )^2 \left (4+x^2\right )^2}+\frac {36 e^6 \left (100+3 e^2\right )}{\left (1600+9 e^4\right )^2 \left (4+x^2\right )}\right ) \, dx\\ &=\frac {720 e^6 \left (100+3 e^2\right )}{\left (1600+9 e^4\right )^2 \left (3 e^2+20 x\right )}+\frac {40 \int \frac {-6 e^2 \left (100+3 e^2\right ) \left (320+3 e^4\right )+e^2 \left (25600+288 e^4-27 e^6\right ) x}{\left (4+x^2\right )^2} \, dx}{\left (1600+9 e^4\right )^2}+\frac {\left (36 e^6 \left (100+3 e^2\right )\right ) \int \frac {1}{4+x^2} \, dx}{\left (1600+9 e^4\right )^2}+\frac {64 \int \frac {4 e^2 \left (100+3 e^2\right )-5 e^2 \left (16-3 e^2\right ) x}{\left (4+x^2\right )^3} \, dx}{1600+9 e^4}\\ &=\frac {720 e^6 \left (100+3 e^2\right )}{\left (1600+9 e^4\right )^2 \left (3 e^2+20 x\right )}+\frac {16 e^2 \left (5 \left (16-3 e^2\right )+\left (100+3 e^2\right ) x\right )}{\left (1600+9 e^4\right ) \left (4+x^2\right )^2}-\frac {10 e^2 \left (2 \left (25600+288 e^4-27 e^6\right )+3 \left (100+3 e^2\right ) \left (320+3 e^4\right ) x\right )}{\left (1600+9 e^4\right )^2 \left (4+x^2\right )}+\frac {18 e^6 \left (100+3 e^2\right ) \tan ^{-1}\left (\frac {x}{2}\right )}{\left (1600+9 e^4\right )^2}-\frac {\left (30 e^2 \left (100+3 e^2\right ) \left (320+3 e^4\right )\right ) \int \frac {1}{4+x^2} \, dx}{\left (1600+9 e^4\right )^2}+\frac {\left (48 e^2 \left (100+3 e^2\right )\right ) \int \frac {1}{\left (4+x^2\right )^2} \, dx}{1600+9 e^4}\\ &=\frac {720 e^6 \left (100+3 e^2\right )}{\left (1600+9 e^4\right )^2 \left (3 e^2+20 x\right )}+\frac {16 e^2 \left (5 \left (16-3 e^2\right )+\left (100+3 e^2\right ) x\right )}{\left (1600+9 e^4\right ) \left (4+x^2\right )^2}+\frac {6 e^2 \left (100+3 e^2\right ) x}{\left (1600+9 e^4\right ) \left (4+x^2\right )}-\frac {10 e^2 \left (2 \left (25600+288 e^4-27 e^6\right )+3 \left (100+3 e^2\right ) \left (320+3 e^4\right ) x\right )}{\left (1600+9 e^4\right )^2 \left (4+x^2\right )}+\frac {18 e^6 \left (100+3 e^2\right ) \tan ^{-1}\left (\frac {x}{2}\right )}{\left (1600+9 e^4\right )^2}-\frac {15 e^2 \left (100+3 e^2\right ) \left (320+3 e^4\right ) \tan ^{-1}\left (\frac {x}{2}\right )}{\left (1600+9 e^4\right )^2}+\frac {\left (6 e^2 \left (100+3 e^2\right )\right ) \int \frac {1}{4+x^2} \, dx}{1600+9 e^4}\\ &=\frac {720 e^6 \left (100+3 e^2\right )}{\left (1600+9 e^4\right )^2 \left (3 e^2+20 x\right )}+\frac {16 e^2 \left (5 \left (16-3 e^2\right )+\left (100+3 e^2\right ) x\right )}{\left (1600+9 e^4\right ) \left (4+x^2\right )^2}+\frac {6 e^2 \left (100+3 e^2\right ) x}{\left (1600+9 e^4\right ) \left (4+x^2\right )}-\frac {10 e^2 \left (2 \left (25600+288 e^4-27 e^6\right )+3 \left (100+3 e^2\right ) \left (320+3 e^4\right ) x\right )}{\left (1600+9 e^4\right )^2 \left (4+x^2\right )}+\frac {18 e^6 \left (100+3 e^2\right ) \tan ^{-1}\left (\frac {x}{2}\right )}{\left (1600+9 e^4\right )^2}-\frac {15 e^2 \left (100+3 e^2\right ) \left (320+3 e^4\right ) \tan ^{-1}\left (\frac {x}{2}\right )}{\left (1600+9 e^4\right )^2}+\frac {3 e^2 \left (100+3 e^2\right ) \tan ^{-1}\left (\frac {x}{2}\right )}{1600+9 e^4}\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.05, size = 29, normalized size = 1.04 \begin {gather*} -\frac {4 e^2 (-5+x) x^2}{\left (3 e^2+20 x\right ) \left (4+x^2\right )^2} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 1.12, size = 43, normalized size = 1.54 \begin {gather*} -\frac {4 \, {\left (x^{3} - 5 \, x^{2}\right )} e^{2}}{20 \, x^{5} + 160 \, x^{3} + 3 \, {\left (x^{4} + 8 \, x^{2} + 16\right )} e^{2} + 320 \, x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: NotImplementedError} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.28, size = 34, normalized size = 1.21
method | result | size |
norman | \(\frac {20 x^{2} {\mathrm e}^{2}-4 x^{3} {\mathrm e}^{2}}{\left (x^{2}+4\right )^{2} \left (3 \,{\mathrm e}^{2}+20 x \right )}\) | \(34\) |
gosper | \(-\frac {4 x^{2} \left (x -5\right ) {\mathrm e}^{2}}{3 x^{4} {\mathrm e}^{2}+20 x^{5}+24 x^{2} {\mathrm e}^{2}+160 x^{3}+48 \,{\mathrm e}^{2}+320 x}\) | \(45\) |
risch | \(\frac {-\frac {4 x^{3} {\mathrm e}^{2}}{3}+\frac {20 x^{2} {\mathrm e}^{2}}{3}}{x^{4} {\mathrm e}^{2}+\frac {20 x^{5}}{3}+8 x^{2} {\mathrm e}^{2}+\frac {160 x^{3}}{3}+16 \,{\mathrm e}^{2}+\frac {320 x}{3}}\) | \(50\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.37, size = 50, normalized size = 1.79 \begin {gather*} -\frac {4 \, {\left (x^{3} e^{2} - 5 \, x^{2} e^{2}\right )}}{20 \, x^{5} + 3 \, x^{4} e^{2} + 160 \, x^{3} + 24 \, x^{2} e^{2} + 320 \, x + 48 \, e^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 2.74, size = 50, normalized size = 1.79 \begin {gather*} \frac {20\,x^2\,{\mathrm {e}}^2-4\,x^3\,{\mathrm {e}}^2}{20\,x^5+3\,{\mathrm {e}}^2\,x^4+160\,x^3+24\,{\mathrm {e}}^2\,x^2+320\,x+48\,{\mathrm {e}}^2} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [B] time = 3.38, size = 51, normalized size = 1.82 \begin {gather*} \frac {- 4 x^{3} e^{2} + 20 x^{2} e^{2}}{20 x^{5} + 3 x^{4} e^{2} + 160 x^{3} + 24 x^{2} e^{2} + 320 x + 48 e^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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