Optimal. Leaf size=29 \[ 1-\left (e^3+\left (-x+\frac {4}{-5+3 x^3}\right )^2\right ) \log (3 x) \]
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Rubi [C] time = 2.38, antiderivative size = 772, normalized size of antiderivative = 26.62, number of steps used = 43, number of rules used = 25, integrand size = 127, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.197, Rules used = {6688, 6742, 1829, 1834, 1875, 31, 634, 617, 204, 628, 2357, 2304, 2338, 266, 44, 199, 200, 2330, 2316, 2315, 2314, 2317, 2391, 27, 12} \begin {gather*} -\frac {48 i \sqrt [6]{3} \text {Li}_2\left (-\sqrt [3]{-\frac {3}{5}} x\right )}{5^{2/3} \left (1+\sqrt [3]{-1}\right )^5}-\frac {16 (-1)^{2/3} \text {Li}_2\left (-\sqrt [3]{-\frac {3}{5}} x\right )}{3 \sqrt [3]{3} 5^{2/3}}+\frac {16 \sqrt [3]{-\frac {1}{3}} \text {Li}_2\left ((-1)^{2/3} \sqrt [3]{\frac {3}{5}} x\right )}{3\ 5^{2/3}}-\frac {32 \text {Li}_2\left (-\frac {1}{2} \sqrt [3]{\frac {3}{5}} \left (1-i \sqrt {3}\right ) x\right )}{3 \sqrt [3]{3} 5^{2/3} \left (1-i \sqrt {3}\right )}-\frac {16 x^3}{25 \left (5-3 x^3\right )}+\frac {16}{15 \left (5-3 x^3\right )}-\frac {16 \log (3 x)}{\left (5-3 x^3\right )^2}-\frac {16}{75} \log \left (5-3 x^3\right )-x^2 \log (3 x)+\frac {4}{225} \left (12-5\ 3^{2/3} \sqrt [3]{5}\right ) \log \left (3^{2/3} x^2+\sqrt [3]{15} x+5^{2/3}\right )-\frac {8 x \log (3 x)}{\sqrt [3]{3} 5^{2/3} \left (3^{2/3} \sqrt [3]{5}-3 x\right )}-\frac {8 \sqrt [3]{-1} x \log (3 x)}{3\ 5^{2/3} \left (\sqrt [3]{3} x+\sqrt [3]{-5}\right )}-\frac {8 x \log (3 x)}{15^{2/3} \left (\sqrt [3]{-1} 3^{2/3} x+\sqrt [3]{15}\right )}+\frac {16 \log (45) \log \left (3^{2/3} \sqrt [3]{5}-3 x\right )}{9 \sqrt [3]{3} 5^{2/3}}-\frac {1}{25} \left (16+25 e^3\right ) \log (x)+\frac {16 \log (x)}{25}+\frac {8 \sqrt [3]{-1} 3^{2/3} \log \left (\sqrt [3]{-3} x+\sqrt [3]{5}\right )}{5^{2/3} \left (1+\sqrt [3]{-1}\right )^4}-\frac {48 i \sqrt [6]{3} \log (3 x) \log \left (\sqrt [3]{-\frac {3}{5}} x+1\right )}{5^{2/3} \left (1+\sqrt [3]{-1}\right )^5}-\frac {16 (-1)^{2/3} \log (3 x) \log \left (\sqrt [3]{-\frac {3}{5}} x+1\right )}{3 \sqrt [3]{3} 5^{2/3}}+\frac {16 \sqrt [3]{-\frac {1}{3}} \log (3 x) \log \left (1-(-1)^{2/3} \sqrt [3]{\frac {3}{5}} x\right )}{3\ 5^{2/3}}-\frac {16 \log (45) \log \left (\sqrt [3]{5}-\sqrt [3]{3} x\right )}{9 \sqrt [3]{3} 5^{2/3}}+\frac {8}{225} \left (6+5\ 3^{2/3} \sqrt [3]{5}\right ) \log \left (\sqrt [3]{5}-\sqrt [3]{3} x\right )-\frac {8 \log \left (\sqrt [3]{5}-\sqrt [3]{3} x\right )}{3 \sqrt [3]{3} 5^{2/3}}+\frac {8 \sqrt [3]{-\frac {1}{3}} \log \left (\sqrt [3]{3} x+\sqrt [3]{-5}\right )}{3\ 5^{2/3}}-\frac {32 \log (3 x) \log \left (1+\frac {1}{2} \sqrt [3]{\frac {3}{5}} \left (1-i \sqrt {3}\right ) x\right )}{3 \sqrt [3]{3} 5^{2/3} \left (1-i \sqrt {3}\right )}-\frac {8 \tan ^{-1}\left (\frac {2 x}{\sqrt [6]{3} \sqrt [3]{5}}+\frac {1}{\sqrt {3}}\right )}{3^{5/6} 5^{2/3}} \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 27
Rule 31
Rule 44
Rule 199
Rule 200
Rule 204
Rule 266
Rule 617
Rule 628
Rule 634
Rule 1829
Rule 1834
Rule 1875
Rule 2304
Rule 2314
Rule 2315
Rule 2316
Rule 2317
Rule 2330
Rule 2338
Rule 2357
Rule 2391
Rule 6688
Rule 6742
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\left (-5+3 x^3\right ) \left (e^3 \left (5-3 x^3\right )^2+\left (4+5 x-3 x^4\right )^2\right )-2 x \left (100+125 x+144 x^2+60 x^3-225 x^4-72 x^6+135 x^7-27 x^{10}\right ) \log (3 x)}{x \left (5-3 x^3\right )^3} \, dx\\ &=\int \left (\frac {-16 \left (1+\frac {25 e^3}{16}\right )-40 x-25 x^2+30 e^3 x^3+24 x^4+30 x^5-9 e^3 x^6-9 x^8}{x \left (5-3 x^3\right )^2}-\frac {2 \left (-4-5 x+3 x^4\right ) \left (25+36 x^2-30 x^3+9 x^6\right ) \log (3 x)}{\left (-5+3 x^3\right )^3}\right ) \, dx\\ &=-\left (2 \int \frac {\left (-4-5 x+3 x^4\right ) \left (25+36 x^2-30 x^3+9 x^6\right ) \log (3 x)}{\left (-5+3 x^3\right )^3} \, dx\right )+\int \frac {-16 \left (1+\frac {25 e^3}{16}\right )-40 x-25 x^2+30 e^3 x^3+24 x^4+30 x^5-9 e^3 x^6-9 x^8}{x \left (5-3 x^3\right )^2} \, dx\\ &=-\frac {16 x^3}{25 \left (5-3 x^3\right )}+\frac {1}{405} \int \frac {-81 \left (16+25 e^3\right )-3240 x-2025 x^2+1215 e^3 x^3+1215 x^5}{x \left (5-3 x^3\right )} \, dx-2 \int \left (x \log (3 x)-\frac {144 x^2 \log (3 x)}{\left (-5+3 x^3\right )^3}+\frac {60 \log (3 x)}{\left (-5+3 x^3\right )^2}+\frac {8 \log (3 x)}{-5+3 x^3}\right ) \, dx\\ &=-\frac {16 x^3}{25 \left (5-3 x^3\right )}+\frac {1}{405} \int \left (-\frac {81 \left (16+25 e^3\right )}{5 x}-405 x+\frac {648 \left (25+6 x^2\right )}{5 \left (-5+3 x^3\right )}\right ) \, dx-2 \int x \log (3 x) \, dx-16 \int \frac {\log (3 x)}{-5+3 x^3} \, dx-120 \int \frac {\log (3 x)}{\left (-5+3 x^3\right )^2} \, dx+288 \int \frac {x^2 \log (3 x)}{\left (-5+3 x^3\right )^3} \, dx\\ &=-\frac {16 x^3}{25 \left (5-3 x^3\right )}-\frac {1}{25} \left (16+25 e^3\right ) \log (x)-x^2 \log (3 x)-\frac {16 \log (3 x)}{\left (5-3 x^3\right )^2}+\frac {8}{25} \int \frac {25+6 x^2}{-5+3 x^3} \, dx+16 \int \frac {1}{x \left (-5+3 x^3\right )^2} \, dx-16 \int \left (-\frac {\log (3 x)}{3\ 5^{2/3} \left (\sqrt [3]{5}+\sqrt [3]{-3} x\right )}-\frac {\log (3 x)}{3\ 5^{2/3} \left (\sqrt [3]{5}-\sqrt [3]{3} x\right )}-\frac {\log (3 x)}{3\ 5^{2/3} \left (\sqrt [3]{5}-(-1)^{2/3} \sqrt [3]{3} x\right )}\right ) \, dx-120 \int \left (\frac {2 \log (3 x)}{15 \sqrt [3]{3} 5^{2/3} \left (3^{2/3} \sqrt [3]{5}-3 x\right )}+\frac {3 (-1)^{2/3} \sqrt [3]{\frac {3}{5}} \log (3 x)}{5 \left (1+\sqrt [3]{-1}\right )^4 \left (3^{2/3} \sqrt [3]{5}+3 \sqrt [3]{-1} x\right )^2}+\frac {6 (-1)^{5/6} \sqrt [6]{3} \log (3 x)}{5\ 5^{2/3} \left (1+\sqrt [3]{-1}\right )^5 \left (3^{2/3} \sqrt [3]{5}+3 \sqrt [3]{-1} x\right )}+\frac {3 \sqrt [3]{\frac {3}{5}} \log (3 x)}{5 \left (-1+\sqrt [3]{-1}\right )^2 \left (1+\sqrt [3]{-1}\right )^4 \left (-3^{2/3} \sqrt [3]{5}+3 (-1)^{2/3} x\right )^2}+\frac {4 \log (3 x)}{15 \sqrt [3]{3} 5^{2/3} \left (2\ 3^{2/3} \sqrt [3]{5}+3 \left (1-i \sqrt {3}\right ) x\right )}-\frac {\log (3 x)}{15\ 3^{2/3} \sqrt [3]{5} \left (-\sqrt [3]{3} 5^{2/3}+2\ 3^{2/3} \sqrt [3]{5} x-3 x^2\right )}\right ) \, dx\\ &=-\frac {16 x^3}{25 \left (5-3 x^3\right )}-\frac {1}{25} \left (16+25 e^3\right ) \log (x)-x^2 \log (3 x)-\frac {16 \log (3 x)}{\left (5-3 x^3\right )^2}+\frac {16}{3} \operatorname {Subst}\left (\int \frac {1}{x (-5+3 x)^2} \, dx,x,x^3\right )-\left (8 \sqrt [3]{\frac {3}{5}}\right ) \int \frac {\log (3 x)}{\left (3^{2/3} \sqrt [3]{5}+3 \sqrt [3]{-1} x\right )^2} \, dx-\left (8 \sqrt [3]{\frac {3}{5}}\right ) \int \frac {\log (3 x)}{\left (-3^{2/3} \sqrt [3]{5}+3 (-1)^{2/3} x\right )^2} \, dx+\frac {16 \int \frac {\log (3 x)}{\sqrt [3]{5}+\sqrt [3]{-3} x} \, dx}{3\ 5^{2/3}}+\frac {16 \int \frac {\log (3 x)}{\sqrt [3]{5}-\sqrt [3]{3} x} \, dx}{3\ 5^{2/3}}+\frac {16 \int \frac {\log (3 x)}{\sqrt [3]{5}-(-1)^{2/3} \sqrt [3]{3} x} \, dx}{3\ 5^{2/3}}-\frac {8 \int \frac {\sqrt [3]{\frac {5}{3}} \left (50-2 \sqrt [3]{3} 5^{2/3}\right )+\left (25-4 \sqrt [3]{3} 5^{2/3}\right ) x}{\left (\frac {5}{3}\right )^{2/3}+\sqrt [3]{\frac {5}{3}} x+x^2} \, dx}{75 \sqrt [3]{3} 5^{2/3}}-\frac {16 \int \frac {\log (3 x)}{3^{2/3} \sqrt [3]{5}-3 x} \, dx}{\sqrt [3]{3} 5^{2/3}}-\frac {32 \int \frac {\log (3 x)}{2\ 3^{2/3} \sqrt [3]{5}+3 \left (1-i \sqrt {3}\right ) x} \, dx}{\sqrt [3]{3} 5^{2/3}}+\frac {8 \int \frac {\log (3 x)}{-\sqrt [3]{3} 5^{2/3}+2\ 3^{2/3} \sqrt [3]{5} x-3 x^2} \, dx}{3^{2/3} \sqrt [3]{5}}-\frac {\left (144 (-1)^{5/6} \sqrt [6]{3}\right ) \int \frac {\log (3 x)}{3^{2/3} \sqrt [3]{5}+3 \sqrt [3]{-1} x} \, dx}{5^{2/3} \left (1+\sqrt [3]{-1}\right )^5}-\frac {1}{225} \left (8 \left (6+5\ 3^{2/3} \sqrt [3]{5}\right )\right ) \int \frac {1}{\sqrt [3]{\frac {5}{3}}-x} \, dx\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}
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Mathematica [A] time = 0.08, size = 35, normalized size = 1.21 \begin {gather*} -e^3 \log (x)-\frac {\left (-4-5 x+3 x^4\right )^2 \log (3 x)}{\left (-5+3 x^3\right )^2} \end {gather*}
Antiderivative was successfully verified.
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fricas [B] time = 0.76, size = 60, normalized size = 2.07 \begin {gather*} -\frac {{\left (9 \, x^{8} - 30 \, x^{5} - 24 \, x^{4} + 25 \, x^{2} + {\left (9 \, x^{6} - 30 \, x^{3} + 25\right )} e^{3} + 40 \, x + 16\right )} \log \left (3 \, x\right )}{9 \, x^{6} - 30 \, x^{3} + 25} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.16, size = 90, normalized size = 3.10 \begin {gather*} -\frac {9 \, x^{8} \log \left (3 \, x\right ) + 9 \, x^{6} e^{3} \log \relax (x) - 30 \, x^{5} \log \left (3 \, x\right ) - 24 \, x^{4} \log \left (3 \, x\right ) - 30 \, x^{3} e^{3} \log \relax (x) + 25 \, x^{2} \log \left (3 \, x\right ) + 40 \, x \log \left (3 \, x\right ) + 25 \, e^{3} \log \relax (x) + 16 \, \log \left (3 \, x\right )}{9 \, x^{6} - 30 \, x^{3} + 25} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.17, size = 53, normalized size = 1.83
method | result | size |
risch | \(-\frac {\left (9 x^{8}-30 x^{5}-24 x^{4}+25 x^{2}+40 x +16\right ) \ln \left (3 x \right )}{9 x^{6}-30 x^{3}+25}-\ln \relax (x ) {\mathrm e}^{3}\) | \(53\) |
derivativedivides | \(-\ln \left (3 x \right ) {\mathrm e}^{3}-\frac {16 \ln \left (3 x \right )}{25}-x^{2} \ln \left (3 x \right )+\frac {72 \ln \left (3 x \right ) x}{27 x^{3}-45}+\frac {432 \ln \left (3 x \right ) x^{3} \left (27 x^{3}-90\right )}{25 \left (27 x^{3}-45\right )^{2}}\) | \(66\) |
default | \(-\ln \left (3 x \right ) {\mathrm e}^{3}-\frac {16 \ln \left (3 x \right )}{25}-x^{2} \ln \left (3 x \right )+\frac {72 \ln \left (3 x \right ) x}{27 x^{3}-45}+\frac {432 \ln \left (3 x \right ) x^{3} \left (27 x^{3}-90\right )}{25 \left (27 x^{3}-45\right )^{2}}\) | \(66\) |
norman | \(\frac {-16 \ln \left (3 x \right )-40 x \ln \left (3 x \right )-25 x^{2} \ln \left (3 x \right )+24 \ln \left (3 x \right ) x^{4}+30 \ln \left (3 x \right ) x^{5}-9 \ln \left (3 x \right ) x^{8}}{\left (3 x^{3}-5\right )^{2}}-\ln \relax (x ) {\mathrm e}^{3}\) | \(68\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 5.65, size = 48, normalized size = 1.66 \begin {gather*} -{\mathrm {e}}^3\,\ln \relax (x)-\frac {\ln \left (3\,x\right )\,\left (x^8-\frac {10\,x^5}{3}-\frac {8\,x^4}{3}+\frac {25\,x^2}{9}+\frac {40\,x}{9}+\frac {16}{9}\right )}{x^6-\frac {10\,x^3}{3}+\frac {25}{9}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [B] time = 0.40, size = 48, normalized size = 1.66 \begin {gather*} - e^{3} \log {\relax (x )} + \frac {\left (- 9 x^{8} + 30 x^{5} + 24 x^{4} - 25 x^{2} - 40 x - 16\right ) \log {\left (3 x \right )}}{9 x^{6} - 30 x^{3} + 25} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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