Optimal. Leaf size=30 \[ \frac {\log \left (3-x+x^2\right )}{5 \left (-x+\frac {x^2}{5+3 x}\right )} \]
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Rubi [A]
time = 0.66, antiderivative size = 37, normalized size of antiderivative = 1.23, number
of steps used = 45, number of rules used = 14, integrand size = 78, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.180, Rules
used = {6873, 12, 6860, 723, 814, 648, 632, 210, 642, 907, 1642, 2608, 2605, 2099}
\begin {gather*} -\frac {\log \left (x^2-x+3\right )}{5 x}-\frac {\log \left (x^2-x+3\right )}{5 (2 x+5)} \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 210
Rule 632
Rule 642
Rule 648
Rule 723
Rule 814
Rule 907
Rule 1642
Rule 2099
Rule 2605
Rule 2608
Rule 6860
Rule 6873
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {25 x-25 x^2-44 x^3-12 x^4+\left (75+35 x+23 x^2+14 x^3+6 x^4\right ) \log \left (3-x+x^2\right )}{5 x^2 (5+2 x)^2 \left (3-x+x^2\right )} \, dx\\ &=\frac {1}{5} \int \frac {25 x-25 x^2-44 x^3-12 x^4+\left (75+35 x+23 x^2+14 x^3+6 x^4\right ) \log \left (3-x+x^2\right )}{x^2 (5+2 x)^2 \left (3-x+x^2\right )} \, dx\\ &=\frac {1}{5} \int \left (-\frac {25}{(5+2 x)^2 \left (3-x+x^2\right )}+\frac {25}{x (5+2 x)^2 \left (3-x+x^2\right )}-\frac {44 x}{(5+2 x)^2 \left (3-x+x^2\right )}-\frac {12 x^2}{(5+2 x)^2 \left (3-x+x^2\right )}+\frac {\left (25+20 x+6 x^2\right ) \log \left (3-x+x^2\right )}{x^2 (5+2 x)^2}\right ) \, dx\\ &=\frac {1}{5} \int \frac {\left (25+20 x+6 x^2\right ) \log \left (3-x+x^2\right )}{x^2 (5+2 x)^2} \, dx-\frac {12}{5} \int \frac {x^2}{(5+2 x)^2 \left (3-x+x^2\right )} \, dx-5 \int \frac {1}{(5+2 x)^2 \left (3-x+x^2\right )} \, dx+5 \int \frac {1}{x (5+2 x)^2 \left (3-x+x^2\right )} \, dx-\frac {44}{5} \int \frac {x}{(5+2 x)^2 \left (3-x+x^2\right )} \, dx\\ &=\frac {10}{47 (5+2 x)}-\frac {5}{47} \int \frac {7-2 x}{(5+2 x) \left (3-x+x^2\right )} \, dx+\frac {1}{5} \int \left (\frac {\log \left (3-x+x^2\right )}{x^2}+\frac {2 \log \left (3-x+x^2\right )}{(5+2 x)^2}\right ) \, dx-\frac {12}{5} \int \left (\frac {25}{47 (5+2 x)^2}-\frac {170}{2209 (5+2 x)}+\frac {-39+85 x}{2209 \left (3-x+x^2\right )}\right ) \, dx+5 \int \left (\frac {1}{75 x}-\frac {8}{235 (5+2 x)^2}-\frac {856}{55225 (5+2 x)}+\frac {-35-37 x}{6627 \left (3-x+x^2\right )}\right ) \, dx-\frac {44}{5} \int \left (-\frac {10}{47 (5+2 x)^2}-\frac {26}{2209 (5+2 x)}+\frac {72+13 x}{2209 \left (3-x+x^2\right )}\right ) \, dx\\ &=\frac {\log (x)}{15}+\frac {1164 \log (5+2 x)}{11045}+\frac {5 \int \frac {-35-37 x}{3-x+x^2} \, dx}{6627}-\frac {12 \int \frac {-39+85 x}{3-x+x^2} \, dx}{11045}-\frac {44 \int \frac {72+13 x}{3-x+x^2} \, dx}{11045}-\frac {5}{47} \int \left (\frac {48}{47 (5+2 x)}+\frac {37-24 x}{47 \left (3-x+x^2\right )}\right ) \, dx+\frac {1}{5} \int \frac {\log \left (3-x+x^2\right )}{x^2} \, dx+\frac {2}{5} \int \frac {\log \left (3-x+x^2\right )}{(5+2 x)^2} \, dx\\ &=\frac {\log (x)}{15}+\frac {12}{235} \log (5+2 x)-\frac {\log \left (3-x+x^2\right )}{5 x}-\frac {\log \left (3-x+x^2\right )}{5 (5+2 x)}-\frac {5 \int \frac {37-24 x}{3-x+x^2} \, dx}{2209}-\frac {42 \int \frac {1}{3-x+x^2} \, dx}{11045}-\frac {185 \int \frac {-1+2 x}{3-x+x^2} \, dx}{13254}-\frac {286 \int \frac {-1+2 x}{3-x+x^2} \, dx}{11045}-\frac {535 \int \frac {1}{3-x+x^2} \, dx}{13254}-\frac {102 \int \frac {-1+2 x}{3-x+x^2} \, dx}{2209}+\frac {1}{5} \int \frac {-1+2 x}{x \left (3-x+x^2\right )} \, dx+\frac {1}{5} \int \frac {-1+2 x}{15+x+3 x^2+2 x^3} \, dx-\frac {3454 \int \frac {1}{3-x+x^2} \, dx}{11045}\\ &=\frac {\log (x)}{15}+\frac {12}{235} \log (5+2 x)-\frac {5701 \log \left (3-x+x^2\right )}{66270}-\frac {\log \left (3-x+x^2\right )}{5 x}-\frac {\log \left (3-x+x^2\right )}{5 (5+2 x)}+\frac {84 \text {Subst}\left (\int \frac {1}{-11-x^2} \, dx,x,-1+2 x\right )}{11045}+\frac {60 \int \frac {-1+2 x}{3-x+x^2} \, dx}{2209}-\frac {125 \int \frac {1}{3-x+x^2} \, dx}{2209}+\frac {535 \text {Subst}\left (\int \frac {1}{-11-x^2} \, dx,x,-1+2 x\right )}{6627}+\frac {1}{5} \int \left (-\frac {1}{3 x}+\frac {5+x}{3 \left (3-x+x^2\right )}\right ) \, dx+\frac {1}{5} \int \left (-\frac {24}{47 (5+2 x)}+\frac {5+12 x}{47 \left (3-x+x^2\right )}\right ) \, dx+\frac {6908 \text {Subst}\left (\int \frac {1}{-11-x^2} \, dx,x,-1+2 x\right )}{11045}\\ &=\frac {2927 \tan ^{-1}\left (\frac {1-2 x}{\sqrt {11}}\right )}{33135 \sqrt {11}}+\frac {628 \sqrt {11} \tan ^{-1}\left (\frac {1-2 x}{\sqrt {11}}\right )}{11045}-\frac {83 \log \left (3-x+x^2\right )}{1410}-\frac {\log \left (3-x+x^2\right )}{5 x}-\frac {\log \left (3-x+x^2\right )}{5 (5+2 x)}+\frac {1}{235} \int \frac {5+12 x}{3-x+x^2} \, dx+\frac {1}{15} \int \frac {5+x}{3-x+x^2} \, dx+\frac {250 \text {Subst}\left (\int \frac {1}{-11-x^2} \, dx,x,-1+2 x\right )}{2209}\\ &=\frac {53}{705} \sqrt {11} \tan ^{-1}\left (\frac {1-2 x}{\sqrt {11}}\right )-\frac {83 \log \left (3-x+x^2\right )}{1410}-\frac {\log \left (3-x+x^2\right )}{5 x}-\frac {\log \left (3-x+x^2\right )}{5 (5+2 x)}+\frac {6}{235} \int \frac {-1+2 x}{3-x+x^2} \, dx+\frac {1}{30} \int \frac {-1+2 x}{3-x+x^2} \, dx+\frac {11}{235} \int \frac {1}{3-x+x^2} \, dx+\frac {11}{30} \int \frac {1}{3-x+x^2} \, dx\\ &=\frac {53}{705} \sqrt {11} \tan ^{-1}\left (\frac {1-2 x}{\sqrt {11}}\right )-\frac {\log \left (3-x+x^2\right )}{5 x}-\frac {\log \left (3-x+x^2\right )}{5 (5+2 x)}-\frac {22}{235} \text {Subst}\left (\int \frac {1}{-11-x^2} \, dx,x,-1+2 x\right )-\frac {11}{15} \text {Subst}\left (\int \frac {1}{-11-x^2} \, dx,x,-1+2 x\right )\\ &=-\frac {\log \left (3-x+x^2\right )}{5 x}-\frac {\log \left (3-x+x^2\right )}{5 (5+2 x)}\\ \end {aligned} \end {gather*}
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Mathematica [A]
time = 0.08, size = 28, normalized size = 0.93 \begin {gather*} -\frac {(5+3 x) \log \left (3-x+x^2\right )}{5 x (5+2 x)} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.09, size = 27, normalized size = 0.90
method | result | size |
risch | \(-\frac {\left (3 x +5\right ) \ln \left (x^{2}-x +3\right )}{5 \left (5+2 x \right ) x}\) | \(27\) |
norman | \(\frac {-\frac {3 \ln \left (x^{2}-x +3\right ) x}{5}-\ln \left (x^{2}-x +3\right )}{\left (5+2 x \right ) x}\) | \(36\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.84, size = 27, normalized size = 0.90 \begin {gather*} -\frac {{\left (3 \, x + 5\right )} \log \left (x^{2} - x + 3\right )}{5 \, {\left (2 \, x^{2} + 5 \, x\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.37, size = 27, normalized size = 0.90 \begin {gather*} -\frac {{\left (3 \, x + 5\right )} \log \left (x^{2} - x + 3\right )}{5 \, {\left (2 \, x^{2} + 5 \, x\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A]
time = 0.09, size = 22, normalized size = 0.73 \begin {gather*} \frac {\left (- 3 x - 5\right ) \log {\left (x^{2} - x + 3 \right )}}{10 x^{2} + 25 x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.40, size = 22, normalized size = 0.73 \begin {gather*} -\frac {1}{5} \, {\left (\frac {1}{2 \, x + 5} + \frac {1}{x}\right )} \log \left (x^{2} - x + 3\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 0.27, size = 26, normalized size = 0.87 \begin {gather*} -\frac {\ln \left (x^2-x+3\right )\,\left (3\,x+5\right )}{5\,x\,\left (2\,x+5\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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