Optimal. Leaf size=51 \[ \frac {1}{10} \tan ^{-1}\left (\frac {5 (x+2)}{2 \sqrt {5 x^2+2 x-7}}\right )+\frac {1}{5} \tanh ^{-1}\left (\frac {5 (x+1)}{\sqrt {5 x^2+2 x-7}}\right ) \]
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Rubi [A] time = 0.07, antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {986, 1029, 203, 207} \begin {gather*} \frac {1}{10} \tan ^{-1}\left (\frac {5 (x+2)}{2 \sqrt {5 x^2+2 x-7}}\right )+\frac {1}{5} \tanh ^{-1}\left (\frac {5 (x+1)}{\sqrt {5 x^2+2 x-7}}\right ) \end {gather*}
Antiderivative was successfully verified.
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Rule 203
Rule 207
Rule 986
Rule 1029
Rubi steps
\begin {align*} \int \frac {1}{\sqrt {-7+2 x+5 x^2} \left (8+12 x+5 x^2\right )} \, dx &=-\left (\frac {1}{50} \int \frac {-100-50 x}{\sqrt {-7+2 x+5 x^2} \left (8+12 x+5 x^2\right )} \, dx\right )+\frac {1}{50} \int \frac {-50-50 x}{\sqrt {-7+2 x+5 x^2} \left (8+12 x+5 x^2\right )} \, dx\\ &=400 \operatorname {Subst}\left (\int \frac {1}{160000+100 x^2} \, dx,x,\frac {200+100 x}{\sqrt {-7+2 x+5 x^2}}\right )+1600 \operatorname {Subst}\left (\int \frac {1}{-640000+100 x^2} \, dx,x,\frac {-400-400 x}{\sqrt {-7+2 x+5 x^2}}\right )\\ &=\frac {1}{10} \tan ^{-1}\left (\frac {5 (2+x)}{2 \sqrt {-7+2 x+5 x^2}}\right )+\frac {1}{5} \tanh ^{-1}\left (\frac {5 (1+x)}{\sqrt {-7+2 x+5 x^2}}\right )\\ \end {align*}
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Mathematica [C] time = 0.04, size = 81, normalized size = 1.59 \begin {gather*} \left (\frac {1}{10}-\frac {i}{20}\right ) \tanh ^{-1}\left (\frac {\left (\frac {1}{100}+\frac {i}{50}\right ) ((100-40 i) x+(164-8 i))}{\sqrt {5 x^2+2 x-7}}\right )-\left (\frac {1}{20}-\frac {i}{10}\right ) \tan ^{-1}\left (\frac {\left (\frac {1}{50}+\frac {i}{100}\right ) ((-100-40 i) x-(164+8 i))}{\sqrt {5 x^2+2 x-7}}\right ) \end {gather*}
Antiderivative was successfully verified.
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IntegrateAlgebraic [A] time = 0.34, size = 53, normalized size = 1.04 \begin {gather*} \frac {1}{10} \tan ^{-1}\left (\frac {\frac {5 x}{2}+5}{\sqrt {5 x^2+2 x-7}}\right )+\frac {1}{5} \tanh ^{-1}\left (\frac {5 x+5}{\sqrt {5 x^2+2 x-7}}\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [B] time = 0.45, size = 154, normalized size = 3.02 \begin {gather*} \frac {1}{20} \, \arctan \left (\frac {27 \, x^{2} + 20 \, \sqrt {5 \, x^{2} + 2 \, x - 7} {\left (x + 2\right )} + 36 \, x}{31 \, x^{2} + 16 \, x - 56}\right ) + \frac {1}{20} \, \arctan \left (-\frac {27 \, x^{2} - 20 \, \sqrt {5 \, x^{2} + 2 \, x - 7} {\left (x + 2\right )} + 36 \, x}{31 \, x^{2} + 16 \, x - 56}\right ) + \frac {1}{20} \, \log \left (\frac {15 \, x^{2} + 5 \, \sqrt {5 \, x^{2} + 2 \, x - 7} {\left (x + 1\right )} + 26 \, x + 9}{x^{2}}\right ) - \frac {1}{20} \, \log \left (\frac {15 \, x^{2} - 5 \, \sqrt {5 \, x^{2} + 2 \, x - 7} {\left (x + 1\right )} + 26 \, x + 9}{x^{2}}\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.28, size = 205, normalized size = 4.02 \begin {gather*} -\frac {1}{10} \, \arctan \left (-\frac {5 \, \sqrt {5} x + 6 \, \sqrt {5} - 5 \, \sqrt {5 \, x^{2} + 2 \, x - 7} + 5}{2 \, {\left (\sqrt {5} + 5\right )}}\right ) - \frac {1}{10} \, \arctan \left (\frac {5 \, \sqrt {5} x + 6 \, \sqrt {5} - 5 \, \sqrt {5 \, x^{2} + 2 \, x - 7} - 5}{2 \, {\left (\sqrt {5} - 5\right )}}\right ) + \frac {1}{10} \, \log \left (5 \, {\left (\sqrt {5} x - \sqrt {5 \, x^{2} + 2 \, x - 7}\right )}^{2} + 2 \, {\left (\sqrt {5} x - \sqrt {5 \, x^{2} + 2 \, x - 7}\right )} {\left (6 \, \sqrt {5} + 5\right )} + 20 \, \sqrt {5} + 65\right ) - \frac {1}{10} \, \log \left (5 \, {\left (\sqrt {5} x - \sqrt {5 \, x^{2} + 2 \, x - 7}\right )}^{2} + 2 \, {\left (\sqrt {5} x - \sqrt {5 \, x^{2} + 2 \, x - 7}\right )} {\left (6 \, \sqrt {5} - 5\right )} - 20 \, \sqrt {5} + 65\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.02, size = 144, normalized size = 2.82 \begin {gather*} -\frac {\sqrt {-\frac {4 \left (x +2\right )^{2}}{\left (-x -1\right )^{2}}+9}\, \left (2 \arctanh \left (\frac {\sqrt {-\frac {4 \left (x +2\right )^{2}}{\left (-x -1\right )^{2}}+9}}{5}\right )+\arctan \left (\frac {5 \sqrt {-\frac {4 \left (x +2\right )^{2}}{\left (-x -1\right )^{2}}+9}\, \left (x +2\right )}{2 \left (\frac {4 \left (x +2\right )^{2}}{\left (-x -1\right )^{2}}-9\right ) \left (-x -1\right )}\right )\right )}{10 \sqrt {-\frac {\frac {4 \left (x +2\right )^{2}}{\left (-x -1\right )^{2}}-9}{\left (1+\frac {x +2}{-x -1}\right )^{2}}}\, \left (1+\frac {x +2}{-x -1}\right )} \end {gather*}
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*} \int \frac {1}{{\left (5 \, x^{2} + 12 \, x + 8\right )} \sqrt {5 \, x^{2} + 2 \, x - 7}}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {1}{\sqrt {5\,x^2+2\,x-7}\,\left (5\,x^2+12\,x+8\right )} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt {\left (x - 1\right ) \left (5 x + 7\right )} \left (5 x^{2} + 12 x + 8\right )}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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