Optimal. Leaf size=62 \[ -\left (\frac {a}{b}\right )^{-1/n} \tan ^{-1}\left (\cot \left (\frac {\pi -2 k \pi }{n}\right )-\left (\frac {a}{b}\right )^{-1/n} x \csc \left (\frac {\pi -2 k \pi }{n}\right )\right ) \csc \left (\frac {\pi -2 k \pi }{n}\right ) \]
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Rubi [A]
time = 0.12, antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps
used = 2, number of rules used = 2, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {632, 210}
\begin {gather*} -\left (\frac {a}{b}\right )^{-1/n} \csc \left (\frac {\pi -2 \pi k}{n}\right ) \text {ArcTan}\left (\cot \left (\frac {\pi -2 \pi k}{n}\right )-x \left (\frac {a}{b}\right )^{-1/n} \csc \left (\frac {\pi -2 \pi k}{n}\right )\right ) \end {gather*}
Antiderivative was successfully verified.
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Rule 210
Rule 632
Rubi steps
\begin {align*} \int \frac {1}{\left (\frac {a}{b}\right )^{2/n}+x^2-2 \left (\frac {a}{b}\right )^{\frac {1}{n}} x \cos \left (\frac {\pi -2 k \pi }{n}\right )} \, dx &=-\left (2 \text {Subst}\left (\int \frac {1}{-x^2-4 \left (\frac {a}{b}\right )^{2/n} \left (1-\cos ^2\left (\frac {\pi -2 k \pi }{n}\right )\right )} \, dx,x,2 x-2 \left (\frac {a}{b}\right )^{\frac {1}{n}} \cos \left (\frac {\pi -2 k \pi }{n}\right )\right )\right )\\ &=-\left (\frac {a}{b}\right )^{-1/n} \tan ^{-1}\left (\cot \left (\frac {\pi -2 k \pi }{n}\right )-\left (\frac {a}{b}\right )^{-1/n} x \csc \left (\frac {\pi -2 k \pi }{n}\right )\right ) \csc \left (\frac {\pi -2 k \pi }{n}\right )\\ \end {align*}
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Mathematica [A]
time = 0.07, size = 65, normalized size = 1.05 \begin {gather*} \left (\frac {a}{b}\right )^{-1/n} \tan ^{-1}\left (\frac {\left (\left (\frac {a}{b}\right )^{\frac {1}{n}}+x\right ) \tan \left (\frac {\pi -2 k \pi }{2 n}\right )}{\left (\frac {a}{b}\right )^{\frac {1}{n}}-x}\right ) \csc \left (\frac {\pi -2 k \pi }{n}\right ) \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 1.62, size = 111, normalized size = 1.79
method | result | size |
default | \(\frac {\arctan \left (\frac {2 x -2 \left (\frac {a}{b}\right )^{\frac {1}{n}} \cos \left (\frac {\pi \left (2 k -1\right )}{n}\right )}{2 \sqrt {-\left (\frac {a}{b}\right )^{\frac {2}{n}} \left (\cos ^{2}\left (\frac {\pi \left (2 k -1\right )}{n}\right )\right )+\left (\frac {a}{b}\right )^{\frac {2}{n}}}}\right )}{\sqrt {-\left (\frac {a}{b}\right )^{\frac {2}{n}} \left (\cos ^{2}\left (\frac {\pi \left (2 k -1\right )}{n}\right )\right )+\left (\frac {a}{b}\right )^{\frac {2}{n}}}}\) | \(111\) |
risch | \(\text {Expression too large to display}\) | \(1343\) |
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: ValueError} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 2.08, size = 89, normalized size = 1.44 \begin {gather*} -\frac {\arctan \left (\frac {\left (\frac {a}{b}\right )^{\left (\frac {1}{n}\right )} \cos \left (\frac {2 \, \pi k}{n} - \frac {\pi }{n}\right ) - x}{\left (\frac {a}{b}\right )^{\left (\frac {1}{n}\right )} \sin \left (\frac {2 \, \pi k}{n} - \frac {\pi }{n}\right )}\right )}{\left (\frac {a}{b}\right )^{\left (\frac {1}{n}\right )} \sin \left (\frac {2 \, \pi k}{n} - \frac {\pi }{n}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 212 vs.
\(2 (46) = 92\).
time = 0.43, size = 212, normalized size = 3.42 \begin {gather*} - \frac {\sqrt {\frac {\left (\frac {a}{b}\right )^{- \frac {2}{n}}}{\cos ^{2}{\left (\frac {\pi \left (2 k - 1\right )}{n} \right )} - 1}} \log {\left (x - \left (\frac {a}{b}\right )^{\frac {1}{n}} \cos {\left (\frac {2 \pi k}{n} - \frac {\pi }{n} \right )} - \frac {\sqrt {\frac {\left (\frac {a}{b}\right )^{- \frac {2}{n}}}{\cos ^{2}{\left (\frac {\pi \left (2 k - 1\right )}{n} \right )} - 1}} \left (- 2 \left (\frac {a}{b}\right )^{\frac {2}{n}} \cos ^{2}{\left (\frac {2 \pi k}{n} - \frac {\pi }{n} \right )} + 2 \left (\frac {a}{b}\right )^{\frac {2}{n}}\right )}{2} \right )}}{2} + \frac {\sqrt {\frac {\left (\frac {a}{b}\right )^{- \frac {2}{n}}}{\cos ^{2}{\left (\frac {\pi \left (2 k - 1\right )}{n} \right )} - 1}} \log {\left (x - \left (\frac {a}{b}\right )^{\frac {1}{n}} \cos {\left (\frac {2 \pi k}{n} - \frac {\pi }{n} \right )} + \frac {\sqrt {\frac {\left (\frac {a}{b}\right )^{- \frac {2}{n}}}{\cos ^{2}{\left (\frac {\pi \left (2 k - 1\right )}{n} \right )} - 1}} \left (- 2 \left (\frac {a}{b}\right )^{\frac {2}{n}} \cos ^{2}{\left (\frac {2 \pi k}{n} - \frac {\pi }{n} \right )} + 2 \left (\frac {a}{b}\right )^{\frac {2}{n}}\right )}{2} \right )}}{2} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.82, size = 100, normalized size = 1.61 \begin {gather*} \frac {\arctan \left (-\frac {\left (\frac {a}{b}\right )^{\left (\frac {1}{n}\right )} \cos \left (-\frac {2 \, \pi k}{n} + \frac {\pi }{n}\right ) - x}{\sqrt {-\cos \left (\frac {2 \, \pi k}{n} - \frac {\pi }{n}\right )^{2} + 1} \left (\frac {a}{b}\right )^{\left (\frac {1}{n}\right )}}\right )}{\sqrt {-\cos \left (\frac {2 \, \pi k}{n} - \frac {\pi }{n}\right )^{2} + 1} \left (\frac {a}{b}\right )^{\left (\frac {1}{n}\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 0.25, size = 110, normalized size = 1.77 \begin {gather*} -\frac {\mathrm {atanh}\left (\frac {x-\cos \left (\frac {\Pi \,\left (2\,k-1\right )}{n}\right )\,{\left (\frac {a}{b}\right )}^{1/n}}{\sqrt {\cos \left (\frac {\Pi \,\left (2\,k-1\right )}{n}\right )-1}\,\sqrt {\cos \left (\frac {\Pi \,\left (2\,k-1\right )}{n}\right )+1}\,{\left (\frac {a}{b}\right )}^{1/n}}\right )}{\sqrt {\cos \left (\frac {\Pi \,\left (2\,k-1\right )}{n}\right )-1}\,\sqrt {\cos \left (\frac {\Pi \,\left (2\,k-1\right )}{n}\right )+1}\,{\left (\frac {a}{b}\right )}^{1/n}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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