Optimal. Leaf size=205 \[ \frac {2 \tan ^{-1}\left (\sqrt {\frac {1-\sqrt [5]{-1}}{1+\sqrt [5]{-1}}} \tan \left (\frac {x}{2}\right )\right )}{5 \sqrt {1-(-1)^{2/5}}}+\frac {2 \tan ^{-1}\left (\sqrt {\frac {1-(-1)^{3/5}}{1+(-1)^{3/5}}} \tan \left (\frac {x}{2}\right )\right )}{5 \sqrt {1+\sqrt [5]{-1}}}-\frac {\sin (x)}{5 (1-\cos (x))}-\frac {2 \tanh ^{-1}\left (\frac {\tan \left (\frac {x}{2}\right )}{\sqrt {-\frac {1-(-1)^{2/5}}{1+(-1)^{2/5}}}}\right )}{5 \sqrt {(-1)^{4/5}-1}}+\frac {2 \tanh ^{-1}\left (\sqrt {-\frac {1+(-1)^{4/5}}{1-(-1)^{4/5}}} \tan \left (\frac {x}{2}\right )\right )}{5 \sqrt {-1-(-1)^{3/5}}} \]
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Rubi [A] time = 0.47, antiderivative size = 205, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 5, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3213, 2648, 2659, 205, 208} \[ \frac {2 \tan ^{-1}\left (\sqrt {\frac {1-\sqrt [5]{-1}}{1+\sqrt [5]{-1}}} \tan \left (\frac {x}{2}\right )\right )}{5 \sqrt {1-(-1)^{2/5}}}+\frac {2 \tan ^{-1}\left (\sqrt {\frac {1-(-1)^{3/5}}{1+(-1)^{3/5}}} \tan \left (\frac {x}{2}\right )\right )}{5 \sqrt {1+\sqrt [5]{-1}}}-\frac {\sin (x)}{5 (1-\cos (x))}-\frac {2 \tanh ^{-1}\left (\frac {\tan \left (\frac {x}{2}\right )}{\sqrt {-\frac {1-(-1)^{2/5}}{1+(-1)^{2/5}}}}\right )}{5 \sqrt {(-1)^{4/5}-1}}+\frac {2 \tanh ^{-1}\left (\sqrt {-\frac {1+(-1)^{4/5}}{1-(-1)^{4/5}}} \tan \left (\frac {x}{2}\right )\right )}{5 \sqrt {-1-(-1)^{3/5}}} \]
Antiderivative was successfully verified.
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Rule 205
Rule 208
Rule 2648
Rule 2659
Rule 3213
Rubi steps
\begin {align*} \int \frac {1}{1-\cos ^5(x)} \, dx &=\int \left (\frac {1}{5 (1-\cos (x))}+\frac {1}{5 \left (1+\sqrt [5]{-1} \cos (x)\right )}+\frac {1}{5 \left (1-(-1)^{2/5} \cos (x)\right )}+\frac {1}{5 \left (1+(-1)^{3/5} \cos (x)\right )}+\frac {1}{5 \left (1-(-1)^{4/5} \cos (x)\right )}\right ) \, dx\\ &=\frac {1}{5} \int \frac {1}{1-\cos (x)} \, dx+\frac {1}{5} \int \frac {1}{1+\sqrt [5]{-1} \cos (x)} \, dx+\frac {1}{5} \int \frac {1}{1-(-1)^{2/5} \cos (x)} \, dx+\frac {1}{5} \int \frac {1}{1+(-1)^{3/5} \cos (x)} \, dx+\frac {1}{5} \int \frac {1}{1-(-1)^{4/5} \cos (x)} \, dx\\ &=-\frac {\sin (x)}{5 (1-\cos (x))}+\frac {2}{5} \operatorname {Subst}\left (\int \frac {1}{1+\sqrt [5]{-1}+\left (1-\sqrt [5]{-1}\right ) x^2} \, dx,x,\tan \left (\frac {x}{2}\right )\right )+\frac {2}{5} \operatorname {Subst}\left (\int \frac {1}{1-(-1)^{2/5}+\left (1+(-1)^{2/5}\right ) x^2} \, dx,x,\tan \left (\frac {x}{2}\right )\right )+\frac {2}{5} \operatorname {Subst}\left (\int \frac {1}{1+(-1)^{3/5}+\left (1-(-1)^{3/5}\right ) x^2} \, dx,x,\tan \left (\frac {x}{2}\right )\right )+\frac {2}{5} \operatorname {Subst}\left (\int \frac {1}{1-(-1)^{4/5}+\left (1+(-1)^{4/5}\right ) x^2} \, dx,x,\tan \left (\frac {x}{2}\right )\right )\\ &=\frac {2 \tan ^{-1}\left (\sqrt {\frac {1-\sqrt [5]{-1}}{1+\sqrt [5]{-1}}} \tan \left (\frac {x}{2}\right )\right )}{5 \sqrt {1-(-1)^{2/5}}}+\frac {2 \tan ^{-1}\left (\sqrt {\frac {1-(-1)^{3/5}}{1+(-1)^{3/5}}} \tan \left (\frac {x}{2}\right )\right )}{5 \sqrt {1+\sqrt [5]{-1}}}-\frac {2 \tanh ^{-1}\left (\frac {\tan \left (\frac {x}{2}\right )}{\sqrt {-\frac {1-(-1)^{2/5}}{1+(-1)^{2/5}}}}\right )}{5 \sqrt {-1+(-1)^{4/5}}}+\frac {2 \tanh ^{-1}\left (\sqrt {-\frac {1+(-1)^{4/5}}{1-(-1)^{4/5}}} \tan \left (\frac {x}{2}\right )\right )}{5 \sqrt {-1-(-1)^{3/5}}}-\frac {\sin (x)}{5 (1-\cos (x))}\\ \end {align*}
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Mathematica [C] time = 0.12, size = 378, normalized size = 1.84 \[ -\frac {1}{5} \cot \left (\frac {x}{2}\right )+\frac {1}{10} \text {RootSum}\left [\text {$\#$1}^8+2 \text {$\#$1}^7+8 \text {$\#$1}^6+14 \text {$\#$1}^5+30 \text {$\#$1}^4+14 \text {$\#$1}^3+8 \text {$\#$1}^2+2 \text {$\#$1}+1\& ,\frac {2 \text {$\#$1}^6 \tan ^{-1}\left (\frac {\sin (x)}{\cos (x)-\text {$\#$1}}\right )+8 \text {$\#$1}^5 \tan ^{-1}\left (\frac {\sin (x)}{\cos (x)-\text {$\#$1}}\right )+30 \text {$\#$1}^4 \tan ^{-1}\left (\frac {\sin (x)}{\cos (x)-\text {$\#$1}}\right )+80 \text {$\#$1}^3 \tan ^{-1}\left (\frac {\sin (x)}{\cos (x)-\text {$\#$1}}\right )-15 i \text {$\#$1}^2 \log \left (\text {$\#$1}^2-2 \text {$\#$1} \cos (x)+1\right )-4 i \text {$\#$1} \log \left (\text {$\#$1}^2-2 \text {$\#$1} \cos (x)+1\right )-i \log \left (\text {$\#$1}^2-2 \text {$\#$1} \cos (x)+1\right )+30 \text {$\#$1}^2 \tan ^{-1}\left (\frac {\sin (x)}{\cos (x)-\text {$\#$1}}\right )-i \text {$\#$1}^6 \log \left (\text {$\#$1}^2-2 \text {$\#$1} \cos (x)+1\right )-4 i \text {$\#$1}^5 \log \left (\text {$\#$1}^2-2 \text {$\#$1} \cos (x)+1\right )-15 i \text {$\#$1}^4 \log \left (\text {$\#$1}^2-2 \text {$\#$1} \cos (x)+1\right )-40 i \text {$\#$1}^3 \log \left (\text {$\#$1}^2-2 \text {$\#$1} \cos (x)+1\right )+8 \text {$\#$1} \tan ^{-1}\left (\frac {\sin (x)}{\cos (x)-\text {$\#$1}}\right )+2 \tan ^{-1}\left (\frac {\sin (x)}{\cos (x)-\text {$\#$1}}\right )}{4 \text {$\#$1}^7+7 \text {$\#$1}^6+24 \text {$\#$1}^5+35 \text {$\#$1}^4+60 \text {$\#$1}^3+21 \text {$\#$1}^2+8 \text {$\#$1}+1}\& \right ] \]
Antiderivative was successfully verified.
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fricas [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {sage}_{0} x \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 0.08, size = 62, normalized size = 0.30 \[ \frac {\left (\munderset {\textit {\_R} =\RootOf \left (\textit {\_Z}^{8}+10 \textit {\_Z}^{4}+5\right )}{\sum }\frac {\left (\textit {\_R}^{6}+5 \textit {\_R}^{4}+5 \textit {\_R}^{2}+5\right ) \ln \left (\tan \left (\frac {x}{2}\right )-\textit {\_R} \right )}{\textit {\_R}^{7}+5 \textit {\_R}^{3}}\right )}{10}-\frac {1}{5 \tan \left (\frac {x}{2}\right )} \]
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 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 2.45, size = 403, normalized size = 1.97 \[ 2\,\mathrm {atanh}\left (\frac {50\,\mathrm {tan}\left (\frac {x}{2}\right )\,\sqrt {\frac {\sqrt {-\frac {2\,\sqrt {5}}{5}-1}}{50}-\frac {1}{50}}-20\,\sqrt {5}\,\mathrm {tan}\left (\frac {x}{2}\right )\,\sqrt {\frac {\sqrt {-\frac {2\,\sqrt {5}}{5}-1}}{50}-\frac {1}{50}}}{5\,\sqrt {5}\,\sqrt {-\frac {2\,\sqrt {5}}{5}-1}+2\,\sqrt {5}-10\,\sqrt {-\frac {2\,\sqrt {5}}{5}-1}-5}\right )\,\sqrt {\frac {\sqrt {-\frac {2\,\sqrt {5}}{5}-1}}{50}-\frac {1}{50}}-2\,\mathrm {atanh}\left (\frac {50\,\mathrm {tan}\left (\frac {x}{2}\right )\,\sqrt {-\frac {\sqrt {-\frac {2\,\sqrt {5}}{5}-1}}{50}-\frac {1}{50}}-20\,\sqrt {5}\,\mathrm {tan}\left (\frac {x}{2}\right )\,\sqrt {-\frac {\sqrt {-\frac {2\,\sqrt {5}}{5}-1}}{50}-\frac {1}{50}}}{5\,\sqrt {5}\,\sqrt {-\frac {2\,\sqrt {5}}{5}-1}-2\,\sqrt {5}-10\,\sqrt {-\frac {2\,\sqrt {5}}{5}-1}+5}\right )\,\sqrt {-\frac {\sqrt {-\frac {2\,\sqrt {5}}{5}-1}}{50}-\frac {1}{50}}-\frac {\mathrm {cot}\left (\frac {x}{2}\right )}{5}+2\,\mathrm {atanh}\left (\frac {50\,\mathrm {tan}\left (\frac {x}{2}\right )\,\sqrt {-\frac {\sqrt {\frac {2\,\sqrt {5}}{5}-1}}{50}-\frac {1}{50}}+20\,\sqrt {5}\,\mathrm {tan}\left (\frac {x}{2}\right )\,\sqrt {-\frac {\sqrt {\frac {2\,\sqrt {5}}{5}-1}}{50}-\frac {1}{50}}}{5\,\sqrt {5}\,\sqrt {\frac {2\,\sqrt {5}}{5}-1}-2\,\sqrt {5}+10\,\sqrt {\frac {2\,\sqrt {5}}{5}-1}-5}\right )\,\sqrt {-\frac {\sqrt {\frac {2\,\sqrt {5}}{5}-1}}{50}-\frac {1}{50}}-2\,\mathrm {atanh}\left (\frac {50\,\mathrm {tan}\left (\frac {x}{2}\right )\,\sqrt {\frac {\sqrt {\frac {2\,\sqrt {5}}{5}-1}}{50}-\frac {1}{50}}+20\,\sqrt {5}\,\mathrm {tan}\left (\frac {x}{2}\right )\,\sqrt {\frac {\sqrt {\frac {2\,\sqrt {5}}{5}-1}}{50}-\frac {1}{50}}}{5\,\sqrt {5}\,\sqrt {\frac {2\,\sqrt {5}}{5}-1}+2\,\sqrt {5}+10\,\sqrt {\frac {2\,\sqrt {5}}{5}-1}+5}\right )\,\sqrt {\frac {\sqrt {\frac {2\,\sqrt {5}}{5}-1}}{50}-\frac {1}{50}} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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