Integrand size = 20, antiderivative size = 114 \[ \int (c+d x) \cos ^2(a+b x) \cot (a+b x) \, dx=\frac {d x}{4 b}-\frac {i (c+d x)^2}{2 d}+\frac {(c+d x) \log \left (1-e^{2 i (a+b x)}\right )}{b}-\frac {i d \operatorname {PolyLog}\left (2,e^{2 i (a+b x)}\right )}{2 b^2}-\frac {d \cos (a+b x) \sin (a+b x)}{4 b^2}-\frac {(c+d x) \sin ^2(a+b x)}{2 b} \]
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Time = 0.15 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {4493, 4489, 2715, 8, 3798, 2221, 2317, 2438} \[ \int (c+d x) \cos ^2(a+b x) \cot (a+b x) \, dx=-\frac {i d \operatorname {PolyLog}\left (2,e^{2 i (a+b x)}\right )}{2 b^2}-\frac {d \sin (a+b x) \cos (a+b x)}{4 b^2}+\frac {(c+d x) \log \left (1-e^{2 i (a+b x)}\right )}{b}-\frac {(c+d x) \sin ^2(a+b x)}{2 b}+\frac {d x}{4 b}-\frac {i (c+d x)^2}{2 d} \]
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Rule 8
Rule 2221
Rule 2317
Rule 2438
Rule 2715
Rule 3798
Rule 4489
Rule 4493
Rubi steps \begin{align*} \text {integral}& = \int (c+d x) \cot (a+b x) \, dx-\int (c+d x) \cos (a+b x) \sin (a+b x) \, dx \\ & = -\frac {i (c+d x)^2}{2 d}-\frac {(c+d x) \sin ^2(a+b x)}{2 b}-2 i \int \frac {e^{2 i (a+b x)} (c+d x)}{1-e^{2 i (a+b x)}} \, dx+\frac {d \int \sin ^2(a+b x) \, dx}{2 b} \\ & = -\frac {i (c+d x)^2}{2 d}+\frac {(c+d x) \log \left (1-e^{2 i (a+b x)}\right )}{b}-\frac {d \cos (a+b x) \sin (a+b x)}{4 b^2}-\frac {(c+d x) \sin ^2(a+b x)}{2 b}+\frac {d \int 1 \, dx}{4 b}-\frac {d \int \log \left (1-e^{2 i (a+b x)}\right ) \, dx}{b} \\ & = \frac {d x}{4 b}-\frac {i (c+d x)^2}{2 d}+\frac {(c+d x) \log \left (1-e^{2 i (a+b x)}\right )}{b}-\frac {d \cos (a+b x) \sin (a+b x)}{4 b^2}-\frac {(c+d x) \sin ^2(a+b x)}{2 b}+\frac {(i d) \text {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^{2 i (a+b x)}\right )}{2 b^2} \\ & = \frac {d x}{4 b}-\frac {i (c+d x)^2}{2 d}+\frac {(c+d x) \log \left (1-e^{2 i (a+b x)}\right )}{b}-\frac {i d \operatorname {PolyLog}\left (2,e^{2 i (a+b x)}\right )}{2 b^2}-\frac {d \cos (a+b x) \sin (a+b x)}{4 b^2}-\frac {(c+d x) \sin ^2(a+b x)}{2 b} \\ \end{align*}
Time = 0.80 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.22 \[ \int (c+d x) \cos ^2(a+b x) \cot (a+b x) \, dx=\frac {d x \cos (2 (a+b x))}{4 b}+\frac {c \log (\sin (a+b x))}{b}-\frac {a d (\log (\cos (a+b x))+\log (\tan (a+b x)))}{b^2}+\frac {d \left ((a+b x) \log \left (1-e^{2 i (a+b x)}\right )-\frac {1}{2} i \left ((a+b x)^2+\operatorname {PolyLog}\left (2,e^{2 i (a+b x)}\right )\right )\right )}{b^2}-\frac {c \sin ^2(a+b x)}{2 b}-\frac {d \sin (2 (a+b x))}{8 b^2} \]
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Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 248 vs. \(2 (98 ) = 196\).
Time = 1.68 (sec) , antiderivative size = 249, normalized size of antiderivative = 2.18
method | result | size |
risch | \(-\frac {i d \,a^{2}}{b^{2}}-\frac {i d \operatorname {polylog}\left (2, {\mathrm e}^{i \left (x b +a \right )}\right )}{b^{2}}+\frac {c \ln \left ({\mathrm e}^{i \left (x b +a \right )}+1\right )}{b}-\frac {2 c \ln \left ({\mathrm e}^{i \left (x b +a \right )}\right )}{b}+\frac {c \ln \left ({\mathrm e}^{i \left (x b +a \right )}-1\right )}{b}-\frac {i d \operatorname {polylog}\left (2, -{\mathrm e}^{i \left (x b +a \right )}\right )}{b^{2}}-\frac {i d \,x^{2}}{2}+i c x +\frac {d \ln \left (1-{\mathrm e}^{i \left (x b +a \right )}\right ) x}{b}+\frac {d \ln \left (1-{\mathrm e}^{i \left (x b +a \right )}\right ) a}{b^{2}}-\frac {2 i d x a}{b}+\frac {d \ln \left ({\mathrm e}^{i \left (x b +a \right )}+1\right ) x}{b}+\frac {2 d a \ln \left ({\mathrm e}^{i \left (x b +a \right )}\right )}{b^{2}}-\frac {d a \ln \left ({\mathrm e}^{i \left (x b +a \right )}-1\right )}{b^{2}}+\frac {\left (d x +c \right ) \cos \left (2 x b +2 a \right )}{4 b}-\frac {d \sin \left (2 x b +2 a \right )}{8 b^{2}}\) | \(249\) |
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Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 292 vs. \(2 (95) = 190\).
Time = 0.27 (sec) , antiderivative size = 292, normalized size of antiderivative = 2.56 \[ \int (c+d x) \cos ^2(a+b x) \cot (a+b x) \, dx=-\frac {b d x - 2 \, {\left (b d x + b c\right )} \cos \left (b x + a\right )^{2} + d \cos \left (b x + a\right ) \sin \left (b x + a\right ) + 2 i \, d {\rm Li}_2\left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right ) - 2 i \, d {\rm Li}_2\left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right ) - 2 i \, d {\rm Li}_2\left (-\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right ) + 2 i \, d {\rm Li}_2\left (-\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right ) - 2 \, {\left (b d x + b c\right )} \log \left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right ) + 1\right ) - 2 \, {\left (b d x + b c\right )} \log \left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right ) + 1\right ) - 2 \, {\left (b c - a d\right )} \log \left (-\frac {1}{2} \, \cos \left (b x + a\right ) + \frac {1}{2} i \, \sin \left (b x + a\right ) + \frac {1}{2}\right ) - 2 \, {\left (b c - a d\right )} \log \left (-\frac {1}{2} \, \cos \left (b x + a\right ) - \frac {1}{2} i \, \sin \left (b x + a\right ) + \frac {1}{2}\right ) - 2 \, {\left (b d x + a d\right )} \log \left (-\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right ) + 1\right ) - 2 \, {\left (b d x + a d\right )} \log \left (-\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right ) + 1\right )}{4 \, b^{2}} \]
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\[ \int (c+d x) \cos ^2(a+b x) \cot (a+b x) \, dx=\int \left (c + d x\right ) \cos ^{2}{\left (a + b x \right )} \cot {\left (a + b x \right )}\, dx \]
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Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 223 vs. \(2 (95) = 190\).
Time = 0.30 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.96 \[ \int (c+d x) \cos ^2(a+b x) \cot (a+b x) \, dx=\frac {-4 i \, b^{2} d x^{2} - 8 i \, b^{2} c x - 8 i \, b d x \arctan \left (\sin \left (b x + a\right ), -\cos \left (b x + a\right ) + 1\right ) + 8 i \, b c \arctan \left (\sin \left (b x + a\right ), \cos \left (b x + a\right ) - 1\right ) - 8 \, {\left (-i \, b d x - i \, b c\right )} \arctan \left (\sin \left (b x + a\right ), \cos \left (b x + a\right ) + 1\right ) + 2 \, {\left (b d x + b c\right )} \cos \left (2 \, b x + 2 \, a\right ) - 8 i \, d {\rm Li}_2\left (-e^{\left (i \, b x + i \, a\right )}\right ) - 8 i \, d {\rm Li}_2\left (e^{\left (i \, b x + i \, a\right )}\right ) + 4 \, {\left (b d x + b c\right )} \log \left (\cos \left (b x + a\right )^{2} + \sin \left (b x + a\right )^{2} + 2 \, \cos \left (b x + a\right ) + 1\right ) + 4 \, {\left (b d x + b c\right )} \log \left (\cos \left (b x + a\right )^{2} + \sin \left (b x + a\right )^{2} - 2 \, \cos \left (b x + a\right ) + 1\right ) - d \sin \left (2 \, b x + 2 \, a\right )}{8 \, b^{2}} \]
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\[ \int (c+d x) \cos ^2(a+b x) \cot (a+b x) \, dx=\int { {\left (d x + c\right )} \cos \left (b x + a\right )^{2} \cot \left (b x + a\right ) \,d x } \]
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Timed out. \[ \int (c+d x) \cos ^2(a+b x) \cot (a+b x) \, dx=\int {\cos \left (a+b\,x\right )}^2\,\mathrm {cot}\left (a+b\,x\right )\,\left (c+d\,x\right ) \,d x \]
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