Integrand size = 19, antiderivative size = 28 \[ \int \cot (e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx=-\frac {b \log (\cos (e+f x))}{f}+\frac {(a+b) \log (\sin (e+f x))}{f} \]
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Time = 0.06 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {4223, 457, 78} \[ \int \cot (e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx=\frac {(a+b) \log (\sin (e+f x))}{f}-\frac {b \log (\cos (e+f x))}{f} \]
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Rule 78
Rule 457
Rule 4223
Rubi steps \begin{align*} \text {integral}& = -\frac {\text {Subst}\left (\int \frac {b+a x^2}{x \left (1-x^2\right )} \, dx,x,\cos (e+f x)\right )}{f} \\ & = -\frac {\text {Subst}\left (\int \frac {b+a x}{(1-x) x} \, dx,x,\cos ^2(e+f x)\right )}{2 f} \\ & = -\frac {\text {Subst}\left (\int \left (\frac {-a-b}{-1+x}+\frac {b}{x}\right ) \, dx,x,\cos ^2(e+f x)\right )}{2 f} \\ & = -\frac {b \log (\cos (e+f x))}{f}+\frac {(a+b) \log (\sin (e+f x))}{f} \\ \end{align*}
Time = 0.03 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.79 \[ \int \cot (e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx=\frac {a \log (\cos (e+f x))}{f}-\frac {b \log (\cos (e+f x))}{f}+\frac {b \log (\sin (e+f x))}{f}+\frac {a \log (\tan (e+f x))}{f} \]
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Time = 0.33 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.86
method | result | size |
derivativedivides | \(\frac {a \ln \left (\sin \left (f x +e \right )\right )+b \ln \left (\tan \left (f x +e \right )\right )}{f}\) | \(24\) |
default | \(\frac {a \ln \left (\sin \left (f x +e \right )\right )+b \ln \left (\tan \left (f x +e \right )\right )}{f}\) | \(24\) |
risch | \(-i a x -\frac {2 i a e}{f}+\frac {\ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right ) a}{f}+\frac {\ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right ) b}{f}-\frac {b \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}{f}\) | \(67\) |
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Time = 0.26 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.25 \[ \int \cot (e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx=-\frac {b \log \left (\cos \left (f x + e\right )^{2}\right ) - {\left (a + b\right )} \log \left (-\frac {1}{4} \, \cos \left (f x + e\right )^{2} + \frac {1}{4}\right )}{2 \, f} \]
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\[ \int \cot (e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx=\int \left (a + b \sec ^{2}{\left (e + f x \right )}\right ) \cot {\left (e + f x \right )}\, dx \]
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none
Time = 0.19 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.18 \[ \int \cot (e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx=-\frac {b \log \left (\sin \left (f x + e\right )^{2} - 1\right ) - {\left (a + b\right )} \log \left (\sin \left (f x + e\right )^{2}\right )}{2 \, f} \]
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Leaf count of result is larger than twice the leaf count of optimal. 98 vs. \(2 (28) = 56\).
Time = 0.28 (sec) , antiderivative size = 98, normalized size of antiderivative = 3.50 \[ \int \cot (e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx=-\frac {a \log \left ({\left | -\frac {\cos \left (f x + e\right ) + 1}{\cos \left (f x + e\right ) - 1} - \frac {\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1} + 2 \right |}\right ) + b \log \left ({\left | -\frac {\cos \left (f x + e\right ) + 1}{\cos \left (f x + e\right ) - 1} - \frac {\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1} - 2 \right |}\right )}{2 \, f} \]
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Time = 19.68 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.14 \[ \int \cot (e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx=\frac {\ln \left (\mathrm {tan}\left (e+f\,x\right )\right )\,\left (a+b\right )}{f}-\frac {a\,\ln \left ({\mathrm {tan}\left (e+f\,x\right )}^2+1\right )}{2\,f} \]
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