Integrand size = 39, antiderivative size = 16 \[ \int \frac {\cos ^2(a+b x)-\sin ^2(a+b x)}{\cos ^2(a+b x)+\sin ^2(a+b x)} \, dx=\frac {\cos (a+b x) \sin (a+b x)}{b} \]
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Time = 0.07 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {4465, 2715, 8} \[ \int \frac {\cos ^2(a+b x)-\sin ^2(a+b x)}{\cos ^2(a+b x)+\sin ^2(a+b x)} \, dx=\frac {\sin (a+b x) \cos (a+b x)}{b} \]
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Rule 8
Rule 2715
Rule 4465
Rubi steps \begin{align*} \text {integral}& = \int \left (\cos ^2(a+b x)-\sin ^2(a+b x)\right ) \, dx \\ & = \int \cos ^2(a+b x) \, dx-\int \sin ^2(a+b x) \, dx \\ & = \frac {\cos (a+b x) \sin (a+b x)}{b} \\ \end{align*}
Leaf count is larger than twice the leaf count of optimal. \(33\) vs. \(2(16)=32\).
Time = 0.01 (sec) , antiderivative size = 33, normalized size of antiderivative = 2.06 \[ \int \frac {\cos ^2(a+b x)-\sin ^2(a+b x)}{\cos ^2(a+b x)+\sin ^2(a+b x)} \, dx=\frac {\cos (2 b x) \sin (2 a)}{2 b}+\frac {\cos (2 a) \sin (2 b x)}{2 b} \]
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Time = 0.62 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.94
method | result | size |
risch | \(\frac {\sin \left (2 x b +2 a \right )}{2 b}\) | \(15\) |
parallelrisch | \(\frac {\sin \left (2 x b +2 a \right )}{2 b}\) | \(15\) |
derivativedivides | \(\frac {\cos \left (x b +a \right ) \sin \left (x b +a \right )}{b}\) | \(17\) |
default | \(\frac {\cos \left (x b +a \right ) \sin \left (x b +a \right )}{b}\) | \(17\) |
norman | \(\frac {\frac {2 \tan \left (\frac {a}{2}+\frac {x b}{2}\right )}{b}+\frac {2 \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{3}}{b}-\frac {2 \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{5}}{b}-\frac {2 \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{7}}{b}}{\left (1+\tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}\right )^{4}}\) | \(80\) |
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none
Time = 0.25 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00 \[ \int \frac {\cos ^2(a+b x)-\sin ^2(a+b x)}{\cos ^2(a+b x)+\sin ^2(a+b x)} \, dx=\frac {\cos \left (b x + a\right ) \sin \left (b x + a\right )}{b} \]
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Leaf count of result is larger than twice the leaf count of optimal. 32 vs. \(2 (14) = 28\).
Time = 0.12 (sec) , antiderivative size = 32, normalized size of antiderivative = 2.00 \[ \int \frac {\cos ^2(a+b x)-\sin ^2(a+b x)}{\cos ^2(a+b x)+\sin ^2(a+b x)} \, dx=\frac {\sin {\left (a + b x \right )} \cos {\left (a + b x \right )}}{b \sin ^{2}{\left (a + b x \right )} + b \cos ^{2}{\left (a + b x \right )}} \]
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Time = 0.22 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.38 \[ \int \frac {\cos ^2(a+b x)-\sin ^2(a+b x)}{\cos ^2(a+b x)+\sin ^2(a+b x)} \, dx=\frac {\tan \left (b x + a\right )}{{\left (\tan \left (b x + a\right )^{2} + 1\right )} b} \]
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Time = 0.30 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.88 \[ \int \frac {\cos ^2(a+b x)-\sin ^2(a+b x)}{\cos ^2(a+b x)+\sin ^2(a+b x)} \, dx=\frac {\sin \left (2 \, b x + 2 \, a\right )}{2 \, b} \]
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Time = 26.99 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.88 \[ \int \frac {\cos ^2(a+b x)-\sin ^2(a+b x)}{\cos ^2(a+b x)+\sin ^2(a+b x)} \, dx=\frac {\sin \left (2\,a+2\,b\,x\right )}{2\,b} \]
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