Integrand size = 24, antiderivative size = 75 \[ \int \frac {1}{(2 a+2 a \cos (d+e x)+2 a \sin (d+e x))^2} \, dx=-\frac {\log \left (1+\tan \left (\frac {1}{2} (d+e x)\right )\right )}{4 a^2 e}-\frac {a \cos (d+e x)-a \sin (d+e x)}{4 e \left (a^3+a^3 \cos (d+e x)+a^3 \sin (d+e x)\right )} \]
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Time = 0.05 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3208, 12, 3203, 31} \[ \int \frac {1}{(2 a+2 a \cos (d+e x)+2 a \sin (d+e x))^2} \, dx=-\frac {a \cos (d+e x)-a \sin (d+e x)}{4 e \left (a^3 \sin (d+e x)+a^3 \cos (d+e x)+a^3\right )}-\frac {\log \left (\tan \left (\frac {1}{2} (d+e x)\right )+1\right )}{4 a^2 e} \]
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Rule 12
Rule 31
Rule 3203
Rule 3208
Rubi steps \begin{align*} \text {integral}& = -\frac {a \cos (d+e x)-a \sin (d+e x)}{4 e \left (a^3+a^3 \cos (d+e x)+a^3 \sin (d+e x)\right )}+\frac {\int -\frac {2 a}{2 a+2 a \cos (d+e x)+2 a \sin (d+e x)} \, dx}{4 a^2} \\ & = -\frac {a \cos (d+e x)-a \sin (d+e x)}{4 e \left (a^3+a^3 \cos (d+e x)+a^3 \sin (d+e x)\right )}-\frac {\int \frac {1}{2 a+2 a \cos (d+e x)+2 a \sin (d+e x)} \, dx}{2 a} \\ & = -\frac {a \cos (d+e x)-a \sin (d+e x)}{4 e \left (a^3+a^3 \cos (d+e x)+a^3 \sin (d+e x)\right )}-\frac {\text {Subst}\left (\int \frac {1}{4 a+4 a x} \, dx,x,\tan \left (\frac {1}{2} (d+e x)\right )\right )}{a e} \\ & = -\frac {\log \left (1+\tan \left (\frac {1}{2} (d+e x)\right )\right )}{4 a^2 e}-\frac {a \cos (d+e x)-a \sin (d+e x)}{4 e \left (a^3+a^3 \cos (d+e x)+a^3 \sin (d+e x)\right )} \\ \end{align*}
Time = 0.25 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.24 \[ \int \frac {1}{(2 a+2 a \cos (d+e x)+2 a \sin (d+e x))^2} \, dx=\frac {2 \log \left (\cos \left (\frac {1}{2} (d+e x)\right )\right )-2 \log \left (\cos \left (\frac {1}{2} (d+e x)\right )+\sin \left (\frac {1}{2} (d+e x)\right )\right )+\frac {2 \sin \left (\frac {1}{2} (d+e x)\right )}{\cos \left (\frac {1}{2} (d+e x)\right )+\sin \left (\frac {1}{2} (d+e x)\right )}+\tan \left (\frac {1}{2} (d+e x)\right )}{8 a^2 e} \]
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Time = 0.92 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.64
method | result | size |
derivativedivides | \(\frac {\tan \left (\frac {e x}{2}+\frac {d}{2}\right )-2 \ln \left (1+\tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )-\frac {2}{1+\tan \left (\frac {e x}{2}+\frac {d}{2}\right )}}{8 e \,a^{2}}\) | \(48\) |
default | \(\frac {\tan \left (\frac {e x}{2}+\frac {d}{2}\right )-2 \ln \left (1+\tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )-\frac {2}{1+\tan \left (\frac {e x}{2}+\frac {d}{2}\right )}}{8 e \,a^{2}}\) | \(48\) |
norman | \(\frac {\frac {\tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2}}{8 a e}-\frac {3}{8 a e}}{a \left (1+\tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )}-\frac {\ln \left (1+\tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )}{4 a^{2} e}\) | \(67\) |
parallelrisch | \(\frac {\left (-2 \tan \left (\frac {e x}{2}+\frac {d}{2}\right )-2\right ) \ln \left (1+\tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )+\tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2}+3 \tan \left (\frac {e x}{2}+\frac {d}{2}\right )}{8 e \,a^{2} \left (1+\tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )}\) | \(71\) |
risch | \(\frac {\left (-\frac {1}{4}+\frac {i}{4}\right ) \left ({\mathrm e}^{i \left (e x +d \right )}+1+i\right )}{a^{2} e \left (i {\mathrm e}^{i \left (e x +d \right )}+{\mathrm e}^{2 i \left (e x +d \right )}+i+{\mathrm e}^{i \left (e x +d \right )}\right )}-\frac {\ln \left ({\mathrm e}^{i \left (e x +d \right )}+i\right )}{4 a^{2} e}+\frac {\ln \left ({\mathrm e}^{i \left (e x +d \right )}+1\right )}{4 a^{2} e}\) | \(101\) |
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Time = 0.25 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.33 \[ \int \frac {1}{(2 a+2 a \cos (d+e x)+2 a \sin (d+e x))^2} \, dx=\frac {{\left (\cos \left (e x + d\right ) + \sin \left (e x + d\right ) + 1\right )} \log \left (\frac {1}{2} \, \cos \left (e x + d\right ) + \frac {1}{2}\right ) - {\left (\cos \left (e x + d\right ) + \sin \left (e x + d\right ) + 1\right )} \log \left (\sin \left (e x + d\right ) + 1\right ) - 2 \, \cos \left (e x + d\right ) + 2 \, \sin \left (e x + d\right )}{8 \, {\left (a^{2} e \cos \left (e x + d\right ) + a^{2} e \sin \left (e x + d\right ) + a^{2} e\right )}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 168 vs. \(2 (65) = 130\).
Time = 0.83 (sec) , antiderivative size = 168, normalized size of antiderivative = 2.24 \[ \int \frac {1}{(2 a+2 a \cos (d+e x)+2 a \sin (d+e x))^2} \, dx=\begin {cases} - \frac {2 \log {\left (\tan {\left (\frac {d}{2} + \frac {e x}{2} \right )} + 1 \right )} \tan {\left (\frac {d}{2} + \frac {e x}{2} \right )}}{8 a^{2} e \tan {\left (\frac {d}{2} + \frac {e x}{2} \right )} + 8 a^{2} e} - \frac {2 \log {\left (\tan {\left (\frac {d}{2} + \frac {e x}{2} \right )} + 1 \right )}}{8 a^{2} e \tan {\left (\frac {d}{2} + \frac {e x}{2} \right )} + 8 a^{2} e} + \frac {\tan ^{2}{\left (\frac {d}{2} + \frac {e x}{2} \right )}}{8 a^{2} e \tan {\left (\frac {d}{2} + \frac {e x}{2} \right )} + 8 a^{2} e} - \frac {3}{8 a^{2} e \tan {\left (\frac {d}{2} + \frac {e x}{2} \right )} + 8 a^{2} e} & \text {for}\: e \neq 0 \\\frac {x}{\left (2 a \sin {\left (d \right )} + 2 a \cos {\left (d \right )} + 2 a\right )^{2}} & \text {otherwise} \end {cases} \]
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Time = 0.22 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.07 \[ \int \frac {1}{(2 a+2 a \cos (d+e x)+2 a \sin (d+e x))^2} \, dx=-\frac {\frac {2}{a^{2} + \frac {a^{2} \sin \left (e x + d\right )}{\cos \left (e x + d\right ) + 1}} + \frac {2 \, \log \left (\frac {\sin \left (e x + d\right )}{\cos \left (e x + d\right ) + 1} + 1\right )}{a^{2}} - \frac {\sin \left (e x + d\right )}{a^{2} {\left (\cos \left (e x + d\right ) + 1\right )}}}{8 \, e} \]
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Time = 0.30 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.87 \[ \int \frac {1}{(2 a+2 a \cos (d+e x)+2 a \sin (d+e x))^2} \, dx=-\frac {\frac {2 \, \log \left ({\left | \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right ) + 1 \right |}\right )}{a^{2}} - \frac {\tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right )}{a^{2}} - \frac {2 \, \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right )}{a^{2} {\left (\tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right ) + 1\right )}}}{8 \, e} \]
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Time = 27.88 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.79 \[ \int \frac {1}{(2 a+2 a \cos (d+e x)+2 a \sin (d+e x))^2} \, dx=\frac {\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}{8\,a^2\,e}-\frac {\ln \left (\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )+1\right )}{4\,a^2\,e}-\frac {1}{4\,a^2\,e\,\left (\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )+1\right )} \]
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