Integrand size = 28, antiderivative size = 75 \[ \int \frac {\sec ^2(c+d x)}{(a \cos (c+d x)+b \sin (c+d x))^2} \, dx=\frac {\frac {1}{a}+\frac {a}{b^2}}{d (b+a \cot (c+d x))}-\frac {2 a \log (b+a \cot (c+d x))}{b^3 d}-\frac {2 a \log (\tan (c+d x))}{b^3 d}+\frac {\tan (c+d x)}{b^2 d} \]
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Time = 0.12 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {3167, 908} \[ \int \frac {\sec ^2(c+d x)}{(a \cos (c+d x)+b \sin (c+d x))^2} \, dx=-\frac {2 a \log (\tan (c+d x))}{b^3 d}-\frac {2 a \log (a \cot (c+d x)+b)}{b^3 d}+\frac {\frac {a}{b^2}+\frac {1}{a}}{d (a \cot (c+d x)+b)}+\frac {\tan (c+d x)}{b^2 d} \]
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Rule 908
Rule 3167
Rubi steps \begin{align*} \text {integral}& = -\frac {\text {Subst}\left (\int \frac {1+x^2}{x^2 (b+a x)^2} \, dx,x,\cot (c+d x)\right )}{d} \\ & = -\frac {\text {Subst}\left (\int \left (\frac {1}{b^2 x^2}-\frac {2 a}{b^3 x}+\frac {a^2+b^2}{b^2 (b+a x)^2}+\frac {2 a^2}{b^3 (b+a x)}\right ) \, dx,x,\cot (c+d x)\right )}{d} \\ & = \frac {\frac {1}{a}+\frac {a}{b^2}}{d (b+a \cot (c+d x))}-\frac {2 a \log (b+a \cot (c+d x))}{b^3 d}-\frac {2 a \log (\tan (c+d x))}{b^3 d}+\frac {\tan (c+d x)}{b^2 d} \\ \end{align*}
Time = 0.27 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.68 \[ \int \frac {\sec ^2(c+d x)}{(a \cos (c+d x)+b \sin (c+d x))^2} \, dx=\frac {-2 a \log (a+b \tan (c+d x))+b \tan (c+d x)-\frac {a^2+b^2}{a+b \tan (c+d x)}}{b^3 d} \]
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Time = 1.08 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.76
method | result | size |
derivativedivides | \(\frac {\frac {\tan \left (d x +c \right )}{b^{2}}-\frac {a^{2}+b^{2}}{b^{3} \left (a +b \tan \left (d x +c \right )\right )}-\frac {2 a \ln \left (a +b \tan \left (d x +c \right )\right )}{b^{3}}}{d}\) | \(57\) |
default | \(\frac {\frac {\tan \left (d x +c \right )}{b^{2}}-\frac {a^{2}+b^{2}}{b^{3} \left (a +b \tan \left (d x +c \right )\right )}-\frac {2 a \ln \left (a +b \tan \left (d x +c \right )\right )}{b^{3}}}{d}\) | \(57\) |
risch | \(-\frac {4 i \left (-i a \,{\mathrm e}^{2 i \left (d x +c \right )}+b -i a \right )}{\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) \left (b \,{\mathrm e}^{2 i \left (d x +c \right )}+i a \,{\mathrm e}^{2 i \left (d x +c \right )}-b +i a \right ) b^{2} d}+\frac {2 a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{b^{3} d}-\frac {2 a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {i b +a}{i b -a}\right )}{b^{3} d}\) | \(136\) |
parallelrisch | \(\frac {-2 a \left (a \cos \left (2 d x +2 c \right )+a +b \sin \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-a \right )+2 a \left (a \cos \left (2 d x +2 c \right )+a +b \sin \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+2 a \left (a \cos \left (2 d x +2 c \right )+a +b \sin \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\left (-2 a^{2}-2 b^{2}\right ) \cos \left (2 d x +2 c \right )-2 a^{2}}{b^{3} d \left (a \cos \left (2 d x +2 c \right )+a +b \sin \left (2 d x +2 c \right )\right )}\) | \(196\) |
norman | \(\frac {\frac {\left (4 a^{2}+6 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{b^{3} d}-\frac {4 a^{2}+2 b^{2}}{2 b^{3} d}-\frac {\left (4 a^{2}+2 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{2 b^{3} d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-a \right )}+\frac {2 a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{b^{3} d}+\frac {2 a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{b^{3} d}-\frac {2 a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-a \right )}{b^{3} d}\) | \(209\) |
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Leaf count of result is larger than twice the leaf count of optimal. 178 vs. \(2 (75) = 150\).
Time = 0.27 (sec) , antiderivative size = 178, normalized size of antiderivative = 2.37 \[ \int \frac {\sec ^2(c+d x)}{(a \cos (c+d x)+b \sin (c+d x))^2} \, dx=-\frac {2 \, b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) - b^{2} + {\left (a^{2} \cos \left (d x + c\right )^{2} + a b \cos \left (d x + c\right ) \sin \left (d x + c\right )\right )} \log \left (2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + b^{2}\right ) - {\left (a^{2} \cos \left (d x + c\right )^{2} + a b \cos \left (d x + c\right ) \sin \left (d x + c\right )\right )} \log \left (\cos \left (d x + c\right )^{2}\right )}{a b^{3} d \cos \left (d x + c\right )^{2} + b^{4} d \cos \left (d x + c\right ) \sin \left (d x + c\right )} \]
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\[ \int \frac {\sec ^2(c+d x)}{(a \cos (c+d x)+b \sin (c+d x))^2} \, dx=\int \frac {\sec ^{2}{\left (c + d x \right )}}{\left (a \cos {\left (c + d x \right )} + b \sin {\left (c + d x \right )}\right )^{2}}\, dx \]
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Time = 0.22 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.80 \[ \int \frac {\sec ^2(c+d x)}{(a \cos (c+d x)+b \sin (c+d x))^2} \, dx=-\frac {\frac {a^{2} + b^{2}}{b^{4} \tan \left (d x + c\right ) + a b^{3}} + \frac {2 \, a \log \left (b \tan \left (d x + c\right ) + a\right )}{b^{3}} - \frac {\tan \left (d x + c\right )}{b^{2}}}{d} \]
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Time = 0.30 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.95 \[ \int \frac {\sec ^2(c+d x)}{(a \cos (c+d x)+b \sin (c+d x))^2} \, dx=-\frac {\frac {2 \, a \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{b^{3}} - \frac {\tan \left (d x + c\right )}{b^{2}} - \frac {2 \, a b \tan \left (d x + c\right ) + a^{2} - b^{2}}{{\left (b \tan \left (d x + c\right ) + a\right )} b^{3}}}{d} \]
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Time = 24.40 (sec) , antiderivative size = 382, normalized size of antiderivative = 5.09 \[ \int \frac {\sec ^2(c+d x)}{(a \cos (c+d x)+b \sin (c+d x))^2} \, dx=\frac {\frac {4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{b}-\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (2\,a^2+b^2\right )}{a\,b^2}+\frac {2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,a^2+b^2\right )}{a\,b^2}}{d\,\left (a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-2\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-2\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a\right )}-\frac {4\,a\,\mathrm {atanh}\left (\frac {64\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{64\,a^3-64\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\frac {128\,a^5}{b^2}-\frac {128\,a^5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{b^2}+\frac {128\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{b}}-\frac {64\,a^3}{64\,a^3-64\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\frac {128\,a^5}{b^2}-\frac {128\,a^5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{b^2}+\frac {128\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{b}}+\frac {128\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{64\,a^3\,b+\frac {128\,a^5}{b}+128\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-\frac {128\,a^5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{b}-64\,a^3\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}\right )}{b^3\,d} \]
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