Integrand size = 14, antiderivative size = 45 \[ \int \frac {1}{a+b \sec ^2(e+f x)} \, dx=\frac {x}{a}+\frac {\sqrt {b} \arctan \left (\frac {\sqrt {a+b} \cot (e+f x)}{\sqrt {b}}\right )}{a \sqrt {a+b} f} \]
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Time = 0.05 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {4212, 3260, 211} \[ \int \frac {1}{a+b \sec ^2(e+f x)} \, dx=\frac {\sqrt {b} \arctan \left (\frac {\sqrt {a+b} \cot (e+f x)}{\sqrt {b}}\right )}{a f \sqrt {a+b}}+\frac {x}{a} \]
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Rule 211
Rule 3260
Rule 4212
Rubi steps \begin{align*} \text {integral}& = \frac {x}{a}-\frac {b \int \frac {1}{b+a \cos ^2(e+f x)} \, dx}{a} \\ & = \frac {x}{a}+\frac {b \text {Subst}\left (\int \frac {1}{b+(a+b) x^2} \, dx,x,\cot (e+f x)\right )}{a f} \\ & = \frac {x}{a}+\frac {\sqrt {b} \arctan \left (\frac {\sqrt {a+b} \cot (e+f x)}{\sqrt {b}}\right )}{a \sqrt {a+b} f} \\ \end{align*}
Result contains complex when optimal does not.
Time = 1.41 (sec) , antiderivative size = 182, normalized size of antiderivative = 4.04 \[ \int \frac {1}{a+b \sec ^2(e+f x)} \, dx=\frac {(a+2 b+a \cos (2 (e+f x))) \sec ^2(e+f x) \left (\sqrt {a+b} f x \sqrt {b (\cos (e)-i \sin (e))^4}+b \arctan \left (\frac {\sec (f x) (\cos (2 e)-i \sin (2 e)) (-((a+2 b) \sin (f x))+a \sin (2 e+f x))}{2 \sqrt {a+b} \sqrt {b (\cos (e)-i \sin (e))^4}}\right ) (\cos (2 e)-i \sin (2 e))\right )}{2 a \sqrt {a+b} f \left (a+b \sec ^2(e+f x)\right ) \sqrt {b (\cos (e)-i \sin (e))^4}} \]
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Time = 0.17 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.02
method | result | size |
derivativedivides | \(\frac {-\frac {b \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {\left (a +b \right ) b}}\right )}{a \sqrt {\left (a +b \right ) b}}+\frac {\arctan \left (\tan \left (f x +e \right )\right )}{a}}{f}\) | \(46\) |
default | \(\frac {-\frac {b \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {\left (a +b \right ) b}}\right )}{a \sqrt {\left (a +b \right ) b}}+\frac {\arctan \left (\tan \left (f x +e \right )\right )}{a}}{f}\) | \(46\) |
risch | \(\frac {x}{a}+\frac {\sqrt {-\left (a +b \right ) b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {-\left (a +b \right ) b}+a +2 b}{a}\right )}{2 \left (a +b \right ) f a}-\frac {\sqrt {-\left (a +b \right ) b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {-\left (a +b \right ) b}-a -2 b}{a}\right )}{2 \left (a +b \right ) f a}\) | \(114\) |
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Time = 0.28 (sec) , antiderivative size = 231, normalized size of antiderivative = 5.13 \[ \int \frac {1}{a+b \sec ^2(e+f x)} \, dx=\left [\frac {4 \, f x + \sqrt {-\frac {b}{a + b}} \log \left (\frac {{\left (a^{2} + 8 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (3 \, a b + 4 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 4 \, {\left ({\left (a^{2} + 3 \, a b + 2 \, b^{2}\right )} \cos \left (f x + e\right )^{3} - {\left (a b + b^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {-\frac {b}{a + b}} \sin \left (f x + e\right ) + b^{2}}{a^{2} \cos \left (f x + e\right )^{4} + 2 \, a b \cos \left (f x + e\right )^{2} + b^{2}}\right )}{4 \, a f}, \frac {2 \, f x + \sqrt {\frac {b}{a + b}} \arctan \left (\frac {{\left ({\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{2} - b\right )} \sqrt {\frac {b}{a + b}}}{2 \, b \cos \left (f x + e\right ) \sin \left (f x + e\right )}\right )}{2 \, a f}\right ] \]
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\[ \int \frac {1}{a+b \sec ^2(e+f x)} \, dx=\int \frac {1}{a + b \sec ^{2}{\left (e + f x \right )}}\, dx \]
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Time = 0.26 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.98 \[ \int \frac {1}{a+b \sec ^2(e+f x)} \, dx=-\frac {\frac {b \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{\sqrt {{\left (a + b\right )} b} a} - \frac {f x + e}{a}}{f} \]
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Time = 0.28 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.44 \[ \int \frac {1}{a+b \sec ^2(e+f x)} \, dx=-\frac {\frac {{\left (\pi \left \lfloor \frac {f x + e}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (b\right ) + \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b + b^{2}}}\right )\right )} b}{\sqrt {a b + b^{2}} a} - \frac {f x + e}{a}}{f} \]
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Time = 18.48 (sec) , antiderivative size = 460, normalized size of antiderivative = 10.22 \[ \int \frac {1}{a+b \sec ^2(e+f x)} \, dx=\frac {x}{a}-\frac {\mathrm {atan}\left (\frac {\frac {\left (2\,b^3\,\mathrm {tan}\left (e+f\,x\right )-\frac {\left (2\,a^2\,b^2-\frac {\mathrm {tan}\left (e+f\,x\right )\,\left (8\,a^3\,b^2+16\,a^2\,b^3\right )\,\sqrt {-b\,\left (a+b\right )}}{4\,\left (a^2+b\,a\right )}\right )\,\sqrt {-b\,\left (a+b\right )}}{2\,\left (a^2+b\,a\right )}\right )\,\sqrt {-b\,\left (a+b\right )}\,1{}\mathrm {i}}{a^2+b\,a}+\frac {\left (2\,b^3\,\mathrm {tan}\left (e+f\,x\right )+\frac {\left (2\,a^2\,b^2+\frac {\mathrm {tan}\left (e+f\,x\right )\,\left (8\,a^3\,b^2+16\,a^2\,b^3\right )\,\sqrt {-b\,\left (a+b\right )}}{4\,\left (a^2+b\,a\right )}\right )\,\sqrt {-b\,\left (a+b\right )}}{2\,\left (a^2+b\,a\right )}\right )\,\sqrt {-b\,\left (a+b\right )}\,1{}\mathrm {i}}{a^2+b\,a}}{\frac {\left (2\,b^3\,\mathrm {tan}\left (e+f\,x\right )-\frac {\left (2\,a^2\,b^2-\frac {\mathrm {tan}\left (e+f\,x\right )\,\left (8\,a^3\,b^2+16\,a^2\,b^3\right )\,\sqrt {-b\,\left (a+b\right )}}{4\,\left (a^2+b\,a\right )}\right )\,\sqrt {-b\,\left (a+b\right )}}{2\,\left (a^2+b\,a\right )}\right )\,\sqrt {-b\,\left (a+b\right )}}{a^2+b\,a}-\frac {\left (2\,b^3\,\mathrm {tan}\left (e+f\,x\right )+\frac {\left (2\,a^2\,b^2+\frac {\mathrm {tan}\left (e+f\,x\right )\,\left (8\,a^3\,b^2+16\,a^2\,b^3\right )\,\sqrt {-b\,\left (a+b\right )}}{4\,\left (a^2+b\,a\right )}\right )\,\sqrt {-b\,\left (a+b\right )}}{2\,\left (a^2+b\,a\right )}\right )\,\sqrt {-b\,\left (a+b\right )}}{a^2+b\,a}}\right )\,\sqrt {-b\,\left (a+b\right )}\,1{}\mathrm {i}}{f\,\left (a^2+b\,a\right )} \]
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