Integrand size = 23, antiderivative size = 46 \[ \int \frac {\tan ^2(e+f x)}{a+b \sec ^2(e+f x)} \, dx=-\frac {x}{a}+\frac {\sqrt {a+b} \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{a \sqrt {b} f} \]
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Time = 0.14 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {4226, 2000, 492, 209, 211} \[ \int \frac {\tan ^2(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\frac {\sqrt {a+b} \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{a \sqrt {b} f}-\frac {x}{a} \]
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Rule 209
Rule 211
Rule 492
Rule 2000
Rule 4226
Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {x^2}{\left (1+x^2\right ) \left (a+b \left (1+x^2\right )\right )} \, dx,x,\tan (e+f x)\right )}{f} \\ & = \frac {\text {Subst}\left (\int \frac {x^2}{\left (1+x^2\right ) \left (a+b+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{f} \\ & = -\frac {\text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\tan (e+f x)\right )}{a f}+\frac {(a+b) \text {Subst}\left (\int \frac {1}{a+b+b x^2} \, dx,x,\tan (e+f x)\right )}{a f} \\ & = -\frac {x}{a}+\frac {\sqrt {a+b} \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{a \sqrt {b} f} \\ \end{align*}
Result contains complex when optimal does not.
Time = 0.76 (sec) , antiderivative size = 184, normalized size of antiderivative = 4.00 \[ \int \frac {\tan ^2(e+f x)}{a+b \sec ^2(e+f x)} \, dx=-\frac {(a+2 b+a \cos (2 (e+f x))) \sec ^2(e+f x) \left (\sqrt {a+b} f x \sqrt {b (\cos (e)-i \sin (e))^4}+(a+b) \arctan \left (\frac {\sec (f x) (\cos (2 e)-i \sin (2 e)) (-((a+2 b) \sin (f x))+a \sin (2 e+f x))}{2 \sqrt {a+b} \sqrt {b (\cos (e)-i \sin (e))^4}}\right ) (\cos (2 e)-i \sin (2 e))\right )}{2 a \sqrt {a+b} f \left (a+b \sec ^2(e+f x)\right ) \sqrt {b (\cos (e)-i \sin (e))^4}} \]
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Time = 0.63 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.04
method | result | size |
derivativedivides | \(\frac {-\frac {\arctan \left (\tan \left (f x +e \right )\right )}{a}+\frac {\left (a +b \right ) \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {\left (a +b \right ) b}}\right )}{a \sqrt {\left (a +b \right ) b}}}{f}\) | \(48\) |
default | \(\frac {-\frac {\arctan \left (\tan \left (f x +e \right )\right )}{a}+\frac {\left (a +b \right ) \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {\left (a +b \right ) b}}\right )}{a \sqrt {\left (a +b \right ) b}}}{f}\) | \(48\) |
risch | \(-\frac {x}{a}-\frac {\sqrt {-\left (a +b \right ) b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {-\left (a +b \right ) b}+a +2 b}{a}\right )}{2 b f a}+\frac {\sqrt {-\left (a +b \right ) b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {-\left (a +b \right ) b}-a -2 b}{a}\right )}{2 b f a}\) | \(111\) |
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Time = 0.28 (sec) , antiderivative size = 226, normalized size of antiderivative = 4.91 \[ \int \frac {\tan ^2(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\left [-\frac {4 \, f x - \sqrt {-\frac {a + b}{b}} \log \left (\frac {{\left (a^{2} + 8 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (3 \, a b + 4 \, b^{2}\right )} \cos \left (f x + e\right )^{2} - 4 \, {\left ({\left (a b + 2 \, b^{2}\right )} \cos \left (f x + e\right )^{3} - b^{2} \cos \left (f x + e\right )\right )} \sqrt {-\frac {a + b}{b}} \sin \left (f x + e\right ) + b^{2}}{a^{2} \cos \left (f x + e\right )^{4} + 2 \, a b \cos \left (f x + e\right )^{2} + b^{2}}\right )}{4 \, a f}, -\frac {2 \, f x + \sqrt {\frac {a + b}{b}} \arctan \left (\frac {{\left ({\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{2} - b\right )} \sqrt {\frac {a + b}{b}}}{2 \, {\left (a + b\right )} \cos \left (f x + e\right ) \sin \left (f x + e\right )}\right )}{2 \, a f}\right ] \]
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\[ \int \frac {\tan ^2(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\int \frac {\tan ^{2}{\left (e + f x \right )}}{a + b \sec ^{2}{\left (e + f x \right )}}\, dx \]
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Time = 0.27 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.98 \[ \int \frac {\tan ^2(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\frac {\frac {{\left (a + b\right )} \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{\sqrt {{\left (a + b\right )} b} a} - \frac {f x + e}{a}}{f} \]
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Time = 0.45 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.43 \[ \int \frac {\tan ^2(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\frac {\frac {{\left (\pi \left \lfloor \frac {f x + e}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (b\right ) + \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b + b^{2}}}\right )\right )} {\left (a + b\right )}}{\sqrt {a b + b^{2}} a} - \frac {f x + e}{a}}{f} \]
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Time = 19.66 (sec) , antiderivative size = 126, normalized size of antiderivative = 2.74 \[ \int \frac {\tan ^2(e+f x)}{a+b \sec ^2(e+f x)} \, dx=-\frac {\mathrm {atan}\left (\frac {2\,a\,b^2\,\mathrm {tan}\left (e+f\,x\right )}{2\,a^2\,b+2\,a\,b^2}+\frac {2\,a^2\,b\,\mathrm {tan}\left (e+f\,x\right )}{2\,a^2\,b+2\,a\,b^2}\right )}{a\,f}-\frac {\mathrm {atanh}\left (\frac {2\,a\,b^2\,\mathrm {tan}\left (e+f\,x\right )\,\sqrt {-b^2-a\,b}}{2\,a^2\,b^2+2\,a\,b^3}\right )\,\sqrt {-b\,\left (a+b\right )}}{a\,b\,f} \]
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