Integrand size = 25, antiderivative size = 86 \[ \int \frac {\tan ^2(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=-\frac {\arctan \left (\frac {\sqrt {a-b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{\sqrt {a-b} f}+\frac {\text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{\sqrt {b} f} \]
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Time = 0.14 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {3751, 494, 223, 212, 385, 209} \[ \int \frac {\tan ^2(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{\sqrt {b} f}-\frac {\arctan \left (\frac {\sqrt {a-b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{f \sqrt {a-b}} \]
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Rule 209
Rule 212
Rule 223
Rule 385
Rule 494
Rule 3751
Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {x^2}{\left (1+x^2\right ) \sqrt {a+b x^2}} \, dx,x,\tan (e+f x)\right )}{f} \\ & = \frac {\text {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,\tan (e+f x)\right )}{f}-\frac {\text {Subst}\left (\int \frac {1}{\left (1+x^2\right ) \sqrt {a+b x^2}} \, dx,x,\tan (e+f x)\right )}{f} \\ & = \frac {\text {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {\tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{f}-\frac {\text {Subst}\left (\int \frac {1}{1-(-a+b) x^2} \, dx,x,\frac {\tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{f} \\ & = -\frac {\arctan \left (\frac {\sqrt {a-b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{\sqrt {a-b} f}+\frac {\text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{\sqrt {b} f} \\ \end{align*}
Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.
Time = 0.89 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.73 \[ \int \frac {\tan ^2(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=\frac {a \csc ^2(e+f x) \operatorname {EllipticPi}\left (-\frac {b}{a-b},\arcsin \left (\frac {\sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}}}{\sqrt {2}}\right ),1\right ) \sqrt {(a+b+(a-b) \cos (2 (e+f x))) \sec ^2(e+f x)} \sin (2 (e+f x))}{2 (a-b) b f \sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}}} \]
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Time = 0.09 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.16
method | result | size |
derivativedivides | \(\frac {\frac {\ln \left (\sqrt {b}\, \tan \left (f x +e \right )+\sqrt {a +b \tan \left (f x +e \right )^{2}}\right )}{\sqrt {b}}-\frac {\sqrt {b^{4} \left (a -b \right )}\, \arctan \left (\frac {b^{2} \left (a -b \right ) \tan \left (f x +e \right )}{\sqrt {b^{4} \left (a -b \right )}\, \sqrt {a +b \tan \left (f x +e \right )^{2}}}\right )}{b^{2} \left (a -b \right )}}{f}\) | \(100\) |
default | \(\frac {\frac {\ln \left (\sqrt {b}\, \tan \left (f x +e \right )+\sqrt {a +b \tan \left (f x +e \right )^{2}}\right )}{\sqrt {b}}-\frac {\sqrt {b^{4} \left (a -b \right )}\, \arctan \left (\frac {b^{2} \left (a -b \right ) \tan \left (f x +e \right )}{\sqrt {b^{4} \left (a -b \right )}\, \sqrt {a +b \tan \left (f x +e \right )^{2}}}\right )}{b^{2} \left (a -b \right )}}{f}\) | \(100\) |
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Time = 0.35 (sec) , antiderivative size = 479, normalized size of antiderivative = 5.57 \[ \int \frac {\tan ^2(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=\left [\frac {{\left (a - b\right )} \sqrt {b} \log \left (2 \, b \tan \left (f x + e\right )^{2} + 2 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {b} \tan \left (f x + e\right ) + a\right ) - \sqrt {-a + b} b \log \left (-\frac {{\left (a - 2 \, b\right )} \tan \left (f x + e\right )^{2} + 2 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {-a + b} \tan \left (f x + e\right ) - a}{\tan \left (f x + e\right )^{2} + 1}\right )}{2 \, {\left (a b - b^{2}\right )} f}, -\frac {2 \, {\left (a - b\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {-b}}{b \tan \left (f x + e\right )}\right ) + \sqrt {-a + b} b \log \left (-\frac {{\left (a - 2 \, b\right )} \tan \left (f x + e\right )^{2} + 2 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {-a + b} \tan \left (f x + e\right ) - a}{\tan \left (f x + e\right )^{2} + 1}\right )}{2 \, {\left (a b - b^{2}\right )} f}, -\frac {2 \, \sqrt {a - b} b \arctan \left (-\frac {\sqrt {b \tan \left (f x + e\right )^{2} + a}}{\sqrt {a - b} \tan \left (f x + e\right )}\right ) - {\left (a - b\right )} \sqrt {b} \log \left (2 \, b \tan \left (f x + e\right )^{2} + 2 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {b} \tan \left (f x + e\right ) + a\right )}{2 \, {\left (a b - b^{2}\right )} f}, -\frac {\sqrt {a - b} b \arctan \left (-\frac {\sqrt {b \tan \left (f x + e\right )^{2} + a}}{\sqrt {a - b} \tan \left (f x + e\right )}\right ) + {\left (a - b\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {-b}}{b \tan \left (f x + e\right )}\right )}{{\left (a b - b^{2}\right )} f}\right ] \]
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\[ \int \frac {\tan ^2(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=\int \frac {\tan ^{2}{\left (e + f x \right )}}{\sqrt {a + b \tan ^{2}{\left (e + f x \right )}}}\, dx \]
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\[ \int \frac {\tan ^2(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=\int { \frac {\tan \left (f x + e\right )^{2}}{\sqrt {b \tan \left (f x + e\right )^{2} + a}} \,d x } \]
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Timed out. \[ \int \frac {\tan ^2(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=\text {Timed out} \]
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Timed out. \[ \int \frac {\tan ^2(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=\int \frac {{\mathrm {tan}\left (e+f\,x\right )}^2}{\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a}} \,d x \]
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