Optimal. Leaf size=62 \[ \frac{\sqrt{a+b \tan ^2(e+f x)}}{f}-\frac{\sqrt{a-b} \tanh ^{-1}\left (\frac{\sqrt{a+b \tan ^2(e+f x)}}{\sqrt{a-b}}\right )}{f} \]
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Rubi [A] time = 0.0710242, antiderivative size = 62, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {3670, 444, 50, 63, 208} \[ \frac{\sqrt{a+b \tan ^2(e+f x)}}{f}-\frac{\sqrt{a-b} \tanh ^{-1}\left (\frac{\sqrt{a+b \tan ^2(e+f x)}}{\sqrt{a-b}}\right )}{f} \]
Antiderivative was successfully verified.
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Rule 3670
Rule 444
Rule 50
Rule 63
Rule 208
Rubi steps
\begin{align*} \int \tan (e+f x) \sqrt{a+b \tan ^2(e+f x)} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x \sqrt{a+b x^2}}{1+x^2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{\operatorname{Subst}\left (\int \frac{\sqrt{a+b x}}{1+x} \, dx,x,\tan ^2(e+f x)\right )}{2 f}\\ &=\frac{\sqrt{a+b \tan ^2(e+f x)}}{f}+\frac{(a-b) \operatorname{Subst}\left (\int \frac{1}{(1+x) \sqrt{a+b x}} \, dx,x,\tan ^2(e+f x)\right )}{2 f}\\ &=\frac{\sqrt{a+b \tan ^2(e+f x)}}{f}+\frac{(a-b) \operatorname{Subst}\left (\int \frac{1}{1-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+b \tan ^2(e+f x)}\right )}{b f}\\ &=-\frac{\sqrt{a-b} \tanh ^{-1}\left (\frac{\sqrt{a+b \tan ^2(e+f x)}}{\sqrt{a-b}}\right )}{f}+\frac{\sqrt{a+b \tan ^2(e+f x)}}{f}\\ \end{align*}
Mathematica [A] time = 0.0504613, size = 59, normalized size = 0.95 \[ \frac{\sqrt{a+b \tan ^2(e+f x)}-\sqrt{a-b} \tanh ^{-1}\left (\frac{\sqrt{a+b \tan ^2(e+f x)}}{\sqrt{a-b}}\right )}{f} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.032, size = 91, normalized size = 1.5 \begin{align*}{\frac{1}{f}\sqrt{a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2}}}-{\frac{b}{f}\arctan \left ({\sqrt{a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2}}{\frac{1}{\sqrt{-a+b}}}} \right ){\frac{1}{\sqrt{-a+b}}}}+{\frac{a}{f}\arctan \left ({\sqrt{a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2}}{\frac{1}{\sqrt{-a+b}}}} \right ){\frac{1}{\sqrt{-a+b}}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{b \tan \left (f x + e\right )^{2} + a} \tan \left (f x + e\right )\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 2.30909, size = 518, normalized size = 8.35 \begin{align*} \left [\frac{\sqrt{a - b} \log \left (-\frac{b^{2} \tan \left (f x + e\right )^{4} + 2 \,{\left (4 \, a b - 3 \, b^{2}\right )} \tan \left (f x + e\right )^{2} - 4 \,{\left (b \tan \left (f x + e\right )^{2} + 2 \, a - b\right )} \sqrt{b \tan \left (f x + e\right )^{2} + a} \sqrt{a - b} + 8 \, a^{2} - 8 \, a b + b^{2}}{\tan \left (f x + e\right )^{4} + 2 \, \tan \left (f x + e\right )^{2} + 1}\right ) + 4 \, \sqrt{b \tan \left (f x + e\right )^{2} + a}}{4 \, f}, \frac{\sqrt{-a + b} \arctan \left (\frac{2 \, \sqrt{b \tan \left (f x + e\right )^{2} + a} \sqrt{-a + b}}{b \tan \left (f x + e\right )^{2} + 2 \, a - b}\right ) + 2 \, \sqrt{b \tan \left (f x + e\right )^{2} + a}}{2 \, f}\right ] \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{a + b \tan ^{2}{\left (e + f x \right )}} \tan{\left (e + f x \right )}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.15546, size = 81, normalized size = 1.31 \begin{align*} \frac{{\left (a - b\right )} \arctan \left (\frac{\sqrt{b \tan \left (f x + e\right )^{2} + a}}{\sqrt{-a + b}}\right )}{\sqrt{-a + b} f} + \frac{\sqrt{b \tan \left (f x + e\right )^{2} + a}}{f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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