Optimal. Leaf size=85 \[ \frac{\sqrt{a-b} \tan ^{-1}\left (\frac{\sqrt{a-b} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)}}\right )}{f}+\frac{\sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)}}\right )}{f} \]
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Rubi [A] time = 0.0543786, antiderivative size = 85, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.375, Rules used = {3661, 402, 217, 206, 377, 203} \[ \frac{\sqrt{a-b} \tan ^{-1}\left (\frac{\sqrt{a-b} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)}}\right )}{f}+\frac{\sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)}}\right )}{f} \]
Antiderivative was successfully verified.
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Rule 3661
Rule 402
Rule 217
Rule 206
Rule 377
Rule 203
Rubi steps
\begin{align*} \int \sqrt{a+b \tan ^2(e+f x)} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\sqrt{a+b x^2}}{1+x^2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{(a-b) \operatorname{Subst}\left (\int \frac{1}{\left (1+x^2\right ) \sqrt{a+b x^2}} \, dx,x,\tan (e+f x)\right )}{f}+\frac{b \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x^2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{(a-b) \operatorname{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{b \operatorname{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{\sqrt{a-b} \tan ^{-1}\left (\frac{\sqrt{a-b} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)}}\right )}{f}+\frac{\sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)}}\right )}{f}\\ \end{align*}
Mathematica [C] time = 0.855987, size = 203, normalized size = 2.39 \[ \frac{-i \sqrt{a-b} \log \left (-\frac{4 i \left (\sqrt{a-b} \sqrt{a+b \tan ^2(e+f x)}+a-i b \tan (e+f x)\right )}{(a-b)^{3/2} (\tan (e+f x)+i)}\right )+i \sqrt{a-b} \log \left (\frac{4 i \left (\sqrt{a-b} \sqrt{a+b \tan ^2(e+f x)}+a+i b \tan (e+f x)\right )}{(a-b)^{3/2} (\tan (e+f x)-i)}\right )+2 \sqrt{b} \log \left (\sqrt{b} \sqrt{a+b \tan ^2(e+f x)}+b \tan (e+f x)\right )}{2 f} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.04, size = 169, normalized size = 2. \begin{align*}{\frac{1}{f}\sqrt{b}\ln \left ( \sqrt{b}\tan \left ( fx+e \right ) +\sqrt{a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2}} \right ) }-{\frac{1}{fb \left ( a-b \right ) }\sqrt{{b}^{4} \left ( a-b \right ) }\arctan \left ({ \left ( a-b \right ){b}^{2}\tan \left ( fx+e \right ){\frac{1}{\sqrt{{b}^{4} \left ( a-b \right ) }}}{\frac{1}{\sqrt{a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2}}}}} \right ) }+{\frac{a}{f{b}^{2} \left ( a-b \right ) }\sqrt{{b}^{4} \left ( a-b \right ) }\arctan \left ({ \left ( a-b \right ){b}^{2}\tan \left ( fx+e \right ){\frac{1}{\sqrt{{b}^{4} \left ( a-b \right ) }}}{\frac{1}{\sqrt{a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2}}}}} \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 2.12744, size = 1046, normalized size = 12.31 \begin{align*} \left [\frac{\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} \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 \, f}, \frac{2 \, \sqrt{a - b} \arctan \left (-\frac{\sqrt{b \tan \left (f x + e\right )^{2} + a}}{\sqrt{a - b} \tan \left (f x + e\right )}\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 \, f}, -\frac{2 \, \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} \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 \, f}, \frac{\sqrt{a - b} \arctan \left (-\frac{\sqrt{b \tan \left (f x + e\right )^{2} + a}}{\sqrt{a - b} \tan \left (f x + e\right )}\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 )}{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 )}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{b \tan \left (f x + e\right )^{2} + a}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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