Optimal. Leaf size=290 \[ -\frac{\sqrt{a-i b} (i A+B-i C) \tanh ^{-1}\left (\frac{\sqrt{c-i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{a-i b} \sqrt{c+d \tan (e+f x)}}\right )}{f \sqrt{c-i d}}+\frac{\sqrt{a+i b} (i A-B-i C) \tanh ^{-1}\left (\frac{\sqrt{c+i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{a+i b} \sqrt{c+d \tan (e+f x)}}\right )}{f \sqrt{c+i d}}-\frac{(-a C d-2 b B d+b c C) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b \tan (e+f x)}}{\sqrt{b} \sqrt{c+d \tan (e+f x)}}\right )}{\sqrt{b} d^{3/2} f}+\frac{C \sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{d f} \]
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Rubi [A] time = 2.55491, antiderivative size = 290, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 8, integrand size = 49, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.163, Rules used = {3647, 3655, 6725, 63, 217, 206, 93, 208} \[ -\frac{\sqrt{a-i b} (i A+B-i C) \tanh ^{-1}\left (\frac{\sqrt{c-i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{a-i b} \sqrt{c+d \tan (e+f x)}}\right )}{f \sqrt{c-i d}}+\frac{\sqrt{a+i b} (i A-B-i C) \tanh ^{-1}\left (\frac{\sqrt{c+i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{a+i b} \sqrt{c+d \tan (e+f x)}}\right )}{f \sqrt{c+i d}}-\frac{(-a C d-2 b B d+b c C) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b \tan (e+f x)}}{\sqrt{b} \sqrt{c+d \tan (e+f x)}}\right )}{\sqrt{b} d^{3/2} f}+\frac{C \sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{d f} \]
Antiderivative was successfully verified.
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Rule 3647
Rule 3655
Rule 6725
Rule 63
Rule 217
Rule 206
Rule 93
Rule 208
Rubi steps
\begin{align*} \int \frac{\sqrt{a+b \tan (e+f x)} \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right )}{\sqrt{c+d \tan (e+f x)}} \, dx &=\frac{C \sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{d f}+\frac{\int \frac{\frac{1}{2} (-b c C+2 a A d-a C d)+(A b+a B-b C) d \tan (e+f x)-\frac{1}{2} (b c C-2 b B d-a C d) \tan ^2(e+f x)}{\sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}} \, dx}{d}\\ &=\frac{C \sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{d f}+\frac{\operatorname{Subst}\left (\int \frac{\frac{1}{2} (-b c C+2 a A d-a C d)+(A b+a B-b C) d x+\frac{1}{2} (-b c C+2 b B d+a C d) x^2}{\sqrt{a+b x} \sqrt{c+d x} \left (1+x^2\right )} \, dx,x,\tan (e+f x)\right )}{d f}\\ &=\frac{C \sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{d f}+\frac{\operatorname{Subst}\left (\int \left (\frac{-b c C+2 b B d+a C d}{2 \sqrt{a+b x} \sqrt{c+d x}}+\frac{-(b B-a (A-C)) d+(A b+a B-b C) d x}{\sqrt{a+b x} \sqrt{c+d x} \left (1+x^2\right )}\right ) \, dx,x,\tan (e+f x)\right )}{d f}\\ &=\frac{C \sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{d f}+\frac{\operatorname{Subst}\left (\int \frac{-(b B-a (A-C)) d+(A b+a B-b C) d x}{\sqrt{a+b x} \sqrt{c+d x} \left (1+x^2\right )} \, dx,x,\tan (e+f x)\right )}{d f}-\frac{(b c C-2 b B d-a C d) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x} \sqrt{c+d x}} \, dx,x,\tan (e+f x)\right )}{2 d f}\\ &=\frac{C \sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{d f}+\frac{\operatorname{Subst}\left (\int \left (\frac{-i (b B-a (A-C)) d-(A b+a B-b C) d}{2 (i-x) \sqrt{a+b x} \sqrt{c+d x}}+\frac{-i (b B-a (A-C)) d+(A b+a B-b C) d}{2 (i+x) \sqrt{a+b x} \sqrt{c+d x}}\right ) \, dx,x,\tan (e+f x)\right )}{d f}-\frac{(b c C-2 b B d-a C d) \operatorname{Subst}\left (\int \frac{1}{\sqrt{c-\frac{a d}{b}+\frac{d x^2}{b}}} \, dx,x,\sqrt{a+b \tan (e+f x)}\right )}{b d f}\\ &=\frac{C \sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{d f}+\frac{((i a+b) (A-i B-C)) \operatorname{Subst}\left (\int \frac{1}{(i+x) \sqrt{a+b x} \sqrt{c+d x}} \, dx,x,\tan (e+f x)\right )}{2 f}+\frac{((i a-b) (A+i B-C)) \operatorname{Subst}\left (\int \frac{1}{(i-x) \sqrt{a+b x} \sqrt{c+d x}} \, dx,x,\tan (e+f x)\right )}{2 f}-\frac{(b c C-2 b B d-a C d) \operatorname{Subst}\left (\int \frac{1}{1-\frac{d x^2}{b}} \, dx,x,\frac{\sqrt{a+b \tan (e+f x)}}{\sqrt{c+d \tan (e+f x)}}\right )}{b d f}\\ &=-\frac{(b c C-2 b B d-a C d) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b \tan (e+f x)}}{\sqrt{b} \sqrt{c+d \tan (e+f x)}}\right )}{\sqrt{b} d^{3/2} f}+\frac{C \sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{d f}+\frac{((i a+b) (A-i B-C)) \operatorname{Subst}\left (\int \frac{1}{-a+i b-(-c+i d) x^2} \, dx,x,\frac{\sqrt{a+b \tan (e+f x)}}{\sqrt{c+d \tan (e+f x)}}\right )}{f}+\frac{((i a-b) (A+i B-C)) \operatorname{Subst}\left (\int \frac{1}{a+i b-(c+i d) x^2} \, dx,x,\frac{\sqrt{a+b \tan (e+f x)}}{\sqrt{c+d \tan (e+f x)}}\right )}{f}\\ &=-\frac{\sqrt{a-i b} (i A+B-i C) \tanh ^{-1}\left (\frac{\sqrt{c-i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{a-i b} \sqrt{c+d \tan (e+f x)}}\right )}{\sqrt{c-i d} f}-\frac{\sqrt{a+i b} (B-i (A-C)) \tanh ^{-1}\left (\frac{\sqrt{c+i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{a+i b} \sqrt{c+d \tan (e+f x)}}\right )}{\sqrt{c+i d} f}-\frac{(b c C-2 b B d-a C d) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b \tan (e+f x)}}{\sqrt{b} \sqrt{c+d \tan (e+f x)}}\right )}{\sqrt{b} d^{3/2} f}+\frac{C \sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{d f}\\ \end{align*}
Mathematica [A] time = 6.85748, size = 456, normalized size = 1.57 \[ \frac{-\frac{d \left (\sqrt{-b^2} (b B-a (A-C))-b (a B+A b-b C)\right ) \tan ^{-1}\left (\frac{\sqrt{\frac{b d}{\sqrt{-b^2}}+c} \sqrt{a+b \tan (e+f x)}}{\sqrt{\sqrt{-b^2}-a} \sqrt{c+d \tan (e+f x)}}\right )}{\sqrt{\sqrt{-b^2}-a} \sqrt{\frac{b d}{\sqrt{-b^2}}+c}}-\frac{d \left (\sqrt{-b^2} (b B-a (A-C))+b (a B+A b-b C)\right ) \tan ^{-1}\left (\frac{\sqrt{-\frac{\sqrt{-b^2} d+b c}{b}} \sqrt{a+b \tan (e+f x)}}{\sqrt{a+\sqrt{-b^2}} \sqrt{c+d \tan (e+f x)}}\right )}{\sqrt{a+\sqrt{-b^2}} \sqrt{-\frac{\sqrt{-b^2} d+b c}{b}}}-\frac{\sqrt{b} \sqrt{c-\frac{a d}{b}} (-a C d-2 b B d+b c C) \sqrt{\frac{b c+b d \tan (e+f x)}{b c-a d}} \sinh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b \tan (e+f x)}}{\sqrt{b} \sqrt{c-\frac{a d}{b}}}\right )}{\sqrt{d} \sqrt{c+d \tan (e+f x)}}}{b d f}+\frac{C \sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{d f} \]
Antiderivative was successfully verified.
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Maple [F] time = 180., size = 0, normalized size = 0. \begin{align*} \int{(A+B\tan \left ( fx+e \right ) +C \left ( \tan \left ( fx+e \right ) \right ) ^{2})\sqrt{a+b\tan \left ( fx+e \right ) }{\frac{1}{\sqrt{c+d\tan \left ( fx+e \right ) }}}}\, dx \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 \frac{{\left (C \tan \left (f x + e\right )^{2} + B \tan \left (f x + e\right ) + A\right )} \sqrt{b \tan \left (f x + e\right ) + a}}{\sqrt{d \tan \left (f x + e\right ) + c}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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 \frac{\sqrt{a + b \tan{\left (e + f x \right )}} \left (A + B \tan{\left (e + f x \right )} + C \tan ^{2}{\left (e + f x \right )}\right )}{\sqrt{c + d \tan{\left (e + f x \right )}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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