Optimal. Leaf size=287 \[ -\frac {\sqrt {c-i d} (i A+B-i C) \tanh ^{-1}\left (\frac {\sqrt {c-i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a-i b} \sqrt {c+d \tan (e+f x)}}\right )}{f \sqrt {a-i b}}-\frac {\sqrt {c+i d} (B-i (A-C)) \tanh ^{-1}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{f \sqrt {a+i b}}+\frac {(-a C d+2 b B d+b c C) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b \tan (e+f x)}}{\sqrt {b} \sqrt {c+d \tan (e+f x)}}\right )}{b^{3/2} \sqrt {d} f}+\frac {C \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{b f} \]
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Rubi [A] time = 2.63, antiderivative size = 287, normalized size of antiderivative = 1.00, 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} \[ -\frac {\sqrt {c-i d} (i A+B-i C) \tanh ^{-1}\left (\frac {\sqrt {c-i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a-i b} \sqrt {c+d \tan (e+f x)}}\right )}{f \sqrt {a-i b}}-\frac {\sqrt {c+i d} (B-i (A-C)) \tanh ^{-1}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{f \sqrt {a+i b}}+\frac {(-a C d+2 b B d+b c C) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b \tan (e+f x)}}{\sqrt {b} \sqrt {c+d \tan (e+f x)}}\right )}{b^{3/2} \sqrt {d} f}+\frac {C \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{b f} \]
Antiderivative was successfully verified.
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Rule 63
Rule 93
Rule 206
Rule 208
Rule 217
Rule 3647
Rule 3655
Rule 6725
Rubi steps
\begin {align*} \int \frac {\sqrt {c+d \tan (e+f x)} \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right )}{\sqrt {a+b \tan (e+f x)}} \, dx &=\frac {C \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{b f}+\frac {\int \frac {\frac {1}{2} (2 A b c-C (b c+a d))+b (B c+(A-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}{b}\\ &=\frac {C \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{b f}+\frac {\operatorname {Subst}\left (\int \frac {\frac {1}{2} (2 A b c-C (b c+a d))+b (B c+(A-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 )}{b f}\\ &=\frac {C \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{b 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 (A c-c C-B d)+b (B c+(A-C) d) x}{\sqrt {a+b x} \sqrt {c+d x} \left (1+x^2\right )}\right ) \, dx,x,\tan (e+f x)\right )}{b f}\\ &=\frac {C \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{b f}+\frac {\operatorname {Subst}\left (\int \frac {b (A c-c C-B d)+b (B c+(A-C) d) x}{\sqrt {a+b x} \sqrt {c+d x} \left (1+x^2\right )} \, dx,x,\tan (e+f x)\right )}{b 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 b f}\\ &=\frac {C \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{b f}+\frac {\operatorname {Subst}\left (\int \left (\frac {i b (A c-c C-B d)-b (B c+(A-C) d)}{2 (i-x) \sqrt {a+b x} \sqrt {c+d x}}+\frac {i b (A c-c C-B d)+b (B c+(A-C) d)}{2 (i+x) \sqrt {a+b x} \sqrt {c+d x}}\right ) \, dx,x,\tan (e+f x)\right )}{b 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^2 f}\\ &=\frac {C \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{b f}+\frac {((A-i B-C) (i c+d)) \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^2 f}+\frac {(i b (A c-c C-B d)-b (B c+(A-C) d)) \operatorname {Subst}\left (\int \frac {1}{(i-x) \sqrt {a+b x} \sqrt {c+d x}} \, dx,x,\tan (e+f x)\right )}{2 b 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 )}{b^{3/2} \sqrt {d} f}+\frac {C \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{b f}+\frac {((A-i B-C) (i c+d)) \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 b (A c-c C-B d)-b (B c+(A-C) d)) \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 )}{b f}\\ &=-\frac {(i A+B-i C) \sqrt {c-i d} \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 {a-i b} f}-\frac {(B-i (A-C)) \sqrt {c+i d} \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 {a+i b} 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 )}{b^{3/2} \sqrt {d} f}+\frac {C \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{b f}\\ \end {align*}
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Mathematica [A] time = 4.24, size = 441, normalized size = 1.54 \[ \frac {\frac {b \left (\sqrt {-b^2} (A c-B d-c C)+b d (A-C)+b B c\right ) \tanh ^{-1}\left (\frac {\sqrt {\frac {\sqrt {-b^2} d}{b}-c} \sqrt {a+b \tan (e+f x)}}{\sqrt {\sqrt {-b^2}-a} \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {\sqrt {-b^2}-a} \sqrt {\frac {\sqrt {-b^2} d}{b}-c}}+\frac {b \left (\sqrt {-b^2} (A c-B d-c C)-b (d (A-C)+B c)\right ) \tanh ^{-1}\left (\frac {\sqrt {\frac {\sqrt {-b^2} d}{b}+c} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+\sqrt {-b^2}} \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {a+\sqrt {-b^2}} \sqrt {\frac {\sqrt {-b^2} d}{b}+c}}+\frac {\sqrt {b} \sqrt {c-\frac {a d}{b}} (-a C d+2 b B d+b c C) \sqrt {\frac {b (c+d \tan (e+f x))}{b c-a d}} \sinh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b \tan (e+f x)}}{\sqrt {b} \sqrt {c-\frac {a d}{b}}}\right )}{\sqrt {d} \sqrt {c+d \tan (e+f x)}}+b C \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{b^2 f} \]
Antiderivative was successfully verified.
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fricas [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F(-1)] time = 180.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {c +d \tan \left (f x +e \right )}\, \left (A +B \tan \left (f x +e \right )+C \left (\tan ^{2}\left (f x +e \right )\right )\right )}{\sqrt {a +b \tan \left (f x +e \right )}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (C \tan \left (f x + e\right )^{2} + B \tan \left (f x + e\right ) + A\right )} \sqrt {d \tan \left (f x + e\right ) + c}}{\sqrt {b \tan \left (f x + e\right ) + a}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F(-1)] time = 0.00, size = -1, normalized size = -0.00 \[ \text {Hanged} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {c + d \tan {\left (e + f x \right )}} \left (A + B \tan {\left (e + f x \right )} + C \tan ^{2}{\left (e + f x \right )}\right )}{\sqrt {a + b \tan {\left (e + f x \right )}}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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