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 = 2.55, antiderivative size = 290, 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} \[ -\frac {\sqrt {a-i b} (i A+B-i C) \tanh ^{-1}\left (\frac {\sqrt {c-i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a-i b} \sqrt {c+d \tan (e+f x)}}\right )}{f \sqrt {c-i d}}+\frac {\sqrt {a+i b} (i A-B-i C) \tanh ^{-1}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{f \sqrt {c+i d}}-\frac {(-a C d-2 b B d+b c C) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b \tan (e+f x)}}{\sqrt {b} \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {b} d^{3/2} f}+\frac {C \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{d f} \]
Antiderivative was successfully verified.
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Rule 63
Rule 93
Rule 206
Rule 208
Rule 217
Rule 3647
Rule 3655
Rule 6725
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*}
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Mathematica [A] time = 6.63, size = 450, normalized size = 1.55 \[ \frac {C \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{d f}-\frac {\frac {d \left (\sqrt {-b^2} (a (C-A)+b B)-b (a B+A b-b C)\right ) \tanh ^{-1}\left (\frac {\sqrt {\frac {\sqrt {-b^2} d}{b}-c} \sqrt {a+b \tan (e+f x)}}{\sqrt {\sqrt {-b^2}-a} \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {\sqrt {-b^2}-a} \sqrt {\frac {\sqrt {-b^2} d}{b}-c}}+\frac {d \left (\sqrt {-b^2} (a (C-A)+b B)+b (a B+A b-b C)\right ) \tanh ^{-1}\left (\frac {\sqrt {\frac {\sqrt {-b^2} d}{b}+c} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+\sqrt {-b^2}} \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {a+\sqrt {-b^2}} \sqrt {\frac {\sqrt {-b^2} d}{b}+c}}+\frac {\sqrt {b} \sqrt {c-\frac {a d}{b}} (-a C d-2 b B d+b c C) \sqrt {\frac {b (c+d \tan (e+f x))}{b c-a d}} \sinh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b \tan (e+f x)}}{\sqrt {b} \sqrt {c-\frac {a d}{b}}}\right )}{\sqrt {d} \sqrt {c+d \tan (e+f x)}}}{b d f} \]
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 \[ \int \frac {\sqrt {a +b \tan \left (f x +e \right )}\, \left (A +B \tan \left (f x +e \right )+C \left (\tan ^{2}\left (f x +e \right )\right )\right )}{\sqrt {c +d \tan \left (f x +e \right )}}\, dx \]
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 \[ \int \frac {{\left (C \tan \left (f x + e\right )^{2} + B \tan \left (f x + e\right ) + A\right )} \sqrt {b \tan \left (f x + e\right ) + a}}{\sqrt {d \tan \left (f x + e\right ) + c}}\,{d x} \]
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 \[ \int \frac {\sqrt {a + b \tan {\left (e + f x \right )}} \left (A + B \tan {\left (e + f x \right )} + C \tan ^{2}{\left (e + f x \right )}\right )}{\sqrt {c + d \tan {\left (e + f x \right )}}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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