Optimal. Leaf size=109 \[ \frac {(-B+i A) \sqrt {c-i c \tan (e+f x)}}{2 a f (1+i \tan (e+f x))}+\frac {\sqrt {c} (3 B+i A) \tanh ^{-1}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{2 \sqrt {2} a f} \]
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Rubi [A] time = 0.19, antiderivative size = 109, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.093, Rules used = {3588, 78, 63, 208} \[ \frac {(-B+i A) \sqrt {c-i c \tan (e+f x)}}{2 a f (1+i \tan (e+f x))}+\frac {\sqrt {c} (3 B+i A) \tanh ^{-1}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{2 \sqrt {2} a f} \]
Antiderivative was successfully verified.
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Rule 63
Rule 78
Rule 208
Rule 3588
Rubi steps
\begin {align*} \int \frac {(A+B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}}{a+i a \tan (e+f x)} \, dx &=\frac {(a c) \operatorname {Subst}\left (\int \frac {A+B x}{(a+i a x)^2 \sqrt {c-i c x}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {(i A-B) \sqrt {c-i c \tan (e+f x)}}{2 a f (1+i \tan (e+f x))}+\frac {((A-3 i B) c) \operatorname {Subst}\left (\int \frac {1}{(a+i a x) \sqrt {c-i c x}} \, dx,x,\tan (e+f x)\right )}{4 f}\\ &=\frac {(i A-B) \sqrt {c-i c \tan (e+f x)}}{2 a f (1+i \tan (e+f x))}+\frac {(i A+3 B) \operatorname {Subst}\left (\int \frac {1}{2 a-\frac {a x^2}{c}} \, dx,x,\sqrt {c-i c \tan (e+f x)}\right )}{2 f}\\ &=\frac {(i A+3 B) \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{2 \sqrt {2} a f}+\frac {(i A-B) \sqrt {c-i c \tan (e+f x)}}{2 a f (1+i \tan (e+f x))}\\ \end {align*}
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Mathematica [A] time = 2.69, size = 168, normalized size = 1.54 \[ \frac {(\cos (f x)+i \sin (f x)) (A+B \tan (e+f x)) \left (2 (A+i B) \cos (e+f x) (\sin (f x)+i \cos (f x)) \sqrt {c-i c \tan (e+f x)}+\sqrt {2} \sqrt {c} (3 B+i A) (\cos (e)+i \sin (e)) \tanh ^{-1}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )\right )}{4 f (a+i a \tan (e+f x)) (A \cos (e+f x)+B \sin (e+f x))} \]
Antiderivative was successfully verified.
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fricas [B] time = 1.99, size = 326, normalized size = 2.99 \[ \frac {{\left (\sqrt {\frac {1}{2}} a f \sqrt {-\frac {{\left (A^{2} - 6 i \, A B - 9 \, B^{2}\right )} c}{a^{2} f^{2}}} e^{\left (2 i \, f x + 2 i \, e\right )} \log \left (\frac {{\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a f e^{\left (2 i \, f x + 2 i \, e\right )} + a f\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {{\left (A^{2} - 6 i \, A B - 9 \, B^{2}\right )} c}{a^{2} f^{2}}} + {\left (i \, A + 3 \, B\right )} c\right )} e^{\left (-i \, f x - i \, e\right )}}{a f}\right ) - \sqrt {\frac {1}{2}} a f \sqrt {-\frac {{\left (A^{2} - 6 i \, A B - 9 \, B^{2}\right )} c}{a^{2} f^{2}}} e^{\left (2 i \, f x + 2 i \, e\right )} \log \left (-\frac {{\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a f e^{\left (2 i \, f x + 2 i \, e\right )} + a f\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {{\left (A^{2} - 6 i \, A B - 9 \, B^{2}\right )} c}{a^{2} f^{2}}} - {\left (i \, A + 3 \, B\right )} c\right )} e^{\left (-i \, f x - i \, e\right )}}{a f}\right ) + \sqrt {2} {\left ({\left (i \, A - B\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + i \, A - B\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{4 \, a f} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B \tan \left (f x + e\right ) + A\right )} \sqrt {-i \, c \tan \left (f x + e\right ) + c}}{i \, a \tan \left (f x + e\right ) + a}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.70, size = 88, normalized size = 0.81 \[ \frac {2 i c \left (\frac {\left (-\frac {A}{4}-\frac {i B}{4}\right ) \sqrt {c -i c \tan \left (f x +e \right )}}{-c -i c \tan \left (f x +e \right )}+\frac {\left (-3 i B +A \right ) \sqrt {2}\, \arctanh \left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{8 \sqrt {c}}\right )}{f a} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.73, size = 116, normalized size = 1.06 \[ -\frac {i \, {\left (\frac {\sqrt {2} {\left (A - 3 i \, B\right )} c^{\frac {3}{2}} \log \left (-\frac {\sqrt {2} \sqrt {c} - \sqrt {-i \, c \tan \left (f x + e\right ) + c}}{\sqrt {2} \sqrt {c} + \sqrt {-i \, c \tan \left (f x + e\right ) + c}}\right )}{a} + \frac {4 \, \sqrt {-i \, c \tan \left (f x + e\right ) + c} {\left (A + i \, B\right )} c^{2}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )} a - 2 \, a c}\right )}}{8 \, c f} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 9.56, size = 159, normalized size = 1.46 \[ -\frac {B\,c\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{2\,\left (a\,c\,f+a\,c\,f\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}+\frac {A\,c\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2\,a\,f\,\left (c+c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}+\frac {\sqrt {2}\,A\,\sqrt {-c}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-c}}\right )\,1{}\mathrm {i}}{4\,a\,f}+\frac {3\,\sqrt {2}\,B\,\sqrt {c}\,\mathrm {atanh}\left (\frac {\sqrt {2}\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {c}}\right )}{4\,a\,f} \]
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 \[ - \frac {i \left (\int \frac {A \sqrt {- i c \tan {\left (e + f x \right )} + c}}{\tan {\left (e + f x \right )} - i}\, dx + \int \frac {B \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )}}{\tan {\left (e + f x \right )} - i}\, dx\right )}{a} \]
Verification of antiderivative is not currently implemented for this CAS.
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