Optimal. Leaf size=160 \[ -\frac {c^{3/2} (7 B+i A) \tanh ^{-1}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{8 \sqrt {2} a^2 f}+\frac {c (7 B+i A) \sqrt {c-i c \tan (e+f x)}}{8 a^2 f (1+i \tan (e+f x))}+\frac {(-B+i A) (c-i c \tan (e+f x))^{3/2}}{4 a^2 f (1+i \tan (e+f x))^2} \]
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Rubi [A] time = 0.23, antiderivative size = 160, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.116, Rules used = {3588, 78, 47, 63, 208} \[ -\frac {c^{3/2} (7 B+i A) \tanh ^{-1}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{8 \sqrt {2} a^2 f}+\frac {c (7 B+i A) \sqrt {c-i c \tan (e+f x)}}{8 a^2 f (1+i \tan (e+f x))}+\frac {(-B+i A) (c-i c \tan (e+f x))^{3/2}}{4 a^2 f (1+i \tan (e+f x))^2} \]
Antiderivative was successfully verified.
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Rule 47
Rule 63
Rule 78
Rule 208
Rule 3588
Rubi steps
\begin {align*} \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}{(a+i a \tan (e+f x))^2} \, dx &=\frac {(a c) \operatorname {Subst}\left (\int \frac {(A+B x) \sqrt {c-i c x}}{(a+i a x)^3} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {(i A-B) (c-i c \tan (e+f x))^{3/2}}{4 a^2 f (1+i \tan (e+f x))^2}+\frac {((A-7 i B) c) \operatorname {Subst}\left (\int \frac {\sqrt {c-i c x}}{(a+i a x)^2} \, dx,x,\tan (e+f x)\right )}{8 f}\\ &=\frac {(i A+7 B) c \sqrt {c-i c \tan (e+f x)}}{8 a^2 f (1+i \tan (e+f x))}+\frac {(i A-B) (c-i c \tan (e+f x))^{3/2}}{4 a^2 f (1+i \tan (e+f x))^2}-\frac {\left ((A-7 i B) c^2\right ) \operatorname {Subst}\left (\int \frac {1}{(a+i a x) \sqrt {c-i c x}} \, dx,x,\tan (e+f x)\right )}{16 a f}\\ &=\frac {(i A+7 B) c \sqrt {c-i c \tan (e+f x)}}{8 a^2 f (1+i \tan (e+f x))}+\frac {(i A-B) (c-i c \tan (e+f x))^{3/2}}{4 a^2 f (1+i \tan (e+f x))^2}-\frac {((i A+7 B) c) \operatorname {Subst}\left (\int \frac {1}{2 a-\frac {a x^2}{c}} \, dx,x,\sqrt {c-i c \tan (e+f x)}\right )}{8 a f}\\ &=-\frac {(i A+7 B) c^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{8 \sqrt {2} a^2 f}+\frac {(i A+7 B) c \sqrt {c-i c \tan (e+f x)}}{8 a^2 f (1+i \tan (e+f x))}+\frac {(i A-B) (c-i c \tan (e+f x))^{3/2}}{4 a^2 f (1+i \tan (e+f x))^2}\\ \end {align*}
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Mathematica [A] time = 4.32, size = 205, normalized size = 1.28 \[ \frac {\sec (e+f x) (\cos (f x)+i \sin (f x))^2 (A+B \tan (e+f x)) \left (\sqrt {2} c^{3/2} (A-7 i B) (\sin (2 e)-i \cos (2 e)) \tanh ^{-1}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )+2 c \cos (e+f x) (\cos (2 f x)-i \sin (2 f x)) \sqrt {c-i c \tan (e+f x)} ((A+9 i B) \sin (e+f x)+(5 B+3 i A) \cos (e+f x))\right )}{16 f (a+i a \tan (e+f x))^2 (A \cos (e+f x)+B \sin (e+f x))} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.81, size = 373, normalized size = 2.33 \[ \frac {{\left (\sqrt {\frac {1}{2}} a^{2} f \sqrt {-\frac {{\left (A^{2} - 14 i \, A B - 49 \, B^{2}\right )} c^{3}}{a^{4} f^{2}}} e^{\left (4 i \, f x + 4 i \, e\right )} \log \left (\frac {{\left ({\left (-i \, A - 7 \, B\right )} c^{2} + \sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{2} f\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {{\left (A^{2} - 14 i \, A B - 49 \, B^{2}\right )} c^{3}}{a^{4} f^{2}}}\right )} e^{\left (-i \, f x - i \, e\right )}}{4 \, a^{2} f}\right ) - \sqrt {\frac {1}{2}} a^{2} f \sqrt {-\frac {{\left (A^{2} - 14 i \, A B - 49 \, B^{2}\right )} c^{3}}{a^{4} f^{2}}} e^{\left (4 i \, f x + 4 i \, e\right )} \log \left (\frac {{\left ({\left (-i \, A - 7 \, B\right )} c^{2} - \sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{2} f\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {{\left (A^{2} - 14 i \, A B - 49 \, B^{2}\right )} c^{3}}{a^{4} f^{2}}}\right )} e^{\left (-i \, f x - i \, e\right )}}{4 \, a^{2} f}\right ) + \sqrt {2} {\left ({\left (i \, A + 7 \, B\right )} c e^{\left (4 i \, f x + 4 i \, e\right )} + {\left (3 i \, A + 5 \, B\right )} c e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (2 i \, A - 2 \, B\right )} c\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-4 i \, f x - 4 i \, e\right )}}{16 \, a^{2} f} \]
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 \[ \int \frac {{\left (B \tan \left (f x + e\right ) + A\right )} {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}}}{{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.64, size = 117, normalized size = 0.73 \[ -\frac {2 i c^{2} \left (\frac {\left (-\frac {9 i B}{16}-\frac {A}{16}\right ) \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}+\left (\frac {7}{8} i B c -\frac {1}{8} c A \right ) \sqrt {c -i c \tan \left (f x +e \right )}}{\left (-c -i c \tan \left (f x +e \right )\right )^{2}}+\frac {\left (-7 i B +A \right ) \sqrt {2}\, \arctanh \left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{32 \sqrt {c}}\right )}{f \,a^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 1.01, size = 166, normalized size = 1.04 \[ \frac {i \, {\left (\frac {\sqrt {2} {\left (A - 7 i \, B\right )} c^{\frac {5}{2}} \log \left (-\frac {\sqrt {2} \sqrt {c} - \sqrt {-i \, c \tan \left (f x + e\right ) + c}}{\sqrt {2} \sqrt {c} + \sqrt {-i \, c \tan \left (f x + e\right ) + c}}\right )}{a^{2}} + \frac {4 \, {\left ({\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} {\left (A + 9 i \, B\right )} c^{3} + 2 \, \sqrt {-i \, c \tan \left (f x + e\right ) + c} {\left (A - 7 i \, B\right )} c^{4}\right )}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{2} a^{2} - 4 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )} a^{2} c + 4 \, a^{2} c^{2}}\right )}}{32 \, c f} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 9.39, size = 267, normalized size = 1.67 \[ \frac {\frac {7\,B\,c^3\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{4}-\frac {9\,B\,c^2\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}{8}}{4\,a^2\,c^2\,f+a^2\,f\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^2-4\,a^2\,c\,f\,\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}+\frac {\frac {A\,c^3\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}\,1{}\mathrm {i}}{4\,a^2\,f}+\frac {A\,c^2\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2}\,1{}\mathrm {i}}{8\,a^2\,f}}{{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^2-4\,c\,\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )+4\,c^2}+\frac {\sqrt {2}\,A\,{\left (-c\right )}^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-c}}\right )\,1{}\mathrm {i}}{16\,a^2\,f}-\frac {7\,\sqrt {2}\,B\,c^{3/2}\,\mathrm {atanh}\left (\frac {\sqrt {2}\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {c}}\right )}{16\,a^2\,f} \]
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 \[ - \frac {\int \frac {A c \sqrt {- i c \tan {\left (e + f x \right )} + c}}{\tan ^{2}{\left (e + f x \right )} - 2 i \tan {\left (e + f x \right )} - 1}\, dx + \int \frac {B c \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )}}{\tan ^{2}{\left (e + f x \right )} - 2 i \tan {\left (e + f x \right )} - 1}\, dx + \int \left (- \frac {i A c \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )}}{\tan ^{2}{\left (e + f x \right )} - 2 i \tan {\left (e + f x \right )} - 1}\right )\, dx + \int \left (- \frac {i B c \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{2}{\left (e + f x \right )}}{\tan ^{2}{\left (e + f x \right )} - 2 i \tan {\left (e + f x \right )} - 1}\right )\, dx}{a^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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