Optimal. Leaf size=195 \[ \frac {3 (5 i A+3 B) \tanh ^{-1}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{32 \sqrt {2} a^2 \sqrt {c} f}-\frac {3 (5 i A+3 B)}{32 a^2 f \sqrt {c-i c \tan (e+f x)}}+\frac {i A-B}{4 a^2 f (1+i \tan (e+f x))^2 \sqrt {c-i c \tan (e+f x)}}+\frac {5 i A+3 B}{16 a^2 f (1+i \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}} \]
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Rubi [A]
time = 0.17, antiderivative size = 195, normalized size of antiderivative = 1.00, number of steps
used = 6, number of rules used = 6, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.140, Rules used = {3669, 79, 44,
53, 65, 214} \begin {gather*} \frac {-B+i A}{4 a^2 f (1+i \tan (e+f x))^2 \sqrt {c-i c \tan (e+f x)}}-\frac {3 (3 B+5 i A)}{32 a^2 f \sqrt {c-i c \tan (e+f x)}}+\frac {3 B+5 i A}{16 a^2 f (1+i \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}}+\frac {3 (3 B+5 i A) \tanh ^{-1}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{32 \sqrt {2} a^2 \sqrt {c} f} \end {gather*}
Antiderivative was successfully verified.
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Rule 44
Rule 53
Rule 65
Rule 79
Rule 214
Rule 3669
Rubi steps
\begin {align*} \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^2 \sqrt {c-i c \tan (e+f x)}} \, dx &=\frac {(a c) \text {Subst}\left (\int \frac {A+B x}{(a+i a x)^3 (c-i c x)^{3/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {i A-B}{4 a^2 f (1+i \tan (e+f x))^2 \sqrt {c-i c \tan (e+f x)}}+\frac {((5 A-3 i B) c) \text {Subst}\left (\int \frac {1}{(a+i a x)^2 (c-i c x)^{3/2}} \, dx,x,\tan (e+f x)\right )}{8 f}\\ &=\frac {i A-B}{4 a^2 f (1+i \tan (e+f x))^2 \sqrt {c-i c \tan (e+f x)}}+\frac {5 i A+3 B}{16 a^2 f (1+i \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}}+\frac {(3 (5 A-3 i B) c) \text {Subst}\left (\int \frac {1}{(a+i a x) (c-i c x)^{3/2}} \, dx,x,\tan (e+f x)\right )}{32 a f}\\ &=-\frac {3 (5 i A+3 B)}{32 a^2 f \sqrt {c-i c \tan (e+f x)}}+\frac {i A-B}{4 a^2 f (1+i \tan (e+f x))^2 \sqrt {c-i c \tan (e+f x)}}+\frac {5 i A+3 B}{16 a^2 f (1+i \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}}+\frac {(3 (5 A-3 i B)) \text {Subst}\left (\int \frac {1}{(a+i a x) \sqrt {c-i c x}} \, dx,x,\tan (e+f x)\right )}{64 a f}\\ &=-\frac {3 (5 i A+3 B)}{32 a^2 f \sqrt {c-i c \tan (e+f x)}}+\frac {i A-B}{4 a^2 f (1+i \tan (e+f x))^2 \sqrt {c-i c \tan (e+f x)}}+\frac {5 i A+3 B}{16 a^2 f (1+i \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}}+\frac {(3 (5 i A+3 B)) \text {Subst}\left (\int \frac {1}{2 a-\frac {a x^2}{c}} \, dx,x,\sqrt {c-i c \tan (e+f x)}\right )}{32 a c f}\\ &=\frac {3 (5 i A+3 B) \tanh ^{-1}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{32 \sqrt {2} a^2 \sqrt {c} f}-\frac {3 (5 i A+3 B)}{32 a^2 f \sqrt {c-i c \tan (e+f x)}}+\frac {i A-B}{4 a^2 f (1+i \tan (e+f x))^2 \sqrt {c-i c \tan (e+f x)}}+\frac {5 i A+3 B}{16 a^2 f (1+i \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}}\\ \end {align*}
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Mathematica [A]
time = 2.14, size = 160, normalized size = 0.82 \begin {gather*} \frac {(i \cos (e+f x)+\sin (e+f x)) \left (3 (5 A-3 i B) e^{i (e+f x)} \sqrt {1+e^{2 i (e+f x)}} \tanh ^{-1}\left (\sqrt {1+e^{2 i (e+f x)}}\right )-2 \cos (e+f x) (-9 A-i B+2 (3 A-5 i B) \cos (2 (e+f x))+2 (5 i A+3 B) \sin (2 (e+f x)))\right ) \sqrt {c-i c \tan (e+f x)}}{64 a^2 c f} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.35, size = 151, normalized size = 0.77
method | result | size |
derivativedivides | \(-\frac {2 i c^{2} \left (-\frac {i B -A}{8 c^{2} \sqrt {c -i c \tan \left (f x +e \right )}}-\frac {\frac {4 \left (\frac {i B}{32}-\frac {7 A}{32}\right ) \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}+4 \left (\frac {9}{16} A c +\frac {1}{16} i B c \right ) \sqrt {c -i c \tan \left (f x +e \right )}}{\left (c +i c \tan \left (f x +e \right )\right )^{2}}+\frac {3 \left (-\frac {3 i B}{4}+\frac {5 A}{4}\right ) \sqrt {2}\, \arctanh \left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{4 \sqrt {c}}}{8 c^{2}}\right )}{f \,a^{2}}\) | \(151\) |
default | \(-\frac {2 i c^{2} \left (-\frac {i B -A}{8 c^{2} \sqrt {c -i c \tan \left (f x +e \right )}}-\frac {\frac {4 \left (\frac {i B}{32}-\frac {7 A}{32}\right ) \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}+4 \left (\frac {9}{16} A c +\frac {1}{16} i B c \right ) \sqrt {c -i c \tan \left (f x +e \right )}}{\left (c +i c \tan \left (f x +e \right )\right )^{2}}+\frac {3 \left (-\frac {3 i B}{4}+\frac {5 A}{4}\right ) \sqrt {2}\, \arctanh \left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{4 \sqrt {c}}}{8 c^{2}}\right )}{f \,a^{2}}\) | \(151\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.54, size = 202, normalized size = 1.04 \begin {gather*} -\frac {i \, {\left (\frac {3 \, \sqrt {2} {\left (5 \, A - 3 i \, B\right )} \sqrt {c} \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 (3 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{2} {\left (5 \, A - 3 i \, B\right )} c - 10 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )} {\left (5 \, A - 3 i \, B\right )} c^{2} + 32 \, {\left (A - i \, B\right )} c^{3}\right )}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}} a^{2} - 4 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} a^{2} c + 4 \, \sqrt {-i \, c \tan \left (f x + e\right ) + c} a^{2} c^{2}}\right )}}{128 \, c f} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice
the leaf count of optimal. 403 vs. \(2 (156) = 312\).
time = 3.92, size = 403, normalized size = 2.07 \begin {gather*} \frac {{\left (3 \, \sqrt {\frac {1}{2}} a^{2} c f \sqrt {-\frac {25 \, A^{2} - 30 i \, A B - 9 \, B^{2}}{a^{4} c f^{2}}} e^{\left (4 i \, f x + 4 i \, e\right )} \log \left (\frac {3 \, {\left (\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 {25 \, A^{2} - 30 i \, A B - 9 \, B^{2}}{a^{4} c f^{2}}} + 5 i \, A + 3 \, B\right )} e^{\left (-i \, f x - i \, e\right )}}{16 \, a^{2} f}\right ) - 3 \, \sqrt {\frac {1}{2}} a^{2} c f \sqrt {-\frac {25 \, A^{2} - 30 i \, A B - 9 \, B^{2}}{a^{4} c f^{2}}} e^{\left (4 i \, f x + 4 i \, e\right )} \log \left (-\frac {3 \, {\left (\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 {25 \, A^{2} - 30 i \, A B - 9 \, B^{2}}{a^{4} c f^{2}}} - 5 i \, A - 3 \, B\right )} e^{\left (-i \, f x - i \, e\right )}}{16 \, a^{2} f}\right ) - \sqrt {2} {\left (8 \, {\left (i \, A + B\right )} e^{\left (6 i \, f x + 6 i \, e\right )} - {\left (i \, A - 9 \, B\right )} e^{\left (4 i \, f x + 4 i \, e\right )} - {\left (11 i \, A - 3 \, B\right )} e^{\left (2 i \, f x + 2 i \, e\right )} - 2 i \, A + 2 \, B\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 )}}{64 \, a^{2} c f} \end {gather*}
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 \begin {gather*} - \frac {\int \frac {A}{\sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{2}{\left (e + f x \right )} - 2 i \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )} - \sqrt {- i c \tan {\left (e + f x \right )} + c}}\, dx + \int \frac {B \tan {\left (e + f x \right )}}{\sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{2}{\left (e + f x \right )} - 2 i \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )} - \sqrt {- i c \tan {\left (e + f x \right )} + c}}\, dx}{a^{2}} \end {gather*}
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 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 9.44, size = 305, normalized size = 1.56 \begin {gather*} -\frac {\frac {A\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^2\,15{}\mathrm {i}}{32\,a^2\,f}+\frac {A\,c^2\,1{}\mathrm {i}}{a^2\,f}-\frac {A\,c\,\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )\,25{}\mathrm {i}}{16\,a^2\,f}}{-4\,c\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2}+{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{5/2}+4\,c^2\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}-\frac {B\,c^2+\frac {9\,B\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^2}{32}-\frac {15\,B\,c\,\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}{16}}{a^2\,f\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{5/2}-4\,a^2\,c\,f\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2}+4\,a^2\,c^2\,f\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}-\frac {\sqrt {2}\,A\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-c}}\right )\,15{}\mathrm {i}}{64\,a^2\,\sqrt {-c}\,f}+\frac {9\,\sqrt {2}\,B\,\mathrm {atanh}\left (\frac {\sqrt {2}\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {c}}\right )}{64\,a^2\,\sqrt {c}\,f} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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