Optimal. Leaf size=159 \[ \frac {(3 i A+5 B) \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{16 \sqrt {2} a^2 f}+\frac {(i A-B) \sqrt {c-i c \tan (e+f x)}}{4 a^2 f (1+i \tan (e+f x))^2}+\frac {(3 i A+5 B) \sqrt {c-i c \tan (e+f x)}}{16 a^2 f (1+i \tan (e+f x))} \]
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Rubi [A]
time = 0.15, antiderivative size = 159, 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 = {3669, 79, 44,
65, 214} \begin {gather*} \frac {(-B+i A) \sqrt {c-i c \tan (e+f x)}}{4 a^2 f (1+i \tan (e+f x))^2}+\frac {(5 B+3 i A) \sqrt {c-i c \tan (e+f x)}}{16 a^2 f (1+i \tan (e+f x))}+\frac {\sqrt {c} (5 B+3 i A) \tanh ^{-1}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{16 \sqrt {2} a^2 f} \end {gather*}
Antiderivative was successfully verified.
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Rule 44
Rule 65
Rule 79
Rule 214
Rule 3669
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))^2} \, dx &=\frac {(a c) \text {Subst}\left (\int \frac {A+B x}{(a+i a x)^3 \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)}}{4 a^2 f (1+i \tan (e+f x))^2}+\frac {((3 A-5 i B) c) \text {Subst}\left (\int \frac {1}{(a+i a x)^2 \sqrt {c-i c x}} \, dx,x,\tan (e+f x)\right )}{8 f}\\ &=\frac {(i A-B) \sqrt {c-i c \tan (e+f x)}}{4 a^2 f (1+i \tan (e+f x))^2}+\frac {(3 i A+5 B) \sqrt {c-i c \tan (e+f x)}}{16 a^2 f (1+i \tan (e+f x))}+\frac {((3 A-5 i B) c) \text {Subst}\left (\int \frac {1}{(a+i a x) \sqrt {c-i c x}} \, dx,x,\tan (e+f x)\right )}{32 a f}\\ &=\frac {(i A-B) \sqrt {c-i c \tan (e+f x)}}{4 a^2 f (1+i \tan (e+f x))^2}+\frac {(3 i A+5 B) \sqrt {c-i c \tan (e+f x)}}{16 a^2 f (1+i \tan (e+f x))}+\frac {(3 i A+5 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 )}{16 a f}\\ &=\frac {(3 i A+5 B) \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{16 \sqrt {2} a^2 f}+\frac {(i A-B) \sqrt {c-i c \tan (e+f x)}}{4 a^2 f (1+i \tan (e+f x))^2}+\frac {(3 i A+5 B) \sqrt {c-i c \tan (e+f x)}}{16 a^2 f (1+i \tan (e+f x))}\\ \end {align*}
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Mathematica [A]
time = 1.10, size = 206, normalized size = 1.30 \begin {gather*} \frac {\sec (e+f x) (\cos (f x)+i \sin (f x))^2 (A+B \tan (e+f x)) \left (\sqrt {2} (3 i A+5 B) \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right ) (\cos (2 e)+i \sin (2 e))+2 \cos (e+f x) (\cos (2 f x)-i \sin (2 f x)) ((7 i A+B) \cos (e+f x)+(-3 A+5 i B) \sin (e+f x)) \sqrt {c-i c \tan (e+f x)}\right )}{32 f (A \cos (e+f x)+B \sin (e+f x)) (a+i a \tan (e+f x))^2} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.40, size = 120, normalized size = 0.75
method | result | size |
derivativedivides | \(-\frac {2 i c^{2} \left (-\frac {4 \left (-\frac {\left (-5 i B +3 A \right ) \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{128 c}+\left (\frac {5 A}{64}-\frac {3 i B}{64}\right ) \sqrt {c -i c \tan \left (f x +e \right )}\right )}{\left (c +i c \tan \left (f x +e \right )\right )^{2}}-\frac {\left (-5 i B +3 A \right ) \sqrt {2}\, \arctanh \left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{64 c^{\frac {3}{2}}}\right )}{f \,a^{2}}\) | \(120\) |
default | \(-\frac {2 i c^{2} \left (-\frac {4 \left (-\frac {\left (-5 i B +3 A \right ) \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{128 c}+\left (\frac {5 A}{64}-\frac {3 i B}{64}\right ) \sqrt {c -i c \tan \left (f x +e \right )}\right )}{\left (c +i c \tan \left (f x +e \right )\right )^{2}}-\frac {\left (-5 i B +3 A \right ) \sqrt {2}\, \arctanh \left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{64 c^{\frac {3}{2}}}\right )}{f \,a^{2}}\) | \(120\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.54, size = 178, normalized size = 1.12 \begin {gather*} -\frac {i \, {\left (\frac {\sqrt {2} {\left (3 \, A - 5 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^{2}} + \frac {4 \, {\left ({\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} {\left (3 \, A - 5 i \, B\right )} c^{2} - 2 \, \sqrt {-i \, c \tan \left (f x + e\right ) + c} {\left (5 \, A - 3 i \, B\right )} c^{3}\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 )}}{64 \, 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. 374 vs. \(2 (127) = 254\).
time = 4.72, size = 374, normalized size = 2.35 \begin {gather*} \frac {{\left (\sqrt {\frac {1}{2}} a^{2} f \sqrt {-\frac {{\left (9 \, A^{2} - 30 i \, A B - 25 \, B^{2}\right )} c}{a^{4} f^{2}}} e^{\left (4 i \, f x + 4 i \, e\right )} \log \left (\frac {{\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 {{\left (9 \, A^{2} - 30 i \, A B - 25 \, B^{2}\right )} c}{a^{4} f^{2}}} + {\left (3 i \, A + 5 \, B\right )} c\right )} e^{\left (-i \, f x - i \, e\right )}}{8 \, a^{2} f}\right ) - \sqrt {\frac {1}{2}} a^{2} f \sqrt {-\frac {{\left (9 \, A^{2} - 30 i \, A B - 25 \, B^{2}\right )} c}{a^{4} f^{2}}} e^{\left (4 i \, f x + 4 i \, e\right )} \log \left (-\frac {{\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 {{\left (9 \, A^{2} - 30 i \, A B - 25 \, B^{2}\right )} c}{a^{4} f^{2}}} - {\left (3 i \, A + 5 \, B\right )} c\right )} e^{\left (-i \, f x - i \, e\right )}}{8 \, a^{2} f}\right ) + \sqrt {2} {\left ({\left (5 i \, A + 3 \, B\right )} e^{\left (4 i \, f x + 4 i \, e\right )} + {\left (7 i \, A + 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 )}}{32 \, a^{2} 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 \tan {\left (e + f x \right )} - 1}\, dx + \int \frac {B \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}{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 = 9.18, size = 264, normalized size = 1.66 \begin {gather*} -\frac {\frac {5\,B\,c\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}{16}-\frac {3\,B\,c^2\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{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^2\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}\,5{}\mathrm {i}}{8\,a^2\,f}-\frac {A\,c\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2}\,3{}\mathrm {i}}{16\,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\,\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 )\,3{}\mathrm {i}}{32\,a^2\,f}+\frac {5\,\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 )}{32\,a^2\,f} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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