Optimal. Leaf size=104 \[ -\frac {2 i \sqrt {a} A \sqrt {c} \text {ArcTan}\left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{f}+\frac {B \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{f} \]
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Rubi [A]
time = 0.14, antiderivative size = 104, normalized size of antiderivative = 1.00, number of steps
used = 5, number of rules used = 5, integrand size = 45, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {3669, 81, 65,
223, 209} \begin {gather*} \frac {B \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{f}-\frac {2 i \sqrt {a} A \sqrt {c} \text {ArcTan}\left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{f} \end {gather*}
Antiderivative was successfully verified.
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Rule 65
Rule 81
Rule 209
Rule 223
Rule 3669
Rubi steps
\begin {align*} \int \sqrt {a+i a \tan (e+f x)} (A+B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)} \, dx &=\frac {(a c) \text {Subst}\left (\int \frac {A+B x}{\sqrt {a+i a x} \sqrt {c-i c x}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {B \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{f}+\frac {(a A c) \text {Subst}\left (\int \frac {1}{\sqrt {a+i a x} \sqrt {c-i c x}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {B \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{f}-\frac {(2 i A c) \text {Subst}\left (\int \frac {1}{\sqrt {2 c-\frac {c x^2}{a}}} \, dx,x,\sqrt {a+i a \tan (e+f x)}\right )}{f}\\ &=\frac {B \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{f}-\frac {(2 i A c) \text {Subst}\left (\int \frac {1}{1+\frac {c x^2}{a}} \, dx,x,\frac {\sqrt {a+i a \tan (e+f x)}}{\sqrt {c-i c \tan (e+f x)}}\right )}{f}\\ &=-\frac {2 i \sqrt {a} A \sqrt {c} \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{f}+\frac {B \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{f}\\ \end {align*}
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Mathematica [A]
time = 1.33, size = 102, normalized size = 0.98 \begin {gather*} \frac {\sqrt {2} e^{-i (e+f x)} \sqrt {\frac {c}{1+e^{2 i (e+f x)}}} \left (B e^{i (e+f x)}-i A \left (1+e^{2 i (e+f x)}\right ) \text {ArcTan}\left (e^{i (e+f x)}\right )\right ) \sqrt {a+i a \tan (e+f x)}}{f} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.39, size = 121, normalized size = 1.16
method | result | size |
derivativedivides | \(\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \left (A \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}}{\sqrt {a c}}\right ) a c +B \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}\right )}{f \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}}\) | \(121\) |
default | \(\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \left (A \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}}{\sqrt {a c}}\right ) a c +B \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}\right )}{f \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}}\) | \(121\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [B] Both result and optimal contain complex but leaf count of result is larger than
twice the leaf count of optimal. 481 vs. \(2 (82) = 164\).
time = 0.63, size = 481, normalized size = 4.62 \begin {gather*} -\frac {{\left (2 \, {\left (A \cos \left (2 \, f x + 2 \, e\right ) + i \, A \sin \left (2 \, f x + 2 \, e\right ) + A\right )} \arctan \left (\cos \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ), \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + 1\right ) + 2 \, {\left (A \cos \left (2 \, f x + 2 \, e\right ) + i \, A \sin \left (2 \, f x + 2 \, e\right ) + A\right )} \arctan \left (\cos \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ), -\sin \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + 1\right ) + 4 i \, B \cos \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) - {\left (-i \, A \cos \left (2 \, f x + 2 \, e\right ) + A \sin \left (2 \, f x + 2 \, e\right ) - i \, A\right )} \log \left (\cos \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right )^{2} + \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right )^{2} + 2 \, \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + 1\right ) - {\left (i \, A \cos \left (2 \, f x + 2 \, e\right ) - A \sin \left (2 \, f x + 2 \, e\right ) + i \, A\right )} \log \left (\cos \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right )^{2} + \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right )^{2} - 2 \, \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + 1\right ) - 4 \, B \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right )\right )} \sqrt {a} \sqrt {c}}{-2 \, f {\left (i \, \cos \left (2 \, f x + 2 \, e\right ) - \sin \left (2 \, f x + 2 \, e\right ) + i\right )}} \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. 308 vs. \(2 (82) = 164\).
time = 5.10, size = 308, normalized size = 2.96 \begin {gather*} \frac {4 \, B \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} e^{\left (i \, f x + i \, e\right )} + \sqrt {\frac {A^{2} a c}{f^{2}}} f \log \left (\frac {4 \, {\left (2 \, {\left (A e^{\left (3 i \, f x + 3 i \, e\right )} + A e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} - \sqrt {\frac {A^{2} a c}{f^{2}}} {\left (i \, f e^{\left (2 i \, f x + 2 i \, e\right )} - i \, f\right )}\right )}}{A e^{\left (2 i \, f x + 2 i \, e\right )} + A}\right ) - \sqrt {\frac {A^{2} a c}{f^{2}}} f \log \left (\frac {4 \, {\left (2 \, {\left (A e^{\left (3 i \, f x + 3 i \, e\right )} + A e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} - \sqrt {\frac {A^{2} a c}{f^{2}}} {\left (-i \, f e^{\left (2 i \, f x + 2 i \, e\right )} + i \, f\right )}\right )}}{A e^{\left (2 i \, f x + 2 i \, e\right )} + A}\right )}{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*} \int \sqrt {i a \left (\tan {\left (e + f x \right )} - i\right )} \sqrt {- i c \left (\tan {\left (e + f x \right )} + i\right )} \left (A + B \tan {\left (e + f x \right )}\right )\, dx \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 = 11.09, size = 133, normalized size = 1.28 \begin {gather*} -\frac {A\,\sqrt {a}\,\sqrt {c}\,\mathrm {atan}\left (\frac {\sqrt {c}\,\left (\sqrt {a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}-\sqrt {a}\right )}{\sqrt {a}\,\left (\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}-\sqrt {c}\right )}\right )\,4{}\mathrm {i}}{f}+\frac {\sqrt {2}\,B\,\sqrt {\frac {c}{2\,{\cos \left (e+f\,x\right )}^2+\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}}}\,\sqrt {\frac {a\,\left (2\,{\cos \left (e+f\,x\right )}^2+\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{2\,{\cos \left (e+f\,x\right )}^2}}}{f} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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