Optimal. Leaf size=60 \[ \frac {2 a (i A+B) \sqrt {c-i c \tan (e+f x)}}{f}-\frac {2 a B (c-i c \tan (e+f x))^{3/2}}{3 c f} \]
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Rubi [A]
time = 0.06, antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps
used = 3, number of rules used = 2, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.049, Rules used = {3669, 45}
\begin {gather*} \frac {2 a (B+i A) \sqrt {c-i c \tan (e+f x)}}{f}-\frac {2 a B (c-i c \tan (e+f x))^{3/2}}{3 c f} \end {gather*}
Antiderivative was successfully verified.
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Rule 45
Rule 3669
Rubi steps
\begin {align*} \int (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 {c-i c x}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {(a c) \text {Subst}\left (\int \left (\frac {A-i B}{\sqrt {c-i c x}}+\frac {i B \sqrt {c-i c x}}{c}\right ) \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {2 a (i A+B) \sqrt {c-i c \tan (e+f x)}}{f}-\frac {2 a B (c-i c \tan (e+f x))^{3/2}}{3 c f}\\ \end {align*}
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Mathematica [A]
time = 0.81, size = 45, normalized size = 0.75 \begin {gather*} \frac {2 a (3 i A+2 B+i B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}}{3 f} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.27, size = 66, normalized size = 1.10
method | result | size |
derivativedivides | \(\frac {2 i a \left (\frac {i B \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}-i B c \sqrt {c -i c \tan \left (f x +e \right )}+A c \sqrt {c -i c \tan \left (f x +e \right )}\right )}{f c}\) | \(66\) |
default | \(\frac {2 i a \left (\frac {i B \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}-i B c \sqrt {c -i c \tan \left (f x +e \right )}+A c \sqrt {c -i c \tan \left (f x +e \right )}\right )}{f c}\) | \(66\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.28, size = 50, normalized size = 0.83 \begin {gather*} \frac {2 i \, {\left (i \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} B a + 3 \, \sqrt {-i \, c \tan \left (f x + e\right ) + c} {\left (A - i \, B\right )} a c\right )}}{3 \, c f} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 1.18, size = 69, normalized size = 1.15 \begin {gather*} -\frac {2 \, \sqrt {2} {\left (3 \, {\left (-i \, A - B\right )} a e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (-3 i \, A - B\right )} a\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{3 \, {\left (f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )}} \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*} i a \left (\int \left (- i A \sqrt {- i c \tan {\left (e + f x \right )} + c}\right )\, dx + \int A \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )}\, dx + \int B \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{2}{\left (e + f x \right )}\, dx + \int \left (- i B \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )}\right )\, dx\right ) \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 = 0.69, size = 102, normalized size = 1.70 \begin {gather*} \frac {a\,\sqrt {-\frac {c\,\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}}\,\left (A\,3{}\mathrm {i}+2\,B+A\,\left (2\,{\cos \left (e+f\,x\right )}^2-1\right )\,3{}\mathrm {i}+2\,B\,\left (2\,{\cos \left (e+f\,x\right )}^2-1\right )+B\,\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{3\,f\,{\cos \left (e+f\,x\right )}^2} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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