Optimal. Leaf size=106 \[ \frac {i a \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{f}-\frac {2 i a^{3/2} \sqrt {c} \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{f} \]
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Rubi [A] time = 0.14, antiderivative size = 106, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3523, 50, 63, 217, 203} \[ \frac {i a \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{f}-\frac {2 i a^{3/2} \sqrt {c} \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{f} \]
Antiderivative was successfully verified.
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Rule 50
Rule 63
Rule 203
Rule 217
Rule 3523
Rubi steps
\begin {align*} \int (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)} \, dx &=\frac {(a c) \operatorname {Subst}\left (\int \frac {\sqrt {a+i a x}}{\sqrt {c-i c x}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {i a \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{f}+\frac {\left (a^2 c\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+i a x} \sqrt {c-i c x}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {i a \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{f}-\frac {(2 i a c) \operatorname {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 {i a \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{f}-\frac {(2 i a c) \operatorname {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 a^{3/2} \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 {i a \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 = 2.70, size = 129, normalized size = 1.22 \[ \frac {a c e^{-\frac {1}{2} i (4 e+f x)} \left (\sin \left (\frac {3 e}{2}\right )-i \cos \left (\frac {3 e}{2}\right )\right ) \sqrt {a+i a \tan (e+f x)} \left (\cos \left (\frac {1}{2} (e+f x)\right )-i \sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (-\sec (e+f x)+2 \tan ^{-1}\left (e^{i (e+f x)}\right )\right )}{\sqrt {2} f \sqrt {\frac {c}{1+e^{2 i (e+f x)}}}} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.46, size = 288, normalized size = 2.72 \[ \frac {8 i \, a \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} e^{\left (i \, f x + i \, e\right )} - 2 \, \sqrt {\frac {a^{3} c}{f^{2}}} f \log \left (\frac {2 \, {\left (4 \, {\left (a e^{\left (3 i \, f x + 3 i \, e\right )} + a e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} + \sqrt {\frac {a^{3} c}{f^{2}}} {\left (2 i \, f e^{\left (2 i \, f x + 2 i \, e\right )} - 2 i \, f\right )}\right )}}{a e^{\left (2 i \, f x + 2 i \, e\right )} + a}\right ) + 2 \, \sqrt {\frac {a^{3} c}{f^{2}}} f \log \left (\frac {2 \, {\left (4 \, {\left (a e^{\left (3 i \, f x + 3 i \, e\right )} + a e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} + \sqrt {\frac {a^{3} c}{f^{2}}} {\left (-2 i \, f e^{\left (2 i \, f x + 2 i \, e\right )} + 2 i \, f\right )}\right )}}{a e^{\left (2 i \, f x + 2 i \, e\right )} + a}\right )}{4 \, f} \]
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 \[ \int {\left (i \, a \tan \left (f x + e\right ) + a\right )}^{\frac {3}{2}} \sqrt {-i \, c \tan \left (f x + e\right ) + c}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.29, size = 122, normalized size = 1.15 \[ \frac {\sqrt {-c \left (-1+i \tan \left (f x +e \right )\right )}\, \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, a \left (i \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}+a c \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}}{\sqrt {c a}}\right )\right )}{f \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 1.15, size = 447, normalized size = 4.22 \[ -\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 \, a \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 i \, a \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}}{f {\left (-2 i \, \cos \left (2 \, f x + 2 \, e\right ) + 2 \, \sin \left (2 \, f x + 2 \, e\right ) - 2 i\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2}\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}} \,d x \]
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 \[ \int \left (i a \left (\tan {\left (e + f x \right )} - i\right )\right )^{\frac {3}{2}} \sqrt {- i c \left (\tan {\left (e + f x \right )} + i\right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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