Optimal. Leaf size=153 \[ \frac {6 i c^{5/2} \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{\sqrt {a} f}+\frac {3 i c^2 \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{a f}+\frac {2 i c (c-i c \tan (e+f x))^{3/2}}{f \sqrt {a+i a \tan (e+f x)}} \]
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Rubi [A] time = 0.16, antiderivative size = 153, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {3523, 47, 50, 63, 217, 203} \[ \frac {6 i c^{5/2} \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{\sqrt {a} f}+\frac {3 i c^2 \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{a f}+\frac {2 i c (c-i c \tan (e+f x))^{3/2}}{f \sqrt {a+i a \tan (e+f x)}} \]
Antiderivative was successfully verified.
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Rule 47
Rule 50
Rule 63
Rule 203
Rule 217
Rule 3523
Rubi steps
\begin {align*} \int \frac {(c-i c \tan (e+f x))^{5/2}}{\sqrt {a+i a \tan (e+f x)}} \, dx &=\frac {(a c) \operatorname {Subst}\left (\int \frac {(c-i c x)^{3/2}}{(a+i a x)^{3/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {2 i c (c-i c \tan (e+f x))^{3/2}}{f \sqrt {a+i a \tan (e+f x)}}-\frac {\left (3 c^2\right ) \operatorname {Subst}\left (\int \frac {\sqrt {c-i c x}}{\sqrt {a+i a x}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {3 i c^2 \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{a f}+\frac {2 i c (c-i c \tan (e+f x))^{3/2}}{f \sqrt {a+i a \tan (e+f x)}}-\frac {\left (3 c^3\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 {3 i c^2 \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{a f}+\frac {2 i c (c-i c \tan (e+f x))^{3/2}}{f \sqrt {a+i a \tan (e+f x)}}+\frac {\left (6 i c^3\right ) \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 )}{a f}\\ &=\frac {3 i c^2 \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{a f}+\frac {2 i c (c-i c \tan (e+f x))^{3/2}}{f \sqrt {a+i a \tan (e+f x)}}+\frac {\left (6 i c^3\right ) \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 )}{a f}\\ &=\frac {6 i c^{5/2} \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 )}{\sqrt {a} f}+\frac {3 i c^2 \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{a f}+\frac {2 i c (c-i c \tan (e+f x))^{3/2}}{f \sqrt {a+i a \tan (e+f x)}}\\ \end {align*}
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Mathematica [A] time = 3.96, size = 113, normalized size = 0.74 \[ \frac {c^3 (\cos (f x)+i \sin (f x)) (\sin (f x)+i \cos (f x)) \left (\tan ^2(e+f x)-4 i \tan (e+f x)+6 \sec (e+f x) \tan ^{-1}(\cos (e+f x)+i \sin (e+f x))+5\right )}{f \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.46, size = 357, normalized size = 2.33 \[ \frac {{\left (3 \, \sqrt {\frac {c^{5}}{a f^{2}}} a f e^{\left (i \, f x + i \, e\right )} \log \left (\frac {2 \, {\left (4 \, {\left (c^{2} e^{\left (3 i \, f x + 3 i \, e\right )} + c^{2} e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} + {\left (2 i \, a f e^{\left (2 i \, f x + 2 i \, e\right )} - 2 i \, a f\right )} \sqrt {\frac {c^{5}}{a f^{2}}}\right )}}{c^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + c^{2}}\right ) - 3 \, \sqrt {\frac {c^{5}}{a f^{2}}} a f e^{\left (i \, f x + i \, e\right )} \log \left (\frac {2 \, {\left (4 \, {\left (c^{2} e^{\left (3 i \, f x + 3 i \, e\right )} + c^{2} e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} + {\left (-2 i \, a f e^{\left (2 i \, f x + 2 i \, e\right )} + 2 i \, a f\right )} \sqrt {\frac {c^{5}}{a f^{2}}}\right )}}{c^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + c^{2}}\right ) + {\left (12 i \, c^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + 8 i \, c^{2}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-i \, f x - i \, e\right )}}{2 \, a f} \]
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 \[ \int \frac {{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}}}{\sqrt {i \, a \tan \left (f x + e\right ) + a}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.30, size = 300, normalized size = 1.96 \[ \frac {i \sqrt {-c \left (-1+i \tan \left (f x +e \right )\right )}\, \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, c^{2} \left (3 i \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}}{\sqrt {c a}}\right ) \left (\tan ^{2}\left (f x +e \right )\right ) a c -3 i \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}}{\sqrt {c a}}\right ) a c -6 i \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}\, \tan \left (f x +e \right )+6 \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}}{\sqrt {c a}}\right ) \tan \left (f x +e \right ) a c +\sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}\, \left (\tan ^{2}\left (f x +e \right )\right )-5 \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}\right )}{f a \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \left (-\tan \left (f x +e \right )+i\right )^{2} \sqrt {c a}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 1.08, size = 185, normalized size = 1.21 \[ -\frac {{\left (6 i \, c^{2} \arctan \left (\cos \left (f x + e\right ), \sin \left (f x + e\right ) + 1\right ) \cos \left (f x + e\right ) + 6 i \, c^{2} \arctan \left (\cos \left (f x + e\right ), -\sin \left (f x + e\right ) + 1\right ) \cos \left (f x + e\right ) - 8 i \, c^{2} \cos \left (f x + e\right )^{2} + 3 \, c^{2} \cos \left (f x + e\right ) \log \left (\cos \left (f x + e\right )^{2} + \sin \left (f x + e\right )^{2} + 2 \, \sin \left (f x + e\right ) + 1\right ) - 3 \, c^{2} \cos \left (f x + e\right ) \log \left (\cos \left (f x + e\right )^{2} + \sin \left (f x + e\right )^{2} - 2 \, \sin \left (f x + e\right ) + 1\right ) - 8 \, c^{2} \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 2 i \, c^{2}\right )} \sqrt {c}}{2 \, \sqrt {a} f \cos \left (f x + e\right )} \]
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 \[ \int \frac {{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}{\sqrt {a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}} \,d x \]
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 \[ \int \frac {\left (- i c \left (\tan {\left (e + f x \right )} + i\right )\right )^{\frac {5}{2}}}{\sqrt {i a \left (\tan {\left (e + f x \right )} - i\right )}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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