Optimal. Leaf size=274 \[ \frac {231 i \tanh ^{-1}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{512 \sqrt {2} a^3 c^{5/2} f}-\frac {231 i}{512 a^3 c^2 f \sqrt {c-i c \tan (e+f x)}}-\frac {77 i}{256 a^3 c f (c-i c \tan (e+f x))^{3/2}}-\frac {231 i}{640 a^3 f (c-i c \tan (e+f x))^{5/2}}+\frac {33 i}{64 a^3 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}+\frac {11 i}{48 a^3 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}}+\frac {i}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{5/2}} \]
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Rubi [A] time = 0.25, antiderivative size = 274, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 5, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {3522, 3487, 51, 63, 206} \[ -\frac {231 i}{512 a^3 c^2 f \sqrt {c-i c \tan (e+f x)}}+\frac {231 i \tanh ^{-1}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{512 \sqrt {2} a^3 c^{5/2} f}-\frac {77 i}{256 a^3 c f (c-i c \tan (e+f x))^{3/2}}-\frac {231 i}{640 a^3 f (c-i c \tan (e+f x))^{5/2}}+\frac {33 i}{64 a^3 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}+\frac {11 i}{48 a^3 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}}+\frac {i}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{5/2}} \]
Antiderivative was successfully verified.
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Rule 51
Rule 63
Rule 206
Rule 3487
Rule 3522
Rubi steps
\begin {align*} \int \frac {1}{(a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^{5/2}} \, dx &=\frac {\int \cos ^6(e+f x) \sqrt {c-i c \tan (e+f x)} \, dx}{a^3 c^3}\\ &=\frac {\left (i c^4\right ) \operatorname {Subst}\left (\int \frac {1}{(c-x)^4 (c+x)^{7/2}} \, dx,x,-i c \tan (e+f x)\right )}{a^3 f}\\ &=\frac {i}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{5/2}}+\frac {\left (11 i c^3\right ) \operatorname {Subst}\left (\int \frac {1}{(c-x)^3 (c+x)^{7/2}} \, dx,x,-i c \tan (e+f x)\right )}{12 a^3 f}\\ &=\frac {i}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{5/2}}+\frac {11 i}{48 a^3 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}}+\frac {\left (33 i c^2\right ) \operatorname {Subst}\left (\int \frac {1}{(c-x)^2 (c+x)^{7/2}} \, dx,x,-i c \tan (e+f x)\right )}{32 a^3 f}\\ &=\frac {i}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{5/2}}+\frac {11 i}{48 a^3 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}}+\frac {33 i}{64 a^3 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}+\frac {(231 i c) \operatorname {Subst}\left (\int \frac {1}{(c-x) (c+x)^{7/2}} \, dx,x,-i c \tan (e+f x)\right )}{128 a^3 f}\\ &=-\frac {231 i}{640 a^3 f (c-i c \tan (e+f x))^{5/2}}+\frac {i}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{5/2}}+\frac {11 i}{48 a^3 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}}+\frac {33 i}{64 a^3 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}+\frac {(231 i) \operatorname {Subst}\left (\int \frac {1}{(c-x) (c+x)^{5/2}} \, dx,x,-i c \tan (e+f x)\right )}{256 a^3 f}\\ &=-\frac {231 i}{640 a^3 f (c-i c \tan (e+f x))^{5/2}}+\frac {i}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{5/2}}+\frac {11 i}{48 a^3 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}}+\frac {33 i}{64 a^3 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}-\frac {77 i}{256 a^3 c f (c-i c \tan (e+f x))^{3/2}}+\frac {(231 i) \operatorname {Subst}\left (\int \frac {1}{(c-x) (c+x)^{3/2}} \, dx,x,-i c \tan (e+f x)\right )}{512 a^3 c f}\\ &=-\frac {231 i}{640 a^3 f (c-i c \tan (e+f x))^{5/2}}+\frac {i}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{5/2}}+\frac {11 i}{48 a^3 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}}+\frac {33 i}{64 a^3 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}-\frac {77 i}{256 a^3 c f (c-i c \tan (e+f x))^{3/2}}-\frac {231 i}{512 a^3 c^2 f \sqrt {c-i c \tan (e+f x)}}+\frac {(231 i) \operatorname {Subst}\left (\int \frac {1}{(c-x) \sqrt {c+x}} \, dx,x,-i c \tan (e+f x)\right )}{1024 a^3 c^2 f}\\ &=-\frac {231 i}{640 a^3 f (c-i c \tan (e+f x))^{5/2}}+\frac {i}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{5/2}}+\frac {11 i}{48 a^3 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}}+\frac {33 i}{64 a^3 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}-\frac {77 i}{256 a^3 c f (c-i c \tan (e+f x))^{3/2}}-\frac {231 i}{512 a^3 c^2 f \sqrt {c-i c \tan (e+f x)}}+\frac {(231 i) \operatorname {Subst}\left (\int \frac {1}{2 c-x^2} \, dx,x,\sqrt {c-i c \tan (e+f x)}\right )}{512 a^3 c^2 f}\\ &=\frac {231 i \tanh ^{-1}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{512 \sqrt {2} a^3 c^{5/2} f}-\frac {231 i}{640 a^3 f (c-i c \tan (e+f x))^{5/2}}+\frac {i}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{5/2}}+\frac {11 i}{48 a^3 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}}+\frac {33 i}{64 a^3 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}-\frac {77 i}{256 a^3 c f (c-i c \tan (e+f x))^{3/2}}-\frac {231 i}{512 a^3 c^2 f \sqrt {c-i c \tan (e+f x)}}\\ \end {align*}
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Mathematica [A] time = 6.36, size = 171, normalized size = 0.62 \[ -\frac {i e^{-6 i (e+f x)} \left (-350 e^{2 i (e+f x)}-1645 e^{4 i (e+f x)}+1433 e^{6 i (e+f x)}+3184 e^{8 i (e+f x)}+464 e^{10 i (e+f x)}+48 e^{12 i (e+f x)}-3465 e^{6 i (e+f x)} \sqrt {1+e^{2 i (e+f x)}} \tanh ^{-1}\left (\sqrt {1+e^{2 i (e+f x)}}\right )-40\right ) \sqrt {c-i c \tan (e+f x)}}{15360 a^3 c^3 f} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.51, size = 350, normalized size = 1.28 \[ \frac {{\left (3465 i \, \sqrt {\frac {1}{2}} a^{3} c^{3} f \sqrt {\frac {1}{a^{6} c^{5} f^{2}}} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (\frac {{\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (59136 i \, a^{3} c^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + 59136 i \, a^{3} c^{2} f\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {1}{a^{6} c^{5} f^{2}}} + 59136 i\right )} e^{\left (-i \, f x - i \, e\right )}}{65536 \, a^{3} c^{2} f}\right ) - 3465 i \, \sqrt {\frac {1}{2}} a^{3} c^{3} f \sqrt {\frac {1}{a^{6} c^{5} f^{2}}} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (\frac {{\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (-59136 i \, a^{3} c^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} - 59136 i \, a^{3} c^{2} f\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {1}{a^{6} c^{5} f^{2}}} + 59136 i\right )} e^{\left (-i \, f x - i \, e\right )}}{65536 \, a^{3} c^{2} f}\right ) + \sqrt {2} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} {\left (-48 i \, e^{\left (12 i \, f x + 12 i \, e\right )} - 464 i \, e^{\left (10 i \, f x + 10 i \, e\right )} - 3184 i \, e^{\left (8 i \, f x + 8 i \, e\right )} - 1433 i \, e^{\left (6 i \, f x + 6 i \, e\right )} + 1645 i \, e^{\left (4 i \, f x + 4 i \, e\right )} + 350 i \, e^{\left (2 i \, f x + 2 i \, e\right )} + 40 i\right )}\right )} e^{\left (-6 i \, f x - 6 i \, e\right )}}{15360 \, a^{3} c^{3} f} \]
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 \[ \int \frac {1}{{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{3} {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.38, size = 178, normalized size = 0.65 \[ \frac {2 i c^{4} \left (-\frac {\frac {\frac {71 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{32}-\frac {59 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}} c}{6}+\frac {89 c^{2} \sqrt {c -i c \tan \left (f x +e \right )}}{8}}{\left (-c -i c \tan \left (f x +e \right )\right )^{3}}-\frac {231 \sqrt {2}\, \arctanh \left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{64 \sqrt {c}}}{32 c^{6}}-\frac {5}{32 c^{6} \sqrt {c -i c \tan \left (f x +e \right )}}-\frac {1}{24 c^{5} \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}-\frac {1}{80 c^{4} \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}\right )}{f \,a^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.60, size = 244, normalized size = 0.89 \[ -\frac {i \, {\left (\frac {4 \, {\left (3465 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{5} - 18480 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{4} c + 30492 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{3} c^{2} - 12672 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{2} c^{3} - 2816 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )} c^{4} - 1536 \, c^{5}\right )}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {11}{2}} a^{3} c - 6 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {9}{2}} a^{3} c^{2} + 12 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {7}{2}} a^{3} c^{3} - 8 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}} a^{3} c^{4}} + \frac {3465 \, \sqrt {2} \log \left (-\frac {\sqrt {2} \sqrt {c} - \sqrt {-i \, c \tan \left (f x + e\right ) + c}}{\sqrt {2} \sqrt {c} + \sqrt {-i \, c \tan \left (f x + e\right ) + c}}\right )}{a^{3} c^{\frac {3}{2}}}\right )}}{30720 \, c f} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 5.14, size = 255, normalized size = 0.93 \[ -\frac {-\frac {{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^3\,2541{}\mathrm {i}}{640\,a^3\,f}+\frac {c^3\,1{}\mathrm {i}}{5\,a^3\,f}+\frac {{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^4\,77{}\mathrm {i}}{32\,a^3\,c\,f}-\frac {{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^5\,231{}\mathrm {i}}{512\,a^3\,c^2\,f}+\frac {c\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^2\,33{}\mathrm {i}}{20\,a^3\,f}+\frac {c^2\,\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )\,11{}\mathrm {i}}{30\,a^3\,f}}{6\,c\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{9/2}-{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{11/2}+8\,c^3\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{5/2}-12\,c^2\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{7/2}}-\frac {\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-c}}\right )\,231{}\mathrm {i}}{1024\,a^3\,{\left (-c\right )}^{5/2}\,f} \]
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 \[ \frac {i \int \frac {1}{- c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{5}{\left (e + f x \right )} + i c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{4}{\left (e + f x \right )} - 2 c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{3}{\left (e + f x \right )} + 2 i c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{2}{\left (e + f x \right )} - c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )} + i c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c}}\, dx}{a^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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