Optimal. Leaf size=188 \[ \frac {7 i \tanh ^{-1}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{16 \sqrt {2} a c^{5/2} f}-\frac {7 i}{16 a c^2 f \sqrt {c-i c \tan (e+f x)}}-\frac {7 i}{24 a c f (c-i c \tan (e+f x))^{3/2}}-\frac {7 i}{20 a f (c-i c \tan (e+f x))^{5/2}}+\frac {i}{2 a f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}} \]
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Rubi [A] time = 0.22, antiderivative size = 188, normalized size of antiderivative = 1.00, number of steps used = 8, 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} \[ -\frac {7 i}{16 a c^2 f \sqrt {c-i c \tan (e+f x)}}+\frac {7 i \tanh ^{-1}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{16 \sqrt {2} a c^{5/2} f}-\frac {7 i}{24 a c f (c-i c \tan (e+f x))^{3/2}}-\frac {7 i}{20 a f (c-i c \tan (e+f x))^{5/2}}+\frac {i}{2 a f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}} \]
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)) (c-i c \tan (e+f x))^{5/2}} \, dx &=\frac {\int \frac {\cos ^2(e+f x)}{(c-i c \tan (e+f x))^{3/2}} \, dx}{a c}\\ &=\frac {\left (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 )}{a f}\\ &=\frac {i}{2 a f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}+\frac {(7 i c) \operatorname {Subst}\left (\int \frac {1}{(c-x) (c+x)^{7/2}} \, dx,x,-i c \tan (e+f x)\right )}{4 a f}\\ &=-\frac {7 i}{20 a f (c-i c \tan (e+f x))^{5/2}}+\frac {i}{2 a f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}+\frac {(7 i) \operatorname {Subst}\left (\int \frac {1}{(c-x) (c+x)^{5/2}} \, dx,x,-i c \tan (e+f x)\right )}{8 a f}\\ &=-\frac {7 i}{20 a f (c-i c \tan (e+f x))^{5/2}}+\frac {i}{2 a f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}-\frac {7 i}{24 a c f (c-i c \tan (e+f x))^{3/2}}+\frac {(7 i) \operatorname {Subst}\left (\int \frac {1}{(c-x) (c+x)^{3/2}} \, dx,x,-i c \tan (e+f x)\right )}{16 a c f}\\ &=-\frac {7 i}{20 a f (c-i c \tan (e+f x))^{5/2}}+\frac {i}{2 a f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}-\frac {7 i}{24 a c f (c-i c \tan (e+f x))^{3/2}}-\frac {7 i}{16 a c^2 f \sqrt {c-i c \tan (e+f x)}}+\frac {(7 i) \operatorname {Subst}\left (\int \frac {1}{(c-x) \sqrt {c+x}} \, dx,x,-i c \tan (e+f x)\right )}{32 a c^2 f}\\ &=-\frac {7 i}{20 a f (c-i c \tan (e+f x))^{5/2}}+\frac {i}{2 a f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}-\frac {7 i}{24 a c f (c-i c \tan (e+f x))^{3/2}}-\frac {7 i}{16 a c^2 f \sqrt {c-i c \tan (e+f x)}}+\frac {(7 i) \operatorname {Subst}\left (\int \frac {1}{2 c-x^2} \, dx,x,\sqrt {c-i c \tan (e+f x)}\right )}{16 a c^2 f}\\ &=\frac {7 i \tanh ^{-1}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{16 \sqrt {2} a c^{5/2} f}-\frac {7 i}{20 a f (c-i c \tan (e+f x))^{5/2}}+\frac {i}{2 a f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}-\frac {7 i}{24 a c f (c-i c \tan (e+f x))^{3/2}}-\frac {7 i}{16 a c^2 f \sqrt {c-i c \tan (e+f x)}}\\ \end {align*}
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Mathematica [A] time = 3.99, size = 149, normalized size = 0.79 \[ \frac {\sqrt {c-i c \tan (e+f x)} (\cos (2 (e+f x))+i \sin (2 (e+f x))) \left (-63 \sin (2 (e+f x))+21 \sin (4 (e+f x))-139 i \cos (2 (e+f x))+9 i \cos (4 (e+f x))+105 i e^{-2 i (e+f x)} \sqrt {1+e^{2 i (e+f x)}} \tanh ^{-1}\left (\sqrt {1+e^{2 i (e+f x)}}\right )-148 i\right )}{480 a c^3 f} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.48, size = 316, normalized size = 1.68 \[ \frac {{\left (105 i \, \sqrt {\frac {1}{2}} a c^{3} f \sqrt {\frac {1}{a^{2} c^{5} f^{2}}} e^{\left (2 i \, f x + 2 i \, e\right )} \log \left (\frac {{\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (56 i \, a c^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + 56 i \, a c^{2} f\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {1}{a^{2} c^{5} f^{2}}} + 56 i\right )} e^{\left (-i \, f x - i \, e\right )}}{64 \, a c^{2} f}\right ) - 105 i \, \sqrt {\frac {1}{2}} a c^{3} f \sqrt {\frac {1}{a^{2} c^{5} f^{2}}} e^{\left (2 i \, f x + 2 i \, e\right )} \log \left (\frac {{\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (-56 i \, a c^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} - 56 i \, a c^{2} f\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {1}{a^{2} c^{5} f^{2}}} + 56 i\right )} e^{\left (-i \, f x - i \, e\right )}}{64 \, a c^{2} f}\right ) + \sqrt {2} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} {\left (-6 i \, e^{\left (8 i \, f x + 8 i \, e\right )} - 38 i \, e^{\left (6 i \, f x + 6 i \, e\right )} - 148 i \, e^{\left (4 i \, f x + 4 i \, e\right )} - 101 i \, e^{\left (2 i \, f x + 2 i \, e\right )} + 15 i\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{480 \, a c^{3} f} \]
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 \[ \int \frac {1}{{\left (i \, a \tan \left (f x + e\right ) + a\right )} {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.28, size = 140, normalized size = 0.74 \[ \frac {2 i c^{2} \left (-\frac {\frac {\sqrt {c -i c \tan \left (f x +e \right )}}{-2 c -2 i c \tan \left (f x +e \right )}-\frac {7 \sqrt {2}\, \arctanh \left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{4 \sqrt {c}}}{16 c^{4}}-\frac {3}{16 c^{4} \sqrt {c -i c \tan \left (f x +e \right )}}-\frac {1}{12 c^{3} \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}-\frac {1}{20 c^{2} \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}\right )}{f a} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.83, size = 162, normalized size = 0.86 \[ -\frac {i \, {\left (\frac {4 \, {\left (105 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{3} - 140 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{2} c - 56 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )} c^{2} - 48 \, c^{3}\right )}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {7}{2}} a c - 2 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}} a c^{2}} + \frac {105 \, \sqrt {2} \log \left (-\frac {\sqrt {2} \sqrt {c} - \sqrt {-i \, c \tan \left (f x + e\right ) + c}}{\sqrt {2} \sqrt {c} + \sqrt {-i \, c \tan \left (f x + e\right ) + c}}\right )}{a c^{\frac {3}{2}}}\right )}}{960 \, c f} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.64, size = 165, normalized size = 0.88 \[ -\frac {\frac {c\,1{}\mathrm {i}}{5\,a\,f}+\frac {\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )\,7{}\mathrm {i}}{30\,a\,f}+\frac {{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^2\,7{}\mathrm {i}}{12\,a\,c\,f}-\frac {{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^3\,7{}\mathrm {i}}{16\,a\,c^2\,f}}{2\,c\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{5/2}-{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{7/2}}-\frac {\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-c}}\right )\,7{}\mathrm {i}}{32\,a\,{\left (-c\right )}^{5/2}\,f} \]
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 \[ - \frac {i \int \frac {1}{- c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{3}{\left (e + f x \right )} - i c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{2}{\left (e + f x \right )} - c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )} - i c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c}}\, dx}{a} \]
Verification of antiderivative is not currently implemented for this CAS.
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