Optimal. Leaf size=273 \[ \frac {7 (-B+9 i A) \tanh ^{-1}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{128 \sqrt {2} a^2 c^{5/2} f}-\frac {7 (-B+9 i A)}{128 a^2 c^2 f \sqrt {c-i c \tan (e+f x)}}+\frac {-B+i A}{4 a^2 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}}-\frac {7 (-B+9 i A)}{192 a^2 c f (c-i c \tan (e+f x))^{3/2}}-\frac {7 (-B+9 i A)}{160 a^2 f (c-i c \tan (e+f x))^{5/2}}+\frac {-B+9 i A}{16 a^2 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}} \]
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Rubi [A] time = 0.32, antiderivative size = 273, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.116, Rules used = {3588, 78, 51, 63, 208} \[ -\frac {7 (-B+9 i A)}{128 a^2 c^2 f \sqrt {c-i c \tan (e+f x)}}+\frac {7 (-B+9 i A) \tanh ^{-1}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{128 \sqrt {2} a^2 c^{5/2} f}+\frac {-B+i A}{4 a^2 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}}-\frac {7 (-B+9 i A)}{192 a^2 c f (c-i c \tan (e+f x))^{3/2}}-\frac {7 (-B+9 i A)}{160 a^2 f (c-i c \tan (e+f x))^{5/2}}+\frac {-B+9 i A}{16 a^2 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}} \]
Antiderivative was successfully verified.
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Rule 51
Rule 63
Rule 78
Rule 208
Rule 3588
Rubi steps
\begin {align*} \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}} \, dx &=\frac {(a c) \operatorname {Subst}\left (\int \frac {A+B x}{(a+i a x)^3 (c-i c x)^{7/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {i A-B}{4 a^2 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}}+\frac {((9 A+i B) c) \operatorname {Subst}\left (\int \frac {1}{(a+i a x)^2 (c-i c x)^{7/2}} \, dx,x,\tan (e+f x)\right )}{8 f}\\ &=\frac {i A-B}{4 a^2 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}}+\frac {9 i A-B}{16 a^2 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}+\frac {(7 (9 A+i B) c) \operatorname {Subst}\left (\int \frac {1}{(a+i a x) (c-i c x)^{7/2}} \, dx,x,\tan (e+f x)\right )}{32 a f}\\ &=-\frac {7 (9 i A-B)}{160 a^2 f (c-i c \tan (e+f x))^{5/2}}+\frac {i A-B}{4 a^2 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}}+\frac {9 i A-B}{16 a^2 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}+\frac {(7 (9 A+i B)) \operatorname {Subst}\left (\int \frac {1}{(a+i a x) (c-i c x)^{5/2}} \, dx,x,\tan (e+f x)\right )}{64 a f}\\ &=-\frac {7 (9 i A-B)}{160 a^2 f (c-i c \tan (e+f x))^{5/2}}+\frac {i A-B}{4 a^2 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}}+\frac {9 i A-B}{16 a^2 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}-\frac {7 (9 i A-B)}{192 a^2 c f (c-i c \tan (e+f x))^{3/2}}+\frac {(7 (9 A+i B)) \operatorname {Subst}\left (\int \frac {1}{(a+i a x) (c-i c x)^{3/2}} \, dx,x,\tan (e+f x)\right )}{128 a c f}\\ &=-\frac {7 (9 i A-B)}{160 a^2 f (c-i c \tan (e+f x))^{5/2}}+\frac {i A-B}{4 a^2 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}}+\frac {9 i A-B}{16 a^2 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}-\frac {7 (9 i A-B)}{192 a^2 c f (c-i c \tan (e+f x))^{3/2}}-\frac {7 (9 i A-B)}{128 a^2 c^2 f \sqrt {c-i c \tan (e+f x)}}+\frac {(7 (9 A+i B)) \operatorname {Subst}\left (\int \frac {1}{(a+i a x) \sqrt {c-i c x}} \, dx,x,\tan (e+f x)\right )}{256 a c^2 f}\\ &=-\frac {7 (9 i A-B)}{160 a^2 f (c-i c \tan (e+f x))^{5/2}}+\frac {i A-B}{4 a^2 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}}+\frac {9 i A-B}{16 a^2 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}-\frac {7 (9 i A-B)}{192 a^2 c f (c-i c \tan (e+f x))^{3/2}}-\frac {7 (9 i A-B)}{128 a^2 c^2 f \sqrt {c-i c \tan (e+f x)}}+\frac {(7 (9 i A-B)) \operatorname {Subst}\left (\int \frac {1}{2 a-\frac {a x^2}{c}} \, dx,x,\sqrt {c-i c \tan (e+f x)}\right )}{128 a c^3 f}\\ &=\frac {7 (9 i A-B) \tanh ^{-1}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{128 \sqrt {2} a^2 c^{5/2} f}-\frac {7 (9 i A-B)}{160 a^2 f (c-i c \tan (e+f x))^{5/2}}+\frac {i A-B}{4 a^2 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}}+\frac {9 i A-B}{16 a^2 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}-\frac {7 (9 i A-B)}{192 a^2 c f (c-i c \tan (e+f x))^{3/2}}-\frac {7 (9 i A-B)}{128 a^2 c^2 f \sqrt {c-i c \tan (e+f x)}}\\ \end {align*}
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Mathematica [A] time = 9.85, size = 209, normalized size = 0.77 \[ \frac {\sqrt {c-i c \tan (e+f x)} (\cos (e+f x)+i \sin (e+f x)) \left (105 i (9 A+i B) e^{-i (e+f x)} \sqrt {1+e^{2 i (e+f x)}} \tanh ^{-1}\left (\sqrt {1+e^{2 i (e+f x)}}\right )+2 \cos (e+f x) ((-223 B+87 i A) \cos (2 (e+f x))+6 i (A+9 i B) \cos (4 (e+f x))+423 A \sin (2 (e+f x))+54 A \sin (4 (e+f x))-864 i A+47 i B \sin (2 (e+f x))+6 i B \sin (4 (e+f x))-64 B)\right )}{3840 a^2 c^3 f} \]
Antiderivative was successfully verified.
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fricas [B] time = 1.19, size = 444, normalized size = 1.63 \[ \frac {{\left (15 \, \sqrt {\frac {1}{2}} a^{2} c^{3} f \sqrt {-\frac {3969 \, A^{2} + 882 i \, A B - 49 \, B^{2}}{a^{4} c^{5} f^{2}}} e^{\left (4 i \, f x + 4 i \, e\right )} \log \left (\frac {{\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{2} c^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{2} c^{2} f\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {3969 \, A^{2} + 882 i \, A B - 49 \, B^{2}}{a^{4} c^{5} f^{2}}} + 63 i \, A - 7 \, B\right )} e^{\left (-i \, f x - i \, e\right )}}{64 \, a^{2} c^{2} f}\right ) - 15 \, \sqrt {\frac {1}{2}} a^{2} c^{3} f \sqrt {-\frac {3969 \, A^{2} + 882 i \, A B - 49 \, B^{2}}{a^{4} c^{5} f^{2}}} e^{\left (4 i \, f x + 4 i \, e\right )} \log \left (-\frac {{\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{2} c^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{2} c^{2} f\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {3969 \, A^{2} + 882 i \, A B - 49 \, B^{2}}{a^{4} c^{5} f^{2}}} - 63 i \, A + 7 \, B\right )} e^{\left (-i \, f x - i \, e\right )}}{64 \, a^{2} c^{2} f}\right ) + \sqrt {2} {\left ({\left (-24 i \, A - 24 \, B\right )} e^{\left (10 i \, f x + 10 i \, e\right )} + {\left (-192 i \, A - 112 \, B\right )} e^{\left (8 i \, f x + 8 i \, e\right )} + {\left (-1032 i \, A - 152 \, B\right )} e^{\left (6 i \, f x + 6 i \, e\right )} + {\left (-609 i \, A - 199 \, B\right )} e^{\left (4 i \, f x + 4 i \, e\right )} + {\left (285 i \, A - 165 \, B\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + 30 i \, A - 30 \, B\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-4 i \, f x - 4 i \, e\right )}}{3840 \, a^{2} c^{3} f} \]
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 \[ \int \frac {B \tan \left (f x + e\right ) + A}{{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{2} {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.44, size = 199, normalized size = 0.73 \[ -\frac {2 i c^{2} \left (\frac {\frac {\left (\frac {7 i B}{16}+\frac {15 A}{16}\right ) \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}+\left (-\frac {9}{8} i B c -\frac {17}{8} c A \right ) \sqrt {c -i c \tan \left (f x +e \right )}}{\left (-c -i c \tan \left (f x +e \right )\right )^{2}}-\frac {7 \left (i B +9 A \right ) \sqrt {2}\, \arctanh \left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{32 \sqrt {c}}}{16 c^{4}}+\frac {3 A}{16 c^{4} \sqrt {c -i c \tan \left (f x +e \right )}}-\frac {i B -3 A}{48 c^{3} \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}-\frac {i B -A}{40 c^{2} \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}\right )}{f \,a^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.97, size = 245, normalized size = 0.90 \[ -\frac {i \, {\left (\frac {4 \, {\left (105 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{4} {\left (9 \, A + i \, B\right )} - 350 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{3} {\left (9 \, A + i \, B\right )} c + 224 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{2} {\left (9 \, A + i \, B\right )} c^{2} + 64 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )} {\left (9 \, A + i \, B\right )} c^{3} + 384 \, {\left (A - i \, B\right )} c^{4}\right )}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {9}{2}} a^{2} c - 4 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {7}{2}} a^{2} c^{2} + 4 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}} a^{2} c^{3}} + \frac {105 \, \sqrt {2} {\left (9 \, A + i \, B\right )} \log \left (-\frac {\sqrt {2} \sqrt {c} - \sqrt {-i \, c \tan \left (f x + e\right ) + c}}{\sqrt {2} \sqrt {c} + \sqrt {-i \, c \tan \left (f x + e\right ) + c}}\right )}{a^{2} c^{\frac {3}{2}}}\right )}}{7680 \, c f} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 10.14, size = 399, normalized size = 1.46 \[ \frac {-\frac {B\,c^2}{5}+\frac {7\,B\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^2}{60}-\frac {35\,B\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^3}{192\,c}+\frac {7\,B\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^4}{128\,c^2}+\frac {B\,c\,\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}{30}}{a^2\,f\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{9/2}-4\,a^2\,c\,f\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{7/2}+4\,a^2\,c^2\,f\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}-\frac {\frac {A\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^2\,21{}\mathrm {i}}{20\,a^2\,f}+\frac {A\,c^2\,1{}\mathrm {i}}{5\,a^2\,f}-\frac {A\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^3\,105{}\mathrm {i}}{64\,a^2\,c\,f}+\frac {A\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^4\,63{}\mathrm {i}}{128\,a^2\,c^2\,f}+\frac {A\,c\,\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )\,3{}\mathrm {i}}{10\,a^2\,f}}{-4\,c\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{7/2}+{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{9/2}+4\,c^2\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}-\frac {\sqrt {2}\,A\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-c}}\right )\,63{}\mathrm {i}}{256\,a^2\,{\left (-c\right )}^{5/2}\,f}-\frac {7\,\sqrt {2}\,B\,\mathrm {atanh}\left (\frac {\sqrt {2}\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {c}}\right )}{256\,a^2\,c^{5/2}\,f} \]
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 \[ - \frac {\int \frac {A}{- c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{4}{\left (e + f x \right )} - 2 c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{2}{\left (e + f x \right )} - c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c}}\, dx + \int \frac {B \tan {\left (e + f x \right )}}{- c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{4}{\left (e + f x \right )} - 2 c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{2}{\left (e + f x \right )} - c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c}}\, dx}{a^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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