Integrand size = 45, antiderivative size = 208 \[ \int \frac {(a+i a \tan (e+f x))^{3/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{9/2}} \, dx=-\frac {(i A+B) (a+i a \tan (e+f x))^{3/2}}{9 f (c-i c \tan (e+f x))^{9/2}}-\frac {(i A-2 B) (a+i a \tan (e+f x))^{3/2}}{21 c f (c-i c \tan (e+f x))^{7/2}}-\frac {2 (i A-2 B) (a+i a \tan (e+f x))^{3/2}}{105 c^2 f (c-i c \tan (e+f x))^{5/2}}-\frac {2 (i A-2 B) (a+i a \tan (e+f x))^{3/2}}{315 c^3 f (c-i c \tan (e+f x))^{3/2}} \]
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Time = 0.33 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.089, Rules used = {3669, 79, 47, 37} \[ \int \frac {(a+i a \tan (e+f x))^{3/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{9/2}} \, dx=-\frac {2 (-2 B+i A) (a+i a \tan (e+f x))^{3/2}}{315 c^3 f (c-i c \tan (e+f x))^{3/2}}-\frac {2 (-2 B+i A) (a+i a \tan (e+f x))^{3/2}}{105 c^2 f (c-i c \tan (e+f x))^{5/2}}-\frac {(-2 B+i A) (a+i a \tan (e+f x))^{3/2}}{21 c f (c-i c \tan (e+f x))^{7/2}}-\frac {(B+i A) (a+i a \tan (e+f x))^{3/2}}{9 f (c-i c \tan (e+f x))^{9/2}} \]
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Rule 37
Rule 47
Rule 79
Rule 3669
Rubi steps \begin{align*} \text {integral}& = \frac {(a c) \text {Subst}\left (\int \frac {\sqrt {a+i a x} (A+B x)}{(c-i c x)^{11/2}} \, dx,x,\tan (e+f x)\right )}{f} \\ & = -\frac {(i A+B) (a+i a \tan (e+f x))^{3/2}}{9 f (c-i c \tan (e+f x))^{9/2}}+\frac {(a (A+2 i B)) \text {Subst}\left (\int \frac {\sqrt {a+i a x}}{(c-i c x)^{9/2}} \, dx,x,\tan (e+f x)\right )}{3 f} \\ & = -\frac {(i A+B) (a+i a \tan (e+f x))^{3/2}}{9 f (c-i c \tan (e+f x))^{9/2}}-\frac {(i A-2 B) (a+i a \tan (e+f x))^{3/2}}{21 c f (c-i c \tan (e+f x))^{7/2}}+\frac {(2 a (A+2 i B)) \text {Subst}\left (\int \frac {\sqrt {a+i a x}}{(c-i c x)^{7/2}} \, dx,x,\tan (e+f x)\right )}{21 c f} \\ & = -\frac {(i A+B) (a+i a \tan (e+f x))^{3/2}}{9 f (c-i c \tan (e+f x))^{9/2}}-\frac {(i A-2 B) (a+i a \tan (e+f x))^{3/2}}{21 c f (c-i c \tan (e+f x))^{7/2}}-\frac {2 (i A-2 B) (a+i a \tan (e+f x))^{3/2}}{105 c^2 f (c-i c \tan (e+f x))^{5/2}}+\frac {(2 a (A+2 i B)) \text {Subst}\left (\int \frac {\sqrt {a+i a x}}{(c-i c x)^{5/2}} \, dx,x,\tan (e+f x)\right )}{105 c^2 f} \\ & = -\frac {(i A+B) (a+i a \tan (e+f x))^{3/2}}{9 f (c-i c \tan (e+f x))^{9/2}}-\frac {(i A-2 B) (a+i a \tan (e+f x))^{3/2}}{21 c f (c-i c \tan (e+f x))^{7/2}}-\frac {2 (i A-2 B) (a+i a \tan (e+f x))^{3/2}}{105 c^2 f (c-i c \tan (e+f x))^{5/2}}-\frac {2 (i A-2 B) (a+i a \tan (e+f x))^{3/2}}{315 c^3 f (c-i c \tan (e+f x))^{3/2}} \\ \end{align*}
Time = 7.06 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.63 \[ \int \frac {(a+i a \tan (e+f x))^{3/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{9/2}} \, dx=-\frac {a^2 (-i+\tan (e+f x))^2 \left (-58 i A+11 B-33 (A+2 i B) \tan (e+f x)+12 i (A+2 i B) \tan ^2(e+f x)+2 (A+2 i B) \tan ^3(e+f x)\right )}{315 c^4 f (i+\tan (e+f x))^4 \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}} \]
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Time = 0.39 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.65
method | result | size |
derivativedivides | \(\frac {i \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, a \left (1+\tan \left (f x +e \right )^{2}\right ) \left (2 i A \tan \left (f x +e \right )^{3}-24 i B \tan \left (f x +e \right )^{2}-4 B \tan \left (f x +e \right )^{3}-33 i A \tan \left (f x +e \right )-12 A \tan \left (f x +e \right )^{2}+11 i B +66 B \tan \left (f x +e \right )+58 A \right )}{315 f \,c^{5} \left (i+\tan \left (f x +e \right )\right )^{6}}\) | \(136\) |
default | \(\frac {i \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, a \left (1+\tan \left (f x +e \right )^{2}\right ) \left (2 i A \tan \left (f x +e \right )^{3}-24 i B \tan \left (f x +e \right )^{2}-4 B \tan \left (f x +e \right )^{3}-33 i A \tan \left (f x +e \right )-12 A \tan \left (f x +e \right )^{2}+11 i B +66 B \tan \left (f x +e \right )+58 A \right )}{315 f \,c^{5} \left (i+\tan \left (f x +e \right )\right )^{6}}\) | \(136\) |
risch | \(-\frac {a \sqrt {\frac {a \,{\mathrm e}^{2 i \left (f x +e \right )}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, \left (35 i A \,{\mathrm e}^{8 i \left (f x +e \right )}+35 B \,{\mathrm e}^{8 i \left (f x +e \right )}+135 i A \,{\mathrm e}^{6 i \left (f x +e \right )}+45 B \,{\mathrm e}^{6 i \left (f x +e \right )}+189 i A \,{\mathrm e}^{4 i \left (f x +e \right )}-63 B \,{\mathrm e}^{4 i \left (f x +e \right )}+105 i A \,{\mathrm e}^{2 i \left (f x +e \right )}-105 B \,{\mathrm e}^{2 i \left (f x +e \right )}\right )}{2520 c^{4} \sqrt {\frac {c}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, f}\) | \(154\) |
parts | \(\frac {i A \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, a \left (1+\tan \left (f x +e \right )^{2}\right ) \left (58-33 i \tan \left (f x +e \right )-12 \tan \left (f x +e \right )^{2}+2 i \tan \left (f x +e \right )^{3}\right )}{315 f \,c^{5} \left (i+\tan \left (f x +e \right )\right )^{6}}-\frac {i B \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, a \left (1+\tan \left (f x +e \right )^{2}\right ) \left (-11 i-66 \tan \left (f x +e \right )+24 i \tan \left (f x +e \right )^{2}+4 \tan \left (f x +e \right )^{3}\right )}{315 f \,c^{5} \left (i+\tan \left (f x +e \right )\right )^{6}}\) | \(192\) |
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Time = 0.25 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.65 \[ \int \frac {(a+i a \tan (e+f x))^{3/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{9/2}} \, dx=-\frac {{\left (35 \, {\left (i \, A + B\right )} a e^{\left (11 i \, f x + 11 i \, e\right )} + 10 \, {\left (17 i \, A + 8 \, B\right )} a e^{\left (9 i \, f x + 9 i \, e\right )} + 18 \, {\left (18 i \, A - B\right )} a e^{\left (7 i \, f x + 7 i \, e\right )} + 42 \, {\left (7 i \, A - 4 \, B\right )} a e^{\left (5 i \, f x + 5 i \, e\right )} + 105 \, {\left (i \, A - B\right )} a e^{\left (3 i \, f x + 3 i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{2520 \, c^{5} f} \]
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Timed out. \[ \int \frac {(a+i a \tan (e+f x))^{3/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{9/2}} \, dx=\text {Timed out} \]
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Time = 0.43 (sec) , antiderivative size = 260, normalized size of antiderivative = 1.25 \[ \int \frac {(a+i a \tan (e+f x))^{3/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{9/2}} \, dx=\frac {{\left (35 \, {\left (-i \, A - B\right )} a \cos \left (\frac {9}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + 45 \, {\left (-3 i \, A - B\right )} a \cos \left (\frac {7}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + 63 \, {\left (-3 i \, A + B\right )} a \cos \left (\frac {5}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + 105 \, {\left (-i \, A + B\right )} a \cos \left (\frac {3}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + 35 \, {\left (A - i \, B\right )} a \sin \left (\frac {9}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + 45 \, {\left (3 \, A - i \, B\right )} a \sin \left (\frac {7}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + 63 \, {\left (3 \, A + i \, B\right )} a \sin \left (\frac {5}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + 105 \, {\left (A + i \, B\right )} a \sin \left (\frac {3}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right )\right )} \sqrt {a}}{2520 \, c^{\frac {9}{2}} f} \]
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\[ \int \frac {(a+i a \tan (e+f x))^{3/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{9/2}} \, dx=\int { \frac {{\left (B \tan \left (f x + e\right ) + A\right )} {\left (i \, a \tan \left (f x + e\right ) + a\right )}^{\frac {3}{2}}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {9}{2}}} \,d x } \]
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Time = 11.05 (sec) , antiderivative size = 290, normalized size of antiderivative = 1.39 \[ \int \frac {(a+i a \tan (e+f x))^{3/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{9/2}} \, dx=-\frac {a\,\sqrt {\frac {a\,\left (\cos \left (2\,e+2\,f\,x\right )+1+\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}\,\left (A\,\cos \left (2\,e+2\,f\,x\right )\,105{}\mathrm {i}+A\,\cos \left (4\,e+4\,f\,x\right )\,189{}\mathrm {i}+A\,\cos \left (6\,e+6\,f\,x\right )\,135{}\mathrm {i}+A\,\cos \left (8\,e+8\,f\,x\right )\,35{}\mathrm {i}-105\,B\,\cos \left (2\,e+2\,f\,x\right )-63\,B\,\cos \left (4\,e+4\,f\,x\right )+45\,B\,\cos \left (6\,e+6\,f\,x\right )+35\,B\,\cos \left (8\,e+8\,f\,x\right )-105\,A\,\sin \left (2\,e+2\,f\,x\right )-189\,A\,\sin \left (4\,e+4\,f\,x\right )-135\,A\,\sin \left (6\,e+6\,f\,x\right )-35\,A\,\sin \left (8\,e+8\,f\,x\right )-B\,\sin \left (2\,e+2\,f\,x\right )\,105{}\mathrm {i}-B\,\sin \left (4\,e+4\,f\,x\right )\,63{}\mathrm {i}+B\,\sin \left (6\,e+6\,f\,x\right )\,45{}\mathrm {i}+B\,\sin \left (8\,e+8\,f\,x\right )\,35{}\mathrm {i}\right )}{2520\,c^4\,f\,\sqrt {\frac {c\,\left (\cos \left (2\,e+2\,f\,x\right )+1-\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}} \]
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