Integrand size = 45, antiderivative size = 218 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{3/2}} \, dx=\frac {-i A-B}{3 f (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{3/2}}+\frac {4 i A+B}{15 c f (a+i a \tan (e+f x))^{5/2} \sqrt {c-i c \tan (e+f x)}}+\frac {4 i A+B}{15 a c f (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}+\frac {2 (4 A-i B) \tan (e+f x)}{15 a^2 c f \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}} \]
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Time = 0.35 (sec) , antiderivative size = 216, normalized size of antiderivative = 0.99, 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, 39} \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{3/2}} \, dx=\frac {2 (4 A-i B) \tan (e+f x)}{15 a^2 c f \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}-\frac {B+i A}{3 f (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{3/2}}+\frac {B+4 i A}{15 a c f (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}+\frac {B+4 i A}{15 c f (a+i a \tan (e+f x))^{5/2} \sqrt {c-i c \tan (e+f x)}} \]
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Rule 39
Rule 47
Rule 79
Rule 3669
Rubi steps \begin{align*} \text {integral}& = \frac {(a c) \text {Subst}\left (\int \frac {A+B x}{(a+i a x)^{7/2} (c-i c x)^{5/2}} \, dx,x,\tan (e+f x)\right )}{f} \\ & = -\frac {i A+B}{3 f (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{3/2}}+\frac {(a (4 A-i B)) \text {Subst}\left (\int \frac {1}{(a+i a x)^{7/2} (c-i c x)^{3/2}} \, dx,x,\tan (e+f x)\right )}{3 f} \\ & = -\frac {i A+B}{3 f (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{3/2}}+\frac {4 i A+B}{15 c f (a+i a \tan (e+f x))^{5/2} \sqrt {c-i c \tan (e+f x)}}+\frac {(4 A-i B) \text {Subst}\left (\int \frac {1}{(a+i a x)^{5/2} (c-i c x)^{3/2}} \, dx,x,\tan (e+f x)\right )}{5 f} \\ & = -\frac {i A+B}{3 f (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{3/2}}+\frac {4 i A+B}{15 c f (a+i a \tan (e+f x))^{5/2} \sqrt {c-i c \tan (e+f x)}}+\frac {4 i A+B}{15 a c f (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}+\frac {(2 (4 A-i B)) \text {Subst}\left (\int \frac {1}{(a+i a x)^{3/2} (c-i c x)^{3/2}} \, dx,x,\tan (e+f x)\right )}{15 a f} \\ & = -\frac {i A+B}{3 f (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{3/2}}+\frac {4 i A+B}{15 c f (a+i a \tan (e+f x))^{5/2} \sqrt {c-i c \tan (e+f x)}}+\frac {4 i A+B}{15 a c f (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}+\frac {2 (4 A-i B) \tan (e+f x)}{15 a^2 c f \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}} \\ \end{align*}
Time = 6.66 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.70 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{3/2}} \, dx=\frac {3 (A+i B)+(-12 i A-3 B) \tan (e+f x)+3 (4 A-i B) \tan ^2(e+f x)+(-8 i A-2 B) \tan ^3(e+f x)+(8 A-2 i B) \tan ^4(e+f x)}{15 a^2 c f (-i+\tan (e+f x))^2 (i+\tan (e+f x)) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}} \]
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Time = 0.39 (sec) , antiderivative size = 199, normalized size of antiderivative = 0.91
method | result | size |
derivativedivides | \(-\frac {i \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \left (8 i A \tan \left (f x +e \right )^{6}-2 i B \tan \left (f x +e \right )^{5}+2 B \tan \left (f x +e \right )^{6}+20 i A \tan \left (f x +e \right )^{4}+8 A \tan \left (f x +e \right )^{5}-5 i B \tan \left (f x +e \right )^{3}+5 B \tan \left (f x +e \right )^{4}+15 i A \tan \left (f x +e \right )^{2}+20 A \tan \left (f x +e \right )^{3}-3 i \tan \left (f x +e \right ) B +3 i A +12 A \tan \left (f x +e \right )-3 B \right )}{15 f \,a^{3} c^{2} \left (i-\tan \left (f x +e \right )\right )^{4} \left (i+\tan \left (f x +e \right )\right )^{3}}\) | \(199\) |
default | \(-\frac {i \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \left (8 i A \tan \left (f x +e \right )^{6}-2 i B \tan \left (f x +e \right )^{5}+2 B \tan \left (f x +e \right )^{6}+20 i A \tan \left (f x +e \right )^{4}+8 A \tan \left (f x +e \right )^{5}-5 i B \tan \left (f x +e \right )^{3}+5 B \tan \left (f x +e \right )^{4}+15 i A \tan \left (f x +e \right )^{2}+20 A \tan \left (f x +e \right )^{3}-3 i \tan \left (f x +e \right ) B +3 i A +12 A \tan \left (f x +e \right )-3 B \right )}{15 f \,a^{3} c^{2} \left (i-\tan \left (f x +e \right )\right )^{4} \left (i+\tan \left (f x +e \right )\right )^{3}}\) | \(199\) |
parts | \(-\frac {A \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \left (8 i \tan \left (f x +e \right )^{5}-8 \tan \left (f x +e \right )^{6}+20 i \tan \left (f x +e \right )^{3}-20 \tan \left (f x +e \right )^{4}+12 i \tan \left (f x +e \right )-15 \tan \left (f x +e \right )^{2}-3\right )}{15 f \,a^{3} c^{2} \left (i+\tan \left (f x +e \right )\right )^{3} \left (i-\tan \left (f x +e \right )\right )^{4}}+\frac {i B \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \left (2 i \tan \left (f x +e \right )^{5}-2 \tan \left (f x +e \right )^{6}+5 i \tan \left (f x +e \right )^{3}-5 \tan \left (f x +e \right )^{4}+3 i \tan \left (f x +e \right )+3\right )}{15 f \,a^{3} c^{2} \left (i+\tan \left (f x +e \right )\right )^{3} \left (i-\tan \left (f x +e \right )\right )^{4}}\) | \(253\) |
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Time = 0.25 (sec) , antiderivative size = 181, normalized size of antiderivative = 0.83 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{3/2}} \, dx=-\frac {{\left (5 \, {\left (i \, A + B\right )} e^{\left (10 i \, f x + 10 i \, e\right )} + 5 \, {\left (13 i \, A + 7 \, B\right )} e^{\left (8 i \, f x + 8 i \, e\right )} + 48 \, {\left (i \, A - B\right )} e^{\left (7 i \, f x + 7 i \, e\right )} + 30 \, {\left (-i \, A + B\right )} e^{\left (6 i \, f x + 6 i \, e\right )} + 48 \, {\left (i \, A - B\right )} e^{\left (5 i \, f x + 5 i \, e\right )} + 10 \, {\left (-11 i \, A + B\right )} e^{\left (4 i \, f x + 4 i \, e\right )} - {\left (23 i \, A - 13 \, B\right )} e^{\left (2 i \, f x + 2 i \, e\right )} - 3 i \, A + 3 \, B\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} e^{\left (-5 i \, f x - 5 i \, e\right )}}{240 \, a^{3} c^{2} f} \]
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\[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{3/2}} \, dx=\int \frac {A + B \tan {\left (e + f x \right )}}{\left (i a \left (\tan {\left (e + f x \right )} - i\right )\right )^{\frac {5}{2}} \left (- i c \left (\tan {\left (e + f x \right )} + i\right )\right )^{\frac {3}{2}}}\, dx \]
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Exception generated. \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{3/2}} \, dx=\text {Exception raised: RuntimeError} \]
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\[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{3/2}} \, dx=\int { \frac {B \tan \left (f x + e\right ) + A}{{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{\frac {5}{2}} {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}}} \,d x } \]
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Time = 9.97 (sec) , antiderivative size = 249, normalized size of antiderivative = 1.14 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{3/2}} \, dx=\frac {\sqrt {\frac {a\,\left (\cos \left (2\,e+2\,f\,x\right )+1+\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}\,\left (95\,A\,\sin \left (2\,e+2\,f\,x\right )-30\,B+A\,\cos \left (2\,e+2\,f\,x\right )\,85{}\mathrm {i}+A\,\cos \left (4\,e+4\,f\,x\right )\,20{}\mathrm {i}+A\,\cos \left (6\,e+6\,f\,x\right )\,3{}\mathrm {i}-5\,B\,\cos \left (2\,e+2\,f\,x\right )-10\,B\,\cos \left (4\,e+4\,f\,x\right )-3\,B\,\cos \left (6\,e+6\,f\,x\right )-A\,60{}\mathrm {i}+20\,A\,\sin \left (4\,e+4\,f\,x\right )+3\,A\,\sin \left (6\,e+6\,f\,x\right )-B\,\sin \left (2\,e+2\,f\,x\right )\,5{}\mathrm {i}+B\,\sin \left (4\,e+4\,f\,x\right )\,10{}\mathrm {i}+B\,\sin \left (6\,e+6\,f\,x\right )\,3{}\mathrm {i}\right )}{240\,a^3\,c\,f\,\sqrt {\frac {c\,\left (\cos \left (2\,e+2\,f\,x\right )+1-\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}} \]
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