Integrand size = 45, antiderivative size = 212 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^{5/2} \sqrt {c-i c \tan (e+f x)}} \, dx=-\frac {i A+B}{f (a+i a \tan (e+f x))^{5/2} \sqrt {c-i c \tan (e+f x)}}+\frac {(3 i A+2 B) \sqrt {c-i c \tan (e+f x)}}{5 c f (a+i a \tan (e+f x))^{5/2}}+\frac {2 (3 i A+2 B) \sqrt {c-i c \tan (e+f x)}}{15 a c f (a+i a \tan (e+f x))^{3/2}}+\frac {2 (3 i A+2 B) \sqrt {c-i c \tan (e+f x)}}{15 a^2 c f \sqrt {a+i a \tan (e+f x)}} \]
[Out]
Time = 0.35 (sec) , antiderivative size = 212, 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} \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^{5/2} \sqrt {c-i c \tan (e+f x)}} \, dx=\frac {2 (2 B+3 i A) \sqrt {c-i c \tan (e+f x)}}{15 a^2 c f \sqrt {a+i a \tan (e+f x)}}-\frac {B+i A}{f (a+i a \tan (e+f x))^{5/2} \sqrt {c-i c \tan (e+f x)}}+\frac {2 (2 B+3 i A) \sqrt {c-i c \tan (e+f x)}}{15 a c f (a+i a \tan (e+f x))^{3/2}}+\frac {(2 B+3 i A) \sqrt {c-i c \tan (e+f x)}}{5 c f (a+i a \tan (e+f x))^{5/2}} \]
[In]
[Out]
Rule 37
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)^{3/2}} \, dx,x,\tan (e+f x)\right )}{f} \\ & = -\frac {i A+B}{f (a+i a \tan (e+f x))^{5/2} \sqrt {c-i c \tan (e+f x)}}+\frac {(a (3 A-2 i B)) \text {Subst}\left (\int \frac {1}{(a+i a x)^{7/2} \sqrt {c-i c x}} \, dx,x,\tan (e+f x)\right )}{f} \\ & = -\frac {i A+B}{f (a+i a \tan (e+f x))^{5/2} \sqrt {c-i c \tan (e+f x)}}+\frac {(3 i A+2 B) \sqrt {c-i c \tan (e+f x)}}{5 c f (a+i a \tan (e+f x))^{5/2}}+\frac {(2 (3 A-2 i B)) \text {Subst}\left (\int \frac {1}{(a+i a x)^{5/2} \sqrt {c-i c x}} \, dx,x,\tan (e+f x)\right )}{5 f} \\ & = -\frac {i A+B}{f (a+i a \tan (e+f x))^{5/2} \sqrt {c-i c \tan (e+f x)}}+\frac {(3 i A+2 B) \sqrt {c-i c \tan (e+f x)}}{5 c f (a+i a \tan (e+f x))^{5/2}}+\frac {2 (3 i A+2 B) \sqrt {c-i c \tan (e+f x)}}{15 a c f (a+i a \tan (e+f x))^{3/2}}+\frac {(2 (3 A-2 i B)) \text {Subst}\left (\int \frac {1}{(a+i a x)^{3/2} \sqrt {c-i c x}} \, dx,x,\tan (e+f x)\right )}{15 a f} \\ & = -\frac {i A+B}{f (a+i a \tan (e+f x))^{5/2} \sqrt {c-i c \tan (e+f x)}}+\frac {(3 i A+2 B) \sqrt {c-i c \tan (e+f x)}}{5 c f (a+i a \tan (e+f x))^{5/2}}+\frac {2 (3 i A+2 B) \sqrt {c-i c \tan (e+f x)}}{15 a c f (a+i a \tan (e+f x))^{3/2}}+\frac {2 (3 i A+2 B) \sqrt {c-i c \tan (e+f x)}}{15 a^2 c f \sqrt {a+i a \tan (e+f x)}} \\ \end{align*}
Time = 4.51 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.54 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^{5/2} \sqrt {c-i c \tan (e+f x)}} \, dx=\frac {-6 i A+B+(-3 A+2 i B) \tan (e+f x)+(-12 i A-8 B) \tan ^2(e+f x)+(6 A-4 i B) \tan ^3(e+f x)}{15 a^2 f (-i+\tan (e+f x))^2 \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}} \]
[In]
[Out]
Time = 0.39 (sec) , antiderivative size = 186, normalized size of antiderivative = 0.88
method | result | size |
derivativedivides | \(-\frac {\sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \left (4 i B \tan \left (f x +e \right )^{5}+12 i A \tan \left (f x +e \right )^{4}-6 A \tan \left (f x +e \right )^{5}+2 i B \tan \left (f x +e \right )^{3}+8 B \tan \left (f x +e \right )^{4}+18 i A \tan \left (f x +e \right )^{2}-3 A \tan \left (f x +e \right )^{3}-2 i \tan \left (f x +e \right ) B +7 B \tan \left (f x +e \right )^{2}+6 i A +3 A \tan \left (f x +e \right )-B \right )}{15 f \,a^{3} c \left (i-\tan \left (f x +e \right )\right )^{4} \left (i+\tan \left (f x +e \right )\right )^{2}}\) | \(186\) |
default | \(-\frac {\sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \left (4 i B \tan \left (f x +e \right )^{5}+12 i A \tan \left (f x +e \right )^{4}-6 A \tan \left (f x +e \right )^{5}+2 i B \tan \left (f x +e \right )^{3}+8 B \tan \left (f x +e \right )^{4}+18 i A \tan \left (f x +e \right )^{2}-3 A \tan \left (f x +e \right )^{3}-2 i \tan \left (f x +e \right ) B +7 B \tan \left (f x +e \right )^{2}+6 i A +3 A \tan \left (f x +e \right )-B \right )}{15 f \,a^{3} c \left (i-\tan \left (f x +e \right )\right )^{4} \left (i+\tan \left (f x +e \right )\right )^{2}}\) | \(186\) |
parts | \(-\frac {A \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \left (4 i \tan \left (f x +e \right )^{4}-2 \tan \left (f x +e \right )^{5}+6 i \tan \left (f x +e \right )^{2}-\tan \left (f x +e \right )^{3}+2 i+\tan \left (f x +e \right )\right )}{5 f \,a^{3} c \left (i-\tan \left (f x +e \right )\right )^{4} \left (i+\tan \left (f x +e \right )\right )^{2}}+\frac {i B \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \left (1+\tan \left (f x +e \right )^{2}\right ) \left (8 i \tan \left (f x +e \right )^{2}-4 \tan \left (f x +e \right )^{3}-i+2 \tan \left (f x +e \right )\right )}{15 f \,a^{3} c \left (i-\tan \left (f x +e \right )\right )^{4} \left (i+\tan \left (f x +e \right )\right )^{2}}\) | \(230\) |
[In]
[Out]
none
Time = 0.28 (sec) , antiderivative size = 161, normalized size of antiderivative = 0.76 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^{5/2} \sqrt {c-i c \tan (e+f x)}} \, dx=-\frac {{\left (15 \, {\left (i \, A + B\right )} e^{\left (8 i \, f x + 8 i \, e\right )} + 8 \, {\left (6 i \, A - B\right )} e^{\left (7 i \, f x + 7 i \, e\right )} - 30 i \, A e^{\left (6 i \, f x + 6 i \, e\right )} + 8 \, {\left (6 i \, A - B\right )} e^{\left (5 i \, f x + 5 i \, e\right )} + 10 \, {\left (-6 i \, A - B\right )} e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, {\left (-9 i \, A + 4 \, B\right )} e^{\left (2 i \, f x + 2 i \, e\right )} - 3 i \, A + 3 \, B\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} e^{\left (-5 i \, f x - 5 i \, e\right )}}{120 \, a^{3} c f} \]
[In]
[Out]
\[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^{5/2} \sqrt {c-i c \tan (e+f x)}} \, dx=\int \frac {A + B \tan {\left (e + f x \right )}}{\left (i a \left (\tan {\left (e + f x \right )} - i\right )\right )^{\frac {5}{2}} \sqrt {- i c \left (\tan {\left (e + f x \right )} + i\right )}}\, dx \]
[In]
[Out]
Exception generated. \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^{5/2} \sqrt {c-i c \tan (e+f x)}} \, dx=\text {Exception raised: RuntimeError} \]
[In]
[Out]
\[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^{5/2} \sqrt {c-i c \tan (e+f x)}} \, dx=\int { \frac {B \tan \left (f x + e\right ) + A}{{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{\frac {5}{2}} \sqrt {-i \, c \tan \left (f x + e\right ) + c}} \,d x } \]
[In]
[Out]
Time = 9.88 (sec) , antiderivative size = 246, normalized size of antiderivative = 1.16 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^{5/2} \sqrt {c-i c \tan (e+f x)}} \, dx=\frac {\sqrt {\frac {a\,\left (\cos \left (2\,e+2\,f\,x\right )+1+\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}\,\left (15\,B\,\cos \left (2\,e+2\,f\,x\right )-15\,B+A\,\cos \left (2\,e+2\,f\,x\right )\,45{}\mathrm {i}+A\,\cos \left (4\,e+4\,f\,x\right )\,15{}\mathrm {i}+A\,\cos \left (6\,e+6\,f\,x\right )\,3{}\mathrm {i}-A\,15{}\mathrm {i}-5\,B\,\cos \left (4\,e+4\,f\,x\right )-3\,B\,\cos \left (6\,e+6\,f\,x\right )+45\,A\,\sin \left (2\,e+2\,f\,x\right )+15\,A\,\sin \left (4\,e+4\,f\,x\right )+3\,A\,\sin \left (6\,e+6\,f\,x\right )-B\,\sin \left (2\,e+2\,f\,x\right )\,15{}\mathrm {i}+B\,\sin \left (4\,e+4\,f\,x\right )\,5{}\mathrm {i}+B\,\sin \left (6\,e+6\,f\,x\right )\,3{}\mathrm {i}\right )}{120\,a^3\,f\,\sqrt {\frac {c\,\left (\cos \left (2\,e+2\,f\,x\right )+1-\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}} \]
[In]
[Out]