Integrand size = 43, antiderivative size = 144 \[ \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}{a+i a \tan (e+f x)} \, dx=-\frac {(i A-5 B) c^{3/2} \text {arctanh}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{\sqrt {2} a f}+\frac {(i A-5 B) c \sqrt {c-i c \tan (e+f x)}}{2 a f}+\frac {(i A-B) (c-i c \tan (e+f x))^{3/2}}{2 a f (1+i \tan (e+f x))} \]
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Time = 0.27 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.116, Rules used = {3669, 79, 52, 65, 214} \[ \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}{a+i a \tan (e+f x)} \, dx=-\frac {c^{3/2} (-5 B+i A) \text {arctanh}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{\sqrt {2} a f}+\frac {c (-5 B+i A) \sqrt {c-i c \tan (e+f x)}}{2 a f}+\frac {(-B+i A) (c-i c \tan (e+f x))^{3/2}}{2 a f (1+i \tan (e+f x))} \]
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Rule 52
Rule 65
Rule 79
Rule 214
Rule 3669
Rubi steps \begin{align*} \text {integral}& = \frac {(a c) \text {Subst}\left (\int \frac {(A+B x) \sqrt {c-i c x}}{(a+i a x)^2} \, dx,x,\tan (e+f x)\right )}{f} \\ & = \frac {(i A-B) (c-i c \tan (e+f x))^{3/2}}{2 a f (1+i \tan (e+f x))}-\frac {((A+5 i B) c) \text {Subst}\left (\int \frac {\sqrt {c-i c x}}{a+i a x} \, dx,x,\tan (e+f x)\right )}{4 f} \\ & = \frac {(i A-5 B) c \sqrt {c-i c \tan (e+f x)}}{2 a f}+\frac {(i A-B) (c-i c \tan (e+f x))^{3/2}}{2 a f (1+i \tan (e+f x))}-\frac {\left ((A+5 i B) c^2\right ) \text {Subst}\left (\int \frac {1}{(a+i a x) \sqrt {c-i c x}} \, dx,x,\tan (e+f x)\right )}{2 f} \\ & = \frac {(i A-5 B) c \sqrt {c-i c \tan (e+f x)}}{2 a f}+\frac {(i A-B) (c-i c \tan (e+f x))^{3/2}}{2 a f (1+i \tan (e+f x))}-\frac {((i A-5 B) c) \text {Subst}\left (\int \frac {1}{2 a-\frac {a x^2}{c}} \, dx,x,\sqrt {c-i c \tan (e+f x)}\right )}{f} \\ & = -\frac {(i A-5 B) c^{3/2} \text {arctanh}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{\sqrt {2} a f}+\frac {(i A-5 B) c \sqrt {c-i c \tan (e+f x)}}{2 a f}+\frac {(i A-B) (c-i c \tan (e+f x))^{3/2}}{2 a f (1+i \tan (e+f x))} \\ \end{align*}
Time = 3.90 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.75 \[ \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}{a+i a \tan (e+f x)} \, dx=\frac {c \left (\sqrt {2} (-i A+5 B) \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )+\frac {2 (A+3 i B-2 B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}}{-i+\tan (e+f x)}\right )}{2 a f} \]
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Time = 0.28 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.77
method | result | size |
derivativedivides | \(\frac {2 i c \left (i \sqrt {c -i c \tan \left (f x +e \right )}\, B +c \left (\frac {\left (\frac {i B}{4}+\frac {A}{4}\right ) \sqrt {c -i c \tan \left (f x +e \right )}}{\frac {c}{2}+\frac {i c \tan \left (f x +e \right )}{2}}-\frac {\left (\frac {A}{2}+\frac {5 i B}{2}\right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{2 \sqrt {c}}\right )\right )}{f a}\) | \(111\) |
default | \(\frac {2 i c \left (i \sqrt {c -i c \tan \left (f x +e \right )}\, B +c \left (\frac {\left (\frac {i B}{4}+\frac {A}{4}\right ) \sqrt {c -i c \tan \left (f x +e \right )}}{\frac {c}{2}+\frac {i c \tan \left (f x +e \right )}{2}}-\frac {\left (\frac {A}{2}+\frac {5 i B}{2}\right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{2 \sqrt {c}}\right )\right )}{f a}\) | \(111\) |
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Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 330 vs. \(2 (111) = 222\).
Time = 0.25 (sec) , antiderivative size = 330, normalized size of antiderivative = 2.29 \[ \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}{a+i a \tan (e+f x)} \, dx=-\frac {{\left (\sqrt {2} a f \sqrt {-\frac {{\left (A^{2} + 10 i \, A B - 25 \, B^{2}\right )} c^{3}}{a^{2} f^{2}}} e^{\left (2 i \, f x + 2 i \, e\right )} \log \left (-\frac {2 \, {\left ({\left (i \, A - 5 \, B\right )} c^{2} + {\left (a f e^{\left (2 i \, f x + 2 i \, e\right )} + a f\right )} \sqrt {-\frac {{\left (A^{2} + 10 i \, A B - 25 \, B^{2}\right )} c^{3}}{a^{2} f^{2}}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-i \, f x - i \, e\right )}}{a f}\right ) - \sqrt {2} a f \sqrt {-\frac {{\left (A^{2} + 10 i \, A B - 25 \, B^{2}\right )} c^{3}}{a^{2} f^{2}}} e^{\left (2 i \, f x + 2 i \, e\right )} \log \left (-\frac {2 \, {\left ({\left (i \, A - 5 \, B\right )} c^{2} - {\left (a f e^{\left (2 i \, f x + 2 i \, e\right )} + a f\right )} \sqrt {-\frac {{\left (A^{2} + 10 i \, A B - 25 \, B^{2}\right )} c^{3}}{a^{2} f^{2}}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-i \, f x - i \, e\right )}}{a f}\right ) + 2 \, \sqrt {2} {\left ({\left (-i \, A + 5 \, B\right )} c e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (-i \, A + B\right )} c\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{4 \, a f} \]
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\[ \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}{a+i a \tan (e+f x)} \, dx=- \frac {i \left (\int \frac {A c \sqrt {- i c \tan {\left (e + f x \right )} + c}}{\tan {\left (e + f x \right )} - i}\, dx + \int \frac {B c \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )}}{\tan {\left (e + f x \right )} - i}\, dx + \int \left (- \frac {i A c \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )}}{\tan {\left (e + f x \right )} - i}\right )\, dx + \int \left (- \frac {i B c \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{2}{\left (e + f x \right )}}{\tan {\left (e + f x \right )} - i}\right )\, dx\right )}{a} \]
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none
Time = 0.32 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.96 \[ \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}{a+i a \tan (e+f x)} \, dx=\frac {i \, {\left (\frac {\sqrt {2} {\left (A + 5 i \, B\right )} c^{\frac {5}{2}} \log \left (-\frac {\sqrt {2} \sqrt {c} - \sqrt {-i \, c \tan \left (f x + e\right ) + c}}{\sqrt {2} \sqrt {c} + \sqrt {-i \, c \tan \left (f x + e\right ) + c}}\right )}{a} - \frac {4 \, \sqrt {-i \, c \tan \left (f x + e\right ) + c} {\left (A + i \, B\right )} c^{3}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )} a - 2 \, a c} + \frac {8 i \, \sqrt {-i \, c \tan \left (f x + e\right ) + c} B c^{2}}{a}\right )}}{4 \, c f} \]
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\[ \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}{a+i a \tan (e+f x)} \, dx=\int { \frac {{\left (B \tan \left (f x + e\right ) + A\right )} {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}}}{i \, a \tan \left (f x + e\right ) + a} \,d x } \]
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Time = 9.20 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.31 \[ \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}{a+i a \tan (e+f x)} \, dx=\frac {B\,c^2\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{a\,f\,\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )-2\,a\,c\,f}-\frac {2\,B\,c\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{a\,f}+\frac {\sqrt {2}\,A\,{\left (-c\right )}^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-c}}\right )\,1{}\mathrm {i}}{2\,a\,f}+\frac {5\,\sqrt {2}\,B\,c^{3/2}\,\mathrm {atanh}\left (\frac {\sqrt {2}\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {c}}\right )}{2\,a\,f}+\frac {A\,c^2\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}\,1{}\mathrm {i}}{a\,f\,\left (c+c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )} \]
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