Integrand size = 33, antiderivative size = 95 \[ \int \frac {\sqrt {c-i c \tan (e+f x)}}{a+i a \tan (e+f x)} \, dx=\frac {i \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{2 \sqrt {2} a f}+\frac {i \sqrt {c-i c \tan (e+f x)}}{2 a f (1+i \tan (e+f x))} \]
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Time = 0.20 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {3603, 3568, 44, 65, 212} \[ \int \frac {\sqrt {c-i c \tan (e+f x)}}{a+i a \tan (e+f x)} \, dx=\frac {i \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{2 \sqrt {2} a f}+\frac {i \sqrt {c-i c \tan (e+f x)}}{2 a f (1+i \tan (e+f x))} \]
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Rule 44
Rule 65
Rule 212
Rule 3568
Rule 3603
Rubi steps \begin{align*} \text {integral}& = \frac {\int \cos ^2(e+f x) (c-i c \tan (e+f x))^{3/2} \, dx}{a c} \\ & = \frac {\left (i c^2\right ) \text {Subst}\left (\int \frac {1}{(c-x)^2 \sqrt {c+x}} \, dx,x,-i c \tan (e+f x)\right )}{a f} \\ & = \frac {i \sqrt {c-i c \tan (e+f x)}}{2 a f (1+i \tan (e+f x))}+\frac {(i c) \text {Subst}\left (\int \frac {1}{(c-x) \sqrt {c+x}} \, dx,x,-i c \tan (e+f x)\right )}{4 a f} \\ & = \frac {i \sqrt {c-i c \tan (e+f x)}}{2 a f (1+i \tan (e+f x))}+\frac {(i c) \text {Subst}\left (\int \frac {1}{2 c-x^2} \, dx,x,\sqrt {c-i c \tan (e+f x)}\right )}{2 a f} \\ & = \frac {i \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{2 \sqrt {2} a f}+\frac {i \sqrt {c-i c \tan (e+f x)}}{2 a f (1+i \tan (e+f x))} \\ \end{align*}
Time = 0.76 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.99 \[ \int \frac {\sqrt {c-i c \tan (e+f x)}}{a+i a \tan (e+f x)} \, dx=\frac {\sqrt {2} \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right ) (1+i \tan (e+f x))+2 \sqrt {c-i c \tan (e+f x)}}{4 a f (-i+\tan (e+f x))} \]
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Time = 0.77 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.82
method | result | size |
derivativedivides | \(\frac {2 i c^{2} \left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}}{4 c \left (c +i c \tan \left (f x +e \right )\right )}+\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{8 c^{\frac {3}{2}}}\right )}{f a}\) | \(78\) |
default | \(\frac {2 i c^{2} \left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}}{4 c \left (c +i c \tan \left (f x +e \right )\right )}+\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{8 c^{\frac {3}{2}}}\right )}{f a}\) | \(78\) |
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Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 250 vs. \(2 (70) = 140\).
Time = 0.25 (sec) , antiderivative size = 250, normalized size of antiderivative = 2.63 \[ \int \frac {\sqrt {c-i c \tan (e+f x)}}{a+i a \tan (e+f x)} \, dx=\frac {{\left (\sqrt {\frac {1}{2}} a f \sqrt {-\frac {c}{a^{2} f^{2}}} e^{\left (2 i \, f x + 2 i \, e\right )} \log \left (\frac {{\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a f e^{\left (2 i \, f x + 2 i \, e\right )} + a f\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {c}{a^{2} f^{2}}} + i \, c\right )} e^{\left (-i \, f x - i \, e\right )}}{a f}\right ) - \sqrt {\frac {1}{2}} a f \sqrt {-\frac {c}{a^{2} f^{2}}} e^{\left (2 i \, f x + 2 i \, e\right )} \log \left (-\frac {{\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a f e^{\left (2 i \, f x + 2 i \, e\right )} + a f\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {c}{a^{2} f^{2}}} - i \, c\right )} e^{\left (-i \, f x - i \, e\right )}}{a f}\right ) + \sqrt {2} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} {\left (i \, e^{\left (2 i \, f x + 2 i \, e\right )} + i\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{4 \, a f} \]
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\[ \int \frac {\sqrt {c-i c \tan (e+f x)}}{a+i a \tan (e+f x)} \, dx=- \frac {i \int \frac {\sqrt {- i c \tan {\left (e + f x \right )} + c}}{\tan {\left (e + f x \right )} - i}\, dx}{a} \]
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none
Time = 0.30 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.12 \[ \int \frac {\sqrt {c-i c \tan (e+f x)}}{a+i a \tan (e+f x)} \, dx=-\frac {i \, {\left (\frac {\sqrt {2} c^{\frac {3}{2}} \log \left (-\frac {\sqrt {2} \sqrt {c} - \sqrt {-i \, c \tan \left (f x + e\right ) + c}}{\sqrt {2} \sqrt {c} + \sqrt {-i \, c \tan \left (f x + e\right ) + c}}\right )}{a} + \frac {4 \, \sqrt {-i \, c \tan \left (f x + e\right ) + c} c^{2}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )} a - 2 \, a c}\right )}}{8 \, c f} \]
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\[ \int \frac {\sqrt {c-i c \tan (e+f x)}}{a+i a \tan (e+f x)} \, dx=\int { \frac {\sqrt {-i \, c \tan \left (f x + e\right ) + c}}{i \, a \tan \left (f x + e\right ) + a} \,d x } \]
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Time = 0.51 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.85 \[ \int \frac {\sqrt {c-i c \tan (e+f x)}}{a+i a \tan (e+f x)} \, dx=\frac {\sqrt {2}\,\sqrt {-c}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-c}}\right )\,1{}\mathrm {i}}{4\,a\,f}+\frac {c\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2\,a\,f\,\left (c+c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )} \]
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