Integrand size = 33, antiderivative size = 138 \[ \int \frac {\sqrt {c-i c \tan (e+f x)}}{(a+i a \tan (e+f x))^2} \, dx=\frac {3 i \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{16 \sqrt {2} a^2 f}+\frac {i \sqrt {c-i c \tan (e+f x)}}{4 a^2 f (1+i \tan (e+f x))^2}+\frac {3 i \sqrt {c-i c \tan (e+f x)}}{16 a^2 f (1+i \tan (e+f x))} \]
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Time = 0.23 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {3603, 3568, 44, 65, 212} \[ \int \frac {\sqrt {c-i c \tan (e+f x)}}{(a+i a \tan (e+f x))^2} \, dx=\frac {3 i \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{16 \sqrt {2} a^2 f}+\frac {3 i \sqrt {c-i c \tan (e+f x)}}{16 a^2 f (1+i \tan (e+f x))}+\frac {i \sqrt {c-i c \tan (e+f x)}}{4 a^2 f (1+i \tan (e+f x))^2} \]
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Rule 44
Rule 65
Rule 212
Rule 3568
Rule 3603
Rubi steps \begin{align*} \text {integral}& = \frac {\int \cos ^4(e+f x) (c-i c \tan (e+f x))^{5/2} \, dx}{a^2 c^2} \\ & = \frac {\left (i c^3\right ) \text {Subst}\left (\int \frac {1}{(c-x)^3 \sqrt {c+x}} \, dx,x,-i c \tan (e+f x)\right )}{a^2 f} \\ & = \frac {i \sqrt {c-i c \tan (e+f x)}}{4 a^2 f (1+i \tan (e+f x))^2}+\frac {\left (3 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 )}{8 a^2 f} \\ & = \frac {i \sqrt {c-i c \tan (e+f x)}}{4 a^2 f (1+i \tan (e+f x))^2}+\frac {3 i \sqrt {c-i c \tan (e+f x)}}{16 a^2 f (1+i \tan (e+f x))}+\frac {(3 i c) \text {Subst}\left (\int \frac {1}{(c-x) \sqrt {c+x}} \, dx,x,-i c \tan (e+f x)\right )}{32 a^2 f} \\ & = \frac {i \sqrt {c-i c \tan (e+f x)}}{4 a^2 f (1+i \tan (e+f x))^2}+\frac {3 i \sqrt {c-i c \tan (e+f x)}}{16 a^2 f (1+i \tan (e+f x))}+\frac {(3 i c) \text {Subst}\left (\int \frac {1}{2 c-x^2} \, dx,x,\sqrt {c-i c \tan (e+f x)}\right )}{16 a^2 f} \\ & = \frac {3 i \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{16 \sqrt {2} a^2 f}+\frac {i \sqrt {c-i c \tan (e+f x)}}{4 a^2 f (1+i \tan (e+f x))^2}+\frac {3 i \sqrt {c-i c \tan (e+f x)}}{16 a^2 f (1+i \tan (e+f x))} \\ \end{align*}
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.78 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.38 \[ \int \frac {\sqrt {c-i c \tan (e+f x)}}{(a+i a \tan (e+f x))^2} \, dx=\frac {i \operatorname {Hypergeometric2F1}\left (\frac {1}{2},3,\frac {3}{2},-\frac {1}{2} i (i+\tan (e+f x))\right ) \sqrt {c-i c \tan (e+f x)}}{4 a^2 f} \]
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Time = 0.72 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.85
method | result | size |
derivativedivides | \(\frac {2 i c^{3} \left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}}{8 c \left (c +i c \tan \left (f x +e \right )\right )^{2}}+\frac {\frac {3 \sqrt {c -i c \tan \left (f x +e \right )}}{32 c \left (c +i c \tan \left (f x +e \right )\right )}+\frac {3 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{64 c^{\frac {3}{2}}}}{c}\right )}{f \,a^{2}}\) | \(117\) |
default | \(\frac {2 i c^{3} \left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}}{8 c \left (c +i c \tan \left (f x +e \right )\right )^{2}}+\frac {\frac {3 \sqrt {c -i c \tan \left (f x +e \right )}}{32 c \left (c +i c \tan \left (f x +e \right )\right )}+\frac {3 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{64 c^{\frac {3}{2}}}}{c}\right )}{f \,a^{2}}\) | \(117\) |
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Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 275 vs. \(2 (103) = 206\).
Time = 0.24 (sec) , antiderivative size = 275, normalized size of antiderivative = 1.99 \[ \int \frac {\sqrt {c-i c \tan (e+f x)}}{(a+i a \tan (e+f x))^2} \, dx=\frac {{\left (3 \, \sqrt {\frac {1}{2}} a^{2} f \sqrt {-\frac {c}{a^{4} f^{2}}} e^{\left (4 i \, f x + 4 i \, e\right )} \log \left (\frac {3 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{2} f\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {c}{a^{4} f^{2}}} + i \, c\right )} e^{\left (-i \, f x - i \, e\right )}}{8 \, a^{2} f}\right ) - 3 \, \sqrt {\frac {1}{2}} a^{2} f \sqrt {-\frac {c}{a^{4} f^{2}}} e^{\left (4 i \, f x + 4 i \, e\right )} \log \left (-\frac {3 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{2} f\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {c}{a^{4} f^{2}}} - i \, c\right )} e^{\left (-i \, f x - i \, e\right )}}{8 \, a^{2} f}\right ) + \sqrt {2} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} {\left (5 i \, e^{\left (4 i \, f x + 4 i \, e\right )} + 7 i \, e^{\left (2 i \, f x + 2 i \, e\right )} + 2 i\right )}\right )} e^{\left (-4 i \, f x - 4 i \, e\right )}}{32 \, a^{2} f} \]
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\[ \int \frac {\sqrt {c-i c \tan (e+f x)}}{(a+i a \tan (e+f x))^2} \, dx=- \frac {\int \frac {\sqrt {- i c \tan {\left (e + f x \right )} + c}}{\tan ^{2}{\left (e + f x \right )} - 2 i \tan {\left (e + f x \right )} - 1}\, dx}{a^{2}} \]
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Time = 0.29 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.11 \[ \int \frac {\sqrt {c-i c \tan (e+f x)}}{(a+i a \tan (e+f x))^2} \, dx=-\frac {i \, {\left (\frac {3 \, \sqrt {2} c^{\frac {3}{2}} \log \left (-\frac {\sqrt {2} \sqrt {c} - \sqrt {-i \, c \tan \left (f x + e\right ) + c}}{\sqrt {2} \sqrt {c} + \sqrt {-i \, c \tan \left (f x + e\right ) + c}}\right )}{a^{2}} + \frac {4 \, {\left (3 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} c^{2} - 10 \, \sqrt {-i \, c \tan \left (f x + e\right ) + c} c^{3}\right )}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{2} a^{2} - 4 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )} a^{2} c + 4 \, a^{2} c^{2}}\right )}}{64 \, c f} \]
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\[ \int \frac {\sqrt {c-i c \tan (e+f x)}}{(a+i a \tan (e+f x))^2} \, dx=\int { \frac {\sqrt {-i \, c \tan \left (f x + e\right ) + c}}{{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{2}} \,d x } \]
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Time = 6.23 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.96 \[ \int \frac {\sqrt {c-i c \tan (e+f x)}}{(a+i a \tan (e+f x))^2} \, dx=\frac {\frac {c^2\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}\,5{}\mathrm {i}}{8\,a^2\,f}-\frac {c\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2}\,3{}\mathrm {i}}{16\,a^2\,f}}{{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^2-4\,c\,\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )+4\,c^2}+\frac {\sqrt {2}\,\sqrt {-c}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-c}}\right )\,3{}\mathrm {i}}{32\,a^2\,f} \]
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