Integrand size = 35, antiderivative size = 43 \[ \int \frac {(c-i c \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^{5/2}} \, dx=\frac {i (c-i c \tan (e+f x))^{5/2}}{5 f (a+i a \tan (e+f x))^{5/2}} \]
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Time = 0.17 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.057, Rules used = {3604, 37} \[ \int \frac {(c-i c \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^{5/2}} \, dx=\frac {i (c-i c \tan (e+f x))^{5/2}}{5 f (a+i a \tan (e+f x))^{5/2}} \]
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Rule 37
Rule 3604
Rubi steps \begin{align*} \text {integral}& = \frac {(a c) \text {Subst}\left (\int \frac {(c-i c x)^{3/2}}{(a+i a x)^{7/2}} \, dx,x,\tan (e+f x)\right )}{f} \\ & = \frac {i (c-i c \tan (e+f x))^{5/2}}{5 f (a+i a \tan (e+f x))^{5/2}} \\ \end{align*}
Time = 2.34 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.00 \[ \int \frac {(c-i c \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^{5/2}} \, dx=\frac {i (c-i c \tan (e+f x))^{5/2}}{5 f (a+i a \tan (e+f x))^{5/2}} \]
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Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 74 vs. \(2 (34 ) = 68\).
Time = 0.84 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.74
method | result | size |
derivativedivides | \(\frac {\sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, c^{2} \left (1+\tan ^{2}\left (f x +e \right )\right ) \left (\tan \left (f x +e \right )+i\right )}{5 f \,a^{3} \left (-\tan \left (f x +e \right )+i\right )^{4}}\) | \(75\) |
default | \(\frac {\sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, c^{2} \left (1+\tan ^{2}\left (f x +e \right )\right ) \left (\tan \left (f x +e \right )+i\right )}{5 f \,a^{3} \left (-\tan \left (f x +e \right )+i\right )^{4}}\) | \(75\) |
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Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 71 vs. \(2 (31) = 62\).
Time = 0.24 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.65 \[ \int \frac {(c-i c \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^{5/2}} \, dx=\frac {{\left (i \, c^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + i \, c^{2}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} e^{\left (-5 i \, f x - 5 i \, e\right )}}{5 \, a^{3} f} \]
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\[ \int \frac {(c-i c \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^{5/2}} \, dx=\int \frac {\left (- i c \left (\tan {\left (e + f x \right )} + i\right )\right )^{\frac {5}{2}}}{\left (i a \left (\tan {\left (e + f x \right )} - i\right )\right )^{\frac {5}{2}}}\, dx \]
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none
Time = 0.37 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.91 \[ \int \frac {(c-i c \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^{5/2}} \, dx=\frac {{\left (i \, c^{2} \cos \left (5 \, f x + 5 \, e\right ) + c^{2} \sin \left (5 \, f x + 5 \, e\right )\right )} \sqrt {c}}{5 \, a^{\frac {5}{2}} f} \]
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\[ \int \frac {(c-i c \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^{5/2}} \, dx=\int { \frac {{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}}}{{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]
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Time = 8.99 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.51 \[ \int \frac {(c-i c \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^{5/2}} \, dx=\frac {c^2\,{\left (\mathrm {tan}\left (e+f\,x\right )+1{}\mathrm {i}\right )}^5\,\sqrt {a\,\left (1+\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}\,\sqrt {-c\,\left (-1+\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}}{5\,a^3\,f\,{\left ({\mathrm {tan}\left (e+f\,x\right )}^2+1\right )}^3} \]
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