Integrand size = 35, antiderivative size = 154 \[ \int \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{5/2} \, dx=-\frac {3 i \sqrt {a} c^{5/2} \arctan \left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{f}-\frac {3 i c^2 \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{2 f}-\frac {i c \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{3/2}}{2 f} \]
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Time = 0.27 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3604, 52, 65, 223, 209} \[ \int \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{5/2} \, dx=-\frac {3 i \sqrt {a} c^{5/2} \arctan \left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{f}-\frac {3 i c^2 \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{2 f}-\frac {i c \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{3/2}}{2 f} \]
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Rule 52
Rule 65
Rule 209
Rule 223
Rule 3604
Rubi steps \begin{align*} \text {integral}& = \frac {(a c) \text {Subst}\left (\int \frac {(c-i c x)^{3/2}}{\sqrt {a+i a x}} \, dx,x,\tan (e+f x)\right )}{f} \\ & = -\frac {i c \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{3/2}}{2 f}+\frac {\left (3 a c^2\right ) \text {Subst}\left (\int \frac {\sqrt {c-i c x}}{\sqrt {a+i a x}} \, dx,x,\tan (e+f x)\right )}{2 f} \\ & = -\frac {3 i c^2 \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{2 f}-\frac {i c \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{3/2}}{2 f}+\frac {\left (3 a c^3\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a+i a x} \sqrt {c-i c x}} \, dx,x,\tan (e+f x)\right )}{2 f} \\ & = -\frac {3 i c^2 \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{2 f}-\frac {i c \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{3/2}}{2 f}-\frac {\left (3 i c^3\right ) \text {Subst}\left (\int \frac {1}{\sqrt {2 c-\frac {c x^2}{a}}} \, dx,x,\sqrt {a+i a \tan (e+f x)}\right )}{f} \\ & = -\frac {3 i c^2 \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{2 f}-\frac {i c \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{3/2}}{2 f}-\frac {\left (3 i c^3\right ) \text {Subst}\left (\int \frac {1}{1+\frac {c x^2}{a}} \, dx,x,\frac {\sqrt {a+i a \tan (e+f x)}}{\sqrt {c-i c \tan (e+f x)}}\right )}{f} \\ & = -\frac {3 i \sqrt {a} c^{5/2} \arctan \left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{f}-\frac {3 i c^2 \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{2 f}-\frac {i c \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{3/2}}{2 f} \\ \end{align*}
Time = 1.46 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.82 \[ \int \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{5/2} \, dx=-\frac {c^2 \sqrt {c-i c \tan (e+f x)} \left (6 i \sqrt {a} \arcsin \left (\frac {\sqrt {a+i a \tan (e+f x)}}{\sqrt {2} \sqrt {a}}\right )+\sqrt {1-i \tan (e+f x)} (4 i+\tan (e+f x)) \sqrt {a+i a \tan (e+f x)}\right )}{2 f \sqrt {1-i \tan (e+f x)}} \]
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Time = 0.81 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.99
method | result | size |
derivativedivides | \(-\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, c^{2} \left (4 i \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}+\tan \left (f x +e \right ) \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}-3 a c \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}}{\sqrt {a c}}\right )\right )}{2 f \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}}\) | \(153\) |
default | \(-\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, c^{2} \left (4 i \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}+\tan \left (f x +e \right ) \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}-3 a c \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}}{\sqrt {a c}}\right )\right )}{2 f \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}}\) | \(153\) |
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Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 361 vs. \(2 (112) = 224\).
Time = 0.27 (sec) , antiderivative size = 361, normalized size of antiderivative = 2.34 \[ \int \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{5/2} \, dx=\frac {3 \, \sqrt {\frac {a c^{5}}{f^{2}}} {\left (f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \log \left (\frac {4 \, {\left (2 \, {\left (c^{2} e^{\left (3 i \, f x + 3 i \, e\right )} + c^{2} e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} - \sqrt {\frac {a c^{5}}{f^{2}}} {\left (i \, f e^{\left (2 i \, f x + 2 i \, e\right )} - i \, f\right )}\right )}}{c^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + c^{2}}\right ) - 3 \, \sqrt {\frac {a c^{5}}{f^{2}}} {\left (f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \log \left (\frac {4 \, {\left (2 \, {\left (c^{2} e^{\left (3 i \, f x + 3 i \, e\right )} + c^{2} e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} - \sqrt {\frac {a c^{5}}{f^{2}}} {\left (-i \, f e^{\left (2 i \, f x + 2 i \, e\right )} + i \, f\right )}\right )}}{c^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + c^{2}}\right ) - 4 \, {\left (3 i \, c^{2} e^{\left (3 i \, f x + 3 i \, e\right )} + 5 i \, c^{2} e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{4 \, {\left (f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )}} \]
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\[ \int \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{5/2} \, dx=\int \sqrt {i a \left (\tan {\left (e + f x \right )} - i\right )} \left (- i c \left (\tan {\left (e + f x \right )} + i\right )\right )^{\frac {5}{2}}\, dx \]
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Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 445 vs. \(2 (112) = 224\).
Time = 0.56 (sec) , antiderivative size = 445, normalized size of antiderivative = 2.89 \[ \int \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{5/2} \, dx=-\frac {{\left (12 \, c^{2} \cos \left (3 \, f x + 3 \, e\right ) + 20 \, c^{2} \cos \left (f x + e\right ) + 12 i \, c^{2} \sin \left (3 \, f x + 3 \, e\right ) + 20 i \, c^{2} \sin \left (f x + e\right ) + 6 \, {\left (c^{2} \cos \left (4 \, f x + 4 \, e\right ) + 2 \, c^{2} \cos \left (2 \, f x + 2 \, e\right ) + i \, c^{2} \sin \left (4 \, f x + 4 \, e\right ) + 2 i \, c^{2} \sin \left (2 \, f x + 2 \, e\right ) + c^{2}\right )} \arctan \left (\cos \left (f x + e\right ), \sin \left (f x + e\right ) + 1\right ) + 6 \, {\left (c^{2} \cos \left (4 \, f x + 4 \, e\right ) + 2 \, c^{2} \cos \left (2 \, f x + 2 \, e\right ) + i \, c^{2} \sin \left (4 \, f x + 4 \, e\right ) + 2 i \, c^{2} \sin \left (2 \, f x + 2 \, e\right ) + c^{2}\right )} \arctan \left (\cos \left (f x + e\right ), -\sin \left (f x + e\right ) + 1\right ) + 3 \, {\left (i \, c^{2} \cos \left (4 \, f x + 4 \, e\right ) + 2 i \, c^{2} \cos \left (2 \, f x + 2 \, e\right ) - c^{2} \sin \left (4 \, f x + 4 \, e\right ) - 2 \, c^{2} \sin \left (2 \, f x + 2 \, e\right ) + i \, c^{2}\right )} \log \left (\cos \left (f x + e\right )^{2} + \sin \left (f x + e\right )^{2} + 2 \, \sin \left (f x + e\right ) + 1\right ) + 3 \, {\left (-i \, c^{2} \cos \left (4 \, f x + 4 \, e\right ) - 2 i \, c^{2} \cos \left (2 \, f x + 2 \, e\right ) + c^{2} \sin \left (4 \, f x + 4 \, e\right ) + 2 \, c^{2} \sin \left (2 \, f x + 2 \, e\right ) - i \, c^{2}\right )} \log \left (\cos \left (f x + e\right )^{2} + \sin \left (f x + e\right )^{2} - 2 \, \sin \left (f x + e\right ) + 1\right )\right )} \sqrt {a} \sqrt {c}}{-4 \, f {\left (i \, \cos \left (4 \, f x + 4 \, e\right ) + 2 i \, \cos \left (2 \, f x + 2 \, e\right ) - \sin \left (4 \, f x + 4 \, e\right ) - 2 \, \sin \left (2 \, f x + 2 \, e\right ) + i\right )}} \]
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Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 300 vs. \(2 (112) = 224\).
Time = 2.27 (sec) , antiderivative size = 300, normalized size of antiderivative = 1.95 \[ \int \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{5/2} \, dx=-\frac {11 \, {\left (a c^{2} - c^{2}\right )} \sqrt {-a c} e^{\left (9 i \, f x + 9 i \, e\right )} + 50 \, {\left (a c^{2} - c^{2}\right )} \sqrt {-a c} e^{\left (7 i \, f x + 7 i \, e\right )} + 84 \, {\left (a c^{2} - c^{2}\right )} \sqrt {-a c} e^{\left (5 i \, f x + 5 i \, e\right )} + 62 \, {\left (a c^{2} - c^{2}\right )} \sqrt {-a c} e^{\left (3 i \, f x + 3 i \, e\right )} + 17 \, {\left (a c^{2} - c^{2}\right )} \sqrt {-a c} e^{\left (i \, f x + i \, e\right )}}{4 \, {\left ({\left (a - 1\right )} f e^{\left (10 i \, f x + 10 i \, e\right )} + 5 \, {\left (a - 1\right )} f e^{\left (8 i \, f x + 8 i \, e\right )} + 10 \, {\left (a - 1\right )} f e^{\left (6 i \, f x + 6 i \, e\right )} + 10 \, {\left (a - 1\right )} f e^{\left (4 i \, f x + 4 i \, e\right )} + 5 \, {\left (a - 1\right )} f e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (a - 1\right )} f\right )}} - \frac {32 i \, \sqrt {a} c^{\frac {5}{2}} \arctan \left (e^{\left (i \, f x + i \, e\right )}\right ) + \frac {i \, {\left (11 \, \sqrt {a} c^{\frac {5}{2}} e^{\left (3 i \, f x + 3 i \, e\right )} + 9 \, \sqrt {a} c^{\frac {5}{2}} e^{\left (i \, f x + i \, e\right )}\right )}}{{\left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )}^{2}}}{4 \, f} \]
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Timed out. \[ \int \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{5/2} \, dx=\int \sqrt {a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{5/2} \,d x \]
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