Integrand size = 28, antiderivative size = 329 \[ \int \frac {(d \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^3} \, dx=\frac {d^{5/2} \arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{8 \sqrt {2} a^3 f}-\frac {d^{5/2} \arctan \left (1+\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{8 \sqrt {2} a^3 f}-\frac {d^{5/2} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{16 \sqrt {2} a^3 f}+\frac {d^{5/2} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{16 \sqrt {2} a^3 f}-\frac {d (d \tan (e+f x))^{3/2}}{6 f (a+i a \tan (e+f x))^3}+\frac {i d^2 \sqrt {d \tan (e+f x)}}{4 a f (a+i a \tan (e+f x))^2}-\frac {i d^2 \sqrt {d \tan (e+f x)}}{4 f \left (a^3+i a^3 \tan (e+f x)\right )} \]
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Time = 0.67 (sec) , antiderivative size = 329, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.464, Rules used = {3639, 3676, 3677, 12, 16, 3557, 335, 303, 1176, 631, 210, 1179, 642} \[ \int \frac {(d \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^3} \, dx=\frac {d^{5/2} \arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{8 \sqrt {2} a^3 f}-\frac {d^{5/2} \arctan \left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{8 \sqrt {2} a^3 f}-\frac {d^{5/2} \log \left (\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}+\sqrt {d}\right )}{16 \sqrt {2} a^3 f}+\frac {d^{5/2} \log \left (\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}+\sqrt {d}\right )}{16 \sqrt {2} a^3 f}-\frac {i d^2 \sqrt {d \tan (e+f x)}}{4 f \left (a^3+i a^3 \tan (e+f x)\right )}+\frac {i d^2 \sqrt {d \tan (e+f x)}}{4 a f (a+i a \tan (e+f x))^2}-\frac {d (d \tan (e+f x))^{3/2}}{6 f (a+i a \tan (e+f x))^3} \]
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Rule 12
Rule 16
Rule 210
Rule 303
Rule 335
Rule 631
Rule 642
Rule 1176
Rule 1179
Rule 3557
Rule 3639
Rule 3676
Rule 3677
Rubi steps \begin{align*} \text {integral}& = -\frac {d (d \tan (e+f x))^{3/2}}{6 f (a+i a \tan (e+f x))^3}-\frac {\int \frac {\sqrt {d \tan (e+f x)} \left (-\frac {3 a d^2}{2}+\frac {9}{2} i a d^2 \tan (e+f x)\right )}{(a+i a \tan (e+f x))^2} \, dx}{6 a^2} \\ & = -\frac {d (d \tan (e+f x))^{3/2}}{6 f (a+i a \tan (e+f x))^3}+\frac {i d^2 \sqrt {d \tan (e+f x)}}{4 a f (a+i a \tan (e+f x))^2}+\frac {\int \frac {-3 i a^2 d^3-9 a^2 d^3 \tan (e+f x)}{\sqrt {d \tan (e+f x)} (a+i a \tan (e+f x))} \, dx}{24 a^4} \\ & = -\frac {d (d \tan (e+f x))^{3/2}}{6 f (a+i a \tan (e+f x))^3}+\frac {i d^2 \sqrt {d \tan (e+f x)}}{4 a f (a+i a \tan (e+f x))^2}-\frac {i d^2 \sqrt {d \tan (e+f x)}}{4 f \left (a^3+i a^3 \tan (e+f x)\right )}+\frac {\int -\frac {6 a^3 d^4 \tan (e+f x)}{\sqrt {d \tan (e+f x)}} \, dx}{48 a^6 d} \\ & = -\frac {d (d \tan (e+f x))^{3/2}}{6 f (a+i a \tan (e+f x))^3}+\frac {i d^2 \sqrt {d \tan (e+f x)}}{4 a f (a+i a \tan (e+f x))^2}-\frac {i d^2 \sqrt {d \tan (e+f x)}}{4 f \left (a^3+i a^3 \tan (e+f x)\right )}-\frac {d^3 \int \frac {\tan (e+f x)}{\sqrt {d \tan (e+f x)}} \, dx}{8 a^3} \\ & = -\frac {d (d \tan (e+f x))^{3/2}}{6 f (a+i a \tan (e+f x))^3}+\frac {i d^2 \sqrt {d \tan (e+f x)}}{4 a f (a+i a \tan (e+f x))^2}-\frac {i d^2 \sqrt {d \tan (e+f x)}}{4 f \left (a^3+i a^3 \tan (e+f x)\right )}-\frac {d^2 \int \sqrt {d \tan (e+f x)} \, dx}{8 a^3} \\ & = -\frac {d (d \tan (e+f x))^{3/2}}{6 f (a+i a \tan (e+f x))^3}+\frac {i d^2 \sqrt {d \tan (e+f x)}}{4 a f (a+i a \tan (e+f x))^2}-\frac {i d^2 \sqrt {d \tan (e+f x)}}{4 f \left (a^3+i a^3 \tan (e+f x)\right )}-\frac {d^3 \text {Subst}\left (\int \frac {\sqrt {x}}{d^2+x^2} \, dx,x,d \tan (e+f x)\right )}{8 a^3 f} \\ & = -\frac {d (d \tan (e+f x))^{3/2}}{6 f (a+i a \tan (e+f x))^3}+\frac {i d^2 \sqrt {d \tan (e+f x)}}{4 a f (a+i a \tan (e+f x))^2}-\frac {i d^2 \sqrt {d \tan (e+f x)}}{4 f \left (a^3+i a^3 \tan (e+f x)\right )}-\frac {d^3 \text {Subst}\left (\int \frac {x^2}{d^2+x^4} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{4 a^3 f} \\ & = -\frac {d (d \tan (e+f x))^{3/2}}{6 f (a+i a \tan (e+f x))^3}+\frac {i d^2 \sqrt {d \tan (e+f x)}}{4 a f (a+i a \tan (e+f x))^2}-\frac {i d^2 \sqrt {d \tan (e+f x)}}{4 f \left (a^3+i a^3 \tan (e+f x)\right )}+\frac {d^3 \text {Subst}\left (\int \frac {d-x^2}{d^2+x^4} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{8 a^3 f}-\frac {d^3 \text {Subst}\left (\int \frac {d+x^2}{d^2+x^4} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{8 a^3 f} \\ & = -\frac {d (d \tan (e+f x))^{3/2}}{6 f (a+i a \tan (e+f x))^3}+\frac {i d^2 \sqrt {d \tan (e+f x)}}{4 a f (a+i a \tan (e+f x))^2}-\frac {i d^2 \sqrt {d \tan (e+f x)}}{4 f \left (a^3+i a^3 \tan (e+f x)\right )}-\frac {d^{5/2} \text {Subst}\left (\int \frac {\sqrt {2} \sqrt {d}+2 x}{-d-\sqrt {2} \sqrt {d} x-x^2} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{16 \sqrt {2} a^3 f}-\frac {d^{5/2} \text {Subst}\left (\int \frac {\sqrt {2} \sqrt {d}-2 x}{-d+\sqrt {2} \sqrt {d} x-x^2} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{16 \sqrt {2} a^3 f}-\frac {d^3 \text {Subst}\left (\int \frac {1}{d-\sqrt {2} \sqrt {d} x+x^2} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{16 a^3 f}-\frac {d^3 \text {Subst}\left (\int \frac {1}{d+\sqrt {2} \sqrt {d} x+x^2} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{16 a^3 f} \\ & = -\frac {d^{5/2} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{16 \sqrt {2} a^3 f}+\frac {d^{5/2} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{16 \sqrt {2} a^3 f}-\frac {d (d \tan (e+f x))^{3/2}}{6 f (a+i a \tan (e+f x))^3}+\frac {i d^2 \sqrt {d \tan (e+f x)}}{4 a f (a+i a \tan (e+f x))^2}-\frac {i d^2 \sqrt {d \tan (e+f x)}}{4 f \left (a^3+i a^3 \tan (e+f x)\right )}-\frac {d^{5/2} \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{8 \sqrt {2} a^3 f}+\frac {d^{5/2} \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{8 \sqrt {2} a^3 f} \\ & = \frac {d^{5/2} \arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{8 \sqrt {2} a^3 f}-\frac {d^{5/2} \arctan \left (1+\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{8 \sqrt {2} a^3 f}-\frac {d^{5/2} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{16 \sqrt {2} a^3 f}+\frac {d^{5/2} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{16 \sqrt {2} a^3 f}-\frac {d (d \tan (e+f x))^{3/2}}{6 f (a+i a \tan (e+f x))^3}+\frac {i d^2 \sqrt {d \tan (e+f x)}}{4 a f (a+i a \tan (e+f x))^2}-\frac {i d^2 \sqrt {d \tan (e+f x)}}{4 f \left (a^3+i a^3 \tan (e+f x)\right )} \\ \end{align*}
Time = 1.56 (sec) , antiderivative size = 195, normalized size of antiderivative = 0.59 \[ \int \frac {(d \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^3} \, dx=\frac {d \sec ^3(e+f x) \left (6 i \left (-d^2\right )^{3/4} \arctan \left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt [4]{-d^2}}\right ) (\cos (3 (e+f x))+i \sin (3 (e+f x)))+6 \left (-d^2\right )^{3/4} \text {arctanh}\left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt [4]{-d^2}}\right ) (-i \cos (3 (e+f x))+\sin (3 (e+f x)))+d \left (-3 \cos (e+f x)+3 \cos (3 (e+f x))+4 i \cos ^2(e+f x) \sin (e+f x)\right ) \sqrt {d \tan (e+f x)}\right )}{48 a^3 f (-i+\tan (e+f x))^3} \]
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Time = 0.78 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.35
method | result | size |
derivativedivides | \(\frac {2 d^{4} \left (-\frac {\arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {i d}}\right )}{16 d \sqrt {i d}}-\frac {\frac {2 \left (d \tan \left (f x +e \right )\right )^{\frac {5}{2}}-\frac {2 i d \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}}{\left (d \tan \left (f x +e \right )-i d \right )^{3}}+\frac {\arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {-i d}}\right )}{\sqrt {-i d}}}{16 d}\right )}{f \,a^{3}}\) | \(116\) |
default | \(\frac {2 d^{4} \left (-\frac {\arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {i d}}\right )}{16 d \sqrt {i d}}-\frac {\frac {2 \left (d \tan \left (f x +e \right )\right )^{\frac {5}{2}}-\frac {2 i d \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}}{\left (d \tan \left (f x +e \right )-i d \right )^{3}}+\frac {\arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {-i d}}\right )}{\sqrt {-i d}}}{16 d}\right )}{f \,a^{3}}\) | \(116\) |
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Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 587 vs. \(2 (255) = 510\).
Time = 0.25 (sec) , antiderivative size = 587, normalized size of antiderivative = 1.78 \[ \int \frac {(d \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^3} \, dx=\frac {{\left (12 \, a^{3} f \sqrt {\frac {i \, d^{5}}{64 \, a^{6} f^{2}}} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (-\frac {2 \, {\left (i \, d^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + 8 \, {\left (i \, a^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + i \, a^{3} f\right )} \sqrt {\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {i \, d^{5}}{64 \, a^{6} f^{2}}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{d^{2}}\right ) - 12 \, a^{3} f \sqrt {\frac {i \, d^{5}}{64 \, a^{6} f^{2}}} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (-\frac {2 \, {\left (i \, d^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + 8 \, {\left (-i \, a^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} - i \, a^{3} f\right )} \sqrt {\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {i \, d^{5}}{64 \, a^{6} f^{2}}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{d^{2}}\right ) + 12 \, a^{3} f \sqrt {-\frac {i \, d^{5}}{64 \, a^{6} f^{2}}} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (\frac {{\left (d^{3} + 8 \, {\left (a^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{3} f\right )} \sqrt {\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {i \, d^{5}}{64 \, a^{6} f^{2}}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{8 \, a^{3} f}\right ) - 12 \, a^{3} f \sqrt {-\frac {i \, d^{5}}{64 \, a^{6} f^{2}}} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (\frac {{\left (d^{3} - 8 \, {\left (a^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{3} f\right )} \sqrt {\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {i \, d^{5}}{64 \, a^{6} f^{2}}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{8 \, a^{3} f}\right ) + {\left (-2 i \, d^{2} e^{\left (6 i \, f x + 6 i \, e\right )} + i \, d^{2} e^{\left (4 i \, f x + 4 i \, e\right )} + 2 i \, d^{2} e^{\left (2 i \, f x + 2 i \, e\right )} - i \, d^{2}\right )} \sqrt {\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-6 i \, f x - 6 i \, e\right )}}{48 \, a^{3} f} \]
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\[ \int \frac {(d \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^3} \, dx=\frac {i \int \frac {\left (d \tan {\left (e + f x \right )}\right )^{\frac {5}{2}}}{\tan ^{3}{\left (e + f x \right )} - 3 i \tan ^{2}{\left (e + f x \right )} - 3 \tan {\left (e + f x \right )} + i}\, dx}{a^{3}} \]
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Exception generated. \[ \int \frac {(d \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^3} \, dx=\text {Exception raised: RuntimeError} \]
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none
Time = 0.83 (sec) , antiderivative size = 212, normalized size of antiderivative = 0.64 \[ \int \frac {(d \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^3} \, dx=-\frac {1}{24} \, d^{4} {\left (\frac {3 \, \sqrt {2} \arctan \left (\frac {8 \, \sqrt {d^{2}} \sqrt {d \tan \left (f x + e\right )}}{4 i \, \sqrt {2} d^{\frac {3}{2}} + 4 \, \sqrt {2} \sqrt {d^{2}} \sqrt {d}}\right )}{a^{3} d^{\frac {3}{2}} f {\left (\frac {i \, d}{\sqrt {d^{2}}} + 1\right )}} + \frac {3 \, \sqrt {2} \arctan \left (\frac {8 \, \sqrt {d^{2}} \sqrt {d \tan \left (f x + e\right )}}{-4 i \, \sqrt {2} d^{\frac {3}{2}} + 4 \, \sqrt {2} \sqrt {d^{2}} \sqrt {d}}\right )}{a^{3} d^{\frac {3}{2}} f {\left (-\frac {i \, d}{\sqrt {d^{2}}} + 1\right )}} + \frac {2 \, {\left (3 \, \sqrt {d \tan \left (f x + e\right )} d^{2} \tan \left (f x + e\right )^{2} - i \, \sqrt {d \tan \left (f x + e\right )} d^{2} \tan \left (f x + e\right )\right )}}{{\left (d \tan \left (f x + e\right ) - i \, d\right )}^{3} a^{3} d f}\right )} \]
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Time = 5.63 (sec) , antiderivative size = 158, normalized size of antiderivative = 0.48 \[ \int \frac {(d \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^3} \, dx=\frac {{\left (-1\right )}^{1/4}\,d^{5/2}\,\mathrm {atanh}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{\sqrt {d}}\right )}{8\,a^3\,f}-\frac {{\left (-1\right )}^{1/4}\,d^{5/2}\,\mathrm {atan}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{\sqrt {d}}\right )}{8\,a^3\,f}-\frac {-\frac {d^3\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{5/2}}{4\,a^3\,f}+\frac {d^4\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}\,1{}\mathrm {i}}{12\,a^3\,f}}{-d^3\,{\mathrm {tan}\left (e+f\,x\right )}^3+d^3\,{\mathrm {tan}\left (e+f\,x\right )}^2\,3{}\mathrm {i}+3\,d^3\,\mathrm {tan}\left (e+f\,x\right )-d^3\,1{}\mathrm {i}} \]
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