Integrand size = 30, antiderivative size = 150 \[ \int (a+i a \tan (e+f x))^3 \sqrt {c+d \tan (e+f x)} \, dx=-\frac {8 i a^3 \sqrt {c-i d} \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{f}+\frac {8 i a^3 \sqrt {c+d \tan (e+f x)}}{f}+\frac {4 a^3 (i c-6 d) (c+d \tan (e+f x))^{3/2}}{15 d^2 f}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right ) (c+d \tan (e+f x))^{3/2}}{5 d f} \]
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Time = 0.51 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3637, 3673, 3609, 3618, 65, 214} \[ \int (a+i a \tan (e+f x))^3 \sqrt {c+d \tan (e+f x)} \, dx=-\frac {8 i a^3 \sqrt {c-i d} \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{f}+\frac {4 a^3 (-6 d+i c) (c+d \tan (e+f x))^{3/2}}{15 d^2 f}+\frac {8 i a^3 \sqrt {c+d \tan (e+f x)}}{f}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right ) (c+d \tan (e+f x))^{3/2}}{5 d f} \]
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Rule 65
Rule 214
Rule 3609
Rule 3618
Rule 3637
Rule 3673
Rubi steps \begin{align*} \text {integral}& = -\frac {2 \left (a^3+i a^3 \tan (e+f x)\right ) (c+d \tan (e+f x))^{3/2}}{5 d f}+\frac {(2 a) \int (a+i a \tan (e+f x)) (a (i c+4 d)+a (c+6 i d) \tan (e+f x)) \sqrt {c+d \tan (e+f x)} \, dx}{5 d} \\ & = \frac {4 a^3 (i c-6 d) (c+d \tan (e+f x))^{3/2}}{15 d^2 f}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right ) (c+d \tan (e+f x))^{3/2}}{5 d f}+\frac {(2 a) \int \sqrt {c+d \tan (e+f x)} \left (10 a^2 d+10 i a^2 d \tan (e+f x)\right ) \, dx}{5 d} \\ & = \frac {8 i a^3 \sqrt {c+d \tan (e+f x)}}{f}+\frac {4 a^3 (i c-6 d) (c+d \tan (e+f x))^{3/2}}{15 d^2 f}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right ) (c+d \tan (e+f x))^{3/2}}{5 d f}+\frac {(2 a) \int \frac {10 a^2 (c-i d) d+10 a^2 d (i c+d) \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}} \, dx}{5 d} \\ & = \frac {8 i a^3 \sqrt {c+d \tan (e+f x)}}{f}+\frac {4 a^3 (i c-6 d) (c+d \tan (e+f x))^{3/2}}{15 d^2 f}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right ) (c+d \tan (e+f x))^{3/2}}{5 d f}+\frac {\left (40 i a^5 (c-i d)^2 d\right ) \text {Subst}\left (\int \frac {1}{\left (100 a^4 d^2 (i c+d)^2+10 a^2 (c-i d) d x\right ) \sqrt {c+\frac {x}{10 a^2 (i c+d)}}} \, dx,x,10 a^2 d (i c+d) \tan (e+f x)\right )}{f} \\ & = \frac {8 i a^3 \sqrt {c+d \tan (e+f x)}}{f}+\frac {4 a^3 (i c-6 d) (c+d \tan (e+f x))^{3/2}}{15 d^2 f}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right ) (c+d \tan (e+f x))^{3/2}}{5 d f}-\frac {\left (800 a^7 (c-i d)^3 d\right ) \text {Subst}\left (\int \frac {1}{-100 a^4 c (c-i d) d (i c+d)+100 a^4 d^2 (i c+d)^2+100 a^4 (c-i d) d (i c+d) x^2} \, dx,x,\sqrt {c+d \tan (e+f x)}\right )}{f} \\ & = -\frac {8 i a^3 \sqrt {c-i d} \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{f}+\frac {8 i a^3 \sqrt {c+d \tan (e+f x)}}{f}+\frac {4 a^3 (i c-6 d) (c+d \tan (e+f x))^{3/2}}{15 d^2 f}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right ) (c+d \tan (e+f x))^{3/2}}{5 d f} \\ \end{align*}
Time = 2.14 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.80 \[ \int (a+i a \tan (e+f x))^3 \sqrt {c+d \tan (e+f x)} \, dx=\frac {i a^2 \left (-8 a \sqrt {c-i d} \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )+8 a \sqrt {c+d \tan (e+f x)}+\frac {2 a (c+3 i d) (c+d \tan (e+f x))^{3/2}}{3 d^2}-\frac {2 a (c+d \tan (e+f x))^{5/2}}{5 d^2}\right )}{f} \]
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Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1271 vs. \(2 (128 ) = 256\).
Time = 1.27 (sec) , antiderivative size = 1272, normalized size of antiderivative = 8.48
method | result | size |
derivativedivides | \(\text {Expression too large to display}\) | \(1272\) |
default | \(\text {Expression too large to display}\) | \(1272\) |
parts | \(\text {Expression too large to display}\) | \(1689\) |
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Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 512 vs. \(2 (124) = 248\).
Time = 0.28 (sec) , antiderivative size = 512, normalized size of antiderivative = 3.41 \[ \int (a+i a \tan (e+f x))^3 \sqrt {c+d \tan (e+f x)} \, dx=\frac {2 \, {\left (15 \, {\left (d^{2} f e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, d^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + d^{2} f\right )} \sqrt {-\frac {a^{6} c - i \, a^{6} d}{f^{2}}} \log \left (\frac {2 \, {\left (a^{3} c + {\left (i \, f e^{\left (2 i \, f x + 2 i \, e\right )} + i \, f\right )} \sqrt {\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {a^{6} c - i \, a^{6} d}{f^{2}}} + {\left (a^{3} c - i \, a^{3} d\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{a^{3}}\right ) - 15 \, {\left (d^{2} f e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, d^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + d^{2} f\right )} \sqrt {-\frac {a^{6} c - i \, a^{6} d}{f^{2}}} \log \left (\frac {2 \, {\left (a^{3} c + {\left (-i \, f e^{\left (2 i \, f x + 2 i \, e\right )} - i \, f\right )} \sqrt {\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {a^{6} c - i \, a^{6} d}{f^{2}}} + {\left (a^{3} c - i \, a^{3} d\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{a^{3}}\right ) - 2 \, {\left (-i \, a^{3} c^{2} + 7 \, a^{3} c d - 24 i \, a^{3} d^{2} + {\left (-i \, a^{3} c^{2} + 8 \, a^{3} c d - 39 i \, a^{3} d^{2}\right )} e^{\left (4 i \, f x + 4 i \, e\right )} + {\left (-2 i \, a^{3} c^{2} + 15 \, a^{3} c d - 57 i \, a^{3} d^{2}\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} \sqrt {\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )}}{15 \, {\left (d^{2} f e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, d^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + d^{2} f\right )}} \]
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\[ \int (a+i a \tan (e+f x))^3 \sqrt {c+d \tan (e+f x)} \, dx=- i a^{3} \left (\int i \sqrt {c + d \tan {\left (e + f x \right )}}\, dx + \int \left (- 3 \sqrt {c + d \tan {\left (e + f x \right )}} \tan {\left (e + f x \right )}\right )\, dx + \int \sqrt {c + d \tan {\left (e + f x \right )}} \tan ^{3}{\left (e + f x \right )}\, dx + \int \left (- 3 i \sqrt {c + d \tan {\left (e + f x \right )}} \tan ^{2}{\left (e + f x \right )}\right )\, dx\right ) \]
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\[ \int (a+i a \tan (e+f x))^3 \sqrt {c+d \tan (e+f x)} \, dx=\int { {\left (i \, a \tan \left (f x + e\right ) + a\right )}^{3} \sqrt {d \tan \left (f x + e\right ) + c} \,d x } \]
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Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 271 vs. \(2 (124) = 248\).
Time = 0.65 (sec) , antiderivative size = 271, normalized size of antiderivative = 1.81 \[ \int (a+i a \tan (e+f x))^3 \sqrt {c+d \tan (e+f x)} \, dx=-\frac {16 \, {\left (-i \, a^{3} c - a^{3} d\right )} \arctan \left (\frac {2 \, {\left (\sqrt {d \tan \left (f x + e\right ) + c} c - \sqrt {c^{2} + d^{2}} \sqrt {d \tan \left (f x + e\right ) + c}\right )}}{c \sqrt {-2 \, c + 2 \, \sqrt {c^{2} + d^{2}}} - i \, \sqrt {-2 \, c + 2 \, \sqrt {c^{2} + d^{2}}} d - \sqrt {c^{2} + d^{2}} \sqrt {-2 \, c + 2 \, \sqrt {c^{2} + d^{2}}}}\right )}{\sqrt {-2 \, c + 2 \, \sqrt {c^{2} + d^{2}}} f {\left (-\frac {i \, d}{c - \sqrt {c^{2} + d^{2}}} + 1\right )}} - \frac {2 \, {\left (3 i \, {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}} a^{3} d^{8} f^{4} - 5 i \, {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} a^{3} c d^{8} f^{4} + 15 \, {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} a^{3} d^{9} f^{4} - 60 i \, \sqrt {d \tan \left (f x + e\right ) + c} a^{3} d^{10} f^{4}\right )}}{15 \, d^{10} f^{5}} \]
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Time = 9.74 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.33 \[ \int (a+i a \tan (e+f x))^3 \sqrt {c+d \tan (e+f x)} \, dx=-\left (\left (c-d\,1{}\mathrm {i}\right )\,\left (\frac {a^3\,\left (c-d\,1{}\mathrm {i}\right )\,2{}\mathrm {i}}{d^2\,f}-\frac {a^3\,\left (c+d\,1{}\mathrm {i}\right )\,4{}\mathrm {i}}{d^2\,f}\right )+\frac {a^3\,{\left (c+d\,1{}\mathrm {i}\right )}^2\,2{}\mathrm {i}}{d^2\,f}\right )\,\sqrt {c+d\,\mathrm {tan}\left (e+f\,x\right )}-\left (\frac {a^3\,\left (c-d\,1{}\mathrm {i}\right )\,2{}\mathrm {i}}{3\,d^2\,f}-\frac {a^3\,\left (c+d\,1{}\mathrm {i}\right )\,4{}\mathrm {i}}{3\,d^2\,f}\right )\,{\left (c+d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}-\frac {a^3\,{\left (c+d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{5/2}\,2{}\mathrm {i}}{5\,d^2\,f}+\frac {\sqrt {16{}\mathrm {i}}\,a^3\,\mathrm {atan}\left (\frac {\sqrt {16{}\mathrm {i}}\,\sqrt {c+d\,\mathrm {tan}\left (e+f\,x\right )}\,1{}\mathrm {i}}{4\,\sqrt {d+c\,1{}\mathrm {i}}}\right )\,\sqrt {d+c\,1{}\mathrm {i}}\,2{}\mathrm {i}}{f} \]
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