Integrand size = 28, antiderivative size = 111 \[ \int \frac {(a+i a \tan (c+d x))^3}{(e \sec (c+d x))^{5/2}} \, dx=\frac {6 i a^3}{5 d e^2 \sqrt {e \sec (c+d x)}}-\frac {6 a^3 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d e^2 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}-\frac {4 i a (a+i a \tan (c+d x))^2}{5 d (e \sec (c+d x))^{5/2}} \]
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Time = 0.14 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3577, 3567, 3856, 2719} \[ \int \frac {(a+i a \tan (c+d x))^3}{(e \sec (c+d x))^{5/2}} \, dx=\frac {6 i a^3}{5 d e^2 \sqrt {e \sec (c+d x)}}-\frac {6 a^3 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d e^2 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}-\frac {4 i a (a+i a \tan (c+d x))^2}{5 d (e \sec (c+d x))^{5/2}} \]
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Rule 2719
Rule 3567
Rule 3577
Rule 3856
Rubi steps \begin{align*} \text {integral}& = -\frac {4 i a (a+i a \tan (c+d x))^2}{5 d (e \sec (c+d x))^{5/2}}-\frac {\left (3 a^2\right ) \int \frac {a+i a \tan (c+d x)}{\sqrt {e \sec (c+d x)}} \, dx}{5 e^2} \\ & = \frac {6 i a^3}{5 d e^2 \sqrt {e \sec (c+d x)}}-\frac {4 i a (a+i a \tan (c+d x))^2}{5 d (e \sec (c+d x))^{5/2}}-\frac {\left (3 a^3\right ) \int \frac {1}{\sqrt {e \sec (c+d x)}} \, dx}{5 e^2} \\ & = \frac {6 i a^3}{5 d e^2 \sqrt {e \sec (c+d x)}}-\frac {4 i a (a+i a \tan (c+d x))^2}{5 d (e \sec (c+d x))^{5/2}}-\frac {\left (3 a^3\right ) \int \sqrt {\cos (c+d x)} \, dx}{5 e^2 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}} \\ & = \frac {6 i a^3}{5 d e^2 \sqrt {e \sec (c+d x)}}-\frac {6 a^3 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d e^2 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}-\frac {4 i a (a+i a \tan (c+d x))^2}{5 d (e \sec (c+d x))^{5/2}} \\ \end{align*}
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 2.40 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.97 \[ \int \frac {(a+i a \tan (c+d x))^3}{(e \sec (c+d x))^{5/2}} \, dx=-\frac {4 i a^3 e^{2 i (c+d x)} \left (1+e^{2 i (c+d x)}-\sqrt {1+e^{2 i (c+d x)}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )\right )}{5 d e^2 \left (1+e^{2 i (c+d x)}\right ) \sqrt {e \sec (c+d x)}} \]
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Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 323 vs. \(2 (120 ) = 240\).
Time = 14.92 (sec) , antiderivative size = 324, normalized size of antiderivative = 2.92
method | result | size |
risch | \(-\frac {2 i \left ({\mathrm e}^{2 i \left (d x +c \right )}-3\right ) a^{3} \sqrt {2}}{5 d \,e^{2} \sqrt {\frac {e \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}}+\frac {3 i \left (-\frac {2 \left (e \,{\mathrm e}^{2 i \left (d x +c \right )}+e \right )}{e \sqrt {{\mathrm e}^{i \left (d x +c \right )} \left (e \,{\mathrm e}^{2 i \left (d x +c \right )}+e \right )}}+\frac {i \sqrt {-i \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}\, \sqrt {2}\, \sqrt {i \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}\, \sqrt {i {\mathrm e}^{i \left (d x +c \right )}}\, \left (-2 i E\left (\sqrt {-i \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}, \frac {\sqrt {2}}{2}\right )+i F\left (\sqrt {-i \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}, \frac {\sqrt {2}}{2}\right )\right )}{\sqrt {e \,{\mathrm e}^{3 i \left (d x +c \right )}+e \,{\mathrm e}^{i \left (d x +c \right )}}}\right ) a^{3} \sqrt {2}\, \sqrt {e \,{\mathrm e}^{i \left (d x +c \right )} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}}{5 d \,e^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) \sqrt {\frac {e \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}}\) | \(324\) |
default | \(\text {Expression too large to display}\) | \(882\) |
parts | \(\text {Expression too large to display}\) | \(1289\) |
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Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.07 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.85 \[ \int \frac {(a+i a \tan (c+d x))^3}{(e \sec (c+d x))^{5/2}} \, dx=-\frac {2 \, {\left (3 i \, \sqrt {2} a^{3} \sqrt {e} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right )\right ) + \sqrt {2} {\left (i \, a^{3} e^{\left (3 i \, d x + 3 i \, c\right )} + i \, a^{3} e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )}\right )}}{5 \, d e^{3}} \]
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\[ \int \frac {(a+i a \tan (c+d x))^3}{(e \sec (c+d x))^{5/2}} \, dx=- i a^{3} \left (\int \frac {i}{\left (e \sec {\left (c + d x \right )}\right )^{\frac {5}{2}}}\, dx + \int \left (- \frac {3 \tan {\left (c + d x \right )}}{\left (e \sec {\left (c + d x \right )}\right )^{\frac {5}{2}}}\right )\, dx + \int \frac {\tan ^{3}{\left (c + d x \right )}}{\left (e \sec {\left (c + d x \right )}\right )^{\frac {5}{2}}}\, dx + \int \left (- \frac {3 i \tan ^{2}{\left (c + d x \right )}}{\left (e \sec {\left (c + d x \right )}\right )^{\frac {5}{2}}}\right )\, dx\right ) \]
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\[ \int \frac {(a+i a \tan (c+d x))^3}{(e \sec (c+d x))^{5/2}} \, dx=\int { \frac {{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3}}{\left (e \sec \left (d x + c\right )\right )^{\frac {5}{2}}} \,d x } \]
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\[ \int \frac {(a+i a \tan (c+d x))^3}{(e \sec (c+d x))^{5/2}} \, dx=\int { \frac {{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3}}{\left (e \sec \left (d x + c\right )\right )^{\frac {5}{2}}} \,d x } \]
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Timed out. \[ \int \frac {(a+i a \tan (c+d x))^3}{(e \sec (c+d x))^{5/2}} \, dx=\int \frac {{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^3}{{\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \]
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