Integrand size = 28, antiderivative size = 132 \[ \int \frac {(e \sec (c+d x))^{5/2}}{(a+i a \tan (c+d x))^3} \, dx=-\frac {2 e^2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {e \sec (c+d x)}}{21 a^3 d}+\frac {4 i e^2 \sqrt {e \sec (c+d x)}}{7 a d (a+i a \tan (c+d x))^2}-\frac {2 i e^2 \sqrt {e \sec (c+d x)}}{21 d \left (a^3+i a^3 \tan (c+d x)\right )} \]
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Time = 0.16 (sec) , antiderivative size = 132, 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 = {3581, 3583, 3856, 2720} \[ \int \frac {(e \sec (c+d x))^{5/2}}{(a+i a \tan (c+d x))^3} \, dx=-\frac {2 i e^2 \sqrt {e \sec (c+d x)}}{21 d \left (a^3+i a^3 \tan (c+d x)\right )}-\frac {2 e^2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {e \sec (c+d x)}}{21 a^3 d}+\frac {4 i e^2 \sqrt {e \sec (c+d x)}}{7 a d (a+i a \tan (c+d x))^2} \]
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Rule 2720
Rule 3581
Rule 3583
Rule 3856
Rubi steps \begin{align*} \text {integral}& = \frac {4 i e^2 \sqrt {e \sec (c+d x)}}{7 a d (a+i a \tan (c+d x))^2}-\frac {e^2 \int \frac {\sqrt {e \sec (c+d x)}}{a+i a \tan (c+d x)} \, dx}{7 a^2} \\ & = \frac {4 i e^2 \sqrt {e \sec (c+d x)}}{7 a d (a+i a \tan (c+d x))^2}-\frac {2 i e^2 \sqrt {e \sec (c+d x)}}{21 d \left (a^3+i a^3 \tan (c+d x)\right )}-\frac {e^2 \int \sqrt {e \sec (c+d x)} \, dx}{21 a^3} \\ & = \frac {4 i e^2 \sqrt {e \sec (c+d x)}}{7 a d (a+i a \tan (c+d x))^2}-\frac {2 i e^2 \sqrt {e \sec (c+d x)}}{21 d \left (a^3+i a^3 \tan (c+d x)\right )}-\frac {\left (e^2 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{21 a^3} \\ & = -\frac {2 e^2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {e \sec (c+d x)}}{21 a^3 d}+\frac {4 i e^2 \sqrt {e \sec (c+d x)}}{7 a d (a+i a \tan (c+d x))^2}-\frac {2 i e^2 \sqrt {e \sec (c+d x)}}{21 d \left (a^3+i a^3 \tan (c+d x)\right )} \\ \end{align*}
Time = 1.57 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.79 \[ \int \frac {(e \sec (c+d x))^{5/2}}{(a+i a \tan (c+d x))^3} \, dx=\frac {(e \sec (c+d x))^{5/2} \left (-5 i-5 i \cos (2 (c+d x))+2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) (\cos (2 (c+d x))+i \sin (2 (c+d x)))-\sin (2 (c+d x))\right )}{21 a^3 d (-i+\tan (c+d x))^2} \]
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Time = 6.61 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.45
method | result | size |
default | \(\frac {2 e^{2} \left (12 i \left (\cos ^{4}\left (d x +c \right )\right )-i \cos \left (d x +c \right ) F\left (i \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right ), i\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}-i F\left (i \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right ), i\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}+12 \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )-7 i \left (\cos ^{2}\left (d x +c \right )\right )-\sin \left (d x +c \right ) \cos \left (d x +c \right )\right ) \sqrt {e \sec \left (d x +c \right )}}{21 a^{3} d}\) | \(191\) |
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Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.07 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.84 \[ \int \frac {(e \sec (c+d x))^{5/2}}{(a+i a \tan (c+d x))^3} \, dx=\frac {{\left (2 i \, \sqrt {2} e^{\frac {5}{2}} e^{\left (4 i \, d x + 4 i \, c\right )} {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right ) + \sqrt {2} {\left (2 i \, e^{2} e^{\left (4 i \, d x + 4 i \, c\right )} + 5 i \, e^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + 3 i \, e^{2}\right )} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )}\right )} e^{\left (-4 i \, d x - 4 i \, c\right )}}{21 \, a^{3} d} \]
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\[ \int \frac {(e \sec (c+d x))^{5/2}}{(a+i a \tan (c+d x))^3} \, dx=\frac {i \int \frac {\left (e \sec {\left (c + d x \right )}\right )^{\frac {5}{2}}}{\tan ^{3}{\left (c + d x \right )} - 3 i \tan ^{2}{\left (c + d x \right )} - 3 \tan {\left (c + d x \right )} + i}\, dx}{a^{3}} \]
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Exception generated. \[ \int \frac {(e \sec (c+d x))^{5/2}}{(a+i a \tan (c+d x))^3} \, dx=\text {Exception raised: RuntimeError} \]
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\[ \int \frac {(e \sec (c+d x))^{5/2}}{(a+i a \tan (c+d x))^3} \, dx=\int { \frac {\left (e \sec \left (d x + c\right )\right )^{\frac {5}{2}}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3}} \,d x } \]
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Timed out. \[ \int \frac {(e \sec (c+d x))^{5/2}}{(a+i a \tan (c+d x))^3} \, dx=\int \frac {{\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^{5/2}}{{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^3} \,d x \]
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