Integrand size = 28, antiderivative size = 191 \[ \int \frac {\sqrt {e \sec (c+d x)}}{(a+i a \tan (c+d x))^4} \, dx=\frac {2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {e \sec (c+d x)}}{33 a^4 d}+\frac {2 e \sin (c+d x)}{33 a^4 d \sqrt {e \sec (c+d x)}}+\frac {2 i \sqrt {e \sec (c+d x)}}{15 d (a+i a \tan (c+d x))^4}+\frac {14 i \sqrt {e \sec (c+d x)}}{165 a d (a+i a \tan (c+d x))^3}+\frac {4 i e^2}{33 d (e \sec (c+d x))^{3/2} \left (a^4+i a^4 \tan (c+d x)\right )} \]
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Time = 0.25 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {3583, 3581, 3854, 3856, 2720} \[ \int \frac {\sqrt {e \sec (c+d x)}}{(a+i a \tan (c+d x))^4} \, dx=\frac {4 i e^2}{33 d \left (a^4+i a^4 \tan (c+d x)\right ) (e \sec (c+d x))^{3/2}}+\frac {2 e \sin (c+d x)}{33 a^4 d \sqrt {e \sec (c+d x)}}+\frac {2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {e \sec (c+d x)}}{33 a^4 d}+\frac {14 i \sqrt {e \sec (c+d x)}}{165 a d (a+i a \tan (c+d x))^3}+\frac {2 i \sqrt {e \sec (c+d x)}}{15 d (a+i a \tan (c+d x))^4} \]
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Rule 2720
Rule 3581
Rule 3583
Rule 3854
Rule 3856
Rubi steps \begin{align*} \text {integral}& = \frac {2 i \sqrt {e \sec (c+d x)}}{15 d (a+i a \tan (c+d x))^4}+\frac {7 \int \frac {\sqrt {e \sec (c+d x)}}{(a+i a \tan (c+d x))^3} \, dx}{15 a} \\ & = \frac {2 i \sqrt {e \sec (c+d x)}}{15 d (a+i a \tan (c+d x))^4}+\frac {14 i \sqrt {e \sec (c+d x)}}{165 a d (a+i a \tan (c+d x))^3}+\frac {7 \int \frac {\sqrt {e \sec (c+d x)}}{(a+i a \tan (c+d x))^2} \, dx}{33 a^2} \\ & = \frac {2 i \sqrt {e \sec (c+d x)}}{15 d (a+i a \tan (c+d x))^4}+\frac {14 i \sqrt {e \sec (c+d x)}}{165 a d (a+i a \tan (c+d x))^3}+\frac {4 i e^2}{33 d (e \sec (c+d x))^{3/2} \left (a^4+i a^4 \tan (c+d x)\right )}+\frac {e^2 \int \frac {1}{(e \sec (c+d x))^{3/2}} \, dx}{11 a^4} \\ & = \frac {2 e \sin (c+d x)}{33 a^4 d \sqrt {e \sec (c+d x)}}+\frac {2 i \sqrt {e \sec (c+d x)}}{15 d (a+i a \tan (c+d x))^4}+\frac {14 i \sqrt {e \sec (c+d x)}}{165 a d (a+i a \tan (c+d x))^3}+\frac {4 i e^2}{33 d (e \sec (c+d x))^{3/2} \left (a^4+i a^4 \tan (c+d x)\right )}+\frac {\int \sqrt {e \sec (c+d x)} \, dx}{33 a^4} \\ & = \frac {2 e \sin (c+d x)}{33 a^4 d \sqrt {e \sec (c+d x)}}+\frac {2 i \sqrt {e \sec (c+d x)}}{15 d (a+i a \tan (c+d x))^4}+\frac {14 i \sqrt {e \sec (c+d x)}}{165 a d (a+i a \tan (c+d x))^3}+\frac {4 i e^2}{33 d (e \sec (c+d x))^{3/2} \left (a^4+i a^4 \tan (c+d x)\right )}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{33 a^4} \\ & = \frac {2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {e \sec (c+d x)}}{33 a^4 d}+\frac {2 e \sin (c+d x)}{33 a^4 d \sqrt {e \sec (c+d x)}}+\frac {2 i \sqrt {e \sec (c+d x)}}{15 d (a+i a \tan (c+d x))^4}+\frac {14 i \sqrt {e \sec (c+d x)}}{165 a d (a+i a \tan (c+d x))^3}+\frac {4 i e^2}{33 d (e \sec (c+d x))^{3/2} \left (a^4+i a^4 \tan (c+d x)\right )} \\ \end{align*}
Time = 1.33 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.72 \[ \int \frac {\sqrt {e \sec (c+d x)}}{(a+i a \tan (c+d x))^4} \, dx=\frac {\sec ^4(c+d x) \sqrt {e \sec (c+d x)} \left (40 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) (\cos (4 (c+d x))+i \sin (4 (c+d x)))+i (64+112 \cos (2 (c+d x))+48 \cos (4 (c+d x))+54 i \sin (2 (c+d x))+37 i \sin (4 (c+d x)))\right )}{660 a^4 d (-i+\tan (c+d x))^4} \]
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Time = 6.55 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.80
method | result | size |
default | \(\frac {2 i \left (\left (5 \cos \left (d x +c \right )+5\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, F\left (i \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right ), i\right )+i \sin \left (d x +c \right ) \cos \left (d x +c \right ) \left (-88 \left (\cos ^{6}\left (d x +c \right )\right )+16 \left (\cos ^{4}\left (d x +c \right )\right )-3 \left (\cos ^{2}\left (d x +c \right )\right )-5\right )+\left (\cos ^{6}\left (d x +c \right )\right ) \left (88 \left (\cos ^{2}\left (d x +c \right )\right )-60\right )\right ) \sqrt {e \sec \left (d x +c \right )}}{165 a^{4} d}\) | \(152\) |
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Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.08 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.64 \[ \int \frac {\sqrt {e \sec (c+d x)}}{(a+i a \tan (c+d x))^4} \, dx=\frac {{\left (\sqrt {2} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (85 i \, e^{\left (8 i \, d x + 8 i \, c\right )} + 166 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 128 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 58 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 11 i\right )} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )} - 80 i \, \sqrt {2} \sqrt {e} e^{\left (8 i \, d x + 8 i \, c\right )} {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right )\right )} e^{\left (-8 i \, d x - 8 i \, c\right )}}{1320 \, a^{4} d} \]
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\[ \int \frac {\sqrt {e \sec (c+d x)}}{(a+i a \tan (c+d x))^4} \, dx=\frac {\int \frac {\sqrt {e \sec {\left (c + d x \right )}}}{\tan ^{4}{\left (c + d x \right )} - 4 i \tan ^{3}{\left (c + d x \right )} - 6 \tan ^{2}{\left (c + d x \right )} + 4 i \tan {\left (c + d x \right )} + 1}\, dx}{a^{4}} \]
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Exception generated. \[ \int \frac {\sqrt {e \sec (c+d x)}}{(a+i a \tan (c+d x))^4} \, dx=\text {Exception raised: RuntimeError} \]
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\[ \int \frac {\sqrt {e \sec (c+d x)}}{(a+i a \tan (c+d x))^4} \, dx=\int { \frac {\sqrt {e \sec \left (d x + c\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{4}} \,d x } \]
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Timed out. \[ \int \frac {\sqrt {e \sec (c+d x)}}{(a+i a \tan (c+d x))^4} \, dx=\int \frac {\sqrt {\frac {e}{\cos \left (c+d\,x\right )}}}{{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^4} \,d x \]
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