Integrand size = 28, antiderivative size = 157 \[ \int \frac {(e \sec (c+d x))^{13/2}}{(a+i a \tan (c+d x))^4} \, dx=-\frac {10 e^6 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {e \sec (c+d x)}}{a^4 d}-\frac {10 e^5 (e \sec (c+d x))^{3/2} \sin (c+d x)}{a^4 d}+\frac {4 i e^2 (e \sec (c+d x))^{9/2}}{3 a d (a+i a \tan (c+d x))^3}+\frac {12 i e^4 (e \sec (c+d x))^{5/2}}{d \left (a^4+i a^4 \tan (c+d x)\right )} \]
[Out]
Time = 0.22 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3581, 3853, 3856, 2720} \[ \int \frac {(e \sec (c+d x))^{13/2}}{(a+i a \tan (c+d x))^4} \, dx=-\frac {10 e^6 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {e \sec (c+d x)}}{a^4 d}-\frac {10 e^5 \sin (c+d x) (e \sec (c+d x))^{3/2}}{a^4 d}+\frac {12 i e^4 (e \sec (c+d x))^{5/2}}{d \left (a^4+i a^4 \tan (c+d x)\right )}+\frac {4 i e^2 (e \sec (c+d x))^{9/2}}{3 a d (a+i a \tan (c+d x))^3} \]
[In]
[Out]
Rule 2720
Rule 3581
Rule 3853
Rule 3856
Rubi steps \begin{align*} \text {integral}& = \frac {4 i e^2 (e \sec (c+d x))^{9/2}}{3 a d (a+i a \tan (c+d x))^3}-\frac {\left (3 e^2\right ) \int \frac {(e \sec (c+d x))^{9/2}}{(a+i a \tan (c+d x))^2} \, dx}{a^2} \\ & = \frac {4 i e^2 (e \sec (c+d x))^{9/2}}{3 a d (a+i a \tan (c+d x))^3}+\frac {12 i e^4 (e \sec (c+d x))^{5/2}}{d \left (a^4+i a^4 \tan (c+d x)\right )}-\frac {\left (15 e^4\right ) \int (e \sec (c+d x))^{5/2} \, dx}{a^4} \\ & = -\frac {10 e^5 (e \sec (c+d x))^{3/2} \sin (c+d x)}{a^4 d}+\frac {4 i e^2 (e \sec (c+d x))^{9/2}}{3 a d (a+i a \tan (c+d x))^3}+\frac {12 i e^4 (e \sec (c+d x))^{5/2}}{d \left (a^4+i a^4 \tan (c+d x)\right )}-\frac {\left (5 e^6\right ) \int \sqrt {e \sec (c+d x)} \, dx}{a^4} \\ & = -\frac {10 e^5 (e \sec (c+d x))^{3/2} \sin (c+d x)}{a^4 d}+\frac {4 i e^2 (e \sec (c+d x))^{9/2}}{3 a d (a+i a \tan (c+d x))^3}+\frac {12 i e^4 (e \sec (c+d x))^{5/2}}{d \left (a^4+i a^4 \tan (c+d x)\right )}-\frac {\left (5 e^6 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{a^4} \\ & = -\frac {10 e^6 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {e \sec (c+d x)}}{a^4 d}-\frac {10 e^5 (e \sec (c+d x))^{3/2} \sin (c+d x)}{a^4 d}+\frac {4 i e^2 (e \sec (c+d x))^{9/2}}{3 a d (a+i a \tan (c+d x))^3}+\frac {12 i e^4 (e \sec (c+d x))^{5/2}}{d \left (a^4+i a^4 \tan (c+d x)\right )} \\ \end{align*}
Time = 1.56 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.85 \[ \int \frac {(e \sec (c+d x))^{13/2}}{(a+i a \tan (c+d x))^4} \, dx=\frac {i e^6 \sec ^5(c+d x) \sqrt {e \sec (c+d x)} \left (21+19 \cos (2 (c+d x))+30 i \cos ^{\frac {3}{2}}(c+d x) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) (\cos (c+d x)+i \sin (c+d x))+11 i \sin (2 (c+d x))\right ) (\cos (3 (c+d x))+i \sin (3 (c+d x)))}{3 a^4 d (-i+\tan (c+d x))^4} \]
[In]
[Out]
Time = 10.10 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.10
method | result | size |
default | \(\frac {2 e^{6} \left (-15 i \cos \left (d x +c \right ) F\left (i \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right ), i\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}-15 i F\left (i \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right ), i\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}+8 \sin \left (d x +c \right ) \cos \left (d x +c \right )+8 i \left (\cos ^{2}\left (d x +c \right )\right )+\tan \left (d x +c \right )+12 i\right ) \sqrt {e \sec \left (d x +c \right )}}{3 a^{4} d}\) | \(172\) |
[In]
[Out]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.08 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.94 \[ \int \frac {(e \sec (c+d x))^{13/2}}{(a+i a \tan (c+d x))^4} \, dx=-\frac {2 \, {\left (\sqrt {2} {\left (-15 i \, e^{6} e^{\left (4 i \, d x + 4 i \, c\right )} - 21 i \, e^{6} e^{\left (2 i \, d x + 2 i \, c\right )} - 4 i \, e^{6}\right )} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )} + 15 \, \sqrt {2} {\left (-i \, e^{6} e^{\left (4 i \, d x + 4 i \, c\right )} - i \, e^{6} e^{\left (2 i \, d x + 2 i \, c\right )}\right )} \sqrt {e} {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right )\right )}}{3 \, {\left (a^{4} d e^{\left (4 i \, d x + 4 i \, c\right )} + a^{4} d e^{\left (2 i \, d x + 2 i \, c\right )}\right )}} \]
[In]
[Out]
Timed out. \[ \int \frac {(e \sec (c+d x))^{13/2}}{(a+i a \tan (c+d x))^4} \, dx=\text {Timed out} \]
[In]
[Out]
Exception generated. \[ \int \frac {(e \sec (c+d x))^{13/2}}{(a+i a \tan (c+d x))^4} \, dx=\text {Exception raised: RuntimeError} \]
[In]
[Out]
\[ \int \frac {(e \sec (c+d x))^{13/2}}{(a+i a \tan (c+d x))^4} \, dx=\int { \frac {\left (e \sec \left (d x + c\right )\right )^{\frac {13}{2}}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{4}} \,d x } \]
[In]
[Out]
Timed out. \[ \int \frac {(e \sec (c+d x))^{13/2}}{(a+i a \tan (c+d x))^4} \, dx=\int \frac {{\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^{13/2}}{{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^4} \,d x \]
[In]
[Out]