Integrand size = 28, antiderivative size = 190 \[ \int \frac {(e \cos (c+d x))^{7/2}}{(a+i a \tan (c+d x))^2} \, dx=\frac {2 (e \cos (c+d x))^{7/2} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{7 a^2 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {2 \cos (c+d x) (e \cos (c+d x))^{7/2} \sin (c+d x)}{15 a^2 d}+\frac {6 (e \cos (c+d x))^{7/2} \tan (c+d x)}{35 a^2 d}+\frac {2 (e \cos (c+d x))^{7/2} \sec ^2(c+d x) \tan (c+d x)}{7 a^2 d}+\frac {4 i \cos ^2(c+d x) (e \cos (c+d x))^{7/2}}{15 d \left (a^2+i a^2 \tan (c+d x)\right )} \]
[Out]
Time = 0.28 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {3596, 3581, 3854, 3856, 2720} \[ \int \frac {(e \cos (c+d x))^{7/2}}{(a+i a \tan (c+d x))^2} \, dx=\frac {2 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) (e \cos (c+d x))^{7/2}}{7 a^2 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {4 i \cos ^2(c+d x) (e \cos (c+d x))^{7/2}}{15 d \left (a^2+i a^2 \tan (c+d x)\right )}+\frac {2 \sin (c+d x) \cos (c+d x) (e \cos (c+d x))^{7/2}}{15 a^2 d}+\frac {6 \tan (c+d x) (e \cos (c+d x))^{7/2}}{35 a^2 d}+\frac {2 \tan (c+d x) \sec ^2(c+d x) (e \cos (c+d x))^{7/2}}{7 a^2 d} \]
[In]
[Out]
Rule 2720
Rule 3581
Rule 3596
Rule 3854
Rule 3856
Rubi steps \begin{align*} \text {integral}& = \left ((e \cos (c+d x))^{7/2} (e \sec (c+d x))^{7/2}\right ) \int \frac {1}{(e \sec (c+d x))^{7/2} (a+i a \tan (c+d x))^2} \, dx \\ & = \frac {4 i \cos ^2(c+d x) (e \cos (c+d x))^{7/2}}{15 d \left (a^2+i a^2 \tan (c+d x)\right )}+\frac {\left (11 e^2 (e \cos (c+d x))^{7/2} (e \sec (c+d x))^{7/2}\right ) \int \frac {1}{(e \sec (c+d x))^{11/2}} \, dx}{15 a^2} \\ & = \frac {2 \cos (c+d x) (e \cos (c+d x))^{7/2} \sin (c+d x)}{15 a^2 d}+\frac {4 i \cos ^2(c+d x) (e \cos (c+d x))^{7/2}}{15 d \left (a^2+i a^2 \tan (c+d x)\right )}+\frac {\left (3 (e \cos (c+d x))^{7/2} (e \sec (c+d x))^{7/2}\right ) \int \frac {1}{(e \sec (c+d x))^{7/2}} \, dx}{5 a^2} \\ & = \frac {2 \cos (c+d x) (e \cos (c+d x))^{7/2} \sin (c+d x)}{15 a^2 d}+\frac {6 (e \cos (c+d x))^{7/2} \tan (c+d x)}{35 a^2 d}+\frac {4 i \cos ^2(c+d x) (e \cos (c+d x))^{7/2}}{15 d \left (a^2+i a^2 \tan (c+d x)\right )}+\frac {\left (3 (e \cos (c+d x))^{7/2} (e \sec (c+d x))^{7/2}\right ) \int \frac {1}{(e \sec (c+d x))^{3/2}} \, dx}{7 a^2 e^2} \\ & = \frac {2 \cos (c+d x) (e \cos (c+d x))^{7/2} \sin (c+d x)}{15 a^2 d}+\frac {6 (e \cos (c+d x))^{7/2} \tan (c+d x)}{35 a^2 d}+\frac {2 (e \cos (c+d x))^{7/2} \sec ^2(c+d x) \tan (c+d x)}{7 a^2 d}+\frac {4 i \cos ^2(c+d x) (e \cos (c+d x))^{7/2}}{15 d \left (a^2+i a^2 \tan (c+d x)\right )}+\frac {\left ((e \cos (c+d x))^{7/2} (e \sec (c+d x))^{7/2}\right ) \int \sqrt {e \sec (c+d x)} \, dx}{7 a^2 e^4} \\ & = \frac {2 \cos (c+d x) (e \cos (c+d x))^{7/2} \sin (c+d x)}{15 a^2 d}+\frac {6 (e \cos (c+d x))^{7/2} \tan (c+d x)}{35 a^2 d}+\frac {2 (e \cos (c+d x))^{7/2} \sec ^2(c+d x) \tan (c+d x)}{7 a^2 d}+\frac {4 i \cos ^2(c+d x) (e \cos (c+d x))^{7/2}}{15 d \left (a^2+i a^2 \tan (c+d x)\right )}+\frac {(e \cos (c+d x))^{7/2} \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{7 a^2 \cos ^{\frac {7}{2}}(c+d x)} \\ & = \frac {2 (e \cos (c+d x))^{7/2} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{7 a^2 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {2 \cos (c+d x) (e \cos (c+d x))^{7/2} \sin (c+d x)}{15 a^2 d}+\frac {6 (e \cos (c+d x))^{7/2} \tan (c+d x)}{35 a^2 d}+\frac {2 (e \cos (c+d x))^{7/2} \sec ^2(c+d x) \tan (c+d x)}{7 a^2 d}+\frac {4 i \cos ^2(c+d x) (e \cos (c+d x))^{7/2}}{15 d \left (a^2+i a^2 \tan (c+d x)\right )} \\ \end{align*}
Time = 2.63 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.82 \[ \int \frac {(e \cos (c+d x))^{7/2}}{(a+i a \tan (c+d x))^2} \, dx=\frac {e^3 \sqrt {e \cos (c+d x)} \left (-240 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) (\cos (2 (c+d x))+i \sin (2 (c+d x)))+\sqrt {\cos (c+d x)} (-296 i \cos (c+d x)+68 i \cos (3 (c+d x))+4 i \cos (5 (c+d x))+134 \sin (c+d x)-117 \sin (3 (c+d x))-11 \sin (5 (c+d x)))\right )}{840 a^2 d \cos ^{\frac {5}{2}}(c+d x) (-i+\tan (c+d x))^2} \]
[In]
[Out]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 386 vs. \(2 (192 ) = 384\).
Time = 9.44 (sec) , antiderivative size = 387, normalized size of antiderivative = 2.04
method | result | size |
default | \(-\frac {2 e^{4} \left (25088 i \left (\sin ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+3584 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{16}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-3584 i \left (\sin ^{17}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-12544 \left (\sin ^{14}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+6272 i \left (\sin ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+19264 \left (\sin ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-14 i \sin \left (\frac {d x}{2}+\frac {c}{2}\right )-16800 \left (\sin ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-1568 i \left (\sin ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+9104 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+224 i \left (\sin ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-3128 \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+14336 i \left (\sin ^{15}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+700 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-15680 i \left (\sin ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-90 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+15 \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}-25088 i \left (\sin ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}{105 a^{2} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e +e}\, d}\) | \(387\) |
[In]
[Out]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.10 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.79 \[ \int \frac {(e \cos (c+d x))^{7/2}}{(a+i a \tan (c+d x))^2} \, dx=\frac {{\left (-480 i \, \sqrt {2} e^{\frac {7}{2}} e^{\left (7 i \, d x + 7 i \, c\right )} {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right ) + \sqrt {\frac {1}{2}} {\left (-15 i \, e^{3} e^{\left (10 i \, d x + 10 i \, c\right )} - 185 i \, e^{3} e^{\left (8 i \, d x + 8 i \, c\right )} + 430 i \, e^{3} e^{\left (6 i \, d x + 6 i \, c\right )} + 162 i \, e^{3} e^{\left (4 i \, d x + 4 i \, c\right )} + 49 i \, e^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + 7 i \, e^{3}\right )} \sqrt {e e^{\left (2 i \, d x + 2 i \, c\right )} + e} e^{\left (-\frac {1}{2} i \, d x - \frac {1}{2} i \, c\right )}\right )} e^{\left (-7 i \, d x - 7 i \, c\right )}}{1680 \, a^{2} d} \]
[In]
[Out]
Timed out. \[ \int \frac {(e \cos (c+d x))^{7/2}}{(a+i a \tan (c+d x))^2} \, dx=\text {Timed out} \]
[In]
[Out]
Exception generated. \[ \int \frac {(e \cos (c+d x))^{7/2}}{(a+i a \tan (c+d x))^2} \, dx=\text {Exception raised: RuntimeError} \]
[In]
[Out]
\[ \int \frac {(e \cos (c+d x))^{7/2}}{(a+i a \tan (c+d x))^2} \, dx=\int { \frac {\left (e \cos \left (d x + c\right )\right )^{\frac {7}{2}}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2}} \,d x } \]
[In]
[Out]
Timed out. \[ \int \frac {(e \cos (c+d x))^{7/2}}{(a+i a \tan (c+d x))^2} \, dx=\int \frac {{\left (e\,\cos \left (c+d\,x\right )\right )}^{7/2}}{{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^2} \,d x \]
[In]
[Out]