Integrand size = 23, antiderivative size = 232 \[ \int \frac {(e \sin (c+d x))^m}{a+b \sec (c+d x)} \, dx=-\frac {b e \operatorname {AppellF1}\left (1-m,\frac {1-m}{2},\frac {1-m}{2},2-m,-\frac {a-b}{b+a \cos (c+d x)},\frac {a+b}{b+a \cos (c+d x)}\right ) \left (-\frac {a (1-\cos (c+d x))}{b+a \cos (c+d x)}\right )^{\frac {1-m}{2}} \left (\frac {a (1+\cos (c+d x))}{b+a \cos (c+d x)}\right )^{\frac {1-m}{2}} (e \sin (c+d x))^{-1+m}}{a^2 d (1-m)}+\frac {\cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},\sin ^2(c+d x)\right ) (e \sin (c+d x))^{1+m}}{a d e (1+m) \sqrt {\cos ^2(c+d x)}} \]
[Out]
Time = 0.31 (sec) , antiderivative size = 232, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3957, 2946, 2722, 2782} \[ \int \frac {(e \sin (c+d x))^m}{a+b \sec (c+d x)} \, dx=\frac {\cos (c+d x) (e \sin (c+d x))^{m+1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+1}{2},\frac {m+3}{2},\sin ^2(c+d x)\right )}{a d e (m+1) \sqrt {\cos ^2(c+d x)}}-\frac {b e (e \sin (c+d x))^{m-1} \left (-\frac {a (1-\cos (c+d x))}{a \cos (c+d x)+b}\right )^{\frac {1-m}{2}} \left (\frac {a (\cos (c+d x)+1)}{a \cos (c+d x)+b}\right )^{\frac {1-m}{2}} \operatorname {AppellF1}\left (1-m,\frac {1-m}{2},\frac {1-m}{2},2-m,-\frac {a-b}{b+a \cos (c+d x)},\frac {a+b}{b+a \cos (c+d x)}\right )}{a^2 d (1-m)} \]
[In]
[Out]
Rule 2722
Rule 2782
Rule 2946
Rule 3957
Rubi steps \begin{align*} \text {integral}& = -\int \frac {\cos (c+d x) (e \sin (c+d x))^m}{-b-a \cos (c+d x)} \, dx \\ & = \frac {\int (e \sin (c+d x))^m \, dx}{a}+\frac {b \int \frac {(e \sin (c+d x))^m}{-b-a \cos (c+d x)} \, dx}{a} \\ & = -\frac {b e \operatorname {AppellF1}\left (1-m,\frac {1-m}{2},\frac {1-m}{2},2-m,-\frac {a-b}{b+a \cos (c+d x)},\frac {a+b}{b+a \cos (c+d x)}\right ) \left (-\frac {a (1-\cos (c+d x))}{b+a \cos (c+d x)}\right )^{\frac {1-m}{2}} \left (\frac {a (1+\cos (c+d x))}{b+a \cos (c+d x)}\right )^{\frac {1-m}{2}} (e \sin (c+d x))^{-1+m}}{a^2 d (1-m)}+\frac {\cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},\sin ^2(c+d x)\right ) (e \sin (c+d x))^{1+m}}{a d e (1+m) \sqrt {\cos ^2(c+d x)}} \\ \end{align*}
Leaf count is larger than twice the leaf count of optimal. \(687\) vs. \(2(232)=464\).
Time = 4.96 (sec) , antiderivative size = 687, normalized size of antiderivative = 2.96 \[ \int \frac {(e \sin (c+d x))^m}{a+b \sec (c+d x)} \, dx=\frac {2 \left (-b \operatorname {AppellF1}\left (\frac {1+m}{2},m,1,\frac {3+m}{2},-\tan ^2\left (\frac {1}{2} (c+d x)\right ),\frac {(a-b) \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}\right )+(a+b) \operatorname {Hypergeometric2F1}\left (\frac {1+m}{2},1+m,\frac {3+m}{2},-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )\right ) (e \sin (c+d x))^m \tan \left (\frac {1}{2} (c+d x)\right )}{d (a+b \sec (c+d x)) \left (\left (-b \operatorname {AppellF1}\left (\frac {1+m}{2},m,1,\frac {3+m}{2},-\tan ^2\left (\frac {1}{2} (c+d x)\right ),\frac {(a-b) \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}\right )+(a+b) \operatorname {Hypergeometric2F1}\left (\frac {1+m}{2},1+m,\frac {3+m}{2},-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )\right ) \sec ^2\left (\frac {1}{2} (c+d x)\right )+2 m \cot (c+d x) \left (-b \operatorname {AppellF1}\left (\frac {1+m}{2},m,1,\frac {3+m}{2},-\tan ^2\left (\frac {1}{2} (c+d x)\right ),\frac {(a-b) \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}\right )+(a+b) \operatorname {Hypergeometric2F1}\left (\frac {1+m}{2},1+m,\frac {3+m}{2},-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )\right ) \tan \left (\frac {1}{2} (c+d x)\right )+2 m \left (-b \operatorname {AppellF1}\left (\frac {1+m}{2},m,1,\frac {3+m}{2},-\tan ^2\left (\frac {1}{2} (c+d x)\right ),\frac {(a-b) \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}\right )+(a+b) \operatorname {Hypergeometric2F1}\left (\frac {1+m}{2},1+m,\frac {3+m}{2},-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right )+\frac {(1+m) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \left (-(a+b)^2 \left (\operatorname {Hypergeometric2F1}\left (\frac {1+m}{2},1+m,\frac {3+m}{2},-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )-\sec ^2\left (\frac {1}{2} (c+d x)\right )^{-1-m}\right )+\frac {2 b \left ((-a+b) \operatorname {AppellF1}\left (\frac {3+m}{2},m,2,\frac {5+m}{2},-\tan ^2\left (\frac {1}{2} (c+d x)\right ),\frac {(a-b) \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}\right )+(a+b) m \operatorname {AppellF1}\left (\frac {3+m}{2},1+m,1,\frac {5+m}{2},-\tan ^2\left (\frac {1}{2} (c+d x)\right ),\frac {(a-b) \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}\right )\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right )}{3+m}\right )}{a+b}\right )} \]
[In]
[Out]
\[\int \frac {\left (e \sin \left (d x +c \right )\right )^{m}}{a +b \sec \left (d x +c \right )}d x\]
[In]
[Out]
\[ \int \frac {(e \sin (c+d x))^m}{a+b \sec (c+d x)} \, dx=\int { \frac {\left (e \sin \left (d x + c\right )\right )^{m}}{b \sec \left (d x + c\right ) + a} \,d x } \]
[In]
[Out]
\[ \int \frac {(e \sin (c+d x))^m}{a+b \sec (c+d x)} \, dx=\int \frac {\left (e \sin {\left (c + d x \right )}\right )^{m}}{a + b \sec {\left (c + d x \right )}}\, dx \]
[In]
[Out]
\[ \int \frac {(e \sin (c+d x))^m}{a+b \sec (c+d x)} \, dx=\int { \frac {\left (e \sin \left (d x + c\right )\right )^{m}}{b \sec \left (d x + c\right ) + a} \,d x } \]
[In]
[Out]
\[ \int \frac {(e \sin (c+d x))^m}{a+b \sec (c+d x)} \, dx=\int { \frac {\left (e \sin \left (d x + c\right )\right )^{m}}{b \sec \left (d x + c\right ) + a} \,d x } \]
[In]
[Out]
Timed out. \[ \int \frac {(e \sin (c+d x))^m}{a+b \sec (c+d x)} \, dx=\int \frac {\cos \left (c+d\,x\right )\,{\left (e\,\sin \left (c+d\,x\right )\right )}^m}{b+a\,\cos \left (c+d\,x\right )} \,d x \]
[In]
[Out]