Integrand size = 25, antiderivative size = 326 \[ \int \frac {(e \tan (c+d x))^{9/2}}{a+a \sec (c+d x)} \, dx=-\frac {e^{9/2} \arctan \left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a d}+\frac {e^{9/2} \arctan \left (1+\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a d}+\frac {e^{9/2} \log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)-\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} a d}-\frac {e^{9/2} \log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)+\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} a d}+\frac {6 e^4 \cos (c+d x) E\left (\left .c-\frac {\pi }{4}+d x\right |2\right ) \sqrt {e \tan (c+d x)}}{5 a d \sqrt {\sin (2 c+2 d x)}}-\frac {6 e^3 \cos (c+d x) (e \tan (c+d x))^{3/2}}{5 a d}-\frac {2 e^3 (5-3 \sec (c+d x)) (e \tan (c+d x))^{3/2}}{15 a d} \]
-1/2*e^(9/2)*arctan(1-2^(1/2)*(e*tan(d*x+c))^(1/2)/e^(1/2))/a/d*2^(1/2)+1/ 2*e^(9/2)*arctan(1+2^(1/2)*(e*tan(d*x+c))^(1/2)/e^(1/2))/a/d*2^(1/2)+1/4*e ^(9/2)*ln(e^(1/2)-2^(1/2)*(e*tan(d*x+c))^(1/2)+e^(1/2)*tan(d*x+c))/a/d*2^( 1/2)-1/4*e^(9/2)*ln(e^(1/2)+2^(1/2)*(e*tan(d*x+c))^(1/2)+e^(1/2)*tan(d*x+c ))/a/d*2^(1/2)-6/5*e^4*cos(d*x+c)*(sin(c+1/4*Pi+d*x)^2)^(1/2)/sin(c+1/4*Pi +d*x)*EllipticE(cos(c+1/4*Pi+d*x),2^(1/2))*(e*tan(d*x+c))^(1/2)/a/d/sin(2* d*x+2*c)^(1/2)-6/5*e^3*cos(d*x+c)*(e*tan(d*x+c))^(3/2)/a/d-2/15*e^3*(5-3*s ec(d*x+c))*(e*tan(d*x+c))^(3/2)/a/d
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 17.80 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.40 \[ \int \frac {(e \tan (c+d x))^{9/2}}{a+a \sec (c+d x)} \, dx=\frac {4 e^3 \cos ^2\left (\frac {1}{2} (c+d x)\right ) \left (-1+\operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {3}{4},\frac {7}{4},-\tan ^2(c+d x)\right )-\operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-\tan ^2(c+d x)\right )+\operatorname {Hypergeometric2F1}\left (\frac {3}{4},1,\frac {7}{4},-\tan ^2(c+d x)\right )\right ) \sec (c+d x) \left (1+\sqrt {\sec ^2(c+d x)}\right ) (e \tan (c+d x))^{3/2}}{3 a d (1+\sec (c+d x))^2} \]
(4*e^3*Cos[(c + d*x)/2]^2*(-1 + Hypergeometric2F1[-1/2, 3/4, 7/4, -Tan[c + d*x]^2] - Hypergeometric2F1[1/2, 3/4, 7/4, -Tan[c + d*x]^2] + Hypergeomet ric2F1[3/4, 1, 7/4, -Tan[c + d*x]^2])*Sec[c + d*x]*(1 + Sqrt[Sec[c + d*x]^ 2])*(e*Tan[c + d*x])^(3/2))/(3*a*d*(1 + Sec[c + d*x])^2)
Time = 1.25 (sec) , antiderivative size = 303, normalized size of antiderivative = 0.93, number of steps used = 27, number of rules used = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.040, Rules used = {3042, 4376, 25, 3042, 4369, 27, 3042, 4372, 3042, 3093, 3042, 3095, 3042, 3052, 3042, 3119, 3957, 266, 826, 1476, 1082, 217, 1479, 25, 27, 1103}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {(e \tan (c+d x))^{9/2}}{a \sec (c+d x)+a} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (-e \cot \left (c+d x+\frac {\pi }{2}\right )\right )^{9/2}}{a \csc \left (c+d x+\frac {\pi }{2}\right )+a}dx\) |
| \(\Big \downarrow \) 4376 |
| \(\displaystyle \frac {e^2 \int -\left ((a-a \sec (c+d x)) (e \tan (c+d x))^{5/2}\right )dx}{a^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {e^2 \int (a-a \sec (c+d x)) (e \tan (c+d x))^{5/2}dx}{a^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {e^2 \int \left (-e \cot \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2} \left (a-a \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx}{a^2}\) |
| \(\Big \downarrow \) 4369 |
| \(\displaystyle -\frac {e^2 \left (\frac {2 e (5 a-3 a \sec (c+d x)) (e \tan (c+d x))^{3/2}}{15 d}-\frac {2}{5} e^2 \int \frac {1}{2} (5 a-3 a \sec (c+d x)) \sqrt {e \tan (c+d x)}dx\right )}{a^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {e^2 \left (\frac {2 e (5 a-3 a \sec (c+d x)) (e \tan (c+d x))^{3/2}}{15 d}-\frac {1}{5} e^2 \int (5 a-3 a \sec (c+d x)) \sqrt {e \tan (c+d x)}dx\right )}{a^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {e^2 \left (\frac {2 e (5 a-3 a \sec (c+d x)) (e \tan (c+d x))^{3/2}}{15 d}-\frac {1}{5} e^2 \int \sqrt {-e \cot \left (c+d x+\frac {\pi }{2}\right )} \left (5 a-3 a \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx\right )}{a^2}\) |
| \(\Big \downarrow \) 4372 |
| \(\displaystyle -\frac {e^2 \left (\frac {2 e (5 a-3 a \sec (c+d x)) (e \tan (c+d x))^{3/2}}{15 d}-\frac {1}{5} e^2 \left (5 a \int \sqrt {e \tan (c+d x)}dx-3 a \int \sec (c+d x) \sqrt {e \tan (c+d x)}dx\right )\right )}{a^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {e^2 \left (\frac {2 e (5 a-3 a \sec (c+d x)) (e \tan (c+d x))^{3/2}}{15 d}-\frac {1}{5} e^2 \left (5 a \int \sqrt {e \tan (c+d x)}dx-3 a \int \sec (c+d x) \sqrt {e \tan (c+d x)}dx\right )\right )}{a^2}\) |
| \(\Big \downarrow \) 3093 |
| \(\displaystyle -\frac {e^2 \left (\frac {2 e (5 a-3 a \sec (c+d x)) (e \tan (c+d x))^{3/2}}{15 d}-\frac {1}{5} e^2 \left (5 a \int \sqrt {e \tan (c+d x)}dx-3 a \left (\frac {2 \cos (c+d x) (e \tan (c+d x))^{3/2}}{d e}-2 \int \cos (c+d x) \sqrt {e \tan (c+d x)}dx\right )\right )\right )}{a^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {e^2 \left (\frac {2 e (5 a-3 a \sec (c+d x)) (e \tan (c+d x))^{3/2}}{15 d}-\frac {1}{5} e^2 \left (5 a \int \sqrt {e \tan (c+d x)}dx-3 a \left (\frac {2 \cos (c+d x) (e \tan (c+d x))^{3/2}}{d e}-2 \int \frac {\sqrt {e \tan (c+d x)}}{\sec (c+d x)}dx\right )\right )\right )}{a^2}\) |
| \(\Big \downarrow \) 3095 |
| \(\displaystyle -\frac {e^2 \left (\frac {2 e (5 a-3 a \sec (c+d x)) (e \tan (c+d x))^{3/2}}{15 d}-\frac {1}{5} e^2 \left (5 a \int \sqrt {e \tan (c+d x)}dx-3 a \left (\frac {2 \cos (c+d x) (e \tan (c+d x))^{3/2}}{d e}-\frac {2 \sqrt {\cos (c+d x)} \sqrt {e \tan (c+d x)} \int \sqrt {\cos (c+d x)} \sqrt {\sin (c+d x)}dx}{\sqrt {\sin (c+d x)}}\right )\right )\right )}{a^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {e^2 \left (\frac {2 e (5 a-3 a \sec (c+d x)) (e \tan (c+d x))^{3/2}}{15 d}-\frac {1}{5} e^2 \left (5 a \int \sqrt {e \tan (c+d x)}dx-3 a \left (\frac {2 \cos (c+d x) (e \tan (c+d x))^{3/2}}{d e}-\frac {2 \sqrt {\cos (c+d x)} \sqrt {e \tan (c+d x)} \int \sqrt {\cos (c+d x)} \sqrt {\sin (c+d x)}dx}{\sqrt {\sin (c+d x)}}\right )\right )\right )}{a^2}\) |
| \(\Big \downarrow \) 3052 |
| \(\displaystyle -\frac {e^2 \left (\frac {2 e (5 a-3 a \sec (c+d x)) (e \tan (c+d x))^{3/2}}{15 d}-\frac {1}{5} e^2 \left (5 a \int \sqrt {e \tan (c+d x)}dx-3 a \left (\frac {2 \cos (c+d x) (e \tan (c+d x))^{3/2}}{d e}-\frac {2 \cos (c+d x) \sqrt {e \tan (c+d x)} \int \sqrt {\sin (2 c+2 d x)}dx}{\sqrt {\sin (2 c+2 d x)}}\right )\right )\right )}{a^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {e^2 \left (\frac {2 e (5 a-3 a \sec (c+d x)) (e \tan (c+d x))^{3/2}}{15 d}-\frac {1}{5} e^2 \left (5 a \int \sqrt {e \tan (c+d x)}dx-3 a \left (\frac {2 \cos (c+d x) (e \tan (c+d x))^{3/2}}{d e}-\frac {2 \cos (c+d x) \sqrt {e \tan (c+d x)} \int \sqrt {\sin (2 c+2 d x)}dx}{\sqrt {\sin (2 c+2 d x)}}\right )\right )\right )}{a^2}\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle -\frac {e^2 \left (\frac {2 e (5 a-3 a \sec (c+d x)) (e \tan (c+d x))^{3/2}}{15 d}-\frac {1}{5} e^2 \left (5 a \int \sqrt {e \tan (c+d x)}dx-3 a \left (\frac {2 \cos (c+d x) (e \tan (c+d x))^{3/2}}{d e}-\frac {2 \cos (c+d x) E\left (\left .c+d x-\frac {\pi }{4}\right |2\right ) \sqrt {e \tan (c+d x)}}{d \sqrt {\sin (2 c+2 d x)}}\right )\right )\right )}{a^2}\) |
| \(\Big \downarrow \) 3957 |
| \(\displaystyle -\frac {e^2 \left (\frac {2 e (5 a-3 a \sec (c+d x)) (e \tan (c+d x))^{3/2}}{15 d}-\frac {1}{5} e^2 \left (\frac {5 a e \int \frac {\sqrt {e \tan (c+d x)}}{\tan ^2(c+d x) e^2+e^2}d(e \tan (c+d x))}{d}-3 a \left (\frac {2 \cos (c+d x) (e \tan (c+d x))^{3/2}}{d e}-\frac {2 \cos (c+d x) E\left (\left .c+d x-\frac {\pi }{4}\right |2\right ) \sqrt {e \tan (c+d x)}}{d \sqrt {\sin (2 c+2 d x)}}\right )\right )\right )}{a^2}\) |
| \(\Big \downarrow \) 266 |
| \(\displaystyle -\frac {e^2 \left (\frac {2 e (5 a-3 a \sec (c+d x)) (e \tan (c+d x))^{3/2}}{15 d}-\frac {1}{5} e^2 \left (\frac {10 a e \int \frac {e^2 \tan ^2(c+d x)}{e^4 \tan ^4(c+d x)+e^2}d\sqrt {e \tan (c+d x)}}{d}-3 a \left (\frac {2 \cos (c+d x) (e \tan (c+d x))^{3/2}}{d e}-\frac {2 \cos (c+d x) E\left (\left .c+d x-\frac {\pi }{4}\right |2\right ) \sqrt {e \tan (c+d x)}}{d \sqrt {\sin (2 c+2 d x)}}\right )\right )\right )}{a^2}\) |
| \(\Big \downarrow \) 826 |
| \(\displaystyle -\frac {e^2 \left (\frac {2 e (5 a-3 a \sec (c+d x)) (e \tan (c+d x))^{3/2}}{15 d}-\frac {1}{5} e^2 \left (\frac {10 a e \left (\frac {1}{2} \int \frac {e^2 \tan ^2(c+d x)+e}{e^4 \tan ^4(c+d x)+e^2}d\sqrt {e \tan (c+d x)}-\frac {1}{2} \int \frac {e-e^2 \tan ^2(c+d x)}{e^4 \tan ^4(c+d x)+e^2}d\sqrt {e \tan (c+d x)}\right )}{d}-3 a \left (\frac {2 \cos (c+d x) (e \tan (c+d x))^{3/2}}{d e}-\frac {2 \cos (c+d x) E\left (\left .c+d x-\frac {\pi }{4}\right |2\right ) \sqrt {e \tan (c+d x)}}{d \sqrt {\sin (2 c+2 d x)}}\right )\right )\right )}{a^2}\) |
| \(\Big \downarrow \) 1476 |
| \(\displaystyle -\frac {e^2 \left (\frac {2 e (5 a-3 a \sec (c+d x)) (e \tan (c+d x))^{3/2}}{15 d}-\frac {1}{5} e^2 \left (\frac {10 a e \left (\frac {1}{2} \left (\frac {1}{2} \int \frac {1}{e^2 \tan ^2(c+d x)-\sqrt {2} e^{3/2} \tan (c+d x)+e}d\sqrt {e \tan (c+d x)}+\frac {1}{2} \int \frac {1}{e^2 \tan ^2(c+d x)+\sqrt {2} e^{3/2} \tan (c+d x)+e}d\sqrt {e \tan (c+d x)}\right )-\frac {1}{2} \int \frac {e-e^2 \tan ^2(c+d x)}{e^4 \tan ^4(c+d x)+e^2}d\sqrt {e \tan (c+d x)}\right )}{d}-3 a \left (\frac {2 \cos (c+d x) (e \tan (c+d x))^{3/2}}{d e}-\frac {2 \cos (c+d x) E\left (\left .c+d x-\frac {\pi }{4}\right |2\right ) \sqrt {e \tan (c+d x)}}{d \sqrt {\sin (2 c+2 d x)}}\right )\right )\right )}{a^2}\) |
| \(\Big \downarrow \) 1082 |
| \(\displaystyle -\frac {e^2 \left (\frac {2 e (5 a-3 a \sec (c+d x)) (e \tan (c+d x))^{3/2}}{15 d}-\frac {1}{5} e^2 \left (\frac {10 a e \left (\frac {1}{2} \left (\frac {\int \frac {1}{-e^2 \tan ^2(c+d x)-1}d\left (1-\sqrt {2} \sqrt {e} \tan (c+d x)\right )}{\sqrt {2} \sqrt {e}}-\frac {\int \frac {1}{-e^2 \tan ^2(c+d x)-1}d\left (\sqrt {2} \sqrt {e} \tan (c+d x)+1\right )}{\sqrt {2} \sqrt {e}}\right )-\frac {1}{2} \int \frac {e-e^2 \tan ^2(c+d x)}{e^4 \tan ^4(c+d x)+e^2}d\sqrt {e \tan (c+d x)}\right )}{d}-3 a \left (\frac {2 \cos (c+d x) (e \tan (c+d x))^{3/2}}{d e}-\frac {2 \cos (c+d x) E\left (\left .c+d x-\frac {\pi }{4}\right |2\right ) \sqrt {e \tan (c+d x)}}{d \sqrt {\sin (2 c+2 d x)}}\right )\right )\right )}{a^2}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle -\frac {e^2 \left (\frac {2 e (5 a-3 a \sec (c+d x)) (e \tan (c+d x))^{3/2}}{15 d}-\frac {1}{5} e^2 \left (\frac {10 a e \left (\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {e} \tan (c+d x)+1\right )}{\sqrt {2} \sqrt {e}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {e} \tan (c+d x)\right )}{\sqrt {2} \sqrt {e}}\right )-\frac {1}{2} \int \frac {e-e^2 \tan ^2(c+d x)}{e^4 \tan ^4(c+d x)+e^2}d\sqrt {e \tan (c+d x)}\right )}{d}-3 a \left (\frac {2 \cos (c+d x) (e \tan (c+d x))^{3/2}}{d e}-\frac {2 \cos (c+d x) E\left (\left .c+d x-\frac {\pi }{4}\right |2\right ) \sqrt {e \tan (c+d x)}}{d \sqrt {\sin (2 c+2 d x)}}\right )\right )\right )}{a^2}\) |
| \(\Big \downarrow \) 1479 |
| \(\displaystyle -\frac {e^2 \left (\frac {2 e (5 a-3 a \sec (c+d x)) (e \tan (c+d x))^{3/2}}{15 d}-\frac {1}{5} e^2 \left (\frac {10 a e \left (\frac {1}{2} \left (\frac {\int -\frac {\sqrt {2} \sqrt {e}-2 \sqrt {e \tan (c+d x)}}{e^2 \tan ^2(c+d x)-\sqrt {2} e^{3/2} \tan (c+d x)+e}d\sqrt {e \tan (c+d x)}}{2 \sqrt {2} \sqrt {e}}+\frac {\int -\frac {\sqrt {2} \left (\sqrt {e}+\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{e^2 \tan ^2(c+d x)+\sqrt {2} e^{3/2} \tan (c+d x)+e}d\sqrt {e \tan (c+d x)}}{2 \sqrt {2} \sqrt {e}}\right )+\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {e} \tan (c+d x)+1\right )}{\sqrt {2} \sqrt {e}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {e} \tan (c+d x)\right )}{\sqrt {2} \sqrt {e}}\right )\right )}{d}-3 a \left (\frac {2 \cos (c+d x) (e \tan (c+d x))^{3/2}}{d e}-\frac {2 \cos (c+d x) E\left (\left .c+d x-\frac {\pi }{4}\right |2\right ) \sqrt {e \tan (c+d x)}}{d \sqrt {\sin (2 c+2 d x)}}\right )\right )\right )}{a^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {e^2 \left (\frac {2 e (5 a-3 a \sec (c+d x)) (e \tan (c+d x))^{3/2}}{15 d}-\frac {1}{5} e^2 \left (\frac {10 a e \left (\frac {1}{2} \left (-\frac {\int \frac {\sqrt {2} \sqrt {e}-2 \sqrt {e \tan (c+d x)}}{e^2 \tan ^2(c+d x)-\sqrt {2} e^{3/2} \tan (c+d x)+e}d\sqrt {e \tan (c+d x)}}{2 \sqrt {2} \sqrt {e}}-\frac {\int \frac {\sqrt {2} \left (\sqrt {e}+\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{e^2 \tan ^2(c+d x)+\sqrt {2} e^{3/2} \tan (c+d x)+e}d\sqrt {e \tan (c+d x)}}{2 \sqrt {2} \sqrt {e}}\right )+\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {e} \tan (c+d x)+1\right )}{\sqrt {2} \sqrt {e}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {e} \tan (c+d x)\right )}{\sqrt {2} \sqrt {e}}\right )\right )}{d}-3 a \left (\frac {2 \cos (c+d x) (e \tan (c+d x))^{3/2}}{d e}-\frac {2 \cos (c+d x) E\left (\left .c+d x-\frac {\pi }{4}\right |2\right ) \sqrt {e \tan (c+d x)}}{d \sqrt {\sin (2 c+2 d x)}}\right )\right )\right )}{a^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {e^2 \left (\frac {2 e (5 a-3 a \sec (c+d x)) (e \tan (c+d x))^{3/2}}{15 d}-\frac {1}{5} e^2 \left (\frac {10 a e \left (\frac {1}{2} \left (-\frac {\int \frac {\sqrt {2} \sqrt {e}-2 \sqrt {e \tan (c+d x)}}{e^2 \tan ^2(c+d x)-\sqrt {2} e^{3/2} \tan (c+d x)+e}d\sqrt {e \tan (c+d x)}}{2 \sqrt {2} \sqrt {e}}-\frac {\int \frac {\sqrt {e}+\sqrt {2} \sqrt {e \tan (c+d x)}}{e^2 \tan ^2(c+d x)+\sqrt {2} e^{3/2} \tan (c+d x)+e}d\sqrt {e \tan (c+d x)}}{2 \sqrt {e}}\right )+\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {e} \tan (c+d x)+1\right )}{\sqrt {2} \sqrt {e}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {e} \tan (c+d x)\right )}{\sqrt {2} \sqrt {e}}\right )\right )}{d}-3 a \left (\frac {2 \cos (c+d x) (e \tan (c+d x))^{3/2}}{d e}-\frac {2 \cos (c+d x) E\left (\left .c+d x-\frac {\pi }{4}\right |2\right ) \sqrt {e \tan (c+d x)}}{d \sqrt {\sin (2 c+2 d x)}}\right )\right )\right )}{a^2}\) |
| \(\Big \downarrow \) 1103 |
| \(\displaystyle -\frac {e^2 \left (\frac {2 e (5 a-3 a \sec (c+d x)) (e \tan (c+d x))^{3/2}}{15 d}-\frac {1}{5} e^2 \left (\frac {10 a e \left (\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {e} \tan (c+d x)+1\right )}{\sqrt {2} \sqrt {e}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {e} \tan (c+d x)\right )}{\sqrt {2} \sqrt {e}}\right )+\frac {1}{2} \left (\frac {\log \left (-\sqrt {2} e^{3/2} \tan (c+d x)+e^2 \tan ^2(c+d x)+e\right )}{2 \sqrt {2} \sqrt {e}}-\frac {\log \left (\sqrt {2} e^{3/2} \tan (c+d x)+e^2 \tan ^2(c+d x)+e\right )}{2 \sqrt {2} \sqrt {e}}\right )\right )}{d}-3 a \left (\frac {2 \cos (c+d x) (e \tan (c+d x))^{3/2}}{d e}-\frac {2 \cos (c+d x) E\left (\left .c+d x-\frac {\pi }{4}\right |2\right ) \sqrt {e \tan (c+d x)}}{d \sqrt {\sin (2 c+2 d x)}}\right )\right )\right )}{a^2}\) |
-((e^2*((2*e*(5*a - 3*a*Sec[c + d*x])*(e*Tan[c + d*x])^(3/2))/(15*d) - (e^ 2*((10*a*e*((-(ArcTan[1 - Sqrt[2]*Sqrt[e]*Tan[c + d*x]]/(Sqrt[2]*Sqrt[e])) + ArcTan[1 + Sqrt[2]*Sqrt[e]*Tan[c + d*x]]/(Sqrt[2]*Sqrt[e]))/2 + (Log[e - Sqrt[2]*e^(3/2)*Tan[c + d*x] + e^2*Tan[c + d*x]^2]/(2*Sqrt[2]*Sqrt[e]) - Log[e + Sqrt[2]*e^(3/2)*Tan[c + d*x] + e^2*Tan[c + d*x]^2]/(2*Sqrt[2]*Sqr t[e]))/2))/d - 3*a*((-2*Cos[c + d*x]*EllipticE[c - Pi/4 + d*x, 2]*Sqrt[e*T an[c + d*x]])/(d*Sqrt[Sin[2*c + 2*d*x]]) + (2*Cos[c + d*x]*(e*Tan[c + d*x] )^(3/2))/(d*e))))/5))/a^2)
3.2.19.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( -1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & & (LtQ[a, 0] || LtQ[b, 0])
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De nominator[m]}, Simp[k/c Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) ^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I ntBinomialQ[a, b, c, 2, m, p, x]
Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]}, Simp[1/(2*s) Int[(r + s*x^2)/(a + b*x^ 4), x], x] - Simp[1/(2*s) Int[(r - s*x^2)/(a + b*x^4), x], x]] /; FreeQ[{ a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ, b]]))
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S implify[a*(c/b^2)]}, Simp[-2/b Subst[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b )], x] /; RationalQ[q] && (EqQ[q^2, 1] || !RationalQ[b^2 - 4*a*c])] /; Fre eQ[{a, b, c}, x]
Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]
Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 2*(d/e), 2]}, Simp[e/(2*c) Int[1/Simp[d/e + q*x + x^2, x], x], x] + Simp[ e/(2*c) Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]
Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ -2*(d/e), 2]}, Simp[e/(2*c*q) Int[(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Simp[e/(2*c*q) Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /; F reeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]
Int[Sqrt[cos[(e_.) + (f_.)*(x_)]*(b_.)]*Sqrt[(a_.)*sin[(e_.) + (f_.)*(x_)]] , x_Symbol] :> Simp[Sqrt[a*Sin[e + f*x]]*(Sqrt[b*Cos[e + f*x]]/Sqrt[Sin[2*e + 2*f*x]]) Int[Sqrt[Sin[2*e + 2*f*x]], x], x] /; FreeQ[{a, b, e, f}, x]
Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^( n_), x_Symbol] :> Simp[a^2*(a*Sec[e + f*x])^(m - 2)*((b*Tan[e + f*x])^(n + 1)/(b*f*(m + n - 1))), x] + Simp[a^2*((m - 2)/(m + n - 1)) Int[(a*Sec[e + f*x])^(m - 2)*(b*Tan[e + f*x])^n, x], x] /; FreeQ[{a, b, e, f, n}, x] && ( GtQ[m, 1] || (EqQ[m, 1] && EqQ[n, 1/2])) && NeQ[m + n - 1, 0] && IntegersQ[ 2*m, 2*n]
Int[Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]]/sec[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[Sqrt[Cos[e + f*x]]*(Sqrt[b*Tan[e + f*x]]/Sqrt[Sin[e + f*x]]) Int[ Sqrt[Cos[e + f*x]]*Sqrt[Sin[e + f*x]], x], x] /; FreeQ[{b, e, f}, x]
Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)* (c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b/d Subst[Int [x^n/(b^2 + x^2), x], x, b*Tan[c + d*x]], x] /; FreeQ[{b, c, d, n}, x] && !IntegerQ[n]
Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + ( a_)), x_Symbol] :> Simp[(-e)*(e*Cot[c + d*x])^(m - 1)*((a*m + b*(m - 1)*Csc [c + d*x])/(d*m*(m - 1))), x] - Simp[e^2/m Int[(e*Cot[c + d*x])^(m - 2)*( a*m + b*(m - 1)*Csc[c + d*x]), x], x] /; FreeQ[{a, b, c, d, e}, x] && GtQ[m , 1]
Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[a Int[(e*Cot[c + d*x])^m, x], x] + Simp[b Int[ (e*Cot[c + d*x])^m*Csc[c + d*x], x], x] /; FreeQ[{a, b, c, d, e, m}, x]
Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + ( a_))^(n_), x_Symbol] :> Simp[a^(2*n)/e^(2*n) Int[(e*Cot[c + d*x])^(m + 2* n)/(-a + b*Csc[c + d*x])^n, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && EqQ[a ^2 - b^2, 0] && ILtQ[n, 0]
Result contains complex when optimal does not.
Time = 3.86 (sec) , antiderivative size = 1166, normalized size of antiderivative = 3.58
1/30/a/d*2^(1/2)*e^4*(-15*I*(cot(d*x+c)-csc(d*x+c)+1)^(1/2)*(cot(d*x+c)-cs c(d*x+c))^(1/2)*(csc(d*x+c)-cot(d*x+c)+1)^(1/2)*EllipticPi((csc(d*x+c)-cot (d*x+c)+1)^(1/2),1/2+1/2*I,1/2*2^(1/2))*cos(d*x+c)^2-15*I*(cot(d*x+c)-csc( d*x+c)+1)^(1/2)*(cot(d*x+c)-csc(d*x+c))^(1/2)*(csc(d*x+c)-cot(d*x+c)+1)^(1 /2)*EllipticPi((csc(d*x+c)-cot(d*x+c)+1)^(1/2),1/2+1/2*I,1/2*2^(1/2))*cos( d*x+c)^3+15*I*(cot(d*x+c)-csc(d*x+c)+1)^(1/2)*(cot(d*x+c)-csc(d*x+c))^(1/2 )*(csc(d*x+c)-cot(d*x+c)+1)^(1/2)*EllipticPi((csc(d*x+c)-cot(d*x+c)+1)^(1/ 2),1/2-1/2*I,1/2*2^(1/2))*cos(d*x+c)^3+15*I*(cot(d*x+c)-csc(d*x+c)+1)^(1/2 )*(cot(d*x+c)-csc(d*x+c))^(1/2)*(csc(d*x+c)-cot(d*x+c)+1)^(1/2)*EllipticPi ((csc(d*x+c)-cot(d*x+c)+1)^(1/2),1/2-1/2*I,1/2*2^(1/2))*cos(d*x+c)^2+18*(c sc(d*x+c)-cot(d*x+c)+1)^(1/2)*(cot(d*x+c)-csc(d*x+c)+1)^(1/2)*(cot(d*x+c)- csc(d*x+c))^(1/2)*EllipticF((csc(d*x+c)-cot(d*x+c)+1)^(1/2),1/2*2^(1/2))*c os(d*x+c)^3-15*(cot(d*x+c)-csc(d*x+c)+1)^(1/2)*(cot(d*x+c)-csc(d*x+c))^(1/ 2)*(csc(d*x+c)-cot(d*x+c)+1)^(1/2)*EllipticPi((csc(d*x+c)-cot(d*x+c)+1)^(1 /2),1/2-1/2*I,1/2*2^(1/2))*cos(d*x+c)^3-15*(cot(d*x+c)-csc(d*x+c)+1)^(1/2) *(cot(d*x+c)-csc(d*x+c))^(1/2)*EllipticPi((csc(d*x+c)-cot(d*x+c)+1)^(1/2), 1/2+1/2*I,1/2*2^(1/2))*(csc(d*x+c)-cot(d*x+c)+1)^(1/2)*cos(d*x+c)^3-36*(cs c(d*x+c)-cot(d*x+c)+1)^(1/2)*(cot(d*x+c)-csc(d*x+c)+1)^(1/2)*(cot(d*x+c)-c sc(d*x+c))^(1/2)*EllipticE((csc(d*x+c)-cot(d*x+c)+1)^(1/2),1/2*2^(1/2))*co s(d*x+c)^3+18*(csc(d*x+c)-cot(d*x+c)+1)^(1/2)*(cot(d*x+c)-csc(d*x+c)+1)...
Timed out. \[ \int \frac {(e \tan (c+d x))^{9/2}}{a+a \sec (c+d x)} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {(e \tan (c+d x))^{9/2}}{a+a \sec (c+d x)} \, dx=\text {Timed out} \]
\[ \int \frac {(e \tan (c+d x))^{9/2}}{a+a \sec (c+d x)} \, dx=\int { \frac {\left (e \tan \left (d x + c\right )\right )^{\frac {9}{2}}}{a \sec \left (d x + c\right ) + a} \,d x } \]
\[ \int \frac {(e \tan (c+d x))^{9/2}}{a+a \sec (c+d x)} \, dx=\int { \frac {\left (e \tan \left (d x + c\right )\right )^{\frac {9}{2}}}{a \sec \left (d x + c\right ) + a} \,d x } \]
Timed out. \[ \int \frac {(e \tan (c+d x))^{9/2}}{a+a \sec (c+d x)} \, dx=\int \frac {\cos \left (c+d\,x\right )\,{\left (e\,\mathrm {tan}\left (c+d\,x\right )\right )}^{9/2}}{a\,\left (\cos \left (c+d\,x\right )+1\right )} \,d x \]