Integrand size = 23, antiderivative size = 310 \[ \int (a+a \sec (c+d x)) (e \tan (c+d x))^{5/2} \, dx=\frac {a e^{5/2} \arctan \left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} d}-\frac {a e^{5/2} \arctan \left (1+\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} d}-\frac {a e^{5/2} \log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)-\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} d}+\frac {a e^{5/2} \log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)+\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} d}+\frac {6 a e^2 \cos (c+d x) E\left (\left .c-\frac {\pi }{4}+d x\right |2\right ) \sqrt {e \tan (c+d x)}}{5 d \sqrt {\sin (2 c+2 d x)}}-\frac {6 a e \cos (c+d x) (e \tan (c+d x))^{3/2}}{5 d}+\frac {2 e (5 a+3 a \sec (c+d x)) (e \tan (c+d x))^{3/2}}{15 d} \]
1/2*a*e^(5/2)*arctan(1-2^(1/2)*(e*tan(d*x+c))^(1/2)/e^(1/2))/d*2^(1/2)-1/2 *a*e^(5/2)*arctan(1+2^(1/2)*(e*tan(d*x+c))^(1/2)/e^(1/2))/d*2^(1/2)-1/4*a* e^(5/2)*ln(e^(1/2)-2^(1/2)*(e*tan(d*x+c))^(1/2)+e^(1/2)*tan(d*x+c))/d*2^(1 /2)+1/4*a*e^(5/2)*ln(e^(1/2)+2^(1/2)*(e*tan(d*x+c))^(1/2)+e^(1/2)*tan(d*x+ c))/d*2^(1/2)-6/5*a*e^2*cos(d*x+c)*(sin(c+1/4*Pi+d*x)^2)^(1/2)/sin(c+1/4*P i+d*x)*EllipticE(cos(c+1/4*Pi+d*x),2^(1/2))*(e*tan(d*x+c))^(1/2)/d/sin(2*d *x+2*c)^(1/2)-6/5*a*e*cos(d*x+c)*(e*tan(d*x+c))^(3/2)/d+2/15*e*(5*a+3*a*se c(d*x+c))*(e*tan(d*x+c))^(3/2)/d
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 2.58 (sec) , antiderivative size = 186, normalized size of antiderivative = 0.60 \[ \int (a+a \sec (c+d x)) (e \tan (c+d x))^{5/2} \, dx=\frac {a (1+\cos (c+d x)) \csc (c+d x) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \left (12+20 \cos (c+d x)-36 \cos ^2(c+d x)+\frac {24 \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {3}{2},\frac {7}{4},-\tan ^2(c+d x)\right )}{\sqrt {\sec ^2(c+d x)}}+15 \arcsin (\cos (c+d x)-\sin (c+d x)) \cot ^2(c+d x) \sqrt {\sin (2 (c+d x))}+15 \cot ^2(c+d x) \log \left (\cos (c+d x)+\sin (c+d x)+\sqrt {\sin (2 (c+d x))}\right ) \sqrt {\sin (2 (c+d x))}\right ) (e \tan (c+d x))^{5/2}}{60 d} \]
(a*(1 + Cos[c + d*x])*Csc[c + d*x]*Sec[(c + d*x)/2]^2*(12 + 20*Cos[c + d*x ] - 36*Cos[c + d*x]^2 + (24*Hypergeometric2F1[3/4, 3/2, 7/4, -Tan[c + d*x] ^2])/Sqrt[Sec[c + d*x]^2] + 15*ArcSin[Cos[c + d*x] - Sin[c + d*x]]*Cot[c + d*x]^2*Sqrt[Sin[2*(c + d*x)]] + 15*Cot[c + d*x]^2*Log[Cos[c + d*x] + Sin[ c + d*x] + Sqrt[Sin[2*(c + d*x)]]]*Sqrt[Sin[2*(c + d*x)]])*(e*Tan[c + d*x] )^(5/2))/(60*d)
Time = 1.09 (sec) , antiderivative size = 295, normalized size of antiderivative = 0.95, number of steps used = 24, number of rules used = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.000, Rules used = {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.
\(\displaystyle \int (a \sec (c+d x)+a) (e \tan (c+d x))^{5/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right ) \left (-e \cot \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2}dx\) |
\(\Big \downarrow \) 4369 |
\(\displaystyle \frac {2 e (3 a \sec (c+d x)+5 a) (e \tan (c+d x))^{3/2}}{15 d}-\frac {2}{5} e^2 \int \frac {1}{2} (3 \sec (c+d x) a+5 a) \sqrt {e \tan (c+d x)}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {2 e (3 a \sec (c+d x)+5 a) (e \tan (c+d x))^{3/2}}{15 d}-\frac {1}{5} e^2 \int (3 \sec (c+d x) a+5 a) \sqrt {e \tan (c+d x)}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 e (3 a \sec (c+d x)+5 a) (e \tan (c+d x))^{3/2}}{15 d}-\frac {1}{5} e^2 \int \sqrt {-e \cot \left (c+d x+\frac {\pi }{2}\right )} \left (3 \csc \left (c+d x+\frac {\pi }{2}\right ) a+5 a\right )dx\) |
\(\Big \downarrow \) 4372 |
\(\displaystyle \frac {2 e (3 a \sec (c+d x)+5 a) (e \tan (c+d x))^{3/2}}{15 d}-\frac {1}{5} e^2 \left (5 a \int \sqrt {e \tan (c+d x)}dx+3 a \int \sec (c+d x) \sqrt {e \tan (c+d x)}dx\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 e (3 a \sec (c+d x)+5 a) (e \tan (c+d x))^{3/2}}{15 d}-\frac {1}{5} e^2 \left (5 a \int \sqrt {e \tan (c+d x)}dx+3 a \int \sec (c+d x) \sqrt {e \tan (c+d x)}dx\right )\) |
\(\Big \downarrow \) 3093 |
\(\displaystyle \frac {2 e (3 a \sec (c+d x)+5 a) (e \tan (c+d x))^{3/2}}{15 d}-\frac {1}{5} e^2 \left (5 a \int \sqrt {e \tan (c+d x)}dx+3 a \left (\frac {2 \cos (c+d x) (e \tan (c+d x))^{3/2}}{d e}-2 \int \cos (c+d x) \sqrt {e \tan (c+d x)}dx\right )\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 e (3 a \sec (c+d x)+5 a) (e \tan (c+d x))^{3/2}}{15 d}-\frac {1}{5} e^2 \left (5 a \int \sqrt {e \tan (c+d x)}dx+3 a \left (\frac {2 \cos (c+d x) (e \tan (c+d x))^{3/2}}{d e}-2 \int \frac {\sqrt {e \tan (c+d x)}}{\sec (c+d x)}dx\right )\right )\) |
\(\Big \downarrow \) 3095 |
\(\displaystyle \frac {2 e (3 a \sec (c+d x)+5 a) (e \tan (c+d x))^{3/2}}{15 d}-\frac {1}{5} e^2 \left (5 a \int \sqrt {e \tan (c+d x)}dx+3 a \left (\frac {2 \cos (c+d x) (e \tan (c+d x))^{3/2}}{d e}-\frac {2 \sqrt {\cos (c+d x)} \sqrt {e \tan (c+d x)} \int \sqrt {\cos (c+d x)} \sqrt {\sin (c+d x)}dx}{\sqrt {\sin (c+d x)}}\right )\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 e (3 a \sec (c+d x)+5 a) (e \tan (c+d x))^{3/2}}{15 d}-\frac {1}{5} e^2 \left (5 a \int \sqrt {e \tan (c+d x)}dx+3 a \left (\frac {2 \cos (c+d x) (e \tan (c+d x))^{3/2}}{d e}-\frac {2 \sqrt {\cos (c+d x)} \sqrt {e \tan (c+d x)} \int \sqrt {\cos (c+d x)} \sqrt {\sin (c+d x)}dx}{\sqrt {\sin (c+d x)}}\right )\right )\) |
\(\Big \downarrow \) 3052 |
\(\displaystyle \frac {2 e (3 a \sec (c+d x)+5 a) (e \tan (c+d x))^{3/2}}{15 d}-\frac {1}{5} e^2 \left (5 a \int \sqrt {e \tan (c+d x)}dx+3 a \left (\frac {2 \cos (c+d x) (e \tan (c+d x))^{3/2}}{d e}-\frac {2 \cos (c+d x) \sqrt {e \tan (c+d x)} \int \sqrt {\sin (2 c+2 d x)}dx}{\sqrt {\sin (2 c+2 d x)}}\right )\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 e (3 a \sec (c+d x)+5 a) (e \tan (c+d x))^{3/2}}{15 d}-\frac {1}{5} e^2 \left (5 a \int \sqrt {e \tan (c+d x)}dx+3 a \left (\frac {2 \cos (c+d x) (e \tan (c+d x))^{3/2}}{d e}-\frac {2 \cos (c+d x) \sqrt {e \tan (c+d x)} \int \sqrt {\sin (2 c+2 d x)}dx}{\sqrt {\sin (2 c+2 d x)}}\right )\right )\) |
\(\Big \downarrow \) 3119 |
\(\displaystyle \frac {2 e (3 a \sec (c+d x)+5 a) (e \tan (c+d x))^{3/2}}{15 d}-\frac {1}{5} e^2 \left (5 a \int \sqrt {e \tan (c+d x)}dx+3 a \left (\frac {2 \cos (c+d x) (e \tan (c+d x))^{3/2}}{d e}-\frac {2 \cos (c+d x) E\left (\left .c+d x-\frac {\pi }{4}\right |2\right ) \sqrt {e \tan (c+d x)}}{d \sqrt {\sin (2 c+2 d x)}}\right )\right )\) |
\(\Big \downarrow \) 3957 |
\(\displaystyle \frac {2 e (3 a \sec (c+d x)+5 a) (e \tan (c+d x))^{3/2}}{15 d}-\frac {1}{5} e^2 \left (\frac {5 a e \int \frac {\sqrt {e \tan (c+d x)}}{\tan ^2(c+d x) e^2+e^2}d(e \tan (c+d x))}{d}+3 a \left (\frac {2 \cos (c+d x) (e \tan (c+d x))^{3/2}}{d e}-\frac {2 \cos (c+d x) E\left (\left .c+d x-\frac {\pi }{4}\right |2\right ) \sqrt {e \tan (c+d x)}}{d \sqrt {\sin (2 c+2 d x)}}\right )\right )\) |
\(\Big \downarrow \) 266 |
\(\displaystyle \frac {2 e (3 a \sec (c+d x)+5 a) (e \tan (c+d x))^{3/2}}{15 d}-\frac {1}{5} e^2 \left (\frac {10 a e \int \frac {e^2 \tan ^2(c+d x)}{e^4 \tan ^4(c+d x)+e^2}d\sqrt {e \tan (c+d x)}}{d}+3 a \left (\frac {2 \cos (c+d x) (e \tan (c+d x))^{3/2}}{d e}-\frac {2 \cos (c+d x) E\left (\left .c+d x-\frac {\pi }{4}\right |2\right ) \sqrt {e \tan (c+d x)}}{d \sqrt {\sin (2 c+2 d x)}}\right )\right )\) |
\(\Big \downarrow \) 826 |
\(\displaystyle \frac {2 e (3 a \sec (c+d x)+5 a) (e \tan (c+d x))^{3/2}}{15 d}-\frac {1}{5} e^2 \left (\frac {10 a e \left (\frac {1}{2} \int \frac {e^2 \tan ^2(c+d x)+e}{e^4 \tan ^4(c+d x)+e^2}d\sqrt {e \tan (c+d x)}-\frac {1}{2} \int \frac {e-e^2 \tan ^2(c+d x)}{e^4 \tan ^4(c+d x)+e^2}d\sqrt {e \tan (c+d x)}\right )}{d}+3 a \left (\frac {2 \cos (c+d x) (e \tan (c+d x))^{3/2}}{d e}-\frac {2 \cos (c+d x) E\left (\left .c+d x-\frac {\pi }{4}\right |2\right ) \sqrt {e \tan (c+d x)}}{d \sqrt {\sin (2 c+2 d x)}}\right )\right )\) |
\(\Big \downarrow \) 1476 |
\(\displaystyle \frac {2 e (3 a \sec (c+d x)+5 a) (e \tan (c+d x))^{3/2}}{15 d}-\frac {1}{5} e^2 \left (\frac {10 a e \left (\frac {1}{2} \left (\frac {1}{2} \int \frac {1}{e^2 \tan ^2(c+d x)-\sqrt {2} e^{3/2} \tan (c+d x)+e}d\sqrt {e \tan (c+d x)}+\frac {1}{2} \int \frac {1}{e^2 \tan ^2(c+d x)+\sqrt {2} e^{3/2} \tan (c+d x)+e}d\sqrt {e \tan (c+d x)}\right )-\frac {1}{2} \int \frac {e-e^2 \tan ^2(c+d x)}{e^4 \tan ^4(c+d x)+e^2}d\sqrt {e \tan (c+d x)}\right )}{d}+3 a \left (\frac {2 \cos (c+d x) (e \tan (c+d x))^{3/2}}{d e}-\frac {2 \cos (c+d x) E\left (\left .c+d x-\frac {\pi }{4}\right |2\right ) \sqrt {e \tan (c+d x)}}{d \sqrt {\sin (2 c+2 d x)}}\right )\right )\) |
\(\Big \downarrow \) 1082 |
\(\displaystyle \frac {2 e (3 a \sec (c+d x)+5 a) (e \tan (c+d x))^{3/2}}{15 d}-\frac {1}{5} e^2 \left (\frac {10 a e \left (\frac {1}{2} \left (\frac {\int \frac {1}{-e^2 \tan ^2(c+d x)-1}d\left (1-\sqrt {2} \sqrt {e} \tan (c+d x)\right )}{\sqrt {2} \sqrt {e}}-\frac {\int \frac {1}{-e^2 \tan ^2(c+d x)-1}d\left (\sqrt {2} \sqrt {e} \tan (c+d x)+1\right )}{\sqrt {2} \sqrt {e}}\right )-\frac {1}{2} \int \frac {e-e^2 \tan ^2(c+d x)}{e^4 \tan ^4(c+d x)+e^2}d\sqrt {e \tan (c+d x)}\right )}{d}+3 a \left (\frac {2 \cos (c+d x) (e \tan (c+d x))^{3/2}}{d e}-\frac {2 \cos (c+d x) E\left (\left .c+d x-\frac {\pi }{4}\right |2\right ) \sqrt {e \tan (c+d x)}}{d \sqrt {\sin (2 c+2 d x)}}\right )\right )\) |
\(\Big \downarrow \) 217 |
\(\displaystyle \frac {2 e (3 a \sec (c+d x)+5 a) (e \tan (c+d x))^{3/2}}{15 d}-\frac {1}{5} e^2 \left (\frac {10 a e \left (\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {e} \tan (c+d x)+1\right )}{\sqrt {2} \sqrt {e}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {e} \tan (c+d x)\right )}{\sqrt {2} \sqrt {e}}\right )-\frac {1}{2} \int \frac {e-e^2 \tan ^2(c+d x)}{e^4 \tan ^4(c+d x)+e^2}d\sqrt {e \tan (c+d x)}\right )}{d}+3 a \left (\frac {2 \cos (c+d x) (e \tan (c+d x))^{3/2}}{d e}-\frac {2 \cos (c+d x) E\left (\left .c+d x-\frac {\pi }{4}\right |2\right ) \sqrt {e \tan (c+d x)}}{d \sqrt {\sin (2 c+2 d x)}}\right )\right )\) |
\(\Big \downarrow \) 1479 |
\(\displaystyle \frac {2 e (3 a \sec (c+d x)+5 a) (e \tan (c+d x))^{3/2}}{15 d}-\frac {1}{5} e^2 \left (\frac {10 a e \left (\frac {1}{2} \left (\frac {\int -\frac {\sqrt {2} \sqrt {e}-2 \sqrt {e \tan (c+d x)}}{e^2 \tan ^2(c+d x)-\sqrt {2} e^{3/2} \tan (c+d x)+e}d\sqrt {e \tan (c+d x)}}{2 \sqrt {2} \sqrt {e}}+\frac {\int -\frac {\sqrt {2} \left (\sqrt {e}+\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{e^2 \tan ^2(c+d x)+\sqrt {2} e^{3/2} \tan (c+d x)+e}d\sqrt {e \tan (c+d x)}}{2 \sqrt {2} \sqrt {e}}\right )+\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {e} \tan (c+d x)+1\right )}{\sqrt {2} \sqrt {e}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {e} \tan (c+d x)\right )}{\sqrt {2} \sqrt {e}}\right )\right )}{d}+3 a \left (\frac {2 \cos (c+d x) (e \tan (c+d x))^{3/2}}{d e}-\frac {2 \cos (c+d x) E\left (\left .c+d x-\frac {\pi }{4}\right |2\right ) \sqrt {e \tan (c+d x)}}{d \sqrt {\sin (2 c+2 d x)}}\right )\right )\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {2 e (3 a \sec (c+d x)+5 a) (e \tan (c+d x))^{3/2}}{15 d}-\frac {1}{5} e^2 \left (\frac {10 a e \left (\frac {1}{2} \left (-\frac {\int \frac {\sqrt {2} \sqrt {e}-2 \sqrt {e \tan (c+d x)}}{e^2 \tan ^2(c+d x)-\sqrt {2} e^{3/2} \tan (c+d x)+e}d\sqrt {e \tan (c+d x)}}{2 \sqrt {2} \sqrt {e}}-\frac {\int \frac {\sqrt {2} \left (\sqrt {e}+\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{e^2 \tan ^2(c+d x)+\sqrt {2} e^{3/2} \tan (c+d x)+e}d\sqrt {e \tan (c+d x)}}{2 \sqrt {2} \sqrt {e}}\right )+\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {e} \tan (c+d x)+1\right )}{\sqrt {2} \sqrt {e}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {e} \tan (c+d x)\right )}{\sqrt {2} \sqrt {e}}\right )\right )}{d}+3 a \left (\frac {2 \cos (c+d x) (e \tan (c+d x))^{3/2}}{d e}-\frac {2 \cos (c+d x) E\left (\left .c+d x-\frac {\pi }{4}\right |2\right ) \sqrt {e \tan (c+d x)}}{d \sqrt {\sin (2 c+2 d x)}}\right )\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {2 e (3 a \sec (c+d x)+5 a) (e \tan (c+d x))^{3/2}}{15 d}-\frac {1}{5} e^2 \left (\frac {10 a e \left (\frac {1}{2} \left (-\frac {\int \frac {\sqrt {2} \sqrt {e}-2 \sqrt {e \tan (c+d x)}}{e^2 \tan ^2(c+d x)-\sqrt {2} e^{3/2} \tan (c+d x)+e}d\sqrt {e \tan (c+d x)}}{2 \sqrt {2} \sqrt {e}}-\frac {\int \frac {\sqrt {e}+\sqrt {2} \sqrt {e \tan (c+d x)}}{e^2 \tan ^2(c+d x)+\sqrt {2} e^{3/2} \tan (c+d x)+e}d\sqrt {e \tan (c+d x)}}{2 \sqrt {e}}\right )+\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {e} \tan (c+d x)+1\right )}{\sqrt {2} \sqrt {e}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {e} \tan (c+d x)\right )}{\sqrt {2} \sqrt {e}}\right )\right )}{d}+3 a \left (\frac {2 \cos (c+d x) (e \tan (c+d x))^{3/2}}{d e}-\frac {2 \cos (c+d x) E\left (\left .c+d x-\frac {\pi }{4}\right |2\right ) \sqrt {e \tan (c+d x)}}{d \sqrt {\sin (2 c+2 d x)}}\right )\right )\) |
\(\Big \downarrow \) 1103 |
\(\displaystyle \frac {2 e (3 a \sec (c+d x)+5 a) (e \tan (c+d x))^{3/2}}{15 d}-\frac {1}{5} e^2 \left (\frac {10 a e \left (\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {e} \tan (c+d x)+1\right )}{\sqrt {2} \sqrt {e}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {e} \tan (c+d x)\right )}{\sqrt {2} \sqrt {e}}\right )+\frac {1}{2} \left (\frac {\log \left (-\sqrt {2} e^{3/2} \tan (c+d x)+e^2 \tan ^2(c+d x)+e\right )}{2 \sqrt {2} \sqrt {e}}-\frac {\log \left (\sqrt {2} e^{3/2} \tan (c+d x)+e^2 \tan ^2(c+d x)+e\right )}{2 \sqrt {2} \sqrt {e}}\right )\right )}{d}+3 a \left (\frac {2 \cos (c+d x) (e \tan (c+d x))^{3/2}}{d e}-\frac {2 \cos (c+d x) E\left (\left .c+d x-\frac {\pi }{4}\right |2\right ) \sqrt {e \tan (c+d x)}}{d \sqrt {\sin (2 c+2 d x)}}\right )\right )\) |
(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])) + ArcTa n[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]*Sqrt[e]))/2 ))/d + 3*a*((-2*Cos[c + d*x]*EllipticE[c - Pi/4 + d*x, 2]*Sqrt[e*Tan[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
3.2.4.3.1 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]
Leaf count of result is larger than twice the leaf count of optimal. \(579\) vs. \(2(274)=548\).
Time = 6.00 (sec) , antiderivative size = 580, normalized size of antiderivative = 1.87
method | result | size |
parts | \(\frac {2 a e \left (\frac {\left (e \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}-\frac {e^{2} \sqrt {2}\, \left (\ln \left (\frac {e \tan \left (d x +c \right )-\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \tan \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}{e \tan \left (d x +c \right )+\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \tan \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {e \tan \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {e \tan \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 \left (e^{2}\right )^{\frac {1}{4}}}\right )}{d}-\frac {a \sqrt {2}\, \left (-6 \sqrt {\csc \left (d x +c \right )-\cot \left (d x +c \right )+1}\, \sqrt {\cot \left (d x +c \right )-\csc \left (d x +c \right )+1}\, \sqrt {\cot \left (d x +c \right )-\csc \left (d x +c \right )}\, \operatorname {EllipticE}\left (\sqrt {\csc \left (d x +c \right )-\cot \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right ) \cos \left (d x +c \right )^{3}+3 \sqrt {\csc \left (d x +c \right )-\cot \left (d x +c \right )+1}\, \sqrt {\cot \left (d x +c \right )-\csc \left (d x +c \right )+1}\, \sqrt {\cot \left (d x +c \right )-\csc \left (d x +c \right )}\, \operatorname {EllipticF}\left (\sqrt {\csc \left (d x +c \right )-\cot \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right ) \cos \left (d x +c \right )^{3}-6 \sqrt {\csc \left (d x +c \right )-\cot \left (d x +c \right )+1}\, \sqrt {\cot \left (d x +c \right )-\csc \left (d x +c \right )+1}\, \sqrt {\cot \left (d x +c \right )-\csc \left (d x +c \right )}\, \operatorname {EllipticE}\left (\sqrt {\csc \left (d x +c \right )-\cot \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right ) \cos \left (d x +c \right )^{2}+3 \sqrt {\csc \left (d x +c \right )-\cot \left (d x +c \right )+1}\, \sqrt {\cot \left (d x +c \right )-\csc \left (d x +c \right )+1}\, \sqrt {\cot \left (d x +c \right )-\csc \left (d x +c \right )}\, \operatorname {EllipticF}\left (\sqrt {\csc \left (d x +c \right )-\cot \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right ) \cos \left (d x +c \right )^{2}+3 \cos \left (d x +c \right )^{3} \sqrt {2}-4 \sqrt {2}\, \cos \left (d x +c \right )^{2}+\sqrt {2}\right ) e^{2} \sqrt {e \tan \left (d x +c \right )}\, \sec \left (d x +c \right ) \tan \left (d x +c \right )}{5 d \left (\cos \left (d x +c \right )^{2}-1\right )}\) | \(580\) |
default | \(\text {Expression too large to display}\) | \(1152\) |
2*a/d*e*(1/3*(e*tan(d*x+c))^(3/2)-1/8*e^2/(e^2)^(1/4)*2^(1/2)*(ln((e*tan(d *x+c)-(e^2)^(1/4)*(e*tan(d*x+c))^(1/2)*2^(1/2)+(e^2)^(1/2))/(e*tan(d*x+c)+ (e^2)^(1/4)*(e*tan(d*x+c))^(1/2)*2^(1/2)+(e^2)^(1/2)))+2*arctan(2^(1/2)/(e ^2)^(1/4)*(e*tan(d*x+c))^(1/2)+1)-2*arctan(-2^(1/2)/(e^2)^(1/4)*(e*tan(d*x +c))^(1/2)+1)))-1/5*a/d*2^(1/2)*(-6*(csc(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)*EllipticE((csc(d*x +c)-cot(d*x+c)+1)^(1/2),1/2*2^(1/2))*cos(d*x+c)^3+3*(csc(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)*El lipticF((csc(d*x+c)-cot(d*x+c)+1)^(1/2),1/2*2^(1/2))*cos(d*x+c)^3-6*(csc(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)*EllipticE((csc(d*x+c)-cot(d*x+c)+1)^(1/2),1/2*2^(1/2))*cos(d *x+c)^2+3*(csc(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))*cos(d*x+c)^2+3*cos(d*x+c)^3*2^(1/2)-4*2^(1/2)*cos(d*x+c)^2+2^(1 /2))*e^2*(e*tan(d*x+c))^(1/2)/(cos(d*x+c)^2-1)*sec(d*x+c)*tan(d*x+c)
Timed out. \[ \int (a+a \sec (c+d x)) (e \tan (c+d x))^{5/2} \, dx=\text {Timed out} \]
\[ \int (a+a \sec (c+d x)) (e \tan (c+d x))^{5/2} \, dx=a \left (\int \left (e \tan {\left (c + d x \right )}\right )^{\frac {5}{2}}\, dx + \int \left (e \tan {\left (c + d x \right )}\right )^{\frac {5}{2}} \sec {\left (c + d x \right )}\, dx\right ) \]
a*(Integral((e*tan(c + d*x))**(5/2), x) + Integral((e*tan(c + d*x))**(5/2) *sec(c + d*x), x))
Exception generated. \[ \int (a+a \sec (c+d x)) (e \tan (c+d x))^{5/2} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(e>0)', see `assume?` for more de tails)Is e
\[ \int (a+a \sec (c+d x)) (e \tan (c+d x))^{5/2} \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )} \left (e \tan \left (d x + c\right )\right )^{\frac {5}{2}} \,d x } \]
Timed out. \[ \int (a+a \sec (c+d x)) (e \tan (c+d x))^{5/2} \, dx=\int {\left (e\,\mathrm {tan}\left (c+d\,x\right )\right )}^{5/2}\,\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right ) \,d x \]