\(\int (a+a \sec (c+d x)) (e \tan (c+d x))^{5/2} \, dx\) [104]

Optimal result

Integrand size = 23, antiderivative size = 253 \[ \int (a+a \sec (c+d x)) (e \tan (c+d x))^{5/2} \, dx=\frac {a e^{5/2} \arctan \left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} d}-\frac {a e^{5/2} \arctan \left (1+\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} d}+\frac {a e^{5/2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}+\sqrt {e} \tan (c+d x)}\right )}{\sqrt {2} d}+\frac {6 a e^2 \cos (c+d x) E\left (\left .c-\frac {\pi }{4}+d x\right |2\right ) \sqrt {e \tan (c+d x)}}{5 d \sqrt {\sin (2 c+2 d x)}}-\frac {6 a e \cos (c+d x) (e \tan (c+d x))^{3/2}}{5 d}+\frac {2 e (5 a+3 a \sec (c+d x)) (e \tan (c+d x))^{3/2}}{15 d} \] Output:

1/2*a*e^(5/2)*arctan(1-2^(1/2)*(e*tan(d*x+c))^(1/2)/e^(1/2))*2^(1/2)/d-1/2 
*a*e^(5/2)*arctan(1+2^(1/2)*(e*tan(d*x+c))^(1/2)/e^(1/2))*2^(1/2)/d+1/2*a* 
e^(5/2)*arctanh(2^(1/2)*(e*tan(d*x+c))^(1/2)/(e^(1/2)+e^(1/2)*tan(d*x+c))) 
*2^(1/2)/d-6/5*a*e^2*cos(d*x+c)*EllipticE(cos(c+1/4*Pi+d*x),2^(1/2))*(e*ta 
n(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*sec(d*x+c))*(e*tan(d*x+c))^(3/2)/d
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 4.94 (sec) , antiderivative size = 186, normalized size of antiderivative = 0.74 \[ \int (a+a \sec (c+d x)) (e \tan (c+d x))^{5/2} \, dx=\frac {a (1+\cos (c+d x)) \csc (c+d x) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \left (12+20 \cos (c+d x)-36 \cos ^2(c+d x)+\frac {24 \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {3}{2},\frac {7}{4},-\tan ^2(c+d x)\right )}{\sqrt {\sec ^2(c+d x)}}+15 \arcsin (\cos (c+d x)-\sin (c+d x)) \cot ^2(c+d x) \sqrt {\sin (2 (c+d x))}+15 \cot ^2(c+d x) \log \left (\cos (c+d x)+\sin (c+d x)+\sqrt {\sin (2 (c+d x))}\right ) \sqrt {\sin (2 (c+d x))}\right ) (e \tan (c+d x))^{5/2}}{60 d} \] Input:

Integrate[(a + a*Sec[c + d*x])*(e*Tan[c + d*x])^(5/2),x]
 

Output:

(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)
 

Rubi [A] (warning: unable to verify)

Time = 1.14 (sec) , antiderivative size = 295, normalized size of antiderivative = 1.17, 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

\(\displaystyle \frac {2 e (3 a \sec (c+d x)+5 a) (e \tan (c+d x))^{3/2}}{15 d}-\frac {1}{5} e^2 \left (\frac {10 a e \left (\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {e} \tan (c+d x)+1\right )}{\sqrt {2} \sqrt {e}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {e} \tan (c+d x)\right )}{\sqrt {2} \sqrt {e}}\right )+\frac {1}{2} \left (\frac {\log \left (-\sqrt {2} e^{3/2} \tan (c+d x)+e^2 \tan ^2(c+d x)+e\right )}{2 \sqrt {2} \sqrt {e}}-\frac {\log \left (\sqrt {2} e^{3/2} \tan (c+d x)+e^2 \tan ^2(c+d x)+e\right )}{2 \sqrt {2} \sqrt {e}}\right )\right )}{d}+3 a \left (\frac {2 \cos (c+d x) (e \tan (c+d x))^{3/2}}{d e}-\frac {2 \cos (c+d x) E\left (\left .c+d x-\frac {\pi }{4}\right |2\right ) \sqrt {e \tan (c+d x)}}{d \sqrt {\sin (2 c+2 d x)}}\right )\right )\)

Input:

Int[(a + a*Sec[c + d*x])*(e*Tan[c + d*x])^(5/2),x]
 

Output:

(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
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 217

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])
 

rule 266

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]
 

rule 826

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]]))
 

rule 1082

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]
 

rule 1103

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]
 

rule 1476

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]
 

rule 1479

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]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3052

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]
 

rule 3093

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]
 

rule 3095

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]
 

rule 3119

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]
 

rule 3957

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]
 

rule 4369

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]
 

rule 4372

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]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(456\) vs. \(2(210)=420\).

Time = 1.97 (sec) , antiderivative size = 457, normalized size of antiderivative = 1.81

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}\, \sqrt {-\frac {2 \sin \left (d x +c \right ) \cos \left (d x +c \right )}{\left (1+\cos \left (d x +c \right )\right )^{2}}}\, \sqrt {e \tan \left (d x +c \right )}\, e^{2} \left (\sqrt {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}\, \sqrt {2 \cot \left (d x +c \right )-2 \csc \left (d x +c \right )+2}\, \sqrt {-\csc \left (d x +c \right )+\cot \left (d x +c \right )}\, \operatorname {EllipticE}\left (\sqrt {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right ) \left (6 \cot \left (d x +c \right )+6 \csc \left (d x +c \right )\right )+\sqrt {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}\, \sqrt {2 \cot \left (d x +c \right )-2 \csc \left (d x +c \right )+2}\, \sqrt {-\csc \left (d x +c \right )+\cot \left (d x +c \right )}\, \operatorname {EllipticF}\left (\sqrt {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right ) \left (-3 \cot \left (d x +c \right )-3 \csc \left (d x +c \right )\right )-6 \cot \left (d x +c \right )+8 \csc \left (d x +c \right )-2 \sec \left (d x +c \right )^{2} \csc \left (d x +c \right )\right )}{10 d \sqrt {-\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{\left (1+\cos \left (d x +c \right )\right )^{2}}}}\)	\(457\)
default	\(-\frac {a \sqrt {2}\, \sqrt {-\frac {2 \sin \left (d x +c \right ) \cos \left (d x +c \right )}{\left (1+\cos \left (d x +c \right )\right )^{2}}}\, \sqrt {e \tan \left (d x +c \right )}\, e^{2} \left (i \sqrt {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}\, \sqrt {2 \cot \left (d x +c \right )-2 \csc \left (d x +c \right )+2}\, \sqrt {-\csc \left (d x +c \right )+\cot \left (d x +c \right )}\, \operatorname {EllipticPi}\left (\sqrt {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}, \frac {1}{2}-\frac {i}{2}, \frac {\sqrt {2}}{2}\right ) \left (15 \cot \left (d x +c \right )+15 \csc \left (d x +c \right )\right )+i \sqrt {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}\, \sqrt {2 \cot \left (d x +c \right )-2 \csc \left (d x +c \right )+2}\, \sqrt {-\csc \left (d x +c \right )+\cot \left (d x +c \right )}\, \operatorname {EllipticPi}\left (\sqrt {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}, \frac {1}{2}+\frac {i}{2}, \frac {\sqrt {2}}{2}\right ) \left (-15 \cot \left (d x +c \right )-15 \csc \left (d x +c \right )\right )+\sqrt {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}\, \sqrt {2 \cot \left (d x +c \right )-2 \csc \left (d x +c \right )+2}\, \sqrt {-\csc \left (d x +c \right )+\cot \left (d x +c \right )}\, \operatorname {EllipticF}\left (\sqrt {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right ) \left (-18 \cot \left (d x +c \right )-18 \csc \left (d x +c \right )\right )+\sqrt {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}\, \sqrt {2 \cot \left (d x +c \right )-2 \csc \left (d x +c \right )+2}\, \sqrt {-\csc \left (d x +c \right )+\cot \left (d x +c \right )}\, \operatorname {EllipticPi}\left (\sqrt {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}, \frac {1}{2}-\frac {i}{2}, \frac {\sqrt {2}}{2}\right ) \left (-15 \cot \left (d x +c \right )-15 \csc \left (d x +c \right )\right )+\sqrt {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}\, \sqrt {2 \cot \left (d x +c \right )-2 \csc \left (d x +c \right )+2}\, \sqrt {-\csc \left (d x +c \right )+\cot \left (d x +c \right )}\, \operatorname {EllipticPi}\left (\sqrt {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}, \frac {1}{2}+\frac {i}{2}, \frac {\sqrt {2}}{2}\right ) \left (-15 \cot \left (d x +c \right )-15 \csc \left (d x +c \right )\right )+\sqrt {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}\, \sqrt {2 \cot \left (d x +c \right )-2 \csc \left (d x +c \right )+2}\, \sqrt {-\csc \left (d x +c \right )+\cot \left (d x +c \right )}\, \operatorname {EllipticE}\left (\sqrt {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right ) \left (36 \cot \left (d x +c \right )+36 \csc \left (d x +c \right )\right )-16 \cot \left (d x +c \right )+48 \csc \left (d x +c \right )-20 \sec \left (d x +c \right ) \csc \left (d x +c \right )-12 \sec \left (d x +c \right )^{2} \csc \left (d x +c \right )\right )}{60 d \sqrt {-\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{\left (1+\cos \left (d x +c \right )\right )^{2}}}}\)	\(720\)

Input:

int((a+a*sec(d*x+c))*(e*tan(d*x+c))^(5/2),x,method=_RETURNVERBOSE)
 

Output:

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)))-1/10*a/d*2^(1/2)*(-2*sin(d*x+c)*cos(d*x+c)/(1+cos(d*x+c))^ 
2)^(1/2)*(e*tan(d*x+c))^(1/2)*e^2/(-sin(d*x+c)*cos(d*x+c)/(1+cos(d*x+c))^2 
)^(1/2)*((-cot(d*x+c)+csc(d*x+c)+1)^(1/2)*(2*cot(d*x+c)-2*csc(d*x+c)+2)^(1 
/2)*(-csc(d*x+c)+cot(d*x+c))^(1/2)*EllipticE((-cot(d*x+c)+csc(d*x+c)+1)^(1 
/2),1/2*2^(1/2))*(6*cot(d*x+c)+6*csc(d*x+c))+(-cot(d*x+c)+csc(d*x+c)+1)^(1 
/2)*(2*cot(d*x+c)-2*csc(d*x+c)+2)^(1/2)*(-csc(d*x+c)+cot(d*x+c))^(1/2)*Ell 
ipticF((-cot(d*x+c)+csc(d*x+c)+1)^(1/2),1/2*2^(1/2))*(-3*cot(d*x+c)-3*csc( 
d*x+c))-6*cot(d*x+c)+8*csc(d*x+c)-2*sec(d*x+c)^2*csc(d*x+c))
 

Fricas [F(-1)]

Timed out. \[ \int (a+a \sec (c+d x)) (e \tan (c+d x))^{5/2} \, dx=\text {Timed out} \] Input:

integrate((a+a*sec(d*x+c))*(e*tan(d*x+c))^(5/2),x, algorithm="fricas")
 

Output:

Timed out
 

Sympy [F]

\[ \int (a+a \sec (c+d x)) (e \tan (c+d x))^{5/2} \, dx=a \left (\int \left (e \tan {\left (c + d x \right )}\right )^{\frac {5}{2}}\, dx + \int \left (e \tan {\left (c + d x \right )}\right )^{\frac {5}{2}} \sec {\left (c + d x \right )}\, dx\right ) \] Input:

integrate((a+a*sec(d*x+c))*(e*tan(d*x+c))**(5/2),x)
 

Output:

a*(Integral((e*tan(c + d*x))**(5/2), x) + Integral((e*tan(c + d*x))**(5/2) 
*sec(c + d*x), x))
 

Maxima [F(-2)]

Exception generated. \[ \int (a+a \sec (c+d x)) (e \tan (c+d x))^{5/2} \, dx=\text {Exception raised: ValueError} \] Input:

integrate((a+a*sec(d*x+c))*(e*tan(d*x+c))^(5/2),x, algorithm="maxima")
 

Output:

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
 

Giac [F]

\[ \int (a+a \sec (c+d x)) (e \tan (c+d x))^{5/2} \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )} \left (e \tan \left (d x + c\right )\right )^{\frac {5}{2}} \,d x } \] Input:

integrate((a+a*sec(d*x+c))*(e*tan(d*x+c))^(5/2),x, algorithm="giac")
 

Output:

integrate((a*sec(d*x + c) + a)*(e*tan(d*x + c))^(5/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int (a+a \sec (c+d x)) (e \tan (c+d x))^{5/2} \, dx=\int {\left (e\,\mathrm {tan}\left (c+d\,x\right )\right )}^{5/2}\,\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right ) \,d x \] Input:

int((e*tan(c + d*x))^(5/2)*(a + a/cos(c + d*x)),x)
 

Output:

int((e*tan(c + d*x))^(5/2)*(a + a/cos(c + d*x)), x)
 

Reduce [F]

\[ \int (a+a \sec (c+d x)) (e \tan (c+d x))^{5/2} \, dx=\sqrt {e}\, a \,e^{2} \left (\int \sqrt {\tan \left (d x +c \right )}\, \tan \left (d x +c \right )^{2}d x +\int \sqrt {\tan \left (d x +c \right )}\, \sec \left (d x +c \right ) \tan \left (d x +c \right )^{2}d x \right ) \] Input:

int((a+a*sec(d*x+c))*(e*tan(d*x+c))^(5/2),x)
 

Output:

sqrt(e)*a*e**2*(int(sqrt(tan(c + d*x))*tan(c + d*x)**2,x) + int(sqrt(tan(c 
 + d*x))*sec(c + d*x)*tan(c + d*x)**2,x))