\(\int \frac {(a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)+C \sec ^2(c+d x))}{\cos ^{\frac {3}{2}}(c+d x)} \, dx\) [1270]

Optimal result

Integrand size = 45, antiderivative size = 301 \[ \int \frac {(a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\frac {a^{5/2} (400 A+326 B+283 C) \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{128 d}+\frac {a^3 (1040 A+950 B+787 C) \sin (c+d x)}{960 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {a^3 (400 A+326 B+283 C) \sin (c+d x)}{128 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {a^2 (80 A+110 B+79 C) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{240 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {a (2 B+C) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{8 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {C (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)} \] Output:

1/128*a^(5/2)*(400*A+326*B+283*C)*arcsinh(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+ 
c))^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/d+1/960*a^3*(1040*A+950*B+787 
*C)*sin(d*x+c)/d/cos(d*x+c)^(5/2)/(a+a*sec(d*x+c))^(1/2)+1/128*a^3*(400*A+ 
326*B+283*C)*sin(d*x+c)/d/cos(d*x+c)^(3/2)/(a+a*sec(d*x+c))^(1/2)+1/240*a^ 
2*(80*A+110*B+79*C)*(a+a*sec(d*x+c))^(1/2)*sin(d*x+c)/d/cos(d*x+c)^(5/2)+1 
/8*a*(2*B+C)*(a+a*sec(d*x+c))^(3/2)*sin(d*x+c)/d/cos(d*x+c)^(5/2)+1/5*C*(a 
+a*sec(d*x+c))^(5/2)*sin(d*x+c)/d/cos(d*x+c)^(5/2)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 10.19 (sec) , antiderivative size = 212, normalized size of antiderivative = 0.70 \[ \int \frac {(a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\frac {a^2 \sec \left (\frac {1}{2} (c+d x)\right ) \sqrt {a (1+\sec (c+d x))} \left (60 \sqrt {2} (400 A+326 B+283 C) \text {arctanh}\left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right ) \cos ^5(c+d x)+(20560 A+22030 B+24863 C+12 (1360 A+1950 B+2343 C) \cos (c+d x)+4 (6640 A+6730 B+6509 C) \cos (2 (c+d x))+5440 A \cos (3 (c+d x))+6520 B \cos (3 (c+d x))+5660 C \cos (3 (c+d x))+6000 A \cos (4 (c+d x))+4890 B \cos (4 (c+d x))+4245 C \cos (4 (c+d x))) \sin \left (\frac {1}{2} (c+d x)\right )\right )}{15360 d \cos ^{\frac {9}{2}}(c+d x)} \] Input:

Integrate[((a + a*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x] 
^2))/Cos[c + d*x]^(3/2),x]
 

Output:

(a^2*Sec[(c + d*x)/2]*Sqrt[a*(1 + Sec[c + d*x])]*(60*Sqrt[2]*(400*A + 326* 
B + 283*C)*ArcTanh[Sqrt[2]*Sin[(c + d*x)/2]]*Cos[c + d*x]^5 + (20560*A + 2 
2030*B + 24863*C + 12*(1360*A + 1950*B + 2343*C)*Cos[c + d*x] + 4*(6640*A 
+ 6730*B + 6509*C)*Cos[2*(c + d*x)] + 5440*A*Cos[3*(c + d*x)] + 6520*B*Cos 
[3*(c + d*x)] + 5660*C*Cos[3*(c + d*x)] + 6000*A*Cos[4*(c + d*x)] + 4890*B 
*Cos[4*(c + d*x)] + 4245*C*Cos[4*(c + d*x)])*Sin[(c + d*x)/2]))/(15360*d*C 
os[c + d*x]^(9/2))
 

Rubi [A] (verified)

Time = 2.04 (sec) , antiderivative size = 312, normalized size of antiderivative = 1.04, number of steps used = 19, number of rules used = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 4753, 3042, 4576, 27, 3042, 4506, 27, 3042, 4506, 27, 3042, 4504, 3042, 4290, 3042, 4288, 222}

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.

\(\displaystyle \int \frac {(a \sec (c+d x)+a)^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(a \sec (c+d x)+a)^{5/2} \left (A+B \sec (c+d x)+C \sec (c+d x)^2\right )}{\cos (c+d x)^{3/2}}dx\)

\(\Big \downarrow \) 4753

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sec ^{\frac {3}{2}}(c+d x) (\sec (c+d x) a+a)^{5/2} \left (C \sec ^2(c+d x)+B \sec (c+d x)+A\right )dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{5/2} \left (C \csc \left (c+d x+\frac {\pi }{2}\right )^2+B \csc \left (c+d x+\frac {\pi }{2}\right )+A\right )dx\)

\(\Big \downarrow \) 4576

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {1}{2} \sec ^{\frac {3}{2}}(c+d x) (\sec (c+d x) a+a)^{5/2} (a (10 A+3 C)+5 a (2 B+C) \sec (c+d x))dx}{5 a}+\frac {C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^{5/2}}{5 d}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \sec ^{\frac {3}{2}}(c+d x) (\sec (c+d x) a+a)^{5/2} (a (10 A+3 C)+5 a (2 B+C) \sec (c+d x))dx}{10 a}+\frac {C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^{5/2}}{5 d}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{5/2} \left (a (10 A+3 C)+5 a (2 B+C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx}{10 a}+\frac {C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^{5/2}}{5 d}\right )\)

\(\Big \downarrow \) 4506

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{4} \int \frac {1}{2} \sec ^{\frac {3}{2}}(c+d x) (\sec (c+d x) a+a)^{3/2} \left ((80 A+30 B+39 C) a^2+(80 A+110 B+79 C) \sec (c+d x) a^2\right )dx+\frac {5 a^2 (2 B+C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{4 d}}{10 a}+\frac {C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^{5/2}}{5 d}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{8} \int \sec ^{\frac {3}{2}}(c+d x) (\sec (c+d x) a+a)^{3/2} \left ((80 A+30 B+39 C) a^2+(80 A+110 B+79 C) \sec (c+d x) a^2\right )dx+\frac {5 a^2 (2 B+C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{4 d}}{10 a}+\frac {C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^{5/2}}{5 d}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{8} \int \csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2} \left ((80 A+30 B+39 C) a^2+(80 A+110 B+79 C) \csc \left (c+d x+\frac {\pi }{2}\right ) a^2\right )dx+\frac {5 a^2 (2 B+C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{4 d}}{10 a}+\frac {C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^{5/2}}{5 d}\right )\)

\(\Big \downarrow \) 4506

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{8} \left (\frac {1}{3} \int \frac {1}{2} \sec ^{\frac {3}{2}}(c+d x) \sqrt {\sec (c+d x) a+a} \left (3 (240 A+170 B+157 C) a^3+(1040 A+950 B+787 C) \sec (c+d x) a^3\right )dx+\frac {a^3 (80 A+110 B+79 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\right )+\frac {5 a^2 (2 B+C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{4 d}}{10 a}+\frac {C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^{5/2}}{5 d}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{8} \left (\frac {1}{6} \int \sec ^{\frac {3}{2}}(c+d x) \sqrt {\sec (c+d x) a+a} \left (3 (240 A+170 B+157 C) a^3+(1040 A+950 B+787 C) \sec (c+d x) a^3\right )dx+\frac {a^3 (80 A+110 B+79 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\right )+\frac {5 a^2 (2 B+C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{4 d}}{10 a}+\frac {C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^{5/2}}{5 d}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{8} \left (\frac {1}{6} \int \csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a} \left (3 (240 A+170 B+157 C) a^3+(1040 A+950 B+787 C) \csc \left (c+d x+\frac {\pi }{2}\right ) a^3\right )dx+\frac {a^3 (80 A+110 B+79 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\right )+\frac {5 a^2 (2 B+C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{4 d}}{10 a}+\frac {C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^{5/2}}{5 d}\right )\)

\(\Big \downarrow \) 4504

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{8} \left (\frac {1}{6} \left (\frac {15}{4} a^3 (400 A+326 B+283 C) \int \sec ^{\frac {3}{2}}(c+d x) \sqrt {\sec (c+d x) a+a}dx+\frac {a^4 (1040 A+950 B+787 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^3 (80 A+110 B+79 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\right )+\frac {5 a^2 (2 B+C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{4 d}}{10 a}+\frac {C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^{5/2}}{5 d}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{8} \left (\frac {1}{6} \left (\frac {15}{4} a^3 (400 A+326 B+283 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {a^4 (1040 A+950 B+787 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^3 (80 A+110 B+79 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\right )+\frac {5 a^2 (2 B+C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{4 d}}{10 a}+\frac {C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^{5/2}}{5 d}\right )\)

\(\Big \downarrow \) 4290

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{8} \left (\frac {1}{6} \left (\frac {15}{4} a^3 (400 A+326 B+283 C) \left (\frac {1}{2} \int \sqrt {\sec (c+d x)} \sqrt {\sec (c+d x) a+a}dx+\frac {a \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^4 (1040 A+950 B+787 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^3 (80 A+110 B+79 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\right )+\frac {5 a^2 (2 B+C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{4 d}}{10 a}+\frac {C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^{5/2}}{5 d}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{8} \left (\frac {1}{6} \left (\frac {15}{4} a^3 (400 A+326 B+283 C) \left (\frac {1}{2} \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {a \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^4 (1040 A+950 B+787 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^3 (80 A+110 B+79 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\right )+\frac {5 a^2 (2 B+C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{4 d}}{10 a}+\frac {C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^{5/2}}{5 d}\right )\)

\(\Big \downarrow \) 4288

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{8} \left (\frac {1}{6} \left (\frac {15}{4} a^3 (400 A+326 B+283 C) \left (\frac {a \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\int \frac {1}{\sqrt {\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+1}}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}\right )+\frac {a^4 (1040 A+950 B+787 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^3 (80 A+110 B+79 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\right )+\frac {5 a^2 (2 B+C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{4 d}}{10 a}+\frac {C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^{5/2}}{5 d}\right )\)

\(\Big \downarrow \) 222

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {5 a^2 (2 B+C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{4 d}+\frac {1}{8} \left (\frac {a^3 (80 A+110 B+79 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}+\frac {1}{6} \left (\frac {a^4 (1040 A+950 B+787 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}+\frac {15}{4} a^3 (400 A+326 B+283 C) \left (\frac {\sqrt {a} \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}+\frac {a \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}\right )\right )\right )}{10 a}+\frac {C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^{5/2}}{5 d}\right )\)

Input:

Int[((a + a*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/C 
os[c + d*x]^(3/2),x]
 

Output:

Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((C*Sec[c + d*x]^(5/2)*(a + a*Sec[c 
+ d*x])^(5/2)*Sin[c + d*x])/(5*d) + ((5*a^2*(2*B + C)*Sec[c + d*x]^(5/2)*( 
a + a*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(4*d) + ((a^3*(80*A + 110*B + 79*C 
)*Sec[c + d*x]^(5/2)*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(3*d) + ((a^4* 
(1040*A + 950*B + 787*C)*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(2*d*Sqrt[a + a* 
Sec[c + d*x]]) + (15*a^3*(400*A + 326*B + 283*C)*((Sqrt[a]*ArcSinh[(Sqrt[a 
]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/d + (a*Sec[c + d*x]^(3/2)*Sin[c 
 + d*x])/(d*Sqrt[a + a*Sec[c + d*x]])))/4)/6)/8)/(10*a))
 

Defintions of rubi rules used

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 222

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[Rt[b, 2]*(x/Sqrt 
[a])]/Rt[b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b]
 

rule 3042

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

rule 4288

Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) 
+ (a_)], x_Symbol] :> Simp[-2*(a/(b*f))*Sqrt[a*(d/b)]   Subst[Int[1/Sqrt[1 
+ x^2/a], x], x, b*(Cot[e + f*x]/Sqrt[a + b*Csc[e + f*x]])], x] /; FreeQ[{a 
, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && GtQ[a*(d/b), 0]
 

rule 4290

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) 
 + (a_)], x_Symbol] :> Simp[-2*b*d*Cot[e + f*x]*((d*Csc[e + f*x])^(n - 1)/( 
f*(2*n - 1)*Sqrt[a + b*Csc[e + f*x]])), x] + Simp[2*a*d*((n - 1)/(b*(2*n - 
1)))   Int[Sqrt[a + b*Csc[e + f*x]]*(d*Csc[e + f*x])^(n - 1), x], x] /; Fre 
eQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && GtQ[n, 1] && IntegerQ[2*n]
 

rule 4504

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) 
 + (a_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[-2*b*B*C 
ot[e + f*x]*((d*Csc[e + f*x])^n/(f*(2*n + 1)*Sqrt[a + b*Csc[e + f*x]])), x] 
 + Simp[(A*b*(2*n + 1) + 2*a*B*n)/(b*(2*n + 1))   Int[Sqrt[a + b*Csc[e + f* 
x]]*(d*Csc[e + f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, A, B, n}, x] && NeQ 
[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && NeQ[A*b*(2*n + 1) + 2*a*B*n, 0] && 
!LtQ[n, 0]
 

rule 4506

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( 
a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-b)*B* 
Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 1)*((d*Csc[e + f*x])^n/(f*(m + n))), 
 x] + Simp[1/(d*(m + n))   Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x] 
)^n*Simp[a*A*d*(m + n) + B*(b*d*n) + (A*b*d*(m + n) + a*B*d*(2*m + n - 1))* 
Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, n}, x] && NeQ[A*b - 
 a*B, 0] && EqQ[a^2 - b^2, 0] && GtQ[m, 1/2] &&  !LtQ[n, -1]
 

rule 4576

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. 
))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a 
_))^(m_), x_Symbol] :> Simp[(-C)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Cs 
c[e + f*x])^n/(f*(m + n + 1))), x] + Simp[1/(b*(m + n + 1))   Int[(a + b*Cs 
c[e + f*x])^m*(d*Csc[e + f*x])^n*Simp[A*b*(m + n + 1) + b*C*n + (a*C*m + b* 
B*(m + n + 1))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, m 
, n}, x] && EqQ[a^2 - b^2, 0] &&  !LtQ[m, -2^(-1)] &&  !LtQ[n, -2^(-1)] && 
NeQ[m + n + 1, 0]
 

rule 4753

Int[(cos[(a_.) + (b_.)*(x_)]*(c_.))^(m_.)*(u_), x_Symbol] :> Simp[(c*Cos[a 
+ b*x])^m*(c*Sec[a + b*x])^m   Int[ActivateTrig[u]/(c*Sec[a + b*x])^m, x], 
x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownSecantIntegrandQ[u, x 
]
 

Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(1017\) vs. \(2(259)=518\).

Time = 4.69 (sec) , antiderivative size = 1018, normalized size of antiderivative = 3.38

method	result	size
default	\(\text {Expression too large to display}\)	\(1018\)

Input:

int((a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c)^(3/2 
),x,method=_RETURNVERBOSE)
 

Output:

4*2^(1/2)*(-1/15360*C/d*(a/(2*cos(1/2*d*x+1/2*c)^2-1)*cos(1/2*d*x+1/2*c)^2 
)^(1/2)*a^2/(2*cos(1/2*d*x+1/2*c)^2-1)^(11/2)*((-271680*cos(1/2*d*x+1/2*c) 
^10+588640*cos(1/2*d*x+1/2*c)^8-534304*cos(1/2*d*x+1/2*c)^6+246960*cos(1/2 
*d*x+1/2*c)^4-57188*cos(1/2*d*x+1/2*c)^2+5342)*tan(1/2*d*x+1/2*c)+2^(1/2)* 
arctanh(1/2*2^(1/2)*(cot(1/2*d*x+1/2*c)-csc(1/2*d*x+1/2*c)-1))*(271680*cos 
(1/2*d*x+1/2*c)^11-815040*cos(1/2*d*x+1/2*c)^9+1018800*cos(1/2*d*x+1/2*c)^ 
7-679200*cos(1/2*d*x+1/2*c)^5+254700*cos(1/2*d*x+1/2*c)^3-50940*cos(1/2*d* 
x+1/2*c)+4245*sec(1/2*d*x+1/2*c))+2^(1/2)*arctanh(1/2*2^(1/2)*(cot(1/2*d*x 
+1/2*c)-csc(1/2*d*x+1/2*c)+1))*(271680*cos(1/2*d*x+1/2*c)^11-815040*cos(1/ 
2*d*x+1/2*c)^9+1018800*cos(1/2*d*x+1/2*c)^7-679200*cos(1/2*d*x+1/2*c)^5+25 
4700*cos(1/2*d*x+1/2*c)^3-50940*cos(1/2*d*x+1/2*c)+4245*sec(1/2*d*x+1/2*c) 
))-1/1536*B/d*(a/(2*cos(1/2*d*x+1/2*c)^2-1)*cos(1/2*d*x+1/2*c)^2)^(1/2)*a^ 
2/(2*cos(1/2*d*x+1/2*c)^2-1)^(9/2)*((-15648*cos(1/2*d*x+1/2*c)^8+26080*cos 
(1/2*d*x+1/2*c)^6-17120*cos(1/2*d*x+1/2*c)^4+5192*cos(1/2*d*x+1/2*c)^2-598 
)*tan(1/2*d*x+1/2*c)+2^(1/2)*arctanh(1/2*2^(1/2)*(cot(1/2*d*x+1/2*c)-csc(1 
/2*d*x+1/2*c)-1))*(15648*cos(1/2*d*x+1/2*c)^9-39120*cos(1/2*d*x+1/2*c)^7+3 
9120*cos(1/2*d*x+1/2*c)^5-19560*cos(1/2*d*x+1/2*c)^3+4890*cos(1/2*d*x+1/2* 
c)-489*sec(1/2*d*x+1/2*c))+2^(1/2)*arctanh(1/2*2^(1/2)*(cot(1/2*d*x+1/2*c) 
-csc(1/2*d*x+1/2*c)+1))*(15648*cos(1/2*d*x+1/2*c)^9-39120*cos(1/2*d*x+1/2* 
c)^7+39120*cos(1/2*d*x+1/2*c)^5-19560*cos(1/2*d*x+1/2*c)^3+4890*cos(1/2...
 

Fricas [A] (verification not implemented)

Time = 0.31 (sec) , antiderivative size = 579, normalized size of antiderivative = 1.92 \[ \int \frac {(a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx =\text {Too large to display} \] Input:

integrate((a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c 
)^(3/2),x, algorithm="fricas")
 

Output:

[1/7680*(4*(15*(400*A + 326*B + 283*C)*a^2*cos(d*x + c)^4 + 10*(272*A + 32 
6*B + 283*C)*a^2*cos(d*x + c)^3 + 8*(80*A + 230*B + 283*C)*a^2*cos(d*x + c 
)^2 + 48*(10*B + 29*C)*a^2*cos(d*x + c) + 384*C*a^2)*sqrt((a*cos(d*x + c) 
+ a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c) + 15*((400*A + 326*B + 
283*C)*a^2*cos(d*x + c)^6 + (400*A + 326*B + 283*C)*a^2*cos(d*x + c)^5)*sq 
rt(a)*log((a*cos(d*x + c)^3 - 4*sqrt(a)*sqrt((a*cos(d*x + c) + a)/cos(d*x 
+ c))*(cos(d*x + c) - 2)*sqrt(cos(d*x + c))*sin(d*x + c) - 7*a*cos(d*x + c 
)^2 + 8*a)/(cos(d*x + c)^3 + cos(d*x + c)^2)))/(d*cos(d*x + c)^6 + d*cos(d 
*x + c)^5), 1/3840*(2*(15*(400*A + 326*B + 283*C)*a^2*cos(d*x + c)^4 + 10* 
(272*A + 326*B + 283*C)*a^2*cos(d*x + c)^3 + 8*(80*A + 230*B + 283*C)*a^2* 
cos(d*x + c)^2 + 48*(10*B + 29*C)*a^2*cos(d*x + c) + 384*C*a^2)*sqrt((a*co 
s(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c) + 15*((400*A 
 + 326*B + 283*C)*a^2*cos(d*x + c)^6 + (400*A + 326*B + 283*C)*a^2*cos(d*x 
 + c)^5)*sqrt(-a)*arctan(2*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c) 
)*sqrt(cos(d*x + c))*sin(d*x + c)/(a*cos(d*x + c)^2 - a*cos(d*x + c) - 2*a 
)))/(d*cos(d*x + c)^6 + d*cos(d*x + c)^5)]
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\text {Timed out} \] Input:

integrate((a+a*sec(d*x+c))**(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)**2)/cos(d*x 
+c)**(3/2),x)
 

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 12710 vs. \(2 (259) = 518\).

Time = 1.29 (sec) , antiderivative size = 12710, normalized size of antiderivative = 42.23 \[ \int \frac {(a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\text {Too large to display} \] Input:

integrate((a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c 
)^(3/2),x, algorithm="maxima")
 

Output:

1/7680*(80*(300*sqrt(2)*a^2*cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2* 
d*x + 3/2*c)))*sin(6*d*x + 6*c) - 28*sqrt(2)*a^2*sin(9/2*d*x + 9/2*c) + 28 
*sqrt(2)*a^2*sin(3/2*d*x + 3/2*c) - 28*(sqrt(2)*a^2*sin(9/2*d*x + 9/2*c) - 
 sqrt(2)*a^2*sin(3/2*d*x + 3/2*c))*cos(6*d*x + 6*c) - 300*(sqrt(2)*a^2*sin 
(6*d*x + 6*c) + 3*sqrt(2)*a^2*sin(8/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/ 
2*d*x + 3/2*c))) + 3*sqrt(2)*a^2*sin(4/3*arctan2(sin(3/2*d*x + 3/2*c), cos 
(3/2*d*x + 3/2*c))))*cos(11/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 
3/2*c))) - 12*(7*sqrt(2)*a^2*sin(9/2*d*x + 9/2*c) - 7*sqrt(2)*a^2*sin(3/2* 
d*x + 3/2*c) - 114*sqrt(2)*a^2*sin(7/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3 
/2*d*x + 3/2*c))) + 114*sqrt(2)*a^2*sin(5/3*arctan2(sin(3/2*d*x + 3/2*c), 
cos(3/2*d*x + 3/2*c))) + 75*sqrt(2)*a^2*sin(1/3*arctan2(sin(3/2*d*x + 3/2* 
c), cos(3/2*d*x + 3/2*c))))*cos(8/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2* 
d*x + 3/2*c))) - 456*(sqrt(2)*a^2*sin(6*d*x + 6*c) + 3*sqrt(2)*a^2*sin(4/3 
*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))))*cos(7/3*arctan2(sin 
(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 456*(sqrt(2)*a^2*sin(6*d*x + 6 
*c) + 3*sqrt(2)*a^2*sin(4/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/ 
2*c))))*cos(5/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) - 12* 
(7*sqrt(2)*a^2*sin(9/2*d*x + 9/2*c) - 7*sqrt(2)*a^2*sin(3/2*d*x + 3/2*c) + 
 75*sqrt(2)*a^2*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c) 
)))*cos(4/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 75*(...
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1692 vs. \(2 (259) = 518\).

Time = 1.21 (sec) , antiderivative size = 1692, normalized size of antiderivative = 5.62 \[ \int \frac {(a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\text {Too large to display} \] Input:

integrate((a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c 
)^(3/2),x, algorithm="giac")
 

Output:

1/3840*(15*(400*A*a^(5/2)*sgn(cos(d*x + c)) + 326*B*a^(5/2)*sgn(cos(d*x + 
c)) + 283*C*a^(5/2)*sgn(cos(d*x + c)))*log(abs((sqrt(a)*tan(1/2*d*x + 1/2* 
c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^2 - a*(2*sqrt(2) + 3))) - 15*(400 
*A*a^(5/2)*sgn(cos(d*x + c)) + 326*B*a^(5/2)*sgn(cos(d*x + c)) + 283*C*a^( 
5/2)*sgn(cos(d*x + c)))*log(abs((sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan 
(1/2*d*x + 1/2*c)^2 + a))^2 + a*(2*sqrt(2) - 3))) + 4*(6000*sqrt(2)*(sqrt( 
a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^18*A*a^(7/2) 
*sgn(cos(d*x + c)) + 4890*sqrt(2)*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*t 
an(1/2*d*x + 1/2*c)^2 + a))^18*B*a^(7/2)*sgn(cos(d*x + c)) + 4245*sqrt(2)* 
(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^18*C*a 
^(7/2)*sgn(cos(d*x + c)) - 162000*sqrt(2)*(sqrt(a)*tan(1/2*d*x + 1/2*c) - 
sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^16*A*a^(9/2)*sgn(cos(d*x + c)) - 13203 
0*sqrt(2)*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + 
a))^16*B*a^(9/2)*sgn(cos(d*x + c)) - 114615*sqrt(2)*(sqrt(a)*tan(1/2*d*x + 
 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^16*C*a^(9/2)*sgn(cos(d*x + c 
)) + 1801920*sqrt(2)*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 
1/2*c)^2 + a))^14*A*a^(11/2)*sgn(cos(d*x + c)) + 1319880*sqrt(2)*(sqrt(a)* 
tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^14*B*a^(11/2)*s 
gn(cos(d*x + c)) + 1298820*sqrt(2)*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a* 
tan(1/2*d*x + 1/2*c)^2 + a))^14*C*a^(11/2)*sgn(cos(d*x + c)) - 9764160*...
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\int \frac {{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )}{{\cos \left (c+d\,x\right )}^{3/2}} \,d x \] Input:

int(((a + a/cos(c + d*x))^(5/2)*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/c 
os(c + d*x)^(3/2),x)
 

Output:

int(((a + a/cos(c + d*x))^(5/2)*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/c 
os(c + d*x)^(3/2), x)
 

Reduce [F]

\[ \int \frac {(a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\sqrt {a}\, a^{2} \left (\left (\int \frac {\sqrt {\sec \left (d x +c \right )+1}\, \sqrt {\cos \left (d x +c \right )}\, \sec \left (d x +c \right )^{4}}{\cos \left (d x +c \right )^{2}}d x \right ) c +\left (\int \frac {\sqrt {\sec \left (d x +c \right )+1}\, \sqrt {\cos \left (d x +c \right )}\, \sec \left (d x +c \right )^{3}}{\cos \left (d x +c \right )^{2}}d x \right ) b +2 \left (\int \frac {\sqrt {\sec \left (d x +c \right )+1}\, \sqrt {\cos \left (d x +c \right )}\, \sec \left (d x +c \right )^{3}}{\cos \left (d x +c \right )^{2}}d x \right ) c +\left (\int \frac {\sqrt {\sec \left (d x +c \right )+1}\, \sqrt {\cos \left (d x +c \right )}\, \sec \left (d x +c \right )^{2}}{\cos \left (d x +c \right )^{2}}d x \right ) a +2 \left (\int \frac {\sqrt {\sec \left (d x +c \right )+1}\, \sqrt {\cos \left (d x +c \right )}\, \sec \left (d x +c \right )^{2}}{\cos \left (d x +c \right )^{2}}d x \right ) b +\left (\int \frac {\sqrt {\sec \left (d x +c \right )+1}\, \sqrt {\cos \left (d x +c \right )}\, \sec \left (d x +c \right )^{2}}{\cos \left (d x +c \right )^{2}}d x \right ) c +2 \left (\int \frac {\sqrt {\sec \left (d x +c \right )+1}\, \sqrt {\cos \left (d x +c \right )}\, \sec \left (d x +c \right )}{\cos \left (d x +c \right )^{2}}d x \right ) a +\left (\int \frac {\sqrt {\sec \left (d x +c \right )+1}\, \sqrt {\cos \left (d x +c \right )}\, \sec \left (d x +c \right )}{\cos \left (d x +c \right )^{2}}d x \right ) b +\left (\int \frac {\sqrt {\sec \left (d x +c \right )+1}\, \sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{2}}d x \right ) a \right ) \] Input:

int((a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c)^(3/2 
),x)
 

Output:

sqrt(a)*a**2*(int((sqrt(sec(c + d*x) + 1)*sqrt(cos(c + d*x))*sec(c + d*x)* 
*4)/cos(c + d*x)**2,x)*c + int((sqrt(sec(c + d*x) + 1)*sqrt(cos(c + d*x))* 
sec(c + d*x)**3)/cos(c + d*x)**2,x)*b + 2*int((sqrt(sec(c + d*x) + 1)*sqrt 
(cos(c + d*x))*sec(c + d*x)**3)/cos(c + d*x)**2,x)*c + int((sqrt(sec(c + d 
*x) + 1)*sqrt(cos(c + d*x))*sec(c + d*x)**2)/cos(c + d*x)**2,x)*a + 2*int( 
(sqrt(sec(c + d*x) + 1)*sqrt(cos(c + d*x))*sec(c + d*x)**2)/cos(c + d*x)** 
2,x)*b + int((sqrt(sec(c + d*x) + 1)*sqrt(cos(c + d*x))*sec(c + d*x)**2)/c 
os(c + d*x)**2,x)*c + 2*int((sqrt(sec(c + d*x) + 1)*sqrt(cos(c + d*x))*sec 
(c + d*x))/cos(c + d*x)**2,x)*a + int((sqrt(sec(c + d*x) + 1)*sqrt(cos(c + 
 d*x))*sec(c + d*x))/cos(c + d*x)**2,x)*b + int((sqrt(sec(c + d*x) + 1)*sq 
rt(cos(c + d*x)))/cos(c + d*x)**2,x)*a)