\(\int \sec ^3(c+d x) (a+a \sec (c+d x))^{5/2} (A+C \sec ^2(c+d x)) \, dx\) [173]

Optimal result

Integrand size = 35, antiderivative size = 273 \[ \int \sec ^3(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {2 a^3 (10439 A+8368 C) \tan (c+d x)}{6435 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^3 (2717 A+2224 C) \sec ^3(c+d x) \tan (c+d x)}{9009 d \sqrt {a+a \sec (c+d x)}}-\frac {4 a^2 (10439 A+8368 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{45045 d}+\frac {2 a^2 (143 A+136 C) \sec ^3(c+d x) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{1287 d}+\frac {2 a (10439 A+8368 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{15015 d}+\frac {10 a C \sec ^3(c+d x) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{143 d}+\frac {2 C \sec ^3(c+d x) (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{13 d} \] Output:

2/6435*a^3*(10439*A+8368*C)*tan(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+2/9009*a^3 
*(2717*A+2224*C)*sec(d*x+c)^3*tan(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)-4/45045* 
a^2*(10439*A+8368*C)*(a+a*sec(d*x+c))^(1/2)*tan(d*x+c)/d+2/1287*a^2*(143*A 
+136*C)*sec(d*x+c)^3*(a+a*sec(d*x+c))^(1/2)*tan(d*x+c)/d+2/15015*a*(10439* 
A+8368*C)*(a+a*sec(d*x+c))^(3/2)*tan(d*x+c)/d+10/143*a*C*sec(d*x+c)^3*(a+a 
*sec(d*x+c))^(3/2)*tan(d*x+c)/d+2/13*C*sec(d*x+c)^3*(a+a*sec(d*x+c))^(5/2) 
*tan(d*x+c)/d
 

Mathematica [A] (verified)

Time = 1.70 (sec) , antiderivative size = 165, normalized size of antiderivative = 0.60 \[ \int \sec ^3(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {a^3 (322751 A+343612 C+1120 (286 A+347 C) \cos (c+d x)+14 (32747 A+30334 C) \cos (2 (c+d x))+141570 A \cos (3 (c+d x))+125520 C \cos (3 (c+d x))+156585 A \cos (4 (c+d x))+125520 C \cos (4 (c+d x))+20878 A \cos (5 (c+d x))+16736 C \cos (5 (c+d x))+20878 A \cos (6 (c+d x))+16736 C \cos (6 (c+d x))) \sec ^6(c+d x) \tan (c+d x)}{180180 d \sqrt {a (1+\sec (c+d x))}} \] Input:

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

Output:

(a^3*(322751*A + 343612*C + 1120*(286*A + 347*C)*Cos[c + d*x] + 14*(32747* 
A + 30334*C)*Cos[2*(c + d*x)] + 141570*A*Cos[3*(c + d*x)] + 125520*C*Cos[3 
*(c + d*x)] + 156585*A*Cos[4*(c + d*x)] + 125520*C*Cos[4*(c + d*x)] + 2087 
8*A*Cos[5*(c + d*x)] + 16736*C*Cos[5*(c + d*x)] + 20878*A*Cos[6*(c + d*x)] 
 + 16736*C*Cos[6*(c + d*x)])*Sec[c + d*x]^6*Tan[c + d*x])/(180180*d*Sqrt[a 
*(1 + Sec[c + d*x])])
 

Rubi [A] (verified)

Time = 1.92 (sec) , antiderivative size = 295, normalized size of antiderivative = 1.08, number of steps used = 18, number of rules used = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.514, Rules used = {3042, 4577, 27, 3042, 4506, 27, 3042, 4506, 27, 3042, 4504, 3042, 4287, 27, 3042, 4489, 3042, 4279}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4577

\(\displaystyle \frac {2 \int \frac {1}{2} \sec ^3(c+d x) (\sec (c+d x) a+a)^{5/2} (a (13 A+6 C)+5 a C \sec (c+d x))dx}{13 a}+\frac {2 C \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^{5/2}}{13 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \sec ^3(c+d x) (\sec (c+d x) a+a)^{5/2} (a (13 A+6 C)+5 a C \sec (c+d x))dx}{13 a}+\frac {2 C \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^{5/2}}{13 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \csc \left (c+d x+\frac {\pi }{2}\right )^3 \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{5/2} \left (a (13 A+6 C)+5 a C \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx}{13 a}+\frac {2 C \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^{5/2}}{13 d}\)

\(\Big \downarrow \) 4506

\(\displaystyle \frac {\frac {2}{11} \int \frac {1}{2} \sec ^3(c+d x) (\sec (c+d x) a+a)^{3/2} \left ((143 A+96 C) a^2+(143 A+136 C) \sec (c+d x) a^2\right )dx+\frac {10 a^2 C \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d}}{13 a}+\frac {2 C \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^{5/2}}{13 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {1}{11} \int \sec ^3(c+d x) (\sec (c+d x) a+a)^{3/2} \left ((143 A+96 C) a^2+(143 A+136 C) \sec (c+d x) a^2\right )dx+\frac {10 a^2 C \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d}}{13 a}+\frac {2 C \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^{5/2}}{13 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{11} \int \csc \left (c+d x+\frac {\pi }{2}\right )^3 \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2} \left ((143 A+96 C) a^2+(143 A+136 C) \csc \left (c+d x+\frac {\pi }{2}\right ) a^2\right )dx+\frac {10 a^2 C \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d}}{13 a}+\frac {2 C \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^{5/2}}{13 d}\)

\(\Big \downarrow \) 4506

\(\displaystyle \frac {\frac {1}{11} \left (\frac {2}{9} \int \frac {1}{2} \sec ^3(c+d x) \sqrt {\sec (c+d x) a+a} \left (15 (143 A+112 C) a^3+(2717 A+2224 C) \sec (c+d x) a^3\right )dx+\frac {2 a^3 (143 A+136 C) \tan (c+d x) \sec ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{9 d}\right )+\frac {10 a^2 C \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d}}{13 a}+\frac {2 C \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^{5/2}}{13 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {1}{11} \left (\frac {1}{9} \int \sec ^3(c+d x) \sqrt {\sec (c+d x) a+a} \left (15 (143 A+112 C) a^3+(2717 A+2224 C) \sec (c+d x) a^3\right )dx+\frac {2 a^3 (143 A+136 C) \tan (c+d x) \sec ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{9 d}\right )+\frac {10 a^2 C \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d}}{13 a}+\frac {2 C \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^{5/2}}{13 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{11} \left (\frac {1}{9} \int \csc \left (c+d x+\frac {\pi }{2}\right )^3 \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a} \left (15 (143 A+112 C) a^3+(2717 A+2224 C) \csc \left (c+d x+\frac {\pi }{2}\right ) a^3\right )dx+\frac {2 a^3 (143 A+136 C) \tan (c+d x) \sec ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{9 d}\right )+\frac {10 a^2 C \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d}}{13 a}+\frac {2 C \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^{5/2}}{13 d}\)

\(\Big \downarrow \) 4504

\(\displaystyle \frac {\frac {1}{11} \left (\frac {1}{9} \left (\frac {3}{7} a^3 (10439 A+8368 C) \int \sec ^3(c+d x) \sqrt {\sec (c+d x) a+a}dx+\frac {2 a^4 (2717 A+2224 C) \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a^3 (143 A+136 C) \tan (c+d x) \sec ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{9 d}\right )+\frac {10 a^2 C \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d}}{13 a}+\frac {2 C \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^{5/2}}{13 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{11} \left (\frac {1}{9} \left (\frac {3}{7} a^3 (10439 A+8368 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3 \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {2 a^4 (2717 A+2224 C) \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a^3 (143 A+136 C) \tan (c+d x) \sec ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{9 d}\right )+\frac {10 a^2 C \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d}}{13 a}+\frac {2 C \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^{5/2}}{13 d}\)

\(\Big \downarrow \) 4287

\(\displaystyle \frac {\frac {1}{11} \left (\frac {1}{9} \left (\frac {3}{7} a^3 (10439 A+8368 C) \left (\frac {2 \int \frac {1}{2} \sec (c+d x) (3 a-2 a \sec (c+d x)) \sqrt {\sec (c+d x) a+a}dx}{5 a}+\frac {2 \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 a d}\right )+\frac {2 a^4 (2717 A+2224 C) \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a^3 (143 A+136 C) \tan (c+d x) \sec ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{9 d}\right )+\frac {10 a^2 C \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d}}{13 a}+\frac {2 C \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^{5/2}}{13 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {1}{11} \left (\frac {1}{9} \left (\frac {3}{7} a^3 (10439 A+8368 C) \left (\frac {\int \sec (c+d x) (3 a-2 a \sec (c+d x)) \sqrt {\sec (c+d x) a+a}dx}{5 a}+\frac {2 \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 a d}\right )+\frac {2 a^4 (2717 A+2224 C) \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a^3 (143 A+136 C) \tan (c+d x) \sec ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{9 d}\right )+\frac {10 a^2 C \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d}}{13 a}+\frac {2 C \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^{5/2}}{13 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{11} \left (\frac {1}{9} \left (\frac {3}{7} a^3 (10439 A+8368 C) \left (\frac {\int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (3 a-2 a \csc \left (c+d x+\frac {\pi }{2}\right )\right ) \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{5 a}+\frac {2 \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 a d}\right )+\frac {2 a^4 (2717 A+2224 C) \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a^3 (143 A+136 C) \tan (c+d x) \sec ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{9 d}\right )+\frac {10 a^2 C \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d}}{13 a}+\frac {2 C \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^{5/2}}{13 d}\)

\(\Big \downarrow \) 4489

\(\displaystyle \frac {\frac {1}{11} \left (\frac {1}{9} \left (\frac {3}{7} a^3 (10439 A+8368 C) \left (\frac {\frac {7}{3} a \int \sec (c+d x) \sqrt {\sec (c+d x) a+a}dx-\frac {4 a \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{5 a}+\frac {2 \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 a d}\right )+\frac {2 a^4 (2717 A+2224 C) \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a^3 (143 A+136 C) \tan (c+d x) \sec ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{9 d}\right )+\frac {10 a^2 C \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d}}{13 a}+\frac {2 C \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^{5/2}}{13 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{11} \left (\frac {1}{9} \left (\frac {3}{7} a^3 (10439 A+8368 C) \left (\frac {\frac {7}{3} a \int \csc \left (c+d x+\frac {\pi }{2}\right ) \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx-\frac {4 a \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{5 a}+\frac {2 \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 a d}\right )+\frac {2 a^4 (2717 A+2224 C) \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a^3 (143 A+136 C) \tan (c+d x) \sec ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{9 d}\right )+\frac {10 a^2 C \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d}}{13 a}+\frac {2 C \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^{5/2}}{13 d}\)

\(\Big \downarrow \) 4279

\(\displaystyle \frac {\frac {10 a^2 C \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d}+\frac {1}{11} \left (\frac {2 a^3 (143 A+136 C) \tan (c+d x) \sec ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{9 d}+\frac {1}{9} \left (\frac {2 a^4 (2717 A+2224 C) \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}+\frac {3}{7} a^3 (10439 A+8368 C) \left (\frac {\frac {14 a^2 \tan (c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}-\frac {4 a \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{5 a}+\frac {2 \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 a d}\right )\right )\right )}{13 a}+\frac {2 C \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^{5/2}}{13 d}\)

Input:

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

Output:

(2*C*Sec[c + d*x]^3*(a + a*Sec[c + d*x])^(5/2)*Tan[c + d*x])/(13*d) + ((10 
*a^2*C*Sec[c + d*x]^3*(a + a*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(11*d) + (( 
2*a^3*(143*A + 136*C)*Sec[c + d*x]^3*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x] 
)/(9*d) + ((2*a^4*(2717*A + 2224*C)*Sec[c + d*x]^3*Tan[c + d*x])/(7*d*Sqrt 
[a + a*Sec[c + d*x]]) + (3*a^3*(10439*A + 8368*C)*((2*(a + a*Sec[c + d*x]) 
^(3/2)*Tan[c + d*x])/(5*a*d) + ((14*a^2*Tan[c + d*x])/(3*d*Sqrt[a + a*Sec[ 
c + d*x]]) - (4*a*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(3*d))/(5*a)))/7) 
/9)/11)/(13*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 3042

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

rule 4279

Int[csc[(e_.) + (f_.)*(x_)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_S 
ymbol] :> Simp[-2*b*(Cot[e + f*x]/(f*Sqrt[a + b*Csc[e + f*x]])), x] /; Free 
Q[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0]
 

rule 4287

Int[csc[(e_.) + (f_.)*(x_)]^3*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), 
x_Symbol] :> Simp[(-Cot[e + f*x])*((a + b*Csc[e + f*x])^(m + 1)/(b*f*(m + 2 
))), x] + Simp[1/(b*(m + 2))   Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*(b*( 
m + 1) - a*Csc[e + f*x]), x], x] /; FreeQ[{a, b, e, f, m}, x] && EqQ[a^2 - 
b^2, 0] &&  !LtQ[m, -2^(-1)]
 

rule 4489

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

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 4577

Int[((A_.) + 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*Csc[e + f*x])^n/(f*(m + n + 1))), 
x] + Simp[1/(b*(m + n + 1))   Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^n 
*Simp[A*b*(m + n + 1) + b*C*n + a*C*m*Csc[e + f*x], x], x], x] /; FreeQ[{a, 
 b, d, e, f, A, C, m, n}, x] && EqQ[a^2 - b^2, 0] &&  !LtQ[m, -2^(-1)] && 
!LtQ[n, -2^(-1)] && NeQ[m + n + 1, 0]
 

Maple [A] (verified)

Time = 0.60 (sec) , antiderivative size = 158, normalized size of antiderivative = 0.58

\[\frac {2 a^{2} \left (\cos \left (d x +c \right )^{2} \left (83512 \cos \left (d x +c \right )^{4}+41756 \cos \left (d x +c \right )^{3}+31317 \cos \left (d x +c \right )^{2}+18590 \cos \left (d x +c \right )+5005\right ) A +\left (66944 \cos \left (d x +c \right )^{6}+33472 \cos \left (d x +c \right )^{5}+25104 \cos \left (d x +c \right )^{4}+20920 \cos \left (d x +c \right )^{3}+18305 \cos \left (d x +c \right )^{2}+11970 \cos \left (d x +c \right )+3465\right ) C \right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right ) \sec \left (d x +c \right )^{5}}{45045 d \left (\cos \left (d x +c \right )+1\right )}\]

Input:

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

Output:

2/45045/d*a^2*(cos(d*x+c)^2*(83512*cos(d*x+c)^4+41756*cos(d*x+c)^3+31317*c 
os(d*x+c)^2+18590*cos(d*x+c)+5005)*A+(66944*cos(d*x+c)^6+33472*cos(d*x+c)^ 
5+25104*cos(d*x+c)^4+20920*cos(d*x+c)^3+18305*cos(d*x+c)^2+11970*cos(d*x+c 
)+3465)*C)*(a*(1+sec(d*x+c)))^(1/2)/(cos(d*x+c)+1)*tan(d*x+c)*sec(d*x+c)^5
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 171, normalized size of antiderivative = 0.63 \[ \int \sec ^3(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {2 \, {\left (8 \, {\left (10439 \, A + 8368 \, C\right )} a^{2} \cos \left (d x + c\right )^{6} + 4 \, {\left (10439 \, A + 8368 \, C\right )} a^{2} \cos \left (d x + c\right )^{5} + 3 \, {\left (10439 \, A + 8368 \, C\right )} a^{2} \cos \left (d x + c\right )^{4} + 10 \, {\left (1859 \, A + 2092 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + 35 \, {\left (143 \, A + 523 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 11970 \, C a^{2} \cos \left (d x + c\right ) + 3465 \, C a^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{45045 \, {\left (d \cos \left (d x + c\right )^{7} + d \cos \left (d x + c\right )^{6}\right )}} \] Input:

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

Output:

2/45045*(8*(10439*A + 8368*C)*a^2*cos(d*x + c)^6 + 4*(10439*A + 8368*C)*a^ 
2*cos(d*x + c)^5 + 3*(10439*A + 8368*C)*a^2*cos(d*x + c)^4 + 10*(1859*A + 
2092*C)*a^2*cos(d*x + c)^3 + 35*(143*A + 523*C)*a^2*cos(d*x + c)^2 + 11970 
*C*a^2*cos(d*x + c) + 3465*C*a^2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))* 
sin(d*x + c)/(d*cos(d*x + c)^7 + d*cos(d*x + c)^6)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F(-1)]

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

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

Output:

Timed out
 

Giac [A] (verification not implemented)

Time = 1.05 (sec) , antiderivative size = 351, normalized size of antiderivative = 1.29 \[ \int \sec ^3(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {8 \, {\left ({\left ({\left ({\left ({\left (4 \, {\left (2 \, \sqrt {2} {\left (1859 \, A a^{9} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 1483 \, C a^{9} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 13 \, \sqrt {2} {\left (1859 \, A a^{9} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 1483 \, C a^{9} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 143 \, \sqrt {2} {\left (1859 \, A a^{9} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 1483 \, C a^{9} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1716 \, \sqrt {2} {\left (228 \, A a^{9} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 181 \, C a^{9} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 6006 \, \sqrt {2} {\left (57 \, A a^{9} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 49 \, C a^{9} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 60060 \, \sqrt {2} {\left (3 \, A a^{9} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 2 \, C a^{9} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 45045 \, \sqrt {2} {\left (A a^{9} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + C a^{9} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{45045 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{6} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} d} \] Input:

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

Output:

8/45045*(((((4*(2*sqrt(2)*(1859*A*a^9*sgn(cos(d*x + c)) + 1483*C*a^9*sgn(c 
os(d*x + c)))*tan(1/2*d*x + 1/2*c)^2 - 13*sqrt(2)*(1859*A*a^9*sgn(cos(d*x 
+ c)) + 1483*C*a^9*sgn(cos(d*x + c))))*tan(1/2*d*x + 1/2*c)^2 + 143*sqrt(2 
)*(1859*A*a^9*sgn(cos(d*x + c)) + 1483*C*a^9*sgn(cos(d*x + c))))*tan(1/2*d 
*x + 1/2*c)^2 - 1716*sqrt(2)*(228*A*a^9*sgn(cos(d*x + c)) + 181*C*a^9*sgn( 
cos(d*x + c))))*tan(1/2*d*x + 1/2*c)^2 + 6006*sqrt(2)*(57*A*a^9*sgn(cos(d* 
x + c)) + 49*C*a^9*sgn(cos(d*x + c))))*tan(1/2*d*x + 1/2*c)^2 - 60060*sqrt 
(2)*(3*A*a^9*sgn(cos(d*x + c)) + 2*C*a^9*sgn(cos(d*x + c))))*tan(1/2*d*x + 
 1/2*c)^2 + 45045*sqrt(2)*(A*a^9*sgn(cos(d*x + c)) + C*a^9*sgn(cos(d*x + c 
))))*tan(1/2*d*x + 1/2*c)/((a*tan(1/2*d*x + 1/2*c)^2 - a)^6*sqrt(-a*tan(1/ 
2*d*x + 1/2*c)^2 + a)*d)
 

Mupad [B] (verification not implemented)

Time = 22.04 (sec) , antiderivative size = 1039, normalized size of antiderivative = 3.81 \[ \int \sec ^3(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\text {Too large to display} \] Input:

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

Output:

((a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*((A*a^2*8i) 
/(3*d) - (a^2*exp(c*1i + d*x*1i)*(10439*A + 8368*C)*8i)/(45045*d)))/((exp( 
c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i) + 1)) + ((a + a/(exp(- c*1i - d*x* 
1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*(exp(c*1i + d*x*1i)*((A*a^2*8i)/d + ( 
a^2*(286*A - 523*C)*32i)/(15015*d)) - (A*a^2*8i)/(5*d) + (a^2*(2*A + C)*32 
i)/(5*d)))/((exp(c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i) + 1)^2) + ((a + a 
/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*(exp(c*1i + d*x*1i 
)*((A*a^2*48i)/(13*d) - (a^2*(3*A + C)*32i)/(13*d) + (a^2*(13*A + 20*C)*16 
i)/(13*d) - (a^2*(A + C)*160i)/(13*d)) - (A*a^2*48i)/(13*d) + (a^2*(3*A + 
C)*32i)/(13*d) - (a^2*(13*A + 20*C)*16i)/(13*d) + (a^2*(A + C)*160i)/(13*d 
)))/((exp(c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i) + 1)^6) - ((a + a/(exp(- 
 c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*(exp(c*1i + d*x*1i)*((A*a 
^2*40i)/(11*d) - (a^2*(A - 16*C)*8i)/(11*d) + (C*a^2*128i)/(143*d) - (a^2* 
(3*A + 4*C)*40i)/(11*d) + (a^2*(11*A + 20*C)*8i)/(11*d)) - (A*a^2*8i)/(11* 
d) - (C*a^2*128i)/(11*d) + (a^2*(11*A + 4*C)*8i)/(11*d) - (a^2*(5*A + 12*C 
)*24i)/(11*d) + (a^2*(5*A - 16*C)*8i)/(11*d)))/((exp(c*1i + d*x*1i) + 1)*( 
exp(c*2i + d*x*2i) + 1)^5) + ((a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + 
d*x*1i)/2))^(1/2)*(exp(c*1i + d*x*1i)*((a^2*(A + 4*C)*40i)/(7*d) - (A*a^2* 
40i)/(7*d) + (a^2*(143*A + 811*C)*32i)/(9009*d)) + (A*a^2*8i)/(7*d) - (a^2 
*(A - 7*C)*32i)/(7*d) - (a^2*(9*A + 4*C)*8i)/(7*d)))/((exp(c*1i + d*x*1...
 

Reduce [F]

\[ \int \sec ^3(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\sqrt {a}\, a^{2} \left (\left (\int \sqrt {\sec \left (d x +c \right )+1}\, \sec \left (d x +c \right )^{7}d x \right ) c +2 \left (\int \sqrt {\sec \left (d x +c \right )+1}\, \sec \left (d x +c \right )^{6}d x \right ) c +\left (\int \sqrt {\sec \left (d x +c \right )+1}\, \sec \left (d x +c \right )^{5}d x \right ) a +\left (\int \sqrt {\sec \left (d x +c \right )+1}\, \sec \left (d x +c \right )^{5}d x \right ) c +2 \left (\int \sqrt {\sec \left (d x +c \right )+1}\, \sec \left (d x +c \right )^{4}d x \right ) a +\left (\int \sqrt {\sec \left (d x +c \right )+1}\, \sec \left (d x +c \right )^{3}d x \right ) a \right ) \] Input:

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

Output:

sqrt(a)*a**2*(int(sqrt(sec(c + d*x) + 1)*sec(c + d*x)**7,x)*c + 2*int(sqrt 
(sec(c + d*x) + 1)*sec(c + d*x)**6,x)*c + int(sqrt(sec(c + d*x) + 1)*sec(c 
 + d*x)**5,x)*a + int(sqrt(sec(c + d*x) + 1)*sec(c + d*x)**5,x)*c + 2*int( 
sqrt(sec(c + d*x) + 1)*sec(c + d*x)**4,x)*a + int(sqrt(sec(c + d*x) + 1)*s 
ec(c + d*x)**3,x)*a)