\(\int (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^{\frac {5}{2}}(c+d x) \, dx\) [1332]

Optimal result

Integrand size = 45, antiderivative size = 253 \[ \int (a+a \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x) \, dx=\frac {a^{5/2} (8 A+20 B+19 C) \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{4 d}-\frac {a^3 (56 A+12 B-27 C) \sin (c+d x)}{12 d \sqrt {a+a \cos (c+d x)} \sqrt {\sec (c+d x)}}-\frac {a^2 (8 A+4 B-C) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{2 d \sqrt {\sec (c+d x)}}+\frac {2 a (5 A+3 B) (a+a \cos (c+d x))^{3/2} \sqrt {\sec (c+d x)} \sin (c+d x)}{3 d}+\frac {2 A (a+a \cos (c+d x))^{5/2} \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 d} \] Output:

1/4*a^(5/2)*(8*A+20*B+19*C)*arcsin(a^(1/2)*sin(d*x+c)/(a+a*cos(d*x+c))^(1/ 
2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/d-1/12*a^3*(56*A+12*B-27*C)*sin(d*x+ 
c)/d/(a+a*cos(d*x+c))^(1/2)/sec(d*x+c)^(1/2)-1/2*a^2*(8*A+4*B-C)*(a+a*cos( 
d*x+c))^(1/2)*sin(d*x+c)/d/sec(d*x+c)^(1/2)+2/3*a*(5*A+3*B)*(a+a*cos(d*x+c 
))^(3/2)*sec(d*x+c)^(1/2)*sin(d*x+c)/d+2/3*A*(a+a*cos(d*x+c))^(5/2)*sec(d* 
x+c)^(3/2)*sin(d*x+c)/d
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 1.23 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.62 \[ \int (a+a \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x) \, dx=\frac {a^2 \sqrt {a (1+\cos (c+d x))} \sec \left (\frac {1}{2} (c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x) \left (6 \sqrt {2} (8 A+20 B+19 C) \arcsin \left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right ) \cos ^{\frac {3}{2}}(c+d x)+2 (16 A+12 B+33 C+(128 A+48 B+9 C) \cos (c+d x)+3 (4 B+11 C) \cos (2 (c+d x))+3 C \cos (3 (c+d x))) \sin \left (\frac {1}{2} (c+d x)\right )\right )}{48 d} \] Input:

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

Output:

(a^2*Sqrt[a*(1 + Cos[c + d*x])]*Sec[(c + d*x)/2]*Sec[c + d*x]^(3/2)*(6*Sqr 
t[2]*(8*A + 20*B + 19*C)*ArcSin[Sqrt[2]*Sin[(c + d*x)/2]]*Cos[c + d*x]^(3/ 
2) + 2*(16*A + 12*B + 33*C + (128*A + 48*B + 9*C)*Cos[c + d*x] + 3*(4*B + 
11*C)*Cos[2*(c + d*x)] + 3*C*Cos[3*(c + d*x)])*Sin[(c + d*x)/2]))/(48*d)
 

Rubi [A] (verified)

Time = 1.69 (sec) , antiderivative size = 263, normalized size of antiderivative = 1.04, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.356, Rules used = {3042, 4709, 3042, 3522, 27, 3042, 3454, 27, 3042, 3455, 27, 3042, 3460, 3042, 3253, 223}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4709

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3522

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 \int \frac {(\cos (c+d x) a+a)^{5/2} (a (5 A+3 B)-a (4 A-3 C) \cos (c+d x))}{2 \cos ^{\frac {3}{2}}(c+d x)}dx}{3 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {(\cos (c+d x) a+a)^{5/2} (a (5 A+3 B)-a (4 A-3 C) \cos (c+d x))}{\cos ^{\frac {3}{2}}(c+d x)}dx}{3 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{5/2} \left (a (5 A+3 B)-a (4 A-3 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx}{3 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 3454

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 \int \frac {(\cos (c+d x) a+a)^{3/2} \left (a^2 (16 A+12 B+3 C)-3 a^2 (8 A+4 B-C) \cos (c+d x)\right )}{2 \sqrt {\cos (c+d x)}}dx+\frac {2 a^2 (5 A+3 B) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{d \sqrt {\cos (c+d x)}}}{3 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {(\cos (c+d x) a+a)^{3/2} \left (a^2 (16 A+12 B+3 C)-3 a^2 (8 A+4 B-C) \cos (c+d x)\right )}{\sqrt {\cos (c+d x)}}dx+\frac {2 a^2 (5 A+3 B) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{d \sqrt {\cos (c+d x)}}}{3 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2} \left (a^2 (16 A+12 B+3 C)-3 a^2 (8 A+4 B-C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 a^2 (5 A+3 B) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{d \sqrt {\cos (c+d x)}}}{3 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 3455

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{2} \int \frac {\sqrt {\cos (c+d x) a+a} \left (a^3 (40 A+36 B+15 C)-a^3 (56 A+12 B-27 C) \cos (c+d x)\right )}{2 \sqrt {\cos (c+d x)}}dx-\frac {3 a^3 (8 A+4 B-C) \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}{2 d}+\frac {2 a^2 (5 A+3 B) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{d \sqrt {\cos (c+d x)}}}{3 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{4} \int \frac {\sqrt {\cos (c+d x) a+a} \left (a^3 (40 A+36 B+15 C)-a^3 (56 A+12 B-27 C) \cos (c+d x)\right )}{\sqrt {\cos (c+d x)}}dx-\frac {3 a^3 (8 A+4 B-C) \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}{2 d}+\frac {2 a^2 (5 A+3 B) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{d \sqrt {\cos (c+d x)}}}{3 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{4} \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a} \left (a^3 (40 A+36 B+15 C)-a^3 (56 A+12 B-27 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {3 a^3 (8 A+4 B-C) \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}{2 d}+\frac {2 a^2 (5 A+3 B) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{d \sqrt {\cos (c+d x)}}}{3 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 3460

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{4} \left (\frac {3}{2} a^3 (8 A+20 B+19 C) \int \frac {\sqrt {\cos (c+d x) a+a}}{\sqrt {\cos (c+d x)}}dx-\frac {a^4 (56 A+12 B-27 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}\right )-\frac {3 a^3 (8 A+4 B-C) \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}{2 d}+\frac {2 a^2 (5 A+3 B) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{d \sqrt {\cos (c+d x)}}}{3 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{4} \left (\frac {3}{2} a^3 (8 A+20 B+19 C) \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {a^4 (56 A+12 B-27 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}\right )-\frac {3 a^3 (8 A+4 B-C) \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}{2 d}+\frac {2 a^2 (5 A+3 B) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{d \sqrt {\cos (c+d x)}}}{3 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 3253

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{4} \left (-\frac {3 a^3 (8 A+20 B+19 C) \int \frac {1}{\sqrt {1-\frac {a \sin ^2(c+d x)}{\cos (c+d x) a+a}}}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x) a+a}}\right )}{d}-\frac {a^4 (56 A+12 B-27 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}\right )-\frac {3 a^3 (8 A+4 B-C) \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}{2 d}+\frac {2 a^2 (5 A+3 B) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{d \sqrt {\cos (c+d x)}}}{3 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 223

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {-\frac {3 a^3 (8 A+4 B-C) \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}{2 d}+\frac {2 a^2 (5 A+3 B) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{d \sqrt {\cos (c+d x)}}+\frac {1}{4} \left (\frac {3 a^{7/2} (8 A+20 B+19 C) \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{d}-\frac {a^4 (56 A+12 B-27 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}\right )}{3 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )\)

Input:

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

Output:

Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((2*A*(a + a*Cos[c + d*x])^(5/2)*Sin 
[c + d*x])/(3*d*Cos[c + d*x]^(3/2)) + ((-3*a^3*(8*A + 4*B - C)*Sqrt[Cos[c 
+ d*x]]*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(2*d) + (2*a^2*(5*A + 3*B)* 
(a + a*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]]) + ((3*a^(7 
/2)*(8*A + 20*B + 19*C)*ArcSin[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d 
*x]]])/d - (a^4*(56*A + 12*B - 27*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(d*S 
qrt[a + a*Cos[c + d*x]]))/4)/(3*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 223

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

rule 3042

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

rule 3253

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(d_.)*sin[(e_.) + (f_.) 
*(x_)]], x_Symbol] :> Simp[-2/f   Subst[Int[1/Sqrt[1 - x^2/a], x], x, b*(Co 
s[e + f*x]/Sqrt[a + b*Sin[e + f*x]])], x] /; FreeQ[{a, b, d, e, f}, x] && E 
qQ[a^2 - b^2, 0] && EqQ[d, a/b]
 

rule 3454

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + 
(f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim 
p[(-b^2)*(B*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((c + d*Sin[ 
e + f*x])^(n + 1)/(d*f*(n + 1)*(b*c + a*d))), x] - Simp[b/(d*(n + 1)*(b*c + 
 a*d))   Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp 
[a*A*d*(m - n - 2) - B*(a*c*(m - 1) + b*d*(n + 1)) - (A*b*d*(m + n + 1) - B 
*(b*c*m - a*d*(n + 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f 
, A, B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] 
&& GtQ[m, 1/2] && LtQ[n, -1] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0 
])
 

rule 3455

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

rule 3460

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + ( 
f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp 
[-2*b*B*Cos[e + f*x]*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(2*n + 3)*Sqrt[a + 
b*Sin[e + f*x]])), x] + Simp[(A*b*d*(2*n + 3) - B*(b*c - 2*a*d*(n + 1)))/(b 
*d*(2*n + 3))   Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^n, x], x] 
 /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - 
 b^2, 0] && NeQ[c^2 - d^2, 0] &&  !LtQ[n, -1]
 

rule 3522

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

rule 4709

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

Maple [A] (verified)

Time = 0.83 (sec) , antiderivative size = 306, normalized size of antiderivative = 1.21

\[\frac {a^{2} \sqrt {\left (1+\cos \left (d x +c \right )\right ) a}\, \sec \left (d x +c \right )^{\frac {5}{2}} \left (24 A \cos \left (d x +c \right )^{3} \arctan \left (\sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \tan \left (d x +c \right )\right )+60 B \cos \left (d x +c \right )^{3} \arctan \left (\sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \tan \left (d x +c \right )\right )+57 C \cos \left (d x +c \right )^{3} \arctan \left (\sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \tan \left (d x +c \right )\right )+\left (64 \cos \left (d x +c \right )+8\right ) \sin \left (d x +c \right ) A \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \cos \left (d x +c \right )+\sin \left (d x +c \right ) \cos \left (d x +c \right )^{2} \left (12 \cos \left (d x +c \right )+24\right ) B \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+\sin \left (d x +c \right ) \cos \left (d x +c \right )^{3} \left (6 \cos \left (d x +c \right )+33\right ) C \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right )}{12 d \left (1+\cos \left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\]

Input:

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

Output:

1/12/d*a^2*((1+cos(d*x+c))*a)^(1/2)*sec(d*x+c)^(5/2)/(1+cos(d*x+c))/(cos(d 
*x+c)/(1+cos(d*x+c)))^(1/2)*(24*A*cos(d*x+c)^3*arctan((cos(d*x+c)/(1+cos(d 
*x+c)))^(1/2)*tan(d*x+c))+60*B*cos(d*x+c)^3*arctan((cos(d*x+c)/(1+cos(d*x+ 
c)))^(1/2)*tan(d*x+c))+57*C*cos(d*x+c)^3*arctan((cos(d*x+c)/(1+cos(d*x+c)) 
)^(1/2)*tan(d*x+c))+(64*cos(d*x+c)+8)*sin(d*x+c)*A*(cos(d*x+c)/(1+cos(d*x+ 
c)))^(1/2)*cos(d*x+c)+sin(d*x+c)*cos(d*x+c)^2*(12*cos(d*x+c)+24)*B*(cos(d* 
x+c)/(1+cos(d*x+c)))^(1/2)+sin(d*x+c)*cos(d*x+c)^3*(6*cos(d*x+c)+33)*C*(co 
s(d*x+c)/(1+cos(d*x+c)))^(1/2))
 

Fricas [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 192, normalized size of antiderivative = 0.76 \[ \int (a+a \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x) \, dx=-\frac {3 \, {\left ({\left (8 \, A + 20 \, B + 19 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + {\left (8 \, A + 20 \, B + 19 \, C\right )} a^{2} \cos \left (d x + c\right )\right )} \sqrt {a} \arctan \left (\frac {\sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right ) - \frac {{\left (6 \, C a^{2} \cos \left (d x + c\right )^{3} + 3 \, {\left (4 \, B + 11 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 8 \, {\left (8 \, A + 3 \, B\right )} a^{2} \cos \left (d x + c\right ) + 8 \, A a^{2}\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{12 \, {\left (d \cos \left (d x + c\right )^{2} + d \cos \left (d x + c\right )\right )}} \] Input:

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

Output:

-1/12*(3*((8*A + 20*B + 19*C)*a^2*cos(d*x + c)^2 + (8*A + 20*B + 19*C)*a^2 
*cos(d*x + c))*sqrt(a)*arctan(sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))/ 
(sqrt(a)*sin(d*x + c))) - (6*C*a^2*cos(d*x + c)^3 + 3*(4*B + 11*C)*a^2*cos 
(d*x + c)^2 + 8*(8*A + 3*B)*a^2*cos(d*x + c) + 8*A*a^2)*sqrt(a*cos(d*x + c 
) + a)*sin(d*x + c)/sqrt(cos(d*x + c)))/(d*cos(d*x + c)^2 + d*cos(d*x + c) 
)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F(-1)]

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

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

Output:

Timed out
 

Giac [F(-1)]

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

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

Output:

Timed out
 

Mupad [F(-1)]

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

int((1/cos(c + d*x))^(5/2)*(a + a*cos(c + d*x))^(5/2)*(A + B*cos(c + d*x) 
+ C*cos(c + d*x)^2),x)
                                                                                    
                                                                                    
 

Output:

int((1/cos(c + d*x))^(5/2)*(a + a*cos(c + d*x))^(5/2)*(A + B*cos(c + d*x) 
+ C*cos(c + d*x)^2), x)
 

Reduce [F]

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

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

Output:

sqrt(a)*a**2*(2*int(sqrt(sec(c + d*x))*sqrt(cos(c + d*x) + 1)*cos(c + d*x) 
*sec(c + d*x)**2,x)*a + int(sqrt(sec(c + d*x))*sqrt(cos(c + d*x) + 1)*cos( 
c + d*x)*sec(c + d*x)**2,x)*b + int(sqrt(sec(c + d*x))*sqrt(cos(c + d*x) + 
 1)*cos(c + d*x)**4*sec(c + d*x)**2,x)*c + int(sqrt(sec(c + d*x))*sqrt(cos 
(c + d*x) + 1)*cos(c + d*x)**3*sec(c + d*x)**2,x)*b + 2*int(sqrt(sec(c + d 
*x))*sqrt(cos(c + d*x) + 1)*cos(c + d*x)**3*sec(c + d*x)**2,x)*c + int(sqr 
t(sec(c + d*x))*sqrt(cos(c + d*x) + 1)*cos(c + d*x)**2*sec(c + d*x)**2,x)* 
a + 2*int(sqrt(sec(c + d*x))*sqrt(cos(c + d*x) + 1)*cos(c + d*x)**2*sec(c 
+ d*x)**2,x)*b + int(sqrt(sec(c + d*x))*sqrt(cos(c + d*x) + 1)*cos(c + d*x 
)**2*sec(c + d*x)**2,x)*c + int(sqrt(sec(c + d*x))*sqrt(cos(c + d*x) + 1)* 
sec(c + d*x)**2,x)*a)