\(\int (a+a \cos (c+d x))^{3/2} (A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^6(c+d x) \, dx\) [390]

Optimal result

Integrand size = 43, antiderivative size = 263 \[ \int (a+a \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\frac {a^{3/2} (133 A+150 B+176 C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{128 d}+\frac {a^2 (133 A+150 B+176 C) \tan (c+d x)}{128 d \sqrt {a+a \cos (c+d x)}}+\frac {a^2 (133 A+150 B+176 C) \sec (c+d x) \tan (c+d x)}{192 d \sqrt {a+a \cos (c+d x)}}+\frac {a^2 (67 A+90 B+80 C) \sec ^2(c+d x) \tan (c+d x)}{240 d \sqrt {a+a \cos (c+d x)}}+\frac {a (3 A+10 B) \sqrt {a+a \cos (c+d x)} \sec ^3(c+d x) \tan (c+d x)}{40 d}+\frac {A (a+a \cos (c+d x))^{3/2} \sec ^4(c+d x) \tan (c+d x)}{5 d} \] Output:

1/128*a^(3/2)*(133*A+150*B+176*C)*arctanh(a^(1/2)*sin(d*x+c)/(a+a*cos(d*x+ 
c))^(1/2))/d+1/128*a^2*(133*A+150*B+176*C)*tan(d*x+c)/d/(a+a*cos(d*x+c))^( 
1/2)+1/192*a^2*(133*A+150*B+176*C)*sec(d*x+c)*tan(d*x+c)/d/(a+a*cos(d*x+c) 
)^(1/2)+1/240*a^2*(67*A+90*B+80*C)*sec(d*x+c)^2*tan(d*x+c)/d/(a+a*cos(d*x+ 
c))^(1/2)+1/40*a*(3*A+10*B)*(a+a*cos(d*x+c))^(1/2)*sec(d*x+c)^3*tan(d*x+c) 
/d+1/5*A*(a+a*cos(d*x+c))^(3/2)*sec(d*x+c)^4*tan(d*x+c)/d
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 3.57 (sec) , antiderivative size = 208, normalized size of antiderivative = 0.79 \[ \int (a+a \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\frac {a \sqrt {a (1+\cos (c+d x))} \sec \left (\frac {1}{2} (c+d x)\right ) \sec ^5(c+d x) \left (60 \sqrt {2} (133 A+150 B+176 C) \text {arctanh}\left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right ) \cos ^5(c+d x)+(13313 A+11550 B+10480 C+12 (1273 A+1070 B+880 C) \cos (c+d x)+4 (3059 A+3450 B+3280 C) \cos (2 (c+d x))+2660 A \cos (3 (c+d x))+3000 B \cos (3 (c+d x))+3520 C \cos (3 (c+d x))+1995 A \cos (4 (c+d x))+2250 B \cos (4 (c+d x))+2640 C \cos (4 (c+d x))) \sin \left (\frac {1}{2} (c+d x)\right )\right )}{15360 d} \] Input:

Integrate[(a + a*Cos[c + d*x])^(3/2)*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^ 
2)*Sec[c + d*x]^6,x]
 

Output:

(a*Sqrt[a*(1 + Cos[c + d*x])]*Sec[(c + d*x)/2]*Sec[c + d*x]^5*(60*Sqrt[2]* 
(133*A + 150*B + 176*C)*ArcTanh[Sqrt[2]*Sin[(c + d*x)/2]]*Cos[c + d*x]^5 + 
 (13313*A + 11550*B + 10480*C + 12*(1273*A + 1070*B + 880*C)*Cos[c + d*x] 
+ 4*(3059*A + 3450*B + 3280*C)*Cos[2*(c + d*x)] + 2660*A*Cos[3*(c + d*x)] 
+ 3000*B*Cos[3*(c + d*x)] + 3520*C*Cos[3*(c + d*x)] + 1995*A*Cos[4*(c + d* 
x)] + 2250*B*Cos[4*(c + d*x)] + 2640*C*Cos[4*(c + d*x)])*Sin[(c + d*x)/2]) 
)/(15360*d)
 

Rubi [A] (verified)

Time = 1.58 (sec) , antiderivative size = 261, normalized size of antiderivative = 0.99, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.349, Rules used = {3042, 3522, 27, 3042, 3454, 27, 3042, 3459, 3042, 3251, 3042, 3251, 3042, 3252, 219}

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.

\(\displaystyle \int \sec ^6(c+d x) (a \cos (c+d x)+a)^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^{3/2} \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^6}dx\)

\(\Big \downarrow \) 3522

\(\displaystyle \frac {\int \frac {1}{2} (\cos (c+d x) a+a)^{3/2} (a (3 A+10 B)+5 a (A+2 C) \cos (c+d x)) \sec ^5(c+d x)dx}{5 a}+\frac {A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int (\cos (c+d x) a+a)^{3/2} (a (3 A+10 B)+5 a (A+2 C) \cos (c+d x)) \sec ^5(c+d x)dx}{10 a}+\frac {A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2} \left (a (3 A+10 B)+5 a (A+2 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^5}dx}{10 a}+\frac {A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d}\)

\(\Big \downarrow \) 3454

\(\displaystyle \frac {\frac {1}{4} \int \frac {1}{2} \sqrt {\cos (c+d x) a+a} \left ((67 A+90 B+80 C) a^2+5 (11 A+10 B+16 C) \cos (c+d x) a^2\right ) \sec ^4(c+d x)dx+\frac {a^2 (3 A+10 B) \tan (c+d x) \sec ^3(c+d x) \sqrt {a \cos (c+d x)+a}}{4 d}}{10 a}+\frac {A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {1}{8} \int \sqrt {\cos (c+d x) a+a} \left ((67 A+90 B+80 C) a^2+5 (11 A+10 B+16 C) \cos (c+d x) a^2\right ) \sec ^4(c+d x)dx+\frac {a^2 (3 A+10 B) \tan (c+d x) \sec ^3(c+d x) \sqrt {a \cos (c+d x)+a}}{4 d}}{10 a}+\frac {A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{8} \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a} \left ((67 A+90 B+80 C) a^2+5 (11 A+10 B+16 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^4}dx+\frac {a^2 (3 A+10 B) \tan (c+d x) \sec ^3(c+d x) \sqrt {a \cos (c+d x)+a}}{4 d}}{10 a}+\frac {A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d}\)

\(\Big \downarrow \) 3459

\(\displaystyle \frac {\frac {1}{8} \left (\frac {5}{6} a^2 (133 A+150 B+176 C) \int \sqrt {\cos (c+d x) a+a} \sec ^3(c+d x)dx+\frac {a^3 (67 A+90 B+80 C) \tan (c+d x) \sec ^2(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^2 (3 A+10 B) \tan (c+d x) \sec ^3(c+d x) \sqrt {a \cos (c+d x)+a}}{4 d}}{10 a}+\frac {A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{8} \left (\frac {5}{6} a^2 (133 A+150 B+176 C) \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )^3}dx+\frac {a^3 (67 A+90 B+80 C) \tan (c+d x) \sec ^2(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^2 (3 A+10 B) \tan (c+d x) \sec ^3(c+d x) \sqrt {a \cos (c+d x)+a}}{4 d}}{10 a}+\frac {A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d}\)

\(\Big \downarrow \) 3251

\(\displaystyle \frac {\frac {1}{8} \left (\frac {5}{6} a^2 (133 A+150 B+176 C) \left (\frac {3}{4} \int \sqrt {\cos (c+d x) a+a} \sec ^2(c+d x)dx+\frac {a \tan (c+d x) \sec (c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^3 (67 A+90 B+80 C) \tan (c+d x) \sec ^2(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^2 (3 A+10 B) \tan (c+d x) \sec ^3(c+d x) \sqrt {a \cos (c+d x)+a}}{4 d}}{10 a}+\frac {A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{8} \left (\frac {5}{6} a^2 (133 A+150 B+176 C) \left (\frac {3}{4} \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )^2}dx+\frac {a \tan (c+d x) \sec (c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^3 (67 A+90 B+80 C) \tan (c+d x) \sec ^2(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^2 (3 A+10 B) \tan (c+d x) \sec ^3(c+d x) \sqrt {a \cos (c+d x)+a}}{4 d}}{10 a}+\frac {A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d}\)

\(\Big \downarrow \) 3251

\(\displaystyle \frac {\frac {1}{8} \left (\frac {5}{6} a^2 (133 A+150 B+176 C) \left (\frac {3}{4} \left (\frac {1}{2} \int \sqrt {\cos (c+d x) a+a} \sec (c+d x)dx+\frac {a \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a \tan (c+d x) \sec (c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^3 (67 A+90 B+80 C) \tan (c+d x) \sec ^2(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^2 (3 A+10 B) \tan (c+d x) \sec ^3(c+d x) \sqrt {a \cos (c+d x)+a}}{4 d}}{10 a}+\frac {A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{8} \left (\frac {5}{6} a^2 (133 A+150 B+176 C) \left (\frac {3}{4} \left (\frac {1}{2} \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {a \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a \tan (c+d x) \sec (c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^3 (67 A+90 B+80 C) \tan (c+d x) \sec ^2(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^2 (3 A+10 B) \tan (c+d x) \sec ^3(c+d x) \sqrt {a \cos (c+d x)+a}}{4 d}}{10 a}+\frac {A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d}\)

\(\Big \downarrow \) 3252

\(\displaystyle \frac {\frac {1}{8} \left (\frac {5}{6} a^2 (133 A+150 B+176 C) \left (\frac {3}{4} \left (\frac {a \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {a \int \frac {1}{a-\frac {a^2 \sin ^2(c+d x)}{\cos (c+d x) a+a}}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x) a+a}}\right )}{d}\right )+\frac {a \tan (c+d x) \sec (c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^3 (67 A+90 B+80 C) \tan (c+d x) \sec ^2(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^2 (3 A+10 B) \tan (c+d x) \sec ^3(c+d x) \sqrt {a \cos (c+d x)+a}}{4 d}}{10 a}+\frac {A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {a^2 (3 A+10 B) \tan (c+d x) \sec ^3(c+d x) \sqrt {a \cos (c+d x)+a}}{4 d}+\frac {1}{8} \left (\frac {a^3 (67 A+90 B+80 C) \tan (c+d x) \sec ^2(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}+\frac {5}{6} a^2 (133 A+150 B+176 C) \left (\frac {3}{4} \left (\frac {\sqrt {a} \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{d}+\frac {a \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a \tan (c+d x) \sec (c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\right )\right )}{10 a}+\frac {A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d}\)

Input:

Int[(a + a*Cos[c + d*x])^(3/2)*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec 
[c + d*x]^6,x]
 

Output:

(A*(a + a*Cos[c + d*x])^(3/2)*Sec[c + d*x]^4*Tan[c + d*x])/(5*d) + ((a^2*( 
3*A + 10*B)*Sqrt[a + a*Cos[c + d*x]]*Sec[c + d*x]^3*Tan[c + d*x])/(4*d) + 
((a^3*(67*A + 90*B + 80*C)*Sec[c + d*x]^2*Tan[c + d*x])/(3*d*Sqrt[a + a*Co 
s[c + d*x]]) + (5*a^2*(133*A + 150*B + 176*C)*((a*Sec[c + d*x]*Tan[c + d*x 
])/(2*d*Sqrt[a + a*Cos[c + d*x]]) + (3*((Sqrt[a]*ArcTanh[(Sqrt[a]*Sin[c + 
d*x])/Sqrt[a + a*Cos[c + d*x]]])/d + (a*Tan[c + d*x])/(d*Sqrt[a + a*Cos[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 219

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

rule 3042

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

rule 3251

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

rule 3252

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

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 3459

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

Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(2682\) vs. \(2(235)=470\).

Time = 0.95 (sec) , antiderivative size = 2683, normalized size of antiderivative = 10.20

method	result	size
parts	\(\text {Expression too large to display}\)	\(2683\)
default	\(\text {Expression too large to display}\)	\(2879\)

Input:

int((a+a*cos(d*x+c))^(3/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^6,x, 
method=_RETURNVERBOSE)
 

Output:

1/120*A*a^(1/2)*cos(1/2*d*x+1/2*c)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)*(-63840* 
a*(ln(-4/(2*cos(1/2*d*x+1/2*c)-2^(1/2))*(a*2^(1/2)*cos(1/2*d*x+1/2*c)-a^(1 
/2)*2^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)-2*a))+ln(4/(2*cos(1/2*d*x+1/2*c 
)+2^(1/2))*(a*2^(1/2)*cos(1/2*d*x+1/2*c)+a^(1/2)*2^(1/2)*(sin(1/2*d*x+1/2* 
c)^2*a)^(1/2)+2*a)))*sin(1/2*d*x+1/2*c)^10+31920*(2*a^(1/2)*2^(1/2)*(sin(1 
/2*d*x+1/2*c)^2*a)^(1/2)+5*ln(-4/(2*cos(1/2*d*x+1/2*c)-2^(1/2))*(a*2^(1/2) 
*cos(1/2*d*x+1/2*c)-a^(1/2)*2^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)-2*a))*a 
+5*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(a*2^(1/2)*cos(1/2*d*x+1/2*c)+a^(1/ 
2)*2^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)+2*a))*a)*sin(1/2*d*x+1/2*c)^8-10 
640*(14*a^(1/2)*2^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)+15*ln(-4/(2*cos(1/2 
*d*x+1/2*c)-2^(1/2))*(a*2^(1/2)*cos(1/2*d*x+1/2*c)-a^(1/2)*2^(1/2)*(sin(1/ 
2*d*x+1/2*c)^2*a)^(1/2)-2*a))*a+15*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(a* 
2^(1/2)*cos(1/2*d*x+1/2*c)+a^(1/2)*2^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)+ 
2*a))*a)*sin(1/2*d*x+1/2*c)^6+1064*(128*a^(1/2)*2^(1/2)*(sin(1/2*d*x+1/2*c 
)^2*a)^(1/2)+75*ln(-4/(2*cos(1/2*d*x+1/2*c)-2^(1/2))*(a*2^(1/2)*cos(1/2*d* 
x+1/2*c)-a^(1/2)*2^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)-2*a))*a+75*ln(4/(2 
*cos(1/2*d*x+1/2*c)+2^(1/2))*(a*2^(1/2)*cos(1/2*d*x+1/2*c)+a^(1/2)*2^(1/2) 
*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)+2*a))*a)*sin(1/2*d*x+1/2*c)^4-190*(316*a^( 
1/2)*2^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)+105*ln(-4/(2*cos(1/2*d*x+1/2*c 
)-2^(1/2))*(a*2^(1/2)*cos(1/2*d*x+1/2*c)-a^(1/2)*2^(1/2)*(sin(1/2*d*x+1...
 

Fricas [A] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 253, normalized size of antiderivative = 0.96 \[ \int (a+a \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\frac {15 \, {\left ({\left (133 \, A + 150 \, B + 176 \, C\right )} a \cos \left (d x + c\right )^{6} + {\left (133 \, A + 150 \, B + 176 \, C\right )} a \cos \left (d x + c\right )^{5}\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - 4 \, \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} {\left (\cos \left (d x + c\right ) - 2\right )} \sin \left (d x + c\right ) + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right ) + 4 \, {\left (15 \, {\left (133 \, A + 150 \, B + 176 \, C\right )} a \cos \left (d x + c\right )^{4} + 10 \, {\left (133 \, A + 150 \, B + 176 \, C\right )} a \cos \left (d x + c\right )^{3} + 8 \, {\left (133 \, A + 150 \, B + 80 \, C\right )} a \cos \left (d x + c\right )^{2} + 48 \, {\left (19 \, A + 10 \, B\right )} a \cos \left (d x + c\right ) + 384 \, A a\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{7680 \, {\left (d \cos \left (d x + c\right )^{6} + d \cos \left (d x + c\right )^{5}\right )}} \] Input:

integrate((a+a*cos(d*x+c))^(3/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c 
)^6,x, algorithm="fricas")
 

Output:

1/7680*(15*((133*A + 150*B + 176*C)*a*cos(d*x + c)^6 + (133*A + 150*B + 17 
6*C)*a*cos(d*x + c)^5)*sqrt(a)*log((a*cos(d*x + c)^3 - 7*a*cos(d*x + c)^2 
- 4*sqrt(a*cos(d*x + c) + a)*sqrt(a)*(cos(d*x + c) - 2)*sin(d*x + c) + 8*a 
)/(cos(d*x + c)^3 + cos(d*x + c)^2)) + 4*(15*(133*A + 150*B + 176*C)*a*cos 
(d*x + c)^4 + 10*(133*A + 150*B + 176*C)*a*cos(d*x + c)^3 + 8*(133*A + 150 
*B + 80*C)*a*cos(d*x + c)^2 + 48*(19*A + 10*B)*a*cos(d*x + c) + 384*A*a)*s 
qrt(a*cos(d*x + c) + a)*sin(d*x + c))/(d*cos(d*x + c)^6 + d*cos(d*x + c)^5 
)
 

Sympy [F(-1)]

Timed out. \[ \int (a+a \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\text {Timed out} \] Input:

integrate((a+a*cos(d*x+c))**(3/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x 
+c)**6,x)
 

Output:

Timed out
 

Maxima [F(-1)]

Timed out. \[ \int (a+a \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\text {Timed out} \] Input:

integrate((a+a*cos(d*x+c))^(3/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c 
)^6,x, algorithm="maxima")
 

Output:

Timed out
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 489 vs. \(2 (235) = 470\).

Time = 0.65 (sec) , antiderivative size = 489, normalized size of antiderivative = 1.86 \[ \int (a+a \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx =\text {Too large to display} \] Input:

integrate((a+a*cos(d*x+c))^(3/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c 
)^6,x, algorithm="giac")
 

Output:

-1/7680*sqrt(2)*(15*sqrt(2)*(133*A*a*sgn(cos(1/2*d*x + 1/2*c)) + 150*B*a*s 
gn(cos(1/2*d*x + 1/2*c)) + 176*C*a*sgn(cos(1/2*d*x + 1/2*c)))*log(abs(-2*s 
qrt(2) + 4*sin(1/2*d*x + 1/2*c))/abs(2*sqrt(2) + 4*sin(1/2*d*x + 1/2*c))) 
+ 4*(31920*A*a*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c)^9 + 36000*B* 
a*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c)^9 + 42240*C*a*sgn(cos(1/2 
*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c)^9 - 74480*A*a*sgn(cos(1/2*d*x + 1/2*c) 
)*sin(1/2*d*x + 1/2*c)^7 - 84000*B*a*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x 
 + 1/2*c)^7 - 98560*C*a*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c)^7 + 
 68096*A*a*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c)^5 + 76800*B*a*sg 
n(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c)^5 + 87040*C*a*sgn(cos(1/2*d*x 
 + 1/2*c))*sin(1/2*d*x + 1/2*c)^5 - 30020*A*a*sgn(cos(1/2*d*x + 1/2*c))*si 
n(1/2*d*x + 1/2*c)^3 - 32760*B*a*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1 
/2*c)^3 - 34240*C*a*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c)^3 + 568 
5*A*a*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c) + 5430*B*a*sgn(cos(1/ 
2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c) + 5040*C*a*sgn(cos(1/2*d*x + 1/2*c))* 
sin(1/2*d*x + 1/2*c))/(2*sin(1/2*d*x + 1/2*c)^2 - 1)^5)*sqrt(a)/d
 

Mupad [F(-1)]

Timed out. \[ \int (a+a \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\int \frac {{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{3/2}\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A\right )}{{\cos \left (c+d\,x\right )}^6} \,d x \] Input:

int(((a + a*cos(c + d*x))^(3/2)*(A + B*cos(c + d*x) + C*cos(c + d*x)^2))/c 
os(c + d*x)^6,x)
 

Output:

int(((a + a*cos(c + d*x))^(3/2)*(A + B*cos(c + d*x) + C*cos(c + d*x)^2))/c 
os(c + d*x)^6, x)
 

Reduce [F]

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

int((a+a*cos(d*x+c))^(3/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^6,x)
 

Output:

sqrt(a)*a*(int(sqrt(cos(c + d*x) + 1)*cos(c + d*x)*sec(c + d*x)**6,x)*a + 
int(sqrt(cos(c + d*x) + 1)*cos(c + d*x)*sec(c + d*x)**6,x)*b + int(sqrt(co 
s(c + d*x) + 1)*cos(c + d*x)**3*sec(c + d*x)**6,x)*c + int(sqrt(cos(c + d* 
x) + 1)*cos(c + d*x)**2*sec(c + d*x)**6,x)*b + int(sqrt(cos(c + d*x) + 1)* 
cos(c + d*x)**2*sec(c + d*x)**6,x)*c + int(sqrt(cos(c + d*x) + 1)*sec(c + 
d*x)**6,x)*a)