\(\int (a+a \cos (c+d x))^{3/2} (A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^5(c+d x) \, dx\) [389]

Optimal result

Integrand size = 43, antiderivative size = 215 \[ \int (a+a \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx=\frac {a^{3/2} (75 A+88 B+112 C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{64 d}+\frac {a^2 (75 A+88 B+112 C) \tan (c+d x)}{64 d \sqrt {a+a \cos (c+d x)}}+\frac {a^2 (39 A+56 B+48 C) \sec (c+d x) \tan (c+d x)}{96 d \sqrt {a+a \cos (c+d x)}}+\frac {a (3 A+8 B) \sqrt {a+a \cos (c+d x)} \sec ^2(c+d x) \tan (c+d x)}{24 d}+\frac {A (a+a \cos (c+d x))^{3/2} \sec ^3(c+d x) \tan (c+d x)}{4 d} \] Output:

1/64*a^(3/2)*(75*A+88*B+112*C)*arctanh(a^(1/2)*sin(d*x+c)/(a+a*cos(d*x+c)) 
^(1/2))/d+1/64*a^2*(75*A+88*B+112*C)*tan(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+1 
/96*a^2*(39*A+56*B+48*C)*sec(d*x+c)*tan(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+1/ 
24*a*(3*A+8*B)*(a+a*cos(d*x+c))^(1/2)*sec(d*x+c)^2*tan(d*x+c)/d+1/4*A*(a+a 
*cos(d*x+c))^(3/2)*sec(d*x+c)^3*tan(d*x+c)/d
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 2.37 (sec) , antiderivative size = 174, normalized size of antiderivative = 0.81 \[ \int (a+a \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx=\frac {a \sqrt {a (1+\cos (c+d x))} \sec \left (\frac {1}{2} (c+d x)\right ) \sec ^4(c+d x) \left (6 \sqrt {2} (75 A+88 B+112 C) \text {arctanh}\left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right ) \cos ^4(c+d x)+(492 A+352 B+192 C+(1155 A+1048 B+1008 C) \cos (c+d x)+4 (75 A+88 B+48 C) \cos (2 (c+d x))+225 A \cos (3 (c+d x))+264 B \cos (3 (c+d x))+336 C \cos (3 (c+d x))) \sin \left (\frac {1}{2} (c+d x)\right )\right )}{768 d} \] 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]^5,x]
 

Output:

(a*Sqrt[a*(1 + Cos[c + d*x])]*Sec[(c + d*x)/2]*Sec[c + d*x]^4*(6*Sqrt[2]*( 
75*A + 88*B + 112*C)*ArcTanh[Sqrt[2]*Sin[(c + d*x)/2]]*Cos[c + d*x]^4 + (4 
92*A + 352*B + 192*C + (1155*A + 1048*B + 1008*C)*Cos[c + d*x] + 4*(75*A + 
 88*B + 48*C)*Cos[2*(c + d*x)] + 225*A*Cos[3*(c + d*x)] + 264*B*Cos[3*(c + 
 d*x)] + 336*C*Cos[3*(c + d*x)])*Sin[(c + d*x)/2]))/(768*d)
 

Rubi [A] (verified)

Time = 1.28 (sec) , antiderivative size = 220, normalized size of antiderivative = 1.02, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.302, Rules used = {3042, 3522, 27, 3042, 3454, 27, 3042, 3459, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3522

\(\displaystyle \frac {\int \frac {1}{2} (\cos (c+d x) a+a)^{3/2} (a (3 A+8 B)+a (3 A+8 C) \cos (c+d x)) \sec ^4(c+d x)dx}{4 a}+\frac {A \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^{3/2}}{4 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int (\cos (c+d x) a+a)^{3/2} (a (3 A+8 B)+a (3 A+8 C) \cos (c+d x)) \sec ^4(c+d x)dx}{8 a}+\frac {A \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^{3/2}}{4 d}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3454

\(\displaystyle \frac {\frac {1}{3} \int \frac {1}{2} \sqrt {\cos (c+d x) a+a} \left ((39 A+56 B+48 C) a^2+3 (9 A+8 B+16 C) \cos (c+d x) a^2\right ) \sec ^3(c+d x)dx+\frac {a^2 (3 A+8 B) \tan (c+d x) \sec ^2(c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}}{8 a}+\frac {A \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^{3/2}}{4 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {1}{6} \int \sqrt {\cos (c+d x) a+a} \left ((39 A+56 B+48 C) a^2+3 (9 A+8 B+16 C) \cos (c+d x) a^2\right ) \sec ^3(c+d x)dx+\frac {a^2 (3 A+8 B) \tan (c+d x) \sec ^2(c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}}{8 a}+\frac {A \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^{3/2}}{4 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{6} \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a} \left ((39 A+56 B+48 C) a^2+3 (9 A+8 B+16 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^3}dx+\frac {a^2 (3 A+8 B) \tan (c+d x) \sec ^2(c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}}{8 a}+\frac {A \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^{3/2}}{4 d}\)

\(\Big \downarrow \) 3459

\(\displaystyle \frac {\frac {1}{6} \left (\frac {3}{4} a^2 (75 A+88 B+112 C) \int \sqrt {\cos (c+d x) a+a} \sec ^2(c+d x)dx+\frac {a^3 (39 A+56 B+48 C) \tan (c+d x) \sec (c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^2 (3 A+8 B) \tan (c+d x) \sec ^2(c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}}{8 a}+\frac {A \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^{3/2}}{4 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{6} \left (\frac {3}{4} a^2 (75 A+88 B+112 C) \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )^2}dx+\frac {a^3 (39 A+56 B+48 C) \tan (c+d x) \sec (c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^2 (3 A+8 B) \tan (c+d x) \sec ^2(c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}}{8 a}+\frac {A \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^{3/2}}{4 d}\)

\(\Big \downarrow \) 3251

\(\displaystyle \frac {\frac {1}{6} \left (\frac {3}{4} a^2 (75 A+88 B+112 C) \left (\frac {1}{2} \int \sqrt {\cos (c+d x) a+a} \sec (c+d x)dx+\frac {a \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^3 (39 A+56 B+48 C) \tan (c+d x) \sec (c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^2 (3 A+8 B) \tan (c+d x) \sec ^2(c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}}{8 a}+\frac {A \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^{3/2}}{4 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{6} \left (\frac {3}{4} a^2 (75 A+88 B+112 C) \left (\frac {1}{2} \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {a \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^3 (39 A+56 B+48 C) \tan (c+d x) \sec (c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^2 (3 A+8 B) \tan (c+d x) \sec ^2(c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}}{8 a}+\frac {A \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^{3/2}}{4 d}\)

\(\Big \downarrow \) 3252

\(\displaystyle \frac {\frac {1}{6} \left (\frac {3}{4} a^2 (75 A+88 B+112 C) \left (\frac {a \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {a \int \frac {1}{a-\frac {a^2 \sin ^2(c+d x)}{\cos (c+d x) a+a}}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x) a+a}}\right )}{d}\right )+\frac {a^3 (39 A+56 B+48 C) \tan (c+d x) \sec (c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^2 (3 A+8 B) \tan (c+d x) \sec ^2(c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}}{8 a}+\frac {A \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^{3/2}}{4 d}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {a^2 (3 A+8 B) \tan (c+d x) \sec ^2(c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}+\frac {1}{6} \left (\frac {a^3 (39 A+56 B+48 C) \tan (c+d x) \sec (c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}+\frac {3}{4} a^2 (75 A+88 B+112 C) \left (\frac {\sqrt {a} \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{d}+\frac {a \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}\right )\right )}{8 a}+\frac {A \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^{3/2}}{4 d}\)

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]^5,x]
 

Output:

(A*(a + a*Cos[c + d*x])^(3/2)*Sec[c + d*x]^3*Tan[c + d*x])/(4*d) + ((a^2*( 
3*A + 8*B)*Sqrt[a + a*Cos[c + d*x]]*Sec[c + d*x]^2*Tan[c + d*x])/(3*d) + ( 
(a^3*(39*A + 56*B + 48*C)*Sec[c + d*x]*Tan[c + d*x])/(2*d*Sqrt[a + a*Cos[c 
 + d*x]]) + (3*a^2*(75*A + 88*B + 112*C)*((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*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. \(2179\) vs. \(2(191)=382\).

Time = 0.90 (sec) , antiderivative size = 2180, normalized size of antiderivative = 10.14

method	result	size
parts	\(\text {Expression too large to display}\)	\(2180\)
default	\(\text {Expression too large to display}\)	\(2400\)

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)^5,x, 
method=_RETURNVERBOSE)
 

Output:

1/8*A*a^(1/2)*cos(1/2*d*x+1/2*c)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)*(1200*a*(l 
n(-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)^8-1200*(a^(1/2)*2^(1/2)*(sin(1/2*d*x+1 
/2*c)^2*a)^(1/2)+2*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+2*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+200*(11*a^ 
(1/2)*2^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)+9*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+9*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+(-1460*a^(1/2)*2^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)-60 
0*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-600*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)^2+362*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*...
 

Fricas [A] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 232, normalized size of antiderivative = 1.08 \[ \int (a+a \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx=\frac {3 \, {\left ({\left (75 \, A + 88 \, B + 112 \, C\right )} a \cos \left (d x + c\right )^{5} + {\left (75 \, A + 88 \, B + 112 \, C\right )} a \cos \left (d x + c\right )^{4}\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - 4 \, \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} {\left (\cos \left (d x + c\right ) - 2\right )} \sin \left (d x + c\right ) + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right ) + 4 \, {\left (3 \, {\left (75 \, A + 88 \, B + 112 \, C\right )} a \cos \left (d x + c\right )^{3} + 2 \, {\left (75 \, A + 88 \, B + 48 \, C\right )} a \cos \left (d x + c\right )^{2} + 8 \, {\left (15 \, A + 8 \, B\right )} a \cos \left (d x + c\right ) + 48 \, A a\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{768 \, {\left (d \cos \left (d x + c\right )^{5} + d \cos \left (d x + c\right )^{4}\right )}} \] 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 
)^5,x, algorithm="fricas")
 

Output:

1/768*(3*((75*A + 88*B + 112*C)*a*cos(d*x + c)^5 + (75*A + 88*B + 112*C)*a 
*cos(d*x + c)^4)*sqrt(a)*log((a*cos(d*x + c)^3 - 7*a*cos(d*x + c)^2 - 4*sq 
rt(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*(3*(75*A + 88*B + 112*C)*a*cos(d*x + c) 
^3 + 2*(75*A + 88*B + 48*C)*a*cos(d*x + c)^2 + 8*(15*A + 8*B)*a*cos(d*x + 
c) + 48*A*a)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c))/(d*cos(d*x + c)^5 + d* 
cos(d*x + c)^4)
 

Sympy [F(-1)]

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

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 12526 vs. \(2 (191) = 382\).

Time = 168.82 (sec) , antiderivative size = 12526, normalized size of antiderivative = 58.26 \[ \int (a+a \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx=\text {Too large to display} \] 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 
)^5,x, algorithm="maxima")
 

Output:

-1/768*(3*(140*a*cos(8*d*x + 8*c)^2*sin(3/2*d*x + 3/2*c) + 2240*a*cos(6*d* 
x + 6*c)^2*sin(3/2*d*x + 3/2*c) + 5040*a*cos(4*d*x + 4*c)^2*sin(3/2*d*x + 
3/2*c) + 2240*a*cos(2*d*x + 2*c)^2*sin(3/2*d*x + 3/2*c) + 140*a*sin(8*d*x 
+ 8*c)^2*sin(3/2*d*x + 3/2*c) + 2240*a*sin(6*d*x + 6*c)^2*sin(3/2*d*x + 3/ 
2*c) + 5040*a*sin(4*d*x + 4*c)^2*sin(3/2*d*x + 3/2*c) + 2240*a*sin(2*d*x + 
 2*c)^2*sin(3/2*d*x + 3/2*c) + 4064*a*cos(7/2*d*x + 7/2*c)*sin(2*d*x + 2*c 
) + 336*a*cos(5/2*d*x + 5/2*c)*sin(2*d*x + 2*c) - 240*a*cos(3/2*d*x + 3/2* 
c)*sin(2*d*x + 2*c) + 1360*a*cos(2*d*x + 2*c)*sin(3/2*d*x + 3/2*c) - 36*(a 
*sin(8*d*x + 8*c) + 4*a*sin(6*d*x + 6*c) + 6*a*sin(4*d*x + 4*c) + 4*a*sin( 
2*d*x + 2*c))*cos(21/2*d*x + 21/2*c) + 140*(a*sin(8*d*x + 8*c) + 4*a*sin(6 
*d*x + 6*c) + 6*a*sin(4*d*x + 4*c) + 4*a*sin(2*d*x + 2*c))*cos(19/2*d*x + 
19/2*c) + 456*(a*sin(8*d*x + 8*c) + 4*a*sin(6*d*x + 6*c) + 6*a*sin(4*d*x + 
 4*c) + 4*a*sin(2*d*x + 2*c))*cos(17/2*d*x + 17/2*c) + 4*(280*a*cos(6*d*x 
+ 6*c)*sin(3/2*d*x + 3/2*c) + 420*a*cos(4*d*x + 4*c)*sin(3/2*d*x + 3/2*c) 
+ 280*a*cos(2*d*x + 2*c)*sin(3/2*d*x + 3/2*c) - 290*a*sin(15/2*d*x + 15/2* 
c) - 596*a*sin(13/2*d*x + 13/2*c) - 780*a*sin(11/2*d*x + 11/2*c) - 750*a*s 
in(9/2*d*x + 9/2*c) - 254*a*sin(7/2*d*x + 7/2*c) - 21*a*sin(5/2*d*x + 5/2* 
c) + 85*a*sin(3/2*d*x + 3/2*c))*cos(8*d*x + 8*c) + 2320*(2*a*sin(6*d*x + 6 
*c) + 3*a*sin(4*d*x + 4*c) + 2*a*sin(2*d*x + 2*c))*cos(15/2*d*x + 15/2*c) 
+ 4768*(2*a*sin(6*d*x + 6*c) + 3*a*sin(4*d*x + 4*c) + 2*a*sin(2*d*x + 2...
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 414 vs. \(2 (191) = 382\).

Time = 0.57 (sec) , antiderivative size = 414, normalized size of antiderivative = 1.93 \[ \int (a+a \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx =\text {Too large to display} \] 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 
)^5,x, algorithm="giac")
 

Output:

-1/768*sqrt(2)*(3*sqrt(2)*(75*A*a*sgn(cos(1/2*d*x + 1/2*c)) + 88*B*a*sgn(c 
os(1/2*d*x + 1/2*c)) + 112*C*a*sgn(cos(1/2*d*x + 1/2*c)))*log(abs(-2*sqrt( 
2) + 4*sin(1/2*d*x + 1/2*c))/abs(2*sqrt(2) + 4*sin(1/2*d*x + 1/2*c))) + 4* 
(1800*A*a*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c)^7 + 2112*B*a*sgn( 
cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c)^7 + 2688*C*a*sgn(cos(1/2*d*x + 
1/2*c))*sin(1/2*d*x + 1/2*c)^7 - 3300*A*a*sgn(cos(1/2*d*x + 1/2*c))*sin(1/ 
2*d*x + 1/2*c)^5 - 3872*B*a*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c) 
^5 - 4416*C*a*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c)^5 + 2190*A*a* 
sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c)^3 + 2416*B*a*sgn(cos(1/2*d* 
x + 1/2*c))*sin(1/2*d*x + 1/2*c)^3 + 2400*C*a*sgn(cos(1/2*d*x + 1/2*c))*si 
n(1/2*d*x + 1/2*c)^3 - 543*A*a*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2 
*c) - 504*B*a*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c) - 432*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 
)^4)*sqrt(a)/d
 

Mupad [F(-1)]

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

Reduce [F]

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

Output:

sqrt(a)*a*(int(sqrt(cos(c + d*x) + 1)*cos(c + d*x)*sec(c + d*x)**5,x)*a + 
int(sqrt(cos(c + d*x) + 1)*cos(c + d*x)*sec(c + d*x)**5,x)*b + int(sqrt(co 
s(c + d*x) + 1)*cos(c + d*x)**3*sec(c + d*x)**5,x)*c + int(sqrt(cos(c + d* 
x) + 1)*cos(c + d*x)**2*sec(c + d*x)**5,x)*b + int(sqrt(cos(c + d*x) + 1)* 
cos(c + d*x)**2*sec(c + d*x)**5,x)*c + int(sqrt(cos(c + d*x) + 1)*sec(c + 
d*x)**5,x)*a)