Integrand size = 43, antiderivative size = 209 \[ \int \sqrt {a+a \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx=\frac {\sqrt {a} (35 A+40 B+48 C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{64 d}+\frac {a (35 A+40 B+48 C) \tan (c+d x)}{64 d \sqrt {a+a \cos (c+d x)}}+\frac {a (35 A+40 B+48 C) \sec (c+d x) \tan (c+d x)}{96 d \sqrt {a+a \cos (c+d x)}}+\frac {a (A+8 B) \sec ^2(c+d x) \tan (c+d x)}{24 d \sqrt {a+a \cos (c+d x)}}+\frac {A \sqrt {a+a \cos (c+d x)} \sec ^3(c+d x) \tan (c+d x)}{4 d} \]
1/64*(35*A+40*B+48*C)*arctanh(sin(d*x+c)*a^(1/2)/(a+a*cos(d*x+c))^(1/2))*a ^(1/2)/d+1/64*a*(35*A+40*B+48*C)*tan(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+1/96* a*(35*A+40*B+48*C)*sec(d*x+c)*tan(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+1/24*a*( A+8*B)*sec(d*x+c)^2*tan(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+1/4*A*sec(d*x+c)^3 *(a+a*cos(d*x+c))^(1/2)*tan(d*x+c)/d
Time = 2.07 (sec) , antiderivative size = 173, normalized size of antiderivative = 0.83 \[ \int \sqrt {a+a \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx=\frac {\sqrt {a (1+\cos (c+d x))} \sec \left (\frac {1}{2} (c+d x)\right ) \sec ^4(c+d x) \left (6 \sqrt {2} (35 A+40 B+48 C) \text {arctanh}\left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right ) \cos ^4(c+d x)+(332 A+160 B+192 C+(539 A+616 B+432 C) \cos (c+d x)+4 (35 A+40 B+48 C) \cos (2 (c+d x))+105 A \cos (3 (c+d x))+120 B \cos (3 (c+d x))+144 C \cos (3 (c+d x))) \sin \left (\frac {1}{2} (c+d x)\right )\right )}{768 d} \]
(Sqrt[a*(1 + Cos[c + d*x])]*Sec[(c + d*x)/2]*Sec[c + d*x]^4*(6*Sqrt[2]*(35 *A + 40*B + 48*C)*ArcTanh[Sqrt[2]*Sin[(c + d*x)/2]]*Cos[c + d*x]^4 + (332* A + 160*B + 192*C + (539*A + 616*B + 432*C)*Cos[c + d*x] + 4*(35*A + 40*B + 48*C)*Cos[2*(c + d*x)] + 105*A*Cos[3*(c + d*x)] + 120*B*Cos[3*(c + d*x)] + 144*C*Cos[3*(c + d*x)])*Sin[(c + d*x)/2]))/(768*d)
Time = 1.07 (sec) , antiderivative size = 204, normalized size of antiderivative = 0.98, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.279, Rules used = {3042, 3522, 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.
| \(\displaystyle \int \sec ^5(c+d x) \sqrt {a \cos (c+d x)+a} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sqrt {a \sin \left (c+d x+\frac {\pi }{2}\right )+a} \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^5}dx\) |
| \(\Big \downarrow \) 3522 |
| \(\displaystyle \frac {\int \frac {1}{2} \sqrt {\cos (c+d x) a+a} (a (A+8 B)+a (5 A+8 C) \cos (c+d x)) \sec ^4(c+d x)dx}{4 a}+\frac {A \tan (c+d x) \sec ^3(c+d x) \sqrt {a \cos (c+d x)+a}}{4 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \sqrt {\cos (c+d x) a+a} (a (A+8 B)+a (5 A+8 C) \cos (c+d x)) \sec ^4(c+d x)dx}{8 a}+\frac {A \tan (c+d x) \sec ^3(c+d x) \sqrt {a \cos (c+d x)+a}}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a} \left (a (A+8 B)+a (5 A+8 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^4}dx}{8 a}+\frac {A \tan (c+d x) \sec ^3(c+d x) \sqrt {a \cos (c+d x)+a}}{4 d}\) |
| \(\Big \downarrow \) 3459 |
| \(\displaystyle \frac {\frac {1}{6} a (35 A+40 B+48 C) \int \sqrt {\cos (c+d x) a+a} \sec ^3(c+d x)dx+\frac {a^2 (A+8 B) \tan (c+d x) \sec ^2(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}}{8 a}+\frac {A \tan (c+d x) \sec ^3(c+d x) \sqrt {a \cos (c+d x)+a}}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{6} a (35 A+40 B+48 C) \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )^3}dx+\frac {a^2 (A+8 B) \tan (c+d x) \sec ^2(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}}{8 a}+\frac {A \tan (c+d x) \sec ^3(c+d x) \sqrt {a \cos (c+d x)+a}}{4 d}\) |
| \(\Big \downarrow \) 3251 |
| \(\displaystyle \frac {\frac {1}{6} a (35 A+40 B+48 C) \left (\frac {3}{4} \int \sqrt {\cos (c+d x) a+a} \sec ^2(c+d x)dx+\frac {a \tan (c+d x) \sec (c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^2 (A+8 B) \tan (c+d x) \sec ^2(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}}{8 a}+\frac {A \tan (c+d x) \sec ^3(c+d x) \sqrt {a \cos (c+d x)+a}}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{6} a (35 A+40 B+48 C) \left (\frac {3}{4} \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )^2}dx+\frac {a \tan (c+d x) \sec (c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^2 (A+8 B) \tan (c+d x) \sec ^2(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}}{8 a}+\frac {A \tan (c+d x) \sec ^3(c+d x) \sqrt {a \cos (c+d x)+a}}{4 d}\) |
| \(\Big \downarrow \) 3251 |
| \(\displaystyle \frac {\frac {1}{6} a (35 A+40 B+48 C) \left (\frac {3}{4} \left (\frac {1}{2} \int \sqrt {\cos (c+d x) a+a} \sec (c+d x)dx+\frac {a \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a \tan (c+d x) \sec (c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^2 (A+8 B) \tan (c+d x) \sec ^2(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}}{8 a}+\frac {A \tan (c+d x) \sec ^3(c+d x) \sqrt {a \cos (c+d x)+a}}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{6} a (35 A+40 B+48 C) \left (\frac {3}{4} \left (\frac {1}{2} \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {a \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a \tan (c+d x) \sec (c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^2 (A+8 B) \tan (c+d x) \sec ^2(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}}{8 a}+\frac {A \tan (c+d x) \sec ^3(c+d x) \sqrt {a \cos (c+d x)+a}}{4 d}\) |
| \(\Big \downarrow \) 3252 |
| \(\displaystyle \frac {\frac {1}{6} a (35 A+40 B+48 C) \left (\frac {3}{4} \left (\frac {a \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {a \int \frac {1}{a-\frac {a^2 \sin ^2(c+d x)}{\cos (c+d x) a+a}}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x) a+a}}\right )}{d}\right )+\frac {a \tan (c+d x) \sec (c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^2 (A+8 B) \tan (c+d x) \sec ^2(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}}{8 a}+\frac {A \tan (c+d x) \sec ^3(c+d x) \sqrt {a \cos (c+d x)+a}}{4 d}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {a^2 (A+8 B) \tan (c+d x) \sec ^2(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}+\frac {1}{6} a (35 A+40 B+48 C) \left (\frac {3}{4} \left (\frac {\sqrt {a} \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{d}+\frac {a \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a \tan (c+d x) \sec (c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\right )}{8 a}+\frac {A \tan (c+d x) \sec ^3(c+d x) \sqrt {a \cos (c+d x)+a}}{4 d}\) |
(A*Sqrt[a + a*Cos[c + d*x]]*Sec[c + d*x]^3*Tan[c + d*x])/(4*d) + ((a^2*(A + 8*B)*Sec[c + d*x]^2*Tan[c + d*x])/(3*d*Sqrt[a + a*Cos[c + d*x]]) + (a*(3 5*A + 40*B + 48*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*a)
3.4.81.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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])
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]
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]
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]
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])
Leaf count of result is larger than twice the leaf count of optimal. \(2178\) vs. \(2(185)=370\).
Time = 16.91 (sec) , antiderivative size = 2179, normalized size of antiderivative = 10.43
| method | result | size |
| parts | \(\text {Expression too large to display}\) | \(2179\) |
| default | \(\text {Expression too large to display}\) | \(2400\) |
int((a+cos(d*x+c)*a)^(1/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^5,x, method=_RETURNVERBOSE)
1/24*A*cos(1/2*d*x+1/2*c)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*(1680*a*(ln(4/(2* cos(1/2*d*x+1/2*c)+2^(1/2))*(2^(1/2)*a*cos(1/2*d*x+1/2*c)+2^(1/2)*(a*sin(1 /2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+2*a))+ln(-4/(2*cos(1/2*d*x+1/2*c)-2^(1/2))* (2^(1/2)*a*cos(1/2*d*x+1/2*c)-2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/ 2)-2*a)))*sin(1/2*d*x+1/2*c)^8-1680*(2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2 )*a^(1/2)+2*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(2^(1/2)*a*cos(1/2*d*x+1/2 *c)+2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+2*a))*a+2*ln(-4/(2*cos( 1/2*d*x+1/2*c)-2^(1/2))*(2^(1/2)*a*cos(1/2*d*x+1/2*c)-2^(1/2)*(a*sin(1/2*d *x+1/2*c)^2)^(1/2)*a^(1/2)-2*a))*a)*sin(1/2*d*x+1/2*c)^6+280*(11*2^(1/2)*( a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+9*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2) )*(2^(1/2)*a*cos(1/2*d*x+1/2*c)+2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^( 1/2)+2*a))*a+9*ln(-4/(2*cos(1/2*d*x+1/2*c)-2^(1/2))*(2^(1/2)*a*cos(1/2*d*x +1/2*c)-2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)-2*a))*a)*sin(1/2*d* x+1/2*c)^4+(-840*ln(-4/(2*cos(1/2*d*x+1/2*c)-2^(1/2))*(2^(1/2)*a*cos(1/2*d *x+1/2*c)-2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)-2*a))*a-840*ln(4/ (2*cos(1/2*d*x+1/2*c)+2^(1/2))*(2^(1/2)*a*cos(1/2*d*x+1/2*c)+2^(1/2)*(a*si n(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+2*a))*a-2044*2^(1/2)*(a*sin(1/2*d*x+1/2* c)^2)^(1/2)*a^(1/2))*sin(1/2*d*x+1/2*c)^2+105*ln(-4/(2*cos(1/2*d*x+1/2*c)- 2^(1/2))*(2^(1/2)*a*cos(1/2*d*x+1/2*c)-2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1 /2)*a^(1/2)-2*a))*a+105*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(2^(1/2)*a*...
Time = 0.42 (sec) , antiderivative size = 226, normalized size of antiderivative = 1.08 \[ \int \sqrt {a+a \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx=\frac {3 \, {\left ({\left (35 \, A + 40 \, B + 48 \, C\right )} \cos \left (d x + c\right )^{5} + {\left (35 \, A + 40 \, B + 48 \, C\right )} \cos \left (d x + c\right )^{4}\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - 4 \, \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} {\left (\cos \left (d x + c\right ) - 2\right )} \sin \left (d x + c\right ) + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right ) + 4 \, {\left (3 \, {\left (35 \, A + 40 \, B + 48 \, C\right )} \cos \left (d x + c\right )^{3} + 2 \, {\left (35 \, A + 40 \, B + 48 \, C\right )} \cos \left (d x + c\right )^{2} + 8 \, {\left (7 \, A + 8 \, B\right )} \cos \left (d x + c\right ) + 48 \, A\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{768 \, {\left (d \cos \left (d x + c\right )^{5} + d \cos \left (d x + c\right )^{4}\right )}} \]
integrate((a+a*cos(d*x+c))^(1/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c )^5,x, algorithm="fricas")
1/768*(3*((35*A + 40*B + 48*C)*cos(d*x + c)^5 + (35*A + 40*B + 48*C)*cos(d *x + c)^4)*sqrt(a)*log((a*cos(d*x + c)^3 - 7*a*cos(d*x + c)^2 - 4*sqrt(a*c os(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*(35*A + 40*B + 48*C)*cos(d*x + c)^3 + 2*(3 5*A + 40*B + 48*C)*cos(d*x + c)^2 + 8*(7*A + 8*B)*cos(d*x + c) + 48*A)*sqr t(a*cos(d*x + c) + a)*sin(d*x + c))/(d*cos(d*x + c)^5 + d*cos(d*x + c)^4)
Timed out. \[ \int \sqrt {a+a \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx=\text {Timed out} \]
Timed out. \[ \int \sqrt {a+a \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx=\text {Timed out} \]
integrate((a+a*cos(d*x+c))^(1/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c )^5,x, algorithm="maxima")
Leaf count of result is larger than twice the leaf count of optimal. 399 vs. \(2 (185) = 370\).
Time = 0.43 (sec) , antiderivative size = 399, normalized size of antiderivative = 1.91 \[ \int \sqrt {a+a \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx=-\frac {\sqrt {2} {\left (3 \, \sqrt {2} {\left (35 \, A \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 40 \, B \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 48 \, C \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \log \left (\frac {{\left | -2 \, \sqrt {2} + 4 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}}{{\left | 2 \, \sqrt {2} + 4 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}}\right ) + \frac {4 \, {\left (840 \, A \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 960 \, B \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 1152 \, C \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 1540 \, A \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 1760 \, B \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 2112 \, C \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 1022 \, A \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 1168 \, B \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 1248 \, C \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 279 \, A \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 264 \, B \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 240 \, C \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (2 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{4}}\right )} \sqrt {a}}{768 \, d} \]
integrate((a+a*cos(d*x+c))^(1/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c )^5,x, algorithm="giac")
-1/768*sqrt(2)*(3*sqrt(2)*(35*A*sgn(cos(1/2*d*x + 1/2*c)) + 40*B*sgn(cos(1 /2*d*x + 1/2*c)) + 48*C*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*(840*A* sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c)^7 + 960*B*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c)^7 + 1152*C*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2 *d*x + 1/2*c)^7 - 1540*A*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c)^5 - 1760*B*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c)^5 - 2112*C*sgn(cos (1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c)^5 + 1022*A*sgn(cos(1/2*d*x + 1/2*c ))*sin(1/2*d*x + 1/2*c)^3 + 1168*B*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c)^3 + 1248*C*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c)^3 - 279* A*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c) - 264*B*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c) - 240*C*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
Timed out. \[ \int \sqrt {a+a \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx=\int \frac {\sqrt {a+a\,\cos \left (c+d\,x\right )}\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A\right )}{{\cos \left (c+d\,x\right )}^5} \,d x \]