Integrand size = 33, antiderivative size = 173 \[ \int \cos (c+d x) (a+a \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {5 a^{5/2} A \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{d}+\frac {A (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{d}+\frac {a^3 (15 A+64 C) \tan (c+d x)}{15 d \sqrt {a+a \sec (c+d x)}}-\frac {a^2 (15 A-16 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{15 d}-\frac {a (5 A-2 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{5 d} \]
5*a^(5/2)*A*arctan(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/d+A*(a+a*sec (d*x+c))^(5/2)*sin(d*x+c)/d-1/5*a*(5*A-2*C)*(a+a*sec(d*x+c))^(3/2)*tan(d*x +c)/d+1/15*a^3*(15*A+64*C)*tan(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)-1/15*a^2*(1 5*A-16*C)*(a+a*sec(d*x+c))^(1/2)*tan(d*x+c)/d
Time = 1.47 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.71 \[ \int \cos (c+d x) (a+a \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {a^3 \left (75 A \text {arctanh}\left (\sqrt {1-\sec (c+d x)}\right ) \tan (c+d x)+\sqrt {1-\sec (c+d x)} \left (15 A \sin (c+d x)+2 \left (15 A+43 C+14 C \sec (c+d x)+3 C \sec ^2(c+d x)\right ) \tan (c+d x)\right )\right )}{15 d \sqrt {1-\sec (c+d x)} \sqrt {a (1+\sec (c+d x))}} \]
(a^3*(75*A*ArcTanh[Sqrt[1 - Sec[c + d*x]]]*Tan[c + d*x] + Sqrt[1 - Sec[c + d*x]]*(15*A*Sin[c + d*x] + 2*(15*A + 43*C + 14*C*Sec[c + d*x] + 3*C*Sec[c + d*x]^2)*Tan[c + d*x])))/(15*d*Sqrt[1 - Sec[c + d*x]]*Sqrt[a*(1 + Sec[c + d*x])])
Time = 1.09 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.10, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.455, Rules used = {3042, 4575, 27, 3042, 4405, 27, 3042, 4405, 27, 3042, 4403, 3042, 4261, 216, 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.
| \(\displaystyle \int \cos (c+d x) (a \sec (c+d x)+a)^{5/2} \left (A+C \sec ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{5/2} \left (A+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )}dx\) |
| \(\Big \downarrow \) 4575 |
| \(\displaystyle \frac {\int \frac {1}{2} (\sec (c+d x) a+a)^{5/2} (5 a A-a (5 A-2 C) \sec (c+d x))dx}{a}+\frac {A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int (\sec (c+d x) a+a)^{5/2} (5 a A-a (5 A-2 C) \sec (c+d x))dx}{2 a}+\frac {A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{5/2} \left (5 a A-a (5 A-2 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx}{2 a}+\frac {A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{d}\) |
| \(\Big \downarrow \) 4405 |
| \(\displaystyle \frac {\frac {2}{5} \int \frac {1}{2} (\sec (c+d x) a+a)^{3/2} \left (25 a^2 A-a^2 (15 A-16 C) \sec (c+d x)\right )dx-\frac {2 a^2 (5 A-2 C) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}}{2 a}+\frac {A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {1}{5} \int (\sec (c+d x) a+a)^{3/2} \left (25 a^2 A-a^2 (15 A-16 C) \sec (c+d x)\right )dx-\frac {2 a^2 (5 A-2 C) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}}{2 a}+\frac {A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{5} \int \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2} \left (25 a^2 A-a^2 (15 A-16 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx-\frac {2 a^2 (5 A-2 C) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}}{2 a}+\frac {A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{d}\) |
| \(\Big \downarrow \) 4405 |
| \(\displaystyle \frac {\frac {1}{5} \left (\frac {2}{3} \int \frac {1}{2} \sqrt {\sec (c+d x) a+a} \left (75 A a^3+(15 A+64 C) \sec (c+d x) a^3\right )dx-\frac {2 a^3 (15 A-16 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\right )-\frac {2 a^2 (5 A-2 C) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}}{2 a}+\frac {A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {1}{5} \left (\frac {1}{3} \int \sqrt {\sec (c+d x) a+a} \left (75 A a^3+(15 A+64 C) \sec (c+d x) a^3\right )dx-\frac {2 a^3 (15 A-16 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\right )-\frac {2 a^2 (5 A-2 C) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}}{2 a}+\frac {A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{5} \left (\frac {1}{3} \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a} \left (75 A a^3+(15 A+64 C) \csc \left (c+d x+\frac {\pi }{2}\right ) a^3\right )dx-\frac {2 a^3 (15 A-16 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\right )-\frac {2 a^2 (5 A-2 C) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}}{2 a}+\frac {A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{d}\) |
| \(\Big \downarrow \) 4403 |
| \(\displaystyle \frac {\frac {1}{5} \left (\frac {1}{3} \left (a^3 (15 A+64 C) \int \sec (c+d x) \sqrt {\sec (c+d x) a+a}dx+75 a^3 A \int \sqrt {\sec (c+d x) a+a}dx\right )-\frac {2 a^3 (15 A-16 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\right )-\frac {2 a^2 (5 A-2 C) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}}{2 a}+\frac {A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{5} \left (\frac {1}{3} \left (a^3 (15 A+64 C) \int \csc \left (c+d x+\frac {\pi }{2}\right ) \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+75 a^3 A \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx\right )-\frac {2 a^3 (15 A-16 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\right )-\frac {2 a^2 (5 A-2 C) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}}{2 a}+\frac {A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{d}\) |
| \(\Big \downarrow \) 4261 |
| \(\displaystyle \frac {\frac {1}{5} \left (\frac {1}{3} \left (a^3 (15 A+64 C) \int \csc \left (c+d x+\frac {\pi }{2}\right ) \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx-\frac {150 a^4 A \int \frac {1}{\frac {a^2 \tan ^2(c+d x)}{\sec (c+d x) a+a}+a}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}\right )-\frac {2 a^3 (15 A-16 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\right )-\frac {2 a^2 (5 A-2 C) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}}{2 a}+\frac {A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{d}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {\frac {1}{5} \left (\frac {1}{3} \left (a^3 (15 A+64 C) \int \csc \left (c+d x+\frac {\pi }{2}\right ) \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {150 a^{7/2} A \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}\right )-\frac {2 a^3 (15 A-16 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\right )-\frac {2 a^2 (5 A-2 C) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}}{2 a}+\frac {A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{d}\) |
| \(\Big \downarrow \) 4279 |
| \(\displaystyle \frac {\frac {1}{5} \left (\frac {1}{3} \left (\frac {150 a^{7/2} A \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}+\frac {2 a^4 (15 A+64 C) \tan (c+d x)}{d \sqrt {a \sec (c+d x)+a}}\right )-\frac {2 a^3 (15 A-16 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\right )-\frac {2 a^2 (5 A-2 C) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}}{2 a}+\frac {A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{d}\) |
(A*(a + a*Sec[c + d*x])^(5/2)*Sin[c + d*x])/d + ((-2*a^2*(5*A - 2*C)*(a + a*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(5*d) + ((-2*a^3*(15*A - 16*C)*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(3*d) + ((150*a^(7/2)*A*ArcTan[(Sqrt[a]*Ta n[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/d + (2*a^4*(15*A + 64*C)*Tan[c + d* x])/(d*Sqrt[a + a*Sec[c + d*x]]))/3)/5)/(2*a)
3.2.77.3.1 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]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[Sqrt[csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[-2*(b/d) Subst[Int[1/(a + x^2), x], x, b*(Cot[c + d*x]/Sqrt[a + b*Csc[c + d*x]])], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]
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]
Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]*(csc[(e_.) + (f_.)*(x_)]*(d_ .) + (c_)), x_Symbol] :> Simp[c Int[Sqrt[a + b*Csc[e + f*x]], x], x] + Si mp[d Int[Sqrt[a + b*Csc[e + f*x]]*Csc[e + f*x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(d _.) + (c_)), x_Symbol] :> Simp[(-b)*d*Cot[e + f*x]*((a + b*Csc[e + f*x])^(m - 1)/(f*m)), x] + Simp[1/m Int[(a + b*Csc[e + f*x])^(m - 1)*Simp[a*c*m + (b*c*m + a*d*(2*m - 1))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && GtQ[m, 1] && EqQ[a^2 - b^2, 0] && IntegerQ[2 *m]
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_. ))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[A*Co t[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(f*n)), x] - Simp[1/( b*d*n) Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[a*A*m - b *(A*(m + n + 1) + C*n)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, C, m}, x] && EqQ[a^2 - b^2, 0] && !LtQ[m, -2^(-1)] && (LtQ[n, -2^(-1)] || EqQ[m + n + 1, 0])
Time = 51.03 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.25
| method | result | size |
| default | \(\frac {a^{2} \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (75 A \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right ) \cos \left (d x +c \right )+75 A \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right )+15 A \cos \left (d x +c \right ) \sin \left (d x +c \right )+30 A \sin \left (d x +c \right )+86 C \sin \left (d x +c \right )+28 C \tan \left (d x +c \right )+6 C \sec \left (d x +c \right ) \tan \left (d x +c \right )\right )}{15 d \left (\cos \left (d x +c \right )+1\right )}\) | \(217\) |
1/15*a^2/d*(a*(1+sec(d*x+c)))^(1/2)/(cos(d*x+c)+1)*(75*A*(-cos(d*x+c)/(cos (d*x+c)+1))^(1/2)*arctanh(sin(d*x+c)/(cos(d*x+c)+1)/(-cos(d*x+c)/(cos(d*x+ c)+1))^(1/2))*cos(d*x+c)+75*A*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctanh(s in(d*x+c)/(cos(d*x+c)+1)/(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2))+15*A*cos(d*x+ c)*sin(d*x+c)+30*A*sin(d*x+c)+86*C*sin(d*x+c)+28*C*tan(d*x+c)+6*C*sec(d*x+ c)*tan(d*x+c))
Time = 0.29 (sec) , antiderivative size = 398, normalized size of antiderivative = 2.30 \[ \int \cos (c+d x) (a+a \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\left [\frac {75 \, {\left (A a^{2} \cos \left (d x + c\right )^{3} + A a^{2} \cos \left (d x + c\right )^{2}\right )} \sqrt {-a} \log \left (\frac {2 \, a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right ) + 1}\right ) + 2 \, {\left (15 \, A a^{2} \cos \left (d x + c\right )^{3} + 2 \, {\left (15 \, A + 43 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 28 \, C a^{2} \cos \left (d x + c\right ) + 6 \, C a^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{30 \, {\left (d \cos \left (d x + c\right )^{3} + d \cos \left (d x + c\right )^{2}\right )}}, -\frac {75 \, {\left (A a^{2} \cos \left (d x + c\right )^{3} + A a^{2} \cos \left (d x + c\right )^{2}\right )} \sqrt {a} \arctan \left (\frac {\sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) - {\left (15 \, A a^{2} \cos \left (d x + c\right )^{3} + 2 \, {\left (15 \, A + 43 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 28 \, C a^{2} \cos \left (d x + c\right ) + 6 \, C a^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{15 \, {\left (d \cos \left (d x + c\right )^{3} + d \cos \left (d x + c\right )^{2}\right )}}\right ] \]
[1/30*(75*(A*a^2*cos(d*x + c)^3 + A*a^2*cos(d*x + c)^2)*sqrt(-a)*log((2*a* cos(d*x + c)^2 - 2*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d* x + c)*sin(d*x + c) + a*cos(d*x + c) - a)/(cos(d*x + c) + 1)) + 2*(15*A*a^ 2*cos(d*x + c)^3 + 2*(15*A + 43*C)*a^2*cos(d*x + c)^2 + 28*C*a^2*cos(d*x + c) + 6*C*a^2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(d*co s(d*x + c)^3 + d*cos(d*x + c)^2), -1/15*(75*(A*a^2*cos(d*x + c)^3 + A*a^2* cos(d*x + c)^2)*sqrt(a)*arctan(sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos (d*x + c)/(sqrt(a)*sin(d*x + c))) - (15*A*a^2*cos(d*x + c)^3 + 2*(15*A + 4 3*C)*a^2*cos(d*x + c)^2 + 28*C*a^2*cos(d*x + c) + 6*C*a^2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^3 + d*cos(d*x + c)^ 2)]
Timed out. \[ \int \cos (c+d x) (a+a \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \]
Leaf count of result is larger than twice the leaf count of optimal. 1384 vs. \(2 (153) = 306\).
Time = 0.43 (sec) , antiderivative size = 1384, normalized size of antiderivative = 8.00 \[ \int \cos (c+d x) (a+a \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\text {Too large to display} \]
1/4*(18*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1) ^(3/4)*a^(5/2)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + 2*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4) *((4*a^2*sin(3*d*x + 3*c) + 5*a^2*sin(2*d*x + 2*c) + 4*a^2*sin(d*x + c))*c os(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + (a^2*cos(2*d*x + 2*c)^2*sin(d*x + c) + a^2*sin(2*d*x + 2*c)^2*sin(d*x + c) + 2*a^2*cos(2*d *x + 2*c)*sin(d*x + c) + a^2*sin(d*x + c))*cos(1/2*arctan2(sin(2*d*x + 2*c ), cos(2*d*x + 2*c) + 1)) - (4*a^2*cos(3*d*x + 3*c) + 5*a^2*cos(2*d*x + 2* c) + 4*a^2*cos(d*x + c) + 5*a^2)*sin(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d *x + 2*c) + 1)) - ((a^2*cos(d*x + c) - a^2)*cos(2*d*x + 2*c)^2 + a^2*cos(d *x + c) + (a^2*cos(d*x + c) - a^2)*sin(2*d*x + 2*c)^2 - a^2 + 2*(a^2*cos(d *x + c) - a^2)*cos(2*d*x + 2*c))*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d *x + 2*c) + 1)))*sqrt(a) + 5*((a^2*cos(2*d*x + 2*c)^2 + a^2*sin(2*d*x + 2* c)^2 + 2*a^2*cos(2*d*x + 2*c) + a^2)*arctan2(-(cos(2*d*x + 2*c)^2 + sin(2* d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*(cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))*sin(d*x + c) - cos(d*x + c)*sin(1/2*arctan2(s in(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))), (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*(cos(d*x + c)*cos(1/2*arctan2(sin (2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + sin(d*x + c)*sin(1/2*arctan2(sin(2 *d*x + 2*c), cos(2*d*x + 2*c) + 1))) + 1) - (a^2*cos(2*d*x + 2*c)^2 + a...
\[ \int \cos (c+d x) (a+a \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \cos \left (d x + c\right ) \,d x } \]
Timed out. \[ \int \cos (c+d x) (a+a \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\int \cos \left (c+d\,x\right )\,\left (A+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2} \,d x \]