Integrand size = 35, antiderivative size = 182 \[ \int (a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {2 a^{5/2} A \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{d}+\frac {2 a^3 (245 A+224 B+160 C) \tan (c+d x)}{105 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (35 A+56 B+40 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{105 d}+\frac {2 a (7 B+5 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{35 d}+\frac {2 C (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{7 d} \]
2*a^(5/2)*A*arctan(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/d+2/35*a*(7* B+5*C)*(a+a*sec(d*x+c))^(3/2)*tan(d*x+c)/d+2/7*C*(a+a*sec(d*x+c))^(5/2)*ta n(d*x+c)/d+2/105*a^3*(245*A+224*B+160*C)*tan(d*x+c)/d/(a+a*sec(d*x+c))^(1/ 2)+2/105*a^2*(35*A+56*B+40*C)*(a+a*sec(d*x+c))^(1/2)*tan(d*x+c)/d
Time = 3.15 (sec) , antiderivative size = 170, normalized size of antiderivative = 0.93 \[ \int (a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {a^2 \sec \left (\frac {1}{2} (c+d x)\right ) \sec ^3(c+d x) \sqrt {a (1+\sec (c+d x))} \left (420 \sqrt {2} A \arcsin \left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right ) \cos ^{\frac {7}{2}}(c+d x)+2 (70 A+196 B+290 C+(840 A+987 B+930 C) \cos (c+d x)+2 (35 A+98 B+115 C) \cos (2 (c+d x))+280 A \cos (3 (c+d x))+301 B \cos (3 (c+d x))+230 C \cos (3 (c+d x))) \sin \left (\frac {1}{2} (c+d x)\right )\right )}{420 d} \]
(a^2*Sec[(c + d*x)/2]*Sec[c + d*x]^3*Sqrt[a*(1 + Sec[c + d*x])]*(420*Sqrt[ 2]*A*ArcSin[Sqrt[2]*Sin[(c + d*x)/2]]*Cos[c + d*x]^(7/2) + 2*(70*A + 196*B + 290*C + (840*A + 987*B + 930*C)*Cos[c + d*x] + 2*(35*A + 98*B + 115*C)* Cos[2*(c + d*x)] + 280*A*Cos[3*(c + d*x)] + 301*B*Cos[3*(c + d*x)] + 230*C *Cos[3*(c + d*x)])*Sin[(c + d*x)/2]))/(420*d)
Time = 1.10 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.10, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {3042, 4542, 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.
| \(\displaystyle \int (a \sec (c+d x)+a)^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{5/2} \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )dx\) |
| \(\Big \downarrow \) 4542 |
| \(\displaystyle \frac {2 \int \frac {1}{2} (\sec (c+d x) a+a)^{5/2} (7 a A+a (7 B+5 C) \sec (c+d x))dx}{7 a}+\frac {2 C \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int (\sec (c+d x) a+a)^{5/2} (7 a A+a (7 B+5 C) \sec (c+d x))dx}{7 a}+\frac {2 C \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{5/2} \left (7 a A+a (7 B+5 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx}{7 a}+\frac {2 C \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 d}\) |
| \(\Big \downarrow \) 4405 |
| \(\displaystyle \frac {\frac {2}{5} \int \frac {1}{2} (\sec (c+d x) a+a)^{3/2} \left (35 A a^2+(35 A+56 B+40 C) \sec (c+d x) a^2\right )dx+\frac {2 a^2 (7 B+5 C) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}}{7 a}+\frac {2 C \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {1}{5} \int (\sec (c+d x) a+a)^{3/2} \left (35 A a^2+(35 A+56 B+40 C) \sec (c+d x) a^2\right )dx+\frac {2 a^2 (7 B+5 C) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}}{7 a}+\frac {2 C \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{5} \int \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2} \left (35 A a^2+(35 A+56 B+40 C) \csc \left (c+d x+\frac {\pi }{2}\right ) a^2\right )dx+\frac {2 a^2 (7 B+5 C) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}}{7 a}+\frac {2 C \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 d}\) |
| \(\Big \downarrow \) 4405 |
| \(\displaystyle \frac {\frac {1}{5} \left (\frac {2}{3} \int \frac {1}{2} \sqrt {\sec (c+d x) a+a} \left (105 A a^3+(245 A+224 B+160 C) \sec (c+d x) a^3\right )dx+\frac {2 a^3 (35 A+56 B+40 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\right )+\frac {2 a^2 (7 B+5 C) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}}{7 a}+\frac {2 C \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {1}{5} \left (\frac {1}{3} \int \sqrt {\sec (c+d x) a+a} \left (105 A a^3+(245 A+224 B+160 C) \sec (c+d x) a^3\right )dx+\frac {2 a^3 (35 A+56 B+40 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\right )+\frac {2 a^2 (7 B+5 C) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}}{7 a}+\frac {2 C \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{5} \left (\frac {1}{3} \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a} \left (105 A a^3+(245 A+224 B+160 C) \csc \left (c+d x+\frac {\pi }{2}\right ) a^3\right )dx+\frac {2 a^3 (35 A+56 B+40 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\right )+\frac {2 a^2 (7 B+5 C) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}}{7 a}+\frac {2 C \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 d}\) |
| \(\Big \downarrow \) 4403 |
| \(\displaystyle \frac {\frac {1}{5} \left (\frac {1}{3} \left (a^3 (245 A+224 B+160 C) \int \sec (c+d x) \sqrt {\sec (c+d x) a+a}dx+105 a^3 A \int \sqrt {\sec (c+d x) a+a}dx\right )+\frac {2 a^3 (35 A+56 B+40 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\right )+\frac {2 a^2 (7 B+5 C) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}}{7 a}+\frac {2 C \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{5} \left (\frac {1}{3} \left (a^3 (245 A+224 B+160 C) \int \csc \left (c+d x+\frac {\pi }{2}\right ) \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+105 a^3 A \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx\right )+\frac {2 a^3 (35 A+56 B+40 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\right )+\frac {2 a^2 (7 B+5 C) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}}{7 a}+\frac {2 C \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 d}\) |
| \(\Big \downarrow \) 4261 |
| \(\displaystyle \frac {\frac {1}{5} \left (\frac {1}{3} \left (a^3 (245 A+224 B+160 C) \int \csc \left (c+d x+\frac {\pi }{2}\right ) \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx-\frac {210 a^4 A \int \frac {1}{\frac {a^2 \tan ^2(c+d x)}{\sec (c+d x) a+a}+a}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}\right )+\frac {2 a^3 (35 A+56 B+40 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\right )+\frac {2 a^2 (7 B+5 C) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}}{7 a}+\frac {2 C \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 d}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {\frac {1}{5} \left (\frac {1}{3} \left (a^3 (245 A+224 B+160 C) \int \csc \left (c+d x+\frac {\pi }{2}\right ) \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {210 a^{7/2} A \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}\right )+\frac {2 a^3 (35 A+56 B+40 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\right )+\frac {2 a^2 (7 B+5 C) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}}{7 a}+\frac {2 C \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 d}\) |
| \(\Big \downarrow \) 4279 |
| \(\displaystyle \frac {\frac {2 a^2 (7 B+5 C) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}+\frac {1}{5} \left (\frac {2 a^3 (35 A+56 B+40 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}+\frac {1}{3} \left (\frac {210 a^{7/2} A \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}+\frac {2 a^4 (245 A+224 B+160 C) \tan (c+d x)}{d \sqrt {a \sec (c+d x)+a}}\right )\right )}{7 a}+\frac {2 C \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 d}\) |
(2*C*(a + a*Sec[c + d*x])^(5/2)*Tan[c + d*x])/(7*d) + ((2*a^2*(7*B + 5*C)* (a + a*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(5*d) + ((2*a^3*(35*A + 56*B + 40 *C)*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(3*d) + ((210*a^(7/2)*A*ArcTan[ (Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/d + (2*a^4*(245*A + 224* B + 160*C)*Tan[c + d*x])/(d*Sqrt[a + a*Sec[c + d*x]]))/3)/5)/(7*a)
3.6.3.3.1 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_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. ))*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Simp[(-C)*Cot [e + f*x]*((a + b*Csc[e + f*x])^m/(f*(m + 1))), x] + Simp[1/(b*(m + 1)) I nt[(a + b*Csc[e + f*x])^m*Simp[A*b*(m + 1) + (a*C*m + b*B*(m + 1))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && EqQ[a^2 - b^2, 0] && !LtQ[m, -2^(-1)]
Time = 10.97 (sec) , antiderivative size = 261, normalized size of antiderivative = 1.43
| method | result | size |
| default | \(\frac {2 a^{2} \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (105 A \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right ) \cos \left (d x +c \right )+105 A \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right )+280 A \sin \left (d x +c \right )+301 B \sin \left (d x +c \right )+230 C \sin \left (d x +c \right )+35 A \tan \left (d x +c \right )+98 B \tan \left (d x +c \right )+115 C \tan \left (d x +c \right )+21 B \tan \left (d x +c \right ) \sec \left (d x +c \right )+60 C \sec \left (d x +c \right ) \tan \left (d x +c \right )+15 C \tan \left (d x +c \right ) \sec \left (d x +c \right )^{2}\right )}{105 d \left (\cos \left (d x +c \right )+1\right )}\) | \(261\) |
| parts | \(\frac {2 A \,a^{2} \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (3 \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right ) \cos \left (d x +c \right )+3 \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right )+8 \sin \left (d x +c \right )+\tan \left (d x +c \right )\right )}{3 d \left (\cos \left (d x +c \right )+1\right )}+\frac {2 B \,a^{2} \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (43 \sin \left (d x +c \right )+14 \tan \left (d x +c \right )+3 \sec \left (d x +c \right ) \tan \left (d x +c \right )\right )}{15 d \left (\cos \left (d x +c \right )+1\right )}+\frac {2 C \,a^{2} \left (46 \cos \left (d x +c \right )^{3}+23 \cos \left (d x +c \right )^{2}+12 \cos \left (d x +c \right )+3\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right ) \sec \left (d x +c \right )^{2}}{21 d \left (\cos \left (d x +c \right )+1\right )}\) | \(311\) |
2/105*a^2/d*(a*(1+sec(d*x+c)))^(1/2)/(cos(d*x+c)+1)*(105*A*(-cos(d*x+c)/(c os(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)+105*A*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctan h(sin(d*x+c)/(cos(d*x+c)+1)/(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2))+280*A*sin( d*x+c)+301*B*sin(d*x+c)+230*C*sin(d*x+c)+35*A*tan(d*x+c)+98*B*tan(d*x+c)+1 15*C*tan(d*x+c)+21*B*tan(d*x+c)*sec(d*x+c)+60*C*sec(d*x+c)*tan(d*x+c)+15*C *tan(d*x+c)*sec(d*x+c)^2)
Time = 0.29 (sec) , antiderivative size = 430, normalized size of antiderivative = 2.36 \[ \int (a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\left [\frac {105 \, {\left (A a^{2} \cos \left (d x + c\right )^{4} + A a^{2} \cos \left (d x + c\right )^{3}\right )} \sqrt {-a} \log \left (\frac {2 \, a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right ) + 1}\right ) + 2 \, {\left ({\left (280 \, A + 301 \, B + 230 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + {\left (35 \, A + 98 \, B + 115 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 3 \, {\left (7 \, B + 20 \, C\right )} a^{2} \cos \left (d x + c\right ) + 15 \, C a^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{105 \, {\left (d \cos \left (d x + c\right )^{4} + d \cos \left (d x + c\right )^{3}\right )}}, -\frac {2 \, {\left (105 \, {\left (A a^{2} \cos \left (d x + c\right )^{4} + A a^{2} \cos \left (d x + c\right )^{3}\right )} \sqrt {a} \arctan \left (\frac {\sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) - {\left ({\left (280 \, A + 301 \, B + 230 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + {\left (35 \, A + 98 \, B + 115 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 3 \, {\left (7 \, B + 20 \, C\right )} a^{2} \cos \left (d x + c\right ) + 15 \, C a^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )\right )}}{105 \, {\left (d \cos \left (d x + c\right )^{4} + d \cos \left (d x + c\right )^{3}\right )}}\right ] \]
[1/105*(105*(A*a^2*cos(d*x + c)^4 + A*a^2*cos(d*x + c)^3)*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*((280* A + 301*B + 230*C)*a^2*cos(d*x + c)^3 + (35*A + 98*B + 115*C)*a^2*cos(d*x + c)^2 + 3*(7*B + 20*C)*a^2*cos(d*x + c) + 15*C*a^2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^4 + d*cos(d*x + c)^3), -2 /105*(105*(A*a^2*cos(d*x + c)^4 + A*a^2*cos(d*x + c)^3)*sqrt(a)*arctan(sqr t((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c))) - ((280*A + 301*B + 230*C)*a^2*cos(d*x + c)^3 + (35*A + 98*B + 115*C)*a^2* cos(d*x + c)^2 + 3*(7*B + 20*C)*a^2*cos(d*x + c) + 15*C*a^2)*sqrt((a*cos(d *x + c) + a)/cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^4 + d*cos(d*x + c )^3)]
\[ \int (a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int \left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {5}{2}} \left (A + B \sec {\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right )\, dx \]
\[ \int (a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \,d x } \]
-1/210*(105*((A*a^2*cos(2*d*x + 2*c)^2 + A*a^2*sin(2*d*x + 2*c)^2 + 2*A*a^ 2*cos(2*d*x + 2*c) + A*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)*sin(1/2*arctan2(sin(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(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + 1) - (A*a^2*cos(2*d*x + 2*c)^2 + A*a^2*sin(2*d*x + 2*c)^2 + 2*A*a^2*cos (2*d*x + 2*c) + A*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)*sin(1/2*arctan2(sin(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(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) - 1) - 2*(A*a^2*d*cos(2*d*x + 2*c)^2 + A*a^2*d*sin(2*d*x + 2*c)^2 + 2*A*a^2*d* cos(2*d*x + 2*c) + A*a^2*d)*integrate((((cos(6*d*x + 6*c)*cos(2*d*x + 2*c) + 2*cos(4*d*x + 4*c)*cos(2*d*x + 2*c) + cos(2*d*x + 2*c)^2 + sin(6*d*x + 6*c)*sin(2*d*x + 2*c) + 2*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + sin(2*d*x + 2*c)^2)*cos(9/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + (cos(2*d*x + 2*c)*sin(6*d*x + 6*c) + 2*cos(2*d*x + 2*c)*sin(4*d*x + 4*c) - cos(6*d*x + 6*c)*sin(2*d*x + 2*c) - 2*cos(4*d*x + 4*c)*sin(2*d*x + 2*c))*sin(9/2*arc tan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))))*cos(5/2*arctan2(sin(2*d*x + 2*c ), cos(2*d*x + 2*c) + 1)) - ((cos(2*d*x + 2*c)*sin(6*d*x + 6*c) + 2*cos(2* d*x + 2*c)*sin(4*d*x + 4*c) - cos(6*d*x + 6*c)*sin(2*d*x + 2*c) - 2*cos...
\[ \int (a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \,d x } \]
Timed out. \[ \int (a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int {\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right ) \,d x \]