Integrand size = 33, antiderivative size = 209 \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\frac {a^{5/2} (163 A+200 B) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{64 d}+\frac {a^3 (163 A+200 B) \sin (c+d x)}{64 d \sqrt {a+a \sec (c+d x)}}+\frac {a^3 (95 A+104 B) \cos (c+d x) \sin (c+d x)}{96 d \sqrt {a+a \sec (c+d x)}}+\frac {a^2 (11 A+8 B) \cos ^2(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{24 d}+\frac {a A \cos ^3(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{4 d} \]
1/64*a^(5/2)*(163*A+200*B)*arctan(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2 ))/d+1/4*a*A*cos(d*x+c)^3*(a+a*sec(d*x+c))^(3/2)*sin(d*x+c)/d+1/64*a^3*(16 3*A+200*B)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+1/96*a^3*(95*A+104*B)*cos(d *x+c)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+1/24*a^2*(11*A+8*B)*cos(d*x+c)^2 *sin(d*x+c)*(a+a*sec(d*x+c))^(1/2)/d
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.91 (sec) , antiderivative size = 366, normalized size of antiderivative = 1.75 \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\frac {a^2 \left (6075 A \text {arctanh}\left (\sqrt {1-\sec (c+d x)}\right )+6600 B \text {arctanh}\left (\sqrt {1-\sec (c+d x)}\right )+2079 A \sqrt {1-\sec (c+d x)}+1240 B \sqrt {1-\sec (c+d x)}+7641 A \cos (c+d x) \sqrt {1-\sec (c+d x)}+6360 B \cos (c+d x) \sqrt {1-\sec (c+d x)}+2097 A \cos (2 (c+d x)) \sqrt {1-\sec (c+d x)}+1240 B \cos (2 (c+d x)) \sqrt {1-\sec (c+d x)}+522 A \cos (3 (c+d x)) \sqrt {1-\sec (c+d x)}-80 B \cos (3 (c+d x)) \sqrt {1-\sec (c+d x)}+18 A \cos (4 (c+d x)) \sqrt {1-\sec (c+d x)}+7680 B \operatorname {Hypergeometric2F1}\left (\frac {1}{2},4,\frac {3}{2},1-\sec (c+d x)\right ) \sqrt {1-\sec (c+d x)}+4608 A \operatorname {Hypergeometric2F1}\left (\frac {1}{2},5,\frac {3}{2},1-\sec (c+d x)\right ) \sqrt {1-\sec (c+d x)}\right ) \sqrt {a (1+\sec (c+d x))} \sin (c+d x)}{2880 d (1+\cos (c+d x)) \sqrt {1-\sec (c+d x)}} \]
(a^2*(6075*A*ArcTanh[Sqrt[1 - Sec[c + d*x]]] + 6600*B*ArcTanh[Sqrt[1 - Sec [c + d*x]]] + 2079*A*Sqrt[1 - Sec[c + d*x]] + 1240*B*Sqrt[1 - Sec[c + d*x] ] + 7641*A*Cos[c + d*x]*Sqrt[1 - Sec[c + d*x]] + 6360*B*Cos[c + d*x]*Sqrt[ 1 - Sec[c + d*x]] + 2097*A*Cos[2*(c + d*x)]*Sqrt[1 - Sec[c + d*x]] + 1240* B*Cos[2*(c + d*x)]*Sqrt[1 - Sec[c + d*x]] + 522*A*Cos[3*(c + d*x)]*Sqrt[1 - Sec[c + d*x]] - 80*B*Cos[3*(c + d*x)]*Sqrt[1 - Sec[c + d*x]] + 18*A*Cos[ 4*(c + d*x)]*Sqrt[1 - Sec[c + d*x]] + 7680*B*Hypergeometric2F1[1/2, 4, 3/2 , 1 - Sec[c + d*x]]*Sqrt[1 - Sec[c + d*x]] + 4608*A*Hypergeometric2F1[1/2, 5, 3/2, 1 - Sec[c + d*x]]*Sqrt[1 - Sec[c + d*x]])*Sqrt[a*(1 + Sec[c + d*x ])]*Sin[c + d*x])/(2880*d*(1 + Cos[c + d*x])*Sqrt[1 - Sec[c + d*x]])
Time = 1.26 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.01, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.394, Rules used = {3042, 4505, 27, 3042, 4505, 27, 3042, 4503, 3042, 4292, 3042, 4261, 216}
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.
| \(\displaystyle \int \cos ^4(c+d x) (a \sec (c+d x)+a)^{5/2} (A+B \sec (c+d x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{5/2} \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^4}dx\) |
| \(\Big \downarrow \) 4505 |
| \(\displaystyle \frac {1}{4} \int \frac {1}{2} \cos ^3(c+d x) (\sec (c+d x) a+a)^{3/2} (a (11 A+8 B)+a (3 A+8 B) \sec (c+d x))dx+\frac {a A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{4 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{8} \int \cos ^3(c+d x) (\sec (c+d x) a+a)^{3/2} (a (11 A+8 B)+a (3 A+8 B) \sec (c+d x))dx+\frac {a A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{8} \int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2} \left (a (11 A+8 B)+a (3 A+8 B) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^3}dx+\frac {a A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{4 d}\) |
| \(\Big \downarrow \) 4505 |
| \(\displaystyle \frac {1}{8} \left (\frac {1}{3} \int \frac {1}{2} \cos ^2(c+d x) \sqrt {\sec (c+d x) a+a} \left ((95 A+104 B) a^2+3 (17 A+24 B) \sec (c+d x) a^2\right )dx+\frac {a^2 (11 A+8 B) \sin (c+d x) \cos ^2(c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\right )+\frac {a A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{4 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{8} \left (\frac {1}{6} \int \cos ^2(c+d x) \sqrt {\sec (c+d x) a+a} \left ((95 A+104 B) a^2+3 (17 A+24 B) \sec (c+d x) a^2\right )dx+\frac {a^2 (11 A+8 B) \sin (c+d x) \cos ^2(c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\right )+\frac {a A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{8} \left (\frac {1}{6} \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a} \left ((95 A+104 B) a^2+3 (17 A+24 B) \csc \left (c+d x+\frac {\pi }{2}\right ) a^2\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^2}dx+\frac {a^2 (11 A+8 B) \sin (c+d x) \cos ^2(c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\right )+\frac {a A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{4 d}\) |
| \(\Big \downarrow \) 4503 |
| \(\displaystyle \frac {1}{8} \left (\frac {1}{6} \left (\frac {3}{4} a^2 (163 A+200 B) \int \cos (c+d x) \sqrt {\sec (c+d x) a+a}dx+\frac {a^3 (95 A+104 B) \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^2 (11 A+8 B) \sin (c+d x) \cos ^2(c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\right )+\frac {a A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{8} \left (\frac {1}{6} \left (\frac {3}{4} a^2 (163 A+200 B) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\csc \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {a^3 (95 A+104 B) \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^2 (11 A+8 B) \sin (c+d x) \cos ^2(c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\right )+\frac {a A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{4 d}\) |
| \(\Big \downarrow \) 4292 |
| \(\displaystyle \frac {1}{8} \left (\frac {1}{6} \left (\frac {3}{4} a^2 (163 A+200 B) \left (\frac {1}{2} \int \sqrt {\sec (c+d x) a+a}dx+\frac {a \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^3 (95 A+104 B) \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^2 (11 A+8 B) \sin (c+d x) \cos ^2(c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\right )+\frac {a A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{8} \left (\frac {1}{6} \left (\frac {3}{4} a^2 (163 A+200 B) \left (\frac {1}{2} \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {a \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^3 (95 A+104 B) \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^2 (11 A+8 B) \sin (c+d x) \cos ^2(c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\right )+\frac {a A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{4 d}\) |
| \(\Big \downarrow \) 4261 |
| \(\displaystyle \frac {1}{8} \left (\frac {1}{6} \left (\frac {3}{4} a^2 (163 A+200 B) \left (\frac {a \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {a \int \frac {1}{\frac {a^2 \tan ^2(c+d x)}{\sec (c+d x) a+a}+a}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}\right )+\frac {a^3 (95 A+104 B) \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^2 (11 A+8 B) \sin (c+d x) \cos ^2(c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\right )+\frac {a A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{4 d}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {1}{8} \left (\frac {a^2 (11 A+8 B) \sin (c+d x) \cos ^2(c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}+\frac {1}{6} \left (\frac {a^3 (95 A+104 B) \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}+\frac {3}{4} a^2 (163 A+200 B) \left (\frac {\sqrt {a} \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}+\frac {a \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}\right )\right )\right )+\frac {a A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{4 d}\) |
(a*A*Cos[c + d*x]^3*(a + a*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(4*d) + ((a^2 *(11*A + 8*B)*Cos[c + d*x]^2*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(3*d) + ((a^3*(95*A + 104*B)*Cos[c + d*x]*Sin[c + d*x])/(2*d*Sqrt[a + a*Sec[c + d*x]]) + (3*a^2*(163*A + 200*B)*((Sqrt[a]*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sq rt[a + a*Sec[c + d*x]]])/d + (a*Sin[c + d*x])/(d*Sqrt[a + a*Sec[c + d*x]]) ))/4)/6)/8
3.2.42.3.1 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_)]*(d_.))^(n_)*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[a*Cot[e + f*x]*((d*Csc[e + f*x])^n/(f*n*Sqrt[a + b*Csc[e + f*x]])), x] + Simp[a*((2*n + 1)/(2*b*d*n)) Int[Sqrt[a + b*Csc [e + f*x]]*(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && LtQ[n, -2^(-1)] && IntegerQ[2*n]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[A*b^2*Co t[e + f*x]*((d*Csc[e + f*x])^n/(a*f*n*Sqrt[a + b*Csc[e + f*x]])), x] + Simp [(A*b*(2*n + 1) + 2*a*B*n)/(2*a*d*n) Int[Sqrt[a + b*Csc[e + f*x]]*(d*Csc[ e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a *B, 0] && EqQ[a^2 - b^2, 0] && NeQ[A*b*(2*n + 1) + 2*a*B*n, 0] && LtQ[n, 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[a*A*Cot [e + f*x]*(a + b*Csc[e + f*x])^(m - 1)*((d*Csc[e + f*x])^n/(f*n)), x] - Sim p[b/(a*d*n) Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^(n + 1)*Sim p[a*A*(m - n - 1) - b*B*n - (a*B*n + A*b*(m + n))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0 ] && GtQ[m, 1/2] && LtQ[n, -1]
Leaf count of result is larger than twice the leaf count of optimal. \(402\) vs. \(2(185)=370\).
Time = 2.12 (sec) , antiderivative size = 403, normalized size of antiderivative = 1.93
\[\frac {a^{2} \left (48 A \cos \left (d x +c \right )^{4} \sin \left (d x +c \right )+184 A \sin \left (d x +c \right ) \cos \left (d x +c \right )^{3}+64 B \cos \left (d x +c \right )^{3} \sin \left (d x +c \right )+489 A \,\operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \cos \left (d x +c \right )+326 A \cos \left (d x +c \right )^{2} \sin \left (d x +c \right )+600 B \,\operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \cos \left (d x +c \right )+272 B \sin \left (d x +c \right ) \cos \left (d x +c \right )^{2}+489 A \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right )+489 A \cos \left (d x +c \right ) \sin \left (d x +c \right )+600 B \,\operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}+600 B \cos \left (d x +c \right ) \sin \left (d x +c \right )\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{192 d \left (\cos \left (d x +c \right )+1\right )}\]
1/192*a^2/d*(48*A*cos(d*x+c)^4*sin(d*x+c)+184*A*sin(d*x+c)*cos(d*x+c)^3+64 *B*cos(d*x+c)^3*sin(d*x+c)+489*A*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)/(cos(d*x+c)+1))^(1/2)*cos(d*x+c) +326*A*cos(d*x+c)^2*sin(d*x+c)+600*B*arctanh(sin(d*x+c)/(cos(d*x+c)+1)/(-c os(d*x+c)/(cos(d*x+c)+1))^(1/2))*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*cos(d* x+c)+272*B*sin(d*x+c)*cos(d*x+c)^2+489*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))+48 9*A*cos(d*x+c)*sin(d*x+c)+600*B*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)/(cos(d*x+c)+1))^(1/2)+600*B*cos(d *x+c)*sin(d*x+c))*(a*(1+sec(d*x+c)))^(1/2)/(cos(d*x+c)+1)
Time = 0.32 (sec) , antiderivative size = 420, normalized size of antiderivative = 2.01 \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\left [\frac {3 \, {\left ({\left (163 \, A + 200 \, B\right )} a^{2} \cos \left (d x + c\right ) + {\left (163 \, A + 200 \, B\right )} a^{2}\right )} \sqrt {-a} \log \left (\frac {2 \, a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right ) + 1}\right ) + 2 \, {\left (48 \, A a^{2} \cos \left (d x + c\right )^{4} + 8 \, {\left (23 \, A + 8 \, B\right )} a^{2} \cos \left (d x + c\right )^{3} + 2 \, {\left (163 \, A + 136 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} + 3 \, {\left (163 \, A + 200 \, B\right )} a^{2} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{384 \, {\left (d \cos \left (d x + c\right ) + d\right )}}, -\frac {3 \, {\left ({\left (163 \, A + 200 \, B\right )} a^{2} \cos \left (d x + c\right ) + {\left (163 \, A + 200 \, B\right )} a^{2}\right )} \sqrt {a} \arctan \left (\frac {\sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) - {\left (48 \, A a^{2} \cos \left (d x + c\right )^{4} + 8 \, {\left (23 \, A + 8 \, B\right )} a^{2} \cos \left (d x + c\right )^{3} + 2 \, {\left (163 \, A + 136 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} + 3 \, {\left (163 \, A + 200 \, B\right )} a^{2} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{192 \, {\left (d \cos \left (d x + c\right ) + d\right )}}\right ] \]
[1/384*(3*((163*A + 200*B)*a^2*cos(d*x + c) + (163*A + 200*B)*a^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*(48*A*a^2*cos(d*x + c)^4 + 8*(23*A + 8*B)*a^2*cos(d*x + c)^3 + 2*(163*A + 136*B)*a^2*cos(d*x + c)^2 + 3*(163*A + 200*B)*a^2*cos(d*x + c))*sqrt((a *cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c) + d), -1/19 2*(3*((163*A + 200*B)*a^2*cos(d*x + c) + (163*A + 200*B)*a^2)*sqrt(a)*arct an(sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c))) - (48*A*a^2*cos(d*x + c)^4 + 8*(23*A + 8*B)*a^2*cos(d*x + c)^3 + 2*( 163*A + 136*B)*a^2*cos(d*x + c)^2 + 3*(163*A + 200*B)*a^2*cos(d*x + c))*sq rt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c) + d)]
Timed out. \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\text {Timed out} \]
Timed out. \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\text {Timed out} \]
\[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\int { {\left (B \sec \left (d x + c\right ) + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \cos \left (d x + c\right )^{4} \,d x } \]
Timed out. \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\int {\cos \left (c+d\,x\right )}^4\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}\right )\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2} \,d x \]