Integrand size = 37, antiderivative size = 266 \[ \int \cos ^{\frac {11}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {16 a^2 (112 A+143 C) \sin (c+d x)}{1155 d \sqrt {\cos (c+d x)} \sqrt {a+a \sec (c+d x)}}+\frac {8 a^2 (112 A+143 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{1155 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (112 A+143 C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{385 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (28 A+33 C) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{231 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a A \cos ^{\frac {7}{2}}(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{33 d}+\frac {2 A \cos ^{\frac {9}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{11 d} \]
2/11*A*cos(d*x+c)^(9/2)*(a+a*sec(d*x+c))^(3/2)*sin(d*x+c)/d+2/385*a^2*(112 *A+143*C)*cos(d*x+c)^(3/2)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+2/231*a^2*( 28*A+33*C)*cos(d*x+c)^(5/2)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+16/1155*a^ 2*(112*A+143*C)*sin(d*x+c)/d/cos(d*x+c)^(1/2)/(a+a*sec(d*x+c))^(1/2)+8/115 5*a^2*(112*A+143*C)*sin(d*x+c)*cos(d*x+c)^(1/2)/d/(a+a*sec(d*x+c))^(1/2)+2 /33*a*A*cos(d*x+c)^(7/2)*sin(d*x+c)*(a+a*sec(d*x+c))^(1/2)/d
Time = 7.29 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.47 \[ \int \cos ^{\frac {11}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {a \sqrt {\cos (c+d x)} (18494 A+21736 C+2 (5789 A+5566 C) \cos (c+d x)+8 (581 A+429 C) \cos (2 (c+d x))+1645 A \cos (3 (c+d x))+660 C \cos (3 (c+d x))+490 A \cos (4 (c+d x))+105 A \cos (5 (c+d x))) \sqrt {a (1+\sec (c+d x))} \tan \left (\frac {1}{2} (c+d x)\right )}{9240 d} \]
(a*Sqrt[Cos[c + d*x]]*(18494*A + 21736*C + 2*(5789*A + 5566*C)*Cos[c + d*x ] + 8*(581*A + 429*C)*Cos[2*(c + d*x)] + 1645*A*Cos[3*(c + d*x)] + 660*C*C os[3*(c + d*x)] + 490*A*Cos[4*(c + d*x)] + 105*A*Cos[5*(c + d*x)])*Sqrt[a* (1 + Sec[c + d*x])]*Tan[(c + d*x)/2])/(9240*d)
Time = 1.70 (sec) , antiderivative size = 295, normalized size of antiderivative = 1.11, number of steps used = 16, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.432, Rules used = {3042, 4753, 3042, 4575, 27, 3042, 4505, 27, 3042, 4503, 3042, 4292, 3042, 4292, 3042, 4291}
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.
\(\displaystyle \int \cos ^{\frac {11}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2} \left (A+C \sec ^2(c+d x)\right ) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \cos (c+d x)^{11/2} (a \sec (c+d x)+a)^{3/2} \left (A+C \sec (c+d x)^2\right )dx\) |
\(\Big \downarrow \) 4753 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {(\sec (c+d x) a+a)^{3/2} \left (C \sec ^2(c+d x)+A\right )}{\sec ^{\frac {11}{2}}(c+d x)}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2} \left (C \csc \left (c+d x+\frac {\pi }{2}\right )^2+A\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{11/2}}dx\) |
\(\Big \downarrow \) 4575 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 \int \frac {(\sec (c+d x) a+a)^{3/2} (3 a A+a (6 A+11 C) \sec (c+d x))}{2 \sec ^{\frac {9}{2}}(c+d x)}dx}{11 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d \sec ^{\frac {9}{2}}(c+d x)}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {(\sec (c+d x) a+a)^{3/2} (3 a A+a (6 A+11 C) \sec (c+d x))}{\sec ^{\frac {9}{2}}(c+d x)}dx}{11 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d \sec ^{\frac {9}{2}}(c+d x)}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2} \left (3 a A+a (6 A+11 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{9/2}}dx}{11 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d \sec ^{\frac {9}{2}}(c+d x)}\right )\) |
\(\Big \downarrow \) 4505 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {2}{9} \int \frac {3 \sqrt {\sec (c+d x) a+a} \left ((28 A+33 C) a^2+3 (8 A+11 C) \sec (c+d x) a^2\right )}{2 \sec ^{\frac {7}{2}}(c+d x)}dx+\frac {2 a^2 A \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d \sec ^{\frac {7}{2}}(c+d x)}}{11 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d \sec ^{\frac {9}{2}}(c+d x)}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{3} \int \frac {\sqrt {\sec (c+d x) a+a} \left ((28 A+33 C) a^2+3 (8 A+11 C) \sec (c+d x) a^2\right )}{\sec ^{\frac {7}{2}}(c+d x)}dx+\frac {2 a^2 A \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d \sec ^{\frac {7}{2}}(c+d x)}}{11 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d \sec ^{\frac {9}{2}}(c+d x)}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{3} \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a} \left ((28 A+33 C) a^2+3 (8 A+11 C) \csc \left (c+d x+\frac {\pi }{2}\right ) a^2\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{7/2}}dx+\frac {2 a^2 A \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d \sec ^{\frac {7}{2}}(c+d x)}}{11 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d \sec ^{\frac {9}{2}}(c+d x)}\right )\) |
\(\Big \downarrow \) 4503 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{3} \left (\frac {3}{7} a^2 (112 A+143 C) \int \frac {\sqrt {\sec (c+d x) a+a}}{\sec ^{\frac {5}{2}}(c+d x)}dx+\frac {2 a^3 (28 A+33 C) \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a^2 A \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d \sec ^{\frac {7}{2}}(c+d x)}}{11 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d \sec ^{\frac {9}{2}}(c+d x)}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{3} \left (\frac {3}{7} a^2 (112 A+143 C) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\csc \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx+\frac {2 a^3 (28 A+33 C) \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a^2 A \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d \sec ^{\frac {7}{2}}(c+d x)}}{11 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d \sec ^{\frac {9}{2}}(c+d x)}\right )\) |
\(\Big \downarrow \) 4292 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{3} \left (\frac {3}{7} a^2 (112 A+143 C) \left (\frac {4}{5} \int \frac {\sqrt {\sec (c+d x) a+a}}{\sec ^{\frac {3}{2}}(c+d x)}dx+\frac {2 a \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a^3 (28 A+33 C) \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a^2 A \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d \sec ^{\frac {7}{2}}(c+d x)}}{11 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d \sec ^{\frac {9}{2}}(c+d x)}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{3} \left (\frac {3}{7} a^2 (112 A+143 C) \left (\frac {4}{5} \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx+\frac {2 a \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a^3 (28 A+33 C) \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a^2 A \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d \sec ^{\frac {7}{2}}(c+d x)}}{11 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d \sec ^{\frac {9}{2}}(c+d x)}\right )\) |
\(\Big \downarrow \) 4292 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{3} \left (\frac {3}{7} a^2 (112 A+143 C) \left (\frac {4}{5} \left (\frac {2}{3} \int \frac {\sqrt {\sec (c+d x) a+a}}{\sqrt {\sec (c+d x)}}dx+\frac {2 a \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a^3 (28 A+33 C) \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a^2 A \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d \sec ^{\frac {7}{2}}(c+d x)}}{11 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d \sec ^{\frac {9}{2}}(c+d x)}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{3} \left (\frac {3}{7} a^2 (112 A+143 C) \left (\frac {4}{5} \left (\frac {2}{3} \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 a \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a^3 (28 A+33 C) \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a^2 A \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d \sec ^{\frac {7}{2}}(c+d x)}}{11 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d \sec ^{\frac {9}{2}}(c+d x)}\right )\) |
\(\Big \downarrow \) 4291 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {2 a^2 A \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d \sec ^{\frac {7}{2}}(c+d x)}+\frac {1}{3} \left (\frac {2 a^3 (28 A+33 C) \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}+\frac {3}{7} a^2 (112 A+143 C) \left (\frac {2 a \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}+\frac {4}{5} \left (\frac {4 a \sin (c+d x) \sqrt {\sec (c+d x)}}{3 d \sqrt {a \sec (c+d x)+a}}+\frac {2 a \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}\right )\right )\right )}{11 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d \sec ^{\frac {9}{2}}(c+d x)}\right )\) |
Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((2*A*(a + a*Sec[c + d*x])^(3/2)*Sin [c + d*x])/(11*d*Sec[c + d*x]^(9/2)) + ((2*a^2*A*Sqrt[a + a*Sec[c + d*x]]* Sin[c + d*x])/(3*d*Sec[c + d*x]^(7/2)) + ((2*a^3*(28*A + 33*C)*Sin[c + d*x ])/(7*d*Sec[c + d*x]^(5/2)*Sqrt[a + a*Sec[c + d*x]]) + (3*a^2*(112*A + 143 *C)*((2*a*Sin[c + d*x])/(5*d*Sec[c + d*x]^(3/2)*Sqrt[a + a*Sec[c + d*x]]) + (4*((2*a*Sin[c + d*x])/(3*d*Sqrt[Sec[c + d*x]]*Sqrt[a + a*Sec[c + d*x]]) + (4*a*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(3*d*Sqrt[a + a*Sec[c + d*x]])))/ 5))/7)/3)/(11*a))
3.12.35.3.1 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[Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]/Sqrt[csc[(e_.) + (f_.)*(x_)] *(d_.)], x_Symbol] :> Simp[-2*a*(Cot[e + f*x]/(f*Sqrt[a + b*Csc[e + f*x]]*S qrt[d*Csc[e + f*x]])), x] /; FreeQ[{a, b, d, e, f}, 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]
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])
Int[(cos[(a_.) + (b_.)*(x_)]*(c_.))^(m_.)*(u_), x_Symbol] :> Simp[(c*Cos[a + b*x])^m*(c*Sec[a + b*x])^m Int[ActivateTrig[u]/(c*Sec[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] && !IntegerQ[m] && KnownSecantIntegrandQ[u, x ]
Time = 0.63 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.48
method | result | size |
default | \(-\frac {2 a \left (\left (105 \cos \left (d x +c \right )^{5}+245 \cos \left (d x +c \right )^{4}+280 \cos \left (d x +c \right )^{3}+336 \cos \left (d x +c \right )^{2}+448 \cos \left (d x +c \right )+896\right ) A +\left (165 \cos \left (d x +c \right )^{3}+429 \cos \left (d x +c \right )^{2}+572 \cos \left (d x +c \right )+1144\right ) C \right ) \sqrt {\cos \left (d x +c \right )}\, \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )}{1155 d}\) | \(127\) |
-2/1155*a/d*((105*cos(d*x+c)^5+245*cos(d*x+c)^4+280*cos(d*x+c)^3+336*cos(d *x+c)^2+448*cos(d*x+c)+896)*A+(165*cos(d*x+c)^3+429*cos(d*x+c)^2+572*cos(d *x+c)+1144)*C)*cos(d*x+c)^(1/2)*(a*(1+sec(d*x+c)))^(1/2)*(cot(d*x+c)-csc(d *x+c))
Time = 0.30 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.51 \[ \int \cos ^{\frac {11}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {2 \, {\left (105 \, A a \cos \left (d x + c\right )^{5} + 245 \, A a \cos \left (d x + c\right )^{4} + 5 \, {\left (56 \, A + 33 \, C\right )} a \cos \left (d x + c\right )^{3} + 3 \, {\left (112 \, A + 143 \, C\right )} a \cos \left (d x + c\right )^{2} + 4 \, {\left (112 \, A + 143 \, C\right )} a \cos \left (d x + c\right ) + 8 \, {\left (112 \, A + 143 \, C\right )} a\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{1155 \, {\left (d \cos \left (d x + c\right ) + d\right )}} \]
2/1155*(105*A*a*cos(d*x + c)^5 + 245*A*a*cos(d*x + c)^4 + 5*(56*A + 33*C)* a*cos(d*x + c)^3 + 3*(112*A + 143*C)*a*cos(d*x + c)^2 + 4*(112*A + 143*C)* a*cos(d*x + c) + 8*(112*A + 143*C)*a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c)/(d*cos(d*x + c) + d)
Timed out. \[ \int \cos ^{\frac {11}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \]
Leaf count of result is larger than twice the leaf count of optimal. 652 vs. \(2 (230) = 460\).
Time = 0.50 (sec) , antiderivative size = 652, normalized size of antiderivative = 2.45 \[ \int \cos ^{\frac {11}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\text {Too large to display} \]
1/36960*(7*sqrt(2)*(3630*a*cos(10/11*arctan2(sin(11/2*d*x + 11/2*c), cos(1 1/2*d*x + 11/2*c)))*sin(11/2*d*x + 11/2*c) + 990*a*cos(8/11*arctan2(sin(11 /2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c)))*sin(11/2*d*x + 11/2*c) + 429*a* cos(6/11*arctan2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c)))*sin(11/2 *d*x + 11/2*c) + 165*a*cos(4/11*arctan2(sin(11/2*d*x + 11/2*c), cos(11/2*d *x + 11/2*c)))*sin(11/2*d*x + 11/2*c) + 55*a*cos(2/11*arctan2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c)))*sin(11/2*d*x + 11/2*c) - 3630*a*cos(1 1/2*d*x + 11/2*c)*sin(10/11*arctan2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c))) - 990*a*cos(11/2*d*x + 11/2*c)*sin(8/11*arctan2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c))) - 429*a*cos(11/2*d*x + 11/2*c)*sin(6/11* arctan2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c))) - 165*a*cos(11/2* d*x + 11/2*c)*sin(4/11*arctan2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2 *c))) - 55*a*cos(11/2*d*x + 11/2*c)*sin(2/11*arctan2(sin(11/2*d*x + 11/2*c ), cos(11/2*d*x + 11/2*c))) + 30*a*sin(11/2*d*x + 11/2*c) + 55*a*sin(9/11* arctan2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c))) + 165*a*sin(7/11* arctan2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c))) + 429*a*sin(5/11* arctan2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c))) + 990*a*sin(3/11* arctan2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c))) + 3630*a*sin(1/11 *arctan2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c))))*A*sqrt(a) - 44* sqrt(2)*(175*a*cos(7/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))*sin...
\[ \int \cos ^{\frac {11}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \cos \left (d x + c\right )^{\frac {11}{2}} \,d x } \]
Timed out. \[ \int \cos ^{\frac {11}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\int {\cos \left (c+d\,x\right )}^{11/2}\,\left (A+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{3/2} \,d x \]