Integrand size = 45, antiderivative size = 243 \[ \int (a+a \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x) \, dx=\frac {a^{5/2} (2 B+5 C) \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{d}-\frac {a^3 (64 A+70 B+15 C) \sin (c+d x)}{15 d \sqrt {a+a \cos (c+d x)} \sqrt {\sec (c+d x)}}+\frac {2 a^2 (8 A+10 B+5 C) \sqrt {a+a \cos (c+d x)} \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 a (A+B) (a+a \cos (c+d x))^{3/2} \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 d}+\frac {2 A (a+a \cos (c+d x))^{5/2} \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d} \]
2/3*a*(A+B)*(a+a*cos(d*x+c))^(3/2)*sec(d*x+c)^(3/2)*sin(d*x+c)/d+2/5*A*(a+ a*cos(d*x+c))^(5/2)*sec(d*x+c)^(5/2)*sin(d*x+c)/d-1/15*a^3*(64*A+70*B+15*C )*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)/sec(d*x+c)^(1/2)+a^(5/2)*(2*B+5*C)*a rcsin(sin(d*x+c)*a^(1/2)/(a+a*cos(d*x+c))^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+ c)^(1/2)/d+2/5*a^2*(8*A+10*B+5*C)*sin(d*x+c)*(a+a*cos(d*x+c))^(1/2)*sec(d* x+c)^(1/2)/d
Time = 0.98 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.64 \[ \int (a+a \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x) \, dx=\frac {a^2 \sqrt {a (1+\cos (c+d x))} \sec \left (\frac {1}{2} (c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x) \left (60 \sqrt {2} (2 B+5 C) \arcsin \left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right ) \cos ^{\frac {5}{2}}(c+d x)+2 (196 A+160 B+60 C+(112 A+40 B+45 C) \cos (c+d x)+4 (43 A+40 B+15 C) \cos (2 (c+d x))+15 C \cos (3 (c+d x))) \sin \left (\frac {1}{2} (c+d x)\right )\right )}{120 d} \]
Integrate[(a + a*Cos[c + d*x])^(5/2)*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^ 2)*Sec[c + d*x]^(7/2),x]
(a^2*Sqrt[a*(1 + Cos[c + d*x])]*Sec[(c + d*x)/2]*Sec[c + d*x]^(5/2)*(60*Sq rt[2]*(2*B + 5*C)*ArcSin[Sqrt[2]*Sin[(c + d*x)/2]]*Cos[c + d*x]^(5/2) + 2* (196*A + 160*B + 60*C + (112*A + 40*B + 45*C)*Cos[c + d*x] + 4*(43*A + 40* B + 15*C)*Cos[2*(c + d*x)] + 15*C*Cos[3*(c + d*x)])*Sin[(c + d*x)/2]))/(12 0*d)
Time = 1.71 (sec) , antiderivative size = 256, normalized size of antiderivative = 1.05, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.356, Rules used = {3042, 4709, 3042, 3522, 27, 3042, 3454, 27, 3042, 3454, 27, 3042, 3460, 3042, 3253, 223}
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.
\(\displaystyle \int \sec ^{\frac {7}{2}}(c+d x) (a \cos (c+d x)+a)^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sec (c+d x)^{7/2} (a \cos (c+d x)+a)^{5/2} \left (A+B \cos (c+d x)+C \cos (c+d x)^2\right )dx\) |
\(\Big \downarrow \) 4709 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {(\cos (c+d x) a+a)^{5/2} \left (C \cos ^2(c+d x)+B \cos (c+d x)+A\right )}{\cos ^{\frac {7}{2}}(c+d x)}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{5/2} \left (C \sin \left (c+d x+\frac {\pi }{2}\right )^2+B \sin \left (c+d x+\frac {\pi }{2}\right )+A\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{7/2}}dx\) |
\(\Big \downarrow \) 3522 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 \int \frac {(\cos (c+d x) a+a)^{5/2} (5 a (A+B)-a (2 A-5 C) \cos (c+d x))}{2 \cos ^{\frac {5}{2}}(c+d x)}dx}{5 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {(\cos (c+d x) a+a)^{5/2} (5 a (A+B)-a (2 A-5 C) \cos (c+d x))}{\cos ^{\frac {5}{2}}(c+d x)}dx}{5 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{5/2} \left (5 a (A+B)-a (2 A-5 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx}{5 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )\) |
\(\Big \downarrow \) 3454 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {2}{3} \int \frac {(\cos (c+d x) a+a)^{3/2} \left (3 a^2 (8 A+10 B+5 C)-a^2 (16 A+10 B-15 C) \cos (c+d x)\right )}{2 \cos ^{\frac {3}{2}}(c+d x)}dx+\frac {10 a^2 (A+B) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{3 d \cos ^{\frac {3}{2}}(c+d x)}}{5 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{3} \int \frac {(\cos (c+d x) a+a)^{3/2} \left (3 a^2 (8 A+10 B+5 C)-a^2 (16 A+10 B-15 C) \cos (c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)}dx+\frac {10 a^2 (A+B) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{3 d \cos ^{\frac {3}{2}}(c+d x)}}{5 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{3} \int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2} \left (3 a^2 (8 A+10 B+5 C)-a^2 (16 A+10 B-15 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx+\frac {10 a^2 (A+B) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{3 d \cos ^{\frac {3}{2}}(c+d x)}}{5 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )\) |
\(\Big \downarrow \) 3454 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{3} \left (2 \int \frac {\sqrt {\cos (c+d x) a+a} \left (a^3 (32 A+50 B+45 C)-a^3 (64 A+70 B+15 C) \cos (c+d x)\right )}{2 \sqrt {\cos (c+d x)}}dx+\frac {6 a^3 (8 A+10 B+5 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{d \sqrt {\cos (c+d x)}}\right )+\frac {10 a^2 (A+B) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{3 d \cos ^{\frac {3}{2}}(c+d x)}}{5 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{3} \left (\int \frac {\sqrt {\cos (c+d x) a+a} \left (a^3 (32 A+50 B+45 C)-a^3 (64 A+70 B+15 C) \cos (c+d x)\right )}{\sqrt {\cos (c+d x)}}dx+\frac {6 a^3 (8 A+10 B+5 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{d \sqrt {\cos (c+d x)}}\right )+\frac {10 a^2 (A+B) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{3 d \cos ^{\frac {3}{2}}(c+d x)}}{5 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{3} \left (\int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a} \left (a^3 (32 A+50 B+45 C)-a^3 (64 A+70 B+15 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {6 a^3 (8 A+10 B+5 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{d \sqrt {\cos (c+d x)}}\right )+\frac {10 a^2 (A+B) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{3 d \cos ^{\frac {3}{2}}(c+d x)}}{5 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )\) |
\(\Big \downarrow \) 3460 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{3} \left (\frac {15}{2} a^3 (2 B+5 C) \int \frac {\sqrt {\cos (c+d x) a+a}}{\sqrt {\cos (c+d x)}}dx-\frac {a^4 (64 A+70 B+15 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}+\frac {6 a^3 (8 A+10 B+5 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{d \sqrt {\cos (c+d x)}}\right )+\frac {10 a^2 (A+B) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{3 d \cos ^{\frac {3}{2}}(c+d x)}}{5 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{3} \left (\frac {15}{2} a^3 (2 B+5 C) \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {a^4 (64 A+70 B+15 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}+\frac {6 a^3 (8 A+10 B+5 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{d \sqrt {\cos (c+d x)}}\right )+\frac {10 a^2 (A+B) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{3 d \cos ^{\frac {3}{2}}(c+d x)}}{5 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )\) |
\(\Big \downarrow \) 3253 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{3} \left (-\frac {15 a^3 (2 B+5 C) \int \frac {1}{\sqrt {1-\frac {a \sin ^2(c+d x)}{\cos (c+d x) a+a}}}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x) a+a}}\right )}{d}-\frac {a^4 (64 A+70 B+15 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}+\frac {6 a^3 (8 A+10 B+5 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{d \sqrt {\cos (c+d x)}}\right )+\frac {10 a^2 (A+B) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{3 d \cos ^{\frac {3}{2}}(c+d x)}}{5 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )\) |
\(\Big \downarrow \) 223 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {10 a^2 (A+B) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{3 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {1}{3} \left (\frac {15 a^{7/2} (2 B+5 C) \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{d}-\frac {a^4 (64 A+70 B+15 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}+\frac {6 a^3 (8 A+10 B+5 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{d \sqrt {\cos (c+d x)}}\right )}{5 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )\) |
Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((2*A*(a + a*Cos[c + d*x])^(5/2)*Sin [c + d*x])/(5*d*Cos[c + d*x]^(5/2)) + ((10*a^2*(A + B)*(a + a*Cos[c + d*x] )^(3/2)*Sin[c + d*x])/(3*d*Cos[c + d*x]^(3/2)) + ((15*a^(7/2)*(2*B + 5*C)* ArcSin[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/d - (a^4*(64*A + 70*B + 15*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(d*Sqrt[a + a*Cos[c + d*x]]) + (6*a^3*(8*A + 10*B + 5*C)*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(d*Sqr t[Cos[c + d*x]]))/3)/(5*a))
3.14.31.3.1 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[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[Rt[-b, 2]*(x/Sqrt [a])]/Rt[-b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && NegQ[b]
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(d_.)*sin[(e_.) + (f_.) *(x_)]], x_Symbol] :> Simp[-2/f Subst[Int[1/Sqrt[1 - x^2/a], x], x, b*(Co s[e + f*x]/Sqrt[a + b*Sin[e + f*x]])], x] /; FreeQ[{a, b, d, e, f}, x] && E qQ[a^2 - b^2, 0] && EqQ[d, a/b]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim p[(-b^2)*(B*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((c + d*Sin[ e + f*x])^(n + 1)/(d*f*(n + 1)*(b*c + a*d))), x] - Simp[b/(d*(n + 1)*(b*c + a*d)) Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp [a*A*d*(m - n - 2) - B*(a*c*(m - 1) + b*d*(n + 1)) - (A*b*d*(m + n + 1) - B *(b*c*m - a*d*(n + 1)))*Sin[e + f*x], x], 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] && GtQ[m, 1/2] && LtQ[n, -1] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 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 [-2*b*B*Cos[e + f*x]*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(2*n + 3)*Sqrt[a + b*Sin[e + f*x]])), x] + Simp[(A*b*d*(2*n + 3) - B*(b*c - 2*a*d*(n + 1)))/(b *d*(2*n + 3)) Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, 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])
Int[(u_)*((c_.)*sec[(a_.) + (b_.)*(x_)])^(m_.), x_Symbol] :> Simp[(c*Sec[a + b*x])^m*(c*Cos[a + b*x])^m Int[ActivateTrig[u]/(c*Cos[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] && !IntegerQ[m] && KnownSineIntegrandQ[u, x]
Time = 1.59 (sec) , antiderivative size = 385, normalized size of antiderivative = 1.58
\[\frac {a^{2} \left (\sec ^{\frac {7}{2}}\left (d x +c \right )\right ) \sqrt {\left (1+\cos \left (d x +c \right )\right ) a}\, \left (30 B \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \left (\cos ^{4}\left (d x +c \right )\right ) \arctan \left (\sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \tan \left (d x +c \right )\right )+75 C \left (\cos ^{4}\left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \tan \left (d x +c \right )\right )+30 B \arctan \left (\sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \tan \left (d x +c \right )\right ) \left (\cos ^{3}\left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+15 C \left (\cos ^{4}\left (d x +c \right )\right ) \sin \left (d x +c \right )+75 C \left (\cos ^{3}\left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \tan \left (d x +c \right )\right )+86 A \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )+80 B \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )+30 C \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )+28 A \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+10 B \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+6 A \cos \left (d x +c \right ) \sin \left (d x +c \right )\right )}{15 d \left (1+\cos \left (d x +c \right )\right )}\]
1/15*a^2/d*sec(d*x+c)^(7/2)*((1+cos(d*x+c))*a)^(1/2)/(1+cos(d*x+c))*(30*B* (cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)^4*arctan((cos(d*x+c)/(1+cos(d *x+c)))^(1/2)*tan(d*x+c))+75*C*cos(d*x+c)^4*(cos(d*x+c)/(1+cos(d*x+c)))^(1 /2)*arctan((cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*tan(d*x+c))+30*B*arctan((cos( d*x+c)/(1+cos(d*x+c)))^(1/2)*tan(d*x+c))*cos(d*x+c)^3*(cos(d*x+c)/(1+cos(d *x+c)))^(1/2)+15*C*cos(d*x+c)^4*sin(d*x+c)+75*C*cos(d*x+c)^3*(cos(d*x+c)/( 1+cos(d*x+c)))^(1/2)*arctan((cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*tan(d*x+c))+ 86*A*cos(d*x+c)^3*sin(d*x+c)+80*B*cos(d*x+c)^3*sin(d*x+c)+30*C*cos(d*x+c)^ 3*sin(d*x+c)+28*A*cos(d*x+c)^2*sin(d*x+c)+10*B*cos(d*x+c)^2*sin(d*x+c)+6*A *cos(d*x+c)*sin(d*x+c))
Time = 0.32 (sec) , antiderivative size = 193, normalized size of antiderivative = 0.79 \[ \int (a+a \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x) \, dx=-\frac {15 \, {\left ({\left (2 \, B + 5 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + {\left (2 \, B + 5 \, C\right )} a^{2} \cos \left (d x + c\right )^{2}\right )} \sqrt {a} \arctan \left (\frac {\sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right ) - \frac {{\left (15 \, C a^{2} \cos \left (d x + c\right )^{3} + 2 \, {\left (43 \, A + 40 \, B + 15 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 2 \, {\left (14 \, A + 5 \, B\right )} a^{2} \cos \left (d x + c\right ) + 6 \, A a^{2}\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{15 \, {\left (d \cos \left (d x + c\right )^{3} + d \cos \left (d x + c\right )^{2}\right )}} \]
integrate((a+a*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c )^(7/2),x, algorithm="fricas")
-1/15*(15*((2*B + 5*C)*a^2*cos(d*x + c)^3 + (2*B + 5*C)*a^2*cos(d*x + c)^2 )*sqrt(a)*arctan(sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))/(sqrt(a)*sin( d*x + c))) - (15*C*a^2*cos(d*x + c)^3 + 2*(43*A + 40*B + 15*C)*a^2*cos(d*x + c)^2 + 2*(14*A + 5*B)*a^2*cos(d*x + c) + 6*A*a^2)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c)/sqrt(cos(d*x + c)))/(d*cos(d*x + c)^3 + d*cos(d*x + c)^2)
Timed out. \[ \int (a+a \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x) \, dx=\text {Timed out} \]
Leaf count of result is larger than twice the leaf count of optimal. 3231 vs. \(2 (209) = 418\).
Time = 0.74 (sec) , antiderivative size = 3231, normalized size of antiderivative = 13.30 \[ \int (a+a \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x) \, dx=\text {Too large to display} \]
integrate((a+a*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c )^(7/2),x, algorithm="maxima")
1/60*(10*(10*sqrt(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*a^(5/2)*sin(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + 3*((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)))* sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) - cos(1/2*arctan2 (sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))*sin(1/2*arctan2(sin(2*d*x + 2*c) , cos(2*d*x + 2*c)))), (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))*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + sin(1/2*arcta n2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))*sin(1/2*arctan2(sin(2*d*x + 2* c), cos(2*d*x + 2*c)))) + 1) - (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)))*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c ) + 1)) - cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))*sin(1/2 *arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))), (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))*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))*si...
Timed out. \[ \int (a+a \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x) \, dx=\text {Timed out} \]
integrate((a+a*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c )^(7/2),x, algorithm="giac")
Timed out. \[ \int (a+a \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x) \, dx=\int {\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{7/2}\,{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{5/2}\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A\right ) \,d x \]
int((1/cos(c + d*x))^(7/2)*(a + a*cos(c + d*x))^(5/2)*(A + B*cos(c + d*x) + C*cos(c + d*x)^2),x)