Integrand size = 43, antiderivative size = 241 \[ \int \sqrt {\cos (c+d x)} (b \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {3 b^2 B x \sqrt {b \cos (c+d x)}}{8 \sqrt {\cos (c+d x)}}+\frac {b^2 (5 A+4 C) \sqrt {b \cos (c+d x)} \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}+\frac {3 b^2 B \sqrt {\cos (c+d x)} \sqrt {b \cos (c+d x)} \sin (c+d x)}{8 d}+\frac {b^2 B \cos ^{\frac {5}{2}}(c+d x) \sqrt {b \cos (c+d x)} \sin (c+d x)}{4 d}+\frac {b^2 C \cos ^{\frac {7}{2}}(c+d x) \sqrt {b \cos (c+d x)} \sin (c+d x)}{5 d}-\frac {b^2 (5 A+4 C) \sqrt {b \cos (c+d x)} \sin ^3(c+d x)}{15 d \sqrt {\cos (c+d x)}} \]
1/4*b^2*B*cos(d*x+c)^(5/2)*sin(d*x+c)*(b*cos(d*x+c))^(1/2)/d+1/5*b^2*C*cos (d*x+c)^(7/2)*sin(d*x+c)*(b*cos(d*x+c))^(1/2)/d+3/8*b^2*B*x*(b*cos(d*x+c)) ^(1/2)/cos(d*x+c)^(1/2)+1/5*b^2*(5*A+4*C)*sin(d*x+c)*(b*cos(d*x+c))^(1/2)/ d/cos(d*x+c)^(1/2)-1/15*b^2*(5*A+4*C)*sin(d*x+c)^3*(b*cos(d*x+c))^(1/2)/d/ cos(d*x+c)^(1/2)+3/8*b^2*B*sin(d*x+c)*cos(d*x+c)^(1/2)*(b*cos(d*x+c))^(1/2 )/d
Time = 0.64 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.45 \[ \int \sqrt {\cos (c+d x)} (b \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {(b \cos (c+d x))^{5/2} (180 B c+180 B d x+60 (6 A+5 C) \sin (c+d x)+120 B \sin (2 (c+d x))+40 A \sin (3 (c+d x))+50 C \sin (3 (c+d x))+15 B \sin (4 (c+d x))+6 C \sin (5 (c+d x)))}{480 d \cos ^{\frac {5}{2}}(c+d x)} \]
Integrate[Sqrt[Cos[c + d*x]]*(b*Cos[c + d*x])^(5/2)*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2),x]
((b*Cos[c + d*x])^(5/2)*(180*B*c + 180*B*d*x + 60*(6*A + 5*C)*Sin[c + d*x] + 120*B*Sin[2*(c + d*x)] + 40*A*Sin[3*(c + d*x)] + 50*C*Sin[3*(c + d*x)] + 15*B*Sin[4*(c + d*x)] + 6*C*Sin[5*(c + d*x)]))/(480*d*Cos[c + d*x]^(5/2) )
Time = 0.63 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.59, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.279, Rules used = {2031, 3042, 3502, 3042, 3227, 3042, 3113, 2009, 3115, 3042, 3115, 24}
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.
| \(\displaystyle \int \sqrt {\cos (c+d x)} (b \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 2031 |
| \(\displaystyle \frac {b^2 \sqrt {b \cos (c+d x)} \int \cos ^3(c+d x) \left (C \cos ^2(c+d x)+B \cos (c+d x)+A\right )dx}{\sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {b^2 \sqrt {b \cos (c+d x)} \int \sin \left (c+d x+\frac {\pi }{2}\right )^3 \left (C \sin \left (c+d x+\frac {\pi }{2}\right )^2+B \sin \left (c+d x+\frac {\pi }{2}\right )+A\right )dx}{\sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 3502 |
| \(\displaystyle \frac {b^2 \sqrt {b \cos (c+d x)} \left (\frac {1}{5} \int \cos ^3(c+d x) (5 A+4 C+5 B \cos (c+d x))dx+\frac {C \sin (c+d x) \cos ^4(c+d x)}{5 d}\right )}{\sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {b^2 \sqrt {b \cos (c+d x)} \left (\frac {1}{5} \int \sin \left (c+d x+\frac {\pi }{2}\right )^3 \left (5 A+4 C+5 B \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx+\frac {C \sin (c+d x) \cos ^4(c+d x)}{5 d}\right )}{\sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 3227 |
| \(\displaystyle \frac {b^2 \sqrt {b \cos (c+d x)} \left (\frac {1}{5} \left ((5 A+4 C) \int \cos ^3(c+d x)dx+5 B \int \cos ^4(c+d x)dx\right )+\frac {C \sin (c+d x) \cos ^4(c+d x)}{5 d}\right )}{\sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {b^2 \sqrt {b \cos (c+d x)} \left (\frac {1}{5} \left ((5 A+4 C) \int \sin \left (c+d x+\frac {\pi }{2}\right )^3dx+5 B \int \sin \left (c+d x+\frac {\pi }{2}\right )^4dx\right )+\frac {C \sin (c+d x) \cos ^4(c+d x)}{5 d}\right )}{\sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 3113 |
| \(\displaystyle \frac {b^2 \sqrt {b \cos (c+d x)} \left (\frac {1}{5} \left (5 B \int \sin \left (c+d x+\frac {\pi }{2}\right )^4dx-\frac {(5 A+4 C) \int \left (1-\sin ^2(c+d x)\right )d(-\sin (c+d x))}{d}\right )+\frac {C \sin (c+d x) \cos ^4(c+d x)}{5 d}\right )}{\sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {b^2 \sqrt {b \cos (c+d x)} \left (\frac {1}{5} \left (5 B \int \sin \left (c+d x+\frac {\pi }{2}\right )^4dx-\frac {(5 A+4 C) \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\right )+\frac {C \sin (c+d x) \cos ^4(c+d x)}{5 d}\right )}{\sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 3115 |
| \(\displaystyle \frac {b^2 \sqrt {b \cos (c+d x)} \left (\frac {1}{5} \left (5 B \left (\frac {3}{4} \int \cos ^2(c+d x)dx+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}\right )-\frac {(5 A+4 C) \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\right )+\frac {C \sin (c+d x) \cos ^4(c+d x)}{5 d}\right )}{\sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {b^2 \sqrt {b \cos (c+d x)} \left (\frac {1}{5} \left (5 B \left (\frac {3}{4} \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}\right )-\frac {(5 A+4 C) \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\right )+\frac {C \sin (c+d x) \cos ^4(c+d x)}{5 d}\right )}{\sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 3115 |
| \(\displaystyle \frac {b^2 \sqrt {b \cos (c+d x)} \left (\frac {1}{5} \left (5 B \left (\frac {3}{4} \left (\frac {\int 1dx}{2}+\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}\right )-\frac {(5 A+4 C) \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\right )+\frac {C \sin (c+d x) \cos ^4(c+d x)}{5 d}\right )}{\sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {b^2 \sqrt {b \cos (c+d x)} \left (\frac {1}{5} \left (5 B \left (\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac {3}{4} \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )\right )-\frac {(5 A+4 C) \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\right )+\frac {C \sin (c+d x) \cos ^4(c+d x)}{5 d}\right )}{\sqrt {\cos (c+d x)}}\) |
(b^2*Sqrt[b*Cos[c + d*x]]*((C*Cos[c + d*x]^4*Sin[c + d*x])/(5*d) + (-(((5* A + 4*C)*(-Sin[c + d*x] + Sin[c + d*x]^3/3))/d) + 5*B*((Cos[c + d*x]^3*Sin [c + d*x])/(4*d) + (3*(x/2 + (Cos[c + d*x]*Sin[c + d*x])/(2*d)))/4))/5))/S qrt[Cos[c + d*x]]
3.4.6.3.1 Defintions of rubi rules used
Int[(Fx_.)*((a_.)*(v_))^(m_)*((b_.)*(v_))^(n_), x_Symbol] :> Simp[a^(m + 1/ 2)*b^(n - 1/2)*(Sqrt[b*v]/Sqrt[a*v]) Int[v^(m + n)*Fx, x], x] /; FreeQ[{a , b, m}, x] && !IntegerQ[m] && IGtQ[n + 1/2, 0] && IntegerQ[m + n]
Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Simp[-d^(-1) Subst[Int[Exp and[(1 - x^2)^((n - 1)/2), x], x], x, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]
Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* x]*((b*Sin[c + d*x])^(n - 1)/(d*n)), x] + Simp[b^2*((n - 1)/n) Int[(b*Sin [c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[ 2*n]
Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x _)]), x_Symbol] :> Simp[c Int[(b*Sin[e + f*x])^m, x], x] + Simp[d/b Int [(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Co s[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Simp[1/(b*(m + 2)) Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && !LtQ[m, -1]
Time = 8.92 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.57
| method | result | size |
| default | \(\frac {b^{2} \sqrt {\cos \left (d x +c \right ) b}\, \left (24 C \sin \left (d x +c \right ) \left (\cos ^{4}\left (d x +c \right )\right )+30 B \sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )+40 A \sin \left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right )+32 C \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+45 B \sin \left (d x +c \right ) \cos \left (d x +c \right )+80 A \sin \left (d x +c \right )+45 B \left (d x +c \right )+64 \sin \left (d x +c \right ) C \right )}{120 d \sqrt {\cos \left (d x +c \right )}}\) | \(137\) |
| parts | \(\frac {A \,b^{2} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right ) \sqrt {\cos \left (d x +c \right ) b}}{3 d \sqrt {\cos \left (d x +c \right )}}+\frac {B \,b^{2} \sqrt {\cos \left (d x +c \right ) b}\, \left (2 \sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )+3 \cos \left (d x +c \right ) \sin \left (d x +c \right )+3 d x +3 c \right )}{8 d \sqrt {\cos \left (d x +c \right )}}+\frac {C \,b^{2} \left (3 \left (\cos ^{4}\left (d x +c \right )\right )+4 \left (\cos ^{2}\left (d x +c \right )\right )+8\right ) \sin \left (d x +c \right ) \sqrt {\cos \left (d x +c \right ) b}}{15 d \sqrt {\cos \left (d x +c \right )}}\) | \(165\) |
| risch | \(\frac {3 b^{2} \sqrt {\cos \left (d x +c \right ) b}\, \left (\sqrt {\cos }\left (d x +c \right )\right ) {\mathrm e}^{i \left (d x +c \right )} x B}{4 \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}-\frac {i b^{2} \sqrt {\cos \left (d x +c \right ) b}\, \left (\sqrt {\cos }\left (d x +c \right )\right ) {\mathrm e}^{6 i \left (d x +c \right )} C}{80 \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) d}-\frac {i b^{2} \sqrt {\cos \left (d x +c \right ) b}\, \left (\sqrt {\cos }\left (d x +c \right )\right ) {\mathrm e}^{5 i \left (d x +c \right )} B}{32 \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) d}-\frac {i b^{2} \sqrt {\cos \left (d x +c \right ) b}\, \left (\sqrt {\cos }\left (d x +c \right )\right ) {\mathrm e}^{2 i \left (d x +c \right )} \left (6 A +5 C \right )}{8 \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) d}+\frac {i b^{2} \sqrt {\cos \left (d x +c \right ) b}\, \left (\sqrt {\cos }\left (d x +c \right )\right ) \left (6 A +5 C \right )}{8 \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) d}+\frac {i b^{2} \sqrt {\cos \left (d x +c \right ) b}\, \left (\sqrt {\cos }\left (d x +c \right )\right ) {\mathrm e}^{-i \left (d x +c \right )} B}{4 \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) d}+\frac {i b^{2} \sqrt {\cos \left (d x +c \right ) b}\, \left (\sqrt {\cos }\left (d x +c \right )\right ) {\mathrm e}^{-2 i \left (d x +c \right )} \left (4 A +5 C \right )}{48 \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) d}-\frac {i b^{2} \sqrt {\cos \left (d x +c \right ) b}\, \left (\sqrt {\cos }\left (d x +c \right )\right ) \left (11 C +10 A \right ) \cos \left (4 d x +4 c \right )}{120 \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) d}+\frac {b^{2} \sqrt {\cos \left (d x +c \right ) b}\, \left (\sqrt {\cos }\left (d x +c \right )\right ) \left (5 A +7 C \right ) \sin \left (4 d x +4 c \right )}{60 \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) d}-\frac {7 i b^{2} \sqrt {\cos \left (d x +c \right ) b}\, \left (\sqrt {\cos }\left (d x +c \right )\right ) B \cos \left (3 d x +3 c \right )}{32 \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) d}+\frac {9 b^{2} \sqrt {\cos \left (d x +c \right ) b}\, \left (\sqrt {\cos }\left (d x +c \right )\right ) B \sin \left (3 d x +3 c \right )}{32 \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) d}\) | \(568\) |
int((cos(d*x+c)*b)^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*cos(d*x+c)^(1/2), x,method=_RETURNVERBOSE)
1/120*b^2/d*(cos(d*x+c)*b)^(1/2)*(24*C*sin(d*x+c)*cos(d*x+c)^4+30*B*sin(d* x+c)*cos(d*x+c)^3+40*A*sin(d*x+c)*cos(d*x+c)^2+32*C*cos(d*x+c)^2*sin(d*x+c )+45*B*sin(d*x+c)*cos(d*x+c)+80*A*sin(d*x+c)+45*B*(d*x+c)+64*sin(d*x+c)*C) /cos(d*x+c)^(1/2)
Time = 0.32 (sec) , antiderivative size = 331, normalized size of antiderivative = 1.37 \[ \int \sqrt {\cos (c+d x)} (b \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\left [\frac {45 \, B \sqrt {-b} b^{2} \cos \left (d x + c\right ) \log \left (2 \, b \cos \left (d x + c\right )^{2} - 2 \, \sqrt {b \cos \left (d x + c\right )} \sqrt {-b} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - b\right ) + 2 \, {\left (24 \, C b^{2} \cos \left (d x + c\right )^{4} + 30 \, B b^{2} \cos \left (d x + c\right )^{3} + 8 \, {\left (5 \, A + 4 \, C\right )} b^{2} \cos \left (d x + c\right )^{2} + 45 \, B b^{2} \cos \left (d x + c\right ) + 16 \, {\left (5 \, A + 4 \, C\right )} b^{2}\right )} \sqrt {b \cos \left (d x + c\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{240 \, d \cos \left (d x + c\right )}, \frac {45 \, B b^{\frac {5}{2}} \arctan \left (\frac {\sqrt {b \cos \left (d x + c\right )} \sin \left (d x + c\right )}{\sqrt {b} \cos \left (d x + c\right )^{\frac {3}{2}}}\right ) \cos \left (d x + c\right ) + {\left (24 \, C b^{2} \cos \left (d x + c\right )^{4} + 30 \, B b^{2} \cos \left (d x + c\right )^{3} + 8 \, {\left (5 \, A + 4 \, C\right )} b^{2} \cos \left (d x + c\right )^{2} + 45 \, B b^{2} \cos \left (d x + c\right ) + 16 \, {\left (5 \, A + 4 \, C\right )} b^{2}\right )} \sqrt {b \cos \left (d x + c\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{120 \, d \cos \left (d x + c\right )}\right ] \]
integrate((b*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*cos(d*x+c)^ (1/2),x, algorithm="fricas")
[1/240*(45*B*sqrt(-b)*b^2*cos(d*x + c)*log(2*b*cos(d*x + c)^2 - 2*sqrt(b*c os(d*x + c))*sqrt(-b)*sqrt(cos(d*x + c))*sin(d*x + c) - b) + 2*(24*C*b^2*c os(d*x + c)^4 + 30*B*b^2*cos(d*x + c)^3 + 8*(5*A + 4*C)*b^2*cos(d*x + c)^2 + 45*B*b^2*cos(d*x + c) + 16*(5*A + 4*C)*b^2)*sqrt(b*cos(d*x + c))*sqrt(c os(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)), 1/120*(45*B*b^(5/2)*arctan(sq rt(b*cos(d*x + c))*sin(d*x + c)/(sqrt(b)*cos(d*x + c)^(3/2)))*cos(d*x + c) + (24*C*b^2*cos(d*x + c)^4 + 30*B*b^2*cos(d*x + c)^3 + 8*(5*A + 4*C)*b^2* cos(d*x + c)^2 + 45*B*b^2*cos(d*x + c) + 16*(5*A + 4*C)*b^2)*sqrt(b*cos(d* x + c))*sqrt(cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c))]
Timed out. \[ \int \sqrt {\cos (c+d x)} (b \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\text {Timed out} \]
Time = 0.52 (sec) , antiderivative size = 185, normalized size of antiderivative = 0.77 \[ \int \sqrt {\cos (c+d x)} (b \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {40 \, {\left (b^{2} \sin \left (3 \, d x + 3 \, c\right ) + 9 \, b^{2} \sin \left (\frac {1}{3} \, \arctan \left (\sin \left (3 \, d x + 3 \, c\right ), \cos \left (3 \, d x + 3 \, c\right )\right )\right )\right )} A \sqrt {b} + 15 \, {\left (12 \, {\left (d x + c\right )} b^{2} + b^{2} \sin \left (4 \, d x + 4 \, c\right ) + 8 \, b^{2} \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (4 \, d x + 4 \, c\right ), \cos \left (4 \, d x + 4 \, c\right )\right )\right )\right )} B \sqrt {b} + 2 \, {\left (3 \, b^{2} \sin \left (5 \, d x + 5 \, c\right ) + 25 \, b^{2} \sin \left (\frac {3}{5} \, \arctan \left (\sin \left (5 \, d x + 5 \, c\right ), \cos \left (5 \, d x + 5 \, c\right )\right )\right ) + 150 \, b^{2} \sin \left (\frac {1}{5} \, \arctan \left (\sin \left (5 \, d x + 5 \, c\right ), \cos \left (5 \, d x + 5 \, c\right )\right )\right )\right )} C \sqrt {b}}{480 \, d} \]
integrate((b*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*cos(d*x+c)^ (1/2),x, algorithm="maxima")
1/480*(40*(b^2*sin(3*d*x + 3*c) + 9*b^2*sin(1/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c))))*A*sqrt(b) + 15*(12*(d*x + c)*b^2 + b^2*sin(4*d*x + 4*c ) + 8*b^2*sin(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c))))*B*sqrt(b) + 2*(3*b^2*sin(5*d*x + 5*c) + 25*b^2*sin(3/5*arctan2(sin(5*d*x + 5*c), cos (5*d*x + 5*c))) + 150*b^2*sin(1/5*arctan2(sin(5*d*x + 5*c), cos(5*d*x + 5* c))))*C*sqrt(b))/d
Leaf count of result is larger than twice the leaf count of optimal. 411 vs. \(2 (205) = 410\).
Time = 5.82 (sec) , antiderivative size = 411, normalized size of antiderivative = 1.71 \[ \int \sqrt {\cos (c+d x)} (b \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {45 \, B b^{\frac {5}{2}} d x \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{10} + 225 \, B b^{\frac {5}{2}} d x \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} + 240 \, A b^{\frac {5}{2}} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 150 \, B b^{\frac {5}{2}} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 240 \, C b^{\frac {5}{2}} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 450 \, B b^{\frac {5}{2}} d x \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 640 \, A b^{\frac {5}{2}} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 60 \, B b^{\frac {5}{2}} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 320 \, C b^{\frac {5}{2}} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 450 \, B b^{\frac {5}{2}} d x \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 800 \, A b^{\frac {5}{2}} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 928 \, C b^{\frac {5}{2}} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 225 \, B b^{\frac {5}{2}} d x \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 640 \, A b^{\frac {5}{2}} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 60 \, B b^{\frac {5}{2}} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 320 \, C b^{\frac {5}{2}} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 45 \, B b^{\frac {5}{2}} d x + 240 \, A b^{\frac {5}{2}} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 150 \, B b^{\frac {5}{2}} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 240 \, C b^{\frac {5}{2}} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{120 \, {\left (d \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{10} + 5 \, d \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} + 10 \, d \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 10 \, d \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 5 \, d \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + d\right )}} \]
integrate((b*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*cos(d*x+c)^ (1/2),x, algorithm="giac")
1/120*(45*B*b^(5/2)*d*x*tan(1/2*d*x + 1/2*c)^10 + 225*B*b^(5/2)*d*x*tan(1/ 2*d*x + 1/2*c)^8 + 240*A*b^(5/2)*tan(1/2*d*x + 1/2*c)^9 - 150*B*b^(5/2)*ta n(1/2*d*x + 1/2*c)^9 + 240*C*b^(5/2)*tan(1/2*d*x + 1/2*c)^9 + 450*B*b^(5/2 )*d*x*tan(1/2*d*x + 1/2*c)^6 + 640*A*b^(5/2)*tan(1/2*d*x + 1/2*c)^7 - 60*B *b^(5/2)*tan(1/2*d*x + 1/2*c)^7 + 320*C*b^(5/2)*tan(1/2*d*x + 1/2*c)^7 + 4 50*B*b^(5/2)*d*x*tan(1/2*d*x + 1/2*c)^4 + 800*A*b^(5/2)*tan(1/2*d*x + 1/2* c)^5 + 928*C*b^(5/2)*tan(1/2*d*x + 1/2*c)^5 + 225*B*b^(5/2)*d*x*tan(1/2*d* x + 1/2*c)^2 + 640*A*b^(5/2)*tan(1/2*d*x + 1/2*c)^3 + 60*B*b^(5/2)*tan(1/2 *d*x + 1/2*c)^3 + 320*C*b^(5/2)*tan(1/2*d*x + 1/2*c)^3 + 45*B*b^(5/2)*d*x + 240*A*b^(5/2)*tan(1/2*d*x + 1/2*c) + 150*B*b^(5/2)*tan(1/2*d*x + 1/2*c) + 240*C*b^(5/2)*tan(1/2*d*x + 1/2*c))/(d*tan(1/2*d*x + 1/2*c)^10 + 5*d*tan (1/2*d*x + 1/2*c)^8 + 10*d*tan(1/2*d*x + 1/2*c)^6 + 10*d*tan(1/2*d*x + 1/2 *c)^4 + 5*d*tan(1/2*d*x + 1/2*c)^2 + d)
Time = 3.60 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.60 \[ \int \sqrt {\cos (c+d x)} (b \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {b^2\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {b\,\cos \left (c+d\,x\right )}\,\left (120\,B\,\sin \left (c+d\,x\right )+400\,A\,\sin \left (2\,c+2\,d\,x\right )+40\,A\,\sin \left (4\,c+4\,d\,x\right )+135\,B\,\sin \left (3\,c+3\,d\,x\right )+15\,B\,\sin \left (5\,c+5\,d\,x\right )+350\,C\,\sin \left (2\,c+2\,d\,x\right )+56\,C\,\sin \left (4\,c+4\,d\,x\right )+6\,C\,\sin \left (6\,c+6\,d\,x\right )+360\,B\,d\,x\,\cos \left (c+d\,x\right )\right )}{480\,d\,\left (\cos \left (2\,c+2\,d\,x\right )+1\right )} \]
(b^2*cos(c + d*x)^(1/2)*(b*cos(c + d*x))^(1/2)*(120*B*sin(c + d*x) + 400*A *sin(2*c + 2*d*x) + 40*A*sin(4*c + 4*d*x) + 135*B*sin(3*c + 3*d*x) + 15*B* sin(5*c + 5*d*x) + 350*C*sin(2*c + 2*d*x) + 56*C*sin(4*c + 4*d*x) + 6*C*si n(6*c + 6*d*x) + 360*B*d*x*cos(c + d*x)))/(480*d*(cos(2*c + 2*d*x) + 1))