Integrand size = 43, antiderivative size = 223 \[ \int \cos ^{\frac {5}{2}}(c+d x) \sqrt {b \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {3 B x \sqrt {b \cos (c+d x)}}{8 \sqrt {\cos (c+d x)}}+\frac {(5 A+4 C) \sqrt {b \cos (c+d x)} \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}+\frac {3 B \sqrt {\cos (c+d x)} \sqrt {b \cos (c+d x)} \sin (c+d x)}{8 d}+\frac {B \cos ^{\frac {5}{2}}(c+d x) \sqrt {b \cos (c+d x)} \sin (c+d x)}{4 d}+\frac {C \cos ^{\frac {7}{2}}(c+d x) \sqrt {b \cos (c+d x)} \sin (c+d x)}{5 d}-\frac {(5 A+4 C) \sqrt {b \cos (c+d x)} \sin ^3(c+d x)}{15 d \sqrt {\cos (c+d x)}} \] Output:
3/8*B*x*(b*cos(d*x+c))^(1/2)/cos(d*x+c)^(1/2)+1/5*(5*A+4*C)*(b*cos(d*x+c)) ^(1/2)*sin(d*x+c)/d/cos(d*x+c)^(1/2)+3/8*B*cos(d*x+c)^(1/2)*(b*cos(d*x+c)) ^(1/2)*sin(d*x+c)/d+1/4*B*cos(d*x+c)^(5/2)*(b*cos(d*x+c))^(1/2)*sin(d*x+c) /d+1/5*C*cos(d*x+c)^(7/2)*(b*cos(d*x+c))^(1/2)*sin(d*x+c)/d-1/15*(5*A+4*C) *(b*cos(d*x+c))^(1/2)*sin(d*x+c)^3/d/cos(d*x+c)^(1/2)
Time = 1.33 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.49 \[ \int \cos ^{\frac {5}{2}}(c+d x) \sqrt {b \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {\sqrt {b \cos (c+d x)} (180 B c+180 B d x+60 (6 A+5 C) \sin (c+d x)+120 B \sin (2 (c+d x))+40 A \sin (3 (c+d x))+50 C \sin (3 (c+d x))+15 B \sin (4 (c+d x))+6 C \sin (5 (c+d x)))}{480 d \sqrt {\cos (c+d x)}} \] Input:
Integrate[Cos[c + d*x]^(5/2)*Sqrt[b*Cos[c + d*x]]*(A + B*Cos[c + d*x] + C* Cos[c + d*x]^2),x]
Output:
(Sqrt[b*Cos[c + d*x]]*(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*Sqrt[Cos[c + d*x]])
Time = 0.64 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.62, 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.
| \(\displaystyle \int \cos ^{\frac {5}{2}}(c+d x) \sqrt {b \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 2031 |
| \(\displaystyle \frac {\sqrt {b \cos (c+d x)} \int \cos ^3(c+d x) \left (C \cos ^2(c+d x)+B \cos (c+d x)+A\right )dx}{\sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\sqrt {b \cos (c+d x)} \int \sin \left (c+d x+\frac {\pi }{2}\right )^3 \left (C \sin \left (c+d x+\frac {\pi }{2}\right )^2+B \sin \left (c+d x+\frac {\pi }{2}\right )+A\right )dx}{\sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 3502 |
| \(\displaystyle \frac {\sqrt {b \cos (c+d x)} \left (\frac {1}{5} \int \cos ^3(c+d x) (5 A+4 C+5 B \cos (c+d x))dx+\frac {C \sin (c+d x) \cos ^4(c+d x)}{5 d}\right )}{\sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\sqrt {b \cos (c+d x)} \left (\frac {1}{5} \int \sin \left (c+d x+\frac {\pi }{2}\right )^3 \left (5 A+4 C+5 B \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx+\frac {C \sin (c+d x) \cos ^4(c+d x)}{5 d}\right )}{\sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 3227 |
| \(\displaystyle \frac {\sqrt {b \cos (c+d x)} \left (\frac {1}{5} \left ((5 A+4 C) \int \cos ^3(c+d x)dx+5 B \int \cos ^4(c+d x)dx\right )+\frac {C \sin (c+d x) \cos ^4(c+d x)}{5 d}\right )}{\sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\sqrt {b \cos (c+d x)} \left (\frac {1}{5} \left ((5 A+4 C) \int \sin \left (c+d x+\frac {\pi }{2}\right )^3dx+5 B \int \sin \left (c+d x+\frac {\pi }{2}\right )^4dx\right )+\frac {C \sin (c+d x) \cos ^4(c+d x)}{5 d}\right )}{\sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 3113 |
| \(\displaystyle \frac {\sqrt {b \cos (c+d x)} \left (\frac {1}{5} \left (5 B \int \sin \left (c+d x+\frac {\pi }{2}\right )^4dx-\frac {(5 A+4 C) \int \left (1-\sin ^2(c+d x)\right )d(-\sin (c+d x))}{d}\right )+\frac {C \sin (c+d x) \cos ^4(c+d x)}{5 d}\right )}{\sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\sqrt {b \cos (c+d x)} \left (\frac {1}{5} \left (5 B \int \sin \left (c+d x+\frac {\pi }{2}\right )^4dx-\frac {(5 A+4 C) \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\right )+\frac {C \sin (c+d x) \cos ^4(c+d x)}{5 d}\right )}{\sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 3115 |
| \(\displaystyle \frac {\sqrt {b \cos (c+d x)} \left (\frac {1}{5} \left (5 B \left (\frac {3}{4} \int \cos ^2(c+d x)dx+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}\right )-\frac {(5 A+4 C) \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\right )+\frac {C \sin (c+d x) \cos ^4(c+d x)}{5 d}\right )}{\sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\sqrt {b \cos (c+d x)} \left (\frac {1}{5} \left (5 B \left (\frac {3}{4} \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}\right )-\frac {(5 A+4 C) \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\right )+\frac {C \sin (c+d x) \cos ^4(c+d x)}{5 d}\right )}{\sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 3115 |
| \(\displaystyle \frac {\sqrt {b \cos (c+d x)} \left (\frac {1}{5} \left (5 B \left (\frac {3}{4} \left (\frac {\int 1dx}{2}+\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}\right )-\frac {(5 A+4 C) \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\right )+\frac {C \sin (c+d x) \cos ^4(c+d x)}{5 d}\right )}{\sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {\sqrt {b \cos (c+d x)} \left (\frac {1}{5} \left (5 B \left (\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac {3}{4} \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )\right )-\frac {(5 A+4 C) \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\right )+\frac {C \sin (c+d x) \cos ^4(c+d x)}{5 d}\right )}{\sqrt {\cos (c+d x)}}\) |
Input:
Int[Cos[c + d*x]^(5/2)*Sqrt[b*Cos[c + d*x]]*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2),x]
Output:
(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))/Sqrt[ Cos[c + d*x]]
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 = 1.10 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.49
| method | result | size |
| default | \(\frac {\left (45 B \left (d x +c \right )+\left (40 \cos \left (d x +c \right )^{2}+80\right ) \sin \left (d x +c \right ) A +\sin \left (d x +c \right ) \cos \left (d x +c \right ) \left (30 \cos \left (d x +c \right )^{2}+45\right ) B +\left (24 \cos \left (d x +c \right )^{4}+32 \cos \left (d x +c \right )^{2}+64\right ) \sin \left (d x +c \right ) C \right ) \sqrt {b \cos \left (d x +c \right )}}{120 d \sqrt {\cos \left (d x +c \right )}}\) | \(109\) |
| parts | \(\frac {A \sin \left (d x +c \right ) \left (2+\cos \left (d x +c \right )^{2}\right ) \sqrt {b \cos \left (d x +c \right )}}{3 d \sqrt {\cos \left (d x +c \right )}}+\frac {B \left (2 \cos \left (d x +c \right )^{3} \sin \left (d x +c \right )+3 \cos \left (d x +c \right ) \sin \left (d x +c \right )+3 d x +3 c \right ) \sqrt {b \cos \left (d x +c \right )}}{8 d \sqrt {\cos \left (d x +c \right )}}+\frac {C \sin \left (d x +c \right ) \left (3 \cos \left (d x +c \right )^{4}+4 \cos \left (d x +c \right )^{2}+8\right ) \sqrt {b \cos \left (d x +c \right )}}{15 d \sqrt {\cos \left (d x +c \right )}}\) | \(156\) |
| risch | \(\frac {3 \sqrt {b \cos \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right )}\, {\mathrm e}^{i \left (d x +c \right )} B x}{4 \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}-\frac {i \sqrt {b \cos \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \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 \sqrt {b \cos \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \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 \sqrt {b \cos \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \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 \sqrt {b \cos \left (d x +c \right )}\, \sqrt {\cos \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 \sqrt {b \cos \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \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 \sqrt {b \cos \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \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 \sqrt {b \cos \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right )}\, \left (10 A +11 C \right ) \cos \left (4 d x +4 c \right )}{120 \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) d}+\frac {\sqrt {b \cos \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \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 \sqrt {b \cos \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \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 \sqrt {b \cos \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right )}\, B \sin \left (3 d x +3 c \right )}{32 \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) d}\) | \(535\) |
Input:
int(cos(d*x+c)^(5/2)*(b*cos(d*x+c))^(1/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2), x,method=_RETURNVERBOSE)
Output:
1/120/d*(45*B*(d*x+c)+(40*cos(d*x+c)^2+80)*sin(d*x+c)*A+sin(d*x+c)*cos(d*x +c)*(30*cos(d*x+c)^2+45)*B+(24*cos(d*x+c)^4+32*cos(d*x+c)^2+64)*sin(d*x+c) *C)*(b*cos(d*x+c))^(1/2)/cos(d*x+c)^(1/2)
Time = 0.12 (sec) , antiderivative size = 292, normalized size of antiderivative = 1.31 \[ \int \cos ^{\frac {5}{2}}(c+d x) \sqrt {b \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\left [\frac {45 \, B \sqrt {-b} \cos \left (d x + c\right ) \log \left (2 \, b \cos \left (d x + c\right )^{2} - 2 \, \sqrt {b \cos \left (d x + c\right )} \sqrt {-b} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - b\right ) + 2 \, {\left (24 \, C \cos \left (d x + c\right )^{4} + 30 \, B \cos \left (d x + c\right )^{3} + 8 \, {\left (5 \, A + 4 \, C\right )} \cos \left (d x + c\right )^{2} + 45 \, B \cos \left (d x + c\right ) + 80 \, A + 64 \, C\right )} \sqrt {b \cos \left (d x + c\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{240 \, d \cos \left (d x + c\right )}, \frac {45 \, B \sqrt {b} \arctan \left (\frac {\sqrt {b \cos \left (d x + c\right )} \sin \left (d x + c\right )}{\sqrt {b} \cos \left (d x + c\right )^{\frac {3}{2}}}\right ) \cos \left (d x + c\right ) + {\left (24 \, C \cos \left (d x + c\right )^{4} + 30 \, B \cos \left (d x + c\right )^{3} + 8 \, {\left (5 \, A + 4 \, C\right )} \cos \left (d x + c\right )^{2} + 45 \, B \cos \left (d x + c\right ) + 80 \, A + 64 \, C\right )} \sqrt {b \cos \left (d x + c\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{120 \, d \cos \left (d x + c\right )}\right ] \] Input:
integrate(cos(d*x+c)^(5/2)*(b*cos(d*x+c))^(1/2)*(A+B*cos(d*x+c)+C*cos(d*x+ c)^2),x, algorithm="fricas")
Output:
[1/240*(45*B*sqrt(-b)*cos(d*x + c)*log(2*b*cos(d*x + c)^2 - 2*sqrt(b*cos(d *x + c))*sqrt(-b)*sqrt(cos(d*x + c))*sin(d*x + c) - b) + 2*(24*C*cos(d*x + c)^4 + 30*B*cos(d*x + c)^3 + 8*(5*A + 4*C)*cos(d*x + c)^2 + 45*B*cos(d*x + c) + 80*A + 64*C)*sqrt(b*cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c))/ (d*cos(d*x + c)), 1/120*(45*B*sqrt(b)*arctan(sqrt(b*cos(d*x + c))*sin(d*x + c)/(sqrt(b)*cos(d*x + c)^(3/2)))*cos(d*x + c) + (24*C*cos(d*x + c)^4 + 3 0*B*cos(d*x + c)^3 + 8*(5*A + 4*C)*cos(d*x + c)^2 + 45*B*cos(d*x + c) + 80 *A + 64*C)*sqrt(b*cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c))/(d*cos(d* x + c))]
Timed out. \[ \int \cos ^{\frac {5}{2}}(c+d x) \sqrt {b \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\text {Timed out} \] Input:
integrate(cos(d*x+c)**(5/2)*(b*cos(d*x+c))**(1/2)*(A+B*cos(d*x+c)+C*cos(d* x+c)**2),x)
Output:
Timed out
Time = 0.32 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.71 \[ \int \cos ^{\frac {5}{2}}(c+d x) \sqrt {b \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {15 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (4 \, d x + 4 \, c\right ), \cos \left (4 \, d x + 4 \, c\right )\right )\right )\right )} B \sqrt {b} + 2 \, C \sqrt {b} {\left (3 \, \sin \left (5 \, d x + 5 \, c\right ) + 25 \, \sin \left (\frac {3}{5} \, \arctan \left (\sin \left (5 \, d x + 5 \, c\right ), \cos \left (5 \, d x + 5 \, c\right )\right )\right ) + 150 \, \sin \left (\frac {1}{5} \, \arctan \left (\sin \left (5 \, d x + 5 \, c\right ), \cos \left (5 \, d x + 5 \, c\right )\right )\right )\right )} + 40 \, A \sqrt {b} {\left (\sin \left (3 \, d x + 3 \, c\right ) + 9 \, \sin \left (\frac {1}{3} \, \arctan \left (\sin \left (3 \, d x + 3 \, c\right ), \cos \left (3 \, d x + 3 \, c\right )\right )\right )\right )}}{480 \, d} \] Input:
integrate(cos(d*x+c)^(5/2)*(b*cos(d*x+c))^(1/2)*(A+B*cos(d*x+c)+C*cos(d*x+ c)^2),x, algorithm="maxima")
Output:
1/480*(15*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c))))*B*sqrt(b) + 2*C*sqrt(b)*(3*sin(5*d*x + 5*c) + 25*sin(3/5*arctan2(sin(5*d*x + 5*c), cos(5*d*x + 5*c))) + 150*sin(1/5*arct an2(sin(5*d*x + 5*c), cos(5*d*x + 5*c)))) + 40*A*sqrt(b)*(sin(3*d*x + 3*c) + 9*sin(1/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c)))))/d
Time = 0.34 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.42 \[ \int \cos ^{\frac {5}{2}}(c+d x) \sqrt {b \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {1}{480} \, {\left (180 \, B x + \frac {6 \, C \sin \left (5 \, d x + 5 \, c\right )}{d} + \frac {15 \, B \sin \left (4 \, d x + 4 \, c\right )}{d} + \frac {10 \, {\left (4 \, A + 5 \, C\right )} \sin \left (3 \, d x + 3 \, c\right )}{d} + \frac {120 \, B \sin \left (2 \, d x + 2 \, c\right )}{d} + \frac {60 \, {\left (6 \, A + 5 \, C\right )} \sin \left (d x + c\right )}{d}\right )} \sqrt {b} \] Input:
integrate(cos(d*x+c)^(5/2)*(b*cos(d*x+c))^(1/2)*(A+B*cos(d*x+c)+C*cos(d*x+ c)^2),x, algorithm="giac")
Output:
1/480*(180*B*x + 6*C*sin(5*d*x + 5*c)/d + 15*B*sin(4*d*x + 4*c)/d + 10*(4* A + 5*C)*sin(3*d*x + 3*c)/d + 120*B*sin(2*d*x + 2*c)/d + 60*(6*A + 5*C)*si n(d*x + c)/d)*sqrt(b)
Time = 43.38 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.63 \[ \int \cos ^{\frac {5}{2}}(c+d x) \sqrt {b \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {b\,\cos \left (c+d\,x\right )}\,\left (120\,B\,\sin \left (c+d\,x\right )+400\,A\,\sin \left (2\,c+2\,d\,x\right )+40\,A\,\sin \left (4\,c+4\,d\,x\right )+135\,B\,\sin \left (3\,c+3\,d\,x\right )+15\,B\,\sin \left (5\,c+5\,d\,x\right )+350\,C\,\sin \left (2\,c+2\,d\,x\right )+56\,C\,\sin \left (4\,c+4\,d\,x\right )+6\,C\,\sin \left (6\,c+6\,d\,x\right )+360\,B\,d\,x\,\cos \left (c+d\,x\right )\right )}{480\,d\,\left (\cos \left (2\,c+2\,d\,x\right )+1\right )} \] Input:
int(cos(c + d*x)^(5/2)*(b*cos(c + d*x))^(1/2)*(A + B*cos(c + d*x) + C*cos( c + d*x)^2),x)
Output:
(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*sin(6* c + 6*d*x) + 360*B*d*x*cos(c + d*x)))/(480*d*(cos(2*c + 2*d*x) + 1))
Time = 0.16 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.43 \[ \int \cos ^{\frac {5}{2}}(c+d x) \sqrt {b \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {\sqrt {b}\, \left (-30 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} b +75 \cos \left (d x +c \right ) \sin \left (d x +c \right ) b +24 \sin \left (d x +c \right )^{5} c -40 \sin \left (d x +c \right )^{3} a -80 \sin \left (d x +c \right )^{3} c +120 \sin \left (d x +c \right ) a +120 \sin \left (d x +c \right ) c +45 b d x \right )}{120 d} \] Input:
int(cos(d*x+c)^(5/2)*(b*cos(d*x+c))^(1/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2), x)
Output:
(sqrt(b)*( - 30*cos(c + d*x)*sin(c + d*x)**3*b + 75*cos(c + d*x)*sin(c + d *x)*b + 24*sin(c + d*x)**5*c - 40*sin(c + d*x)**3*a - 80*sin(c + d*x)**3*c + 120*sin(c + d*x)*a + 120*sin(c + d*x)*c + 45*b*d*x))/(120*d)