\(\int \frac {(b \cos (c+d x))^{5/2} (A+B \cos (c+d x)+C \cos ^2(c+d x))}{\sqrt {\cos (c+d x)}} \, dx\) [307]

Optimal result

Integrand size = 43, antiderivative size = 199 \[ \int \frac {(b \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx=\frac {b^2 (4 A+3 C) x \sqrt {b \cos (c+d x)}}{8 \sqrt {\cos (c+d x)}}+\frac {b^2 B \sqrt {b \cos (c+d x)} \sin (c+d x)}{d \sqrt {\cos (c+d x)}}+\frac {b^2 (4 A+3 C) \sqrt {\cos (c+d x)} \sqrt {b \cos (c+d x)} \sin (c+d x)}{8 d}+\frac {b^2 C \cos ^{\frac {5}{2}}(c+d x) \sqrt {b \cos (c+d x)} \sin (c+d x)}{4 d}-\frac {b^2 B \sqrt {b \cos (c+d x)} \sin ^3(c+d x)}{3 d \sqrt {\cos (c+d x)}} \] Output:

1/8*b^2*(4*A+3*C)*x*(b*cos(d*x+c))^(1/2)/cos(d*x+c)^(1/2)+b^2*B*(b*cos(d*x 
+c))^(1/2)*sin(d*x+c)/d/cos(d*x+c)^(1/2)+1/8*b^2*(4*A+3*C)*cos(d*x+c)^(1/2 
)*(b*cos(d*x+c))^(1/2)*sin(d*x+c)/d+1/4*b^2*C*cos(d*x+c)^(5/2)*(b*cos(d*x+ 
c))^(1/2)*sin(d*x+c)/d-1/3*b^2*B*(b*cos(d*x+c))^(1/2)*sin(d*x+c)^3/d/cos(d 
*x+c)^(1/2)
 

Mathematica [A] (verified)

Time = 1.38 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.46 \[ \int \frac {(b \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx=\frac {(b \cos (c+d x))^{5/2} (48 A c+36 c C+48 A d x+36 C d x+72 B \sin (c+d x)+24 (A+C) \sin (2 (c+d x))+8 B \sin (3 (c+d x))+3 C \sin (4 (c+d x)))}{96 d \cos ^{\frac {5}{2}}(c+d x)} \] Input:

Integrate[((b*Cos[c + d*x])^(5/2)*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)) 
/Sqrt[Cos[c + d*x]],x]
 

Output:

((b*Cos[c + d*x])^(5/2)*(48*A*c + 36*c*C + 48*A*d*x + 36*C*d*x + 72*B*Sin[ 
c + d*x] + 24*(A + C)*Sin[2*(c + d*x)] + 8*B*Sin[3*(c + d*x)] + 3*C*Sin[4* 
(c + d*x)]))/(96*d*Cos[c + d*x]^(5/2))
 

Rubi [A] (verified)

Time = 0.52 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.57, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.233, Rules used = {2031, 3042, 3502, 3042, 3227, 3042, 3113, 2009, 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.

\(\displaystyle \int \frac {(b \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx\)

\(\Big \downarrow \) 2031

\(\displaystyle \frac {b^2 \sqrt {b \cos (c+d x)} \int \cos ^2(c+d x) \left (C \cos ^2(c+d x)+B \cos (c+d x)+A\right )dx}{\sqrt {\cos (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {b^2 \sqrt {b \cos (c+d x)} \int \sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (C \sin \left (c+d x+\frac {\pi }{2}\right )^2+B \sin \left (c+d x+\frac {\pi }{2}\right )+A\right )dx}{\sqrt {\cos (c+d x)}}\)

\(\Big \downarrow \) 3502

\(\displaystyle \frac {b^2 \sqrt {b \cos (c+d x)} \left (\frac {1}{4} \int \cos ^2(c+d x) (4 A+3 C+4 B \cos (c+d x))dx+\frac {C \sin (c+d x) \cos ^3(c+d x)}{4 d}\right )}{\sqrt {\cos (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {b^2 \sqrt {b \cos (c+d x)} \left (\frac {1}{4} \int \sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (4 A+3 C+4 B \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx+\frac {C \sin (c+d x) \cos ^3(c+d x)}{4 d}\right )}{\sqrt {\cos (c+d x)}}\)

\(\Big \downarrow \) 3227

\(\displaystyle \frac {b^2 \sqrt {b \cos (c+d x)} \left (\frac {1}{4} \left ((4 A+3 C) \int \cos ^2(c+d x)dx+4 B \int \cos ^3(c+d x)dx\right )+\frac {C \sin (c+d x) \cos ^3(c+d x)}{4 d}\right )}{\sqrt {\cos (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {b^2 \sqrt {b \cos (c+d x)} \left (\frac {1}{4} \left ((4 A+3 C) \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx+4 B \int \sin \left (c+d x+\frac {\pi }{2}\right )^3dx\right )+\frac {C \sin (c+d x) \cos ^3(c+d x)}{4 d}\right )}{\sqrt {\cos (c+d x)}}\)

\(\Big \downarrow \) 3113

\(\displaystyle \frac {b^2 \sqrt {b \cos (c+d x)} \left (\frac {1}{4} \left ((4 A+3 C) \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx-\frac {4 B \int \left (1-\sin ^2(c+d x)\right )d(-\sin (c+d x))}{d}\right )+\frac {C \sin (c+d x) \cos ^3(c+d x)}{4 d}\right )}{\sqrt {\cos (c+d x)}}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {b^2 \sqrt {b \cos (c+d x)} \left (\frac {1}{4} \left ((4 A+3 C) \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx-\frac {4 B \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\right )+\frac {C \sin (c+d x) \cos ^3(c+d x)}{4 d}\right )}{\sqrt {\cos (c+d x)}}\)

\(\Big \downarrow \) 3115

\(\displaystyle \frac {b^2 \sqrt {b \cos (c+d x)} \left (\frac {1}{4} \left ((4 A+3 C) \left (\frac {\int 1dx}{2}+\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )-\frac {4 B \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\right )+\frac {C \sin (c+d x) \cos ^3(c+d x)}{4 d}\right )}{\sqrt {\cos (c+d x)}}\)

\(\Big \downarrow \) 24

\(\displaystyle \frac {b^2 \sqrt {b \cos (c+d x)} \left (\frac {1}{4} \left ((4 A+3 C) \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )-\frac {4 B \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\right )+\frac {C \sin (c+d x) \cos ^3(c+d x)}{4 d}\right )}{\sqrt {\cos (c+d x)}}\)

Input:

Int[((b*Cos[c + d*x])^(5/2)*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2))/Sqrt[ 
Cos[c + d*x]],x]
 

Output:

(b^2*Sqrt[b*Cos[c + d*x]]*((C*Cos[c + d*x]^3*Sin[c + d*x])/(4*d) + ((4*A + 
 3*C)*(x/2 + (Cos[c + d*x]*Sin[c + d*x])/(2*d)) - (4*B*(-Sin[c + d*x] + Si 
n[c + d*x]^3/3))/d)/4))/Sqrt[Cos[c + d*x]]
 

Defintions of rubi rules used

rule 24

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2031

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]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3113

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]
 

rule 3115

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]
 

rule 3227

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]
 

rule 3502

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]
 

Maple [A] (verified)

Time = 0.38 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.53

method	result	size
default	\(\frac {b^{2} \left (12 A \left (d x +c \right )+9 C \left (d x +c \right )+12 A \cos \left (d x +c \right ) \sin \left (d x +c \right )+\left (8 \cos \left (d x +c \right )^{2}+16\right ) \sin \left (d x +c \right ) B +\sin \left (d x +c \right ) \cos \left (d x +c \right ) \left (6 \cos \left (d x +c \right )^{2}+9\right ) C \right ) \sqrt {b \cos \left (d x +c \right )}}{24 d \sqrt {\cos \left (d x +c \right )}}\)	\(105\)
parts	\(\frac {A \left (\cos \left (d x +c \right ) \sin \left (d x +c \right )+d x +c \right ) b^{2} \sqrt {b \cos \left (d x +c \right )}}{2 d \sqrt {\cos \left (d x +c \right )}}+\frac {B \sin \left (d x +c \right ) \left (2+\cos \left (d x +c \right )^{2}\right ) b^{2} \sqrt {b \cos \left (d x +c \right )}}{3 d \sqrt {\cos \left (d x +c \right )}}+\frac {C \left (2 \cos \left (d x +c \right )^{3} \sin \left (d x +c \right )+3 \cos \left (d x +c \right ) \sin \left (d x +c \right )+3 d x +3 c \right ) b^{2} \sqrt {b \cos \left (d x +c \right )}}{8 d \sqrt {\cos \left (d x +c \right )}}\)	\(155\)
risch	\(\frac {b^{2} \sqrt {b \cos \left (d x +c \right )}\, \left (8 A +6 C \right ) x}{16 \sqrt {\cos \left (d x +c \right )}}+\frac {3 b^{2} B \sqrt {b \cos \left (d x +c \right )}\, \sin \left (d x +c \right )}{4 d \sqrt {\cos \left (d x +c \right )}}+\frac {b^{2} \sqrt {b \cos \left (d x +c \right )}\, C \sin \left (4 d x +4 c \right )}{32 \sqrt {\cos \left (d x +c \right )}\, d}+\frac {b^{2} \sqrt {b \cos \left (d x +c \right )}\, B \sin \left (3 d x +3 c \right )}{12 \sqrt {\cos \left (d x +c \right )}\, d}+\frac {b^{2} \sqrt {b \cos \left (d x +c \right )}\, \left (A +C \right ) \sin \left (2 d x +2 c \right )}{4 \sqrt {\cos \left (d x +c \right )}\, d}\)	\(176\)

Input:

int((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,method=_RETURNVERBOSE)
 

Output:

1/24*b^2/d*(12*A*(d*x+c)+9*C*(d*x+c)+12*A*cos(d*x+c)*sin(d*x+c)+(8*cos(d*x 
+c)^2+16)*sin(d*x+c)*B+sin(d*x+c)*cos(d*x+c)*(6*cos(d*x+c)^2+9)*C)*(b*cos( 
d*x+c))^(1/2)/cos(d*x+c)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 303, normalized size of antiderivative = 1.52 \[ \int \frac {(b \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx=\left [\frac {3 \, {\left (4 \, A + 3 \, C\right )} \sqrt {-b} b^{2} \cos \left (d x + c\right ) \log \left (2 \, b \cos \left (d x + c\right )^{2} - 2 \, \sqrt {b \cos \left (d x + c\right )} \sqrt {-b} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - b\right ) + 2 \, {\left (6 \, C b^{2} \cos \left (d x + c\right )^{3} + 8 \, B b^{2} \cos \left (d x + c\right )^{2} + 3 \, {\left (4 \, A + 3 \, C\right )} b^{2} \cos \left (d x + c\right ) + 16 \, B b^{2}\right )} \sqrt {b \cos \left (d x + c\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )}, \frac {3 \, {\left (4 \, A + 3 \, C\right )} b^{\frac {5}{2}} \arctan \left (\frac {\sqrt {b \cos \left (d x + c\right )} \sin \left (d x + c\right )}{\sqrt {b} \cos \left (d x + c\right )^{\frac {3}{2}}}\right ) \cos \left (d x + c\right ) + {\left (6 \, C b^{2} \cos \left (d x + c\right )^{3} + 8 \, B b^{2} \cos \left (d x + c\right )^{2} + 3 \, {\left (4 \, A + 3 \, C\right )} b^{2} \cos \left (d x + c\right ) + 16 \, B b^{2}\right )} \sqrt {b \cos \left (d x + c\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{24 \, d \cos \left (d x + c\right )}\right ] \] Input:

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")
 

Output:

[1/48*(3*(4*A + 3*C)*sqrt(-b)*b^2*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*(6* 
C*b^2*cos(d*x + c)^3 + 8*B*b^2*cos(d*x + c)^2 + 3*(4*A + 3*C)*b^2*cos(d*x 
+ c) + 16*B*b^2)*sqrt(b*cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c))/(d* 
cos(d*x + c)), 1/24*(3*(4*A + 3*C)*b^(5/2)*arctan(sqrt(b*cos(d*x + c))*sin 
(d*x + c)/(sqrt(b)*cos(d*x + c)^(3/2)))*cos(d*x + c) + (6*C*b^2*cos(d*x + 
c)^3 + 8*B*b^2*cos(d*x + c)^2 + 3*(4*A + 3*C)*b^2*cos(d*x + c) + 16*B*b^2) 
*sqrt(b*cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c))]
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(b \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx=\text {Timed out} \] Input:

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)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.33 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.70 \[ \int \frac {(b \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx=\frac {24 \, {\left (2 \, {\left (d x + c\right )} b^{2} + b^{2} \sin \left (2 \, d x + 2 \, c\right )\right )} A \sqrt {b} + 8 \, {\left (b^{2} \sin \left (3 \, d x + 3 \, c\right ) + 9 \, b^{2} \sin \left (\frac {1}{3} \, \arctan \left (\sin \left (3 \, d x + 3 \, c\right ), \cos \left (3 \, d x + 3 \, c\right )\right )\right )\right )} B \sqrt {b} + 3 \, {\left (12 \, {\left (d x + c\right )} b^{2} + b^{2} \sin \left (4 \, d x + 4 \, c\right ) + 8 \, b^{2} \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (4 \, d x + 4 \, c\right ), \cos \left (4 \, d x + 4 \, c\right )\right )\right )\right )} C \sqrt {b}}{96 \, d} \] Input:

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")
 

Output:

1/96*(24*(2*(d*x + c)*b^2 + b^2*sin(2*d*x + 2*c))*A*sqrt(b) + 8*(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))))* 
B*sqrt(b) + 3*(12*(d*x + c)*b^2 + b^2*sin(4*d*x + 4*c) + 8*b^2*sin(1/2*arc 
tan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c))))*C*sqrt(b))/d
 

Giac [A] (verification not implemented)

Time = 0.35 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.49 \[ \int \frac {(b \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx=\frac {1}{96} \, {\left (\frac {3 \, C b^{2} \sin \left (4 \, d x + 4 \, c\right )}{d} + \frac {8 \, B b^{2} \sin \left (3 \, d x + 3 \, c\right )}{d} + \frac {72 \, B b^{2} \sin \left (d x + c\right )}{d} + 12 \, {\left (4 \, A b^{2} + 3 \, C b^{2}\right )} x + \frac {24 \, {\left (A b^{2} + C b^{2}\right )} \sin \left (2 \, d x + 2 \, c\right )}{d}\right )} \sqrt {b} \] Input:

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")
 

Output:

1/96*(3*C*b^2*sin(4*d*x + 4*c)/d + 8*B*b^2*sin(3*d*x + 3*c)/d + 72*B*b^2*s 
in(d*x + c)/d + 12*(4*A*b^2 + 3*C*b^2)*x + 24*(A*b^2 + C*b^2)*sin(2*d*x + 
2*c)/d)*sqrt(b)
 

Mupad [B] (verification not implemented)

Time = 1.24 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.47 \[ \int \frac {(b \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx=\frac {b^2\,\sqrt {b\,\cos \left (c+d\,x\right )}\,\left (72\,B\,\sin \left (c+d\,x\right )+24\,A\,\sin \left (2\,c+2\,d\,x\right )+8\,B\,\sin \left (3\,c+3\,d\,x\right )+24\,C\,\sin \left (2\,c+2\,d\,x\right )+3\,C\,\sin \left (4\,c+4\,d\,x\right )+48\,A\,d\,x+36\,C\,d\,x\right )}{96\,d\,\sqrt {\cos \left (c+d\,x\right )}} \] Input:

int(((b*cos(c + d*x))^(5/2)*(A + B*cos(c + d*x) + C*cos(c + d*x)^2))/cos(c 
 + d*x)^(1/2),x)
 

Output:

(b^2*(b*cos(c + d*x))^(1/2)*(72*B*sin(c + d*x) + 24*A*sin(2*c + 2*d*x) + 8 
*B*sin(3*c + 3*d*x) + 24*C*sin(2*c + 2*d*x) + 3*C*sin(4*c + 4*d*x) + 48*A* 
d*x + 36*C*d*x))/(96*d*cos(c + d*x)^(1/2))
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.44 \[ \int \frac {(b \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx=\frac {\sqrt {b}\, b^{2} \left (-6 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} c +12 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a +15 \cos \left (d x +c \right ) \sin \left (d x +c \right ) c -8 \sin \left (d x +c \right )^{3} b +24 \sin \left (d x +c \right ) b +12 a d x +9 c d x \right )}{24 d} \] Input:

int((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)
 

Output:

(sqrt(b)*b**2*( - 6*cos(c + d*x)*sin(c + d*x)**3*c + 12*cos(c + d*x)*sin(c 
 + d*x)*a + 15*cos(c + d*x)*sin(c + d*x)*c - 8*sin(c + d*x)**3*b + 24*sin( 
c + d*x)*b + 12*a*d*x + 9*c*d*x))/(24*d)