\(\int \cos ^{\frac {5}{2}}(c+d x) (b \cos (c+d x))^n (B \cos (c+d x)+C \cos ^2(c+d x)) \, dx\) [224]

Optimal result

Integrand size = 40, antiderivative size = 163 \[ \int \cos ^{\frac {5}{2}}(c+d x) (b \cos (c+d x))^n \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=-\frac {2 B \cos ^{\frac {9}{2}}(c+d x) (b \cos (c+d x))^n \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (9+2 n),\frac {1}{4} (13+2 n),\cos ^2(c+d x)\right ) \sin (c+d x)}{d (9+2 n) \sqrt {\sin ^2(c+d x)}}-\frac {2 C \cos ^{\frac {11}{2}}(c+d x) (b \cos (c+d x))^n \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (11+2 n),\frac {1}{4} (15+2 n),\cos ^2(c+d x)\right ) \sin (c+d x)}{d (11+2 n) \sqrt {\sin ^2(c+d x)}} \] Output:

-2*B*cos(d*x+c)^(9/2)*(b*cos(d*x+c))^n*hypergeom([1/2, 9/4+1/2*n],[13/4+1/ 
2*n],cos(d*x+c)^2)*sin(d*x+c)/d/(9+2*n)/(sin(d*x+c)^2)^(1/2)-2*C*cos(d*x+c 
)^(11/2)*(b*cos(d*x+c))^n*hypergeom([1/2, 11/4+1/2*n],[15/4+1/2*n],cos(d*x 
+c)^2)*sin(d*x+c)/d/(11+2*n)/(sin(d*x+c)^2)^(1/2)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.30 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.85 \[ \int \cos ^{\frac {5}{2}}(c+d x) (b \cos (c+d x))^n \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=-\frac {2 \cos ^{\frac {9}{2}}(c+d x) (b \cos (c+d x))^n \csc (c+d x) \left (B (11+2 n) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (9+2 n),\frac {1}{4} (13+2 n),\cos ^2(c+d x)\right )+C (9+2 n) \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (11+2 n),\frac {1}{4} (15+2 n),\cos ^2(c+d x)\right )\right ) \sqrt {\sin ^2(c+d x)}}{d (9+2 n) (11+2 n)} \] Input:

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

Output:

(-2*Cos[c + d*x]^(9/2)*(b*Cos[c + d*x])^n*Csc[c + d*x]*(B*(11 + 2*n)*Hyper 
geometric2F1[1/2, (9 + 2*n)/4, (13 + 2*n)/4, Cos[c + d*x]^2] + C*(9 + 2*n) 
*Cos[c + d*x]*Hypergeometric2F1[1/2, (11 + 2*n)/4, (15 + 2*n)/4, Cos[c + d 
*x]^2])*Sqrt[Sin[c + d*x]^2])/(d*(9 + 2*n)*(11 + 2*n))
 

Rubi [A] (verified)

Time = 0.56 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.03, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.175, Rules used = {2034, 3042, 3489, 3042, 3227, 3042, 3122}

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.

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

\(\Big \downarrow \) 2034

\(\displaystyle \cos ^{-n}(c+d x) (b \cos (c+d x))^n \int \cos ^{n+\frac {5}{2}}(c+d x) \left (C \cos ^2(c+d x)+B \cos (c+d x)\right )dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \cos ^{-n}(c+d x) (b \cos (c+d x))^n \int \sin \left (c+d x+\frac {\pi }{2}\right )^{n+\frac {5}{2}} \left (C \sin \left (c+d x+\frac {\pi }{2}\right )^2+B \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx\)

\(\Big \downarrow \) 3489

\(\displaystyle \cos ^{-n}(c+d x) (b \cos (c+d x))^n \int \cos ^{n+\frac {7}{2}}(c+d x) (B+C \cos (c+d x))dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \cos ^{-n}(c+d x) (b \cos (c+d x))^n \int \sin \left (c+d x+\frac {\pi }{2}\right )^{n+\frac {7}{2}} \left (B+C \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx\)

\(\Big \downarrow \) 3227

\(\displaystyle \cos ^{-n}(c+d x) (b \cos (c+d x))^n \left (B \int \cos ^{n+\frac {7}{2}}(c+d x)dx+C \int \cos ^{n+\frac {9}{2}}(c+d x)dx\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \cos ^{-n}(c+d x) (b \cos (c+d x))^n \left (B \int \sin \left (c+d x+\frac {\pi }{2}\right )^{n+\frac {7}{2}}dx+C \int \sin \left (c+d x+\frac {\pi }{2}\right )^{n+\frac {9}{2}}dx\right )\)

\(\Big \downarrow \) 3122

\(\displaystyle \cos ^{-n}(c+d x) (b \cos (c+d x))^n \left (-\frac {2 B \sin (c+d x) \cos ^{n+\frac {9}{2}}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (2 n+9),\frac {1}{4} (2 n+13),\cos ^2(c+d x)\right )}{d (2 n+9) \sqrt {\sin ^2(c+d x)}}-\frac {2 C \sin (c+d x) \cos ^{n+\frac {11}{2}}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (2 n+11),\frac {1}{4} (2 n+15),\cos ^2(c+d x)\right )}{d (2 n+11) \sqrt {\sin ^2(c+d x)}}\right )\)

Input:

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

Output:

((b*Cos[c + d*x])^n*((-2*B*Cos[c + d*x]^(9/2 + n)*Hypergeometric2F1[1/2, ( 
9 + 2*n)/4, (13 + 2*n)/4, Cos[c + d*x]^2]*Sin[c + d*x])/(d*(9 + 2*n)*Sqrt[ 
Sin[c + d*x]^2]) - (2*C*Cos[c + d*x]^(11/2 + n)*Hypergeometric2F1[1/2, (11 
 + 2*n)/4, (15 + 2*n)/4, Cos[c + d*x]^2]*Sin[c + d*x])/(d*(11 + 2*n)*Sqrt[ 
Sin[c + d*x]^2])))/Cos[c + d*x]^n
 

Defintions of rubi rules used

rule 2034

Int[(Fx_.)*((a_.)*(v_))^(m_)*((b_.)*(v_))^(n_), x_Symbol] :> Simp[b^IntPart 
[n]*((b*v)^FracPart[n]/(a^IntPart[n]*(a*v)^FracPart[n]))   Int[(a*v)^(m + n 
)*Fx, x], x] /; FreeQ[{a, b, m, n}, x] &&  !IntegerQ[m] &&  !IntegerQ[n] && 
  !IntegerQ[m + n]
 

rule 3042

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

rule 3122

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*(( 
b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1)*Sqrt[Cos[c + d*x]^2]))*Hypergeometric2 
F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2], x] /; FreeQ[{b, c, d, n}, x] 
 &&  !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 3489

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((B_.)*sin[(e_.) + (f_.)*(x_)] + 
(C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[1/b   Int[(b*Sin[e + f* 
x])^(m + 1)*(B + C*Sin[e + f*x]), x], x] /; FreeQ[{b, e, f, B, C, m}, x]
 

Maple [F]

\[\int \cos \left (d x +c \right )^{\frac {5}{2}} \left (b \cos \left (d x +c \right )\right )^{n} \left (B \cos \left (d x +c \right )+C \cos \left (d x +c \right )^{2}\right )d x\]

Input:

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

Output:

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

Fricas [F]

\[ \int \cos ^{\frac {5}{2}}(c+d x) (b \cos (c+d x))^n \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right )\right )} \left (b \cos \left (d x + c\right )\right )^{n} \cos \left (d x + c\right )^{\frac {5}{2}} \,d x } \] Input:

integrate(cos(d*x+c)^(5/2)*(b*cos(d*x+c))^n*(B*cos(d*x+c)+C*cos(d*x+c)^2), 
x, algorithm="fricas")
 

Output:

integral((C*cos(d*x + c)^4 + B*cos(d*x + c)^3)*(b*cos(d*x + c))^n*sqrt(cos 
(d*x + c)), x)
 

Sympy [F(-1)]

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

integrate(cos(d*x+c)**(5/2)*(b*cos(d*x+c))**n*(B*cos(d*x+c)+C*cos(d*x+c)** 
2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \cos ^{\frac {5}{2}}(c+d x) (b \cos (c+d x))^n \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right )\right )} \left (b \cos \left (d x + c\right )\right )^{n} \cos \left (d x + c\right )^{\frac {5}{2}} \,d x } \] Input:

integrate(cos(d*x+c)^(5/2)*(b*cos(d*x+c))^n*(B*cos(d*x+c)+C*cos(d*x+c)^2), 
x, algorithm="maxima")
 

Output:

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c))*(b*cos(d*x + c))^n*cos(d*x + 
 c)^(5/2), x)
 

Giac [F]

\[ \int \cos ^{\frac {5}{2}}(c+d x) (b \cos (c+d x))^n \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right )\right )} \left (b \cos \left (d x + c\right )\right )^{n} \cos \left (d x + c\right )^{\frac {5}{2}} \,d x } \] Input:

integrate(cos(d*x+c)^(5/2)*(b*cos(d*x+c))^n*(B*cos(d*x+c)+C*cos(d*x+c)^2), 
x, algorithm="giac")
 

Output:

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c))*(b*cos(d*x + c))^n*cos(d*x + 
 c)^(5/2), x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \cos ^{\frac {5}{2}}(c+d x) (b \cos (c+d x))^n \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=b^{n} \left (\left (\int \cos \left (d x +c \right )^{n +\frac {1}{2}} \cos \left (d x +c \right )^{4}d x \right ) c +\left (\int \cos \left (d x +c \right )^{n +\frac {1}{2}} \cos \left (d x +c \right )^{3}d x \right ) b \right ) \] Input:

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

Output:

b**n*(int(cos(c + d*x)**((2*n + 1)/2)*cos(c + d*x)**4,x)*c + int(cos(c + d 
*x)**((2*n + 1)/2)*cos(c + d*x)**3,x)*b)