\(\int \cos ^{\frac {3}{2}}(c+d x) (b \cos (c+d x))^n (A+C \cos ^2(c+d x)) \, dx\) [191]

Optimal result

Integrand size = 33, antiderivative size = 142 \[ \int \cos ^{\frac {3}{2}}(c+d x) (b \cos (c+d x))^n \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {2 C \cos ^{\frac {5}{2}}(c+d x) (b \cos (c+d x))^n \sin (c+d x)}{d (7+2 n)}-\frac {2 (C (5+2 n)+A (7+2 n)) \cos ^{\frac {5}{2}}(c+d x) (b \cos (c+d x))^n \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (5+2 n),\frac {1}{4} (9+2 n),\cos ^2(c+d x)\right ) \sin (c+d x)}{d (5+2 n) (7+2 n) \sqrt {\sin ^2(c+d x)}} \] Output:

2*C*cos(d*x+c)^(5/2)*(b*cos(d*x+c))^n*sin(d*x+c)/d/(7+2*n)-2*(C*(5+2*n)+A* 
(7+2*n))*cos(d*x+c)^(5/2)*(b*cos(d*x+c))^n*hypergeom([1/2, 5/4+1/2*n],[9/4 
+1/2*n],cos(d*x+c)^2)*sin(d*x+c)/d/(5+2*n)/(7+2*n)/(sin(d*x+c)^2)^(1/2)
 

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.99 \[ \int \cos ^{\frac {3}{2}}(c+d x) (b \cos (c+d x))^n \left (A+C \cos ^2(c+d x)\right ) \, dx=-\frac {2 \cos ^{\frac {5}{2}}(c+d x) (b \cos (c+d x))^n \csc (c+d x) \left (A (9+2 n) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (5+2 n),\frac {1}{4} (9+2 n),\cos ^2(c+d x)\right )+C (5+2 n) \cos ^2(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (9+2 n),\frac {1}{4} (13+2 n),\cos ^2(c+d x)\right )\right ) \sqrt {\sin ^2(c+d x)}}{d (5+2 n) (9+2 n)} \] Input:

Integrate[Cos[c + d*x]^(3/2)*(b*Cos[c + d*x])^n*(A + C*Cos[c + d*x]^2),x]
 

Output:

(-2*Cos[c + d*x]^(5/2)*(b*Cos[c + d*x])^n*Csc[c + d*x]*(A*(9 + 2*n)*Hyperg 
eometric2F1[1/2, (5 + 2*n)/4, (9 + 2*n)/4, Cos[c + d*x]^2] + C*(5 + 2*n)*C 
os[c + d*x]^2*Hypergeometric2F1[1/2, (9 + 2*n)/4, (13 + 2*n)/4, Cos[c + d* 
x]^2])*Sqrt[Sin[c + d*x]^2])/(d*(5 + 2*n)*(9 + 2*n))
 

Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.04, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {2034, 3042, 3493, 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.

\(\displaystyle \int \cos ^{\frac {3}{2}}(c+d x) \left (A+C \cos ^2(c+d x)\right ) (b \cos (c+d x))^n \, dx\)

\(\Big \downarrow \) 2034

\(\displaystyle \cos ^{-n}(c+d x) (b \cos (c+d x))^n \int \cos ^{n+\frac {3}{2}}(c+d x) \left (C \cos ^2(c+d x)+A\right )dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3493

\(\displaystyle \cos ^{-n}(c+d x) (b \cos (c+d x))^n \left (\frac {(A (2 n+7)+C (2 n+5)) \int \cos ^{n+\frac {3}{2}}(c+d x)dx}{2 n+7}+\frac {2 C \sin (c+d x) \cos ^{n+\frac {5}{2}}(c+d x)}{d (2 n+7)}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \cos ^{-n}(c+d x) (b \cos (c+d x))^n \left (\frac {(A (2 n+7)+C (2 n+5)) \int \sin \left (c+d x+\frac {\pi }{2}\right )^{n+\frac {3}{2}}dx}{2 n+7}+\frac {2 C \sin (c+d x) \cos ^{n+\frac {5}{2}}(c+d x)}{d (2 n+7)}\right )\)

\(\Big \downarrow \) 3122

\(\displaystyle \cos ^{-n}(c+d x) (b \cos (c+d x))^n \left (\frac {2 C \sin (c+d x) \cos ^{n+\frac {5}{2}}(c+d x)}{d (2 n+7)}-\frac {2 (A (2 n+7)+C (2 n+5)) \sin (c+d x) \cos ^{n+\frac {5}{2}}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (2 n+5),\frac {1}{4} (2 n+9),\cos ^2(c+d x)\right )}{d (2 n+5) (2 n+7) \sqrt {\sin ^2(c+d x)}}\right )\)

Input:

Int[Cos[c + d*x]^(3/2)*(b*Cos[c + d*x])^n*(A + C*Cos[c + d*x]^2),x]
 

Output:

((b*Cos[c + d*x])^n*((2*C*Cos[c + d*x]^(5/2 + n)*Sin[c + d*x])/(d*(7 + 2*n 
)) - (2*(C*(5 + 2*n) + A*(7 + 2*n))*Cos[c + d*x]^(5/2 + n)*Hypergeometric2 
F1[1/2, (5 + 2*n)/4, (9 + 2*n)/4, Cos[c + d*x]^2]*Sin[c + d*x])/(d*(5 + 2* 
n)*(7 + 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 3493

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_) + (C_.)*sin[(e_.) + (f_.)*( 
x_)]^2), x_Symbol] :> Simp[(-C)*Cos[e + f*x]*((b*Sin[e + f*x])^(m + 1)/(b*f 
*(m + 2))), x] + Simp[(A*(m + 2) + C*(m + 1))/(m + 2)   Int[(b*Sin[e + f*x] 
)^m, x], x] /; FreeQ[{b, e, f, A, C, m}, x] &&  !LtQ[m, -1]
 

Maple [F]

\[\int \cos \left (d x +c \right )^{\frac {3}{2}} \left (b \cos \left (d x +c \right )\right )^{n} \left (A +C \cos \left (d x +c \right )^{2}\right )d x\]

Input:

int(cos(d*x+c)^(3/2)*(b*cos(d*x+c))^n*(A+C*cos(d*x+c)^2),x)
 

Output:

int(cos(d*x+c)^(3/2)*(b*cos(d*x+c))^n*(A+C*cos(d*x+c)^2),x)
 

Fricas [F]

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

integrate(cos(d*x+c)^(3/2)*(b*cos(d*x+c))^n*(A+C*cos(d*x+c)^2),x, algorith 
m="fricas")
 

Output:

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

Sympy [F(-1)]

Timed out. \[ \int \cos ^{\frac {3}{2}}(c+d x) (b \cos (c+d x))^n \left (A+C \cos ^2(c+d x)\right ) \, dx=\text {Timed out} \] Input:

integrate(cos(d*x+c)**(3/2)*(b*cos(d*x+c))**n*(A+C*cos(d*x+c)**2),x)
 

Output:

Timed out
 

Maxima [F]

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

integrate(cos(d*x+c)^(3/2)*(b*cos(d*x+c))^n*(A+C*cos(d*x+c)^2),x, algorith 
m="maxima")
 

Output:

integrate((C*cos(d*x + c)^2 + A)*(b*cos(d*x + c))^n*cos(d*x + c)^(3/2), x)
 

Giac [F]

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

integrate(cos(d*x+c)^(3/2)*(b*cos(d*x+c))^n*(A+C*cos(d*x+c)^2),x, algorith 
m="giac")
 

Output:

integrate((C*cos(d*x + c)^2 + A)*(b*cos(d*x + c))^n*cos(d*x + c)^(3/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \cos ^{\frac {3}{2}}(c+d x) (b \cos (c+d x))^n \left (A+C \cos ^2(c+d x)\right ) \, dx=\int {\cos \left (c+d\,x\right )}^{3/2}\,\left (C\,{\cos \left (c+d\,x\right )}^2+A\right )\,{\left (b\,\cos \left (c+d\,x\right )\right )}^n \,d x \] Input:

int(cos(c + d*x)^(3/2)*(A + C*cos(c + d*x)^2)*(b*cos(c + d*x))^n,x)
 

Output:

int(cos(c + d*x)^(3/2)*(A + C*cos(c + d*x)^2)*(b*cos(c + d*x))^n, x)
 

Reduce [F]

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

int(cos(d*x+c)^(3/2)*(b*cos(d*x+c))^n*(A+C*cos(d*x+c)^2),x)
 

Output:

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