3.4.79 \(\int \frac {(b \cos (c+d x))^n (A+B \cos (c+d x)+C \cos ^2(c+d x))}{\sqrt {\cos (c+d x)}} \, dx\) [379]

3.4.79.1 Optimal result

Integrand size = 41, antiderivative size = 221 \[ \int \frac {(b \cos (c+d x))^n \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx=\frac {2 C \sqrt {\cos (c+d x)} (b \cos (c+d x))^n \sin (c+d x)}{d (3+2 n)}-\frac {2 (C+2 C n+A (3+2 n)) \sqrt {\cos (c+d x)} (b \cos (c+d x))^n \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (1+2 n),\frac {1}{4} (5+2 n),\cos ^2(c+d x)\right ) \sin (c+d x)}{d (1+2 n) (3+2 n) \sqrt {\sin ^2(c+d x)}}-\frac {2 B \cos ^{\frac {3}{2}}(c+d x) (b \cos (c+d x))^n \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (3+2 n),\frac {1}{4} (7+2 n),\cos ^2(c+d x)\right ) \sin (c+d x)}{d (3+2 n) \sqrt {\sin ^2(c+d x)}} \]

output

2*C*(b*cos(d*x+c))^n*sin(d*x+c)*cos(d*x+c)^(1/2)/d/(3+2*n)-2*B*cos(d*x+c)^ 
(3/2)*(b*cos(d*x+c))^n*hypergeom([1/2, 3/4+1/2*n],[7/4+1/2*n],cos(d*x+c)^2 
)*sin(d*x+c)/d/(3+2*n)/(sin(d*x+c)^2)^(1/2)-2*(C+2*C*n+A*(3+2*n))*(b*cos(d 
*x+c))^n*hypergeom([1/2, 1/4+1/2*n],[5/4+1/2*n],cos(d*x+c)^2)*sin(d*x+c)*c 
os(d*x+c)^(1/2)/d/(4*n^2+8*n+3)/(sin(d*x+c)^2)^(1/2)
 

3.4.79.2 Mathematica [A] (verified)

Time = 0.30 (sec) , antiderivative size = 171, normalized size of antiderivative = 0.77 \[ \int \frac {(b \cos (c+d x))^n \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx=\frac {2 \sqrt {\cos (c+d x)} (b \cos (c+d x))^n \csc (c+d x) \left (-\left ((C+2 C n+A (3+2 n)) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (1+2 n),\frac {1}{4} (5+2 n),\cos ^2(c+d x)\right ) \sqrt {\sin ^2(c+d x)}\right )+(1+2 n) \left (C \sin ^2(c+d x)-B \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (3+2 n),\frac {1}{4} (7+2 n),\cos ^2(c+d x)\right ) \sqrt {\sin ^2(c+d x)}\right )\right )}{d (1+2 n) (3+2 n)} \]

input

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

output

(2*Sqrt[Cos[c + d*x]]*(b*Cos[c + d*x])^n*Csc[c + d*x]*(-((C + 2*C*n + A*(3 
 + 2*n))*Hypergeometric2F1[1/2, (1 + 2*n)/4, (5 + 2*n)/4, Cos[c + d*x]^2]* 
Sqrt[Sin[c + d*x]^2]) + (1 + 2*n)*(C*Sin[c + d*x]^2 - B*Cos[c + d*x]*Hyper 
geometric2F1[1/2, (3 + 2*n)/4, (7 + 2*n)/4, Cos[c + d*x]^2]*Sqrt[Sin[c + d 
*x]^2])))/(d*(1 + 2*n)*(3 + 2*n))
 

3.4.79.3 Rubi [A] (verified)

Time = 0.61 (sec) , antiderivative size = 213, normalized size of antiderivative = 0.96, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.195, Rules used = {2034, 3042, 3502, 27, 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.

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

\(\Big \downarrow \) 2034

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3502

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3227

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3122

\(\displaystyle \cos ^{-n}(c+d x) (b \cos (c+d x))^n \left (\frac {-\frac {2 (A (2 n+3)+2 C n+C) \sin (c+d x) \cos ^{n+\frac {1}{2}}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (2 n+1),\frac {1}{4} (2 n+5),\cos ^2(c+d x)\right )}{d (2 n+1) \sqrt {\sin ^2(c+d x)}}-\frac {2 B \sin (c+d x) \cos ^{n+\frac {3}{2}}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (2 n+3),\frac {1}{4} (2 n+7),\cos ^2(c+d x)\right )}{d \sqrt {\sin ^2(c+d x)}}}{2 n+3}+\frac {2 C \sin (c+d x) \cos ^{n+\frac {1}{2}}(c+d x)}{d (2 n+3)}\right )\)

input

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

output

((b*Cos[c + d*x])^n*((2*C*Cos[c + d*x]^(1/2 + n)*Sin[c + d*x])/(d*(3 + 2*n 
)) + ((-2*(C + 2*C*n + A*(3 + 2*n))*Cos[c + d*x]^(1/2 + n)*Hypergeometric2 
F1[1/2, (1 + 2*n)/4, (5 + 2*n)/4, Cos[c + d*x]^2]*Sin[c + d*x])/(d*(1 + 2* 
n)*Sqrt[Sin[c + d*x]^2]) - (2*B*Cos[c + d*x]^(3/2 + n)*Hypergeometric2F1[1 
/2, (3 + 2*n)/4, (7 + 2*n)/4, Cos[c + d*x]^2]*Sin[c + d*x])/(d*Sqrt[Sin[c 
+ d*x]^2]))/(3 + 2*n)))/Cos[c + d*x]^n
 

3.4.79.3.1 Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

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 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]
 

3.4.79.4 Maple [F]

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

input

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

output

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

3.4.79.5 Fricas [F]

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

input

integrate((b*cos(d*x+c))^n*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(1/2 
),x, algorithm="fricas")
 

output

integral((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*(b*cos(d*x + c))^n/sqrt(c 
os(d*x + c)), x)
 

3.4.79.6 Sympy [F]

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

input

integrate((b*cos(d*x+c))**n*(A+B*cos(d*x+c)+C*cos(d*x+c)**2)/cos(d*x+c)**( 
1/2),x)
 

output

Integral((b*cos(c + d*x))**n*(A + B*cos(c + d*x) + C*cos(c + d*x)**2)/sqrt 
(cos(c + d*x)), x)
 

3.4.79.7 Maxima [F]

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

input

integrate((b*cos(d*x+c))^n*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(1/2 
),x, algorithm="maxima")
 

output

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

3.4.79.8 Giac [F]

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

input

integrate((b*cos(d*x+c))^n*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(1/2 
),x, algorithm="giac")
 

output

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

3.4.79.9 Mupad [F(-1)]

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

input

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

output

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