\(\int \frac {(b \cos (c+d x))^n (B \cos (c+d x)+C \cos ^2(c+d x))}{\sqrt {\cos (c+d x)}} \, dx\) [227]

Optimal result

Integrand size = 40, antiderivative size = 163 \[ \int \frac {(b \cos (c+d x))^n \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx=-\frac {2 B \cos ^{\frac {3}{2}}(c+d x) (b \cos (c+d x))^n \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (3+2 n),\frac {1}{4} (7+2 n),\cos ^2(c+d x)\right ) \sin (c+d x)}{d (3+2 n) \sqrt {\sin ^2(c+d x)}}-\frac {2 C \cos ^{\frac {5}{2}}(c+d x) (b \cos (c+d x))^n \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (5+2 n),\frac {1}{4} (9+2 n),\cos ^2(c+d x)\right ) \sin (c+d x)}{d (5+2 n) \sqrt {\sin ^2(c+d x)}} \] Output:

-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*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)/(sin(d*x+c)^2)^(1/2)
 

Mathematica [A] (verified)

Time = 0.25 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.85 \[ \int \frac {(b \cos (c+d x))^n \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx=-\frac {2 \cos ^{\frac {3}{2}}(c+d x) (b \cos (c+d x))^n \csc (c+d x) \left (B (5+2 n) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (3+2 n),\frac {1}{4} (7+2 n),\cos ^2(c+d x)\right )+C (3+2 n) \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (5+2 n),\frac {1}{4} (9+2 n),\cos ^2(c+d x)\right )\right ) \sqrt {\sin ^2(c+d x)}}{d (3+2 n) (5+2 n)} \] Input:

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

Output:

(-2*Cos[c + d*x]^(3/2)*(b*Cos[c + d*x])^n*Csc[c + d*x]*(B*(5 + 2*n)*Hyperg 
eometric2F1[1/2, (3 + 2*n)/4, (7 + 2*n)/4, Cos[c + d*x]^2] + C*(3 + 2*n)*C 
os[c + d*x]*Hypergeometric2F1[1/2, (5 + 2*n)/4, (9 + 2*n)/4, Cos[c + d*x]^ 
2])*Sqrt[Sin[c + d*x]^2])/(d*(3 + 2*n)*(5 + 2*n))
 

Rubi [A] (verified)

Time = 0.51 (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.

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

\(\Big \downarrow \) 2034

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3489

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3227

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3122

\(\displaystyle \cos ^{-n}(c+d x) (b \cos (c+d x))^n \left (-\frac {2 B \sin (c+d x) \cos ^{n+\frac {3}{2}}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (2 n+3),\frac {1}{4} (2 n+7),\cos ^2(c+d x)\right )}{d (2 n+3) \sqrt {\sin ^2(c+d x)}}-\frac {2 C \sin (c+d x) \cos ^{n+\frac {5}{2}}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (2 n+5),\frac {1}{4} (2 n+9),\cos ^2(c+d x)\right )}{d (2 n+5) \sqrt {\sin ^2(c+d x)}}\right )\)

Input:

Int[((b*Cos[c + d*x])^n*(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*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*(3 + 2*n)*Sqrt[S 
in[c + d*x]^2]) - (2*C*Cos[c + d*x]^(5/2 + n)*Hypergeometric2F1[1/2, (5 + 
2*n)/4, (9 + 2*n)/4, Cos[c + d*x]^2]*Sin[c + d*x])/(d*(5 + 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]

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

Input:

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

Output:

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

Fricas [F]

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

integrate((b*cos(d*x+c))^n*(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) + B)*(b*cos(d*x + c))^n*sqrt(cos(d*x + c)), x)
 

Sympy [F]

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

integrate((b*cos(d*x+c))**n*(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*(B + C*cos(c + d*x))*sqrt(cos(c + d*x)), x)
 

Maxima [F]

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

integrate((b*cos(d*x+c))^n*(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))*(b*cos(d*x + c))^n/sqrt(cos( 
d*x + c)), x)
 

Giac [F]

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

integrate((b*cos(d*x+c))^n*(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))*(b*cos(d*x + c))^n/sqrt(cos( 
d*x + c)), x)
 

Mupad [F(-1)]

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

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

Reduce [F]

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

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

Output:

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