3.4.81 \(\int \frac {(b \cos (c+d x))^n (A+B \cos (c+d x)+C \cos ^2(c+d x))}{\cos ^{\frac {5}{2}}(c+d x)} \, dx\) [381]

3.4.81.1 Optimal result

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

output

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

3.4.81.2 Mathematica [A] (verified)

Time = 0.31 (sec) , antiderivative size = 173, normalized size of antiderivative = 0.78 \[ \int \frac {(b \cos (c+d x))^n \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\frac {2 (b \cos (c+d x))^n \csc (c+d x) \left (-\left ((C (-3+2 n)+A (-1+2 n)) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (-3+2 n),\frac {1}{4} (1+2 n),\cos ^2(c+d x)\right ) \sqrt {\sin ^2(c+d x)}\right )+(-3+2 n) \left (C \sin ^2(c+d x)-B \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (-1+2 n),\frac {1}{4} (3+2 n),\cos ^2(c+d x)\right ) \sqrt {\sin ^2(c+d x)}\right )\right )}{d (-3+2 n) (-1+2 n) \cos ^{\frac {3}{2}}(c+d x)} \]

input

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

output

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

3.4.81.3 Rubi [A] (verified)

Time = 0.64 (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.

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

\(\Big \downarrow \) 2034

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3502

\(\displaystyle \cos ^{-n}(c+d x) (b \cos (c+d x))^n \left (-\frac {2 \int -\frac {1}{2} \cos ^{n-\frac {5}{2}}(c+d x) \left (2 A \left (\frac {1}{2}-n\right )+2 C \left (\frac {3}{2}-n\right )+B (1-2 n) \cos (c+d x)\right )dx}{1-2 n}-\frac {2 C \sin (c+d x) \cos ^{n-\frac {3}{2}}(c+d x)}{d (1-2 n)}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \cos ^{-n}(c+d x) (b \cos (c+d x))^n \left (\frac {\int \cos ^{n-\frac {5}{2}}(c+d x) (-2 n A+A+C (3-2 n)+B (1-2 n) \cos (c+d x))dx}{1-2 n}-\frac {2 C \sin (c+d x) \cos ^{n-\frac {3}{2}}(c+d x)}{d (1-2 n)}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \cos ^{-n}(c+d x) (b \cos (c+d x))^n \left (\frac {\int \sin \left (c+d x+\frac {\pi }{2}\right )^{n-\frac {5}{2}} \left (-2 n A+A+C (3-2 n)+B (1-2 n) \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx}{1-2 n}-\frac {2 C \sin (c+d x) \cos ^{n-\frac {3}{2}}(c+d x)}{d (1-2 n)}\right )\)

\(\Big \downarrow \) 3227

\(\displaystyle \cos ^{-n}(c+d x) (b \cos (c+d x))^n \left (\frac {(-2 A n+A+C (3-2 n)) \int \cos ^{n-\frac {5}{2}}(c+d x)dx+B (1-2 n) \int \cos ^{n-\frac {3}{2}}(c+d x)dx}{1-2 n}-\frac {2 C \sin (c+d x) \cos ^{n-\frac {3}{2}}(c+d x)}{d (1-2 n)}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \cos ^{-n}(c+d x) (b \cos (c+d x))^n \left (\frac {(-2 A n+A+C (3-2 n)) \int \sin \left (c+d x+\frac {\pi }{2}\right )^{n-\frac {5}{2}}dx+B (1-2 n) \int \sin \left (c+d x+\frac {\pi }{2}\right )^{n-\frac {3}{2}}dx}{1-2 n}-\frac {2 C \sin (c+d x) \cos ^{n-\frac {3}{2}}(c+d x)}{d (1-2 n)}\right )\)

\(\Big \downarrow \) 3122

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

input

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

output

((b*Cos[c + d*x])^n*((-2*C*Cos[c + d*x]^(-3/2 + n)*Sin[c + d*x])/(d*(1 - 2 
*n)) + ((2*(A + C*(3 - 2*n) - 2*A*n)*Cos[c + d*x]^(-3/2 + n)*Hypergeometri 
c2F1[1/2, (-3 + 2*n)/4, (1 + 2*n)/4, Cos[c + d*x]^2]*Sin[c + d*x])/(d*(3 - 
 2*n)*Sqrt[Sin[c + d*x]^2]) + (2*B*Cos[c + d*x]^(-1/2 + n)*Hypergeometric2 
F1[1/2, (-1 + 2*n)/4, (3 + 2*n)/4, Cos[c + d*x]^2]*Sin[c + d*x])/(d*Sqrt[S 
in[c + d*x]^2]))/(1 - 2*n)))/Cos[c + d*x]^n
 

3.4.81.3.1 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]
 

3.4.81.4 Maple [F]

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

input

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

3.4.81.5 Fricas [F]

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

input

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

output

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

3.4.81.6 Sympy [F(-1)]

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

input

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

output

Timed out
 

3.4.81.7 Maxima [F]

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

input

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

output

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

3.4.81.8 Giac [F]

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

input

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

output

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

3.4.81.9 Mupad [F(-1)]

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

input

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