\(\int \frac {\cos ^m(c+d x) (A+B \cos (c+d x)+C \cos ^2(c+d x))}{(a+b \cos (c+d x))^2} \, dx\) [1159]

Optimal result

Integrand size = 41, antiderivative size = 564 \[ \int \frac {\cos ^m(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^2} \, dx=\frac {\left (A b^4 m+a^3 b B m-a b^3 B (1+m)-a^4 C (1+m)+a^2 b^2 (A-A m+C (2+m))\right ) \operatorname {AppellF1}\left (\frac {1}{2},\frac {1-m}{2},1,\frac {3}{2},\sin ^2(c+d x),-\frac {b^2 \sin ^2(c+d x)}{a^2-b^2}\right ) \cos ^{-1+m}(c+d x) \cos ^2(c+d x)^{\frac {1-m}{2}} \sin (c+d x)}{b^2 \left (a^2-b^2\right )^2 d}-\frac {\left (A b^4 m+a^3 b B m-a b^3 B (1+m)-a^4 C (1+m)+a^2 b^2 (A-A m+C (2+m))\right ) \operatorname {AppellF1}\left (\frac {1}{2},-\frac {m}{2},1,\frac {3}{2},\sin ^2(c+d x),-\frac {b^2 \sin ^2(c+d x)}{a^2-b^2}\right ) \cos ^m(c+d x) \cos ^2(c+d x)^{-m/2} \sin (c+d x)}{a b \left (a^2-b^2\right )^2 d}+\frac {\left (A b^2-a (b B-a C)\right ) \cos ^{1+m}(c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \cos (c+d x))}+\frac {\left (a b B m-a^2 C (1+m)+b^2 (C-A m)\right ) \cos ^{1+m}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},\cos ^2(c+d x)\right ) \sin (c+d x)}{b^2 \left (a^2-b^2\right ) d (1+m) \sqrt {\sin ^2(c+d x)}}+\frac {\left (A b^2-a (b B-a C)\right ) (1+m) \cos ^{2+m}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+m}{2},\frac {4+m}{2},\cos ^2(c+d x)\right ) \sin (c+d x)}{a b \left (a^2-b^2\right ) d (2+m) \sqrt {\sin ^2(c+d x)}} \] Output:

(A*b^4*m+a^3*b*B*m-a*b^3*B*(1+m)-a^4*C*(1+m)+a^2*b^2*(A-A*m+C*(2+m)))*Appe 
llF1(1/2,1/2-1/2*m,1,3/2,sin(d*x+c)^2,-b^2*sin(d*x+c)^2/(a^2-b^2))*cos(d*x 
+c)^(-1+m)*(cos(d*x+c)^2)^(1/2-1/2*m)*sin(d*x+c)/b^2/(a^2-b^2)^2/d-(A*b^4* 
m+a^3*b*B*m-a*b^3*B*(1+m)-a^4*C*(1+m)+a^2*b^2*(A-A*m+C*(2+m)))*AppellF1(1/ 
2,-1/2*m,1,3/2,sin(d*x+c)^2,-b^2*sin(d*x+c)^2/(a^2-b^2))*cos(d*x+c)^m*sin( 
d*x+c)/a/b/(a^2-b^2)^2/d/((cos(d*x+c)^2)^(1/2*m))+(A*b^2-a*(B*b-C*a))*cos( 
d*x+c)^(1+m)*sin(d*x+c)/a/(a^2-b^2)/d/(a+b*cos(d*x+c))+(a*b*B*m-a^2*C*(1+m 
)+b^2*(-A*m+C))*cos(d*x+c)^(1+m)*hypergeom([1/2, 1/2+1/2*m],[3/2+1/2*m],co 
s(d*x+c)^2)*sin(d*x+c)/b^2/(a^2-b^2)/d/(1+m)/(sin(d*x+c)^2)^(1/2)+(A*b^2-a 
*(B*b-C*a))*(1+m)*cos(d*x+c)^(2+m)*hypergeom([1/2, 1+1/2*m],[2+1/2*m],cos( 
d*x+c)^2)*sin(d*x+c)/a/b/(a^2-b^2)/d/(2+m)/(sin(d*x+c)^2)^(1/2)
 

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(12349\) vs. \(2(564)=1128\).

Time = 51.84 (sec) , antiderivative size = 12349, normalized size of antiderivative = 21.90 \[ \int \frac {\cos ^m(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^2} \, dx=\text {Result too large to show} \] Input:

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

Output:

Result too large to show
 

Rubi [A] (verified)

Time = 1.70 (sec) , antiderivative size = 503, normalized size of antiderivative = 0.89, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.244, Rules used = {3042, 3534, 3042, 3542, 3042, 3122, 3302, 3042, 3668, 333}

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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^m \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2}dx\)

\(\Big \downarrow \) 3534

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3542

\(\displaystyle \frac {-\frac {a \left (a^2 (-C) (m+1)+a b B m+b^2 (C-A m)\right ) \int \cos ^m(c+d x)dx}{b^2}+\frac {\left (a^4 (-C) (m+1)+a^3 b B m+a^2 b^2 (A (-m)+A+C (m+2))-a b^3 B (m+1)+A b^4 m\right ) \int \frac {\cos ^m(c+d x)}{a+b \cos (c+d x)}dx}{b^2}-\frac {(m+1) \left (A b^2-a (b B-a C)\right ) \int \cos ^{m+1}(c+d x)dx}{b}}{a \left (a^2-b^2\right )}+\frac {\sin (c+d x) \left (A b^2-a (b B-a C)\right ) \cos ^{m+1}(c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {a \left (a^2 (-C) (m+1)+a b B m+b^2 (C-A m)\right ) \int \sin \left (c+d x+\frac {\pi }{2}\right )^mdx}{b^2}+\frac {\left (a^4 (-C) (m+1)+a^3 b B m+a^2 b^2 (A (-m)+A+C (m+2))-a b^3 B (m+1)+A b^4 m\right ) \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^m}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b^2}-\frac {(m+1) \left (A b^2-a (b B-a C)\right ) \int \sin \left (c+d x+\frac {\pi }{2}\right )^{m+1}dx}{b}}{a \left (a^2-b^2\right )}+\frac {\sin (c+d x) \left (A b^2-a (b B-a C)\right ) \cos ^{m+1}(c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\)

\(\Big \downarrow \) 3122

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

\(\Big \downarrow \) 3302

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3668

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

\(\Big \downarrow \) 333

\(\displaystyle \frac {\sin (c+d x) \left (A b^2-a (b B-a C)\right ) \cos ^{m+1}(c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}+\frac {\frac {a \sin (c+d x) \cos ^{m+1}(c+d x) \left (a^2 (-C) (m+1)+a b B m+b^2 (C-A m)\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+1}{2},\frac {m+3}{2},\cos ^2(c+d x)\right )}{b^2 d (m+1) \sqrt {\sin ^2(c+d x)}}+\frac {\left (a^4 (-C) (m+1)+a^3 b B m+a^2 b^2 (A (-m)+A+C (m+2))-a b^3 B (m+1)+A b^4 m\right ) \left (\frac {a \sin (c+d x) \cos ^{m-1}(c+d x) \cos ^2(c+d x)^{\frac {1-m}{2}} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1-m}{2},1,\frac {3}{2},\sin ^2(c+d x),-\frac {b^2 \sin ^2(c+d x)}{a^2-b^2}\right )}{d \left (a^2-b^2\right )}-\frac {b \sin (c+d x) \cos ^m(c+d x) \cos ^2(c+d x)^{-m/2} \operatorname {AppellF1}\left (\frac {1}{2},-\frac {m}{2},1,\frac {3}{2},\sin ^2(c+d x),-\frac {b^2 \sin ^2(c+d x)}{a^2-b^2}\right )}{d \left (a^2-b^2\right )}\right )}{b^2}+\frac {(m+1) \sin (c+d x) \left (A b^2-a (b B-a C)\right ) \cos ^{m+2}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+2}{2},\frac {m+4}{2},\cos ^2(c+d x)\right )}{b d (m+2) \sqrt {\sin ^2(c+d x)}}}{a \left (a^2-b^2\right )}\)

Input:

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

Output:

((A*b^2 - a*(b*B - a*C))*Cos[c + d*x]^(1 + m)*Sin[c + d*x])/(a*(a^2 - b^2) 
*d*(a + b*Cos[c + d*x])) + ((a*(a*b*B*m - a^2*C*(1 + m) + b^2*(C - A*m))*C 
os[c + d*x]^(1 + m)*Hypergeometric2F1[1/2, (1 + m)/2, (3 + m)/2, Cos[c + d 
*x]^2]*Sin[c + d*x])/(b^2*d*(1 + m)*Sqrt[Sin[c + d*x]^2]) + ((A*b^2 - a*(b 
*B - a*C))*(1 + m)*Cos[c + d*x]^(2 + m)*Hypergeometric2F1[1/2, (2 + m)/2, 
(4 + m)/2, Cos[c + d*x]^2]*Sin[c + d*x])/(b*d*(2 + m)*Sqrt[Sin[c + d*x]^2] 
) + ((A*b^4*m + a^3*b*B*m - a*b^3*B*(1 + m) - a^4*C*(1 + m) + a^2*b^2*(A - 
 A*m + C*(2 + m)))*((a*AppellF1[1/2, (1 - m)/2, 1, 3/2, Sin[c + d*x]^2, -( 
(b^2*Sin[c + d*x]^2)/(a^2 - b^2))]*Cos[c + d*x]^(-1 + m)*(Cos[c + d*x]^2)^ 
((1 - m)/2)*Sin[c + d*x])/((a^2 - b^2)*d) - (b*AppellF1[1/2, -1/2*m, 1, 3/ 
2, Sin[c + d*x]^2, -((b^2*Sin[c + d*x]^2)/(a^2 - b^2))]*Cos[c + d*x]^m*Sin 
[c + d*x])/((a^2 - b^2)*d*(Cos[c + d*x]^2)^(m/2))))/b^2)/(a*(a^2 - b^2))
 

Defintions of rubi rules used

rule 333

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim 
p[a^p*c^q*x*AppellF1[1/2, -p, -q, 3/2, (-b)*(x^2/a), (-d)*(x^2/c)], x] /; F 
reeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[p] || GtQ[a, 
0]) && (IntegerQ[q] || GtQ[c, 0])
 

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 3302

Int[((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)/((a_) + (b_.)*sin[(e_.) + (f_.)*( 
x_)]), x_Symbol] :> Simp[a   Int[(d*Sin[e + f*x])^n/(a^2 - b^2*Sin[e + f*x] 
^2), x], x] - Simp[b/d   Int[(d*Sin[e + f*x])^(n + 1)/(a^2 - b^2*Sin[e + f* 
x]^2), x], x] /; FreeQ[{a, b, d, e, f, n}, x] && NeQ[a^2 - b^2, 0]
 

rule 3534

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
 (f_.)*(x_)])^(n_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) 
 + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(A*b^2 - a*b*B + a^2*C))*Cos[e + f*x 
]*(a + b*Sin[e + f*x])^(m + 1)*((c + d*Sin[e + f*x])^(n + 1)/(f*(m + 1)*(b* 
c - a*d)*(a^2 - b^2))), x] + Simp[1/((m + 1)*(b*c - a*d)*(a^2 - b^2))   Int 
[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[(m + 1)*(b*c - a* 
d)*(a*A - b*B + a*C) + d*(A*b^2 - a*b*B + a^2*C)*(m + n + 2) - (c*(A*b^2 - 
a*b*B + a^2*C) + (m + 1)*(b*c - a*d)*(A*b - a*B + b*C))*Sin[e + f*x] - d*(A 
*b^2 - a*b*B + a^2*C)*(m + n + 3)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b 
, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && 
NeQ[c^2 - d^2, 0] && LtQ[m, -1] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ 
[n]) ||  !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] &&  !IntegerQ[m]) | 
| EqQ[a, 0])))
 

rule 3542

Int[(((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)*sin[(e_.) + (f_.) 
*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2))/((a_) + (b_.)*sin[(e_.) + (f_.)* 
(x_)]), x_Symbol] :> Simp[(b*B - a*C)/b^2   Int[(d*Sin[e + f*x])^n, x], x] 
+ (Simp[(A*b^2 - a*b*B + a^2*C)/b^2   Int[(d*Sin[e + f*x])^n/(a + b*Sin[e + 
 f*x]), x], x] + Simp[C/(b*d)   Int[(d*Sin[e + f*x])^(n + 1), x], x]) /; Fr 
eeQ[{a, b, d, e, f, A, B, C, n}, x] && NeQ[a^2 - b^2, 0]
 

rule 3668

Int[((d_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)]^2)^(p_.), x_Symbol] :> With[{ff = FreeFactors[Cos[e + f*x], x]}, Simp[( 
-ff)*d^(2*IntPart[(m - 1)/2] + 1)*((d*Sin[e + f*x])^(2*FracPart[(m - 1)/2]) 
/(f*(Sin[e + f*x]^2)^FracPart[(m - 1)/2]))   Subst[Int[(1 - ff^2*x^2)^((m - 
 1)/2)*(a + b - b*ff^2*x^2)^p, x], x, Cos[e + f*x]/ff], x]] /; FreeQ[{a, b, 
 d, e, f, m, p}, x] &&  !IntegerQ[m]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

int(cos(c + d*x)**m/(cos(c + d*x)**2*b**2 + 2*cos(c + d*x)*a*b + a**2),x)* 
a + int((cos(c + d*x)**m*cos(c + d*x))/(cos(c + d*x)**2*b**2 + 2*cos(c + d 
*x)*a*b + a**2),x)*b + int((cos(c + d*x)**m*cos(c + d*x)**2)/(cos(c + d*x) 
**2*b**2 + 2*cos(c + d*x)*a*b + a**2),x)*c