Integrand size = 33, antiderivative size = 514 \[ \int \frac {\cos ^m(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^2} \, dx=\frac {\left (A b^4 m-a^4 C (1+m)+a^2 b^2 (A-A m+C (2+m))\right ) \operatorname {AppellF1}\left (\frac {1}{2},\frac {1-m}{2},1,\frac {3}{2},\sin ^2(c+d x),-\frac {b^2 \sin ^2(c+d x)}{a^2-b^2}\right ) \cos ^{-1+m}(c+d x) \cos ^2(c+d x)^{\frac {1-m}{2}} \sin (c+d x)}{b^2 \left (a^2-b^2\right )^2 d}-\frac {\left (A b^4 m-a^4 C (1+m)+a^2 b^2 (A-A m+C (2+m))\right ) \operatorname {AppellF1}\left (\frac {1}{2},-\frac {m}{2},1,\frac {3}{2},\sin ^2(c+d x),-\frac {b^2 \sin ^2(c+d x)}{a^2-b^2}\right ) \cos ^m(c+d x) \cos ^2(c+d x)^{-m/2} \sin (c+d x)}{a b \left (a^2-b^2\right )^2 d}+\frac {\left (A b^2+a^2 C\right ) \cos ^{1+m}(c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \cos (c+d x))}-\frac {\left (a^2 C (1+m)-b^2 (C-A m)\right ) \cos ^{1+m}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},\cos ^2(c+d x)\right ) \sin (c+d x)}{b^2 \left (a^2-b^2\right ) d (1+m) \sqrt {\sin ^2(c+d x)}}+\frac {\left (A b^2+a^2 C\right ) (1+m) \cos ^{2+m}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+m}{2},\frac {4+m}{2},\cos ^2(c+d x)\right ) \sin (c+d x)}{a b \left (a^2-b^2\right ) d (2+m) \sqrt {\sin ^2(c+d x)}} \]
(A*b^4*m-a^4*C*(1+m)+a^2*b^2*(A-A*m+C*(2+m)))*AppellF1(1/2,-1/2*m+1/2,1,3/ 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*m+1/2)*sin(d*x+c)/b^2/(a^2-b^2)^2/d-(A*b^4*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+C*a^2)*cos(d*x+c)^(1+m)*sin(d*x+c)/a/(a^2-b^2)/d/(a+b*cos(d*x+c ))-(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],cos(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+C*a^2)*(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)
Leaf count is larger than twice the leaf count of optimal. \(14082\) vs. \(2(514)=1028\).
Time = 51.73 (sec) , antiderivative size = 14082, normalized size of antiderivative = 27.40 \[ \int \frac {\cos ^m(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^2} \, dx=\text {Result too large to show} \]
Time = 1.50 (sec) , antiderivative size = 467, normalized size of antiderivative = 0.91, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.303, Rules used = {3042, 3535, 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.
| \(\displaystyle \int \frac {\cos ^m(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^m \left (A+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2}dx\) |
| \(\Big \downarrow \) 3535 |
| \(\displaystyle \frac {\int \frac {\cos ^m(c+d x) \left ((A+C+C m) a^2-b (A+C) \cos (c+d x) a-\left (C a^2+A b^2\right ) (m+1) \cos ^2(c+d x)+A b^2 m\right )}{a+b \cos (c+d x)}dx}{a \left (a^2-b^2\right )}+\frac {\left (a^2 C+A b^2\right ) \sin (c+d x) \cos ^{m+1}(c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^m \left ((A+C+C m) a^2-b (A+C) \sin \left (c+d x+\frac {\pi }{2}\right ) a-\left (C a^2+A b^2\right ) (m+1) \sin \left (c+d x+\frac {\pi }{2}\right )^2+A b^2 m\right )}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{a \left (a^2-b^2\right )}+\frac {\left (a^2 C+A b^2\right ) \sin (c+d x) \cos ^{m+1}(c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
| \(\Big \downarrow \) 3542 |
| \(\displaystyle \frac {-\frac {(m+1) \left (a^2 C+A b^2\right ) \int \cos ^{m+1}(c+d x)dx}{b}+a \left (A m-C \left (1-\frac {a^2 (m+1)}{b^2}\right )\right ) \int \cos ^m(c+d x)dx+\frac {\left (a^4 (-C) (m+1)+a^2 b^2 (A (-m)+A+C (m+2))+A b^4 m\right ) \int \frac {\cos ^m(c+d x)}{a+b \cos (c+d x)}dx}{b^2}}{a \left (a^2-b^2\right )}+\frac {\left (a^2 C+A b^2\right ) \sin (c+d x) \cos ^{m+1}(c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {a \left (A m-C \left (1-\frac {a^2 (m+1)}{b^2}\right )\right ) \int \sin \left (c+d x+\frac {\pi }{2}\right )^mdx-\frac {(m+1) \left (a^2 C+A b^2\right ) \int \sin \left (c+d x+\frac {\pi }{2}\right )^{m+1}dx}{b}+\frac {\left (a^4 (-C) (m+1)+a^2 b^2 (A (-m)+A+C (m+2))+A b^4 m\right ) \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^m}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b^2}}{a \left (a^2-b^2\right )}+\frac {\left (a^2 C+A b^2\right ) \sin (c+d x) \cos ^{m+1}(c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
| \(\Big \downarrow \) 3122 |
| \(\displaystyle \frac {\frac {\left (a^4 (-C) (m+1)+a^2 b^2 (A (-m)+A+C (m+2))+A b^4 m\right ) \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^m}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b^2}-\frac {a \sin (c+d x) \left (A m-C \left (1-\frac {a^2 (m+1)}{b^2}\right )\right ) \cos ^{m+1}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+1}{2},\frac {m+3}{2},\cos ^2(c+d x)\right )}{d (m+1) \sqrt {\sin ^2(c+d x)}}+\frac {(m+1) \left (a^2 C+A b^2\right ) \sin (c+d x) \cos ^{m+2}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+2}{2},\frac {m+4}{2},\cos ^2(c+d x)\right )}{b d (m+2) \sqrt {\sin ^2(c+d x)}}}{a \left (a^2-b^2\right )}+\frac {\left (a^2 C+A b^2\right ) \sin (c+d x) \cos ^{m+1}(c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
| \(\Big \downarrow \) 3302 |
| \(\displaystyle \frac {\frac {\left (a^4 (-C) (m+1)+a^2 b^2 (A (-m)+A+C (m+2))+A b^4 m\right ) \left (a \int \frac {\cos ^m(c+d x)}{a^2-b^2 \cos ^2(c+d x)}dx-b \int \frac {\cos ^{m+1}(c+d x)}{a^2-b^2 \cos ^2(c+d x)}dx\right )}{b^2}-\frac {a \sin (c+d x) \left (A m-C \left (1-\frac {a^2 (m+1)}{b^2}\right )\right ) \cos ^{m+1}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+1}{2},\frac {m+3}{2},\cos ^2(c+d x)\right )}{d (m+1) \sqrt {\sin ^2(c+d x)}}+\frac {(m+1) \left (a^2 C+A b^2\right ) \sin (c+d x) \cos ^{m+2}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+2}{2},\frac {m+4}{2},\cos ^2(c+d x)\right )}{b d (m+2) \sqrt {\sin ^2(c+d x)}}}{a \left (a^2-b^2\right )}+\frac {\left (a^2 C+A b^2\right ) \sin (c+d x) \cos ^{m+1}(c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\left (a^4 (-C) (m+1)+a^2 b^2 (A (-m)+A+C (m+2))+A b^4 m\right ) \left (a \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^m}{a^2-b^2 \sin \left (c+d x+\frac {\pi }{2}\right )^2}dx-b \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^{m+1}}{a^2-b^2 \sin \left (c+d x+\frac {\pi }{2}\right )^2}dx\right )}{b^2}-\frac {a \sin (c+d x) \left (A m-C \left (1-\frac {a^2 (m+1)}{b^2}\right )\right ) \cos ^{m+1}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+1}{2},\frac {m+3}{2},\cos ^2(c+d x)\right )}{d (m+1) \sqrt {\sin ^2(c+d x)}}+\frac {(m+1) \left (a^2 C+A b^2\right ) \sin (c+d x) \cos ^{m+2}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+2}{2},\frac {m+4}{2},\cos ^2(c+d x)\right )}{b d (m+2) \sqrt {\sin ^2(c+d x)}}}{a \left (a^2-b^2\right )}+\frac {\left (a^2 C+A b^2\right ) \sin (c+d x) \cos ^{m+1}(c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
| \(\Big \downarrow \) 3668 |
| \(\displaystyle \frac {\frac {\left (a^4 (-C) (m+1)+a^2 b^2 (A (-m)+A+C (m+2))+A b^4 m\right ) \left (\frac {a \cos ^{m-1}(c+d x) \cos ^2(c+d x)^{\frac {1-m}{2}} \int \frac {\left (1-\sin ^2(c+d x)\right )^{\frac {m-1}{2}}}{a^2-b^2+b^2 \sin ^2(c+d x)}d\sin (c+d x)}{d}-\frac {b \cos ^m(c+d x) \cos ^2(c+d x)^{-m/2} \int \frac {\left (1-\sin ^2(c+d x)\right )^{m/2}}{a^2-b^2+b^2 \sin ^2(c+d x)}d\sin (c+d x)}{d}\right )}{b^2}-\frac {a \sin (c+d x) \left (A m-C \left (1-\frac {a^2 (m+1)}{b^2}\right )\right ) \cos ^{m+1}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+1}{2},\frac {m+3}{2},\cos ^2(c+d x)\right )}{d (m+1) \sqrt {\sin ^2(c+d x)}}+\frac {(m+1) \left (a^2 C+A b^2\right ) \sin (c+d x) \cos ^{m+2}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+2}{2},\frac {m+4}{2},\cos ^2(c+d x)\right )}{b d (m+2) \sqrt {\sin ^2(c+d x)}}}{a \left (a^2-b^2\right )}+\frac {\left (a^2 C+A b^2\right ) \sin (c+d x) \cos ^{m+1}(c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
| \(\Big \downarrow \) 333 |
| \(\displaystyle \frac {\left (a^2 C+A b^2\right ) \sin (c+d x) \cos ^{m+1}(c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}+\frac {-\frac {a \sin (c+d x) \left (A m-C \left (1-\frac {a^2 (m+1)}{b^2}\right )\right ) \cos ^{m+1}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+1}{2},\frac {m+3}{2},\cos ^2(c+d x)\right )}{d (m+1) \sqrt {\sin ^2(c+d x)}}+\frac {(m+1) \left (a^2 C+A b^2\right ) \sin (c+d x) \cos ^{m+2}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+2}{2},\frac {m+4}{2},\cos ^2(c+d x)\right )}{b d (m+2) \sqrt {\sin ^2(c+d x)}}+\frac {\left (a^4 (-C) (m+1)+a^2 b^2 (A (-m)+A+C (m+2))+A b^4 m\right ) \left (\frac {a \sin (c+d x) \cos ^{m-1}(c+d x) \cos ^2(c+d x)^{\frac {1-m}{2}} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1-m}{2},1,\frac {3}{2},\sin ^2(c+d x),-\frac {b^2 \sin ^2(c+d x)}{a^2-b^2}\right )}{d \left (a^2-b^2\right )}-\frac {b \sin (c+d x) \cos ^m(c+d x) \cos ^2(c+d x)^{-m/2} \operatorname {AppellF1}\left (\frac {1}{2},-\frac {m}{2},1,\frac {3}{2},\sin ^2(c+d x),-\frac {b^2 \sin ^2(c+d x)}{a^2-b^2}\right )}{d \left (a^2-b^2\right )}\right )}{b^2}}{a \left (a^2-b^2\right )}\) |
((A*b^2 + a^2*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*m - C*(1 - (a^2*(1 + m))/b^2))*Cos[c + d*x]^(1 + m)*Hypergeometric2F1[1/2, (1 + m)/2, (3 + m)/2, Cos[c + d*x]^2]*Sin[c + d*x])/(d*(1 + m)*Sqrt[Sin[c + d*x]^2])) + ((A*b^2 + a^2*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^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))
3.8.69.3.1 Defintions of rubi rules used
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])
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]
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]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(A*b^2 + a^2*C))*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*((c + d*S in[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[a*(m + 1)*(b*c - a*d)*(A + C) + d*(A*b^2 + a^2*C)*(m + n + 2) - (c*(A*b^2 + a^2*C) + b*(m + 1)*(b*c - a*d)*(A + C))*Sin[e + f*x] - d *(A*b^2 + a^2*C)*(m + n + 3)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, 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])))
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]
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]
\[\int \frac {\left (\cos ^{m}\left (d x +c \right )\right ) \left (A +C \left (\cos ^{2}\left (d x +c \right )\right )\right )}{\left (a +\cos \left (d x +c \right ) b \right )^{2}}d x\]
\[ \int \frac {\cos ^m(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^2} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} \cos \left (d x + c\right )^{m}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{2}} \,d x } \]
integral((C*cos(d*x + c)^2 + A)*cos(d*x + c)^m/(b^2*cos(d*x + c)^2 + 2*a*b *cos(d*x + c) + a^2), x)
Timed out. \[ \int \frac {\cos ^m(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^2} \, dx=\text {Timed out} \]
\[ \int \frac {\cos ^m(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^2} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} \cos \left (d x + c\right )^{m}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{2}} \,d x } \]
\[ \int \frac {\cos ^m(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^2} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} \cos \left (d x + c\right )^{m}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {\cos ^m(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^2} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^m\,\left (C\,{\cos \left (c+d\,x\right )}^2+A\right )}{{\left (a+b\,\cos \left (c+d\,x\right )\right )}^2} \,d x \]