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

Optimal result

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

[Out]

(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(1/2,1/2-1/2*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*s
in(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],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-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)

Rubi [A] (verified)

Time = 1.08 (sec) , antiderivative size = 564, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.146, Rules used = {3134, 3142, 2722, 2902, 3268, 440} \[ \int \frac {\cos ^m(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^2} \, dx=\frac {\sin (c+d x) \cos ^{m+1}(c+d x) \left (a^2 (-C) (m+1)+a b B m+b^2 (C-A m)\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+1}{2},\frac {m+3}{2},\cos ^2(c+d x)\right )}{b^2 d (m+1) \left (a^2-b^2\right ) \sqrt {\sin ^2(c+d x)}}+\frac {(m+1) \sin (c+d x) \left (A b^2-a (b B-a C)\right ) \cos ^{m+2}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+2}{2},\frac {m+4}{2},\cos ^2(c+d x)\right )}{a b d (m+2) \left (a^2-b^2\right ) \sqrt {\sin ^2(c+d x)}}+\frac {\sin (c+d x) \left (A b^2-a (b B-a C)\right ) \cos ^{m+1}(c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}+\frac {\sin (c+d x) \cos ^{m-1}(c+d x) \cos ^2(c+d x)^{\frac {1-m}{2}} \left (a^4 (-C) (m+1)+a^3 b B m+a^2 b^2 (A (-m)+A+C (m+2))-a b^3 B (m+1)+A b^4 m\right ) \operatorname {AppellF1}\left (\frac {1}{2},\frac {1-m}{2},1,\frac {3}{2},\sin ^2(c+d x),-\frac {b^2 \sin ^2(c+d x)}{a^2-b^2}\right )}{b^2 d \left (a^2-b^2\right )^2}-\frac {\sin (c+d x) \cos ^m(c+d x) \cos ^2(c+d x)^{-m/2} \left (a^4 (-C) (m+1)+a^3 b B m+a^2 b^2 (A (-m)+A+C (m+2))-a b^3 B (m+1)+A b^4 m\right ) \operatorname {AppellF1}\left (\frac {1}{2},-\frac {m}{2},1,\frac {3}{2},\sin ^2(c+d x),-\frac {b^2 \sin ^2(c+d x)}{a^2-b^2}\right )}{a b d \left (a^2-b^2\right )^2} \]

[In]

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]

[Out]

((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 - 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])/(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[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*b*(a^2 - b^2)^2*d*(Cos[c + d*x]^2)^(m/2)) + ((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*b*B*m - a^2*C*(1 + m) + b^2*(C - A*m))*Cos[
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*(a^2 - 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])/(a*b*(a^2 - b^2)*d*(2 + m)*Sqrt[Sin[c + d*x]^2])

Rule 440

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

Rule 2722

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]))*Hypergeometric2F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2], x] /; FreeQ[{b, c, d, n
}, x] &&  !IntegerQ[2*n]

Rule 2902

Int[((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[a, Int[(d*
Sin[e + f*x])^n/(a^2 - b^2*Sin[e + f*x]^2), x], x] - Dist[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 3134

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*s
in[(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] + D
ist[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 3142

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] :> Dist[(b*B - a*C)/b^2, Int[(d*Sin[e + f*x])^n, x],
x] + (Dist[(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] + Dist[C/(b*d), I
nt[(d*Sin[e + f*x])^(n + 1), x], x]) /; FreeQ[{a, b, d, e, f, A, B, C, n}, x] && NeQ[a^2 - b^2, 0]

Rule 3268

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]}, Dist[(-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]

Rubi steps \begin{align*} \text {integral}& = \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 {\int \frac {\cos ^m(c+d x) \left (A b^2 m-a b B (1+m)+a^2 (A+C+C m)-a (A b-a B+b C) \cos (c+d x)-\left (A b^2-a (b B-a C)\right ) (1+m) \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx}{a \left (a^2-b^2\right )} \\ & = \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 (\left (A b^2-a (b B-a C)\right ) (1+m)\right ) \int \cos ^{1+m}(c+d x) \, dx}{a b \left (a^2-b^2\right )}-\frac {\left (a b B m-a^2 C (1+m)+b^2 (C-A m)\right ) \int \cos ^m(c+d x) \, dx}{b^2 \left (a^2-b^2\right )}+\frac {\left (a^2 b (A b-a B+b C)-a^2 \left (A b^2-a (b B-a C)\right ) (1+m)+b^2 \left (A b^2 m-a b B (1+m)+a^2 (A+C+C m)\right )\right ) \int \frac {\cos ^m(c+d x)}{a+b \cos (c+d x)} \, dx}{a b^2 \left (a^2-b^2\right )} \\ & = \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)}}+\frac {\left (a^2 b (A b-a B+b C)-a^2 \left (A b^2-a (b B-a C)\right ) (1+m)+b^2 \left (A b^2 m-a b B (1+m)+a^2 (A+C+C m)\right )\right ) \int \frac {\cos ^m(c+d x)}{a^2-b^2 \cos ^2(c+d x)} \, dx}{b^2 \left (a^2-b^2\right )}-\frac {\left (a^2 b (A b-a B+b C)-a^2 \left (A b^2-a (b B-a C)\right ) (1+m)+b^2 \left (A b^2 m-a b B (1+m)+a^2 (A+C+C m)\right )\right ) \int \frac {\cos ^{1+m}(c+d x)}{a^2-b^2 \cos ^2(c+d x)} \, dx}{a b \left (a^2-b^2\right )} \\ & = \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)}}+\frac {\left (\left (a^2 b (A b-a B+b C)-a^2 \left (A b^2-a (b B-a C)\right ) (1+m)+b^2 \left (A b^2 m-a b B (1+m)+a^2 (A+C+C m)\right )\right ) \cos ^{2 \left (-\frac {1}{2}+\frac {m}{2}\right )}(c+d x) \cos ^2(c+d x)^{\frac {1}{2}-\frac {m}{2}}\right ) \text {Subst}\left (\int \frac {\left (1-x^2\right )^{\frac {1}{2} (-1+m)}}{a^2-b^2+b^2 x^2} \, dx,x,\sin (c+d x)\right )}{b^2 \left (a^2-b^2\right ) d}-\frac {\left (\left (a^2 b (A b-a B+b C)-a^2 \left (A b^2-a (b B-a C)\right ) (1+m)+b^2 \left (A b^2 m-a b B (1+m)+a^2 (A+C+C m)\right )\right ) \cos ^m(c+d x) \cos ^2(c+d x)^{-m/2}\right ) \text {Subst}\left (\int \frac {\left (1-x^2\right )^{m/2}}{a^2-b^2+b^2 x^2} \, dx,x,\sin (c+d x)\right )}{a b \left (a^2-b^2\right ) 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 {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 (1-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)}} \\ \end{align*}

Mathematica [B] (warning: unable to verify)

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

Time = 50.71 (sec) , antiderivative size = 12349, normalized size of antiderivative = 21.90 \[ \int \frac {\cos ^m(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^2} \, dx=\text {Result too large to show} \]

[In]

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]

[Out]

Result too large to show

Maple [F]

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

[In]

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)

[Out]

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]

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

[In]

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")

[Out]

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

[In]

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)

[Out]

Timed out

Maxima [F]

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

[In]

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")

[Out]

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]

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

[In]

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")

[Out]

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

[In]

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)

[Out]

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)