\(\int \frac {\cos ^m(c+d x) (A+C \cos ^2(c+d x))}{a+b \cos (c+d x)} \, dx\) [768]

Optimal result

Integrand size = 33, antiderivative size = 353 \[ \int \frac {\cos ^m(c+d x) \left (A+C \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx=\frac {a \left (A b^2+a^2 C\right ) \operatorname {AppellF1}\left (\frac {1}{2},\frac {1-m}{2},1,\frac {3}{2},\sin ^2(c+d x),-\frac {b^2 \sin ^2(c+d x)}{a^2-b^2}\right ) \cos ^{-1+m}(c+d x) \cos ^2(c+d x)^{\frac {1-m}{2}} \sin (c+d x)}{b^2 \left (a^2-b^2\right ) d}-\frac {\left (A b^2+a^2 C\right ) \operatorname {AppellF1}\left (\frac {1}{2},-\frac {m}{2},1,\frac {3}{2},\sin ^2(c+d x),-\frac {b^2 \sin ^2(c+d x)}{a^2-b^2}\right ) \cos ^m(c+d x) \cos ^2(c+d x)^{-m/2} \sin (c+d x)}{b \left (a^2-b^2\right ) d}+\frac {a C \cos ^{1+m}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},\cos ^2(c+d x)\right ) \sin (c+d x)}{b^2 d (1+m) \sqrt {\sin ^2(c+d x)}}-\frac {C \cos ^{2+m}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+m}{2},\frac {4+m}{2},\cos ^2(c+d x)\right ) \sin (c+d x)}{b d (2+m) \sqrt {\sin ^2(c+d x)}} \] Output:

a*(A*b^2+C*a^2)*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)/d-(A*b^2+C*a^2)*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)/b/(a^2-b^2)/d/((cos(d*x+c)^2)^( 
1/2*m))+a*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/d/(1+m)/(sin(d*x+c)^2)^(1/2)-C*cos(d*x+c)^(2+m)*hyp 
ergeom([1/2, 1+1/2*m],[2+1/2*m],cos(d*x+c)^2)*sin(d*x+c)/b/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. \(10459\) vs. \(2(353)=706\).

Time = 37.51 (sec) , antiderivative size = 10459, normalized size of antiderivative = 29.63 \[ \int \frac {\cos ^m(c+d x) \left (A+C \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx=\text {Result too large to show} \] Input:

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

Output:

Result too large to show
 

Rubi [A] (verified)

Time = 0.94 (sec) , antiderivative size = 338, normalized size of antiderivative = 0.96, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.242, Rules used = {3042, 3543, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3543

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3122

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

\(\Big \downarrow \) 3302

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3668

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

\(\Big \downarrow \) 333

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

Input:

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

Output:

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

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 3543

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

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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