\(\int \frac {\cos (c+d x) (A+B \cos (c+d x))}{a+a \cos (c+d x)} \, dx\) [41]

Optimal result

Integrand size = 29, antiderivative size = 54 \[ \int \frac {\cos (c+d x) (A+B \cos (c+d x))}{a+a \cos (c+d x)} \, dx=\frac {(A-B) x}{a}+\frac {B \sin (c+d x)}{a d}-\frac {(A-B) \sin (c+d x)}{a d (1+\cos (c+d x))} \]

[Out]

(A-B)*x/a+B*sin(d*x+c)/a/d-(A-B)*sin(d*x+c)/a/d/(1+cos(d*x+c))

Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {3047, 3102, 12, 2814, 2727} \[ \int \frac {\cos (c+d x) (A+B \cos (c+d x))}{a+a \cos (c+d x)} \, dx=-\frac {(A-B) \sin (c+d x)}{a d (\cos (c+d x)+1)}+\frac {x (A-B)}{a}+\frac {B \sin (c+d x)}{a d} \]

[In]

Int[(Cos[c + d*x]*(A + B*Cos[c + d*x]))/(a + a*Cos[c + d*x]),x]

[Out]

((A - B)*x)/a + (B*Sin[c + d*x])/(a*d) - ((A - B)*Sin[c + d*x])/(a*d*(1 + Cos[c + d*x]))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2727

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> Simp[-Cos[c + d*x]/(d*(b + a*Sin[c + d*x])), x]
/; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 2814

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

Rule 3047

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(
e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x
]^2), x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]

Rule 3102

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Dist[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]

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

Mathematica [A] (verified)

Time = 0.37 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.72 \[ \int \frac {\cos (c+d x) (A+B \cos (c+d x))}{a+a \cos (c+d x)} \, dx=\frac {B \sin (c+d x)}{a d}+(A-B) \left (-\frac {\sin (c+d x)}{a d (1+\cos (c+d x))}-\frac {\arcsin (\cos (c+d x)) \sin (c+d x)}{a d \sqrt {1-\cos (c+d x)} \sqrt {1+\cos (c+d x)}}\right ) \]

[In]

Integrate[(Cos[c + d*x]*(A + B*Cos[c + d*x]))/(a + a*Cos[c + d*x]),x]

[Out]

(B*Sin[c + d*x])/(a*d) + (A - B)*(-(Sin[c + d*x]/(a*d*(1 + Cos[c + d*x]))) - (ArcSin[Cos[c + d*x]]*Sin[c + d*x
])/(a*d*Sqrt[1 - Cos[c + d*x]]*Sqrt[1 + Cos[c + d*x]]))

Maple [A] (verified)

Time = 0.84 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.78

method	result	size
parallelrisch	\(\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (B \cos \left (d x +c \right )-A +2 B \right )+d x \left (A -B \right )}{a d}\)	\(42\)
derivativedivides	\(\frac {-A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\frac {2 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}+2 \left (A -B \right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}\)	\(76\)
default	\(\frac {-A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\frac {2 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}+2 \left (A -B \right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}\)	\(76\)
risch	\(\frac {x A}{a}-\frac {B x}{a}-\frac {i {\mathrm e}^{i \left (d x +c \right )} B}{2 a d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} B}{2 a d}-\frac {2 i A}{d a \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}+\frac {2 i B}{d a \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}\)	\(99\)
norman	\(\frac {\frac {\left (A -B \right ) x}{a}+\frac {\left (A -B \right ) x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {\left (A -3 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d}-\frac {2 \left (A -2 B \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}-\frac {\left (A -B \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}+\frac {2 \left (A -B \right ) x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}\)	\(141\)

[In]

int(cos(d*x+c)*(A+B*cos(d*x+c))/(a+cos(d*x+c)*a),x,method=_RETURNVERBOSE)

[Out]

(tan(1/2*d*x+1/2*c)*(B*cos(d*x+c)-A+2*B)+d*x*(A-B))/a/d

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.13 \[ \int \frac {\cos (c+d x) (A+B \cos (c+d x))}{a+a \cos (c+d x)} \, dx=\frac {{\left (A - B\right )} d x \cos \left (d x + c\right ) + {\left (A - B\right )} d x + {\left (B \cos \left (d x + c\right ) - A + 2 \, B\right )} \sin \left (d x + c\right )}{a d \cos \left (d x + c\right ) + a d} \]

[In]

integrate(cos(d*x+c)*(A+B*cos(d*x+c))/(a+a*cos(d*x+c)),x, algorithm="fricas")

[Out]

((A - B)*d*x*cos(d*x + c) + (A - B)*d*x + (B*cos(d*x + c) - A + 2*B)*sin(d*x + c))/(a*d*cos(d*x + c) + a*d)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 264 vs. \(2 (39) = 78\).

Time = 0.57 (sec) , antiderivative size = 264, normalized size of antiderivative = 4.89 \[ \int \frac {\cos (c+d x) (A+B \cos (c+d x))}{a+a \cos (c+d x)} \, dx=\begin {cases} \frac {A d x \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a d} + \frac {A d x}{a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a d} - \frac {A \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a d} - \frac {A \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a d} - \frac {B d x \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a d} - \frac {B d x}{a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a d} + \frac {B \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a d} + \frac {3 B \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a d} & \text {for}\: d \neq 0 \\\frac {x \left (A + B \cos {\left (c \right )}\right ) \cos {\left (c \right )}}{a \cos {\left (c \right )} + a} & \text {otherwise} \end {cases} \]

[In]

integrate(cos(d*x+c)*(A+B*cos(d*x+c))/(a+a*cos(d*x+c)),x)

[Out]

Piecewise((A*d*x*tan(c/2 + d*x/2)**2/(a*d*tan(c/2 + d*x/2)**2 + a*d) + A*d*x/(a*d*tan(c/2 + d*x/2)**2 + a*d) -
A*tan(c/2 + d*x/2)**3/(a*d*tan(c/2 + d*x/2)**2 + a*d) - A*tan(c/2 + d*x/2)/(a*d*tan(c/2 + d*x/2)**2 + a*d) -
B*d*x*tan(c/2 + d*x/2)**2/(a*d*tan(c/2 + d*x/2)**2 + a*d) - B*d*x/(a*d*tan(c/2 + d*x/2)**2 + a*d) + B*tan(c/2
+ d*x/2)**3/(a*d*tan(c/2 + d*x/2)**2 + a*d) + 3*B*tan(c/2 + d*x/2)/(a*d*tan(c/2 + d*x/2)**2 + a*d), Ne(d, 0)),
(x*(A + B*cos(c))*cos(c)/(a*cos(c) + a), True))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 143 vs. \(2 (54) = 108\).

Time = 0.30 (sec) , antiderivative size = 143, normalized size of antiderivative = 2.65 \[ \int \frac {\cos (c+d x) (A+B \cos (c+d x))}{a+a \cos (c+d x)} \, dx=-\frac {B {\left (\frac {2 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a} - \frac {2 \, \sin \left (d x + c\right )}{{\left (a + \frac {a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}} - \frac {\sin \left (d x + c\right )}{a {\left (\cos \left (d x + c\right ) + 1\right )}}\right )} - A {\left (\frac {2 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a} - \frac {\sin \left (d x + c\right )}{a {\left (\cos \left (d x + c\right ) + 1\right )}}\right )}}{d} \]

[In]

integrate(cos(d*x+c)*(A+B*cos(d*x+c))/(a+a*cos(d*x+c)),x, algorithm="maxima")

[Out]

-(B*(2*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a - 2*sin(d*x + c)/((a + a*sin(d*x + c)^2/(cos(d*x + c) + 1)^2)
*(cos(d*x + c) + 1)) - sin(d*x + c)/(a*(cos(d*x + c) + 1))) - A*(2*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a -
sin(d*x + c)/(a*(cos(d*x + c) + 1))))/d

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.44 \[ \int \frac {\cos (c+d x) (A+B \cos (c+d x))}{a+a \cos (c+d x)} \, dx=\frac {\frac {{\left (d x + c\right )} {\left (A - B\right )}}{a} - \frac {A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a} + \frac {2 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )} a}}{d} \]

[In]

integrate(cos(d*x+c)*(A+B*cos(d*x+c))/(a+a*cos(d*x+c)),x, algorithm="giac")

[Out]

((d*x + c)*(A - B)/a - (A*tan(1/2*d*x + 1/2*c) - B*tan(1/2*d*x + 1/2*c))/a + 2*B*tan(1/2*d*x + 1/2*c)/((tan(1/
2*d*x + 1/2*c)^2 + 1)*a))/d

Mupad [B] (verification not implemented)

Time = 0.32 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.20 \[ \int \frac {\cos (c+d x) (A+B \cos (c+d x))}{a+a \cos (c+d x)} \, dx=\frac {x\,\left (A-B\right )}{a}+\frac {2\,B\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a\right )}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (A-B\right )}{a\,d} \]

[In]

int((cos(c + d*x)*(A + B*cos(c + d*x)))/(a + a*cos(c + d*x)),x)

[Out]

(x*(A - B))/a + (2*B*tan(c/2 + (d*x)/2))/(d*(a + a*tan(c/2 + (d*x)/2)^2)) - (tan(c/2 + (d*x)/2)*(A - B))/(a*d)