\(\int \frac {\cos ^2(c+d x)}{a+a \cos (c+d x)} \, dx\) [46]

Optimal result

Integrand size = 21, antiderivative size = 43 \[ \int \frac {\cos ^2(c+d x)}{a+a \cos (c+d x)} \, dx=-\frac {x}{a}+\frac {\sin (c+d x)}{a d}+\frac {\sin (c+d x)}{a d (1+\cos (c+d x))} \]

[Out]

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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {2825, 12, 2814, 2727} \[ \int \frac {\cos ^2(c+d x)}{a+a \cos (c+d x)} \, dx=\frac {\sin (c+d x)}{a d}+\frac {\sin (c+d x)}{a d (\cos (c+d x)+1)}-\frac {x}{a} \]

[In]

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

[Out]

-(x/a) + Sin[c + d*x]/(a*d) + 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 2825

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

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

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(89\) vs. \(2(43)=86\).

Time = 0.47 (sec) , antiderivative size = 89, normalized size of antiderivative = 2.07 \[ \int \frac {\cos ^2(c+d x)}{a+a \cos (c+d x)} \, dx=\frac {\sec \left (\frac {c}{2}\right ) \sec \left (\frac {1}{2} (c+d x)\right ) \left (-2 d x \cos \left (\frac {d x}{2}\right )-2 d x \cos \left (c+\frac {d x}{2}\right )+5 \sin \left (\frac {d x}{2}\right )+\sin \left (c+\frac {d x}{2}\right )+\sin \left (c+\frac {3 d x}{2}\right )+\sin \left (2 c+\frac {3 d x}{2}\right )\right )}{4 a d} \]

[In]

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

[Out]

(Sec[c/2]*Sec[(c + d*x)/2]*(-2*d*x*Cos[(d*x)/2] - 2*d*x*Cos[c + (d*x)/2] + 5*Sin[(d*x)/2] + Sin[c + (d*x)/2] +
Sin[c + (3*d*x)/2] + Sin[2*c + (3*d*x)/2]))/(4*a*d)

Maple [A] (verified)

Time = 0.74 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.72

method	result	size
parallelrisch	\(\frac {-d x +\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\cos \left (d x +c \right )+2\right )}{a d}\)	\(31\)
derivativedivides	\(\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}-2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}\)	\(56\)
default	\(\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}-2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}\)	\(56\)
risch	\(-\frac {x}{a}-\frac {i {\mathrm e}^{i \left (d x +c \right )}}{2 a d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )}}{2 a d}+\frac {2 i}{d a \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}\)	\(66\)
norman	\(\frac {\frac {\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )}{d a}-\frac {x}{a}+\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d a}+\frac {4 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {2 x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}\)	\(112\)

[In]

int(cos(d*x+c)^2/(a+cos(d*x+c)*a),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.07 \[ \int \frac {\cos ^2(c+d x)}{a+a \cos (c+d x)} \, dx=-\frac {d x \cos \left (d x + c\right ) + d x - {\left (\cos \left (d x + c\right ) + 2\right )} \sin \left (d x + c\right )}{a d \cos \left (d x + c\right ) + a d} \]

[In]

integrate(cos(d*x+c)^2/(a+a*cos(d*x+c)),x, algorithm="fricas")

[Out]

-(d*x*cos(d*x + c) + d*x - (cos(d*x + c) + 2)*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. 129 vs. \(2 (31) = 62\).

Time = 0.56 (sec) , antiderivative size = 129, normalized size of antiderivative = 3.00 \[ \int \frac {\cos ^2(c+d x)}{a+a \cos (c+d x)} \, dx=\begin {cases} - \frac {d x \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a d} - \frac {d x}{a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a d} + \frac {\tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a d} + \frac {3 \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a d} & \text {for}\: d \neq 0 \\\frac {x \cos ^{2}{\left (c \right )}}{a \cos {\left (c \right )} + a} & \text {otherwise} \end {cases} \]

[In]

integrate(cos(d*x+c)**2/(a+a*cos(d*x+c)),x)

[Out]

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 92 vs. \(2 (43) = 86\).

Time = 0.31 (sec) , antiderivative size = 92, normalized size of antiderivative = 2.14 \[ \int \frac {\cos ^2(c+d x)}{a+a \cos (c+d x)} \, dx=-\frac {\frac {2 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a} - \frac {2 \, \sin \left (d x + c\right )}{{\left (a + \frac {a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}} - \frac {\sin \left (d x + c\right )}{a {\left (\cos \left (d x + c\right ) + 1\right )}}}{d} \]

[In]

integrate(cos(d*x+c)^2/(a+a*cos(d*x+c)),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.38 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.35 \[ \int \frac {\cos ^2(c+d x)}{a+a \cos (c+d x)} \, dx=-\frac {\frac {d x + c}{a} - \frac {\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a} - \frac {2 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )} a}}{d} \]

[In]

integrate(cos(d*x+c)^2/(a+a*cos(d*x+c)),x, algorithm="giac")

[Out]

-((d*x + c)/a - tan(1/2*d*x + 1/2*c)/a - 2*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 = 14.92 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.53 \[ \int \frac {\cos ^2(c+d x)}{a+a \cos (c+d x)} \, dx=\frac {2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\left (-c-d\,x\right )\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )+\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{a\,d\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )} \]

[In]

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

[Out]

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