\(\int \frac {A+C \cos ^2(c+d x)}{a+a \cos (c+d x)} \, dx\) [42]

Optimal result

Integrand size = 25, antiderivative size = 48 \[ \int \frac {A+C \cos ^2(c+d x)}{a+a \cos (c+d x)} \, dx=-\frac {C x}{a}+\frac {C \sin (c+d x)}{a d}+\frac {(A+C) \sin (c+d x)}{a d (1+\cos (c+d x))} \]

[Out]

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

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {3103, 2814, 2727} \[ \int \frac {A+C \cos ^2(c+d x)}{a+a \cos (c+d x)} \, dx=\frac {(A+C) \sin (c+d x)}{a d (\cos (c+d x)+1)}+\frac {C \sin (c+d x)}{a d}-\frac {C x}{a} \]

[In]

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

[Out]

-((C*x)/a) + (C*Sin[c + d*x])/(a*d) + ((A + C)*Sin[c + d*x])/(a*d*(1 + Cos[c + d*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 3103

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (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) - a*C*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, C, m}, x] &&  !Lt
Q[m, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {C \sin (c+d x)}{a d}+\frac {\int \frac {a A-a C \cos (c+d x)}{a+a \cos (c+d x)} \, dx}{a} \\ & = -\frac {C x}{a}+\frac {C \sin (c+d x)}{a d}+(A+C) \int \frac {1}{a+a \cos (c+d x)} \, dx \\ & = -\frac {C x}{a}+\frac {C \sin (c+d x)}{a d}+\frac {(A+C) \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. \(108\) vs. \(2(48)=96\).

Time = 0.46 (sec) , antiderivative size = 108, normalized size of antiderivative = 2.25 \[ \int \frac {A+C \cos ^2(c+d x)}{a+a \cos (c+d x)} \, dx=\frac {\sec \left (\frac {c}{2}\right ) \sec \left (\frac {1}{2} (c+d x)\right ) \left (-2 C d x \cos \left (\frac {d x}{2}\right )-2 C d x \cos \left (c+\frac {d x}{2}\right )+4 A \sin \left (\frac {d x}{2}\right )+5 C \sin \left (\frac {d x}{2}\right )+C \sin \left (c+\frac {d x}{2}\right )+C \sin \left (c+\frac {3 d x}{2}\right )+C \sin \left (2 c+\frac {3 d x}{2}\right )\right )}{4 a d} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 1.59 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.77

method	result	size
parallelrisch	\(\frac {-d x C +\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (C \cos \left (d x +c \right )+A +2 C \right )}{a d}\)	\(37\)
derivativedivides	\(\frac {A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C -4 C \left (-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+\frac {\arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}\right )}{d a}\)	\(73\)
default	\(\frac {A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C -4 C \left (-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+\frac {\arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}\right )}{d a}\)	\(73\)
risch	\(-\frac {C x}{a}-\frac {i {\mathrm e}^{i \left (d x +c \right )} C}{2 a d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} C}{2 a d}+\frac {2 i A}{d a \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}+\frac {2 i C}{d a \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}\)	\(93\)
norman	\(\frac {\frac {\left (A +C \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}+\frac {\left (A +3 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d}-\frac {C x}{a}-\frac {2 C x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {C x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}+\frac {2 \left (A +2 C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}\)	\(127\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.10 \[ \int \frac {A+C \cos ^2(c+d x)}{a+a \cos (c+d x)} \, dx=-\frac {C d x \cos \left (d x + c\right ) + C d x - {\left (C \cos \left (d x + c\right ) + A + 2 \, C\right )} \sin \left (d x + c\right )}{a d \cos \left (d x + c\right ) + a d} \]

[In]

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

[Out]

-(C*d*x*cos(d*x + c) + C*d*x - (C*cos(d*x + c) + A + 2*C)*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. 202 vs. \(2 (37) = 74\).

Time = 0.54 (sec) , antiderivative size = 202, normalized size of antiderivative = 4.21 \[ \int \frac {A+C \cos ^2(c+d x)}{a+a \cos (c+d x)} \, dx=\begin {cases} \frac {A \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a d} + \frac {A \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a d} - \frac {C d x \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a d} - \frac {C d x}{a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a d} + \frac {C \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a d} + \frac {3 C \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a d} & \text {for}\: d \neq 0 \\\frac {x \left (A + C \cos ^{2}{\left (c \right )}\right )}{a \cos {\left (c \right )} + a} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((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) - C*d*x*tan(c/2 + d*x/2)**2/(a*d*tan(c/2 + d*x/2)**2 + a*d) - C*d*x/(a*d*tan(c/2 + d*x/2)**2 + a*d) +
C*tan(c/2 + d*x/2)**3/(a*d*tan(c/2 + d*x/2)**2 + a*d) + 3*C*tan(c/2 + d*x/2)/(a*d*tan(c/2 + d*x/2)**2 + a*d),
Ne(d, 0)), (x*(A + C*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. 117 vs. \(2 (48) = 96\).

Time = 0.33 (sec) , antiderivative size = 117, normalized size of antiderivative = 2.44 \[ \int \frac {A+C \cos ^2(c+d x)}{a+a \cos (c+d x)} \, dx=-\frac {C {\left (\frac {2 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a} - \frac {2 \, \sin \left (d x + c\right )}{{\left (a + \frac {a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}} - \frac {\sin \left (d x + c\right )}{a {\left (\cos \left (d x + c\right ) + 1\right )}}\right )} - \frac {A \sin \left (d x + c\right )}{a {\left (\cos \left (d x + c\right ) + 1\right )}}}{d} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.54 \[ \int \frac {A+C \cos ^2(c+d x)}{a+a \cos (c+d x)} \, dx=-\frac {\frac {{\left (d x + c\right )} C}{a} - \frac {A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a} - \frac {2 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )} a}}{d} \]

[In]

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

[Out]

-((d*x + c)*C/a - (A*tan(1/2*d*x + 1/2*c) + C*tan(1/2*d*x + 1/2*c))/a - 2*C*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 = 1.05 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.23 \[ \int \frac {A+C \cos ^2(c+d x)}{a+a \cos (c+d x)} \, dx=\frac {2\,C\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a\right )}-\frac {C\,x}{a}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (A+C\right )}{a\,d} \]

[In]

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

[Out]

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