\(\int \frac {A+B \cos (c+d x)}{(a+a \cos (c+d x))^2} \, dx\) [51]

Optimal result

Integrand size = 23, antiderivative size = 65 \[ \int \frac {A+B \cos (c+d x)}{(a+a \cos (c+d x))^2} \, dx=\frac {(A-B) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2}+\frac {(A+2 B) \sin (c+d x)}{3 d \left (a^2+a^2 \cos (c+d x)\right )} \]

[Out]

1/3*(A-B)*sin(d*x+c)/d/(a+a*cos(d*x+c))^2+1/3*(A+2*B)*sin(d*x+c)/d/(a^2+a^2*cos(d*x+c))

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {2829, 2727} \[ \int \frac {A+B \cos (c+d x)}{(a+a \cos (c+d x))^2} \, dx=\frac {(A+2 B) \sin (c+d x)}{3 d \left (a^2 \cos (c+d x)+a^2\right )}+\frac {(A-B) \sin (c+d x)}{3 d (a \cos (c+d x)+a)^2} \]

[In]

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

[Out]

((A - B)*Sin[c + d*x])/(3*d*(a + a*Cos[c + d*x])^2) + ((A + 2*B)*Sin[c + d*x])/(3*d*(a^2 + a^2*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 2829

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

Rubi steps \begin{align*} \text {integral}& = \frac {(A-B) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2}+\frac {(A+2 B) \int \frac {1}{a+a \cos (c+d x)} \, dx}{3 a} \\ & = \frac {(A-B) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2}+\frac {(A+2 B) \sin (c+d x)}{3 d \left (a^2+a^2 \cos (c+d x)\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.66 \[ \int \frac {A+B \cos (c+d x)}{(a+a \cos (c+d x))^2} \, dx=\frac {(2 A+B+(A+2 B) \cos (c+d x)) \sin (c+d x)}{3 a^2 d (1+\cos (c+d x))^2} \]

[In]

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

[Out]

((2*A + B + (A + 2*B)*Cos[c + d*x])*Sin[c + d*x])/(3*a^2*d*(1 + Cos[c + d*x])^2)

Maple [A] (verified)

Time = 0.90 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.65

method	result	size
parallelrisch	\(\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (3 A +3 B +\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (A -B \right )\right )}{6 a^{2} d}\)	\(42\)
derivativedivides	\(\frac {\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{3}-\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{3}+A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d \,a^{2}}\)	\(60\)
default	\(\frac {\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{3}-\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{3}+A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d \,a^{2}}\)	\(60\)
risch	\(\frac {2 i \left (3 B \,{\mathrm e}^{2 i \left (d x +c \right )}+3 A \,{\mathrm e}^{i \left (d x +c \right )}+3 B \,{\mathrm e}^{i \left (d x +c \right )}+A +2 B \right )}{3 d \,a^{2} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{3}}\)	\(64\)
norman	\(\frac {\frac {\left (A -B \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 a d}+\frac {\left (A +B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a d}+\frac {\left (2 A +B \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 a d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a}\)	\(89\)

[In]

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

[Out]

1/6*tan(1/2*d*x+1/2*c)*(3*A+3*B+tan(1/2*d*x+1/2*c)^2*(A-B))/a^2/d

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.89 \[ \int \frac {A+B \cos (c+d x)}{(a+a \cos (c+d x))^2} \, dx=\frac {{\left ({\left (A + 2 \, B\right )} \cos \left (d x + c\right ) + 2 \, A + B\right )} \sin \left (d x + c\right )}{3 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \]

[In]

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

[Out]

1/3*((A + 2*B)*cos(d*x + c) + 2*A + B)*sin(d*x + c)/(a^2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a^2*d)

Sympy [A] (verification not implemented)

Time = 0.50 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.45 \[ \int \frac {A+B \cos (c+d x)}{(a+a \cos (c+d x))^2} \, dx=\begin {cases} \frac {A \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a^{2} d} + \frac {A \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{2 a^{2} d} - \frac {B \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a^{2} d} + \frac {B \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{2 a^{2} d} & \text {for}\: d \neq 0 \\\frac {x \left (A + B \cos {\left (c \right )}\right )}{\left (a \cos {\left (c \right )} + a\right )^{2}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((A*tan(c/2 + d*x/2)**3/(6*a**2*d) + A*tan(c/2 + d*x/2)/(2*a**2*d) - B*tan(c/2 + d*x/2)**3/(6*a**2*d)
+ B*tan(c/2 + d*x/2)/(2*a**2*d), Ne(d, 0)), (x*(A + B*cos(c))/(a*cos(c) + a)**2, True))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.43 \[ \int \frac {A+B \cos (c+d x)}{(a+a \cos (c+d x))^2} \, dx=\frac {\frac {A {\left (\frac {3 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{2}} + \frac {B {\left (\frac {3 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{2}}}{6 \, d} \]

[In]

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

[Out]

1/6*(A*(3*sin(d*x + c)/(cos(d*x + c) + 1) + sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/a^2 + B*(3*sin(d*x + c)/(cos(
d*x + c) + 1) - sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/a^2)/d

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.92 \[ \int \frac {A+B \cos (c+d x)}{(a+a \cos (c+d x))^2} \, dx=\frac {A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{6 \, a^{2} d} \]

[In]

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

[Out]

1/6*(A*tan(1/2*d*x + 1/2*c)^3 - B*tan(1/2*d*x + 1/2*c)^3 + 3*A*tan(1/2*d*x + 1/2*c) + 3*B*tan(1/2*d*x + 1/2*c)
)/(a^2*d)

Mupad [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.69 \[ \int \frac {A+B \cos (c+d x)}{(a+a \cos (c+d x))^2} \, dx=\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (A-B\right )}{6\,a^2\,d}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (A+B\right )}{2\,a^2\,d} \]

[In]

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

[Out]

(tan(c/2 + (d*x)/2)^3*(A - B))/(6*a^2*d) + (tan(c/2 + (d*x)/2)*(A + B))/(2*a^2*d)