\(\int \frac {\cos (c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^2} \, dx\) [50]

Optimal result

Integrand size = 29, antiderivative size = 70 \[ \int \frac {\cos (c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^2} \, dx=\frac {B x}{a^2}+\frac {(2 A-5 B) \sin (c+d x)}{3 a^2 d (1+\cos (c+d x))}-\frac {(A-B) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2} \]

[Out]

B*x/a^2+1/3*(2*A-5*B)*sin(d*x+c)/a^2/d/(1+cos(d*x+c))-1/3*(A-B)*sin(d*x+c)/d/(a+a*cos(d*x+c))^2

Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {3047, 3098, 2814, 2727} \[ \int \frac {\cos (c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^2} \, dx=\frac {(2 A-5 B) \sin (c+d x)}{3 a^2 d (\cos (c+d x)+1)}+\frac {B x}{a^2}-\frac {(A-B) \sin (c+d x)}{3 d (a \cos (c+d x)+a)^2} \]

[In]

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

[Out]

(B*x)/a^2 + ((2*A - 5*B)*Sin[c + d*x])/(3*a^2*d*(1 + Cos[c + d*x])) - ((A - B)*Sin[c + d*x])/(3*d*(a + a*Cos[c
+ d*x])^2)

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 3098

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_
.)*(x_)]^2), x_Symbol] :> Simp[(A*b - a*B + b*C)*Cos[e + f*x]*((a + b*Sin[e + f*x])^m/(a*f*(2*m + 1))), x] + D
ist[1/(a^2*(2*m + 1)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[a*A*(m + 1) + m*(b*B - a*C) + b*C*(2*m + 1)*Sin[e
+ f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && LtQ[m, -1] && EqQ[a^2 - b^2, 0]

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))^2} \, dx \\ & = -\frac {(A-B) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2}-\frac {\int \frac {-2 a (A-B)-3 a B \cos (c+d x)}{a+a \cos (c+d x)} \, dx}{3 a^2} \\ & = \frac {B x}{a^2}-\frac {(A-B) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2}+\frac {(2 A-5 B) \int \frac {1}{a+a \cos (c+d x)} \, dx}{3 a} \\ & = \frac {B x}{a^2}-\frac {(A-B) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2}+\frac {(2 A-5 B) \sin (c+d x)}{3 d \left (a^2+a^2 \cos (c+d x)\right )} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(153\) vs. \(2(70)=140\).

Time = 0.75 (sec) , antiderivative size = 153, normalized size of antiderivative = 2.19 \[ \int \frac {\cos (c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^2} \, dx=\frac {\sec \left (\frac {c}{2}\right ) \sec ^3\left (\frac {1}{2} (c+d x)\right ) \left (9 B d x \cos \left (\frac {d x}{2}\right )+9 B d x \cos \left (c+\frac {d x}{2}\right )+3 B d x \cos \left (c+\frac {3 d x}{2}\right )+3 B d x \cos \left (2 c+\frac {3 d x}{2}\right )+6 A \sin \left (\frac {d x}{2}\right )-18 B \sin \left (\frac {d x}{2}\right )-6 A \sin \left (c+\frac {d x}{2}\right )+12 B \sin \left (c+\frac {d x}{2}\right )+4 A \sin \left (c+\frac {3 d x}{2}\right )-10 B \sin \left (c+\frac {3 d x}{2}\right )\right )}{24 a^2 d} \]

[In]

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

[Out]

(Sec[c/2]*Sec[(c + d*x)/2]^3*(9*B*d*x*Cos[(d*x)/2] + 9*B*d*x*Cos[c + (d*x)/2] + 3*B*d*x*Cos[c + (3*d*x)/2] + 3
*B*d*x*Cos[2*c + (3*d*x)/2] + 6*A*Sin[(d*x)/2] - 18*B*Sin[(d*x)/2] - 6*A*Sin[c + (d*x)/2] + 12*B*Sin[c + (d*x)
/2] + 4*A*Sin[c + (3*d*x)/2] - 10*B*Sin[c + (3*d*x)/2]))/(24*a^2*d)

Maple [A] (verified)

Time = 0.87 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.70

method	result	size
parallelrisch	\(\frac {\left (-A +B \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (3 A -9 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+6 d x B}{6 a^{2} d}\)	\(49\)
derivativedivides	\(\frac {-\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{3}+\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{3}+A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-3 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+4 B \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d \,a^{2}}\)	\(74\)
default	\(\frac {-\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{3}+\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{3}+A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-3 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+4 B \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d \,a^{2}}\)	\(74\)
risch	\(\frac {B x}{a^{2}}+\frac {2 i \left (3 A \,{\mathrm e}^{2 i \left (d x +c \right )}-6 B \,{\mathrm e}^{2 i \left (d x +c \right )}+3 A \,{\mathrm e}^{i \left (d x +c \right )}-9 B \,{\mathrm e}^{i \left (d x +c \right )}+2 A -5 B \right )}{3 d \,a^{2} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{3}}\)	\(85\)
norman	\(\frac {\frac {B x}{a}+\frac {B x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}+\frac {2 B x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}+\frac {\left (A -7 B \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 a d}+\frac {\left (A -3 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a d}-\frac {\left (A -B \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 a d}+\frac {\left (5 A -17 B \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 a d}}{a \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}\)	\(158\)

[In]

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

[Out]

1/6*((-A+B)*tan(1/2*d*x+1/2*c)^3+(3*A-9*B)*tan(1/2*d*x+1/2*c)+6*d*x*B)/a^2/d

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.30 \[ \int \frac {\cos (c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^2} \, dx=\frac {3 \, B d x \cos \left (d x + c\right )^{2} + 6 \, B d x \cos \left (d x + c\right ) + 3 \, B d x + {\left ({\left (2 \, A - 5 \, B\right )} \cos \left (d x + c\right ) + A - 4 \, B\right )} \sin \left (d x + c\right )}{3 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \]

[In]

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

[Out]

1/3*(3*B*d*x*cos(d*x + c)^2 + 6*B*d*x*cos(d*x + c) + 3*B*d*x + ((2*A - 5*B)*cos(d*x + c) + A - 4*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 = 0.68 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.50 \[ \int \frac {\cos (c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^2} \, dx=\begin {cases} - \frac {A \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a^{2} d} + \frac {A \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{2 a^{2} d} + \frac {B x}{a^{2}} + \frac {B \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a^{2} d} - \frac {3 B \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{2 a^{2} d} & \text {for}\: d \neq 0 \\\frac {x \left (A + B \cos {\left (c \right )}\right ) \cos {\left (c \right )}}{\left (a \cos {\left (c \right )} + a\right )^{2}} & \text {otherwise} \end {cases} \]

[In]

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

Maxima [A] (verification not implemented)

none

Time = 0.41 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.71 \[ \int \frac {\cos (c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^2} \, dx=-\frac {B {\left (\frac {\frac {9 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac {12 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}\right )} - \frac {A {\left (\frac {3 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{2}}}{6 \, d} \]

[In]

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

[Out]

-1/6*(B*((9*sin(d*x + c)/(cos(d*x + c) + 1) - sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/a^2 - 12*arctan(sin(d*x + c
)/(cos(d*x + c) + 1))/a^2) - A*(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 = 0.30 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.23 \[ \int \frac {\cos (c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^2} \, dx=\frac {\frac {6 \, {\left (d x + c\right )} B}{a^{2}} - \frac {A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 9 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{6}}}{6 \, d} \]

[In]

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

[Out]

1/6*(6*(d*x + c)*B/a^2 - (A*a^4*tan(1/2*d*x + 1/2*c)^3 - B*a^4*tan(1/2*d*x + 1/2*c)^3 - 3*A*a^4*tan(1/2*d*x +
1/2*c) + 9*B*a^4*tan(1/2*d*x + 1/2*c))/a^6)/d

Mupad [B] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.93 \[ \int \frac {\cos (c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^2} \, dx=\frac {3\,A\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-9\,B\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-A\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+B\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+6\,B\,d\,x}{6\,a^2\,d} \]

[In]

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

[Out]

(3*A*tan(c/2 + (d*x)/2) - 9*B*tan(c/2 + (d*x)/2) - A*tan(c/2 + (d*x)/2)^3 + B*tan(c/2 + (d*x)/2)^3 + 6*B*d*x)/
(6*a^2*d)