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

Optimal. Leaf size=70 \[ \frac{(2 A-5 B) \sin (c+d x)}{3 a^2 d (\cos (c+d x)+1)}+\frac{B x}{a^2}-\frac{(A-B) \sin (c+d x)}{3 d (a \cos (c+d x)+a)^2} \]

[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)

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Rubi [A]  time = 0.155169, antiderivative size = 70, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.138, Rules used = {2968, 3019, 2735, 2648} \[ \frac{(2 A-5 B) \sin (c+d x)}{3 a^2 d (\cos (c+d x)+1)}+\frac{B x}{a^2}-\frac{(A-B) \sin (c+d x)}{3 d (a \cos (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[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 2968

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 3019

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]

Rule 2735

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 2648

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]

Rubi steps

\begin{align*} \int \frac{\cos (c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^2} \, dx &=\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]  time = 0.333466, size = 153, normalized size = 2.19 \[ \frac{\sec \left (\frac{c}{2}\right ) \sec ^3\left (\frac{1}{2} (c+d x)\right ) \left (-6 A \sin \left (c+\frac{d x}{2}\right )+4 A \sin \left (c+\frac{3 d x}{2}\right )+6 A \sin \left (\frac{d x}{2}\right )+12 B \sin \left (c+\frac{d x}{2}\right )-10 B \sin \left (c+\frac{3 d x}{2}\right )+9 B d x \cos \left (c+\frac{d x}{2}\right )+3 B d x \cos \left (c+\frac{3 d x}{2}\right )+3 B d x \cos \left (2 c+\frac{3 d x}{2}\right )-18 B \sin \left (\frac{d x}{2}\right )+9 B d x \cos \left (\frac{d x}{2}\right )\right )}{24 a^2 d} \]

Antiderivative was successfully verified.

[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)

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Maple [A]  time = 0.052, size = 97, normalized size = 1.4 \begin{align*} -{\frac{A}{6\,{a}^{2}d} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}+{\frac{B}{6\,{a}^{2}d} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}+{\frac{A}{2\,{a}^{2}d}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-{\frac{3\,B}{2\,{a}^{2}d}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+2\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) B}{{a}^{2}d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.8467, size = 162, normalized size = 2.31 \begin{align*} -\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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Fricas [A]  time = 1.35865, size = 228, normalized size = 3.26 \begin{align*} \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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [A]  time = 3.51063, size = 105, normalized size = 1.5 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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Giac [A]  time = 1.2418, size = 116, normalized size = 1.66 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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