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

Optimal. Leaf size=54 \[ -\frac{(A-B) \sin (c+d x)}{a d (\cos (c+d x)+1)}+\frac{x (A-B)}{a}+\frac{B \sin (c+d x)}{a d} \]

[Out]

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

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Rubi [A]  time = 0.138921, antiderivative size = 54, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.172, Rules used = {2968, 3023, 12, 2735, 2648} \[ -\frac{(A-B) \sin (c+d x)}{a d (\cos (c+d x)+1)}+\frac{x (A-B)}{a}+\frac{B \sin (c+d x)}{a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 3023

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (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) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x]
, x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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

Mathematica [B]  time = 0.231352, size = 126, normalized size = 2.33 \[ \frac{\sec \left (\frac{c}{2}\right ) \cos \left (\frac{1}{2} (c+d x)\right ) \left (2 d x (A-B) \cos \left (c+\frac{d x}{2}\right )+2 d x (A-B) \cos \left (\frac{d x}{2}\right )-4 A \sin \left (\frac{d x}{2}\right )+B \sin \left (c+\frac{d x}{2}\right )+B \sin \left (c+\frac{3 d x}{2}\right )+B \sin \left (2 c+\frac{3 d x}{2}\right )+5 B \sin \left (\frac{d x}{2}\right )\right )}{2 a d (\cos (c+d x)+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[(c + d*x)/2]*Sec[c/2]*(2*(A - B)*d*x*Cos[(d*x)/2] + 2*(A - B)*d*x*Cos[c + (d*x)/2] - 4*A*Sin[(d*x)/2] + 5
*B*Sin[(d*x)/2] + B*Sin[c + (d*x)/2] + B*Sin[c + (3*d*x)/2] + B*Sin[2*c + (3*d*x)/2]))/(2*a*d*(1 + Cos[c + d*x
]))

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Maple [A]  time = 0.061, size = 108, normalized size = 2. \begin{align*} -{\frac{A}{da}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+{\frac{B}{da}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+2\,{\frac{B\tan \left ( 1/2\,dx+c/2 \right ) }{da \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) }}+2\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) A}{da}}-2\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) B}{da}} \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),x)

[Out]

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

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Maxima [B]  time = 1.61438, size = 193, normalized size = 3.57 \begin{align*} -\frac{B{\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 )} - A{\left (\frac{2 \, \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a} - \frac{\sin \left (d x + c\right )}{a{\left (\cos \left (d x + c\right ) + 1\right )}}\right )}}{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)),x, algorithm="maxima")

[Out]

-(B*(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*(2*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a -
sin(d*x + c)/(a*(cos(d*x + c) + 1))))/d

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Fricas [A]  time = 1.33714, size = 147, normalized size = 2.72 \begin{align*} \frac{{\left (A - B\right )} d x \cos \left (d x + c\right ) +{\left (A - B\right )} d x +{\left (B \cos \left (d x + c\right ) - A + 2 \, B\right )} \sin \left (d x + c\right )}{a d \cos \left (d x + c\right ) + a 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)),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 2.18575, size = 264, normalized size = 4.89 \begin{align*} \begin{cases} \frac{A 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{A d x}{a d \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + a d} - \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{B 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{B d x}{a d \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + a d} + \frac{B \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 B \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 + B \cos{\left (c \right )}\right ) \cos{\left (c \right )}}{a \cos{\left (c \right )} + a} & \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)),x)

[Out]

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

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Giac [A]  time = 1.21337, size = 105, normalized size = 1.94 \begin{align*} \frac{\frac{{\left (d x + c\right )}{\left (A - B\right )}}{a} - \frac{A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a} + \frac{2 \, B \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} \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)),x, algorithm="giac")

[Out]

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