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

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

[Out]

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

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Rubi [A]  time = 0.088818, antiderivative size = 54, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {3023, 12, 2735, 2648} \[ -\frac{(B-C) \sin (c+d x)}{a d (\cos (c+d x)+1)}+\frac{x (B-C)}{a}+\frac{C \sin (c+d x)}{a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [B]  time = 0.23507, size = 126, normalized size = 2.33 \[ \frac{\sec \left (\frac{c}{2}\right ) \cos \left (\frac{1}{2} (c+d x)\right ) \left (2 d x (B-C) \cos \left (c+\frac{d x}{2}\right )+2 d x (B-C) \cos \left (\frac{d x}{2}\right )-4 B \sin \left (\frac{d x}{2}\right )+C \sin \left (c+\frac{d x}{2}\right )+C \sin \left (c+\frac{3 d x}{2}\right )+C \sin \left (2 c+\frac{3 d x}{2}\right )+5 C \sin \left (\frac{d x}{2}\right )\right )}{2 a d (\cos (c+d x)+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[(c + d*x)/2]*Sec[c/2]*(2*(B - C)*d*x*Cos[(d*x)/2] + 2*(B - C)*d*x*Cos[c + (d*x)/2] - 4*B*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]))/(2*a*d*(1 + Cos[c + d*x
]))

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Maple [A]  time = 0.027, size = 108, normalized size = 2. \begin{align*} -{\frac{B}{ad}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+{\frac{C}{ad}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+2\,{\frac{C\tan \left ( 1/2\,dx+c/2 \right ) }{ad \left ( \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1 \right ) }}+2\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) B}{ad}}-2\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) C}{ad}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.87203, size = 193, normalized size = 3.57 \begin{align*} -\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 )} - B{\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((B*cos(d*x+c)+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))) - B*(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 = 1.58746, size = 147, normalized size = 2.72 \begin{align*} \frac{{\left (B - C\right )} d x \cos \left (d x + c\right ) +{\left (B - C\right )} d x +{\left (C \cos \left (d x + c\right ) - B + 2 \, C\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((B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+a*cos(d*x+c)),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 3.59394, size = 265, normalized size = 4.91 \begin{align*} \begin{cases} \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{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} - \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 (B \cos{\left (c \right )} + C \cos ^{2}{\left (c \right )}\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((B*cos(d*x+c)+C*cos(d*x+c)**2)/(a+a*cos(d*x+c)),x)

[Out]

Piecewise((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) - B*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*(B*cos(c) + C*cos(c)**2)/(a*cos(c) + a), True))

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Giac [A]  time = 1.6102, size = 105, normalized size = 1.94 \begin{align*} \frac{\frac{{\left (d x + c\right )}{\left (B - C\right )}}{a} - \frac{B \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((d*x + c)*(B - C)/a - (B*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