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

Optimal. Leaf size=34 \[ \frac{(A-B) \sin (c+d x)}{d (a \cos (c+d x)+a)}+\frac{B x}{a} \]

[Out]

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

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Rubi [A]  time = 0.0498775, antiderivative size = 34, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.087, Rules used = {2735, 2648} \[ \frac{(A-B) \sin (c+d x)}{d (a \cos (c+d x)+a)}+\frac{B x}{a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(B*x)/a + ((A - B)*Sin[c + d*x])/(d*(a + a*Cos[c + d*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{A+B \cos (c+d x)}{a+a \cos (c+d x)} \, dx &=\frac{B x}{a}-(-A+B) \int \frac{1}{a+a \cos (c+d x)} \, dx\\ &=\frac{B x}{a}+\frac{(A-B) \sin (c+d x)}{d (a+a \cos (c+d x))}\\ \end{align*}

Mathematica [B]  time = 0.11739, size = 72, normalized size = 2.12 \[ \frac{\sec \left (\frac{c}{2}\right ) \cos \left (\frac{1}{2} (c+d x)\right ) \left (2 (A-B) \sin \left (\frac{d x}{2}\right )+B d x \cos \left (c+\frac{d x}{2}\right )+B d x \cos \left (\frac{d x}{2}\right )\right )}{a d (\cos (c+d x)+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.053, size = 56, normalized size = 1.7 \begin{align*}{\frac{A}{da}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+2\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) B}{da}}-{\frac{B}{da}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Maxima [B]  time = 1.50053, size = 99, normalized size = 2.91 \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{\sin \left (d x + c\right )}{a{\left (\cos \left (d x + c\right ) + 1\right )}}\right )} + \frac{A \sin \left (d x + c\right )}{a{\left (\cos \left (d x + c\right ) + 1\right )}}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((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 - sin(d*x + c)/(a*(cos(d*x + c) + 1))) + A*sin(d*x + c)/(a*(co
s(d*x + c) + 1)))/d

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Fricas [A]  time = 1.31193, size = 105, normalized size = 3.09 \begin{align*} \frac{B d x \cos \left (d x + c\right ) + B d x +{\left (A - 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((A+B*cos(d*x+c))/(a+a*cos(d*x+c)),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 1.10818, size = 49, normalized size = 1.44 \begin{align*} \begin{cases} \frac{A \tan{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{a d} + \frac{B x}{a} - \frac{B \tan{\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 )}{a \cos{\left (c \right )} + a} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.16723, size = 58, normalized size = 1.71 \begin{align*} \frac{\frac{{\left (d x + c\right )} B}{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}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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