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

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Rubi [A]  time = 0.09, antiderivative size = 54, normalized size of antiderivative = 1.00, 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 12

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

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]

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 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]

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*}

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Mathematica [B]  time = 0.24, 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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fricas [A]  time = 0.48, size = 61, normalized size = 1.13 \[ \frac {{\left (B - C\right )} d x \cos \left (d x + c\right ) + {\left (B - C\right )} d x + {\left (C \cos \left (d x + c\right ) - B + 2 \, C\right )} \sin \left (d x + c\right )}{a d \cos \left (d x + c\right ) + a d} \]

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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giac [A]  time = 0.40, size = 78, normalized size = 1.44 \[ \frac {\frac {{\left (d x + c\right )} {\left (B - C\right )}}{a} - \frac {B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a} + \frac {2 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )} a}}{d} \]

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

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maple [A]  time = 0.11, size = 108, normalized size = 2.00 \[ -\frac {B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d}+\frac {C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d}+\frac {2 C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{a d}-\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C}{a d} \]

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)/(1+tan(1/2*d*x+1/2*c)^2)+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 = 0.80, size = 143, normalized size = 2.65 \[ -\frac {C {\left (\frac {2 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a} - \frac {2 \, \sin \left (d x + c\right )}{{\left (a + \frac {a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}} - \frac {\sin \left (d x + c\right )}{a {\left (\cos \left (d x + c\right ) + 1\right )}}\right )} - B {\left (\frac {2 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a} - \frac {\sin \left (d x + c\right )}{a {\left (\cos \left (d x + c\right ) + 1\right )}}\right )}}{d} \]

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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mupad [B]  time = 1.13, size = 65, normalized size = 1.20 \[ \frac {x\,\left (B-C\right )}{a}+\frac {2\,C\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a\right )}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (B-C\right )}{a\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 2.30, size = 265, normalized size = 4.91 \[ \begin {cases} \frac {B d x \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a d} + \frac {B d x}{a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a d} - \frac {B \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a d} - \frac {B \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a d} - \frac {C d x \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a d} - \frac {C d x}{a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a d} + \frac {C \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a d} + \frac {3 C \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a d} & \text {for}\: d \neq 0 \\\frac {x \left (B \cos {\relax (c )} + C \cos ^{2}{\relax (c )}\right )}{a \cos {\relax (c )} + a} & \text {otherwise} \end {cases} \]

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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