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

Optimal. Leaf size=99 \[ \frac {2 (B-C) \sin (c+d x)}{a d}+\frac {(B-C) \sin (c+d x) \cos ^2(c+d x)}{d (a \cos (c+d x)+a)}-\frac {(2 B-3 C) \sin (c+d x) \cos (c+d x)}{2 a d}-\frac {x (2 B-3 C)}{2 a} \]

[Out]

-1/2*(2*B-3*C)*x/a+2*(B-C)*sin(d*x+c)/a/d-1/2*(2*B-3*C)*cos(d*x+c)*sin(d*x+c)/a/d+(B-C)*cos(d*x+c)^2*sin(d*x+c
)/d/(a+a*cos(d*x+c))

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Rubi [A]  time = 0.17, antiderivative size = 99, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.079, Rules used = {3029, 2977, 2734} \[ \frac {2 (B-C) \sin (c+d x)}{a d}+\frac {(B-C) \sin (c+d x) \cos ^2(c+d x)}{d (a \cos (c+d x)+a)}-\frac {(2 B-3 C) \sin (c+d x) \cos (c+d x)}{2 a d}-\frac {x (2 B-3 C)}{2 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((2*B - 3*C)*x)/(2*a) + (2*(B - C)*Sin[c + d*x])/(a*d) - ((2*B - 3*C)*Cos[c + d*x]*Sin[c + d*x])/(2*a*d) + ((
B - C)*Cos[c + d*x]^2*Sin[c + d*x])/(d*(a + a*Cos[c + d*x]))

Rule 2734

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[((2*a*c
+ b*d)*x)/2, x] + (-Simp[((b*c + a*d)*Cos[e + f*x])/f, x] - Simp[(b*d*Cos[e + f*x]*Sin[e + f*x])/(2*f), x]) /;
FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]

Rule 2977

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[((A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x]
)^n)/(a*f*(2*m + 1)), x] - Dist[1/(a*b*(2*m + 1)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n -
1)*Simp[A*(a*d*n - b*c*(m + 1)) - B*(a*c*m + b*d*n) - d*(a*B*(m - n) + A*b*(m + n + 1))*Sin[e + f*x], x], x],
x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ
[m, -2^(-1)] && GtQ[n, 0] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0])

Rule 3029

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)
*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Dist[1/b^2, Int[(a + b*Sin[e + f*x])
^(m + 1)*(c + d*Sin[e + f*x])^n*(b*B - a*C + b*C*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m,
n}, x] && NeQ[b*c - a*d, 0] && EqQ[A*b^2 - a*b*B + a^2*C, 0]

Rubi steps

\begin {align*} \int \frac {\cos (c+d x) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{a+a \cos (c+d x)} \, dx &=\int \frac {\cos ^2(c+d x) (B+C \cos (c+d x))}{a+a \cos (c+d x)} \, dx\\ &=\frac {(B-C) \cos ^2(c+d x) \sin (c+d x)}{d (a+a \cos (c+d x))}+\frac {\int \cos (c+d x) (2 a (B-C)-a (2 B-3 C) \cos (c+d x)) \, dx}{a^2}\\ &=-\frac {(2 B-3 C) x}{2 a}+\frac {2 (B-C) \sin (c+d x)}{a d}-\frac {(2 B-3 C) \cos (c+d x) \sin (c+d x)}{2 a d}+\frac {(B-C) \cos ^2(c+d x) \sin (c+d x)}{d (a+a \cos (c+d x))}\\ \end {align*}

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Mathematica [A]  time = 0.44, size = 197, normalized size = 1.99 \[ \frac {\sec \left (\frac {c}{2}\right ) \cos \left (\frac {1}{2} (c+d x)\right ) \left (-4 d x (2 B-3 C) \cos \left (c+\frac {d x}{2}\right )+4 B \sin \left (c+\frac {d x}{2}\right )+4 B \sin \left (c+\frac {3 d x}{2}\right )+4 B \sin \left (2 c+\frac {3 d x}{2}\right )-4 d x (2 B-3 C) \cos \left (\frac {d x}{2}\right )+20 B \sin \left (\frac {d x}{2}\right )-4 C \sin \left (c+\frac {d x}{2}\right )-3 C \sin \left (c+\frac {3 d x}{2}\right )-3 C \sin \left (2 c+\frac {3 d x}{2}\right )+C \sin \left (2 c+\frac {5 d x}{2}\right )+C \sin \left (3 c+\frac {5 d x}{2}\right )-20 C \sin \left (\frac {d x}{2}\right )\right )}{8 a d (\cos (c+d x)+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]*(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]*(-4*(2*B - 3*C)*d*x*Cos[(d*x)/2] - 4*(2*B - 3*C)*d*x*Cos[c + (d*x)/2] + 20*B*Sin[(d
*x)/2] - 20*C*Sin[(d*x)/2] + 4*B*Sin[c + (d*x)/2] - 4*C*Sin[c + (d*x)/2] + 4*B*Sin[c + (3*d*x)/2] - 3*C*Sin[c
+ (3*d*x)/2] + 4*B*Sin[2*c + (3*d*x)/2] - 3*C*Sin[2*c + (3*d*x)/2] + C*Sin[2*c + (5*d*x)/2] + C*Sin[3*c + (5*d
*x)/2]))/(8*a*d*(1 + Cos[c + d*x]))

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fricas [A]  time = 0.45, size = 83, normalized size = 0.84 \[ -\frac {{\left (2 \, B - 3 \, C\right )} d x \cos \left (d x + c\right ) + {\left (2 \, B - 3 \, C\right )} d x - {\left (C \cos \left (d x + c\right )^{2} + {\left (2 \, B - C\right )} \cos \left (d x + c\right ) + 4 \, B - 4 \, C\right )} \sin \left (d x + c\right )}{2 \, {\left (a d \cos \left (d x + c\right ) + a d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*((2*B - 3*C)*d*x*cos(d*x + c) + (2*B - 3*C)*d*x - (C*cos(d*x + c)^2 + (2*B - C)*cos(d*x + c) + 4*B - 4*C)
*sin(d*x + c))/(a*d*cos(d*x + c) + a*d)

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giac [A]  time = 0.32, size = 124, normalized size = 1.25 \[ -\frac {\frac {{\left (d x + c\right )} {\left (2 \, B - 3 \, C\right )}}{a} - \frac {2 \, {\left (B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{a} - \frac {2 \, {\left (2 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2} a}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.12, size = 211, normalized size = 2.13 \[ \frac {B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d}-\frac {C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d}-\frac {3 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C}{a d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {2 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{a d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-\frac {C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {2 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{a d}+\frac {3 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C}{a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)*(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)-3/a/d/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2*c)^3*C+
2/a/d/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2*c)^3*B-1/a/d/(1+tan(1/2*d*x+1/2*c)^2)^2*C*tan(1/2*d*x+1/2*c)+
2/a/d/(1+tan(1/2*d*x+1/2*c)^2)^2*B*tan(1/2*d*x+1/2*c)-2/a/d*arctan(tan(1/2*d*x+1/2*c))*B+3/a/d*arctan(tan(1/2*
d*x+1/2*c))*C

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maxima [B]  time = 0.93, size = 225, normalized size = 2.27 \[ -\frac {C {\left (\frac {\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {3 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a + \frac {2 \, a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} - \frac {3 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a} + \frac {\sin \left (d x + c\right )}{a {\left (\cos \left (d x + c\right ) + 1\right )}}\right )} + B {\left (\frac {2 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a} - \frac {2 \, \sin \left (d x + c\right )}{{\left (a + \frac {a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}} - \frac {\sin \left (d x + c\right )}{a {\left (\cos \left (d x + c\right ) + 1\right )}}\right )}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(C*((sin(d*x + c)/(cos(d*x + c) + 1) + 3*sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/(a + 2*a*sin(d*x + c)^2/(cos(d*
x + c) + 1)^2 + a*sin(d*x + c)^4/(cos(d*x + c) + 1)^4) - 3*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a + sin(d*x
+ c)/(a*(cos(d*x + c) + 1))) + 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))))/d

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mupad [B]  time = 1.37, size = 107, normalized size = 1.08 \[ \frac {\left (2\,B-3\,C\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (2\,B-C\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+2\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a\right )}-\frac {x\,\left (2\,B-3\,C\right )}{2\,a}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (B-C\right )}{a\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(tan(c/2 + (d*x)/2)^3*(2*B - 3*C) + tan(c/2 + (d*x)/2)*(2*B - C))/(d*(a + 2*a*tan(c/2 + (d*x)/2)^2 + a*tan(c/2
+ (d*x)/2)^4)) - (x*(2*B - 3*C))/(2*a) + (tan(c/2 + (d*x)/2)*(B - C))/(a*d)

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sympy [A]  time = 3.98, size = 668, normalized size = 6.75 \[ \begin {cases} - \frac {2 B d x \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{2 a d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 4 a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 2 a d} - \frac {4 B d x \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{2 a d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 4 a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 2 a d} - \frac {2 B d x}{2 a d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 4 a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 2 a d} + \frac {2 B \tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{2 a d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 4 a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 2 a d} + \frac {8 B \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{2 a d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 4 a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 2 a d} + \frac {6 B \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{2 a d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 4 a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 2 a d} + \frac {3 C d x \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{2 a d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 4 a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 2 a d} + \frac {6 C d x \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{2 a d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 4 a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 2 a d} + \frac {3 C d x}{2 a d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 4 a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 2 a d} - \frac {2 C \tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{2 a d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 4 a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 2 a d} - \frac {10 C \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{2 a d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 4 a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 2 a d} - \frac {4 C \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{2 a d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 4 a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 2 a d} & \text {for}\: d \neq 0 \\\frac {x \left (B \cos {\relax (c )} + C \cos ^{2}{\relax (c )}\right ) \cos {\relax (c )}}{a \cos {\relax (c )} + a} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-2*B*d*x*tan(c/2 + d*x/2)**4/(2*a*d*tan(c/2 + d*x/2)**4 + 4*a*d*tan(c/2 + d*x/2)**2 + 2*a*d) - 4*B*
d*x*tan(c/2 + d*x/2)**2/(2*a*d*tan(c/2 + d*x/2)**4 + 4*a*d*tan(c/2 + d*x/2)**2 + 2*a*d) - 2*B*d*x/(2*a*d*tan(c
/2 + d*x/2)**4 + 4*a*d*tan(c/2 + d*x/2)**2 + 2*a*d) + 2*B*tan(c/2 + d*x/2)**5/(2*a*d*tan(c/2 + d*x/2)**4 + 4*a
*d*tan(c/2 + d*x/2)**2 + 2*a*d) + 8*B*tan(c/2 + d*x/2)**3/(2*a*d*tan(c/2 + d*x/2)**4 + 4*a*d*tan(c/2 + d*x/2)*
*2 + 2*a*d) + 6*B*tan(c/2 + d*x/2)/(2*a*d*tan(c/2 + d*x/2)**4 + 4*a*d*tan(c/2 + d*x/2)**2 + 2*a*d) + 3*C*d*x*t
an(c/2 + d*x/2)**4/(2*a*d*tan(c/2 + d*x/2)**4 + 4*a*d*tan(c/2 + d*x/2)**2 + 2*a*d) + 6*C*d*x*tan(c/2 + d*x/2)*
*2/(2*a*d*tan(c/2 + d*x/2)**4 + 4*a*d*tan(c/2 + d*x/2)**2 + 2*a*d) + 3*C*d*x/(2*a*d*tan(c/2 + d*x/2)**4 + 4*a*
d*tan(c/2 + d*x/2)**2 + 2*a*d) - 2*C*tan(c/2 + d*x/2)**5/(2*a*d*tan(c/2 + d*x/2)**4 + 4*a*d*tan(c/2 + d*x/2)**
2 + 2*a*d) - 10*C*tan(c/2 + d*x/2)**3/(2*a*d*tan(c/2 + d*x/2)**4 + 4*a*d*tan(c/2 + d*x/2)**2 + 2*a*d) - 4*C*ta
n(c/2 + d*x/2)/(2*a*d*tan(c/2 + d*x/2)**4 + 4*a*d*tan(c/2 + d*x/2)**2 + 2*a*d), Ne(d, 0)), (x*(B*cos(c) + C*co
s(c)**2)*cos(c)/(a*cos(c) + a), True))

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