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

Optimal. Leaf size=99 \[ -\frac{(B-4 C) \sin (c+d x)}{3 a^2 d}-\frac{(B-2 C) \sin (c+d x)}{a^2 d (\cos (c+d x)+1)}+\frac{x (B-2 C)}{a^2}+\frac{(B-C) \sin (c+d x) \cos ^2(c+d x)}{3 d (a \cos (c+d x)+a)^2} \]

[Out]

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

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Rubi [A]  time = 0.314939, antiderivative size = 99, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.184, Rules used = {3029, 2977, 2968, 3023, 12, 2735, 2648} \[ -\frac{(B-4 C) \sin (c+d x)}{3 a^2 d}-\frac{(B-2 C) \sin (c+d x)}{a^2 d (\cos (c+d x)+1)}+\frac{x (B-2 C)}{a^2}+\frac{(B-C) \sin (c+d x) \cos ^2(c+d x)}{3 d (a \cos (c+d x)+a)^2} \]

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])^2,x]

[Out]

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

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]

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 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) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^2} \, dx &=\int \frac{\cos ^2(c+d x) (B+C \cos (c+d x))}{(a+a \cos (c+d x))^2} \, dx\\ &=\frac{(B-C) \cos ^2(c+d x) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2}+\frac{\int \frac{\cos (c+d x) (2 a (B-C)-a (B-4 C) \cos (c+d x))}{a+a \cos (c+d x)} \, dx}{3 a^2}\\ &=\frac{(B-C) \cos ^2(c+d x) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2}+\frac{\int \frac{2 a (B-C) \cos (c+d x)-a (B-4 C) \cos ^2(c+d x)}{a+a \cos (c+d x)} \, dx}{3 a^2}\\ &=-\frac{(B-4 C) \sin (c+d x)}{3 a^2 d}+\frac{(B-C) \cos ^2(c+d x) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2}+\frac{\int \frac{3 a^2 (B-2 C) \cos (c+d x)}{a+a \cos (c+d x)} \, dx}{3 a^3}\\ &=-\frac{(B-4 C) \sin (c+d x)}{3 a^2 d}+\frac{(B-C) \cos ^2(c+d x) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2}+\frac{(B-2 C) \int \frac{\cos (c+d x)}{a+a \cos (c+d x)} \, dx}{a}\\ &=\frac{(B-2 C) x}{a^2}-\frac{(B-4 C) \sin (c+d x)}{3 a^2 d}+\frac{(B-C) \cos ^2(c+d x) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2}-\frac{(B-2 C) \int \frac{1}{a+a \cos (c+d x)} \, dx}{a}\\ &=\frac{(B-2 C) x}{a^2}-\frac{(B-4 C) \sin (c+d x)}{3 a^2 d}+\frac{(B-C) \cos ^2(c+d x) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2}-\frac{(B-2 C) \sin (c+d x)}{d \left (a^2+a^2 \cos (c+d x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.679917, size = 137, normalized size = 1.38 \[ \frac{2 \cos \left (\frac{1}{2} (c+d x)\right ) \left (6 \cos ^3\left (\frac{1}{2} (c+d x)\right ) (d x (B-2 C)+C \sin (c+d x))+(B-C) \tan \left (\frac{c}{2}\right ) \cos \left (\frac{1}{2} (c+d x)\right )+(B-C) \sec \left (\frac{c}{2}\right ) \sin \left (\frac{d x}{2}\right )-2 (5 B-8 C) \sec \left (\frac{c}{2}\right ) \sin \left (\frac{d x}{2}\right ) \cos ^2\left (\frac{1}{2} (c+d x)\right )\right )}{3 a^2 d (\cos (c+d x)+1)^2} \]

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])^2,x]

[Out]

(2*Cos[(c + d*x)/2]*((B - C)*Sec[c/2]*Sin[(d*x)/2] - 2*(5*B - 8*C)*Cos[(c + d*x)/2]^2*Sec[c/2]*Sin[(d*x)/2] +
6*Cos[(c + d*x)/2]^3*((B - 2*C)*d*x + C*Sin[c + d*x]) + (B - C)*Cos[(c + d*x)/2]*Tan[c/2]))/(3*a^2*d*(1 + Cos[
c + d*x])^2)

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Maple [A]  time = 0.031, size = 149, normalized size = 1.5 \begin{align*}{\frac{B}{6\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}-{\frac{C}{6\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}-{\frac{3\,B}{2\,d{a}^{2}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+{\frac{5\,C}{2\,d{a}^{2}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+2\,{\frac{C\tan \left ( 1/2\,dx+c/2 \right ) }{d{a}^{2} \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}{d{a}^{2}}}-4\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) C}{d{a}^{2}}} \end{align*}

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))^2,x)

[Out]

1/6/d/a^2*tan(1/2*d*x+1/2*c)^3*B-1/6/d/a^2*C*tan(1/2*d*x+1/2*c)^3-3/2/d/a^2*B*tan(1/2*d*x+1/2*c)+5/2/d/a^2*C*t
an(1/2*d*x+1/2*c)+2/d/a^2*C*tan(1/2*d*x+1/2*c)/(tan(1/2*d*x+1/2*c)^2+1)+2/d/a^2*arctan(tan(1/2*d*x+1/2*c))*B-4
/d/a^2*arctan(tan(1/2*d*x+1/2*c))*C

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Maxima [B]  time = 1.99488, size = 258, normalized size = 2.61 \begin{align*} \frac{C{\left (\frac{\frac{15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac{24 \, \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}} + \frac{12 \, \sin \left (d x + c\right )}{{\left (a^{2} + \frac{a^{2} \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 )}}\right )} - B{\left (\frac{\frac{9 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac{12 \, \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}\right )}}{6 \, d} \end{align*}

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))^2,x, algorithm="maxima")

[Out]

1/6*(C*((15*sin(d*x + c)/(cos(d*x + c) + 1) - sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/a^2 - 24*arctan(sin(d*x + c
)/(cos(d*x + c) + 1))/a^2 + 12*sin(d*x + c)/((a^2 + a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2)*(cos(d*x + c) + 1
))) - B*((9*sin(d*x + c)/(cos(d*x + c) + 1) - sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/a^2 - 12*arctan(sin(d*x + c
)/(cos(d*x + c) + 1))/a^2))/d

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Fricas [A]  time = 1.6701, size = 294, normalized size = 2.97 \begin{align*} \frac{3 \,{\left (B - 2 \, C\right )} d x \cos \left (d x + c\right )^{2} + 6 \,{\left (B - 2 \, C\right )} d x \cos \left (d x + c\right ) + 3 \,{\left (B - 2 \, C\right )} d x +{\left (3 \, C \cos \left (d x + c\right )^{2} -{\left (5 \, B - 14 \, C\right )} \cos \left (d x + c\right ) - 4 \, B + 10 \, C\right )} \sin \left (d x + c\right )}{3 \,{\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \end{align*}

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))^2,x, algorithm="fricas")

[Out]

1/3*(3*(B - 2*C)*d*x*cos(d*x + c)^2 + 6*(B - 2*C)*d*x*cos(d*x + c) + 3*(B - 2*C)*d*x + (3*C*cos(d*x + c)^2 - (
5*B - 14*C)*cos(d*x + c) - 4*B + 10*C)*sin(d*x + c))/(a^2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a^2*d)

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Sympy [A]  time = 10.3865, size = 415, normalized size = 4.19 \begin{align*} \begin{cases} \frac{6 B d x \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{6 a^{2} d \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 6 a^{2} d} + \frac{6 B d x}{6 a^{2} d \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 6 a^{2} d} + \frac{B \tan ^{5}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{6 a^{2} d \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 6 a^{2} d} - \frac{8 B \tan ^{3}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{6 a^{2} d \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 6 a^{2} d} - \frac{9 B \tan{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{6 a^{2} d \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 6 a^{2} d} - \frac{12 C d x \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{6 a^{2} d \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 6 a^{2} d} - \frac{12 C d x}{6 a^{2} d \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 6 a^{2} d} - \frac{C \tan ^{5}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{6 a^{2} d \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 6 a^{2} d} + \frac{14 C \tan ^{3}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{6 a^{2} d \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 6 a^{2} d} + \frac{27 C \tan{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{6 a^{2} d \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 6 a^{2} d} & \text{for}\: d \neq 0 \\\frac{x \left (B \cos{\left (c \right )} + C \cos ^{2}{\left (c \right )}\right ) \cos{\left (c \right )}}{\left (a \cos{\left (c \right )} + a\right )^{2}} & \text{otherwise} \end{cases} \end{align*}

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))**2,x)

[Out]

Piecewise((6*B*d*x*tan(c/2 + d*x/2)**2/(6*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) + 6*B*d*x/(6*a**2*d*tan(c/2 +
d*x/2)**2 + 6*a**2*d) + B*tan(c/2 + d*x/2)**5/(6*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) - 8*B*tan(c/2 + d*x/2
)**3/(6*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) - 9*B*tan(c/2 + d*x/2)/(6*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d
) - 12*C*d*x*tan(c/2 + d*x/2)**2/(6*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) - 12*C*d*x/(6*a**2*d*tan(c/2 + d*x/
2)**2 + 6*a**2*d) - C*tan(c/2 + d*x/2)**5/(6*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) + 14*C*tan(c/2 + d*x/2)**3
/(6*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) + 27*C*tan(c/2 + d*x/2)/(6*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d),
Ne(d, 0)), (x*(B*cos(c) + C*cos(c)**2)*cos(c)/(a*cos(c) + a)**2, True))

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Giac [A]  time = 1.39172, size = 161, normalized size = 1.63 \begin{align*} \frac{\frac{6 \,{\left (d x + c\right )}{\left (B - 2 \, C\right )}}{a^{2}} + \frac{12 \, 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^{2}} + \frac{B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - C a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 9 \, B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 15 \, C a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{6}}}{6 \, d} \end{align*}

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))^2,x, algorithm="giac")

[Out]

1/6*(6*(d*x + c)*(B - 2*C)/a^2 + 12*C*tan(1/2*d*x + 1/2*c)/((tan(1/2*d*x + 1/2*c)^2 + 1)*a^2) + (B*a^4*tan(1/2
*d*x + 1/2*c)^3 - C*a^4*tan(1/2*d*x + 1/2*c)^3 - 9*B*a^4*tan(1/2*d*x + 1/2*c) + 15*C*a^4*tan(1/2*d*x + 1/2*c))
/a^6)/d