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

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

[Out]

((A - 2*B)*x)/a^2 - ((A - 4*B)*Sin[c + d*x])/(3*a^2*d) - ((A - 2*B)*Sin[c + d*x])/(a^2*d*(1 + Cos[c + d*x])) +
 ((A - B)*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 = 0.275746, antiderivative size = 99, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.194, Rules used = {2977, 2968, 3023, 12, 2735, 2648} \[ -\frac{(A-4 B) \sin (c+d x)}{3 a^2 d}-\frac{(A-2 B) \sin (c+d x)}{a^2 d (\cos (c+d x)+1)}+\frac{x (A-2 B)}{a^2}+\frac{(A-B) \sin (c+d x) \cos ^2(c+d x)}{3 d (a \cos (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.677156, size = 137, normalized size = 1.38 \[ \frac{2 \cos \left (\frac{1}{2} (c+d x)\right ) \left (6 \cos ^3\left (\frac{1}{2} (c+d x)\right ) (d x (A-2 B)+B \sin (c+d x))+(A-B) \tan \left (\frac{c}{2}\right ) \cos \left (\frac{1}{2} (c+d x)\right )+(A-B) \sec \left (\frac{c}{2}\right ) \sin \left (\frac{d x}{2}\right )-2 (5 A-8 B) \sec \left (\frac{c}{2}\right ) \sin \left (\frac{d x}{2}\right ) \cos ^2\left (\frac{1}{2} (c+d x)\right )\right )}{3 a^2 d (\cos (c+d x)+1)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Cos[(c + d*x)/2]*((A - B)*Sec[c/2]*Sin[(d*x)/2] - 2*(5*A - 8*B)*Cos[(c + d*x)/2]^2*Sec[c/2]*Sin[(d*x)/2] +
6*Cos[(c + d*x)/2]^3*((A - 2*B)*d*x + B*Sin[c + d*x]) + (A - B)*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 = 0.079, size = 149, normalized size = 1.5 \begin{align*}{\frac{A}{6\,{a}^{2}d} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}-{\frac{B}{6\,{a}^{2}d} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}-{\frac{3\,A}{2\,{a}^{2}d}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+{\frac{5\,B}{2\,{a}^{2}d}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+2\,{\frac{B\tan \left ( 1/2\,dx+c/2 \right ) }{{a}^{2}d \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) }}+2\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) A}{{a}^{2}d}}-4\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) B}{{a}^{2}d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.89457, size = 258, normalized size = 2.61 \begin{align*} \frac{B{\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 )} - A{\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)^2*(A+B*cos(d*x+c))/(a+a*cos(d*x+c))^2,x, algorithm="maxima")

[Out]

1/6*(B*((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
))) - A*((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 = 1.36591, size = 294, normalized size = 2.97 \begin{align*} \frac{3 \,{\left (A - 2 \, B\right )} d x \cos \left (d x + c\right )^{2} + 6 \,{\left (A - 2 \, B\right )} d x \cos \left (d x + c\right ) + 3 \,{\left (A - 2 \, B\right )} d x +{\left (3 \, B \cos \left (d x + c\right )^{2} -{\left (5 \, A - 14 \, B\right )} \cos \left (d x + c\right ) - 4 \, A + 10 \, B\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)^2*(A+B*cos(d*x+c))/(a+a*cos(d*x+c))^2,x, algorithm="fricas")

[Out]

1/3*(3*(A - 2*B)*d*x*cos(d*x + c)^2 + 6*(A - 2*B)*d*x*cos(d*x + c) + 3*(A - 2*B)*d*x + (3*B*cos(d*x + c)^2 - (
5*A - 14*B)*cos(d*x + c) - 4*A + 10*B)*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 = 7.01014, size = 411, normalized size = 4.15 \begin{align*} \begin{cases} \frac{6 A 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 A d x}{6 a^{2} d \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 6 a^{2} d} + \frac{A \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 A \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 A \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 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{12 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{14 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{27 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} & \text{for}\: d \neq 0 \\\frac{x \left (A + B \cos{\left (c \right )}\right ) \cos ^{2}{\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)**2*(A+B*cos(d*x+c))/(a+a*cos(d*x+c))**2,x)

[Out]

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

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Giac [A]  time = 1.26666, size = 161, normalized size = 1.63 \begin{align*} \frac{\frac{6 \,{\left (d x + c\right )}{\left (A - 2 \, B\right )}}{a^{2}} + \frac{12 \, B \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{A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 9 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 15 \, B 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)^2*(A+B*cos(d*x+c))/(a+a*cos(d*x+c))^2,x, algorithm="giac")

[Out]

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