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

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Rubi [A]  time = 0.31, antiderivative size = 99, normalized size of antiderivative = 1.00, 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 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 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 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 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 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))^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*}

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Mathematica [A]  time = 0.70, 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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fricas [A]  time = 0.43, size = 117, normalized size = 1.18 \[ \frac {3 \, {\left (B - 2 \, C\right )} d x \cos \left (d x + c\right )^{2} + 6 \, {\left (B - 2 \, C\right )} d x \cos \left (d x + c\right ) + 3 \, {\left (B - 2 \, C\right )} d x + {\left (3 \, C \cos \left (d x + c\right )^{2} - {\left (5 \, B - 14 \, C\right )} \cos \left (d x + c\right ) - 4 \, B + 10 \, C\right )} \sin \left (d x + c\right )}{3 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \]

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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giac [A]  time = 0.49, size = 119, normalized size = 1.20 \[ \frac {\frac {6 \, {\left (d x + c\right )} {\left (B - 2 \, C\right )}}{a^{2}} + \frac {12 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )} a^{2}} + \frac {B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 9 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 15 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{6}}}{6 \, d} \]

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

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maple [A]  time = 0.12, size = 149, normalized size = 1.51 \[ \frac {B \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d \,a^{2}}-\frac {C \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d \,a^{2}}-\frac {3 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d \,a^{2}}+\frac {5 C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d \,a^{2}}+\frac {2 C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{d \,a^{2}}-\frac {4 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C}{d \,a^{2}} \]

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*B*tan(1/2*d*x+1/2*c)^3-1/6/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)/(1+tan(1/2*d*x+1/2*c)^2)+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 = 1.02, size = 191, normalized size = 1.93 \[ \frac {C {\left (\frac {\frac {15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac {24 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}} + \frac {12 \, \sin \left (d x + c\right )}{{\left (a^{2} + \frac {a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}}\right )} - B {\left (\frac {\frac {9 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac {12 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}\right )}}{6 \, d} \]

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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mupad [B]  time = 1.12, size = 105, normalized size = 1.06 \[ \frac {x\,\left (B-2\,C\right )}{a^2}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {B-C}{a^2}+\frac {B-3\,C}{2\,a^2}\right )}{d}+\frac {2\,C\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a^2\right )}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (B-C\right )}{6\,a^2\,d} \]

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))^2,x)

[Out]

(x*(B - 2*C))/a^2 - (tan(c/2 + (d*x)/2)*((B - C)/a^2 + (B - 3*C)/(2*a^2)))/d + (2*C*tan(c/2 + (d*x)/2))/(d*(a^
2*tan(c/2 + (d*x)/2)^2 + a^2)) + (tan(c/2 + (d*x)/2)^3*(B - C))/(6*a^2*d)

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sympy [A]  time = 6.01, size = 415, normalized size = 4.19 \[ \begin {cases} \frac {6 B d x \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a^{2} d} + \frac {6 B d x}{6 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a^{2} d} + \frac {B \tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a^{2} d} - \frac {8 B \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a^{2} d} - \frac {9 B \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a^{2} d} - \frac {12 C d x \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a^{2} d} - \frac {12 C d x}{6 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a^{2} d} - \frac {C \tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a^{2} d} + \frac {14 C \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a^{2} d} + \frac {27 C \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a^{2} d} & \text {for}\: d \neq 0 \\\frac {x \left (B \cos {\relax (c )} + C \cos ^{2}{\relax (c )}\right ) \cos {\relax (c )}}{\left (a \cos {\relax (c )} + a\right )^{2}} & \text {otherwise} \end {cases} \]

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))**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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