3.271 \(\int \frac{\cos (c+d x) (B \cos (c+d x)+C \cos ^2(c+d x))}{(a+a \cos (c+d x))^3} \, dx\)

Optimal. Leaf size=116 \[ \frac{(4 B-29 C) \sin (c+d x)}{15 d \left (a^3 \cos (c+d x)+a^3\right )}+\frac{C x}{a^3}+\frac{(B-C) \sin (c+d x) \cos ^2(c+d x)}{5 d (a \cos (c+d x)+a)^3}-\frac{(2 B-7 C) \sin (c+d x)}{15 a d (a \cos (c+d x)+a)^2} \]

[Out]

(C*x)/a^3 + ((B - C)*Cos[c + d*x]^2*Sin[c + d*x])/(5*d*(a + a*Cos[c + d*x])^3) - ((2*B - 7*C)*Sin[c + d*x])/(1
5*a*d*(a + a*Cos[c + d*x])^2) + ((4*B - 29*C)*Sin[c + d*x])/(15*d*(a^3 + a^3*Cos[c + d*x]))

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Rubi [A]  time = 0.338189, antiderivative size = 116, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {3029, 2977, 2968, 3019, 2735, 2648} \[ \frac{(4 B-29 C) \sin (c+d x)}{15 d \left (a^3 \cos (c+d x)+a^3\right )}+\frac{C x}{a^3}+\frac{(B-C) \sin (c+d x) \cos ^2(c+d x)}{5 d (a \cos (c+d x)+a)^3}-\frac{(2 B-7 C) \sin (c+d x)}{15 a d (a \cos (c+d x)+a)^2} \]

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])^3,x]

[Out]

(C*x)/a^3 + ((B - C)*Cos[c + d*x]^2*Sin[c + d*x])/(5*d*(a + a*Cos[c + d*x])^3) - ((2*B - 7*C)*Sin[c + d*x])/(1
5*a*d*(a + a*Cos[c + d*x])^2) + ((4*B - 29*C)*Sin[c + d*x])/(15*d*(a^3 + a^3*Cos[c + d*x]))

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 3019

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_
.)*(x_)]^2), x_Symbol] :> Simp[((A*b - a*B + b*C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m)/(a*f*(2*m + 1)), x] + D
ist[1/(a^2*(2*m + 1)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[a*A*(m + 1) + m*(b*B - a*C) + b*C*(2*m + 1)*Sin[e
+ f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && LtQ[m, -1] && 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 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))^3} \, dx &=\int \frac{\cos ^2(c+d x) (B+C \cos (c+d x))}{(a+a \cos (c+d x))^3} \, dx\\ &=\frac{(B-C) \cos ^2(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac{\int \frac{\cos (c+d x) (2 a (B-C)+5 a C \cos (c+d x))}{(a+a \cos (c+d x))^2} \, dx}{5 a^2}\\ &=\frac{(B-C) \cos ^2(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac{\int \frac{2 a (B-C) \cos (c+d x)+5 a C \cos ^2(c+d x)}{(a+a \cos (c+d x))^2} \, dx}{5 a^2}\\ &=\frac{(B-C) \cos ^2(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac{(2 B-7 C) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac{\int \frac{-2 a^2 (2 B-7 C)-15 a^2 C \cos (c+d x)}{a+a \cos (c+d x)} \, dx}{15 a^4}\\ &=\frac{C x}{a^3}+\frac{(B-C) \cos ^2(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac{(2 B-7 C) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac{(4 B-29 C) \int \frac{1}{a+a \cos (c+d x)} \, dx}{15 a^2}\\ &=\frac{C x}{a^3}+\frac{(B-C) \cos ^2(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac{(2 B-7 C) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac{(4 B-29 C) \sin (c+d x)}{15 d \left (a^3+a^3 \cos (c+d x)\right )}\\ \end{align*}

Mathematica [B]  time = 0.53862, size = 241, normalized size = 2.08 \[ \frac{\sec \left (\frac{c}{2}\right ) \sec ^5\left (\frac{1}{2} (c+d x)\right ) \left (-60 B \sin \left (c+\frac{d x}{2}\right )+40 B \sin \left (c+\frac{3 d x}{2}\right )-30 B \sin \left (2 c+\frac{3 d x}{2}\right )+14 B \sin \left (2 c+\frac{5 d x}{2}\right )+80 B \sin \left (\frac{d x}{2}\right )+270 C \sin \left (c+\frac{d x}{2}\right )-230 C \sin \left (c+\frac{3 d x}{2}\right )+90 C \sin \left (2 c+\frac{3 d x}{2}\right )-64 C \sin \left (2 c+\frac{5 d x}{2}\right )+150 C d x \cos \left (c+\frac{d x}{2}\right )+75 C d x \cos \left (c+\frac{3 d x}{2}\right )+75 C d x \cos \left (2 c+\frac{3 d x}{2}\right )+15 C d x \cos \left (2 c+\frac{5 d x}{2}\right )+15 C d x \cos \left (3 c+\frac{5 d x}{2}\right )-370 C \sin \left (\frac{d x}{2}\right )+150 C d x \cos \left (\frac{d x}{2}\right )\right )}{480 a^3 d} \]

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])^3,x]

[Out]

(Sec[c/2]*Sec[(c + d*x)/2]^5*(150*C*d*x*Cos[(d*x)/2] + 150*C*d*x*Cos[c + (d*x)/2] + 75*C*d*x*Cos[c + (3*d*x)/2
] + 75*C*d*x*Cos[2*c + (3*d*x)/2] + 15*C*d*x*Cos[2*c + (5*d*x)/2] + 15*C*d*x*Cos[3*c + (5*d*x)/2] + 80*B*Sin[(
d*x)/2] - 370*C*Sin[(d*x)/2] - 60*B*Sin[c + (d*x)/2] + 270*C*Sin[c + (d*x)/2] + 40*B*Sin[c + (3*d*x)/2] - 230*
C*Sin[c + (3*d*x)/2] - 30*B*Sin[2*c + (3*d*x)/2] + 90*C*Sin[2*c + (3*d*x)/2] + 14*B*Sin[2*c + (5*d*x)/2] - 64*
C*Sin[2*c + (5*d*x)/2]))/(480*a^3*d)

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Maple [A]  time = 0.026, size = 137, normalized size = 1.2 \begin{align*}{\frac{B}{20\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}-{\frac{C}{20\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}-{\frac{B}{6\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}+{\frac{C}{3\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}+{\frac{B}{4\,d{a}^{3}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-{\frac{7\,C}{4\,d{a}^{3}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+2\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) C}{d{a}^{3}}} \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))^3,x)

[Out]

1/20/d/a^3*B*tan(1/2*d*x+1/2*c)^5-1/20/d/a^3*C*tan(1/2*d*x+1/2*c)^5-1/6/d/a^3*tan(1/2*d*x+1/2*c)^3*B+1/3/d/a^3
*C*tan(1/2*d*x+1/2*c)^3+1/4/d/a^3*B*tan(1/2*d*x+1/2*c)-7/4/d/a^3*C*tan(1/2*d*x+1/2*c)+2/d/a^3*arctan(tan(1/2*d
*x+1/2*c))*C

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Maxima [A]  time = 1.92624, size = 216, normalized size = 1.86 \begin{align*} -\frac{C{\left (\frac{\frac{105 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{20 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac{120 \, \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}\right )} - \frac{B{\left (\frac{15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a^{3}}}{60 \, 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))^3,x, algorithm="maxima")

[Out]

-1/60*(C*((105*sin(d*x + c)/(cos(d*x + c) + 1) - 20*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 3*sin(d*x + c)^5/(co
s(d*x + c) + 1)^5)/a^3 - 120*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^3) - B*(15*sin(d*x + c)/(cos(d*x + c) +
1) - 10*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 3*sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/a^3)/d

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Fricas [A]  time = 1.64559, size = 351, normalized size = 3.03 \begin{align*} \frac{15 \, C d x \cos \left (d x + c\right )^{3} + 45 \, C d x \cos \left (d x + c\right )^{2} + 45 \, C d x \cos \left (d x + c\right ) + 15 \, C d x +{\left ({\left (7 \, B - 32 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \,{\left (2 \, B - 17 \, C\right )} \cos \left (d x + c\right ) + 2 \, B - 22 \, C\right )} \sin \left (d x + c\right )}{15 \,{\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} 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))^3,x, algorithm="fricas")

[Out]

1/15*(15*C*d*x*cos(d*x + c)^3 + 45*C*d*x*cos(d*x + c)^2 + 45*C*d*x*cos(d*x + c) + 15*C*d*x + ((7*B - 32*C)*cos
(d*x + c)^2 + 3*(2*B - 17*C)*cos(d*x + c) + 2*B - 22*C)*sin(d*x + c))/(a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x
+ c)^2 + 3*a^3*d*cos(d*x + c) + a^3*d)

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Sympy [A]  time = 15.3748, size = 151, normalized size = 1.3 \begin{align*} \begin{cases} \frac{B \tan ^{5}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{20 a^{3} d} - \frac{B \tan ^{3}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{6 a^{3} d} + \frac{B \tan{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{4 a^{3} d} + \frac{C x}{a^{3}} - \frac{C \tan ^{5}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{20 a^{3} d} + \frac{C \tan ^{3}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{3 a^{3} d} - \frac{7 C \tan{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{4 a^{3} 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 )^{3}} & \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))**3,x)

[Out]

Piecewise((B*tan(c/2 + d*x/2)**5/(20*a**3*d) - B*tan(c/2 + d*x/2)**3/(6*a**3*d) + B*tan(c/2 + d*x/2)/(4*a**3*d
) + C*x/a**3 - C*tan(c/2 + d*x/2)**5/(20*a**3*d) + C*tan(c/2 + d*x/2)**3/(3*a**3*d) - 7*C*tan(c/2 + d*x/2)/(4*
a**3*d), Ne(d, 0)), (x*(B*cos(c) + C*cos(c)**2)*cos(c)/(a*cos(c) + a)**3, True))

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Giac [A]  time = 1.46601, size = 162, normalized size = 1.4 \begin{align*} \frac{\frac{60 \,{\left (d x + c\right )} C}{a^{3}} + \frac{3 \, B a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 3 \, C a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 10 \, B a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 20 \, C a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 15 \, B a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 105 \, C a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{15}}}{60 \, 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))^3,x, algorithm="giac")

[Out]

1/60*(60*(d*x + c)*C/a^3 + (3*B*a^12*tan(1/2*d*x + 1/2*c)^5 - 3*C*a^12*tan(1/2*d*x + 1/2*c)^5 - 10*B*a^12*tan(
1/2*d*x + 1/2*c)^3 + 20*C*a^12*tan(1/2*d*x + 1/2*c)^3 + 15*B*a^12*tan(1/2*d*x + 1/2*c) - 105*C*a^12*tan(1/2*d*
x + 1/2*c))/a^15)/d