3.60 \(\int \frac{\cos (c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^3} \, dx\)

Optimal. Leaf size=102 \[ \frac{(3 A+7 B) \sin (c+d x)}{15 d \left (a^3 \cos (c+d x)+a^3\right )}+\frac{(3 A-8 B) \sin (c+d x)}{15 a d (a \cos (c+d x)+a)^2}-\frac{(A-B) \sin (c+d x)}{5 d (a \cos (c+d x)+a)^3} \]

[Out]

-((A - B)*Sin[c + d*x])/(5*d*(a + a*Cos[c + d*x])^3) + ((3*A - 8*B)*Sin[c + d*x])/(15*a*d*(a + a*Cos[c + d*x])
^2) + ((3*A + 7*B)*Sin[c + d*x])/(15*d*(a^3 + a^3*Cos[c + d*x]))

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Rubi [A]  time = 0.188036, antiderivative size = 102, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.138, Rules used = {2968, 3019, 2750, 2648} \[ \frac{(3 A+7 B) \sin (c+d x)}{15 d \left (a^3 \cos (c+d x)+a^3\right )}+\frac{(3 A-8 B) \sin (c+d x)}{15 a d (a \cos (c+d x)+a)^2}-\frac{(A-B) \sin (c+d x)}{5 d (a \cos (c+d x)+a)^3} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]*(A + B*Cos[c + d*x]))/(a + a*Cos[c + d*x])^3,x]

[Out]

-((A - B)*Sin[c + d*x])/(5*d*(a + a*Cos[c + d*x])^3) + ((3*A - 8*B)*Sin[c + d*x])/(15*a*d*(a + a*Cos[c + d*x])
^2) + ((3*A + 7*B)*Sin[c + d*x])/(15*d*(a^3 + a^3*Cos[c + d*x]))

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 2750

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

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) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^3} \, dx &=\int \frac{A \cos (c+d x)+B \cos ^2(c+d x)}{(a+a \cos (c+d x))^3} \, dx\\ &=-\frac{(A-B) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac{\int \frac{-3 a (A-B)-5 a B \cos (c+d x)}{(a+a \cos (c+d x))^2} \, dx}{5 a^2}\\ &=-\frac{(A-B) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac{(3 A-8 B) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac{(3 A+7 B) \int \frac{1}{a+a \cos (c+d x)} \, dx}{15 a^2}\\ &=-\frac{(A-B) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac{(3 A-8 B) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac{(3 A+7 B) \sin (c+d x)}{15 d \left (a^3+a^3 \cos (c+d x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.319653, size = 135, normalized size = 1.32 \[ \frac{\sec \left (\frac{c}{2}\right ) \cos \left (\frac{1}{2} (c+d x)\right ) \left (-15 (A+2 B) \sin \left (c+\frac{d x}{2}\right )+5 (3 A+8 B) \sin \left (\frac{d x}{2}\right )+15 A \sin \left (c+\frac{3 d x}{2}\right )+3 A \sin \left (2 c+\frac{5 d x}{2}\right )+20 B \sin \left (c+\frac{3 d x}{2}\right )-15 B \sin \left (2 c+\frac{3 d x}{2}\right )+7 B \sin \left (2 c+\frac{5 d x}{2}\right )\right )}{30 a^3 d (\cos (c+d x)+1)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]*(A + B*Cos[c + d*x]))/(a + a*Cos[c + d*x])^3,x]

[Out]

(Cos[(c + d*x)/2]*Sec[c/2]*(5*(3*A + 8*B)*Sin[(d*x)/2] - 15*(A + 2*B)*Sin[c + (d*x)/2] + 15*A*Sin[c + (3*d*x)/
2] + 20*B*Sin[c + (3*d*x)/2] - 15*B*Sin[2*c + (3*d*x)/2] + 3*A*Sin[2*c + (5*d*x)/2] + 7*B*Sin[2*c + (5*d*x)/2]
))/(30*a^3*d*(1 + Cos[c + d*x])^3)

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Maple [A]  time = 0.05, size = 64, normalized size = 0.6 \begin{align*}{\frac{1}{4\,d{a}^{3}} \left ({\frac{-A+B}{5} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}-{\frac{2\,B}{3} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}+A\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +B\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)*(A+B*cos(d*x+c))/(a+cos(d*x+c)*a)^3,x)

[Out]

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

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Maxima [A]  time = 1.01729, size = 155, normalized size = 1.52 \begin{align*} \frac{\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}} + \frac{3 \, A{\left (\frac{5 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{\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)*(A+B*cos(d*x+c))/(a+a*cos(d*x+c))^3,x, algorithm="maxima")

[Out]

1/60*(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 + 3*A*(5*sin(d*x + c)/(cos(d*x + c) + 1) - sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/a^3)/d

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Fricas [A]  time = 1.31214, size = 227, normalized size = 2.23 \begin{align*} \frac{{\left ({\left (3 \, A + 7 \, B\right )} \cos \left (d x + c\right )^{2} + 3 \,{\left (3 \, A + 2 \, B\right )} \cos \left (d x + c\right ) + 3 \, A + 2 \, B\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)*(A+B*cos(d*x+c))/(a+a*cos(d*x+c))^3,x, algorithm="fricas")

[Out]

1/15*((3*A + 7*B)*cos(d*x + c)^2 + 3*(3*A + 2*B)*cos(d*x + c) + 3*A + 2*B)*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 = 5.70016, size = 117, normalized size = 1.15 \begin{align*} \begin{cases} - \frac{A \tan ^{5}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{20 a^{3} d} + \frac{A \tan{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{4 a^{3} d} + \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} & \text{for}\: d \neq 0 \\\frac{x \left (A + B \cos{\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)*(A+B*cos(d*x+c))/(a+a*cos(d*x+c))**3,x)

[Out]

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

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Giac [A]  time = 1.21715, size = 101, normalized size = 0.99 \begin{align*} -\frac{3 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 3 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 10 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 15 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 15 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{60 \, a^{3} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*(A+B*cos(d*x+c))/(a+a*cos(d*x+c))^3,x, algorithm="giac")

[Out]

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