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

Optimal. Leaf size=109 \[ \frac{(2 A+3 B+7 C) \sin (c+d x)}{15 d \left (a^3 \cos (c+d x)+a^3\right )}+\frac{(2 A+3 B-8 C) \sin (c+d x)}{15 a d (a \cos (c+d x)+a)^2}+\frac{(A-B+C) \sin (c+d x)}{5 d (a \cos (c+d x)+a)^3} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.132216, antiderivative size = 109, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.091, Rules used = {3019, 2750, 2648} \[ \frac{(2 A+3 B+7 C) \sin (c+d x)}{15 d \left (a^3 \cos (c+d x)+a^3\right )}+\frac{(2 A+3 B-8 C) \sin (c+d x)}{15 a d (a \cos (c+d x)+a)^2}+\frac{(A-B+C) \sin (c+d x)}{5 d (a \cos (c+d x)+a)^3} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)/(a + a*Cos[c + d*x])^3,x]

[Out]

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

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

Mathematica [A]  time = 0.401919, size = 164, normalized size = 1.5 \[ \frac{\sec \left (\frac{c}{2}\right ) \cos \left (\frac{1}{2} (c+d x)\right ) \left (5 (4 A+3 B+8 C) \sin \left (\frac{d x}{2}\right )+10 A \sin \left (c+\frac{3 d x}{2}\right )+2 A \sin \left (2 c+\frac{5 d x}{2}\right )-15 (B+2 C) \sin \left (c+\frac{d x}{2}\right )+15 B \sin \left (c+\frac{3 d x}{2}\right )+3 B \sin \left (2 c+\frac{5 d x}{2}\right )+20 C \sin \left (c+\frac{3 d x}{2}\right )-15 C \sin \left (2 c+\frac{3 d x}{2}\right )+7 C \sin \left (2 c+\frac{5 d x}{2}\right )\right )}{30 a^3 d (\cos (c+d x)+1)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)/(a + a*Cos[c + d*x])^3,x]

[Out]

(Cos[(c + d*x)/2]*Sec[c/2]*(5*(4*A + 3*B + 8*C)*Sin[(d*x)/2] - 15*(B + 2*C)*Sin[c + (d*x)/2] + 10*A*Sin[c + (3
*d*x)/2] + 15*B*Sin[c + (3*d*x)/2] + 20*C*Sin[c + (3*d*x)/2] - 15*C*Sin[2*c + (3*d*x)/2] + 2*A*Sin[2*c + (5*d*
x)/2] + 3*B*Sin[2*c + (5*d*x)/2] + 7*C*Sin[2*c + (5*d*x)/2]))/(30*a^3*d*(1 + Cos[c + d*x])^3)

________________________________________________________________________________________

Maple [A]  time = 0.024, size = 113, normalized size = 1. \begin{align*}{\frac{1}{4\,d{a}^{3}} \left ({\frac{A}{5} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}-{\frac{B}{5} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}+{\frac{C}{5} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}+{\frac{2\,A}{3} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}-{\frac{2\,C}{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 ) +C\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^3,x)

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 1.0728, size = 242, normalized size = 2.22 \begin{align*} \frac{\frac{A{\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{C{\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 \, B{\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((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^3,x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 1.77728, size = 251, normalized size = 2.3 \begin{align*} \frac{{\left ({\left (2 \, A + 3 \, B + 7 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \,{\left (2 \, A + 3 \, B + 2 \, C\right )} \cos \left (d x + c\right ) + 7 \, A + 3 \, B + 2 \, 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((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^3,x, algorithm="fricas")

[Out]

1/15*((2*A + 3*B + 7*C)*cos(d*x + c)^2 + 3*(2*A + 3*B + 2*C)*cos(d*x + c) + 7*A + 3*B + 2*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)

________________________________________________________________________________________

Sympy [A]  time = 9.51399, size = 180, normalized size = 1.65 \begin{align*} \begin{cases} \frac{A \tan ^{5}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{20 a^{3} d} + \frac{A \tan ^{3}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{6 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{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{4 a^{3} d} + \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 )}}{6 a^{3} d} + \frac{C \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 )} + C \cos ^{2}{\left (c \right )}\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((A+B*cos(d*x+c)+C*cos(d*x+c)**2)/(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)**3/(6*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)/(4*a**3*d) + C*tan(c/2 + d*x/2)**5/(20*a**3*d) - C*
tan(c/2 + d*x/2)**3/(6*a**3*d) + C*tan(c/2 + d*x/2)/(4*a**3*d), Ne(d, 0)), (x*(A + B*cos(c) + C*cos(c)**2)/(a*
cos(c) + a)**3, True))

________________________________________________________________________________________

Giac [A]  time = 1.2074, size = 155, normalized size = 1.42 \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} + 3 \, C \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 10 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 10 \, C \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 ) + 15 \, C \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((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(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 + 3*C*tan(1/2*d*x + 1/2*c)^5 + 10*A*tan(1/2*d*x
+ 1/2*c)^3 - 10*C*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) + 15*C*tan(1/
2*d*x + 1/2*c))/(a^3*d)