3.68 \(\int \frac{\cos ^2(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^4} \, dx\)

Optimal. Leaf size=136 \[ \frac{(13 A+36 B) \sin (c+d x)}{105 a^4 d (\cos (c+d x)+1)}-\frac{2 (A+27 B) \sin (c+d x)}{105 a^4 d (\cos (c+d x)+1)^2}+\frac{(A-B) \sin (c+d x) \cos ^2(c+d x)}{7 d (a \cos (c+d x)+a)^4}-\frac{(A-8 B) \sin (c+d x)}{35 a d (a \cos (c+d x)+a)^3} \]

[Out]

(-2*(A + 27*B)*Sin[c + d*x])/(105*a^4*d*(1 + Cos[c + d*x])^2) + ((13*A + 36*B)*Sin[c + d*x])/(105*a^4*d*(1 + C
os[c + d*x])) + ((A - B)*Cos[c + d*x]^2*Sin[c + d*x])/(7*d*(a + a*Cos[c + d*x])^4) - ((A - 8*B)*Sin[c + d*x])/
(35*a*d*(a + a*Cos[c + d*x])^3)

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Rubi [A]  time = 0.347609, antiderivative size = 136, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.161, Rules used = {2977, 2968, 3019, 2750, 2648} \[ \frac{(13 A+36 B) \sin (c+d x)}{105 a^4 d (\cos (c+d x)+1)}-\frac{2 (A+27 B) \sin (c+d x)}{105 a^4 d (\cos (c+d x)+1)^2}+\frac{(A-B) \sin (c+d x) \cos ^2(c+d x)}{7 d (a \cos (c+d x)+a)^4}-\frac{(A-8 B) \sin (c+d x)}{35 a d (a \cos (c+d x)+a)^3} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^2*(A + B*Cos[c + d*x]))/(a + a*Cos[c + d*x])^4,x]

[Out]

(-2*(A + 27*B)*Sin[c + d*x])/(105*a^4*d*(1 + Cos[c + d*x])^2) + ((13*A + 36*B)*Sin[c + d*x])/(105*a^4*d*(1 + C
os[c + d*x])) + ((A - B)*Cos[c + d*x]^2*Sin[c + d*x])/(7*d*(a + a*Cos[c + d*x])^4) - ((A - 8*B)*Sin[c + d*x])/
(35*a*d*(a + a*Cos[c + d*x])^3)

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 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 ^2(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^4} \, dx &=\frac{(A-B) \cos ^2(c+d x) \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}+\frac{\int \frac{\cos (c+d x) (2 a (A-B)+a (A+6 B) \cos (c+d x))}{(a+a \cos (c+d x))^3} \, dx}{7 a^2}\\ &=\frac{(A-B) \cos ^2(c+d x) \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}+\frac{\int \frac{2 a (A-B) \cos (c+d x)+a (A+6 B) \cos ^2(c+d x)}{(a+a \cos (c+d x))^3} \, dx}{7 a^2}\\ &=\frac{(A-B) \cos ^2(c+d x) \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac{(A-8 B) \sin (c+d x)}{35 a d (a+a \cos (c+d x))^3}-\frac{\int \frac{-3 a^2 (A-8 B)-5 a^2 (A+6 B) \cos (c+d x)}{(a+a \cos (c+d x))^2} \, dx}{35 a^4}\\ &=-\frac{2 (A+27 B) \sin (c+d x)}{105 a^4 d (1+\cos (c+d x))^2}+\frac{(A-B) \cos ^2(c+d x) \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac{(A-8 B) \sin (c+d x)}{35 a d (a+a \cos (c+d x))^3}+\frac{(13 A+36 B) \int \frac{1}{a+a \cos (c+d x)} \, dx}{105 a^3}\\ &=-\frac{2 (A+27 B) \sin (c+d x)}{105 a^4 d (1+\cos (c+d x))^2}+\frac{(A-B) \cos ^2(c+d x) \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac{(A-8 B) \sin (c+d x)}{35 a d (a+a \cos (c+d x))^3}+\frac{(13 A+36 B) \sin (c+d x)}{105 d \left (a^4+a^4 \cos (c+d x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.43541, size = 193, normalized size = 1.42 \[ \frac{\sec \left (\frac{c}{2}\right ) \cos \left (\frac{1}{2} (c+d x)\right ) \left (-35 (5 A+18 B) \sin \left (c+\frac{d x}{2}\right )+70 (4 A+9 B) \sin \left (\frac{d x}{2}\right )+168 A \sin \left (c+\frac{3 d x}{2}\right )-105 A \sin \left (2 c+\frac{3 d x}{2}\right )+91 A \sin \left (2 c+\frac{5 d x}{2}\right )+13 A \sin \left (3 c+\frac{7 d x}{2}\right )+441 B \sin \left (c+\frac{3 d x}{2}\right )-315 B \sin \left (2 c+\frac{3 d x}{2}\right )+147 B \sin \left (2 c+\frac{5 d x}{2}\right )-105 B \sin \left (3 c+\frac{5 d x}{2}\right )+36 B \sin \left (3 c+\frac{7 d x}{2}\right )\right )}{420 a^4 d (\cos (c+d x)+1)^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^2*(A + B*Cos[c + d*x]))/(a + a*Cos[c + d*x])^4,x]

[Out]

(Cos[(c + d*x)/2]*Sec[c/2]*(70*(4*A + 9*B)*Sin[(d*x)/2] - 35*(5*A + 18*B)*Sin[c + (d*x)/2] + 168*A*Sin[c + (3*
d*x)/2] + 441*B*Sin[c + (3*d*x)/2] - 105*A*Sin[2*c + (3*d*x)/2] - 315*B*Sin[2*c + (3*d*x)/2] + 91*A*Sin[2*c +
(5*d*x)/2] + 147*B*Sin[2*c + (5*d*x)/2] - 105*B*Sin[3*c + (5*d*x)/2] + 13*A*Sin[3*c + (7*d*x)/2] + 36*B*Sin[3*
c + (7*d*x)/2]))/(420*a^4*d*(1 + Cos[c + d*x])^4)

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Maple [A]  time = 0.053, size = 90, normalized size = 0.7 \begin{align*}{\frac{1}{8\,d{a}^{4}} \left ({\frac{A-B}{7} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{7}}+{\frac{-A+3\,B}{5} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}+{\frac{-A-3\,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)^2*(A+B*cos(d*x+c))/(a+cos(d*x+c)*a)^4,x)

[Out]

1/8/d/a^4*(1/7*(A-B)*tan(1/2*d*x+1/2*c)^7+1/5*(-A+3*B)*tan(1/2*d*x+1/2*c)^5+1/3*(-A-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 = 1.00878, size = 236, normalized size = 1.74 \begin{align*} \frac{\frac{A{\left (\frac{105 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{35 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac{21 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac{15 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}\right )}}{a^{4}} + \frac{3 \, B{\left (\frac{35 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{35 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{21 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac{5 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}\right )}}{a^{4}}}{840 \, 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))^4,x, algorithm="maxima")

[Out]

1/840*(A*(105*sin(d*x + c)/(cos(d*x + c) + 1) - 35*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - 21*sin(d*x + c)^5/(co
s(d*x + c) + 1)^5 + 15*sin(d*x + c)^7/(cos(d*x + c) + 1)^7)/a^4 + 3*B*(35*sin(d*x + c)/(cos(d*x + c) + 1) - 35
*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 21*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 5*sin(d*x + c)^7/(cos(d*x + c)
+ 1)^7)/a^4)/d

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Fricas [A]  time = 1.36699, size = 308, normalized size = 2.26 \begin{align*} \frac{{\left ({\left (13 \, A + 36 \, B\right )} \cos \left (d x + c\right )^{3} + 13 \,{\left (4 \, A + 3 \, B\right )} \cos \left (d x + c\right )^{2} + 8 \,{\left (4 \, A + 3 \, B\right )} \cos \left (d x + c\right ) + 8 \, A + 6 \, B\right )} \sin \left (d x + c\right )}{105 \,{\left (a^{4} d \cos \left (d x + c\right )^{4} + 4 \, a^{4} d \cos \left (d x + c\right )^{3} + 6 \, a^{4} d \cos \left (d x + c\right )^{2} + 4 \, a^{4} d \cos \left (d x + c\right ) + a^{4} 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))^4,x, algorithm="fricas")

[Out]

1/105*((13*A + 36*B)*cos(d*x + c)^3 + 13*(4*A + 3*B)*cos(d*x + c)^2 + 8*(4*A + 3*B)*cos(d*x + c) + 8*A + 6*B)*
sin(d*x + c)/(a^4*d*cos(d*x + c)^4 + 4*a^4*d*cos(d*x + c)^3 + 6*a^4*d*cos(d*x + c)^2 + 4*a^4*d*cos(d*x + c) +
a^4*d)

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Sympy [A]  time = 16.207, size = 182, normalized size = 1.34 \begin{align*} \begin{cases} \frac{A \tan ^{7}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{56 a^{4} d} - \frac{A \tan ^{5}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{40 a^{4} d} - \frac{A \tan ^{3}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{24 a^{4} d} + \frac{A \tan{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{8 a^{4} d} - \frac{B \tan ^{7}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{56 a^{4} d} + \frac{3 B \tan ^{5}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{40 a^{4} d} - \frac{B \tan ^{3}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{8 a^{4} d} + \frac{B \tan{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{8 a^{4} 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 )^{4}} & \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))**4,x)

[Out]

Piecewise((A*tan(c/2 + d*x/2)**7/(56*a**4*d) - A*tan(c/2 + d*x/2)**5/(40*a**4*d) - A*tan(c/2 + d*x/2)**3/(24*a
**4*d) + A*tan(c/2 + d*x/2)/(8*a**4*d) - B*tan(c/2 + d*x/2)**7/(56*a**4*d) + 3*B*tan(c/2 + d*x/2)**5/(40*a**4*
d) - B*tan(c/2 + d*x/2)**3/(8*a**4*d) + B*tan(c/2 + d*x/2)/(8*a**4*d), Ne(d, 0)), (x*(A + B*cos(c))*cos(c)**2/
(a*cos(c) + a)**4, True))

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Giac [A]  time = 1.18538, size = 158, normalized size = 1.16 \begin{align*} \frac{15 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 15 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 21 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 63 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 35 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 105 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 105 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 105 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{840 \, a^{4} 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))^4,x, algorithm="giac")

[Out]

1/840*(15*A*tan(1/2*d*x + 1/2*c)^7 - 15*B*tan(1/2*d*x + 1/2*c)^7 - 21*A*tan(1/2*d*x + 1/2*c)^5 + 63*B*tan(1/2*
d*x + 1/2*c)^5 - 35*A*tan(1/2*d*x + 1/2*c)^3 - 105*B*tan(1/2*d*x + 1/2*c)^3 + 105*A*tan(1/2*d*x + 1/2*c) + 105
*B*tan(1/2*d*x + 1/2*c))/(a^4*d)