3.67 \(\int \frac{\cos ^3(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^4} \, dx\)

Optimal. Leaf size=154 \[ \frac{(12 A-215 B) \sin (c+d x)}{105 a^4 d (\cos (c+d x)+1)}-\frac{(6 A-55 B) \sin (c+d x)}{105 a^4 d (\cos (c+d x)+1)^2}+\frac{B x}{a^4}+\frac{(A-B) \sin (c+d x) \cos ^3(c+d x)}{7 d (a \cos (c+d x)+a)^4}+\frac{(3 A-10 B) \sin (c+d x) \cos ^2(c+d x)}{35 a d (a \cos (c+d x)+a)^3} \]

[Out]

(B*x)/a^4 - ((6*A - 55*B)*Sin[c + d*x])/(105*a^4*d*(1 + Cos[c + d*x])^2) + ((12*A - 215*B)*Sin[c + d*x])/(105*
a^4*d*(1 + Cos[c + d*x])) + ((A - B)*Cos[c + d*x]^3*Sin[c + d*x])/(7*d*(a + a*Cos[c + d*x])^4) + ((3*A - 10*B)
*Cos[c + d*x]^2*Sin[c + d*x])/(35*a*d*(a + a*Cos[c + d*x])^3)

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Rubi [A]  time = 0.498107, antiderivative size = 154, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.161, Rules used = {2977, 2968, 3019, 2735, 2648} \[ \frac{(12 A-215 B) \sin (c+d x)}{105 a^4 d (\cos (c+d x)+1)}-\frac{(6 A-55 B) \sin (c+d x)}{105 a^4 d (\cos (c+d x)+1)^2}+\frac{B x}{a^4}+\frac{(A-B) \sin (c+d x) \cos ^3(c+d x)}{7 d (a \cos (c+d x)+a)^4}+\frac{(3 A-10 B) \sin (c+d x) \cos ^2(c+d x)}{35 a d (a \cos (c+d x)+a)^3} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^3*(A + B*Cos[c + d*x]))/(a + a*Cos[c + d*x])^4,x]

[Out]

(B*x)/a^4 - ((6*A - 55*B)*Sin[c + d*x])/(105*a^4*d*(1 + Cos[c + d*x])^2) + ((12*A - 215*B)*Sin[c + d*x])/(105*
a^4*d*(1 + Cos[c + d*x])) + ((A - B)*Cos[c + d*x]^3*Sin[c + d*x])/(7*d*(a + a*Cos[c + d*x])^4) + ((3*A - 10*B)
*Cos[c + d*x]^2*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 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 ^3(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^4} \, dx &=\frac{(A-B) \cos ^3(c+d x) \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}+\frac{\int \frac{\cos ^2(c+d x) (3 a (A-B)+7 a B \cos (c+d x))}{(a+a \cos (c+d x))^3} \, dx}{7 a^2}\\ &=\frac{(A-B) \cos ^3(c+d x) \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}+\frac{(3 A-10 B) \cos ^2(c+d x) \sin (c+d x)}{35 a d (a+a \cos (c+d x))^3}+\frac{\int \frac{\cos (c+d x) \left (2 a^2 (3 A-10 B)+35 a^2 B \cos (c+d x)\right )}{(a+a \cos (c+d x))^2} \, dx}{35 a^4}\\ &=\frac{(A-B) \cos ^3(c+d x) \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}+\frac{(3 A-10 B) \cos ^2(c+d x) \sin (c+d x)}{35 a d (a+a \cos (c+d x))^3}+\frac{\int \frac{2 a^2 (3 A-10 B) \cos (c+d x)+35 a^2 B \cos ^2(c+d x)}{(a+a \cos (c+d x))^2} \, dx}{35 a^4}\\ &=-\frac{(6 A-55 B) \sin (c+d x)}{105 a^4 d (1+\cos (c+d x))^2}+\frac{(A-B) \cos ^3(c+d x) \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}+\frac{(3 A-10 B) \cos ^2(c+d x) \sin (c+d x)}{35 a d (a+a \cos (c+d x))^3}-\frac{\int \frac{-2 a^3 (6 A-55 B)-105 a^3 B \cos (c+d x)}{a+a \cos (c+d x)} \, dx}{105 a^6}\\ &=\frac{B x}{a^4}-\frac{(6 A-55 B) \sin (c+d x)}{105 a^4 d (1+\cos (c+d x))^2}+\frac{(A-B) \cos ^3(c+d x) \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}+\frac{(3 A-10 B) \cos ^2(c+d x) \sin (c+d x)}{35 a d (a+a \cos (c+d x))^3}+\frac{(12 A-215 B) \int \frac{1}{a+a \cos (c+d x)} \, dx}{105 a^3}\\ &=\frac{B x}{a^4}-\frac{(6 A-55 B) \sin (c+d x)}{105 a^4 d (1+\cos (c+d x))^2}+\frac{(A-B) \cos ^3(c+d x) \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}+\frac{(3 A-10 B) \cos ^2(c+d x) \sin (c+d x)}{35 a d (a+a \cos (c+d x))^3}+\frac{(12 A-215 B) \sin (c+d x)}{105 d \left (a^4+a^4 \cos (c+d x)\right )}\\ \end{align*}

Mathematica [B]  time = 0.741084, size = 329, normalized size = 2.14 \[ \frac{\sec \left (\frac{c}{2}\right ) \sec ^7\left (\frac{1}{2} (c+d x)\right ) \left (-1260 A \sin \left (c+\frac{d x}{2}\right )+882 A \sin \left (c+\frac{3 d x}{2}\right )-630 A \sin \left (2 c+\frac{3 d x}{2}\right )+294 A \sin \left (2 c+\frac{5 d x}{2}\right )-210 A \sin \left (3 c+\frac{5 d x}{2}\right )+72 A \sin \left (3 c+\frac{7 d x}{2}\right )+1260 A \sin \left (\frac{d x}{2}\right )+8260 B \sin \left (c+\frac{d x}{2}\right )-7140 B \sin \left (c+\frac{3 d x}{2}\right )+3780 B \sin \left (2 c+\frac{3 d x}{2}\right )-2800 B \sin \left (2 c+\frac{5 d x}{2}\right )+840 B \sin \left (3 c+\frac{5 d x}{2}\right )-520 B \sin \left (3 c+\frac{7 d x}{2}\right )+3675 B d x \cos \left (c+\frac{d x}{2}\right )+2205 B d x \cos \left (c+\frac{3 d x}{2}\right )+2205 B d x \cos \left (2 c+\frac{3 d x}{2}\right )+735 B d x \cos \left (2 c+\frac{5 d x}{2}\right )+735 B d x \cos \left (3 c+\frac{5 d x}{2}\right )+105 B d x \cos \left (3 c+\frac{7 d x}{2}\right )+105 B d x \cos \left (4 c+\frac{7 d x}{2}\right )-9940 B \sin \left (\frac{d x}{2}\right )+3675 B d x \cos \left (\frac{d x}{2}\right )\right )}{13440 a^4 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^3*(A + B*Cos[c + d*x]))/(a + a*Cos[c + d*x])^4,x]

[Out]

(Sec[c/2]*Sec[(c + d*x)/2]^7*(3675*B*d*x*Cos[(d*x)/2] + 3675*B*d*x*Cos[c + (d*x)/2] + 2205*B*d*x*Cos[c + (3*d*
x)/2] + 2205*B*d*x*Cos[2*c + (3*d*x)/2] + 735*B*d*x*Cos[2*c + (5*d*x)/2] + 735*B*d*x*Cos[3*c + (5*d*x)/2] + 10
5*B*d*x*Cos[3*c + (7*d*x)/2] + 105*B*d*x*Cos[4*c + (7*d*x)/2] + 1260*A*Sin[(d*x)/2] - 9940*B*Sin[(d*x)/2] - 12
60*A*Sin[c + (d*x)/2] + 8260*B*Sin[c + (d*x)/2] + 882*A*Sin[c + (3*d*x)/2] - 7140*B*Sin[c + (3*d*x)/2] - 630*A
*Sin[2*c + (3*d*x)/2] + 3780*B*Sin[2*c + (3*d*x)/2] + 294*A*Sin[2*c + (5*d*x)/2] - 2800*B*Sin[2*c + (5*d*x)/2]
- 210*A*Sin[3*c + (5*d*x)/2] + 840*B*Sin[3*c + (5*d*x)/2] + 72*A*Sin[3*c + (7*d*x)/2] - 520*B*Sin[3*c + (7*d*
x)/2]))/(13440*a^4*d)

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Maple [A]  time = 0.059, size = 177, normalized size = 1.2 \begin{align*} -{\frac{A}{56\,d{a}^{4}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{7}}+{\frac{B}{56\,d{a}^{4}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{7}}+{\frac{3\,A}{40\,d{a}^{4}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}-{\frac{B}{8\,d{a}^{4}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}-{\frac{A}{8\,d{a}^{4}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}+{\frac{11\,B}{24\,d{a}^{4}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}+{\frac{A}{8\,d{a}^{4}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-{\frac{15\,B}{8\,d{a}^{4}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+2\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) B}{d{a}^{4}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^3*(A+B*cos(d*x+c))/(a+cos(d*x+c)*a)^4,x)

[Out]

-1/56/d/a^4*tan(1/2*d*x+1/2*c)^7*A+1/56/d/a^4*B*tan(1/2*d*x+1/2*c)^7+3/40/d/a^4*A*tan(1/2*d*x+1/2*c)^5-1/8/d/a
^4*B*tan(1/2*d*x+1/2*c)^5-1/8/d/a^4*tan(1/2*d*x+1/2*c)^3*A+11/24/d/a^4*B*tan(1/2*d*x+1/2*c)^3+1/8/d/a^4*A*tan(
1/2*d*x+1/2*c)-15/8/d/a^4*B*tan(1/2*d*x+1/2*c)+2/d/a^4*arctan(tan(1/2*d*x+1/2*c))*B

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Maxima [A]  time = 1.47903, size = 271, normalized size = 1.76 \begin{align*} -\frac{5 \, B{\left (\frac{\frac{315 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{77 \, \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{3 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}}{a^{4}} - \frac{336 \, \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{4}}\right )} - \frac{3 \, A{\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)^3*(A+B*cos(d*x+c))/(a+a*cos(d*x+c))^4,x, algorithm="maxima")

[Out]

-1/840*(5*B*((315*sin(d*x + c)/(cos(d*x + c) + 1) - 77*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 21*sin(d*x + c)^5
/(cos(d*x + c) + 1)^5 - 3*sin(d*x + c)^7/(cos(d*x + c) + 1)^7)/a^4 - 336*arctan(sin(d*x + c)/(cos(d*x + c) + 1
))/a^4) - 3*A*(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 = 1.3857, size = 475, normalized size = 3.08 \begin{align*} \frac{105 \, B d x \cos \left (d x + c\right )^{4} + 420 \, B d x \cos \left (d x + c\right )^{3} + 630 \, B d x \cos \left (d x + c\right )^{2} + 420 \, B d x \cos \left (d x + c\right ) + 105 \, B d x +{\left (4 \,{\left (9 \, A - 65 \, B\right )} \cos \left (d x + c\right )^{3} +{\left (39 \, A - 620 \, B\right )} \cos \left (d x + c\right )^{2} +{\left (24 \, A - 535 \, B\right )} \cos \left (d x + c\right ) + 6 \, A - 160 \, 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)^3*(A+B*cos(d*x+c))/(a+a*cos(d*x+c))^4,x, algorithm="fricas")

[Out]

1/105*(105*B*d*x*cos(d*x + c)^4 + 420*B*d*x*cos(d*x + c)^3 + 630*B*d*x*cos(d*x + c)^2 + 420*B*d*x*cos(d*x + c)
+ 105*B*d*x + (4*(9*A - 65*B)*cos(d*x + c)^3 + (39*A - 620*B)*cos(d*x + c)^2 + (24*A - 535*B)*cos(d*x + c) +
6*A - 160*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*c
os(d*x + c) + a^4*d)

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Sympy [A]  time = 24.3107, size = 192, normalized size = 1.25 \begin{align*} \begin{cases} - \frac{A \tan ^{7}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{56 a^{4} d} + \frac{3 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 )}}{8 a^{4} d} + \frac{A \tan{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{8 a^{4} d} + \frac{B x}{a^{4}} + \frac{B \tan ^{7}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{56 a^{4} d} - \frac{B \tan ^{5}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{8 a^{4} d} + \frac{11 B \tan ^{3}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{24 a^{4} d} - \frac{15 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 ^{3}{\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)**3*(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) + 3*A*tan(c/2 + d*x/2)**5/(40*a**4*d) - A*tan(c/2 + d*x/2)**3/(8
*a**4*d) + A*tan(c/2 + d*x/2)/(8*a**4*d) + B*x/a**4 + B*tan(c/2 + d*x/2)**7/(56*a**4*d) - B*tan(c/2 + d*x/2)**
5/(8*a**4*d) + 11*B*tan(c/2 + d*x/2)**3/(24*a**4*d) - 15*B*tan(c/2 + d*x/2)/(8*a**4*d), Ne(d, 0)), (x*(A + B*c
os(c))*cos(c)**3/(a*cos(c) + a)**4, True))

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Giac [A]  time = 1.21546, size = 209, normalized size = 1.36 \begin{align*} \frac{\frac{840 \,{\left (d x + c\right )} B}{a^{4}} - \frac{15 \, A a^{24} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 15 \, B a^{24} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 63 \, A a^{24} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 105 \, B a^{24} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 105 \, A a^{24} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 385 \, B a^{24} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 105 \, A a^{24} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1575 \, B a^{24} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{28}}}{840 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(A+B*cos(d*x+c))/(a+a*cos(d*x+c))^4,x, algorithm="giac")

[Out]

1/840*(840*(d*x + c)*B/a^4 - (15*A*a^24*tan(1/2*d*x + 1/2*c)^7 - 15*B*a^24*tan(1/2*d*x + 1/2*c)^7 - 63*A*a^24*
tan(1/2*d*x + 1/2*c)^5 + 105*B*a^24*tan(1/2*d*x + 1/2*c)^5 + 105*A*a^24*tan(1/2*d*x + 1/2*c)^3 - 385*B*a^24*ta
n(1/2*d*x + 1/2*c)^3 - 105*A*a^24*tan(1/2*d*x + 1/2*c) + 1575*B*a^24*tan(1/2*d*x + 1/2*c))/a^28)/d