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

Optimal. Leaf size=138 \[ \frac{4 (2 A+9 C) \sin (c+d x)}{105 a^4 d (\cos (c+d x)+1)}+\frac{(23 A-54 C) \sin (c+d x)}{105 a^4 d (\cos (c+d x)+1)^2}-\frac{(A+C) \sin (c+d x) \cos ^2(c+d x)}{7 d (a \cos (c+d x)+a)^4}-\frac{2 (3 A-4 C) \sin (c+d x)}{35 a d (a \cos (c+d x)+a)^3} \]

[Out]

((23*A - 54*C)*Sin[c + d*x])/(105*a^4*d*(1 + Cos[c + d*x])^2) + (4*(2*A + 9*C)*Sin[c + d*x])/(105*a^4*d*(1 + C
os[c + d*x])) - ((A + C)*Cos[c + d*x]^2*Sin[c + d*x])/(7*d*(a + a*Cos[c + d*x])^4) - (2*(3*A - 4*C)*Sin[c + d*
x])/(35*a*d*(a + a*Cos[c + d*x])^3)

________________________________________________________________________________________

Rubi [A]  time = 0.349296, antiderivative size = 138, 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 = {3042, 2968, 3019, 2750, 2648} \[ \frac{4 (2 A+9 C) \sin (c+d x)}{105 a^4 d (\cos (c+d x)+1)}+\frac{(23 A-54 C) \sin (c+d x)}{105 a^4 d (\cos (c+d x)+1)^2}-\frac{(A+C) \sin (c+d x) \cos ^2(c+d x)}{7 d (a \cos (c+d x)+a)^4}-\frac{2 (3 A-4 C) \sin (c+d x)}{35 a d (a \cos (c+d x)+a)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((23*A - 54*C)*Sin[c + d*x])/(105*a^4*d*(1 + Cos[c + d*x])^2) + (4*(2*A + 9*C)*Sin[c + d*x])/(105*a^4*d*(1 + C
os[c + d*x])) - ((A + C)*Cos[c + d*x]^2*Sin[c + d*x])/(7*d*(a + a*Cos[c + d*x])^4) - (2*(3*A - 4*C)*Sin[c + d*
x])/(35*a*d*(a + a*Cos[c + d*x])^3)

Rule 3042

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

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) \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^4} \, dx &=-\frac{(A+C) \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) (a (5 A-2 C)-a (A-6 C) \cos (c+d x))}{(a+a \cos (c+d x))^3} \, dx}{7 a^2}\\ &=-\frac{(A+C) \cos ^2(c+d x) \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}+\frac{\int \frac{a (5 A-2 C) \cos (c+d x)-a (A-6 C) \cos ^2(c+d x)}{(a+a \cos (c+d x))^3} \, dx}{7 a^2}\\ &=-\frac{(A+C) \cos ^2(c+d x) \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac{2 (3 A-4 C) \sin (c+d x)}{35 a d (a+a \cos (c+d x))^3}-\frac{\int \frac{-6 a^2 (3 A-4 C)+5 a^2 (A-6 C) \cos (c+d x)}{(a+a \cos (c+d x))^2} \, dx}{35 a^4}\\ &=\frac{(23 A-54 C) \sin (c+d x)}{105 a^4 d (1+\cos (c+d x))^2}-\frac{(A+C) \cos ^2(c+d x) \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac{2 (3 A-4 C) \sin (c+d x)}{35 a d (a+a \cos (c+d x))^3}+\frac{(4 (2 A+9 C)) \int \frac{1}{a+a \cos (c+d x)} \, dx}{105 a^3}\\ &=\frac{(23 A-54 C) \sin (c+d x)}{105 a^4 d (1+\cos (c+d x))^2}-\frac{(A+C) \cos ^2(c+d x) \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac{2 (3 A-4 C) \sin (c+d x)}{35 a d (a+a \cos (c+d x))^3}+\frac{4 (2 A+9 C) \sin (c+d x)}{105 d \left (a^4+a^4 \cos (c+d x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.391696, size = 179, normalized size = 1.3 \[ \frac{\sec \left (\frac{c}{2}\right ) \cos \left (\frac{1}{2} (c+d x)\right ) \left (-70 (2 A+9 C) \sin \left (c+\frac{d x}{2}\right )+168 A \sin \left (c+\frac{3 d x}{2}\right )+56 A \sin \left (2 c+\frac{5 d x}{2}\right )+8 A \sin \left (3 c+\frac{7 d x}{2}\right )+70 (2 A+9 C) \sin \left (\frac{d x}{2}\right )+441 C \sin \left (c+\frac{3 d x}{2}\right )-315 C \sin \left (2 c+\frac{3 d x}{2}\right )+147 C \sin \left (2 c+\frac{5 d x}{2}\right )-105 C \sin \left (3 c+\frac{5 d x}{2}\right )+36 C \sin \left (3 c+\frac{7 d x}{2}\right )\right )}{420 a^4 d (\cos (c+d x)+1)^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[(c + d*x)/2]*Sec[c/2]*(70*(2*A + 9*C)*Sin[(d*x)/2] - 70*(2*A + 9*C)*Sin[c + (d*x)/2] + 168*A*Sin[c + (3*d
*x)/2] + 441*C*Sin[c + (3*d*x)/2] - 315*C*Sin[2*c + (3*d*x)/2] + 56*A*Sin[2*c + (5*d*x)/2] + 147*C*Sin[2*c + (
5*d*x)/2] - 105*C*Sin[3*c + (5*d*x)/2] + 8*A*Sin[3*c + (7*d*x)/2] + 36*C*Sin[3*c + (7*d*x)/2]))/(420*a^4*d*(1
+ Cos[c + d*x])^4)

________________________________________________________________________________________

Maple [A]  time = 0.026, size = 90, normalized size = 0.7 \begin{align*}{\frac{1}{8\,d{a}^{4}} \left ({\frac{-A-C}{7} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{7}}+{\frac{-A+3\,C}{5} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}+{\frac{A-3\,C}{3} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}+A\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(cos(d*x+c)*(A+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^4,x)

[Out]

1/8/d/a^4*(1/7*(-A-C)*tan(1/2*d*x+1/2*c)^7+1/5*(-A+3*C)*tan(1/2*d*x+1/2*c)^5+1/3*(A-3*C)*tan(1/2*d*x+1/2*c)^3+
A*tan(1/2*d*x+1/2*c)+C*tan(1/2*d*x+1/2*c))

________________________________________________________________________________________

Maxima [A]  time = 1.0281, size = 236, normalized size = 1.71 \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 \, C{\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)*(A+C*cos(d*x+c)^2)/(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*C*(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

________________________________________________________________________________________

Fricas [A]  time = 1.33755, size = 309, normalized size = 2.24 \begin{align*} \frac{{\left (4 \,{\left (2 \, A + 9 \, C\right )} \cos \left (d x + c\right )^{3} +{\left (32 \, A + 39 \, C\right )} \cos \left (d x + c\right )^{2} + 4 \,{\left (13 \, A + 6 \, C\right )} \cos \left (d x + c\right ) + 13 \, A + 6 \, C\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)*(A+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^4,x, algorithm="fricas")

[Out]

1/105*(4*(2*A + 9*C)*cos(d*x + c)^3 + (32*A + 39*C)*cos(d*x + c)^2 + 4*(13*A + 6*C)*cos(d*x + c) + 13*A + 6*C)
*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)

________________________________________________________________________________________

Sympy [A]  time = 16.4021, size = 182, normalized size = 1.32 \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{C \tan ^{7}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{56 a^{4} d} + \frac{3 C \tan ^{5}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{40 a^{4} d} - \frac{C \tan ^{3}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{8 a^{4} d} + \frac{C \tan{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{8 a^{4} d} & \text{for}\: d \neq 0 \\\frac{x \left (A + C \cos ^{2}{\left (c \right )}\right ) \cos{\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)*(A+C*cos(d*x+c)**2)/(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) - C*tan(c/2 + d*x/2)**7/(56*a**4*d) + 3*C*tan(c/2 + d*x/2)**5/(40*a**4
*d) - C*tan(c/2 + d*x/2)**3/(8*a**4*d) + C*tan(c/2 + d*x/2)/(8*a**4*d), Ne(d, 0)), (x*(A + C*cos(c)**2)*cos(c)
/(a*cos(c) + a)**4, True))

________________________________________________________________________________________

Giac [A]  time = 1.16239, size = 158, normalized size = 1.14 \begin{align*} -\frac{15 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 15 \, C \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 \, C \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 \, C \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 \, C \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)*(A+C*cos(d*x+c)^2)/(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*C*tan(1/2*d*x + 1/2*c)^7 + 21*A*tan(1/2*d*x + 1/2*c)^5 - 63*C*tan(1/2
*d*x + 1/2*c)^5 - 35*A*tan(1/2*d*x + 1/2*c)^3 + 105*C*tan(1/2*d*x + 1/2*c)^3 - 105*A*tan(1/2*d*x + 1/2*c) - 10
5*C*tan(1/2*d*x + 1/2*c))/(a^4*d)