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

Optimal. Leaf size=164 \[ \frac {(16 A+12 B-215 C) \sin (c+d x)}{105 a^4 d (\cos (c+d x)+1)}-\frac {(8 A+6 B-55 C) \sin (c+d x)}{105 a^4 d (\cos (c+d x)+1)^2}+\frac {C x}{a^4}-\frac {(A-B+C) \sin (c+d x) \cos ^3(c+d x)}{7 d (a \cos (c+d x)+a)^4}+\frac {(4 A+3 B-10 C) \sin (c+d x) \cos ^2(c+d x)}{35 a d (a \cos (c+d x)+a)^3} \]

[Out]

C*x/a^4-1/105*(8*A+6*B-55*C)*sin(d*x+c)/a^4/d/(1+cos(d*x+c))^2+1/105*(16*A+12*B-215*C)*sin(d*x+c)/a^4/d/(1+cos
(d*x+c))-1/7*(A-B+C)*cos(d*x+c)^3*sin(d*x+c)/d/(a+a*cos(d*x+c))^4+1/35*(4*A+3*B-10*C)*cos(d*x+c)^2*sin(d*x+c)/
a/d/(a+a*cos(d*x+c))^3

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Rubi [A]  time = 0.48, antiderivative size = 164, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.146, Rules used = {3041, 2977, 2968, 3019, 2735, 2648} \[ \frac {(16 A+12 B-215 C) \sin (c+d x)}{105 a^4 d (\cos (c+d x)+1)}-\frac {(8 A+6 B-55 C) \sin (c+d x)}{105 a^4 d (\cos (c+d x)+1)^2}+\frac {C x}{a^4}-\frac {(A-B+C) \sin (c+d x) \cos ^3(c+d x)}{7 d (a \cos (c+d x)+a)^4}+\frac {(4 A+3 B-10 C) \sin (c+d x) \cos ^2(c+d x)}{35 a d (a \cos (c+d x)+a)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(C*x)/a^4 - ((8*A + 6*B - 55*C)*Sin[c + d*x])/(105*a^4*d*(1 + Cos[c + d*x])^2) + ((16*A + 12*B - 215*C)*Sin[c
+ d*x])/(105*a^4*d*(1 + Cos[c + d*x])) - ((A - B + C)*Cos[c + d*x]^3*Sin[c + d*x])/(7*d*(a + a*Cos[c + d*x])^4
) + ((4*A + 3*B - 10*C)*Cos[c + d*x]^2*Sin[c + d*x])/(35*a*d*(a + a*Cos[c + d*x])^3)

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]

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 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 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 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 3041

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)*s
in[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[((a*A - b*B + 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)) + B*(
b*c*m + a*d*(n + 1)) - C*(a*c*m + b*d*(n + 1)) + (d*(a*A - b*B)*(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, B, 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)]

Rubi steps

\begin {align*} \int \frac {\cos ^2(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^4} \, dx &=-\frac {(A-B+C) \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) (a (4 A+3 B-3 C)+7 a C \cos (c+d x))}{(a+a \cos (c+d x))^3} \, dx}{7 a^2}\\ &=-\frac {(A-B+C) \cos ^3(c+d x) \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}+\frac {(4 A+3 B-10 C) \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 (4 A+3 B-10 C)+35 a^2 C \cos (c+d x)\right )}{(a+a \cos (c+d x))^2} \, dx}{35 a^4}\\ &=-\frac {(A-B+C) \cos ^3(c+d x) \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}+\frac {(4 A+3 B-10 C) \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 (4 A+3 B-10 C) \cos (c+d x)+35 a^2 C \cos ^2(c+d x)}{(a+a \cos (c+d x))^2} \, dx}{35 a^4}\\ &=-\frac {(8 A+6 B-55 C) \sin (c+d x)}{105 a^4 d (1+\cos (c+d x))^2}-\frac {(A-B+C) \cos ^3(c+d x) \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}+\frac {(4 A+3 B-10 C) \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 (8 A+6 B-55 C)-105 a^3 C \cos (c+d x)}{a+a \cos (c+d x)} \, dx}{105 a^6}\\ &=\frac {C x}{a^4}-\frac {(8 A+6 B-55 C) \sin (c+d x)}{105 a^4 d (1+\cos (c+d x))^2}-\frac {(A-B+C) \cos ^3(c+d x) \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}+\frac {(4 A+3 B-10 C) \cos ^2(c+d x) \sin (c+d x)}{35 a d (a+a \cos (c+d x))^3}+\frac {(16 A+12 B-215 C) \int \frac {1}{a+a \cos (c+d x)} \, dx}{105 a^3}\\ &=\frac {C x}{a^4}-\frac {(8 A+6 B-55 C) \sin (c+d x)}{105 a^4 d (1+\cos (c+d x))^2}-\frac {(A-B+C) \cos ^3(c+d x) \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}+\frac {(4 A+3 B-10 C) \cos ^2(c+d x) \sin (c+d x)}{35 a d (a+a \cos (c+d x))^3}+\frac {(16 A+12 B-215 C) \sin (c+d x)}{105 d \left (a^4+a^4 \cos (c+d x)\right )}\\ \end {align*}

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Mathematica [B]  time = 0.93, size = 405, normalized size = 2.47 \[ \frac {\sec \left (\frac {c}{2}\right ) \sec ^7\left (\frac {1}{2} (c+d x)\right ) \left (-350 A \sin \left (c+\frac {d x}{2}\right )+336 A \sin \left (c+\frac {3 d x}{2}\right )-210 A \sin \left (2 c+\frac {3 d x}{2}\right )+182 A \sin \left (2 c+\frac {5 d x}{2}\right )+26 A \sin \left (3 c+\frac {7 d x}{2}\right )+560 A \sin \left (\frac {d x}{2}\right )-1260 B \sin \left (c+\frac {d x}{2}\right )+882 B \sin \left (c+\frac {3 d x}{2}\right )-630 B \sin \left (2 c+\frac {3 d x}{2}\right )+294 B \sin \left (2 c+\frac {5 d x}{2}\right )-210 B \sin \left (3 c+\frac {5 d x}{2}\right )+72 B \sin \left (3 c+\frac {7 d x}{2}\right )+1260 B \sin \left (\frac {d x}{2}\right )+8260 C \sin \left (c+\frac {d x}{2}\right )-7140 C \sin \left (c+\frac {3 d x}{2}\right )+3780 C \sin \left (2 c+\frac {3 d x}{2}\right )-2800 C \sin \left (2 c+\frac {5 d x}{2}\right )+840 C \sin \left (3 c+\frac {5 d x}{2}\right )-520 C \sin \left (3 c+\frac {7 d x}{2}\right )+3675 C d x \cos \left (c+\frac {d x}{2}\right )+2205 C d x \cos \left (c+\frac {3 d x}{2}\right )+2205 C d x \cos \left (2 c+\frac {3 d x}{2}\right )+735 C d x \cos \left (2 c+\frac {5 d x}{2}\right )+735 C d x \cos \left (3 c+\frac {5 d x}{2}\right )+105 C d x \cos \left (3 c+\frac {7 d x}{2}\right )+105 C d x \cos \left (4 c+\frac {7 d x}{2}\right )-9940 C \sin \left (\frac {d x}{2}\right )+3675 C d x \cos \left (\frac {d x}{2}\right )\right )}{13440 a^4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[c/2]*Sec[(c + d*x)/2]^7*(3675*C*d*x*Cos[(d*x)/2] + 3675*C*d*x*Cos[c + (d*x)/2] + 2205*C*d*x*Cos[c + (3*d*
x)/2] + 2205*C*d*x*Cos[2*c + (3*d*x)/2] + 735*C*d*x*Cos[2*c + (5*d*x)/2] + 735*C*d*x*Cos[3*c + (5*d*x)/2] + 10
5*C*d*x*Cos[3*c + (7*d*x)/2] + 105*C*d*x*Cos[4*c + (7*d*x)/2] + 560*A*Sin[(d*x)/2] + 1260*B*Sin[(d*x)/2] - 994
0*C*Sin[(d*x)/2] - 350*A*Sin[c + (d*x)/2] - 1260*B*Sin[c + (d*x)/2] + 8260*C*Sin[c + (d*x)/2] + 336*A*Sin[c +
(3*d*x)/2] + 882*B*Sin[c + (3*d*x)/2] - 7140*C*Sin[c + (3*d*x)/2] - 210*A*Sin[2*c + (3*d*x)/2] - 630*B*Sin[2*c
+ (3*d*x)/2] + 3780*C*Sin[2*c + (3*d*x)/2] + 182*A*Sin[2*c + (5*d*x)/2] + 294*B*Sin[2*c + (5*d*x)/2] - 2800*C
*Sin[2*c + (5*d*x)/2] - 210*B*Sin[3*c + (5*d*x)/2] + 840*C*Sin[3*c + (5*d*x)/2] + 26*A*Sin[3*c + (7*d*x)/2] +
72*B*Sin[3*c + (7*d*x)/2] - 520*C*Sin[3*c + (7*d*x)/2]))/(13440*a^4*d)

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fricas [A]  time = 0.42, size = 191, normalized size = 1.16 \[ \frac {105 \, C d x \cos \left (d x + c\right )^{4} + 420 \, C d x \cos \left (d x + c\right )^{3} + 630 \, C d x \cos \left (d x + c\right )^{2} + 420 \, C d x \cos \left (d x + c\right ) + 105 \, C d x + {\left ({\left (13 \, A + 36 \, B - 260 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (52 \, A + 39 \, B - 620 \, C\right )} \cos \left (d x + c\right )^{2} + {\left (32 \, A + 24 \, B - 535 \, C\right )} \cos \left (d x + c\right ) + 8 \, A + 6 \, B - 160 \, C\right )} \sin \left (d x + c\right )}{105 \, {\left (a^{4} d \cos \left (d x + c\right )^{4} + 4 \, a^{4} d \cos \left (d x + c\right )^{3} + 6 \, a^{4} d \cos \left (d x + c\right )^{2} + 4 \, a^{4} d \cos \left (d x + c\right ) + a^{4} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^4,x, algorithm="fricas")

[Out]

1/105*(105*C*d*x*cos(d*x + c)^4 + 420*C*d*x*cos(d*x + c)^3 + 630*C*d*x*cos(d*x + c)^2 + 420*C*d*x*cos(d*x + c)
+ 105*C*d*x + ((13*A + 36*B - 260*C)*cos(d*x + c)^3 + (52*A + 39*B - 620*C)*cos(d*x + c)^2 + (32*A + 24*B - 5
35*C)*cos(d*x + c) + 8*A + 6*B - 160*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)

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giac [A]  time = 0.57, size = 220, normalized size = 1.34 \[ \frac {\frac {840 \, {\left (d x + c\right )} C}{a^{4}} + \frac {15 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 15 \, B a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 15 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 21 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 63 \, B a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 105 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 35 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 105 \, B a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 385 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 105 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 105 \, B a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1575 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{28}}}{840 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^4,x, algorithm="giac")

[Out]

1/840*(840*(d*x + c)*C/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 + 15*C*a^24*
tan(1/2*d*x + 1/2*c)^7 - 21*A*a^24*tan(1/2*d*x + 1/2*c)^5 + 63*B*a^24*tan(1/2*d*x + 1/2*c)^5 - 105*C*a^24*tan(
1/2*d*x + 1/2*c)^5 - 35*A*a^24*tan(1/2*d*x + 1/2*c)^3 - 105*B*a^24*tan(1/2*d*x + 1/2*c)^3 + 385*C*a^24*tan(1/2
*d*x + 1/2*c)^3 + 105*A*a^24*tan(1/2*d*x + 1/2*c) + 105*B*a^24*tan(1/2*d*x + 1/2*c) - 1575*C*a^24*tan(1/2*d*x
+ 1/2*c))/a^28)/d

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maple [A]  time = 0.13, size = 255, normalized size = 1.55 \[ \frac {\left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{56 d \,a^{4}}-\frac {B \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{56 d \,a^{4}}+\frac {C \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{56 d \,a^{4}}-\frac {A \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{40 d \,a^{4}}+\frac {3 B \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{40 d \,a^{4}}-\frac {C \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d \,a^{4}}-\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{24 d \,a^{4}}-\frac {B \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d \,a^{4}}+\frac {11 C \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 d \,a^{4}}+\frac {A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d \,a^{4}}+\frac {B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d \,a^{4}}-\frac {15 C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d \,a^{4}}+\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C}{d \,a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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maxima [A]  time = 0.45, size = 286, normalized size = 1.74 \[ -\frac {5 \, C {\left (\frac {\frac {315 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {77 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {21 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {3 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}}{a^{4}} - \frac {336 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{4}}\right )} - \frac {A {\left (\frac {105 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {35 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {21 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {15 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}\right )}}{a^{4}} - \frac {3 \, B {\left (\frac {35 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {35 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {21 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {5 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}\right )}}{a^{4}}}{840 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^4,x, algorithm="maxima")

[Out]

-1/840*(5*C*((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) - 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/
(cos(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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mupad [B]  time = 1.61, size = 229, normalized size = 1.40 \[ \frac {C\,x}{a^4}+\frac {{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (\frac {A\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8}+\frac {B\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8}-\frac {15\,C\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8}\right )-{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (\frac {A\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{24}+\frac {B\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{8}-\frac {11\,C\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{24}\right )-{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {A\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{40}-\frac {3\,B\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{40}+\frac {C\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{8}\right )+\frac {A\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{56}-\frac {B\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{56}+\frac {C\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{56}}{a^4\,d\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(c + d*x)^2*(A + B*cos(c + d*x) + C*cos(c + d*x)^2))/(a + a*cos(c + d*x))^4,x)

[Out]

(C*x)/a^4 + (cos(c/2 + (d*x)/2)^6*((A*sin(c/2 + (d*x)/2))/8 + (B*sin(c/2 + (d*x)/2))/8 - (15*C*sin(c/2 + (d*x)
/2))/8) - cos(c/2 + (d*x)/2)^4*((A*sin(c/2 + (d*x)/2)^3)/24 + (B*sin(c/2 + (d*x)/2)^3)/8 - (11*C*sin(c/2 + (d*
x)/2)^3)/24) - cos(c/2 + (d*x)/2)^2*((A*sin(c/2 + (d*x)/2)^5)/40 - (3*B*sin(c/2 + (d*x)/2)^5)/40 + (C*sin(c/2
+ (d*x)/2)^5)/8) + (A*sin(c/2 + (d*x)/2)^7)/56 - (B*sin(c/2 + (d*x)/2)^7)/56 + (C*sin(c/2 + (d*x)/2)^7)/56)/(a
^4*d*cos(c/2 + (d*x)/2)^7)

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sympy [A]  time = 22.67, size = 279, normalized size = 1.70 \[ \begin {cases} \frac {A \tan ^{7}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{56 a^{4} d} - \frac {A \tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{40 a^{4} d} - \frac {A \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{24 a^{4} d} + \frac {A \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{8 a^{4} d} - \frac {B \tan ^{7}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{56 a^{4} d} + \frac {3 B \tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{40 a^{4} d} - \frac {B \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{8 a^{4} d} + \frac {B \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{8 a^{4} d} + \frac {C x}{a^{4}} + \frac {C \tan ^{7}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{56 a^{4} d} - \frac {C \tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{8 a^{4} d} + \frac {11 C \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{24 a^{4} d} - \frac {15 C \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{8 a^{4} d} & \text {for}\: d \neq 0 \\\frac {x \left (A + B \cos {\relax (c )} + C \cos ^{2}{\relax (c )}\right ) \cos ^{2}{\relax (c )}}{\left (a \cos {\relax (c )} + a\right )^{4}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**2*(A+B*cos(d*x+c)+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) - 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) + C*x/a**4 + C*tan(c/2 + d*x/2)**7/(56*a
**4*d) - C*tan(c/2 + d*x/2)**5/(8*a**4*d) + 11*C*tan(c/2 + d*x/2)**3/(24*a**4*d) - 15*C*tan(c/2 + d*x/2)/(8*a*
*4*d), Ne(d, 0)), (x*(A + B*cos(c) + C*cos(c)**2)*cos(c)**2/(a*cos(c) + a)**4, True))

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