3.365 \(\int \frac {\cos ^3(c+d x) (A+B \cos (c+d x)+C \cos ^2(c+d x))}{(a+a \cos (c+d x))^4} \, dx\)

Optimal. Leaf size=195 \[ \frac {(6 A-55 B+244 C) \sin (c+d x)}{105 a^4 d}+\frac {(3 A+25 B-88 C) \sin (c+d x) \cos ^2(c+d x)}{105 a^4 d (\cos (c+d x)+1)^2}-\frac {(B-4 C) \sin (c+d x)}{a^4 d (\cos (c+d x)+1)}+\frac {x (B-4 C)}{a^4}-\frac {(A-B+C) \sin (c+d x) \cos ^4(c+d x)}{7 d (a \cos (c+d x)+a)^4}+\frac {(2 A+5 B-12 C) \sin (c+d x) \cos ^3(c+d x)}{35 a d (a \cos (c+d x)+a)^3} \]

[Out]

(B-4*C)*x/a^4+1/105*(6*A-55*B+244*C)*sin(d*x+c)/a^4/d+1/105*(3*A+25*B-88*C)*cos(d*x+c)^2*sin(d*x+c)/a^4/d/(1+c
os(d*x+c))^2-(B-4*C)*sin(d*x+c)/a^4/d/(1+cos(d*x+c))-1/7*(A-B+C)*cos(d*x+c)^4*sin(d*x+c)/d/(a+a*cos(d*x+c))^4+
1/35*(2*A+5*B-12*C)*cos(d*x+c)^3*sin(d*x+c)/a/d/(a+a*cos(d*x+c))^3

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Rubi [A]  time = 0.65, antiderivative size = 195, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {3041, 2977, 2968, 3023, 12, 2735, 2648} \[ \frac {(6 A-55 B+244 C) \sin (c+d x)}{105 a^4 d}+\frac {(3 A+25 B-88 C) \sin (c+d x) \cos ^2(c+d x)}{105 a^4 d (\cos (c+d x)+1)^2}-\frac {(B-4 C) \sin (c+d x)}{a^4 d (\cos (c+d x)+1)}+\frac {x (B-4 C)}{a^4}-\frac {(A-B+C) \sin (c+d x) \cos ^4(c+d x)}{7 d (a \cos (c+d x)+a)^4}+\frac {(2 A+5 B-12 C) \sin (c+d x) \cos ^3(c+d x)}{35 a d (a \cos (c+d x)+a)^3} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^3*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2))/(a + a*Cos[c + d*x])^4,x]

[Out]

((B - 4*C)*x)/a^4 + ((6*A - 55*B + 244*C)*Sin[c + d*x])/(105*a^4*d) + ((3*A + 25*B - 88*C)*Cos[c + d*x]^2*Sin[
c + d*x])/(105*a^4*d*(1 + Cos[c + d*x])^2) - ((B - 4*C)*Sin[c + d*x])/(a^4*d*(1 + Cos[c + d*x])) - ((A - B + C
)*Cos[c + d*x]^4*Sin[c + d*x])/(7*d*(a + a*Cos[c + d*x])^4) + ((2*A + 5*B - 12*C)*Cos[c + d*x]^3*Sin[c + d*x])
/(35*a*d*(a + a*Cos[c + d*x])^3)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 3023

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*
(m + 2)), Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x]
, x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

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 ^3(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 ^4(c+d x) \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}+\frac {\int \frac {\cos ^3(c+d x) (a (3 A+4 B-4 C)+a (A-B+8 C) \cos (c+d x))}{(a+a \cos (c+d x))^3} \, dx}{7 a^2}\\ &=-\frac {(A-B+C) \cos ^4(c+d x) \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}+\frac {(2 A+5 B-12 C) \cos ^3(c+d x) \sin (c+d x)}{35 a d (a+a \cos (c+d x))^3}+\frac {\int \frac {\cos ^2(c+d x) \left (3 a^2 (2 A+5 B-12 C)+a^2 (3 A-10 B+52 C) \cos (c+d x)\right )}{(a+a \cos (c+d x))^2} \, dx}{35 a^4}\\ &=\frac {(3 A+25 B-88 C) \cos ^2(c+d x) \sin (c+d x)}{105 a^4 d (1+\cos (c+d x))^2}-\frac {(A-B+C) \cos ^4(c+d x) \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}+\frac {(2 A+5 B-12 C) \cos ^3(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^3 (3 A+25 B-88 C)+a^3 (6 A-55 B+244 C) \cos (c+d x)\right )}{a+a \cos (c+d x)} \, dx}{105 a^6}\\ &=\frac {(3 A+25 B-88 C) \cos ^2(c+d x) \sin (c+d x)}{105 a^4 d (1+\cos (c+d x))^2}-\frac {(A-B+C) \cos ^4(c+d x) \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}+\frac {(2 A+5 B-12 C) \cos ^3(c+d x) \sin (c+d x)}{35 a d (a+a \cos (c+d x))^3}+\frac {\int \frac {2 a^3 (3 A+25 B-88 C) \cos (c+d x)+a^3 (6 A-55 B+244 C) \cos ^2(c+d x)}{a+a \cos (c+d x)} \, dx}{105 a^6}\\ &=\frac {(6 A-55 B+244 C) \sin (c+d x)}{105 a^4 d}+\frac {(3 A+25 B-88 C) \cos ^2(c+d x) \sin (c+d x)}{105 a^4 d (1+\cos (c+d x))^2}-\frac {(A-B+C) \cos ^4(c+d x) \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}+\frac {(2 A+5 B-12 C) \cos ^3(c+d x) \sin (c+d x)}{35 a d (a+a \cos (c+d x))^3}+\frac {\int \frac {105 a^4 (B-4 C) \cos (c+d x)}{a+a \cos (c+d x)} \, dx}{105 a^7}\\ &=\frac {(6 A-55 B+244 C) \sin (c+d x)}{105 a^4 d}+\frac {(3 A+25 B-88 C) \cos ^2(c+d x) \sin (c+d x)}{105 a^4 d (1+\cos (c+d x))^2}-\frac {(A-B+C) \cos ^4(c+d x) \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}+\frac {(2 A+5 B-12 C) \cos ^3(c+d x) \sin (c+d x)}{35 a d (a+a \cos (c+d x))^3}+\frac {(B-4 C) \int \frac {\cos (c+d x)}{a+a \cos (c+d x)} \, dx}{a^3}\\ &=\frac {(B-4 C) x}{a^4}+\frac {(6 A-55 B+244 C) \sin (c+d x)}{105 a^4 d}+\frac {(3 A+25 B-88 C) \cos ^2(c+d x) \sin (c+d x)}{105 a^4 d (1+\cos (c+d x))^2}-\frac {(A-B+C) \cos ^4(c+d x) \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}+\frac {(2 A+5 B-12 C) \cos ^3(c+d x) \sin (c+d x)}{35 a d (a+a \cos (c+d x))^3}-\frac {(B-4 C) \int \frac {1}{a+a \cos (c+d x)} \, dx}{a^3}\\ &=\frac {(B-4 C) x}{a^4}+\frac {(6 A-55 B+244 C) \sin (c+d x)}{105 a^4 d}+\frac {(3 A+25 B-88 C) \cos ^2(c+d x) \sin (c+d x)}{105 a^4 d (1+\cos (c+d x))^2}-\frac {(A-B+C) \cos ^4(c+d x) \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}+\frac {(2 A+5 B-12 C) \cos ^3(c+d x) \sin (c+d x)}{35 a d (a+a \cos (c+d x))^3}-\frac {(B-4 C) \sin (c+d x)}{d \left (a^4+a^4 \cos (c+d x)\right )}\\ \end {align*}

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Mathematica [B]  time = 1.14, size = 571, normalized size = 2.93 \[ \frac {\sec \left (\frac {c}{2}\right ) \cos \left (\frac {1}{2} (c+d x)\right ) \left (-2520 A \sin \left (c+\frac {d x}{2}\right )+1764 A \sin \left (c+\frac {3 d x}{2}\right )-1260 A \sin \left (2 c+\frac {3 d x}{2}\right )+588 A \sin \left (2 c+\frac {5 d x}{2}\right )-420 A \sin \left (3 c+\frac {5 d x}{2}\right )+144 A \sin \left (3 c+\frac {7 d x}{2}\right )+2520 A \sin \left (\frac {d x}{2}\right )+7350 d x (B-4 C) \cos \left (c+\frac {d x}{2}\right )+16520 B \sin \left (c+\frac {d x}{2}\right )-14280 B \sin \left (c+\frac {3 d x}{2}\right )+7560 B \sin \left (2 c+\frac {3 d x}{2}\right )-5600 B \sin \left (2 c+\frac {5 d x}{2}\right )+1680 B \sin \left (3 c+\frac {5 d x}{2}\right )-1040 B \sin \left (3 c+\frac {7 d x}{2}\right )+4410 B d x \cos \left (c+\frac {3 d x}{2}\right )+4410 B d x \cos \left (2 c+\frac {3 d x}{2}\right )+1470 B d x \cos \left (2 c+\frac {5 d x}{2}\right )+1470 B d x \cos \left (3 c+\frac {5 d x}{2}\right )+210 B d x \cos \left (3 c+\frac {7 d x}{2}\right )+210 B d x \cos \left (4 c+\frac {7 d x}{2}\right )+7350 d x (B-4 C) \cos \left (\frac {d x}{2}\right )-19880 B \sin \left (\frac {d x}{2}\right )-46130 C \sin \left (c+\frac {d x}{2}\right )+46116 C \sin \left (c+\frac {3 d x}{2}\right )-18060 C \sin \left (2 c+\frac {3 d x}{2}\right )+19292 C \sin \left (2 c+\frac {5 d x}{2}\right )-2100 C \sin \left (3 c+\frac {5 d x}{2}\right )+3791 C \sin \left (3 c+\frac {7 d x}{2}\right )+735 C \sin \left (4 c+\frac {7 d x}{2}\right )+105 C \sin \left (4 c+\frac {9 d x}{2}\right )+105 C \sin \left (5 c+\frac {9 d x}{2}\right )-17640 C d x \cos \left (c+\frac {3 d x}{2}\right )-17640 C d x \cos \left (2 c+\frac {3 d x}{2}\right )-5880 C d x \cos \left (2 c+\frac {5 d x}{2}\right )-5880 C d x \cos \left (3 c+\frac {5 d x}{2}\right )-840 C d x \cos \left (3 c+\frac {7 d x}{2}\right )-840 C d x \cos \left (4 c+\frac {7 d x}{2}\right )+60830 C \sin \left (\frac {d x}{2}\right )\right )}{1680 a^4 d (\cos (c+d x)+1)^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^3*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2))/(a + a*Cos[c + d*x])^4,x]

[Out]

(Cos[(c + d*x)/2]*Sec[c/2]*(7350*(B - 4*C)*d*x*Cos[(d*x)/2] + 7350*(B - 4*C)*d*x*Cos[c + (d*x)/2] + 4410*B*d*x
*Cos[c + (3*d*x)/2] - 17640*C*d*x*Cos[c + (3*d*x)/2] + 4410*B*d*x*Cos[2*c + (3*d*x)/2] - 17640*C*d*x*Cos[2*c +
(3*d*x)/2] + 1470*B*d*x*Cos[2*c + (5*d*x)/2] - 5880*C*d*x*Cos[2*c + (5*d*x)/2] + 1470*B*d*x*Cos[3*c + (5*d*x)
/2] - 5880*C*d*x*Cos[3*c + (5*d*x)/2] + 210*B*d*x*Cos[3*c + (7*d*x)/2] - 840*C*d*x*Cos[3*c + (7*d*x)/2] + 210*
B*d*x*Cos[4*c + (7*d*x)/2] - 840*C*d*x*Cos[4*c + (7*d*x)/2] + 2520*A*Sin[(d*x)/2] - 19880*B*Sin[(d*x)/2] + 608
30*C*Sin[(d*x)/2] - 2520*A*Sin[c + (d*x)/2] + 16520*B*Sin[c + (d*x)/2] - 46130*C*Sin[c + (d*x)/2] + 1764*A*Sin
[c + (3*d*x)/2] - 14280*B*Sin[c + (3*d*x)/2] + 46116*C*Sin[c + (3*d*x)/2] - 1260*A*Sin[2*c + (3*d*x)/2] + 7560
*B*Sin[2*c + (3*d*x)/2] - 18060*C*Sin[2*c + (3*d*x)/2] + 588*A*Sin[2*c + (5*d*x)/2] - 5600*B*Sin[2*c + (5*d*x)
/2] + 19292*C*Sin[2*c + (5*d*x)/2] - 420*A*Sin[3*c + (5*d*x)/2] + 1680*B*Sin[3*c + (5*d*x)/2] - 2100*C*Sin[3*c
+ (5*d*x)/2] + 144*A*Sin[3*c + (7*d*x)/2] - 1040*B*Sin[3*c + (7*d*x)/2] + 3791*C*Sin[3*c + (7*d*x)/2] + 735*C
*Sin[4*c + (7*d*x)/2] + 105*C*Sin[4*c + (9*d*x)/2] + 105*C*Sin[5*c + (9*d*x)/2]))/(1680*a^4*d*(1 + Cos[c + d*x
])^4)

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fricas [A]  time = 0.42, size = 223, normalized size = 1.14 \[ \frac {105 \, {\left (B - 4 \, C\right )} d x \cos \left (d x + c\right )^{4} + 420 \, {\left (B - 4 \, C\right )} d x \cos \left (d x + c\right )^{3} + 630 \, {\left (B - 4 \, C\right )} d x \cos \left (d x + c\right )^{2} + 420 \, {\left (B - 4 \, C\right )} d x \cos \left (d x + c\right ) + 105 \, {\left (B - 4 \, C\right )} d x + {\left (105 \, C \cos \left (d x + c\right )^{4} + 4 \, {\left (9 \, A - 65 \, B + 296 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (39 \, A - 620 \, B + 2636 \, C\right )} \cos \left (d x + c\right )^{2} + {\left (24 \, A - 535 \, B + 2236 \, C\right )} \cos \left (d x + c\right ) + 6 \, A - 160 \, B + 664 \, C\right )} \sin \left (d x + c\right )}{105 \, {\left (a^{4} d \cos \left (d x + c\right )^{4} + 4 \, a^{4} d \cos \left (d x + c\right )^{3} + 6 \, a^{4} d \cos \left (d x + c\right )^{2} + 4 \, a^{4} d \cos \left (d x + c\right ) + a^{4} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(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*(B - 4*C)*d*x*cos(d*x + c)^4 + 420*(B - 4*C)*d*x*cos(d*x + c)^3 + 630*(B - 4*C)*d*x*cos(d*x + c)^2
+ 420*(B - 4*C)*d*x*cos(d*x + c) + 105*(B - 4*C)*d*x + (105*C*cos(d*x + c)^4 + 4*(9*A - 65*B + 296*C)*cos(d*x
+ c)^3 + (39*A - 620*B + 2636*C)*cos(d*x + c)^2 + (24*A - 535*B + 2236*C)*cos(d*x + c) + 6*A - 160*B + 664*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 = 0.52, size = 255, normalized size = 1.31 \[ \frac {\frac {840 \, {\left (d x + c\right )} {\left (B - 4 \, C\right )}}{a^{4}} + \frac {1680 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )} a^{4}} - \frac {15 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 15 \, B a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 15 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 63 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 105 \, B a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 147 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 105 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 385 \, B a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 805 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 105 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1575 \, B a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 5145 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{28}}}{840 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(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)*(B - 4*C)/a^4 + 1680*C*tan(1/2*d*x + 1/2*c)/((tan(1/2*d*x + 1/2*c)^2 + 1)*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 - 63*A*a^24*ta
n(1/2*d*x + 1/2*c)^5 + 105*B*a^24*tan(1/2*d*x + 1/2*c)^5 - 147*C*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*tan(1/2*d*x + 1/2*c)^3 + 805*C*a^24*tan(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) - 5145*C*a^24*tan(1/2*d*x + 1/2*c))/a^28)/d

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maple [A]  time = 0.14, size = 307, normalized size = 1.57 \[ -\frac {\left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{56 d \,a^{4}}+\frac {B \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{56 d \,a^{4}}-\frac {C \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{56 d \,a^{4}}+\frac {3 A \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{40 d \,a^{4}}-\frac {B \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d \,a^{4}}+\frac {7 C \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{40 d \,a^{4}}-\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{8 d \,a^{4}}+\frac {11 B \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 d \,a^{4}}-\frac {23 C \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 d \,a^{4}}+\frac {A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d \,a^{4}}-\frac {15 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d \,a^{4}}+\frac {49 C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d \,a^{4}}+\frac {2 C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \,a^{4} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{d \,a^{4}}-\frac {8 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C}{d \,a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^3*(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+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+7/40/d/a^4*C*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)^3-23/24/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)-15/8/d/a^4*B*tan(1/2*d*x+1/2*c)+49/8/d/a^4*C*tan(1/2*d*x+1/2*c)+2/d/a^4*C*tan(1/2*d*x+1/2*c)/(1+tan(1/
2*d*x+1/2*c)^2)+2/d/a^4*arctan(tan(1/2*d*x+1/2*c))*B-8/d/a^4*arctan(tan(1/2*d*x+1/2*c))*C

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maxima [A]  time = 0.45, size = 356, normalized size = 1.83 \[ \frac {C {\left (\frac {1680 \, \sin \left (d x + c\right )}{{\left (a^{4} + \frac {a^{4} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}} + \frac {\frac {5145 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {805 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {147 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {15 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}}{a^{4}} - \frac {6720 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{4}}\right )} - 5 \, B {\left (\frac {\frac {315 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {77 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {21 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {3 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}}{a^{4}} - \frac {336 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{4}}\right )} + \frac {3 \, A {\left (\frac {35 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {35 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {21 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {5 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}\right )}}{a^{4}}}{840 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(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*(C*(1680*sin(d*x + c)/((a^4 + a^4*sin(d*x + c)^2/(cos(d*x + c) + 1)^2)*(cos(d*x + c) + 1)) + (5145*sin(d
*x + c)/(cos(d*x + c) + 1) - 805*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 147*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 - 6720*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^4) - 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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mupad [B]  time = 1.37, size = 248, normalized size = 1.27 \[ \frac {B\,d\,x-4\,C\,d\,x}{a^4\,d}+\frac {\left (\frac {12\,A\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{35}-\frac {52\,B\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{21}+\frac {764\,C\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{105}\right )\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+\left (\frac {16\,B\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{21}-\frac {23\,A\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{70}-\frac {143\,C\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{105}\right )\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\left (\frac {9\,A\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{70}-\frac {5\,B\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{28}+\frac {8\,C\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{35}\right )\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-\frac {A\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{56}+\frac {B\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{56}-\frac {C\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{56}}{a^4\,d\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}+\frac {2\,C\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{a^4\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(c + d*x)^3*(A + B*cos(c + d*x) + C*cos(c + d*x)^2))/(a + a*cos(c + d*x))^4,x)

[Out]

(B*d*x - 4*C*d*x)/(a^4*d) + (cos(c/2 + (d*x)/2)^2*((9*A*sin(c/2 + (d*x)/2))/70 - (5*B*sin(c/2 + (d*x)/2))/28 +
(8*C*sin(c/2 + (d*x)/2))/35) - cos(c/2 + (d*x)/2)^4*((23*A*sin(c/2 + (d*x)/2))/70 - (16*B*sin(c/2 + (d*x)/2))
/21 + (143*C*sin(c/2 + (d*x)/2))/105) + cos(c/2 + (d*x)/2)^6*((12*A*sin(c/2 + (d*x)/2))/35 - (52*B*sin(c/2 + (
d*x)/2))/21 + (764*C*sin(c/2 + (d*x)/2))/105) - (A*sin(c/2 + (d*x)/2))/56 + (B*sin(c/2 + (d*x)/2))/56 - (C*sin
(c/2 + (d*x)/2))/56)/(a^4*d*cos(c/2 + (d*x)/2)^7) + (2*C*cos(c/2 + (d*x)/2)*sin(c/2 + (d*x)/2))/(a^4*d)

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sympy [A]  time = 35.08, size = 746, normalized size = 3.83 \[ \begin {cases} - \frac {15 A \tan ^{9}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{840 a^{4} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 840 a^{4} d} + \frac {48 A \tan ^{7}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{840 a^{4} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 840 a^{4} d} - \frac {42 A \tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{840 a^{4} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 840 a^{4} d} + \frac {105 A \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{840 a^{4} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 840 a^{4} d} + \frac {840 B d x \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{840 a^{4} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 840 a^{4} d} + \frac {840 B d x}{840 a^{4} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 840 a^{4} d} + \frac {15 B \tan ^{9}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{840 a^{4} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 840 a^{4} d} - \frac {90 B \tan ^{7}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{840 a^{4} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 840 a^{4} d} + \frac {280 B \tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{840 a^{4} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 840 a^{4} d} - \frac {1190 B \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{840 a^{4} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 840 a^{4} d} - \frac {1575 B \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{840 a^{4} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 840 a^{4} d} - \frac {3360 C d x \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{840 a^{4} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 840 a^{4} d} - \frac {3360 C d x}{840 a^{4} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 840 a^{4} d} - \frac {15 C \tan ^{9}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{840 a^{4} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 840 a^{4} d} + \frac {132 C \tan ^{7}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{840 a^{4} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 840 a^{4} d} - \frac {658 C \tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{840 a^{4} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 840 a^{4} d} + \frac {4340 C \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{840 a^{4} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 840 a^{4} d} + \frac {6825 C \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{840 a^{4} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 840 a^{4} d} & \text {for}\: d \neq 0 \\\frac {x \left (A + B \cos {\relax (c )} + C \cos ^{2}{\relax (c )}\right ) \cos ^{3}{\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)**3*(A+B*cos(d*x+c)+C*cos(d*x+c)**2)/(a+a*cos(d*x+c))**4,x)

[Out]

Piecewise((-15*A*tan(c/2 + d*x/2)**9/(840*a**4*d*tan(c/2 + d*x/2)**2 + 840*a**4*d) + 48*A*tan(c/2 + d*x/2)**7/
(840*a**4*d*tan(c/2 + d*x/2)**2 + 840*a**4*d) - 42*A*tan(c/2 + d*x/2)**5/(840*a**4*d*tan(c/2 + d*x/2)**2 + 840
*a**4*d) + 105*A*tan(c/2 + d*x/2)/(840*a**4*d*tan(c/2 + d*x/2)**2 + 840*a**4*d) + 840*B*d*x*tan(c/2 + d*x/2)**
2/(840*a**4*d*tan(c/2 + d*x/2)**2 + 840*a**4*d) + 840*B*d*x/(840*a**4*d*tan(c/2 + d*x/2)**2 + 840*a**4*d) + 15
*B*tan(c/2 + d*x/2)**9/(840*a**4*d*tan(c/2 + d*x/2)**2 + 840*a**4*d) - 90*B*tan(c/2 + d*x/2)**7/(840*a**4*d*ta
n(c/2 + d*x/2)**2 + 840*a**4*d) + 280*B*tan(c/2 + d*x/2)**5/(840*a**4*d*tan(c/2 + d*x/2)**2 + 840*a**4*d) - 11
90*B*tan(c/2 + d*x/2)**3/(840*a**4*d*tan(c/2 + d*x/2)**2 + 840*a**4*d) - 1575*B*tan(c/2 + d*x/2)/(840*a**4*d*t
an(c/2 + d*x/2)**2 + 840*a**4*d) - 3360*C*d*x*tan(c/2 + d*x/2)**2/(840*a**4*d*tan(c/2 + d*x/2)**2 + 840*a**4*d
) - 3360*C*d*x/(840*a**4*d*tan(c/2 + d*x/2)**2 + 840*a**4*d) - 15*C*tan(c/2 + d*x/2)**9/(840*a**4*d*tan(c/2 +
d*x/2)**2 + 840*a**4*d) + 132*C*tan(c/2 + d*x/2)**7/(840*a**4*d*tan(c/2 + d*x/2)**2 + 840*a**4*d) - 658*C*tan(
c/2 + d*x/2)**5/(840*a**4*d*tan(c/2 + d*x/2)**2 + 840*a**4*d) + 4340*C*tan(c/2 + d*x/2)**3/(840*a**4*d*tan(c/2
+ d*x/2)**2 + 840*a**4*d) + 6825*C*tan(c/2 + d*x/2)/(840*a**4*d*tan(c/2 + d*x/2)**2 + 840*a**4*d), Ne(d, 0)),
(x*(A + B*cos(c) + C*cos(c)**2)*cos(c)**3/(a*cos(c) + a)**4, True))

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