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

Optimal. Leaf size=248 \[ -\frac{8 (216 A-83 B+20 C) \tan (c+d x)}{105 a^4 d}+\frac{(21 A-8 B+2 C) \tanh ^{-1}(\sin (c+d x))}{2 a^4 d}+\frac{(21 A-8 B+2 C) \tan (c+d x) \sec (c+d x)}{2 a^4 d}-\frac{4 (216 A-83 B+20 C) \tan (c+d x) \sec (c+d x)}{105 a^4 d (\cos (c+d x)+1)}-\frac{(129 A-52 B+10 C) \tan (c+d x) \sec (c+d x)}{105 a^4 d (\cos (c+d x)+1)^2}-\frac{(A-B+C) \tan (c+d x) \sec (c+d x)}{7 d (a \cos (c+d x)+a)^4}-\frac{(2 A-B) \tan (c+d x) \sec (c+d x)}{5 a d (a \cos (c+d x)+a)^3} \]

[Out]

((21*A - 8*B + 2*C)*ArcTanh[Sin[c + d*x]])/(2*a^4*d) - (8*(216*A - 83*B + 20*C)*Tan[c + d*x])/(105*a^4*d) + ((
21*A - 8*B + 2*C)*Sec[c + d*x]*Tan[c + d*x])/(2*a^4*d) - ((129*A - 52*B + 10*C)*Sec[c + d*x]*Tan[c + d*x])/(10
5*a^4*d*(1 + Cos[c + d*x])^2) - (4*(216*A - 83*B + 20*C)*Sec[c + d*x]*Tan[c + d*x])/(105*a^4*d*(1 + Cos[c + d*
x])) - ((A - B + C)*Sec[c + d*x]*Tan[c + d*x])/(7*d*(a + a*Cos[c + d*x])^4) - ((2*A - B)*Sec[c + d*x]*Tan[c +
d*x])/(5*a*d*(a + a*Cos[c + d*x])^3)

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Rubi [A]  time = 0.761248, antiderivative size = 248, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.171, Rules used = {3041, 2978, 2748, 3768, 3770, 3767, 8} \[ -\frac{8 (216 A-83 B+20 C) \tan (c+d x)}{105 a^4 d}+\frac{(21 A-8 B+2 C) \tanh ^{-1}(\sin (c+d x))}{2 a^4 d}+\frac{(21 A-8 B+2 C) \tan (c+d x) \sec (c+d x)}{2 a^4 d}-\frac{4 (216 A-83 B+20 C) \tan (c+d x) \sec (c+d x)}{105 a^4 d (\cos (c+d x)+1)}-\frac{(129 A-52 B+10 C) \tan (c+d x) \sec (c+d x)}{105 a^4 d (\cos (c+d x)+1)^2}-\frac{(A-B+C) \tan (c+d x) \sec (c+d x)}{7 d (a \cos (c+d x)+a)^4}-\frac{(2 A-B) \tan (c+d x) \sec (c+d x)}{5 a d (a \cos (c+d x)+a)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((21*A - 8*B + 2*C)*ArcTanh[Sin[c + d*x]])/(2*a^4*d) - (8*(216*A - 83*B + 20*C)*Tan[c + d*x])/(105*a^4*d) + ((
21*A - 8*B + 2*C)*Sec[c + d*x]*Tan[c + d*x])/(2*a^4*d) - ((129*A - 52*B + 10*C)*Sec[c + d*x]*Tan[c + d*x])/(10
5*a^4*d*(1 + Cos[c + d*x])^2) - (4*(216*A - 83*B + 20*C)*Sec[c + d*x]*Tan[c + d*x])/(105*a^4*d*(1 + Cos[c + d*
x])) - ((A - B + C)*Sec[c + d*x]*Tan[c + d*x])/(7*d*(a + a*Cos[c + d*x])^4) - ((2*A - B)*Sec[c + d*x]*Tan[c +
d*x])/(5*a*d*(a + a*Cos[c + d*x])^3)

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)]

Rule 2978

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*
x])^(n + 1))/(a*f*(2*m + 1)*(b*c - a*d)), x] + Dist[1/(a*(2*m + 1)*(b*c - a*d)), Int[(a + b*Sin[e + f*x])^(m +
1)*(c + d*Sin[e + f*x])^n*Simp[B*(a*c*m + b*d*(n + 1)) + A*(b*c*(m + 1) - a*d*(2*m + n + 2)) + d*(A*b - a*B)*
(m + n + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, 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 2748

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

Rule 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

\begin{align*} \int \frac{\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(a+a \cos (c+d x))^4} \, dx &=-\frac{(A-B+C) \sec (c+d x) \tan (c+d x)}{7 d (a+a \cos (c+d x))^4}+\frac{\int \frac{(a (9 A-2 B+2 C)-a (5 A-5 B-2 C) \cos (c+d x)) \sec ^3(c+d x)}{(a+a \cos (c+d x))^3} \, dx}{7 a^2}\\ &=-\frac{(A-B+C) \sec (c+d x) \tan (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac{(2 A-B) \sec (c+d x) \tan (c+d x)}{5 a d (a+a \cos (c+d x))^3}+\frac{\int \frac{\left (a^2 (73 A-24 B+10 C)-28 a^2 (2 A-B) \cos (c+d x)\right ) \sec ^3(c+d x)}{(a+a \cos (c+d x))^2} \, dx}{35 a^4}\\ &=-\frac{(129 A-52 B+10 C) \sec (c+d x) \tan (c+d x)}{105 a^4 d (1+\cos (c+d x))^2}-\frac{(A-B+C) \sec (c+d x) \tan (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac{(2 A-B) \sec (c+d x) \tan (c+d x)}{5 a d (a+a \cos (c+d x))^3}+\frac{\int \frac{\left (a^3 (477 A-176 B+50 C)-3 a^3 (129 A-52 B+10 C) \cos (c+d x)\right ) \sec ^3(c+d x)}{a+a \cos (c+d x)} \, dx}{105 a^6}\\ &=-\frac{(129 A-52 B+10 C) \sec (c+d x) \tan (c+d x)}{105 a^4 d (1+\cos (c+d x))^2}-\frac{(A-B+C) \sec (c+d x) \tan (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac{(2 A-B) \sec (c+d x) \tan (c+d x)}{5 a d (a+a \cos (c+d x))^3}-\frac{4 (216 A-83 B+20 C) \sec (c+d x) \tan (c+d x)}{105 d \left (a^4+a^4 \cos (c+d x)\right )}+\frac{\int \left (105 a^4 (21 A-8 B+2 C)-8 a^4 (216 A-83 B+20 C) \cos (c+d x)\right ) \sec ^3(c+d x) \, dx}{105 a^8}\\ &=-\frac{(129 A-52 B+10 C) \sec (c+d x) \tan (c+d x)}{105 a^4 d (1+\cos (c+d x))^2}-\frac{(A-B+C) \sec (c+d x) \tan (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac{(2 A-B) \sec (c+d x) \tan (c+d x)}{5 a d (a+a \cos (c+d x))^3}-\frac{4 (216 A-83 B+20 C) \sec (c+d x) \tan (c+d x)}{105 d \left (a^4+a^4 \cos (c+d x)\right )}+\frac{(21 A-8 B+2 C) \int \sec ^3(c+d x) \, dx}{a^4}-\frac{(8 (216 A-83 B+20 C)) \int \sec ^2(c+d x) \, dx}{105 a^4}\\ &=\frac{(21 A-8 B+2 C) \sec (c+d x) \tan (c+d x)}{2 a^4 d}-\frac{(129 A-52 B+10 C) \sec (c+d x) \tan (c+d x)}{105 a^4 d (1+\cos (c+d x))^2}-\frac{(A-B+C) \sec (c+d x) \tan (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac{(2 A-B) \sec (c+d x) \tan (c+d x)}{5 a d (a+a \cos (c+d x))^3}-\frac{4 (216 A-83 B+20 C) \sec (c+d x) \tan (c+d x)}{105 d \left (a^4+a^4 \cos (c+d x)\right )}+\frac{(21 A-8 B+2 C) \int \sec (c+d x) \, dx}{2 a^4}+\frac{(8 (216 A-83 B+20 C)) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{105 a^4 d}\\ &=\frac{(21 A-8 B+2 C) \tanh ^{-1}(\sin (c+d x))}{2 a^4 d}-\frac{8 (216 A-83 B+20 C) \tan (c+d x)}{105 a^4 d}+\frac{(21 A-8 B+2 C) \sec (c+d x) \tan (c+d x)}{2 a^4 d}-\frac{(129 A-52 B+10 C) \sec (c+d x) \tan (c+d x)}{105 a^4 d (1+\cos (c+d x))^2}-\frac{(A-B+C) \sec (c+d x) \tan (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac{(2 A-B) \sec (c+d x) \tan (c+d x)}{5 a d (a+a \cos (c+d x))^3}-\frac{4 (216 A-83 B+20 C) \sec (c+d x) \tan (c+d x)}{105 d \left (a^4+a^4 \cos (c+d x)\right )}\\ \end{align*}

Mathematica [A]  time = 1.47684, size = 271, normalized size = 1.09 \[ -\frac{13440 (21 A-8 B+2 C) \cos ^8\left (\frac{1}{2} (c+d x)\right ) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )+2 \sin \left (\frac{1}{2} (c+d x)\right ) \cos \left (\frac{1}{2} (c+d x)\right ) \sec ^2(c+d x) (8 (12813 A-4994 B+1130 C) \cos (c+d x)+60 (1177 A-456 B+106 C) \cos (2 (c+d x))+35928 A \cos (3 (c+d x))+11619 A \cos (4 (c+d x))+1728 A \cos (5 (c+d x))+58161 A-13864 B \cos (3 (c+d x))-4472 B \cos (4 (c+d x))-664 B \cos (5 (c+d x))-22888 B+3280 C \cos (3 (c+d x))+1070 C \cos (4 (c+d x))+160 C \cos (5 (c+d x))+5290 C)}{1680 a^4 d (\cos (c+d x)+1)^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(13440*(21*A - 8*B + 2*C)*Cos[(c + d*x)/2]^8*(Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2]
+ Sin[(c + d*x)/2]]) + 2*Cos[(c + d*x)/2]*(58161*A - 22888*B + 5290*C + 8*(12813*A - 4994*B + 1130*C)*Cos[c +
d*x] + 60*(1177*A - 456*B + 106*C)*Cos[2*(c + d*x)] + 35928*A*Cos[3*(c + d*x)] - 13864*B*Cos[3*(c + d*x)] + 3
280*C*Cos[3*(c + d*x)] + 11619*A*Cos[4*(c + d*x)] - 4472*B*Cos[4*(c + d*x)] + 1070*C*Cos[4*(c + d*x)] + 1728*A
*Cos[5*(c + d*x)] - 664*B*Cos[5*(c + d*x)] + 160*C*Cos[5*(c + d*x)])*Sec[c + d*x]^2*Sin[(c + d*x)/2])/(1680*a^
4*d*(1 + Cos[c + d*x])^4)

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Maple [B]  time = 0.086, size = 493, normalized size = 2. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^3/(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*tan(1/2*d*x+1/2*c)^7*B-1/56/d/a^4*tan(1/2*d*x+1/2*c)^7*C-9/40/d/
a^4*A*tan(1/2*d*x+1/2*c)^5+7/40/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)^5-13/8/d/a^4*tan(1
/2*d*x+1/2*c)^3*A+23/24/d/a^4*tan(1/2*d*x+1/2*c)^3*B-11/24/d/a^4*C*tan(1/2*d*x+1/2*c)^3-111/8/d/a^4*A*tan(1/2*
d*x+1/2*c)+49/8/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)+9/2/d/a^4*A/(tan(1/2*d*x+1/2*c)-1)-
1/d/a^4/(tan(1/2*d*x+1/2*c)-1)*B-21/2/d/a^4*A*ln(tan(1/2*d*x+1/2*c)-1)+4/d/a^4*ln(tan(1/2*d*x+1/2*c)-1)*B-1/d/
a^4*ln(tan(1/2*d*x+1/2*c)-1)*C+1/2/d/a^4*A/(tan(1/2*d*x+1/2*c)-1)^2+9/2/d/a^4*A/(tan(1/2*d*x+1/2*c)+1)-1/d/a^4
/(tan(1/2*d*x+1/2*c)+1)*B+21/2/d/a^4*A*ln(tan(1/2*d*x+1/2*c)+1)-4/d/a^4*ln(tan(1/2*d*x+1/2*c)+1)*B+1/d/a^4*ln(
tan(1/2*d*x+1/2*c)+1)*C-1/2/d/a^4*A/(tan(1/2*d*x+1/2*c)+1)^2

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Maxima [B]  time = 1.4234, size = 751, normalized size = 3.03 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^3/(a+a*cos(d*x+c))^4,x, algorithm="maxima")

[Out]

-1/840*(3*A*(280*(7*sin(d*x + c)/(cos(d*x + c) + 1) - 9*sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/(a^4 - 2*a^4*sin(
d*x + c)^2/(cos(d*x + c) + 1)^2 + a^4*sin(d*x + c)^4/(cos(d*x + c) + 1)^4) + (3885*sin(d*x + c)/(cos(d*x + c)
+ 1) + 455*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 63*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 5*sin(d*x + c)^7/(co
s(d*x + c) + 1)^7)/a^4 - 2940*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^4 + 2940*log(sin(d*x + c)/(cos(d*x +
c) + 1) - 1)/a^4) - B*(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 - 3360*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^
4 + 3360*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a^4) + 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 - 168*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^4 + 168*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a^4))/
d

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Fricas [A]  time = 2.16491, size = 1062, normalized size = 4.28 \begin{align*} \frac{105 \,{\left ({\left (21 \, A - 8 \, B + 2 \, C\right )} \cos \left (d x + c\right )^{6} + 4 \,{\left (21 \, A - 8 \, B + 2 \, C\right )} \cos \left (d x + c\right )^{5} + 6 \,{\left (21 \, A - 8 \, B + 2 \, C\right )} \cos \left (d x + c\right )^{4} + 4 \,{\left (21 \, A - 8 \, B + 2 \, C\right )} \cos \left (d x + c\right )^{3} +{\left (21 \, A - 8 \, B + 2 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 105 \,{\left ({\left (21 \, A - 8 \, B + 2 \, C\right )} \cos \left (d x + c\right )^{6} + 4 \,{\left (21 \, A - 8 \, B + 2 \, C\right )} \cos \left (d x + c\right )^{5} + 6 \,{\left (21 \, A - 8 \, B + 2 \, C\right )} \cos \left (d x + c\right )^{4} + 4 \,{\left (21 \, A - 8 \, B + 2 \, C\right )} \cos \left (d x + c\right )^{3} +{\left (21 \, A - 8 \, B + 2 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \,{\left (16 \,{\left (216 \, A - 83 \, B + 20 \, C\right )} \cos \left (d x + c\right )^{5} +{\left (11619 \, A - 4472 \, B + 1070 \, C\right )} \cos \left (d x + c\right )^{4} + 4 \,{\left (3411 \, A - 1318 \, B + 310 \, C\right )} \cos \left (d x + c\right )^{3} + 4 \,{\left (1509 \, A - 592 \, B + 130 \, C\right )} \cos \left (d x + c\right )^{2} + 210 \,{\left (2 \, A - B\right )} \cos \left (d x + c\right ) - 105 \, A\right )} \sin \left (d x + c\right )}{420 \,{\left (a^{4} d \cos \left (d x + c\right )^{6} + 4 \, a^{4} d \cos \left (d x + c\right )^{5} + 6 \, a^{4} d \cos \left (d x + c\right )^{4} + 4 \, a^{4} d \cos \left (d x + c\right )^{3} + a^{4} d \cos \left (d x + c\right )^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^3/(a+a*cos(d*x+c))^4,x, algorithm="fricas")

[Out]

1/420*(105*((21*A - 8*B + 2*C)*cos(d*x + c)^6 + 4*(21*A - 8*B + 2*C)*cos(d*x + c)^5 + 6*(21*A - 8*B + 2*C)*cos
(d*x + c)^4 + 4*(21*A - 8*B + 2*C)*cos(d*x + c)^3 + (21*A - 8*B + 2*C)*cos(d*x + c)^2)*log(sin(d*x + c) + 1) -
105*((21*A - 8*B + 2*C)*cos(d*x + c)^6 + 4*(21*A - 8*B + 2*C)*cos(d*x + c)^5 + 6*(21*A - 8*B + 2*C)*cos(d*x +
c)^4 + 4*(21*A - 8*B + 2*C)*cos(d*x + c)^3 + (21*A - 8*B + 2*C)*cos(d*x + c)^2)*log(-sin(d*x + c) + 1) - 2*(1
6*(216*A - 83*B + 20*C)*cos(d*x + c)^5 + (11619*A - 4472*B + 1070*C)*cos(d*x + c)^4 + 4*(3411*A - 1318*B + 310
*C)*cos(d*x + c)^3 + 4*(1509*A - 592*B + 130*C)*cos(d*x + c)^2 + 210*(2*A - B)*cos(d*x + c) - 105*A)*sin(d*x +
c))/(a^4*d*cos(d*x + c)^6 + 4*a^4*d*cos(d*x + c)^5 + 6*a^4*d*cos(d*x + c)^4 + 4*a^4*d*cos(d*x + c)^3 + a^4*d*
cos(d*x + c)^2)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 1.25808, size = 458, normalized size = 1.85 \begin{align*} \frac{\frac{420 \,{\left (21 \, A - 8 \, B + 2 \, C\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right )}{a^{4}} - \frac{420 \,{\left (21 \, A - 8 \, B + 2 \, C\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right )}{a^{4}} + \frac{840 \,{\left (9 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 2 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 7 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 2 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}^{2} 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} + 189 \, A a^{24} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 147 \, 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} + 1365 \, A a^{24} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 805 \, 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} + 11655 \, A a^{24} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 5145 \, 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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^3/(a+a*cos(d*x+c))^4,x, algorithm="giac")

[Out]

1/840*(420*(21*A - 8*B + 2*C)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^4 - 420*(21*A - 8*B + 2*C)*log(abs(tan(1/2*
d*x + 1/2*c) - 1))/a^4 + 840*(9*A*tan(1/2*d*x + 1/2*c)^3 - 2*B*tan(1/2*d*x + 1/2*c)^3 - 7*A*tan(1/2*d*x + 1/2*
c) + 2*B*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 - 1)^2*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 + 189*A*a^24*tan(1/2*d*x + 1/2*c)^5 - 147*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 + 1365*A*a^24*tan(1/2*d*x + 1/2*c)^3 - 805*B*a^2
4*tan(1/2*d*x + 1/2*c)^3 + 385*C*a^24*tan(1/2*d*x + 1/2*c)^3 + 11655*A*a^24*tan(1/2*d*x + 1/2*c) - 5145*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