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

Optimal. Leaf size=246 \[ \frac{4 (34 A-19 B+9 C) \tan ^3(c+d x)}{15 a^3 d}+\frac{4 (34 A-19 B+9 C) \tan (c+d x)}{5 a^3 d}-\frac{(23 A-13 B+6 C) \tanh ^{-1}(\sin (c+d x))}{2 a^3 d}-\frac{(23 A-13 B+6 C) \tan (c+d x) \sec (c+d x)}{2 a^3 d}-\frac{(23 A-13 B+6 C) \tan (c+d x) \sec ^2(c+d x)}{3 d \left (a^3 \cos (c+d x)+a^3\right )}-\frac{(13 A-8 B+3 C) \tan (c+d x) \sec ^2(c+d x)}{15 a d (a \cos (c+d x)+a)^2}-\frac{(A-B+C) \tan (c+d x) \sec ^2(c+d x)}{5 d (a \cos (c+d x)+a)^3} \]

[Out]

-((23*A - 13*B + 6*C)*ArcTanh[Sin[c + d*x]])/(2*a^3*d) + (4*(34*A - 19*B + 9*C)*Tan[c + d*x])/(5*a^3*d) - ((23
*A - 13*B + 6*C)*Sec[c + d*x]*Tan[c + d*x])/(2*a^3*d) - ((A - B + C)*Sec[c + d*x]^2*Tan[c + d*x])/(5*d*(a + a*
Cos[c + d*x])^3) - ((13*A - 8*B + 3*C)*Sec[c + d*x]^2*Tan[c + d*x])/(15*a*d*(a + a*Cos[c + d*x])^2) - ((23*A -
13*B + 6*C)*Sec[c + d*x]^2*Tan[c + d*x])/(3*d*(a^3 + a^3*Cos[c + d*x])) + (4*(34*A - 19*B + 9*C)*Tan[c + d*x]
^3)/(15*a^3*d)

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Rubi [A]  time = 0.57554, antiderivative size = 246, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.146, Rules used = {3041, 2978, 2748, 3767, 3768, 3770} \[ \frac{4 (34 A-19 B+9 C) \tan ^3(c+d x)}{15 a^3 d}+\frac{4 (34 A-19 B+9 C) \tan (c+d x)}{5 a^3 d}-\frac{(23 A-13 B+6 C) \tanh ^{-1}(\sin (c+d x))}{2 a^3 d}-\frac{(23 A-13 B+6 C) \tan (c+d x) \sec (c+d x)}{2 a^3 d}-\frac{(23 A-13 B+6 C) \tan (c+d x) \sec ^2(c+d x)}{3 d \left (a^3 \cos (c+d x)+a^3\right )}-\frac{(13 A-8 B+3 C) \tan (c+d x) \sec ^2(c+d x)}{15 a d (a \cos (c+d x)+a)^2}-\frac{(A-B+C) \tan (c+d x) \sec ^2(c+d x)}{5 d (a \cos (c+d x)+a)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((23*A - 13*B + 6*C)*ArcTanh[Sin[c + d*x]])/(2*a^3*d) + (4*(34*A - 19*B + 9*C)*Tan[c + d*x])/(5*a^3*d) - ((23
*A - 13*B + 6*C)*Sec[c + d*x]*Tan[c + d*x])/(2*a^3*d) - ((A - B + C)*Sec[c + d*x]^2*Tan[c + d*x])/(5*d*(a + a*
Cos[c + d*x])^3) - ((13*A - 8*B + 3*C)*Sec[c + d*x]^2*Tan[c + d*x])/(15*a*d*(a + a*Cos[c + d*x])^2) - ((23*A -
13*B + 6*C)*Sec[c + d*x]^2*Tan[c + d*x])/(3*d*(a^3 + a^3*Cos[c + d*x])) + (4*(34*A - 19*B + 9*C)*Tan[c + d*x]
^3)/(15*a^3*d)

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 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 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]

Rubi steps

\begin{align*} \int \frac{\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{(a+a \cos (c+d x))^3} \, dx &=-\frac{(A-B+C) \sec ^2(c+d x) \tan (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac{\int \frac{(a (8 A-3 B+3 C)-5 a (A-B) \cos (c+d x)) \sec ^4(c+d x)}{(a+a \cos (c+d x))^2} \, dx}{5 a^2}\\ &=-\frac{(A-B+C) \sec ^2(c+d x) \tan (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac{(13 A-8 B+3 C) \sec ^2(c+d x) \tan (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac{\int \frac{\left (3 a^2 (21 A-11 B+6 C)-4 a^2 (13 A-8 B+3 C) \cos (c+d x)\right ) \sec ^4(c+d x)}{a+a \cos (c+d x)} \, dx}{15 a^4}\\ &=-\frac{(A-B+C) \sec ^2(c+d x) \tan (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac{(13 A-8 B+3 C) \sec ^2(c+d x) \tan (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac{(23 A-13 B+6 C) \sec ^2(c+d x) \tan (c+d x)}{3 d \left (a^3+a^3 \cos (c+d x)\right )}+\frac{\int \left (12 a^3 (34 A-19 B+9 C)-15 a^3 (23 A-13 B+6 C) \cos (c+d x)\right ) \sec ^4(c+d x) \, dx}{15 a^6}\\ &=-\frac{(A-B+C) \sec ^2(c+d x) \tan (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac{(13 A-8 B+3 C) \sec ^2(c+d x) \tan (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac{(23 A-13 B+6 C) \sec ^2(c+d x) \tan (c+d x)}{3 d \left (a^3+a^3 \cos (c+d x)\right )}-\frac{(23 A-13 B+6 C) \int \sec ^3(c+d x) \, dx}{a^3}+\frac{(4 (34 A-19 B+9 C)) \int \sec ^4(c+d x) \, dx}{5 a^3}\\ &=-\frac{(23 A-13 B+6 C) \sec (c+d x) \tan (c+d x)}{2 a^3 d}-\frac{(A-B+C) \sec ^2(c+d x) \tan (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac{(13 A-8 B+3 C) \sec ^2(c+d x) \tan (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac{(23 A-13 B+6 C) \sec ^2(c+d x) \tan (c+d x)}{3 d \left (a^3+a^3 \cos (c+d x)\right )}-\frac{(23 A-13 B+6 C) \int \sec (c+d x) \, dx}{2 a^3}-\frac{(4 (34 A-19 B+9 C)) \operatorname{Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{5 a^3 d}\\ &=-\frac{(23 A-13 B+6 C) \tanh ^{-1}(\sin (c+d x))}{2 a^3 d}+\frac{4 (34 A-19 B+9 C) \tan (c+d x)}{5 a^3 d}-\frac{(23 A-13 B+6 C) \sec (c+d x) \tan (c+d x)}{2 a^3 d}-\frac{(A-B+C) \sec ^2(c+d x) \tan (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac{(13 A-8 B+3 C) \sec ^2(c+d x) \tan (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac{(23 A-13 B+6 C) \sec ^2(c+d x) \tan (c+d x)}{3 d \left (a^3+a^3 \cos (c+d x)\right )}+\frac{4 (34 A-19 B+9 C) \tan ^3(c+d x)}{15 a^3 d}\\ \end{align*}

Mathematica [A]  time = 0.989298, size = 270, normalized size = 1.1 \[ \frac{960 (23 A-13 B+6 C) \cos ^6\left (\frac{1}{2} (c+d x)\right ) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )+2 \sin \left (\frac{1}{2} (c+d x)\right ) \cos \left (\frac{1}{2} (c+d x)\right ) \sec ^3(c+d x) ((7814 A-4274 B+2124 C) \cos (c+d x)+8 (691 A-381 B+186 C) \cos (2 (c+d x))+3098 A \cos (3 (c+d x))+1287 A \cos (4 (c+d x))+272 A \cos (5 (c+d x))+4321 A-1718 B \cos (3 (c+d x))-717 B \cos (4 (c+d x))-152 B \cos (5 (c+d x))-2331 B+828 C \cos (3 (c+d x))+342 C \cos (4 (c+d x))+72 C \cos (5 (c+d x))+1146 C)}{240 a^3 d (\cos (c+d x)+1)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(960*(23*A - 13*B + 6*C)*Cos[(c + d*x)/2]^6*(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]*(4321*A - 2331*B + 1146*C + (7814*A - 4274*B + 2124*C)*Cos[c + d*x] +
8*(691*A - 381*B + 186*C)*Cos[2*(c + d*x)] + 3098*A*Cos[3*(c + d*x)] - 1718*B*Cos[3*(c + d*x)] + 828*C*Cos[3*
(c + d*x)] + 1287*A*Cos[4*(c + d*x)] - 717*B*Cos[4*(c + d*x)] + 342*C*Cos[4*(c + d*x)] + 272*A*Cos[5*(c + d*x)
] - 152*B*Cos[5*(c + d*x)] + 72*C*Cos[5*(c + d*x)])*Sec[c + d*x]^3*Sin[(c + d*x)/2])/(240*a^3*d*(1 + Cos[c + d
*x])^3)

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Maple [B]  time = 0.085, size = 566, normalized size = 2.3 \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)^4/(a+a*cos(d*x+c))^3,x)

[Out]

1/2/d/a^3*B/(tan(1/2*d*x+1/2*c)-1)^2-1/2/d/a^3*B/(tan(1/2*d*x+1/2*c)+1)^2+7/2/d/a^3*B/(tan(1/2*d*x+1/2*c)-1)+7
/2/d/a^3*B/(tan(1/2*d*x+1/2*c)+1)+13/2/d/a^3*B*ln(tan(1/2*d*x+1/2*c)+1)-13/2/d/a^3*B*ln(tan(1/2*d*x+1/2*c)-1)-
1/d/a^3/(tan(1/2*d*x+1/2*c)-1)*C-1/3/d/a^3*A/(tan(1/2*d*x+1/2*c)-1)^3-1/d/a^3/(tan(1/2*d*x+1/2*c)+1)*C-1/3/d/a
^3*A/(tan(1/2*d*x+1/2*c)+1)^3+3/d/a^3*ln(tan(1/2*d*x+1/2*c)-1)*C-2/d/a^3*A/(tan(1/2*d*x+1/2*c)-1)^2-3/d/a^3*ln
(tan(1/2*d*x+1/2*c)+1)*C+2/d/a^3*A/(tan(1/2*d*x+1/2*c)+1)^2-17/2/d/a^3*A/(tan(1/2*d*x+1/2*c)-1)-17/2/d/a^3*A/(
tan(1/2*d*x+1/2*c)+1)+23/2/d/a^3*A*ln(tan(1/2*d*x+1/2*c)-1)-23/2/d/a^3*A*ln(tan(1/2*d*x+1/2*c)+1)-2/3/d/a^3*ta
n(1/2*d*x+1/2*c)^3*B-31/4/d/a^3*B*tan(1/2*d*x+1/2*c)-1/20/d/a^3*B*tan(1/2*d*x+1/2*c)^5+1/20/d/a^3*A*tan(1/2*d*
x+1/2*c)^5+1/20/d/a^3*C*tan(1/2*d*x+1/2*c)^5+5/6/d/a^3*tan(1/2*d*x+1/2*c)^3*A+1/2/d/a^3*C*tan(1/2*d*x+1/2*c)^3
+49/4/d/a^3*A*tan(1/2*d*x+1/2*c)+17/4/d/a^3*C*tan(1/2*d*x+1/2*c)

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Maxima [B]  time = 1.09323, size = 851, normalized size = 3.46 \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)^4/(a+a*cos(d*x+c))^3,x, algorithm="maxima")

[Out]

1/60*(A*(20*(33*sin(d*x + c)/(cos(d*x + c) + 1) - 76*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 51*sin(d*x + c)^5/(
cos(d*x + c) + 1)^5)/(a^3 - 3*a^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 3*a^3*sin(d*x + c)^4/(cos(d*x + c) + 1
)^4 - a^3*sin(d*x + c)^6/(cos(d*x + c) + 1)^6) + (735*sin(d*x + c)/(cos(d*x + c) + 1) + 50*sin(d*x + c)^3/(cos
(d*x + c) + 1)^3 + 3*sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/a^3 - 690*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a
^3 + 690*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a^3) - B*(60*(5*sin(d*x + c)/(cos(d*x + c) + 1) - 7*sin(d*x
+ c)^3/(cos(d*x + c) + 1)^3)/(a^3 - 2*a^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + a^3*sin(d*x + c)^4/(cos(d*x +
c) + 1)^4) + (465*sin(d*x + c)/(cos(d*x + c) + 1) + 40*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 3*sin(d*x + c)^5/
(cos(d*x + c) + 1)^5)/a^3 - 390*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^3 + 390*log(sin(d*x + c)/(cos(d*x +
c) + 1) - 1)/a^3) + 3*C*(40*sin(d*x + c)/((a^3 - a^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2)*(cos(d*x + c) + 1))
+ (85*sin(d*x + c)/(cos(d*x + c) + 1) + 10*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + sin(d*x + c)^5/(cos(d*x + c)
+ 1)^5)/a^3 - 60*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^3 + 60*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a
^3))/d

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Fricas [A]  time = 2.11088, size = 910, normalized size = 3.7 \begin{align*} -\frac{15 \,{\left ({\left (23 \, A - 13 \, B + 6 \, C\right )} \cos \left (d x + c\right )^{6} + 3 \,{\left (23 \, A - 13 \, B + 6 \, C\right )} \cos \left (d x + c\right )^{5} + 3 \,{\left (23 \, A - 13 \, B + 6 \, C\right )} \cos \left (d x + c\right )^{4} +{\left (23 \, A - 13 \, B + 6 \, C\right )} \cos \left (d x + c\right )^{3}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \,{\left ({\left (23 \, A - 13 \, B + 6 \, C\right )} \cos \left (d x + c\right )^{6} + 3 \,{\left (23 \, A - 13 \, B + 6 \, C\right )} \cos \left (d x + c\right )^{5} + 3 \,{\left (23 \, A - 13 \, B + 6 \, C\right )} \cos \left (d x + c\right )^{4} +{\left (23 \, A - 13 \, B + 6 \, C\right )} \cos \left (d x + c\right )^{3}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \,{\left (16 \,{\left (34 \, A - 19 \, B + 9 \, C\right )} \cos \left (d x + c\right )^{5} + 3 \,{\left (429 \, A - 239 \, B + 114 \, C\right )} \cos \left (d x + c\right )^{4} +{\left (869 \, A - 479 \, B + 234 \, C\right )} \cos \left (d x + c\right )^{3} + 5 \,{\left (19 \, A - 9 \, B + 6 \, C\right )} \cos \left (d x + c\right )^{2} - 15 \,{\left (A - B\right )} \cos \left (d x + c\right ) + 10 \, A\right )} \sin \left (d x + c\right )}{60 \,{\left (a^{3} d \cos \left (d x + c\right )^{6} + 3 \, a^{3} d \cos \left (d x + c\right )^{5} + 3 \, a^{3} d \cos \left (d x + c\right )^{4} + a^{3} d \cos \left (d x + c\right )^{3}\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)^4/(a+a*cos(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/60*(15*((23*A - 13*B + 6*C)*cos(d*x + c)^6 + 3*(23*A - 13*B + 6*C)*cos(d*x + c)^5 + 3*(23*A - 13*B + 6*C)*c
os(d*x + c)^4 + (23*A - 13*B + 6*C)*cos(d*x + c)^3)*log(sin(d*x + c) + 1) - 15*((23*A - 13*B + 6*C)*cos(d*x +
c)^6 + 3*(23*A - 13*B + 6*C)*cos(d*x + c)^5 + 3*(23*A - 13*B + 6*C)*cos(d*x + c)^4 + (23*A - 13*B + 6*C)*cos(d
*x + c)^3)*log(-sin(d*x + c) + 1) - 2*(16*(34*A - 19*B + 9*C)*cos(d*x + c)^5 + 3*(429*A - 239*B + 114*C)*cos(d
*x + c)^4 + (869*A - 479*B + 234*C)*cos(d*x + c)^3 + 5*(19*A - 9*B + 6*C)*cos(d*x + c)^2 - 15*(A - B)*cos(d*x
+ c) + 10*A)*sin(d*x + c))/(a^3*d*cos(d*x + c)^6 + 3*a^3*d*cos(d*x + c)^5 + 3*a^3*d*cos(d*x + c)^4 + a^3*d*cos
(d*x + c)^3)

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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)**4/(a+a*cos(d*x+c))**3,x)

[Out]

Timed out

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Giac [A]  time = 1.24013, size = 481, normalized size = 1.96 \begin{align*} -\frac{\frac{30 \,{\left (23 \, A - 13 \, B + 6 \, C\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} - \frac{30 \,{\left (23 \, A - 13 \, B + 6 \, C\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right )}{a^{3}} + \frac{20 \,{\left (51 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 21 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 6 \, C \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 76 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 36 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 12 \, C \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 33 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 15 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 6 \, C \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 )}^{3} a^{3}} - \frac{3 \, A a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 3 \, B a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 3 \, C a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 50 \, A a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 40 \, B a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 30 \, C a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 735 \, A a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 465 \, B a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 255 \, C a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{15}}}{60 \, 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)^4/(a+a*cos(d*x+c))^3,x, algorithm="giac")

[Out]

-1/60*(30*(23*A - 13*B + 6*C)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^3 - 30*(23*A - 13*B + 6*C)*log(abs(tan(1/2*
d*x + 1/2*c) - 1))/a^3 + 20*(51*A*tan(1/2*d*x + 1/2*c)^5 - 21*B*tan(1/2*d*x + 1/2*c)^5 + 6*C*tan(1/2*d*x + 1/2
*c)^5 - 76*A*tan(1/2*d*x + 1/2*c)^3 + 36*B*tan(1/2*d*x + 1/2*c)^3 - 12*C*tan(1/2*d*x + 1/2*c)^3 + 33*A*tan(1/2
*d*x + 1/2*c) - 15*B*tan(1/2*d*x + 1/2*c) + 6*C*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 - 1)^3*a^3) - (
3*A*a^12*tan(1/2*d*x + 1/2*c)^5 - 3*B*a^12*tan(1/2*d*x + 1/2*c)^5 + 3*C*a^12*tan(1/2*d*x + 1/2*c)^5 + 50*A*a^1
2*tan(1/2*d*x + 1/2*c)^3 - 40*B*a^12*tan(1/2*d*x + 1/2*c)^3 + 30*C*a^12*tan(1/2*d*x + 1/2*c)^3 + 735*A*a^12*ta
n(1/2*d*x + 1/2*c) - 465*B*a^12*tan(1/2*d*x + 1/2*c) + 255*C*a^12*tan(1/2*d*x + 1/2*c))/a^15)/d