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

Optimal. Leaf size=194 \[ \frac{(12 A-8 B+5 C) \tan ^3(c+d x)}{3 a^2 d}+\frac{(12 A-8 B+5 C) \tan (c+d x)}{a^2 d}-\frac{(10 A-7 B+4 C) \tanh ^{-1}(\sin (c+d x))}{2 a^2 d}-\frac{(10 A-7 B+4 C) \tan (c+d x) \sec (c+d x)}{2 a^2 d}-\frac{(10 A-7 B+4 C) \tan (c+d x) \sec ^2(c+d x)}{3 a^2 d (\cos (c+d x)+1)}-\frac{(A-B+C) \tan (c+d x) \sec ^2(c+d x)}{3 d (a \cos (c+d x)+a)^2} \]

[Out]

-((10*A - 7*B + 4*C)*ArcTanh[Sin[c + d*x]])/(2*a^2*d) + ((12*A - 8*B + 5*C)*Tan[c + d*x])/(a^2*d) - ((10*A - 7
*B + 4*C)*Sec[c + d*x]*Tan[c + d*x])/(2*a^2*d) - ((10*A - 7*B + 4*C)*Sec[c + d*x]^2*Tan[c + d*x])/(3*a^2*d*(1
+ Cos[c + d*x])) - ((A - B + C)*Sec[c + d*x]^2*Tan[c + d*x])/(3*d*(a + a*Cos[c + d*x])^2) + ((12*A - 8*B + 5*C
)*Tan[c + d*x]^3)/(3*a^2*d)

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Rubi [A]  time = 0.387534, antiderivative size = 194, normalized size of antiderivative = 1., number of steps used = 7, 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} \[ \frac{(12 A-8 B+5 C) \tan ^3(c+d x)}{3 a^2 d}+\frac{(12 A-8 B+5 C) \tan (c+d x)}{a^2 d}-\frac{(10 A-7 B+4 C) \tanh ^{-1}(\sin (c+d x))}{2 a^2 d}-\frac{(10 A-7 B+4 C) \tan (c+d x) \sec (c+d x)}{2 a^2 d}-\frac{(10 A-7 B+4 C) \tan (c+d x) \sec ^2(c+d x)}{3 a^2 d (\cos (c+d x)+1)}-\frac{(A-B+C) \tan (c+d x) \sec ^2(c+d x)}{3 d (a \cos (c+d x)+a)^2} \]

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])^2,x]

[Out]

-((10*A - 7*B + 4*C)*ArcTanh[Sin[c + d*x]])/(2*a^2*d) + ((12*A - 8*B + 5*C)*Tan[c + d*x])/(a^2*d) - ((10*A - 7
*B + 4*C)*Sec[c + d*x]*Tan[c + d*x])/(2*a^2*d) - ((10*A - 7*B + 4*C)*Sec[c + d*x]^2*Tan[c + d*x])/(3*a^2*d*(1
+ Cos[c + d*x])) - ((A - B + C)*Sec[c + d*x]^2*Tan[c + d*x])/(3*d*(a + a*Cos[c + d*x])^2) + ((12*A - 8*B + 5*C
)*Tan[c + d*x]^3)/(3*a^2*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))^2} \, dx &=-\frac{(A-B+C) \sec ^2(c+d x) \tan (c+d x)}{3 d (a+a \cos (c+d x))^2}+\frac{\int \frac{(3 a (2 A-B+C)-a (4 A-4 B+C) \cos (c+d x)) \sec ^4(c+d x)}{a+a \cos (c+d x)} \, dx}{3 a^2}\\ &=-\frac{(10 A-7 B+4 C) \sec ^2(c+d x) \tan (c+d x)}{3 a^2 d (1+\cos (c+d x))}-\frac{(A-B+C) \sec ^2(c+d x) \tan (c+d x)}{3 d (a+a \cos (c+d x))^2}+\frac{\int \left (3 a^2 (12 A-8 B+5 C)-3 a^2 (10 A-7 B+4 C) \cos (c+d x)\right ) \sec ^4(c+d x) \, dx}{3 a^4}\\ &=-\frac{(10 A-7 B+4 C) \sec ^2(c+d x) \tan (c+d x)}{3 a^2 d (1+\cos (c+d x))}-\frac{(A-B+C) \sec ^2(c+d x) \tan (c+d x)}{3 d (a+a \cos (c+d x))^2}-\frac{(10 A-7 B+4 C) \int \sec ^3(c+d x) \, dx}{a^2}+\frac{(12 A-8 B+5 C) \int \sec ^4(c+d x) \, dx}{a^2}\\ &=-\frac{(10 A-7 B+4 C) \sec (c+d x) \tan (c+d x)}{2 a^2 d}-\frac{(10 A-7 B+4 C) \sec ^2(c+d x) \tan (c+d x)}{3 a^2 d (1+\cos (c+d x))}-\frac{(A-B+C) \sec ^2(c+d x) \tan (c+d x)}{3 d (a+a \cos (c+d x))^2}-\frac{(10 A-7 B+4 C) \int \sec (c+d x) \, dx}{2 a^2}-\frac{(12 A-8 B+5 C) \operatorname{Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{a^2 d}\\ &=-\frac{(10 A-7 B+4 C) \tanh ^{-1}(\sin (c+d x))}{2 a^2 d}+\frac{(12 A-8 B+5 C) \tan (c+d x)}{a^2 d}-\frac{(10 A-7 B+4 C) \sec (c+d x) \tan (c+d x)}{2 a^2 d}-\frac{(10 A-7 B+4 C) \sec ^2(c+d x) \tan (c+d x)}{3 a^2 d (1+\cos (c+d x))}-\frac{(A-B+C) \sec ^2(c+d x) \tan (c+d x)}{3 d (a+a \cos (c+d x))^2}+\frac{(12 A-8 B+5 C) \tan ^3(c+d x)}{3 a^2 d}\\ \end{align*}

Mathematica [B]  time = 6.21015, size = 763, normalized size = 3.93 \[ \frac{4 \cos ^4\left (\frac{c}{2}+\frac{d x}{2}\right ) \left (11 A \sin \left (\frac{1}{2} (c+d x)\right )-6 B \sin \left (\frac{1}{2} (c+d x)\right )+3 C \sin \left (\frac{1}{2} (c+d x)\right )\right )}{3 d (a \cos (c+d x)+a)^2 \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}+\frac{4 \cos ^4\left (\frac{c}{2}+\frac{d x}{2}\right ) \left (11 A \sin \left (\frac{1}{2} (c+d x)\right )-6 B \sin \left (\frac{1}{2} (c+d x)\right )+3 C \sin \left (\frac{1}{2} (c+d x)\right )\right )}{3 d (a \cos (c+d x)+a)^2 \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}+\frac{2 (10 A-7 B+4 C) \cos ^4\left (\frac{c}{2}+\frac{d x}{2}\right ) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}{d (a \cos (c+d x)+a)^2}-\frac{2 (10 A-7 B+4 C) \cos ^4\left (\frac{c}{2}+\frac{d x}{2}\right ) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}{d (a \cos (c+d x)+a)^2}+\frac{2 \cos ^4\left (\frac{c}{2}+\frac{d x}{2}\right ) \sec ^3\left (\frac{1}{2} (c+d x)\right ) \left (A \sin \left (\frac{1}{2} (c+d x)\right )-B \sin \left (\frac{1}{2} (c+d x)\right )+C \sin \left (\frac{1}{2} (c+d x)\right )\right )}{3 d (a \cos (c+d x)+a)^2}+\frac{4 \cos ^4\left (\frac{c}{2}+\frac{d x}{2}\right ) \sec \left (\frac{1}{2} (c+d x)\right ) \left (13 A \sin \left (\frac{1}{2} (c+d x)\right )-10 B \sin \left (\frac{1}{2} (c+d x)\right )+7 C \sin \left (\frac{1}{2} (c+d x)\right )\right )}{3 d (a \cos (c+d x)+a)^2}+\frac{(3 B-5 A) \cos ^4\left (\frac{c}{2}+\frac{d x}{2}\right )}{3 d (a \cos (c+d x)+a)^2 \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}+\frac{(5 A-3 B) \cos ^4\left (\frac{c}{2}+\frac{d x}{2}\right )}{3 d (a \cos (c+d x)+a)^2 \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2}+\frac{2 A \sin \left (\frac{1}{2} (c+d x)\right ) \cos ^4\left (\frac{c}{2}+\frac{d x}{2}\right )}{3 d (a \cos (c+d x)+a)^2 \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^3}+\frac{2 A \sin \left (\frac{1}{2} (c+d x)\right ) \cos ^4\left (\frac{c}{2}+\frac{d x}{2}\right )}{3 d (a \cos (c+d x)+a)^2 \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^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])^2,x]

[Out]

(2*(10*A - 7*B + 4*C)*Cos[c/2 + (d*x)/2]^4*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]])/(d*(a + a*Cos[c + d*x])^2
) - (2*(10*A - 7*B + 4*C)*Cos[c/2 + (d*x)/2]^4*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]])/(d*(a + a*Cos[c + d*x
])^2) + ((-5*A + 3*B)*Cos[c/2 + (d*x)/2]^4)/(3*d*(a + a*Cos[c + d*x])^2*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^
2) + (2*A*Cos[c/2 + (d*x)/2]^4*Sin[(c + d*x)/2])/(3*d*(a + a*Cos[c + d*x])^2*(Cos[(c + d*x)/2] - Sin[(c + d*x)
/2])^3) + (2*A*Cos[c/2 + (d*x)/2]^4*Sin[(c + d*x)/2])/(3*d*(a + a*Cos[c + d*x])^2*(Cos[(c + d*x)/2] + Sin[(c +
d*x)/2])^3) + ((5*A - 3*B)*Cos[c/2 + (d*x)/2]^4)/(3*d*(a + a*Cos[c + d*x])^2*(Cos[(c + d*x)/2] + Sin[(c + d*x
)/2])^2) + (2*Cos[c/2 + (d*x)/2]^4*Sec[(c + d*x)/2]^3*(A*Sin[(c + d*x)/2] - B*Sin[(c + d*x)/2] + C*Sin[(c + d*
x)/2]))/(3*d*(a + a*Cos[c + d*x])^2) + (4*Cos[c/2 + (d*x)/2]^4*(11*A*Sin[(c + d*x)/2] - 6*B*Sin[(c + d*x)/2] +
3*C*Sin[(c + d*x)/2]))/(3*d*(a + a*Cos[c + d*x])^2*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])) + (4*Cos[c/2 + (d*x
)/2]^4*(11*A*Sin[(c + d*x)/2] - 6*B*Sin[(c + d*x)/2] + 3*C*Sin[(c + d*x)/2]))/(3*d*(a + a*Cos[c + d*x])^2*(Cos
[(c + d*x)/2] + Sin[(c + d*x)/2])) + (4*Cos[c/2 + (d*x)/2]^4*Sec[(c + d*x)/2]*(13*A*Sin[(c + d*x)/2] - 10*B*Si
n[(c + d*x)/2] + 7*C*Sin[(c + d*x)/2]))/(3*d*(a + a*Cos[c + d*x])^2)

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Maple [B]  time = 0.078, size = 506, normalized size = 2.6 \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))^2,x)

[Out]

1/6/d/a^2*tan(1/2*d*x+1/2*c)^3*A-1/6/d/a^2*tan(1/2*d*x+1/2*c)^3*B+1/6/d/a^2*C*tan(1/2*d*x+1/2*c)^3+9/2/d/a^2*A
*tan(1/2*d*x+1/2*c)-7/2/d/a^2*B*tan(1/2*d*x+1/2*c)+5/2/d/a^2*C*tan(1/2*d*x+1/2*c)-3/2/d/a^2*A/(tan(1/2*d*x+1/2
*c)-1)^2+1/2/d/a^2*B/(tan(1/2*d*x+1/2*c)-1)^2+5/d/a^2*A*ln(tan(1/2*d*x+1/2*c)-1)-7/2/d/a^2*B*ln(tan(1/2*d*x+1/
2*c)-1)+2/d/a^2*ln(tan(1/2*d*x+1/2*c)-1)*C-5/d/a^2*A/(tan(1/2*d*x+1/2*c)-1)+5/2/d/a^2*B/(tan(1/2*d*x+1/2*c)-1)
-1/d/a^2/(tan(1/2*d*x+1/2*c)-1)*C-1/3/d/a^2*A/(tan(1/2*d*x+1/2*c)-1)^3+3/2/d/a^2*A/(tan(1/2*d*x+1/2*c)+1)^2-1/
2/d/a^2*B/(tan(1/2*d*x+1/2*c)+1)^2-5/d/a^2*A*ln(tan(1/2*d*x+1/2*c)+1)+7/2/d/a^2*B*ln(tan(1/2*d*x+1/2*c)+1)-2/d
/a^2*ln(tan(1/2*d*x+1/2*c)+1)*C-5/d/a^2*A/(tan(1/2*d*x+1/2*c)+1)+5/2/d/a^2*B/(tan(1/2*d*x+1/2*c)+1)-1/d/a^2/(t
an(1/2*d*x+1/2*c)+1)*C-1/3/d/a^2*A/(tan(1/2*d*x+1/2*c)+1)^3

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Maxima [B]  time = 1.04391, size = 765, normalized size = 3.94 \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))^2,x, algorithm="maxima")

[Out]

1/6*(A*(4*(9*sin(d*x + c)/(cos(d*x + c) + 1) - 20*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 15*sin(d*x + c)^5/(cos
(d*x + c) + 1)^5)/(a^2 - 3*a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 3*a^2*sin(d*x + c)^4/(cos(d*x + c) + 1)^4
- a^2*sin(d*x + c)^6/(cos(d*x + c) + 1)^6) + (27*sin(d*x + c)/(cos(d*x + c) + 1) + sin(d*x + c)^3/(cos(d*x +
c) + 1)^3)/a^2 - 30*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^2 + 30*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)
/a^2) - B*(6*(3*sin(d*x + c)/(cos(d*x + c) + 1) - 5*sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/(a^2 - 2*a^2*sin(d*x
+ c)^2/(cos(d*x + c) + 1)^2 + a^2*sin(d*x + c)^4/(cos(d*x + c) + 1)^4) + (21*sin(d*x + c)/(cos(d*x + c) + 1) +
sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/a^2 - 21*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^2 + 21*log(sin(d*x +
c)/(cos(d*x + c) + 1) - 1)/a^2) + C*((15*sin(d*x + c)/(cos(d*x + c) + 1) + sin(d*x + c)^3/(cos(d*x + c) + 1)^
3)/a^2 - 12*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^2 + 12*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a^2 + 1
2*sin(d*x + c)/((a^2 - a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2)*(cos(d*x + c) + 1))))/d

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Fricas [A]  time = 2.02162, size = 689, normalized size = 3.55 \begin{align*} -\frac{3 \,{\left ({\left (10 \, A - 7 \, B + 4 \, C\right )} \cos \left (d x + c\right )^{5} + 2 \,{\left (10 \, A - 7 \, B + 4 \, C\right )} \cos \left (d x + c\right )^{4} +{\left (10 \, A - 7 \, B + 4 \, C\right )} \cos \left (d x + c\right )^{3}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left ({\left (10 \, A - 7 \, B + 4 \, C\right )} \cos \left (d x + c\right )^{5} + 2 \,{\left (10 \, A - 7 \, B + 4 \, C\right )} \cos \left (d x + c\right )^{4} +{\left (10 \, A - 7 \, B + 4 \, C\right )} \cos \left (d x + c\right )^{3}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \,{\left (4 \,{\left (12 \, A - 8 \, B + 5 \, C\right )} \cos \left (d x + c\right )^{4} +{\left (66 \, A - 43 \, B + 28 \, C\right )} \cos \left (d x + c\right )^{3} + 6 \,{\left (2 \, A - B + C\right )} \cos \left (d x + c\right )^{2} -{\left (2 \, A - 3 \, B\right )} \cos \left (d x + c\right ) + 2 \, A\right )} \sin \left (d x + c\right )}{12 \,{\left (a^{2} d \cos \left (d x + c\right )^{5} + 2 \, a^{2} d \cos \left (d x + c\right )^{4} + a^{2} 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))^2,x, algorithm="fricas")

[Out]

-1/12*(3*((10*A - 7*B + 4*C)*cos(d*x + c)^5 + 2*(10*A - 7*B + 4*C)*cos(d*x + c)^4 + (10*A - 7*B + 4*C)*cos(d*x
+ c)^3)*log(sin(d*x + c) + 1) - 3*((10*A - 7*B + 4*C)*cos(d*x + c)^5 + 2*(10*A - 7*B + 4*C)*cos(d*x + c)^4 +
(10*A - 7*B + 4*C)*cos(d*x + c)^3)*log(-sin(d*x + c) + 1) - 2*(4*(12*A - 8*B + 5*C)*cos(d*x + c)^4 + (66*A - 4
3*B + 28*C)*cos(d*x + c)^3 + 6*(2*A - B + C)*cos(d*x + c)^2 - (2*A - 3*B)*cos(d*x + c) + 2*A)*sin(d*x + c))/(a
^2*d*cos(d*x + c)^5 + 2*a^2*d*cos(d*x + c)^4 + a^2*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))**2,x)

[Out]

Timed out

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Giac [A]  time = 1.21765, size = 409, normalized size = 2.11 \begin{align*} -\frac{\frac{3 \,{\left (10 \, A - 7 \, B + 4 \, C\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right )}{a^{2}} - \frac{3 \,{\left (10 \, A - 7 \, B + 4 \, C\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right )}{a^{2}} + \frac{2 \,{\left (30 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 15 \, 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} - 40 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 24 \, 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} + 18 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 9 \, 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^{2}} - \frac{A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + C a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 27 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 21 \, B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 15 \, C a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{6}}}{6 \, 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))^2,x, algorithm="giac")

[Out]

-1/6*(3*(10*A - 7*B + 4*C)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^2 - 3*(10*A - 7*B + 4*C)*log(abs(tan(1/2*d*x +
1/2*c) - 1))/a^2 + 2*(30*A*tan(1/2*d*x + 1/2*c)^5 - 15*B*tan(1/2*d*x + 1/2*c)^5 + 6*C*tan(1/2*d*x + 1/2*c)^5
- 40*A*tan(1/2*d*x + 1/2*c)^3 + 24*B*tan(1/2*d*x + 1/2*c)^3 - 12*C*tan(1/2*d*x + 1/2*c)^3 + 18*A*tan(1/2*d*x +
1/2*c) - 9*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^2) - (A*a^4*t
an(1/2*d*x + 1/2*c)^3 - B*a^4*tan(1/2*d*x + 1/2*c)^3 + C*a^4*tan(1/2*d*x + 1/2*c)^3 + 27*A*a^4*tan(1/2*d*x + 1
/2*c) - 21*B*a^4*tan(1/2*d*x + 1/2*c) + 15*C*a^4*tan(1/2*d*x + 1/2*c))/a^6)/d