3.336 \(\int (a+a \cos (c+d x))^4 (A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^6(c+d x) \, dx\)

Optimal. Leaf size=225 \[ \frac{a^4 (28 A+35 B+40 C) \tan (c+d x)}{8 d}+\frac{a^4 (28 A+35 B+48 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{(28 A+35 B+20 C) \tan (c+d x) \sec ^2(c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{60 d}+\frac{(28 A+35 B+32 C) \tan (c+d x) \sec (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{24 d}+a^4 C x+\frac{a (4 A+5 B) \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^3}{20 d}+\frac{A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^4}{5 d} \]

[Out]

a^4*C*x + (a^4*(28*A + 35*B + 48*C)*ArcTanh[Sin[c + d*x]])/(8*d) + (a^4*(28*A + 35*B + 40*C)*Tan[c + d*x])/(8*
d) + ((28*A + 35*B + 32*C)*(a^4 + a^4*Cos[c + d*x])*Sec[c + d*x]*Tan[c + d*x])/(24*d) + ((28*A + 35*B + 20*C)*
(a^2 + a^2*Cos[c + d*x])^2*Sec[c + d*x]^2*Tan[c + d*x])/(60*d) + (a*(4*A + 5*B)*(a + a*Cos[c + d*x])^3*Sec[c +
d*x]^3*Tan[c + d*x])/(20*d) + (A*(a + a*Cos[c + d*x])^4*Sec[c + d*x]^4*Tan[c + d*x])/(5*d)

________________________________________________________________________________________

Rubi [A]  time = 0.691009, antiderivative size = 225, 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 = {3043, 2975, 2968, 3021, 2735, 3770} \[ \frac{a^4 (28 A+35 B+40 C) \tan (c+d x)}{8 d}+\frac{a^4 (28 A+35 B+48 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{(28 A+35 B+20 C) \tan (c+d x) \sec ^2(c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{60 d}+\frac{(28 A+35 B+32 C) \tan (c+d x) \sec (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{24 d}+a^4 C x+\frac{a (4 A+5 B) \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^3}{20 d}+\frac{A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^4}{5 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Cos[c + d*x])^4*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^6,x]

[Out]

a^4*C*x + (a^4*(28*A + 35*B + 48*C)*ArcTanh[Sin[c + d*x]])/(8*d) + (a^4*(28*A + 35*B + 40*C)*Tan[c + d*x])/(8*
d) + ((28*A + 35*B + 32*C)*(a^4 + a^4*Cos[c + d*x])*Sec[c + d*x]*Tan[c + d*x])/(24*d) + ((28*A + 35*B + 20*C)*
(a^2 + a^2*Cos[c + d*x])^2*Sec[c + d*x]^2*Tan[c + d*x])/(60*d) + (a*(4*A + 5*B)*(a + a*Cos[c + d*x])^3*Sec[c +
d*x]^3*Tan[c + d*x])/(20*d) + (A*(a + a*Cos[c + d*x])^4*Sec[c + d*x]^4*Tan[c + d*x])/(5*d)

Rule 3043

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[((c^2*C - B*c*d + A*d^2)*Cos[e +
f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(n + 1)*(c^2 - d^2)), x] + Dist[1/(b*d*(n + 1)
*(c^2 - d^2)), Int[(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(a*d*m + b*c*(n + 1)) + (c*C -
B*d)*(a*c*m + b*d*(n + 1)) + b*(d*(B*c - A*d)*(m + n + 2) - C*(c^2*(m + 1) + d^2*(n + 1)))*Sin[e + f*x], x],
x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2,
0] &&  !LtQ[m, -2^(-1)] && (LtQ[n, -1] || EqQ[m + n + 2, 0])

Rule 2975

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^2*(B*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*(c + d*S
in[e + f*x])^(n + 1))/(d*f*(n + 1)*(b*c + a*d)), x] - Dist[b/(d*(n + 1)*(b*c + a*d)), Int[(a + b*Sin[e + f*x])
^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp[a*A*d*(m - n - 2) - B*(a*c*(m - 1) + b*d*(n + 1)) - (A*b*d*(m + n +
1) - B*(b*c*m - a*d*(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] && GtQ[m, 1/2] && LtQ[n, -1] && IntegerQ[2*m] && (IntegerQ[2*n]
|| EqQ[c, 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 3021

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^2 - a*b*B + a^2*C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m +
1)*(a^2 - b^2)), x] + Dist[1/(b*(m + 1)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*(a*A - b*B + a*
C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A*b - a*B + b*C)*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e,
f, A, B, C}, x] && LtQ[m, -1] && NeQ[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 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 (a+a \cos (c+d x))^4 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx &=\frac{A (a+a \cos (c+d x))^4 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac{\int (a+a \cos (c+d x))^4 (a (4 A+5 B)+5 a C \cos (c+d x)) \sec ^5(c+d x) \, dx}{5 a}\\ &=\frac{a (4 A+5 B) (a+a \cos (c+d x))^3 \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac{A (a+a \cos (c+d x))^4 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac{\int (a+a \cos (c+d x))^3 \left (a^2 (28 A+35 B+20 C)+20 a^2 C \cos (c+d x)\right ) \sec ^4(c+d x) \, dx}{20 a}\\ &=\frac{(28 A+35 B+20 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^2(c+d x) \tan (c+d x)}{60 d}+\frac{a (4 A+5 B) (a+a \cos (c+d x))^3 \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac{A (a+a \cos (c+d x))^4 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac{\int (a+a \cos (c+d x))^2 \left (5 a^3 (28 A+35 B+32 C)+60 a^3 C \cos (c+d x)\right ) \sec ^3(c+d x) \, dx}{60 a}\\ &=\frac{(28 A+35 B+32 C) \left (a^4+a^4 \cos (c+d x)\right ) \sec (c+d x) \tan (c+d x)}{24 d}+\frac{(28 A+35 B+20 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^2(c+d x) \tan (c+d x)}{60 d}+\frac{a (4 A+5 B) (a+a \cos (c+d x))^3 \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac{A (a+a \cos (c+d x))^4 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac{\int (a+a \cos (c+d x)) \left (15 a^4 (28 A+35 B+40 C)+120 a^4 C \cos (c+d x)\right ) \sec ^2(c+d x) \, dx}{120 a}\\ &=\frac{(28 A+35 B+32 C) \left (a^4+a^4 \cos (c+d x)\right ) \sec (c+d x) \tan (c+d x)}{24 d}+\frac{(28 A+35 B+20 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^2(c+d x) \tan (c+d x)}{60 d}+\frac{a (4 A+5 B) (a+a \cos (c+d x))^3 \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac{A (a+a \cos (c+d x))^4 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac{\int \left (15 a^5 (28 A+35 B+40 C)+\left (120 a^5 C+15 a^5 (28 A+35 B+40 C)\right ) \cos (c+d x)+120 a^5 C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx}{120 a}\\ &=\frac{a^4 (28 A+35 B+40 C) \tan (c+d x)}{8 d}+\frac{(28 A+35 B+32 C) \left (a^4+a^4 \cos (c+d x)\right ) \sec (c+d x) \tan (c+d x)}{24 d}+\frac{(28 A+35 B+20 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^2(c+d x) \tan (c+d x)}{60 d}+\frac{a (4 A+5 B) (a+a \cos (c+d x))^3 \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac{A (a+a \cos (c+d x))^4 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac{\int \left (15 a^5 (28 A+35 B+48 C)+120 a^5 C \cos (c+d x)\right ) \sec (c+d x) \, dx}{120 a}\\ &=a^4 C x+\frac{a^4 (28 A+35 B+40 C) \tan (c+d x)}{8 d}+\frac{(28 A+35 B+32 C) \left (a^4+a^4 \cos (c+d x)\right ) \sec (c+d x) \tan (c+d x)}{24 d}+\frac{(28 A+35 B+20 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^2(c+d x) \tan (c+d x)}{60 d}+\frac{a (4 A+5 B) (a+a \cos (c+d x))^3 \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac{A (a+a \cos (c+d x))^4 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac{1}{8} \left (a^4 (28 A+35 B+48 C)\right ) \int \sec (c+d x) \, dx\\ &=a^4 C x+\frac{a^4 (28 A+35 B+48 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{a^4 (28 A+35 B+40 C) \tan (c+d x)}{8 d}+\frac{(28 A+35 B+32 C) \left (a^4+a^4 \cos (c+d x)\right ) \sec (c+d x) \tan (c+d x)}{24 d}+\frac{(28 A+35 B+20 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^2(c+d x) \tan (c+d x)}{60 d}+\frac{a (4 A+5 B) (a+a \cos (c+d x))^3 \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac{A (a+a \cos (c+d x))^4 \sec ^4(c+d x) \tan (c+d x)}{5 d}\\ \end{align*}

Mathematica [B]  time = 6.20683, size = 971, normalized size = 4.32 \[ \frac{C (c+d x) (\cos (c+d x) a+a)^4 \sec ^8\left (\frac{c}{2}+\frac{d x}{2}\right )}{16 d}+\frac{(-28 A-35 B-48 C) (\cos (c+d x) a+a)^4 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right ) \sec ^8\left (\frac{c}{2}+\frac{d x}{2}\right )}{128 d}+\frac{(28 A+35 B+48 C) (\cos (c+d x) a+a)^4 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )+\sin \left (\frac{1}{2} (c+d x)\right )\right ) \sec ^8\left (\frac{c}{2}+\frac{d x}{2}\right )}{128 d}+\frac{A (\cos (c+d x) a+a)^4 \sin \left (\frac{1}{2} (c+d x)\right ) \sec ^8\left (\frac{c}{2}+\frac{d x}{2}\right )}{320 d \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^5}+\frac{(\cos (c+d x) a+a)^4 \left (139 A \sin \left (\frac{1}{2} (c+d x)\right )+80 B \sin \left (\frac{1}{2} (c+d x)\right )+20 C \sin \left (\frac{1}{2} (c+d x)\right )\right ) \sec ^8\left (\frac{c}{2}+\frac{d x}{2}\right )}{1920 d \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^3}+\frac{(\cos (c+d x) a+a)^4 \left (139 A \sin \left (\frac{1}{2} (c+d x)\right )+80 B \sin \left (\frac{1}{2} (c+d x)\right )+20 C \sin \left (\frac{1}{2} (c+d x)\right )\right ) \sec ^8\left (\frac{c}{2}+\frac{d x}{2}\right )}{1920 d \left (\cos \left (\frac{1}{2} (c+d x)\right )+\sin \left (\frac{1}{2} (c+d x)\right )\right )^3}+\frac{(\cos (c+d x) a+a)^4 \left (83 A \sin \left (\frac{1}{2} (c+d x)\right )+100 B \sin \left (\frac{1}{2} (c+d x)\right )+100 C \sin \left (\frac{1}{2} (c+d x)\right )\right ) \sec ^8\left (\frac{c}{2}+\frac{d x}{2}\right )}{240 d \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}+\frac{(\cos (c+d x) a+a)^4 \left (83 A \sin \left (\frac{1}{2} (c+d x)\right )+100 B \sin \left (\frac{1}{2} (c+d x)\right )+100 C \sin \left (\frac{1}{2} (c+d x)\right )\right ) \sec ^8\left (\frac{c}{2}+\frac{d x}{2}\right )}{240 d \left (\cos \left (\frac{1}{2} (c+d x)\right )+\sin \left (\frac{1}{2} (c+d x)\right )\right )}+\frac{(559 A+485 B+260 C) (\cos (c+d x) a+a)^4 \sec ^8\left (\frac{c}{2}+\frac{d x}{2}\right )}{3840 d \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}+\frac{(-559 A-485 B-260 C) (\cos (c+d x) a+a)^4 \sec ^8\left (\frac{c}{2}+\frac{d x}{2}\right )}{3840 d \left (\cos \left (\frac{1}{2} (c+d x)\right )+\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}+\frac{(22 A+5 B) (\cos (c+d x) a+a)^4 \sec ^8\left (\frac{c}{2}+\frac{d x}{2}\right )}{1280 d \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^4}+\frac{(-22 A-5 B) (\cos (c+d x) a+a)^4 \sec ^8\left (\frac{c}{2}+\frac{d x}{2}\right )}{1280 d \left (\cos \left (\frac{1}{2} (c+d x)\right )+\sin \left (\frac{1}{2} (c+d x)\right )\right )^4}+\frac{A (\cos (c+d x) a+a)^4 \sin \left (\frac{1}{2} (c+d x)\right ) \sec ^8\left (\frac{c}{2}+\frac{d x}{2}\right )}{320 d \left (\cos \left (\frac{1}{2} (c+d x)\right )+\sin \left (\frac{1}{2} (c+d x)\right )\right )^5} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Cos[c + d*x])^4*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^6,x]

[Out]

(C*(c + d*x)*(a + a*Cos[c + d*x])^4*Sec[c/2 + (d*x)/2]^8)/(16*d) + ((-28*A - 35*B - 48*C)*(a + a*Cos[c + d*x])
^4*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]*Sec[c/2 + (d*x)/2]^8)/(128*d) + ((28*A + 35*B + 48*C)*(a + a*Cos[c
+ d*x])^4*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]*Sec[c/2 + (d*x)/2]^8)/(128*d) + ((22*A + 5*B)*(a + a*Cos[c
+ d*x])^4*Sec[c/2 + (d*x)/2]^8)/(1280*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^4) + ((559*A + 485*B + 260*C)*(
a + a*Cos[c + d*x])^4*Sec[c/2 + (d*x)/2]^8)/(3840*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2) + (A*(a + a*Cos[c
+ d*x])^4*Sec[c/2 + (d*x)/2]^8*Sin[(c + d*x)/2])/(320*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^5) + (A*(a + a*
Cos[c + d*x])^4*Sec[c/2 + (d*x)/2]^8*Sin[(c + d*x)/2])/(320*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^5) + ((-22
*A - 5*B)*(a + a*Cos[c + d*x])^4*Sec[c/2 + (d*x)/2]^8)/(1280*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^4) + ((-5
59*A - 485*B - 260*C)*(a + a*Cos[c + d*x])^4*Sec[c/2 + (d*x)/2]^8)/(3840*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2
])^2) + ((a + a*Cos[c + d*x])^4*Sec[c/2 + (d*x)/2]^8*(139*A*Sin[(c + d*x)/2] + 80*B*Sin[(c + d*x)/2] + 20*C*Si
n[(c + d*x)/2]))/(1920*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^3) + ((a + a*Cos[c + d*x])^4*Sec[c/2 + (d*x)/2]
^8*(139*A*Sin[(c + d*x)/2] + 80*B*Sin[(c + d*x)/2] + 20*C*Sin[(c + d*x)/2]))/(1920*d*(Cos[(c + d*x)/2] + Sin[(
c + d*x)/2])^3) + ((a + a*Cos[c + d*x])^4*Sec[c/2 + (d*x)/2]^8*(83*A*Sin[(c + d*x)/2] + 100*B*Sin[(c + d*x)/2]
+ 100*C*Sin[(c + d*x)/2]))/(240*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])) + ((a + a*Cos[c + d*x])^4*Sec[c/2 +
(d*x)/2]^8*(83*A*Sin[(c + d*x)/2] + 100*B*Sin[(c + d*x)/2] + 100*C*Sin[(c + d*x)/2]))/(240*d*(Cos[(c + d*x)/2]
+ Sin[(c + d*x)/2]))

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Maple [A]  time = 0.101, size = 331, normalized size = 1.5 \begin{align*}{\frac{83\,A{a}^{4}\tan \left ( dx+c \right ) }{15\,d}}+{\frac{A{a}^{4}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{4}}{5\,d}}+{\frac{34\,A{a}^{4}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{15\,d}}+{\frac{{a}^{4}B\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{4\,d}}+{\frac{27\,{a}^{4}B\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{8\,d}}+{\frac{35\,{a}^{4}B\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}}+{\frac{20\,{a}^{4}C\tan \left ( dx+c \right ) }{3\,d}}+{\frac{{a}^{4}C\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3\,d}}+{\frac{A{a}^{4}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{d}}+{\frac{7\,A{a}^{4}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{7\,A{a}^{4}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+{\frac{20\,{a}^{4}B\tan \left ( dx+c \right ) }{3\,d}}+{\frac{4\,{a}^{4}B\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3\,d}}+2\,{\frac{{a}^{4}C\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{d}}+6\,{\frac{{a}^{4}C\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{a}^{4}Cx+{\frac{C{a}^{4}c}{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*cos(d*x+c))^4*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^6,x)

[Out]

83/15/d*A*a^4*tan(d*x+c)+1/5/d*A*a^4*tan(d*x+c)*sec(d*x+c)^4+34/15/d*A*a^4*tan(d*x+c)*sec(d*x+c)^2+1/4/d*a^4*B
*tan(d*x+c)*sec(d*x+c)^3+27/8/d*a^4*B*sec(d*x+c)*tan(d*x+c)+35/8/d*a^4*B*ln(sec(d*x+c)+tan(d*x+c))+20/3/d*a^4*
C*tan(d*x+c)+1/3/d*a^4*C*tan(d*x+c)*sec(d*x+c)^2+1/d*A*a^4*tan(d*x+c)*sec(d*x+c)^3+7/2/d*A*a^4*sec(d*x+c)*tan(
d*x+c)+7/2/d*A*a^4*ln(sec(d*x+c)+tan(d*x+c))+20/3/d*a^4*B*tan(d*x+c)+4/3/d*a^4*B*tan(d*x+c)*sec(d*x+c)^2+2/d*a
^4*C*sec(d*x+c)*tan(d*x+c)+6/d*a^4*C*ln(sec(d*x+c)+tan(d*x+c))+a^4*C*x+1/d*a^4*C*c

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Maxima [B]  time = 1.03938, size = 670, normalized size = 2.98 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^4*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^6,x, algorithm="maxima")

[Out]

1/240*(16*(3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c))*A*a^4 + 480*(tan(d*x + c)^3 + 3*tan(d*x + c
))*A*a^4 + 320*(tan(d*x + c)^3 + 3*tan(d*x + c))*B*a^4 + 80*(tan(d*x + c)^3 + 3*tan(d*x + c))*C*a^4 + 240*(d*x
+ c)*C*a^4 - 60*A*a^4*(2*(3*sin(d*x + c)^3 - 5*sin(d*x + c))/(sin(d*x + c)^4 - 2*sin(d*x + c)^2 + 1) - 3*log(
sin(d*x + c) + 1) + 3*log(sin(d*x + c) - 1)) - 15*B*a^4*(2*(3*sin(d*x + c)^3 - 5*sin(d*x + c))/(sin(d*x + c)^4
- 2*sin(d*x + c)^2 + 1) - 3*log(sin(d*x + c) + 1) + 3*log(sin(d*x + c) - 1)) - 240*A*a^4*(2*sin(d*x + c)/(sin
(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) - 360*B*a^4*(2*sin(d*x + c)/(sin(d*x + c)^2
- 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) - 240*C*a^4*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(s
in(d*x + c) + 1) + log(sin(d*x + c) - 1)) + 120*B*a^4*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 480*C*
a^4*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 240*A*a^4*tan(d*x + c) + 960*B*a^4*tan(d*x + c) + 1440*C
*a^4*tan(d*x + c))/d

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Fricas [A]  time = 2.16467, size = 521, normalized size = 2.32 \begin{align*} \frac{240 \, C a^{4} d x \cos \left (d x + c\right )^{5} + 15 \,{\left (28 \, A + 35 \, B + 48 \, C\right )} a^{4} \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \,{\left (28 \, A + 35 \, B + 48 \, C\right )} a^{4} \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (8 \,{\left (83 \, A + 100 \, B + 100 \, C\right )} a^{4} \cos \left (d x + c\right )^{4} + 15 \,{\left (28 \, A + 27 \, B + 16 \, C\right )} a^{4} \cos \left (d x + c\right )^{3} + 8 \,{\left (34 \, A + 20 \, B + 5 \, C\right )} a^{4} \cos \left (d x + c\right )^{2} + 30 \,{\left (4 \, A + B\right )} a^{4} \cos \left (d x + c\right ) + 24 \, A a^{4}\right )} \sin \left (d x + c\right )}{240 \, d \cos \left (d x + c\right )^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^4*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^6,x, algorithm="fricas")

[Out]

1/240*(240*C*a^4*d*x*cos(d*x + c)^5 + 15*(28*A + 35*B + 48*C)*a^4*cos(d*x + c)^5*log(sin(d*x + c) + 1) - 15*(2
8*A + 35*B + 48*C)*a^4*cos(d*x + c)^5*log(-sin(d*x + c) + 1) + 2*(8*(83*A + 100*B + 100*C)*a^4*cos(d*x + c)^4
+ 15*(28*A + 27*B + 16*C)*a^4*cos(d*x + c)^3 + 8*(34*A + 20*B + 5*C)*a^4*cos(d*x + c)^2 + 30*(4*A + B)*a^4*cos
(d*x + c) + 24*A*a^4)*sin(d*x + c))/(d*cos(d*x + c)^5)

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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+a*cos(d*x+c))**4*(A+B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)**6,x)

[Out]

Timed out

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Giac [A]  time = 1.24557, size = 475, normalized size = 2.11 \begin{align*} \frac{120 \,{\left (d x + c\right )} C a^{4} + 15 \,{\left (28 \, A a^{4} + 35 \, B a^{4} + 48 \, C a^{4}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 15 \,{\left (28 \, A a^{4} + 35 \, B a^{4} + 48 \, C a^{4}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) - \frac{2 \,{\left (420 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} + 525 \, B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} + 600 \, C a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} - 1960 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 2450 \, B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 2720 \, C a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 3584 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 4480 \, B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 4720 \, C a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 3160 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 3950 \, B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 3680 \, C a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 1500 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1395 \, B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1080 \, C a^{4} \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 )}^{5}}}{120 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^4*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^6,x, algorithm="giac")

[Out]

1/120*(120*(d*x + c)*C*a^4 + 15*(28*A*a^4 + 35*B*a^4 + 48*C*a^4)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 15*(28*A
*a^4 + 35*B*a^4 + 48*C*a^4)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(420*A*a^4*tan(1/2*d*x + 1/2*c)^9 + 525*B*a
^4*tan(1/2*d*x + 1/2*c)^9 + 600*C*a^4*tan(1/2*d*x + 1/2*c)^9 - 1960*A*a^4*tan(1/2*d*x + 1/2*c)^7 - 2450*B*a^4*
tan(1/2*d*x + 1/2*c)^7 - 2720*C*a^4*tan(1/2*d*x + 1/2*c)^7 + 3584*A*a^4*tan(1/2*d*x + 1/2*c)^5 + 4480*B*a^4*ta
n(1/2*d*x + 1/2*c)^5 + 4720*C*a^4*tan(1/2*d*x + 1/2*c)^5 - 3160*A*a^4*tan(1/2*d*x + 1/2*c)^3 - 3950*B*a^4*tan(
1/2*d*x + 1/2*c)^3 - 3680*C*a^4*tan(1/2*d*x + 1/2*c)^3 + 1500*A*a^4*tan(1/2*d*x + 1/2*c) + 1395*B*a^4*tan(1/2*
d*x + 1/2*c) + 1080*C*a^4*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^5)/d