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

[Out]

a^4*C*x+1/8*a^4*(28*A+35*B+48*C)*arctanh(sin(d*x+c))/d+1/8*a^4*(28*A+35*B+40*C)*tan(d*x+c)/d+1/24*(28*A+35*B+3
2*C)*(a^4+a^4*cos(d*x+c))*sec(d*x+c)*tan(d*x+c)/d+1/60*(28*A+35*B+20*C)*(a^2+a^2*cos(d*x+c))^2*sec(d*x+c)^2*ta
n(d*x+c)/d+1/20*a*(4*A+5*B)*(a+a*cos(d*x+c))^3*sec(d*x+c)^3*tan(d*x+c)/d+1/5*A*(a+a*cos(d*x+c))^4*sec(d*x+c)^4
*tan(d*x+c)/d

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Rubi [A]  time = 0.69, antiderivative size = 225, normalized size of antiderivative = 1.00, 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 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 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 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 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 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*}

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Mathematica [B]  time = 6.22, 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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fricas [A]  time = 0.61, size = 196, normalized size = 0.87 \[ \frac {240 \, C a^{4} d x \cos \left (d x + c\right )^{5} + 15 \, {\left (28 \, A + 35 \, B + 48 \, C\right )} a^{4} \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (28 \, A + 35 \, B + 48 \, C\right )} a^{4} \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (8 \, {\left (83 \, A + 100 \, B + 100 \, C\right )} a^{4} \cos \left (d x + c\right )^{4} + 15 \, {\left (28 \, A + 27 \, B + 16 \, C\right )} a^{4} \cos \left (d x + c\right )^{3} + 8 \, {\left (34 \, A + 20 \, B + 5 \, C\right )} a^{4} \cos \left (d x + c\right )^{2} + 30 \, {\left (4 \, A + B\right )} a^{4} \cos \left (d x + c\right ) + 24 \, A a^{4}\right )} \sin \left (d x + c\right )}{240 \, d \cos \left (d x + c\right )^{5}} \]

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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giac [A]  time = 0.69, size = 352, normalized size = 1.56 \[ \frac {120 \, {\left (d x + c\right )} C a^{4} + 15 \, {\left (28 \, A a^{4} + 35 \, B a^{4} + 48 \, C a^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 15 \, {\left (28 \, A a^{4} + 35 \, B a^{4} + 48 \, C a^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (420 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 525 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 600 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 1960 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 2450 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 2720 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 3584 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 4480 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 4720 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3160 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3950 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3680 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 1500 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1395 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1080 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{5}}}{120 \, d} \]

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

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maple [A]  time = 0.55, size = 331, normalized size = 1.47 \[ \frac {83 A \,a^{4} \tan \left (d x +c \right )}{15 d}+\frac {35 a^{4} B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d}+a^{4} C x +\frac {a^{4} C c}{d}+\frac {7 A \,a^{4} \sec \left (d x +c \right ) \tan \left (d x +c \right )}{2 d}+\frac {7 A \,a^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {20 a^{4} B \tan \left (d x +c \right )}{3 d}+\frac {6 a^{4} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {34 A \,a^{4} \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{15 d}+\frac {27 a^{4} B \sec \left (d x +c \right ) \tan \left (d x +c \right )}{8 d}+\frac {20 a^{4} C \tan \left (d x +c \right )}{3 d}+\frac {A \,a^{4} \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )}{d}+\frac {4 a^{4} B \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{3 d}+\frac {2 a^{4} C \sec \left (d x +c \right ) \tan \left (d x +c \right )}{d}+\frac {A \,a^{4} \tan \left (d x +c \right ) \left (\sec ^{4}\left (d x +c \right )\right )}{5 d}+\frac {a^{4} B \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )}{4 d}+\frac {a^{4} C \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{3 d} \]

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

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maxima [B]  time = 0.39, size = 496, normalized size = 2.20 \[ \frac {16 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} A a^{4} + 480 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a^{4} + 320 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a^{4} + 80 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a^{4} + 240 \, {\left (d x + c\right )} C a^{4} - 60 \, A a^{4} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 15 \, B a^{4} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 240 \, A a^{4} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 360 \, B a^{4} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 240 \, C a^{4} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 120 \, B a^{4} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 480 \, C a^{4} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 240 \, A a^{4} \tan \left (d x + c\right ) + 960 \, B a^{4} \tan \left (d x + c\right ) + 1440 \, C a^{4} \tan \left (d x + c\right )}{240 \, d} \]

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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mupad [B]  time = 2.80, size = 995, normalized size = 4.42 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((11*A*a^4*sin(2*c + 2*d*x))/8 + (77*A*a^4*sin(3*c + 3*d*x))/48 + (7*A*a^4*sin(4*c + 4*d*x))/16 + (83*A*a^4*si
n(5*c + 5*d*x))/240 + (31*B*a^4*sin(2*c + 2*d*x))/32 + (19*B*a^4*sin(3*c + 3*d*x))/12 + (27*B*a^4*sin(4*c + 4*
d*x))/64 + (5*B*a^4*sin(5*c + 5*d*x))/12 + (C*a^4*sin(2*c + 2*d*x))/2 + (4*C*a^4*sin(3*c + 3*d*x))/3 + (C*a^4*
sin(4*c + 4*d*x))/4 + (5*C*a^4*sin(5*c + 5*d*x))/12 + (35*A*a^4*sin(c + d*x))/24 + (7*B*a^4*sin(c + d*x))/6 +
(11*C*a^4*sin(c + d*x))/12 + (5*C*a^4*atan((784*A^2*sin(c/2 + (d*x)/2) + 1225*B^2*sin(c/2 + (d*x)/2) + 2368*C^
2*sin(c/2 + (d*x)/2) + 1960*A*B*sin(c/2 + (d*x)/2) + 2688*A*C*sin(c/2 + (d*x)/2) + 3360*B*C*sin(c/2 + (d*x)/2)
)/(cos(c/2 + (d*x)/2)*(784*A^2 + 1225*B^2 + 2368*C^2 + 1960*A*B + 2688*A*C + 3360*B*C)))*cos(c + d*x))/4 + (5*
C*a^4*atan((784*A^2*sin(c/2 + (d*x)/2) + 1225*B^2*sin(c/2 + (d*x)/2) + 2368*C^2*sin(c/2 + (d*x)/2) + 1960*A*B*
sin(c/2 + (d*x)/2) + 2688*A*C*sin(c/2 + (d*x)/2) + 3360*B*C*sin(c/2 + (d*x)/2))/(cos(c/2 + (d*x)/2)*(784*A^2 +
1225*B^2 + 2368*C^2 + 1960*A*B + 2688*A*C + 3360*B*C)))*cos(3*c + 3*d*x))/8 + (C*a^4*atan((784*A^2*sin(c/2 +
(d*x)/2) + 1225*B^2*sin(c/2 + (d*x)/2) + 2368*C^2*sin(c/2 + (d*x)/2) + 1960*A*B*sin(c/2 + (d*x)/2) + 2688*A*C*
sin(c/2 + (d*x)/2) + 3360*B*C*sin(c/2 + (d*x)/2))/(cos(c/2 + (d*x)/2)*(784*A^2 + 1225*B^2 + 2368*C^2 + 1960*A*
B + 2688*A*C + 3360*B*C)))*cos(5*c + 5*d*x))/8 + (35*A*a^4*cos(c + d*x)*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*
x)/2)))/8 + (175*B*a^4*cos(c + d*x)*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/32 + (15*C*a^4*cos(c + d*x)*
atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/2 + (35*A*a^4*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(3
*c + 3*d*x))/16 + (7*A*a^4*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(5*c + 5*d*x))/16 + (175*B*a^4*atan
h(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(3*c + 3*d*x))/64 + (35*B*a^4*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (
d*x)/2))*cos(5*c + 5*d*x))/64 + (15*C*a^4*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(3*c + 3*d*x))/4 + (
3*C*a^4*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(5*c + 5*d*x))/4)/(d*((5*cos(c + d*x))/8 + (5*cos(3*c
+ 3*d*x))/16 + cos(5*c + 5*d*x)/16))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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