3.335 \(\int (a+a \cos (c+d x))^4 (A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^5(c+d x) \, dx\)

Optimal. Leaf size=217 \[ -\frac {5 a^4 (7 A+8 B+4 C) \sin (c+d x)}{8 d}+\frac {a^4 (35 A+48 B+52 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {(35 A+44 B+36 C) \tan (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{12 d}+a^4 x (B+4 C)+\frac {(7 A+8 B+4 C) \tan (c+d x) \sec (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{8 d}+\frac {a (A+B) \tan (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+a)^3}{3 d}+\frac {A \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^4}{4 d} \]

[Out]

a^4*(B+4*C)*x+1/8*a^4*(35*A+48*B+52*C)*arctanh(sin(d*x+c))/d-5/8*a^4*(7*A+8*B+4*C)*sin(d*x+c)/d+1/12*(35*A+44*
B+36*C)*(a^4+a^4*cos(d*x+c))*tan(d*x+c)/d+1/8*(7*A+8*B+4*C)*(a^2+a^2*cos(d*x+c))^2*sec(d*x+c)*tan(d*x+c)/d+1/3
*a*(A+B)*(a+a*cos(d*x+c))^3*sec(d*x+c)^2*tan(d*x+c)/d+1/4*A*(a+a*cos(d*x+c))^4*sec(d*x+c)^3*tan(d*x+c)/d

________________________________________________________________________________________

Rubi [A]  time = 0.74, antiderivative size = 217, 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, 3023, 2735, 3770} \[ -\frac {5 a^4 (7 A+8 B+4 C) \sin (c+d x)}{8 d}+\frac {a^4 (35 A+48 B+52 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {(35 A+44 B+36 C) \tan (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{12 d}+\frac {(7 A+8 B+4 C) \tan (c+d x) \sec (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{8 d}+a^4 x (B+4 C)+\frac {a (A+B) \tan (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+a)^3}{3 d}+\frac {A \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^4}{4 d} \]

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]^5,x]

[Out]

a^4*(B + 4*C)*x + (a^4*(35*A + 48*B + 52*C)*ArcTanh[Sin[c + d*x]])/(8*d) - (5*a^4*(7*A + 8*B + 4*C)*Sin[c + d*
x])/(8*d) + ((35*A + 44*B + 36*C)*(a^4 + a^4*Cos[c + d*x])*Tan[c + d*x])/(12*d) + ((7*A + 8*B + 4*C)*(a^2 + a^
2*Cos[c + d*x])^2*Sec[c + d*x]*Tan[c + d*x])/(8*d) + (a*(A + B)*(a + a*Cos[c + d*x])^3*Sec[c + d*x]^2*Tan[c +
d*x])/(3*d) + (A*(a + a*Cos[c + d*x])^4*Sec[c + d*x]^3*Tan[c + d*x])/(4*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 3023

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*
(m + 2)), Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x]
, x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

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 ^5(c+d x) \, dx &=\frac {A (a+a \cos (c+d x))^4 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {\int (a+a \cos (c+d x))^4 (4 a (A+B)-a (A-4 C) \cos (c+d x)) \sec ^4(c+d x) \, dx}{4 a}\\ &=\frac {a (A+B) (a+a \cos (c+d x))^3 \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac {A (a+a \cos (c+d x))^4 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {\int (a+a \cos (c+d x))^3 \left (3 a^2 (7 A+8 B+4 C)-a^2 (7 A+4 B-12 C) \cos (c+d x)\right ) \sec ^3(c+d x) \, dx}{12 a}\\ &=\frac {(7 A+8 B+4 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a (A+B) (a+a \cos (c+d x))^3 \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac {A (a+a \cos (c+d x))^4 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {\int (a+a \cos (c+d x))^2 \left (2 a^3 (35 A+44 B+36 C)-a^3 (35 A+32 B-12 C) \cos (c+d x)\right ) \sec ^2(c+d x) \, dx}{24 a}\\ &=\frac {(35 A+44 B+36 C) \left (a^4+a^4 \cos (c+d x)\right ) \tan (c+d x)}{12 d}+\frac {(7 A+8 B+4 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a (A+B) (a+a \cos (c+d x))^3 \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac {A (a+a \cos (c+d x))^4 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {\int (a+a \cos (c+d x)) \left (3 a^4 (35 A+48 B+52 C)-15 a^4 (7 A+8 B+4 C) \cos (c+d x)\right ) \sec (c+d x) \, dx}{24 a}\\ &=\frac {(35 A+44 B+36 C) \left (a^4+a^4 \cos (c+d x)\right ) \tan (c+d x)}{12 d}+\frac {(7 A+8 B+4 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a (A+B) (a+a \cos (c+d x))^3 \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac {A (a+a \cos (c+d x))^4 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {\int \left (3 a^5 (35 A+48 B+52 C)+\left (-15 a^5 (7 A+8 B+4 C)+3 a^5 (35 A+48 B+52 C)\right ) \cos (c+d x)-15 a^5 (7 A+8 B+4 C) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx}{24 a}\\ &=-\frac {5 a^4 (7 A+8 B+4 C) \sin (c+d x)}{8 d}+\frac {(35 A+44 B+36 C) \left (a^4+a^4 \cos (c+d x)\right ) \tan (c+d x)}{12 d}+\frac {(7 A+8 B+4 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a (A+B) (a+a \cos (c+d x))^3 \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac {A (a+a \cos (c+d x))^4 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {\int \left (3 a^5 (35 A+48 B+52 C)+24 a^5 (B+4 C) \cos (c+d x)\right ) \sec (c+d x) \, dx}{24 a}\\ &=a^4 (B+4 C) x-\frac {5 a^4 (7 A+8 B+4 C) \sin (c+d x)}{8 d}+\frac {(35 A+44 B+36 C) \left (a^4+a^4 \cos (c+d x)\right ) \tan (c+d x)}{12 d}+\frac {(7 A+8 B+4 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a (A+B) (a+a \cos (c+d x))^3 \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac {A (a+a \cos (c+d x))^4 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {1}{8} \left (a^4 (35 A+48 B+52 C)\right ) \int \sec (c+d x) \, dx\\ &=a^4 (B+4 C) x+\frac {a^4 (35 A+48 B+52 C) \tanh ^{-1}(\sin (c+d x))}{8 d}-\frac {5 a^4 (7 A+8 B+4 C) \sin (c+d x)}{8 d}+\frac {(35 A+44 B+36 C) \left (a^4+a^4 \cos (c+d x)\right ) \tan (c+d x)}{12 d}+\frac {(7 A+8 B+4 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a (A+B) (a+a \cos (c+d x))^3 \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac {A (a+a \cos (c+d x))^4 \sec ^3(c+d x) \tan (c+d x)}{4 d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [B]  time = 6.20, size = 838, normalized size = 3.86 \[ \frac {(B+4 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 {(-35 A-48 B-52 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 {(35 A+48 B+52 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 {(\cos (c+d x) a+a)^4 \left (4 A \sin \left (\frac {1}{2} (c+d x)\right )+B \sin \left (\frac {1}{2} (c+d x)\right )\right ) \sec ^8\left (\frac {c}{2}+\frac {d x}{2}\right )}{96 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 (4 A \sin \left (\frac {1}{2} (c+d x)\right )+B \sin \left (\frac {1}{2} (c+d x)\right )\right ) \sec ^8\left (\frac {c}{2}+\frac {d x}{2}\right )}{96 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 (5 A \sin \left (\frac {1}{2} (c+d x)\right )+5 B \sin \left (\frac {1}{2} (c+d x)\right )+3 C \sin \left (\frac {1}{2} (c+d x)\right )\right ) \sec ^8\left (\frac {c}{2}+\frac {d x}{2}\right )}{12 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 (5 A \sin \left (\frac {1}{2} (c+d x)\right )+5 B \sin \left (\frac {1}{2} (c+d x)\right )+3 C \sin \left (\frac {1}{2} (c+d x)\right )\right ) \sec ^8\left (\frac {c}{2}+\frac {d x}{2}\right )}{12 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {C (\cos (c+d x) a+a)^4 \sin (c+d x) \sec ^8\left (\frac {c}{2}+\frac {d x}{2}\right )}{16 d}+\frac {(97 A+52 B+12 C) (\cos (c+d x) a+a)^4 \sec ^8\left (\frac {c}{2}+\frac {d x}{2}\right )}{768 d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {(-97 A-52 B-12 C) (\cos (c+d x) a+a)^4 \sec ^8\left (\frac {c}{2}+\frac {d x}{2}\right )}{768 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {A (\cos (c+d x) a+a)^4 \sec ^8\left (\frac {c}{2}+\frac {d x}{2}\right )}{256 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 \sec ^8\left (\frac {c}{2}+\frac {d x}{2}\right )}{256 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^4} \]

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]^5,x]

[Out]

((B + 4*C)*(c + d*x)*(a + a*Cos[c + d*x])^4*Sec[c/2 + (d*x)/2]^8)/(16*d) + ((-35*A - 48*B - 52*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) + ((35*A + 48*B + 52*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) + (A*(a + a*Cos[c +
d*x])^4*Sec[c/2 + (d*x)/2]^8)/(256*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^4) + ((97*A + 52*B + 12*C)*(a + a*C
os[c + d*x])^4*Sec[c/2 + (d*x)/2]^8)/(768*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)/(256*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^4) + ((-97*A - 52*B - 12*C)*(a + a*Cos[c
+ d*x])^4*Sec[c/2 + (d*x)/2]^8)/(768*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*(4*A*Sin[(c + d*x)/2] + B*Sin[(c + d*x)/2]))/(96*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*(4*A*Sin[(c + d*x)/2] + B*Sin[(c + d*x)/2]))/(96*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*(5*A*Sin[(c + d*x)/2] + 5*B*Sin[(c
+ d*x)/2] + 3*C*Sin[(c + d*x)/2]))/(12*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*(5*A*Sin[(c + d*x)/2] + 5*B*Sin[(c + d*x)/2] + 3*C*Sin[(c + d*x)/2]))/(12*d*(Cos[(c + d*x)/2
] + Sin[(c + d*x)/2])) + (C*(a + a*Cos[c + d*x])^4*Sec[c/2 + (d*x)/2]^8*Sin[c + d*x])/(16*d)

________________________________________________________________________________________

fricas [A]  time = 0.46, size = 191, normalized size = 0.88 \[ \frac {48 \, {\left (B + 4 \, C\right )} a^{4} d x \cos \left (d x + c\right )^{4} + 3 \, {\left (35 \, A + 48 \, B + 52 \, C\right )} a^{4} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (35 \, A + 48 \, B + 52 \, C\right )} a^{4} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (24 \, C a^{4} \cos \left (d x + c\right )^{4} + 32 \, {\left (5 \, A + 5 \, B + 3 \, C\right )} a^{4} \cos \left (d x + c\right )^{3} + 3 \, {\left (27 \, A + 16 \, B + 4 \, C\right )} a^{4} \cos \left (d x + c\right )^{2} + 8 \, {\left (4 \, A + B\right )} a^{4} \cos \left (d x + c\right ) + 6 \, A a^{4}\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )^{4}} \]

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)^5,x, algorithm="fricas")

[Out]

1/48*(48*(B + 4*C)*a^4*d*x*cos(d*x + c)^4 + 3*(35*A + 48*B + 52*C)*a^4*cos(d*x + c)^4*log(sin(d*x + c) + 1) -
3*(35*A + 48*B + 52*C)*a^4*cos(d*x + c)^4*log(-sin(d*x + c) + 1) + 2*(24*C*a^4*cos(d*x + c)^4 + 32*(5*A + 5*B
+ 3*C)*a^4*cos(d*x + c)^3 + 3*(27*A + 16*B + 4*C)*a^4*cos(d*x + c)^2 + 8*(4*A + B)*a^4*cos(d*x + c) + 6*A*a^4)
*sin(d*x + c))/(d*cos(d*x + c)^4)

________________________________________________________________________________________

giac [A]  time = 0.70, size = 339, normalized size = 1.56 \[ \frac {\frac {48 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1} + 24 \, {\left (B a^{4} + 4 \, C a^{4}\right )} {\left (d x + c\right )} + 3 \, {\left (35 \, A a^{4} + 48 \, B a^{4} + 52 \, C a^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (35 \, A a^{4} + 48 \, B a^{4} + 52 \, C a^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (105 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 120 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 84 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 385 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 424 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 276 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 511 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 520 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 300 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 279 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 216 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 108 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{4}}}{24 \, d} \]

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)^5,x, algorithm="giac")

[Out]

1/24*(48*C*a^4*tan(1/2*d*x + 1/2*c)/(tan(1/2*d*x + 1/2*c)^2 + 1) + 24*(B*a^4 + 4*C*a^4)*(d*x + c) + 3*(35*A*a^
4 + 48*B*a^4 + 52*C*a^4)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(35*A*a^4 + 48*B*a^4 + 52*C*a^4)*log(abs(tan(1
/2*d*x + 1/2*c) - 1)) - 2*(105*A*a^4*tan(1/2*d*x + 1/2*c)^7 + 120*B*a^4*tan(1/2*d*x + 1/2*c)^7 + 84*C*a^4*tan(
1/2*d*x + 1/2*c)^7 - 385*A*a^4*tan(1/2*d*x + 1/2*c)^5 - 424*B*a^4*tan(1/2*d*x + 1/2*c)^5 - 276*C*a^4*tan(1/2*d
*x + 1/2*c)^5 + 511*A*a^4*tan(1/2*d*x + 1/2*c)^3 + 520*B*a^4*tan(1/2*d*x + 1/2*c)^3 + 300*C*a^4*tan(1/2*d*x +
1/2*c)^3 - 279*A*a^4*tan(1/2*d*x + 1/2*c) - 216*B*a^4*tan(1/2*d*x + 1/2*c) - 108*C*a^4*tan(1/2*d*x + 1/2*c))/(
tan(1/2*d*x + 1/2*c)^2 - 1)^4)/d

________________________________________________________________________________________

maple [A]  time = 0.48, size = 294, normalized size = 1.35 \[ \frac {35 A \,a^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d}+a^{4} B x +\frac {a^{4} B c}{d}+\frac {a^{4} C \sin \left (d x +c \right )}{d}+\frac {20 A \,a^{4} \tan \left (d x +c \right )}{3 d}+\frac {6 a^{4} B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+4 a^{4} C x +\frac {4 a^{4} C c}{d}+\frac {27 A \,a^{4} \sec \left (d x +c \right ) \tan \left (d x +c \right )}{8 d}+\frac {20 a^{4} B \tan \left (d x +c \right )}{3 d}+\frac {13 a^{4} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {4 A \,a^{4} \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{3 d}+\frac {2 a^{4} B \sec \left (d x +c \right ) \tan \left (d x +c \right )}{d}+\frac {4 a^{4} C \tan \left (d x +c \right )}{d}+\frac {A \,a^{4} \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )}{4 d}+\frac {a^{4} B \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{3 d}+\frac {a^{4} C \sec \left (d x +c \right ) \tan \left (d x +c \right )}{2 d} \]

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)^5,x)

[Out]

35/8/d*A*a^4*ln(sec(d*x+c)+tan(d*x+c))+a^4*B*x+1/d*a^4*B*c+1/d*a^4*C*sin(d*x+c)+20/3/d*A*a^4*tan(d*x+c)+6/d*a^
4*B*ln(sec(d*x+c)+tan(d*x+c))+4*a^4*C*x+4/d*a^4*C*c+27/8/d*A*a^4*sec(d*x+c)*tan(d*x+c)+20/3/d*a^4*B*tan(d*x+c)
+13/2/d*a^4*C*ln(sec(d*x+c)+tan(d*x+c))+4/3/d*A*a^4*tan(d*x+c)*sec(d*x+c)^2+2/d*a^4*B*sec(d*x+c)*tan(d*x+c)+4/
d*a^4*C*tan(d*x+c)+1/4/d*A*a^4*tan(d*x+c)*sec(d*x+c)^3+1/3/d*a^4*B*tan(d*x+c)*sec(d*x+c)^2+1/2/d*a^4*C*sec(d*x
+c)*tan(d*x+c)

________________________________________________________________________________________

maxima [B]  time = 0.38, size = 416, normalized size = 1.92 \[ \frac {64 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a^{4} + 16 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a^{4} + 48 \, {\left (d x + c\right )} B a^{4} + 192 \, {\left (d x + c\right )} C a^{4} - 3 \, A a^{4} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 72 \, A a^{4} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 48 \, B a^{4} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 12 \, C a^{4} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 24 \, A a^{4} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 96 \, B a^{4} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 144 \, C a^{4} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 48 \, C a^{4} \sin \left (d x + c\right ) + 192 \, A a^{4} \tan \left (d x + c\right ) + 288 \, B a^{4} \tan \left (d x + c\right ) + 192 \, C a^{4} \tan \left (d x + c\right )}{48 \, d} \]

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)^5,x, algorithm="maxima")

[Out]

1/48*(64*(tan(d*x + c)^3 + 3*tan(d*x + c))*A*a^4 + 16*(tan(d*x + c)^3 + 3*tan(d*x + c))*B*a^4 + 48*(d*x + c)*B
*a^4 + 192*(d*x + c)*C*a^4 - 3*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)) - 72*A*a^4*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - l
og(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) - 48*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)) - 12*C*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)) + 24*A*a^4*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 96*B*a^4*(log(sin(d*x + c) +
1) - log(sin(d*x + c) - 1)) + 144*C*a^4*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 48*C*a^4*sin(d*x + c
) + 192*A*a^4*tan(d*x + c) + 288*B*a^4*tan(d*x + c) + 192*C*a^4*tan(d*x + c))/d

________________________________________________________________________________________

mupad [B]  time = 2.97, size = 1342, normalized size = 6.18 \[ \text {result too large to display} \]

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)^5,x)

[Out]

((105*A*a^4*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/32 + (9*B*a^4*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*
x)/2)))/2 + (39*C*a^4*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/8 + (7*A*a^4*sin(2*c + 2*d*x))/3 + (27*A*a
^4*sin(3*c + 3*d*x))/32 + (5*A*a^4*sin(4*c + 4*d*x))/6 + (11*B*a^4*sin(2*c + 2*d*x))/6 + (B*a^4*sin(3*c + 3*d*
x))/2 + (5*B*a^4*sin(4*c + 4*d*x))/6 + C*a^4*sin(2*c + 2*d*x) + (5*C*a^4*sin(3*c + 3*d*x))/16 + (C*a^4*sin(4*c
+ 4*d*x))/2 + (C*a^4*sin(5*c + 5*d*x))/16 + (3*B*a^4*atan((1225*A^2*sin(c/2 + (d*x)/2) + 2368*B^2*sin(c/2 + (
d*x)/2) + 3728*C^2*sin(c/2 + (d*x)/2) + 3360*A*B*sin(c/2 + (d*x)/2) + 3640*A*C*sin(c/2 + (d*x)/2) + 5504*B*C*s
in(c/2 + (d*x)/2))/(cos(c/2 + (d*x)/2)*(1225*A^2 + 2368*B^2 + 3728*C^2 + 3360*A*B + 3640*A*C + 5504*B*C))))/4
+ 3*C*a^4*atan((1225*A^2*sin(c/2 + (d*x)/2) + 2368*B^2*sin(c/2 + (d*x)/2) + 3728*C^2*sin(c/2 + (d*x)/2) + 3360
*A*B*sin(c/2 + (d*x)/2) + 3640*A*C*sin(c/2 + (d*x)/2) + 5504*B*C*sin(c/2 + (d*x)/2))/(cos(c/2 + (d*x)/2)*(1225
*A^2 + 2368*B^2 + 3728*C^2 + 3360*A*B + 3640*A*C + 5504*B*C))) + (35*A*a^4*sin(c + d*x))/32 + (B*a^4*sin(c + d
*x))/2 + (C*a^4*sin(c + d*x))/4 + B*a^4*atan((1225*A^2*sin(c/2 + (d*x)/2) + 2368*B^2*sin(c/2 + (d*x)/2) + 3728
*C^2*sin(c/2 + (d*x)/2) + 3360*A*B*sin(c/2 + (d*x)/2) + 3640*A*C*sin(c/2 + (d*x)/2) + 5504*B*C*sin(c/2 + (d*x)
/2))/(cos(c/2 + (d*x)/2)*(1225*A^2 + 2368*B^2 + 3728*C^2 + 3360*A*B + 3640*A*C + 5504*B*C)))*cos(2*c + 2*d*x)
+ (B*a^4*atan((1225*A^2*sin(c/2 + (d*x)/2) + 2368*B^2*sin(c/2 + (d*x)/2) + 3728*C^2*sin(c/2 + (d*x)/2) + 3360*
A*B*sin(c/2 + (d*x)/2) + 3640*A*C*sin(c/2 + (d*x)/2) + 5504*B*C*sin(c/2 + (d*x)/2))/(cos(c/2 + (d*x)/2)*(1225*
A^2 + 2368*B^2 + 3728*C^2 + 3360*A*B + 3640*A*C + 5504*B*C)))*cos(4*c + 4*d*x))/4 + 4*C*a^4*atan((1225*A^2*sin
(c/2 + (d*x)/2) + 2368*B^2*sin(c/2 + (d*x)/2) + 3728*C^2*sin(c/2 + (d*x)/2) + 3360*A*B*sin(c/2 + (d*x)/2) + 36
40*A*C*sin(c/2 + (d*x)/2) + 5504*B*C*sin(c/2 + (d*x)/2))/(cos(c/2 + (d*x)/2)*(1225*A^2 + 2368*B^2 + 3728*C^2 +
3360*A*B + 3640*A*C + 5504*B*C)))*cos(2*c + 2*d*x) + C*a^4*atan((1225*A^2*sin(c/2 + (d*x)/2) + 2368*B^2*sin(c
/2 + (d*x)/2) + 3728*C^2*sin(c/2 + (d*x)/2) + 3360*A*B*sin(c/2 + (d*x)/2) + 3640*A*C*sin(c/2 + (d*x)/2) + 5504
*B*C*sin(c/2 + (d*x)/2))/(cos(c/2 + (d*x)/2)*(1225*A^2 + 2368*B^2 + 3728*C^2 + 3360*A*B + 3640*A*C + 5504*B*C)
))*cos(4*c + 4*d*x) + (35*A*a^4*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(2*c + 2*d*x))/8 + (35*A*a^4*a
tanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(4*c + 4*d*x))/32 + 6*B*a^4*atanh(sin(c/2 + (d*x)/2)/cos(c/2 +
(d*x)/2))*cos(2*c + 2*d*x) + (3*B*a^4*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(4*c + 4*d*x))/2 + (13*C
*a^4*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(2*c + 2*d*x))/2 + (13*C*a^4*atanh(sin(c/2 + (d*x)/2)/cos
(c/2 + (d*x)/2))*cos(4*c + 4*d*x))/8)/(d*(cos(2*c + 2*d*x)/2 + cos(4*c + 4*d*x)/8 + 3/8))

________________________________________________________________________________________

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)**5,x)

[Out]

Timed out

________________________________________________________________________________________