3.338 \(\int (a+a \cos (c+d x))^4 (A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^8(c+d x) \, dx\)

Optimal. Leaf size=287 \[ \frac{a^4 (454 A+504 B+581 C) \tan (c+d x)}{105 d}+\frac{a^4 (44 A+49 B+56 C) \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac{a^4 (988 A+1113 B+1232 C) \tan (c+d x) \sec ^2(c+d x)}{840 d}+\frac{a^4 (44 A+49 B+56 C) \tan (c+d x) \sec (c+d x)}{16 d}+\frac{(16 A+21 B+14 C) \tan (c+d x) \sec ^4(c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{70 d}+\frac{(436 A+511 B+504 C) \tan (c+d x) \sec ^3(c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{840 d}+\frac{a (4 A+7 B) \tan (c+d x) \sec ^5(c+d x) (a \cos (c+d x)+a)^3}{42 d}+\frac{A \tan (c+d x) \sec ^6(c+d x) (a \cos (c+d x)+a)^4}{7 d} \]

[Out]

(a^4*(44*A + 49*B + 56*C)*ArcTanh[Sin[c + d*x]])/(16*d) + (a^4*(454*A + 504*B + 581*C)*Tan[c + d*x])/(105*d) +
(a^4*(44*A + 49*B + 56*C)*Sec[c + d*x]*Tan[c + d*x])/(16*d) + (a^4*(988*A + 1113*B + 1232*C)*Sec[c + d*x]^2*T
an[c + d*x])/(840*d) + ((436*A + 511*B + 504*C)*(a^4 + a^4*Cos[c + d*x])*Sec[c + d*x]^3*Tan[c + d*x])/(840*d)
+ ((16*A + 21*B + 14*C)*(a^2 + a^2*Cos[c + d*x])^2*Sec[c + d*x]^4*Tan[c + d*x])/(70*d) + (a*(4*A + 7*B)*(a + a
*Cos[c + d*x])^3*Sec[c + d*x]^5*Tan[c + d*x])/(42*d) + (A*(a + a*Cos[c + d*x])^4*Sec[c + d*x]^6*Tan[c + d*x])/
(7*d)

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Rubi [A]  time = 0.868061, antiderivative size = 287, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 9, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.22, Rules used = {3043, 2975, 2968, 3021, 2748, 3768, 3770, 3767, 8} \[ \frac{a^4 (454 A+504 B+581 C) \tan (c+d x)}{105 d}+\frac{a^4 (44 A+49 B+56 C) \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac{a^4 (988 A+1113 B+1232 C) \tan (c+d x) \sec ^2(c+d x)}{840 d}+\frac{a^4 (44 A+49 B+56 C) \tan (c+d x) \sec (c+d x)}{16 d}+\frac{(16 A+21 B+14 C) \tan (c+d x) \sec ^4(c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{70 d}+\frac{(436 A+511 B+504 C) \tan (c+d x) \sec ^3(c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{840 d}+\frac{a (4 A+7 B) \tan (c+d x) \sec ^5(c+d x) (a \cos (c+d x)+a)^3}{42 d}+\frac{A \tan (c+d x) \sec ^6(c+d x) (a \cos (c+d x)+a)^4}{7 d} \]

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]^8,x]

[Out]

(a^4*(44*A + 49*B + 56*C)*ArcTanh[Sin[c + d*x]])/(16*d) + (a^4*(454*A + 504*B + 581*C)*Tan[c + d*x])/(105*d) +
(a^4*(44*A + 49*B + 56*C)*Sec[c + d*x]*Tan[c + d*x])/(16*d) + (a^4*(988*A + 1113*B + 1232*C)*Sec[c + d*x]^2*T
an[c + d*x])/(840*d) + ((436*A + 511*B + 504*C)*(a^4 + a^4*Cos[c + d*x])*Sec[c + d*x]^3*Tan[c + d*x])/(840*d)
+ ((16*A + 21*B + 14*C)*(a^2 + a^2*Cos[c + d*x])^2*Sec[c + d*x]^4*Tan[c + d*x])/(70*d) + (a*(4*A + 7*B)*(a + a
*Cos[c + d*x])^3*Sec[c + d*x]^5*Tan[c + d*x])/(42*d) + (A*(a + a*Cos[c + d*x])^4*Sec[c + d*x]^6*Tan[c + d*x])/
(7*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 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 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]

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 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, 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 ^8(c+d x) \, dx &=\frac{A (a+a \cos (c+d x))^4 \sec ^6(c+d x) \tan (c+d x)}{7 d}+\frac{\int (a+a \cos (c+d x))^4 (a (4 A+7 B)+a (2 A+7 C) \cos (c+d x)) \sec ^7(c+d x) \, dx}{7 a}\\ &=\frac{a (4 A+7 B) (a+a \cos (c+d x))^3 \sec ^5(c+d x) \tan (c+d x)}{42 d}+\frac{A (a+a \cos (c+d x))^4 \sec ^6(c+d x) \tan (c+d x)}{7 d}+\frac{\int (a+a \cos (c+d x))^3 \left (3 a^2 (16 A+21 B+14 C)+2 a^2 (10 A+7 B+21 C) \cos (c+d x)\right ) \sec ^6(c+d x) \, dx}{42 a}\\ &=\frac{(16 A+21 B+14 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^4(c+d x) \tan (c+d x)}{70 d}+\frac{a (4 A+7 B) (a+a \cos (c+d x))^3 \sec ^5(c+d x) \tan (c+d x)}{42 d}+\frac{A (a+a \cos (c+d x))^4 \sec ^6(c+d x) \tan (c+d x)}{7 d}+\frac{\int (a+a \cos (c+d x))^2 \left (a^3 (436 A+511 B+504 C)+98 a^3 (2 A+2 B+3 C) \cos (c+d x)\right ) \sec ^5(c+d x) \, dx}{210 a}\\ &=\frac{(436 A+511 B+504 C) \left (a^4+a^4 \cos (c+d x)\right ) \sec ^3(c+d x) \tan (c+d x)}{840 d}+\frac{(16 A+21 B+14 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^4(c+d x) \tan (c+d x)}{70 d}+\frac{a (4 A+7 B) (a+a \cos (c+d x))^3 \sec ^5(c+d x) \tan (c+d x)}{42 d}+\frac{A (a+a \cos (c+d x))^4 \sec ^6(c+d x) \tan (c+d x)}{7 d}+\frac{\int (a+a \cos (c+d x)) \left (3 a^4 (988 A+1113 B+1232 C)+6 a^4 (276 A+301 B+364 C) \cos (c+d x)\right ) \sec ^4(c+d x) \, dx}{840 a}\\ &=\frac{(436 A+511 B+504 C) \left (a^4+a^4 \cos (c+d x)\right ) \sec ^3(c+d x) \tan (c+d x)}{840 d}+\frac{(16 A+21 B+14 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^4(c+d x) \tan (c+d x)}{70 d}+\frac{a (4 A+7 B) (a+a \cos (c+d x))^3 \sec ^5(c+d x) \tan (c+d x)}{42 d}+\frac{A (a+a \cos (c+d x))^4 \sec ^6(c+d x) \tan (c+d x)}{7 d}+\frac{\int \left (3 a^5 (988 A+1113 B+1232 C)+\left (6 a^5 (276 A+301 B+364 C)+3 a^5 (988 A+1113 B+1232 C)\right ) \cos (c+d x)+6 a^5 (276 A+301 B+364 C) \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx}{840 a}\\ &=\frac{a^4 (988 A+1113 B+1232 C) \sec ^2(c+d x) \tan (c+d x)}{840 d}+\frac{(436 A+511 B+504 C) \left (a^4+a^4 \cos (c+d x)\right ) \sec ^3(c+d x) \tan (c+d x)}{840 d}+\frac{(16 A+21 B+14 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^4(c+d x) \tan (c+d x)}{70 d}+\frac{a (4 A+7 B) (a+a \cos (c+d x))^3 \sec ^5(c+d x) \tan (c+d x)}{42 d}+\frac{A (a+a \cos (c+d x))^4 \sec ^6(c+d x) \tan (c+d x)}{7 d}+\frac{\int \left (315 a^5 (44 A+49 B+56 C)+24 a^5 (454 A+504 B+581 C) \cos (c+d x)\right ) \sec ^3(c+d x) \, dx}{2520 a}\\ &=\frac{a^4 (988 A+1113 B+1232 C) \sec ^2(c+d x) \tan (c+d x)}{840 d}+\frac{(436 A+511 B+504 C) \left (a^4+a^4 \cos (c+d x)\right ) \sec ^3(c+d x) \tan (c+d x)}{840 d}+\frac{(16 A+21 B+14 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^4(c+d x) \tan (c+d x)}{70 d}+\frac{a (4 A+7 B) (a+a \cos (c+d x))^3 \sec ^5(c+d x) \tan (c+d x)}{42 d}+\frac{A (a+a \cos (c+d x))^4 \sec ^6(c+d x) \tan (c+d x)}{7 d}+\frac{1}{8} \left (a^4 (44 A+49 B+56 C)\right ) \int \sec ^3(c+d x) \, dx+\frac{1}{105} \left (a^4 (454 A+504 B+581 C)\right ) \int \sec ^2(c+d x) \, dx\\ &=\frac{a^4 (44 A+49 B+56 C) \sec (c+d x) \tan (c+d x)}{16 d}+\frac{a^4 (988 A+1113 B+1232 C) \sec ^2(c+d x) \tan (c+d x)}{840 d}+\frac{(436 A+511 B+504 C) \left (a^4+a^4 \cos (c+d x)\right ) \sec ^3(c+d x) \tan (c+d x)}{840 d}+\frac{(16 A+21 B+14 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^4(c+d x) \tan (c+d x)}{70 d}+\frac{a (4 A+7 B) (a+a \cos (c+d x))^3 \sec ^5(c+d x) \tan (c+d x)}{42 d}+\frac{A (a+a \cos (c+d x))^4 \sec ^6(c+d x) \tan (c+d x)}{7 d}+\frac{1}{16} \left (a^4 (44 A+49 B+56 C)\right ) \int \sec (c+d x) \, dx-\frac{\left (a^4 (454 A+504 B+581 C)\right ) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{105 d}\\ &=\frac{a^4 (44 A+49 B+56 C) \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac{a^4 (454 A+504 B+581 C) \tan (c+d x)}{105 d}+\frac{a^4 (44 A+49 B+56 C) \sec (c+d x) \tan (c+d x)}{16 d}+\frac{a^4 (988 A+1113 B+1232 C) \sec ^2(c+d x) \tan (c+d x)}{840 d}+\frac{(436 A+511 B+504 C) \left (a^4+a^4 \cos (c+d x)\right ) \sec ^3(c+d x) \tan (c+d x)}{840 d}+\frac{(16 A+21 B+14 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^4(c+d x) \tan (c+d x)}{70 d}+\frac{a (4 A+7 B) (a+a \cos (c+d x))^3 \sec ^5(c+d x) \tan (c+d x)}{42 d}+\frac{A (a+a \cos (c+d x))^4 \sec ^6(c+d x) \tan (c+d x)}{7 d}\\ \end{align*}

Mathematica [A]  time = 3.44124, size = 298, normalized size = 1.04 \[ -\frac{a^4 (\cos (c+d x)+1)^4 \sec ^8\left (\frac{1}{2} (c+d x)\right ) \sec ^7(c+d x) \left (3360 (44 A+49 B+56 C) \cos ^7(c+d x) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )-2 \sin (c+d x) (70 (1444 A+1291 B+1128 C) \cos (c+d x)+8 (12746 A+12936 B+12859 C) \cos (2 (c+d x))+35420 A \cos (3 (c+d x))+29056 A \cos (4 (c+d x))+4620 A \cos (5 (c+d x))+3632 A \cos (6 (c+d x))+80384 A+37205 B \cos (3 (c+d x))+32256 B \cos (4 (c+d x))+5145 B \cos (5 (c+d x))+4032 B \cos (6 (c+d x))+75264 B+36120 C \cos (3 (c+d x))+35504 C \cos (4 (c+d x))+5880 C \cos (5 (c+d x))+4648 C \cos (6 (c+d x))+72016 C)\right )}{860160 d} \]

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]^8,x]

[Out]

-(a^4*(1 + Cos[c + d*x])^4*Sec[(c + d*x)/2]^8*Sec[c + d*x]^7*(3360*(44*A + 49*B + 56*C)*Cos[c + d*x]^7*(Log[Co
s[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) - 2*(80384*A + 75264*B + 72016*
C + 70*(1444*A + 1291*B + 1128*C)*Cos[c + d*x] + 8*(12746*A + 12936*B + 12859*C)*Cos[2*(c + d*x)] + 35420*A*Co
s[3*(c + d*x)] + 37205*B*Cos[3*(c + d*x)] + 36120*C*Cos[3*(c + d*x)] + 29056*A*Cos[4*(c + d*x)] + 32256*B*Cos[
4*(c + d*x)] + 35504*C*Cos[4*(c + d*x)] + 4620*A*Cos[5*(c + d*x)] + 5145*B*Cos[5*(c + d*x)] + 5880*C*Cos[5*(c
+ d*x)] + 3632*A*Cos[6*(c + d*x)] + 4032*B*Cos[6*(c + d*x)] + 4648*C*Cos[6*(c + d*x)])*Sin[c + d*x]))/(860160*
d)

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Maple [A]  time = 0.114, size = 454, normalized size = 1.6 \begin{align*}{\frac{454\,A{a}^{4}\tan \left ( dx+c \right ) }{105\,d}}+{\frac{A{a}^{4}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{6}}{7\,d}}+{\frac{48\,A{a}^{4}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{4}}{35\,d}}+{\frac{227\,A{a}^{4}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{105\,d}}+{\frac{{a}^{4}C\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{4}}{5\,d}}+{\frac{34\,{a}^{4}C\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{15\,d}}+{\frac{24\,{a}^{4}B\tan \left ( dx+c \right ) }{5\,d}}+{\frac{4\,{a}^{4}B\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{4}}{5\,d}}+{\frac{12\,{a}^{4}B\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{5\,d}}+{\frac{{a}^{4}B\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{5}}{6\,d}}+{\frac{41\,{a}^{4}B\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{24\,d}}+{\frac{49\,{a}^{4}B\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{16\,d}}+{\frac{2\,A{a}^{4}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{5}}{3\,d}}+{\frac{11\,A{a}^{4}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{6\,d}}+{\frac{11\,A{a}^{4}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{4\,d}}+{\frac{{a}^{4}C\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{d}}+{\frac{7\,{a}^{4}C\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{49\,{a}^{4}B\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{16\,d}}+{\frac{11\,A{a}^{4}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{4\,d}}+{\frac{7\,{a}^{4}C\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+{\frac{83\,{a}^{4}C\tan \left ( dx+c \right ) }{15\,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)^8,x)

[Out]

454/105/d*A*a^4*tan(d*x+c)+1/7/d*A*a^4*tan(d*x+c)*sec(d*x+c)^6+48/35/d*A*a^4*tan(d*x+c)*sec(d*x+c)^4+227/105/d
*A*a^4*tan(d*x+c)*sec(d*x+c)^2+1/5/d*a^4*C*tan(d*x+c)*sec(d*x+c)^4+34/15/d*a^4*C*tan(d*x+c)*sec(d*x+c)^2+24/5/
d*a^4*B*tan(d*x+c)+4/5/d*a^4*B*tan(d*x+c)*sec(d*x+c)^4+12/5/d*a^4*B*tan(d*x+c)*sec(d*x+c)^2+1/6/d*a^4*B*tan(d*
x+c)*sec(d*x+c)^5+41/24/d*a^4*B*tan(d*x+c)*sec(d*x+c)^3+49/16/d*a^4*B*sec(d*x+c)*tan(d*x+c)+2/3/d*A*a^4*tan(d*
x+c)*sec(d*x+c)^5+11/6/d*A*a^4*tan(d*x+c)*sec(d*x+c)^3+11/4/d*A*a^4*sec(d*x+c)*tan(d*x+c)+1/d*a^4*C*tan(d*x+c)
*sec(d*x+c)^3+7/2/d*a^4*C*sec(d*x+c)*tan(d*x+c)+49/16/d*a^4*B*ln(sec(d*x+c)+tan(d*x+c))+11/4/d*A*a^4*ln(sec(d*
x+c)+tan(d*x+c))+7/2/d*a^4*C*ln(sec(d*x+c)+tan(d*x+c))+83/15/d*a^4*C*tan(d*x+c)

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Maxima [B]  time = 1.0551, size = 987, normalized size = 3.44 \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)^8,x, algorithm="maxima")

[Out]

1/3360*(96*(5*tan(d*x + c)^7 + 21*tan(d*x + c)^5 + 35*tan(d*x + c)^3 + 35*tan(d*x + c))*A*a^4 + 1344*(3*tan(d*
x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c))*A*a^4 + 1120*(tan(d*x + c)^3 + 3*tan(d*x + c))*A*a^4 + 896*(3*
tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c))*B*a^4 + 4480*(tan(d*x + c)^3 + 3*tan(d*x + c))*B*a^4 + 2
24*(3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c))*C*a^4 + 6720*(tan(d*x + c)^3 + 3*tan(d*x + c))*C*a
^4 - 140*A*a^4*(2*(15*sin(d*x + c)^5 - 40*sin(d*x + c)^3 + 33*sin(d*x + c))/(sin(d*x + c)^6 - 3*sin(d*x + c)^4
+ 3*sin(d*x + c)^2 - 1) - 15*log(sin(d*x + c) + 1) + 15*log(sin(d*x + c) - 1)) - 35*B*a^4*(2*(15*sin(d*x + c)
^5 - 40*sin(d*x + c)^3 + 33*sin(d*x + c))/(sin(d*x + c)^6 - 3*sin(d*x + c)^4 + 3*sin(d*x + c)^2 - 1) - 15*log(
sin(d*x + c) + 1) + 15*log(sin(d*x + c) - 1)) - 840*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)) - 1260*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)) - 840*C*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)) - 840*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)) - 3360*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)) + 3360*C*a^4*tan(d*x + c))/d

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Fricas [A]  time = 2.09729, size = 617, normalized size = 2.15 \begin{align*} \frac{105 \,{\left (44 \, A + 49 \, B + 56 \, C\right )} a^{4} \cos \left (d x + c\right )^{7} \log \left (\sin \left (d x + c\right ) + 1\right ) - 105 \,{\left (44 \, A + 49 \, B + 56 \, C\right )} a^{4} \cos \left (d x + c\right )^{7} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (16 \,{\left (454 \, A + 504 \, B + 581 \, C\right )} a^{4} \cos \left (d x + c\right )^{6} + 105 \,{\left (44 \, A + 49 \, B + 56 \, C\right )} a^{4} \cos \left (d x + c\right )^{5} + 16 \,{\left (227 \, A + 252 \, B + 238 \, C\right )} a^{4} \cos \left (d x + c\right )^{4} + 70 \,{\left (44 \, A + 41 \, B + 24 \, C\right )} a^{4} \cos \left (d x + c\right )^{3} + 48 \,{\left (48 \, A + 28 \, B + 7 \, C\right )} a^{4} \cos \left (d x + c\right )^{2} + 280 \,{\left (4 \, A + B\right )} a^{4} \cos \left (d x + c\right ) + 240 \, A a^{4}\right )} \sin \left (d x + c\right )}{3360 \, d \cos \left (d x + c\right )^{7}} \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)^8,x, algorithm="fricas")

[Out]

1/3360*(105*(44*A + 49*B + 56*C)*a^4*cos(d*x + c)^7*log(sin(d*x + c) + 1) - 105*(44*A + 49*B + 56*C)*a^4*cos(d
*x + c)^7*log(-sin(d*x + c) + 1) + 2*(16*(454*A + 504*B + 581*C)*a^4*cos(d*x + c)^6 + 105*(44*A + 49*B + 56*C)
*a^4*cos(d*x + c)^5 + 16*(227*A + 252*B + 238*C)*a^4*cos(d*x + c)^4 + 70*(44*A + 41*B + 24*C)*a^4*cos(d*x + c)
^3 + 48*(48*A + 28*B + 7*C)*a^4*cos(d*x + c)^2 + 280*(4*A + B)*a^4*cos(d*x + c) + 240*A*a^4)*sin(d*x + c))/(d*
cos(d*x + c)^7)

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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)**8,x)

[Out]

Timed out

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Giac [A]  time = 1.28688, size = 598, normalized size = 2.08 \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)^8,x, algorithm="giac")

[Out]

1/1680*(105*(44*A*a^4 + 49*B*a^4 + 56*C*a^4)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 105*(44*A*a^4 + 49*B*a^4 + 5
6*C*a^4)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(4620*A*a^4*tan(1/2*d*x + 1/2*c)^13 + 5145*B*a^4*tan(1/2*d*x +
1/2*c)^13 + 5880*C*a^4*tan(1/2*d*x + 1/2*c)^13 - 30800*A*a^4*tan(1/2*d*x + 1/2*c)^11 - 34300*B*a^4*tan(1/2*d*
x + 1/2*c)^11 - 39200*C*a^4*tan(1/2*d*x + 1/2*c)^11 + 87164*A*a^4*tan(1/2*d*x + 1/2*c)^9 + 97069*B*a^4*tan(1/2
*d*x + 1/2*c)^9 + 110936*C*a^4*tan(1/2*d*x + 1/2*c)^9 - 135168*A*a^4*tan(1/2*d*x + 1/2*c)^7 - 150528*B*a^4*tan
(1/2*d*x + 1/2*c)^7 - 172032*C*a^4*tan(1/2*d*x + 1/2*c)^7 + 126084*A*a^4*tan(1/2*d*x + 1/2*c)^5 + 134099*B*a^4
*tan(1/2*d*x + 1/2*c)^5 + 159656*C*a^4*tan(1/2*d*x + 1/2*c)^5 - 58800*A*a^4*tan(1/2*d*x + 1/2*c)^3 - 73220*B*a
^4*tan(1/2*d*x + 1/2*c)^3 - 86240*C*a^4*tan(1/2*d*x + 1/2*c)^3 + 22260*A*a^4*tan(1/2*d*x + 1/2*c) + 21735*B*a^
4*tan(1/2*d*x + 1/2*c) + 21000*C*a^4*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^7)/d