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

Optimal. Leaf size=355 \[ \frac {a b \left (a^2 (5 A+6 C)+2 b^2 (3 A+4 C)\right ) \tanh ^{-1}(\sin (c+d x))}{4 d}+\frac {a b \left (a^2 (103 A+126 C)+6 A b^2\right ) \tan (c+d x) \sec ^3(c+d x)}{210 d}+\frac {a b \left (a^2 (5 A+6 C)+2 b^2 (3 A+4 C)\right ) \tan (c+d x) \sec (c+d x)}{4 d}+\frac {\left (a^2 (6 A+7 C)+2 A b^2\right ) \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^2}{35 d}+\frac {\left (8 a^4 (6 A+7 C)+84 a^2 b^2 (4 A+5 C)+35 b^4 (2 A+3 C)\right ) \tan (c+d x)}{105 d}+\frac {\left (4 a^4 (6 A+7 C)+3 a^2 b^2 (50 A+63 C)+4 A b^4\right ) \tan (c+d x) \sec ^2(c+d x)}{105 d}+\frac {A \tan (c+d x) \sec ^6(c+d x) (a+b \cos (c+d x))^4}{7 d}+\frac {2 A b \tan (c+d x) \sec ^5(c+d x) (a+b \cos (c+d x))^3}{21 d} \]

[Out]

1/4*a*b*(2*b^2*(3*A+4*C)+a^2*(5*A+6*C))*arctanh(sin(d*x+c))/d+1/105*(35*b^4*(2*A+3*C)+84*a^2*b^2*(4*A+5*C)+8*a
^4*(6*A+7*C))*tan(d*x+c)/d+1/4*a*b*(2*b^2*(3*A+4*C)+a^2*(5*A+6*C))*sec(d*x+c)*tan(d*x+c)/d+1/105*(4*A*b^4+4*a^
4*(6*A+7*C)+3*a^2*b^2*(50*A+63*C))*sec(d*x+c)^2*tan(d*x+c)/d+1/210*a*b*(6*A*b^2+a^2*(103*A+126*C))*sec(d*x+c)^
3*tan(d*x+c)/d+1/35*(2*A*b^2+a^2*(6*A+7*C))*(a+b*cos(d*x+c))^2*sec(d*x+c)^4*tan(d*x+c)/d+2/21*A*b*(a+b*cos(d*x
+c))^3*sec(d*x+c)^5*tan(d*x+c)/d+1/7*A*(a+b*cos(d*x+c))^4*sec(d*x+c)^6*tan(d*x+c)/d

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Rubi [A]  time = 1.24, antiderivative size = 355, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 9, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {3048, 3047, 3031, 3021, 2748, 3768, 3770, 3767, 8} \[ \frac {\left (84 a^2 b^2 (4 A+5 C)+8 a^4 (6 A+7 C)+35 b^4 (2 A+3 C)\right ) \tan (c+d x)}{105 d}+\frac {a b \left (a^2 (5 A+6 C)+2 b^2 (3 A+4 C)\right ) \tanh ^{-1}(\sin (c+d x))}{4 d}+\frac {a b \left (a^2 (103 A+126 C)+6 A b^2\right ) \tan (c+d x) \sec ^3(c+d x)}{210 d}+\frac {\left (3 a^2 b^2 (50 A+63 C)+4 a^4 (6 A+7 C)+4 A b^4\right ) \tan (c+d x) \sec ^2(c+d x)}{105 d}+\frac {a b \left (a^2 (5 A+6 C)+2 b^2 (3 A+4 C)\right ) \tan (c+d x) \sec (c+d x)}{4 d}+\frac {\left (a^2 (6 A+7 C)+2 A b^2\right ) \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^2}{35 d}+\frac {A \tan (c+d x) \sec ^6(c+d x) (a+b \cos (c+d x))^4}{7 d}+\frac {2 A b \tan (c+d x) \sec ^5(c+d x) (a+b \cos (c+d x))^3}{21 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Cos[c + d*x])^4*(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^8,x]

[Out]

(a*b*(2*b^2*(3*A + 4*C) + a^2*(5*A + 6*C))*ArcTanh[Sin[c + d*x]])/(4*d) + ((35*b^4*(2*A + 3*C) + 84*a^2*b^2*(4
*A + 5*C) + 8*a^4*(6*A + 7*C))*Tan[c + d*x])/(105*d) + (a*b*(2*b^2*(3*A + 4*C) + a^2*(5*A + 6*C))*Sec[c + d*x]
*Tan[c + d*x])/(4*d) + ((4*A*b^4 + 4*a^4*(6*A + 7*C) + 3*a^2*b^2*(50*A + 63*C))*Sec[c + d*x]^2*Tan[c + d*x])/(
105*d) + (a*b*(6*A*b^2 + a^2*(103*A + 126*C))*Sec[c + d*x]^3*Tan[c + d*x])/(210*d) + ((2*A*b^2 + a^2*(6*A + 7*
C))*(a + b*Cos[c + d*x])^2*Sec[c + d*x]^4*Tan[c + d*x])/(35*d) + (2*A*b*(a + b*Cos[c + d*x])^3*Sec[c + d*x]^5*
Tan[c + d*x])/(21*d) + (A*(a + b*Cos[c + d*x])^4*Sec[c + d*x]^6*Tan[c + d*x])/(7*d)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

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 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 3031

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])*((A_.) + (B_.)*sin[(e
_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((b*c - a*d)*(A*b^2 - a*b*B + a^2*C)*
Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b^2*f*(m + 1)*(a^2 - b^2)), x] - Dist[1/(b^2*(m + 1)*(a^2 - b^2)),
Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*(m + 1)*((b*B - a*C)*(b*c - a*d) - A*b*(a*c - b*d)) + (b*B*(a^2*d + b
^2*d*(m + 1) - a*b*c*(m + 2)) + (b*c - a*d)*(A*b^2*(m + 2) + C*(a^2 + b^2*(m + 1))))*Sin[e + f*x] - b*C*d*(m +
1)*(a^2 - b^2)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && Ne
Q[a^2 - b^2, 0] && LtQ[m, -1]

Rule 3047

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

Rule 3048

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*s
in[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((c^2*C + 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/(d*(n + 1)*(c^2 - d^2)), Int[(a + b*Sin[e + f*x])^(m
- 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(b*d*m + a*c*(n + 1)) + c*C*(b*c*m + a*d*(n + 1)) - (A*d*(a*d*(n +
2) - b*c*(n + 1)) - C*(b*c*d*(n + 1) - a*(c^2 + d^2*(n + 1))))*Sin[e + f*x] - b*(A*d^2*(m + n + 2) + C*(c^2*(
m + 1) + d^2*(n + 1)))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C}, x] && NeQ[b*c - a*d, 0] &
& NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 0] && LtQ[n, -1]

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin {align*} \int (a+b \cos (c+d x))^4 \left (A+C \cos ^2(c+d x)\right ) \sec ^8(c+d x) \, dx &=\frac {A (a+b \cos (c+d x))^4 \sec ^6(c+d x) \tan (c+d x)}{7 d}+\frac {1}{7} \int (a+b \cos (c+d x))^3 \left (4 A b+a (6 A+7 C) \cos (c+d x)+b (2 A+7 C) \cos ^2(c+d x)\right ) \sec ^7(c+d x) \, dx\\ &=\frac {2 A b (a+b \cos (c+d x))^3 \sec ^5(c+d x) \tan (c+d x)}{21 d}+\frac {A (a+b \cos (c+d x))^4 \sec ^6(c+d x) \tan (c+d x)}{7 d}+\frac {1}{42} \int (a+b \cos (c+d x))^2 \left (6 \left (2 A b^2+a^2 (6 A+7 C)\right )+4 a b (17 A+21 C) \cos (c+d x)+2 b^2 (10 A+21 C) \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx\\ &=\frac {\left (2 A b^2+a^2 (6 A+7 C)\right ) (a+b \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{35 d}+\frac {2 A b (a+b \cos (c+d x))^3 \sec ^5(c+d x) \tan (c+d x)}{21 d}+\frac {A (a+b \cos (c+d x))^4 \sec ^6(c+d x) \tan (c+d x)}{7 d}+\frac {1}{210} \int (a+b \cos (c+d x)) \left (4 b \left (6 A b^2+a^2 (103 A+126 C)\right )+2 a \left (12 a^2 (6 A+7 C)+b^2 (244 A+315 C)\right ) \cos (c+d x)+2 b \left (6 a^2 (6 A+7 C)+b^2 (62 A+105 C)\right ) \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx\\ &=\frac {a b \left (6 A b^2+a^2 (103 A+126 C)\right ) \sec ^3(c+d x) \tan (c+d x)}{210 d}+\frac {\left (2 A b^2+a^2 (6 A+7 C)\right ) (a+b \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{35 d}+\frac {2 A b (a+b \cos (c+d x))^3 \sec ^5(c+d x) \tan (c+d x)}{21 d}+\frac {A (a+b \cos (c+d x))^4 \sec ^6(c+d x) \tan (c+d x)}{7 d}-\frac {1}{840} \int \left (-24 \left (4 A b^4+4 a^4 (6 A+7 C)+3 a^2 b^2 (50 A+63 C)\right )-420 a b \left (2 b^2 (3 A+4 C)+a^2 (5 A+6 C)\right ) \cos (c+d x)-8 b^2 \left (6 a^2 (6 A+7 C)+b^2 (62 A+105 C)\right ) \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx\\ &=\frac {\left (4 A b^4+4 a^4 (6 A+7 C)+3 a^2 b^2 (50 A+63 C)\right ) \sec ^2(c+d x) \tan (c+d x)}{105 d}+\frac {a b \left (6 A b^2+a^2 (103 A+126 C)\right ) \sec ^3(c+d x) \tan (c+d x)}{210 d}+\frac {\left (2 A b^2+a^2 (6 A+7 C)\right ) (a+b \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{35 d}+\frac {2 A b (a+b \cos (c+d x))^3 \sec ^5(c+d x) \tan (c+d x)}{21 d}+\frac {A (a+b \cos (c+d x))^4 \sec ^6(c+d x) \tan (c+d x)}{7 d}-\frac {\int \left (-1260 a b \left (2 b^2 (3 A+4 C)+a^2 (5 A+6 C)\right )-24 \left (35 b^4 (2 A+3 C)+84 a^2 b^2 (4 A+5 C)+8 a^4 (6 A+7 C)\right ) \cos (c+d x)\right ) \sec ^3(c+d x) \, dx}{2520}\\ &=\frac {\left (4 A b^4+4 a^4 (6 A+7 C)+3 a^2 b^2 (50 A+63 C)\right ) \sec ^2(c+d x) \tan (c+d x)}{105 d}+\frac {a b \left (6 A b^2+a^2 (103 A+126 C)\right ) \sec ^3(c+d x) \tan (c+d x)}{210 d}+\frac {\left (2 A b^2+a^2 (6 A+7 C)\right ) (a+b \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{35 d}+\frac {2 A b (a+b \cos (c+d x))^3 \sec ^5(c+d x) \tan (c+d x)}{21 d}+\frac {A (a+b \cos (c+d x))^4 \sec ^6(c+d x) \tan (c+d x)}{7 d}+\frac {1}{2} \left (a b \left (2 b^2 (3 A+4 C)+a^2 (5 A+6 C)\right )\right ) \int \sec ^3(c+d x) \, dx-\frac {1}{105} \left (-35 b^4 (2 A+3 C)-84 a^2 b^2 (4 A+5 C)-8 a^4 (6 A+7 C)\right ) \int \sec ^2(c+d x) \, dx\\ &=\frac {a b \left (2 b^2 (3 A+4 C)+a^2 (5 A+6 C)\right ) \sec (c+d x) \tan (c+d x)}{4 d}+\frac {\left (4 A b^4+4 a^4 (6 A+7 C)+3 a^2 b^2 (50 A+63 C)\right ) \sec ^2(c+d x) \tan (c+d x)}{105 d}+\frac {a b \left (6 A b^2+a^2 (103 A+126 C)\right ) \sec ^3(c+d x) \tan (c+d x)}{210 d}+\frac {\left (2 A b^2+a^2 (6 A+7 C)\right ) (a+b \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{35 d}+\frac {2 A b (a+b \cos (c+d x))^3 \sec ^5(c+d x) \tan (c+d x)}{21 d}+\frac {A (a+b \cos (c+d x))^4 \sec ^6(c+d x) \tan (c+d x)}{7 d}+\frac {1}{4} \left (a b \left (2 b^2 (3 A+4 C)+a^2 (5 A+6 C)\right )\right ) \int \sec (c+d x) \, dx-\frac {\left (35 b^4 (2 A+3 C)+84 a^2 b^2 (4 A+5 C)+8 a^4 (6 A+7 C)\right ) \operatorname {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{105 d}\\ &=\frac {a b \left (2 b^2 (3 A+4 C)+a^2 (5 A+6 C)\right ) \tanh ^{-1}(\sin (c+d x))}{4 d}+\frac {\left (35 b^4 (2 A+3 C)+84 a^2 b^2 (4 A+5 C)+8 a^4 (6 A+7 C)\right ) \tan (c+d x)}{105 d}+\frac {a b \left (2 b^2 (3 A+4 C)+a^2 (5 A+6 C)\right ) \sec (c+d x) \tan (c+d x)}{4 d}+\frac {\left (4 A b^4+4 a^4 (6 A+7 C)+3 a^2 b^2 (50 A+63 C)\right ) \sec ^2(c+d x) \tan (c+d x)}{105 d}+\frac {a b \left (6 A b^2+a^2 (103 A+126 C)\right ) \sec ^3(c+d x) \tan (c+d x)}{210 d}+\frac {\left (2 A b^2+a^2 (6 A+7 C)\right ) (a+b \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{35 d}+\frac {2 A b (a+b \cos (c+d x))^3 \sec ^5(c+d x) \tan (c+d x)}{21 d}+\frac {A (a+b \cos (c+d x))^4 \sec ^6(c+d x) \tan (c+d x)}{7 d}\\ \end {align*}

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Mathematica [A]  time = 2.17, size = 233, normalized size = 0.66 \[ \frac {105 a b \left (a^2 (5 A+6 C)+2 b^2 (3 A+4 C)\right ) \tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \left (60 a^4 A \tan ^6(c+d x)+280 a^3 A b \sec ^5(c+d x)+84 a^2 \left (a^2 (3 A+C)+6 A b^2\right ) \tan ^4(c+d x)+70 a b \left (a^2 (5 A+6 C)+6 A b^2\right ) \sec ^3(c+d x)+105 a b \left (a^2 (5 A+6 C)+2 b^2 (3 A+4 C)\right ) \sec (c+d x)+140 \left (a^4 (3 A+2 C)+6 a^2 b^2 (2 A+C)+A b^4\right ) \tan ^2(c+d x)+420 \left (a^4+6 a^2 b^2+b^4\right ) (A+C)\right )}{420 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Cos[c + d*x])^4*(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^8,x]

[Out]

(105*a*b*(2*b^2*(3*A + 4*C) + a^2*(5*A + 6*C))*ArcTanh[Sin[c + d*x]] + Tan[c + d*x]*(420*(a^4 + 6*a^2*b^2 + b^
4)*(A + C) + 105*a*b*(2*b^2*(3*A + 4*C) + a^2*(5*A + 6*C))*Sec[c + d*x] + 70*a*b*(6*A*b^2 + a^2*(5*A + 6*C))*S
ec[c + d*x]^3 + 280*a^3*A*b*Sec[c + d*x]^5 + 140*(A*b^4 + 6*a^2*b^2*(2*A + C) + a^4*(3*A + 2*C))*Tan[c + d*x]^
2 + 84*a^2*(6*A*b^2 + a^2*(3*A + C))*Tan[c + d*x]^4 + 60*a^4*A*Tan[c + d*x]^6))/(420*d)

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fricas [A]  time = 0.95, size = 325, normalized size = 0.92 \[ \frac {105 \, {\left ({\left (5 \, A + 6 \, C\right )} a^{3} b + 2 \, {\left (3 \, A + 4 \, C\right )} a b^{3}\right )} \cos \left (d x + c\right )^{7} \log \left (\sin \left (d x + c\right ) + 1\right ) - 105 \, {\left ({\left (5 \, A + 6 \, C\right )} a^{3} b + 2 \, {\left (3 \, A + 4 \, C\right )} a b^{3}\right )} \cos \left (d x + c\right )^{7} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (4 \, {\left (8 \, {\left (6 \, A + 7 \, C\right )} a^{4} + 84 \, {\left (4 \, A + 5 \, C\right )} a^{2} b^{2} + 35 \, {\left (2 \, A + 3 \, C\right )} b^{4}\right )} \cos \left (d x + c\right )^{6} + 280 \, A a^{3} b \cos \left (d x + c\right ) + 105 \, {\left ({\left (5 \, A + 6 \, C\right )} a^{3} b + 2 \, {\left (3 \, A + 4 \, C\right )} a b^{3}\right )} \cos \left (d x + c\right )^{5} + 60 \, A a^{4} + 4 \, {\left (4 \, {\left (6 \, A + 7 \, C\right )} a^{4} + 42 \, {\left (4 \, A + 5 \, C\right )} a^{2} b^{2} + 35 \, A b^{4}\right )} \cos \left (d x + c\right )^{4} + 70 \, {\left ({\left (5 \, A + 6 \, C\right )} a^{3} b + 6 \, A a b^{3}\right )} \cos \left (d x + c\right )^{3} + 12 \, {\left ({\left (6 \, A + 7 \, C\right )} a^{4} + 42 \, A a^{2} b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{840 \, d \cos \left (d x + c\right )^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^4*(A+C*cos(d*x+c)^2)*sec(d*x+c)^8,x, algorithm="fricas")

[Out]

1/840*(105*((5*A + 6*C)*a^3*b + 2*(3*A + 4*C)*a*b^3)*cos(d*x + c)^7*log(sin(d*x + c) + 1) - 105*((5*A + 6*C)*a
^3*b + 2*(3*A + 4*C)*a*b^3)*cos(d*x + c)^7*log(-sin(d*x + c) + 1) + 2*(4*(8*(6*A + 7*C)*a^4 + 84*(4*A + 5*C)*a
^2*b^2 + 35*(2*A + 3*C)*b^4)*cos(d*x + c)^6 + 280*A*a^3*b*cos(d*x + c) + 105*((5*A + 6*C)*a^3*b + 2*(3*A + 4*C
)*a*b^3)*cos(d*x + c)^5 + 60*A*a^4 + 4*(4*(6*A + 7*C)*a^4 + 42*(4*A + 5*C)*a^2*b^2 + 35*A*b^4)*cos(d*x + c)^4
+ 70*((5*A + 6*C)*a^3*b + 6*A*a*b^3)*cos(d*x + c)^3 + 12*((6*A + 7*C)*a^4 + 42*A*a^2*b^2)*cos(d*x + c)^2)*sin(
d*x + c))/(d*cos(d*x + c)^7)

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giac [B]  time = 0.81, size = 1280, normalized size = 3.61 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^4*(A+C*cos(d*x+c)^2)*sec(d*x+c)^8,x, algorithm="giac")

[Out]

1/420*(105*(5*A*a^3*b + 6*C*a^3*b + 6*A*a*b^3 + 8*C*a*b^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 105*(5*A*a^3*b
+ 6*C*a^3*b + 6*A*a*b^3 + 8*C*a*b^3)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(420*A*a^4*tan(1/2*d*x + 1/2*c)^1
3 + 420*C*a^4*tan(1/2*d*x + 1/2*c)^13 - 1155*A*a^3*b*tan(1/2*d*x + 1/2*c)^13 - 1050*C*a^3*b*tan(1/2*d*x + 1/2*
c)^13 + 2520*A*a^2*b^2*tan(1/2*d*x + 1/2*c)^13 + 2520*C*a^2*b^2*tan(1/2*d*x + 1/2*c)^13 - 1050*A*a*b^3*tan(1/2
*d*x + 1/2*c)^13 - 840*C*a*b^3*tan(1/2*d*x + 1/2*c)^13 + 420*A*b^4*tan(1/2*d*x + 1/2*c)^13 + 420*C*b^4*tan(1/2
*d*x + 1/2*c)^13 - 840*A*a^4*tan(1/2*d*x + 1/2*c)^11 - 1400*C*a^4*tan(1/2*d*x + 1/2*c)^11 + 980*A*a^3*b*tan(1/
2*d*x + 1/2*c)^11 + 2520*C*a^3*b*tan(1/2*d*x + 1/2*c)^11 - 8400*A*a^2*b^2*tan(1/2*d*x + 1/2*c)^11 - 11760*C*a^
2*b^2*tan(1/2*d*x + 1/2*c)^11 + 2520*A*a*b^3*tan(1/2*d*x + 1/2*c)^11 + 3360*C*a*b^3*tan(1/2*d*x + 1/2*c)^11 -
1960*A*b^4*tan(1/2*d*x + 1/2*c)^11 - 2520*C*b^4*tan(1/2*d*x + 1/2*c)^11 + 3612*A*a^4*tan(1/2*d*x + 1/2*c)^9 +
3164*C*a^4*tan(1/2*d*x + 1/2*c)^9 - 2975*A*a^3*b*tan(1/2*d*x + 1/2*c)^9 - 1890*C*a^3*b*tan(1/2*d*x + 1/2*c)^9
+ 18984*A*a^2*b^2*tan(1/2*d*x + 1/2*c)^9 + 24360*C*a^2*b^2*tan(1/2*d*x + 1/2*c)^9 - 1890*A*a*b^3*tan(1/2*d*x +
1/2*c)^9 - 4200*C*a*b^3*tan(1/2*d*x + 1/2*c)^9 + 4060*A*b^4*tan(1/2*d*x + 1/2*c)^9 + 6300*C*b^4*tan(1/2*d*x +
1/2*c)^9 - 2544*A*a^4*tan(1/2*d*x + 1/2*c)^7 - 4368*C*a^4*tan(1/2*d*x + 1/2*c)^7 - 26208*A*a^2*b^2*tan(1/2*d*
x + 1/2*c)^7 - 30240*C*a^2*b^2*tan(1/2*d*x + 1/2*c)^7 - 5040*A*b^4*tan(1/2*d*x + 1/2*c)^7 - 8400*C*b^4*tan(1/2
*d*x + 1/2*c)^7 + 3612*A*a^4*tan(1/2*d*x + 1/2*c)^5 + 3164*C*a^4*tan(1/2*d*x + 1/2*c)^5 + 2975*A*a^3*b*tan(1/2
*d*x + 1/2*c)^5 + 1890*C*a^3*b*tan(1/2*d*x + 1/2*c)^5 + 18984*A*a^2*b^2*tan(1/2*d*x + 1/2*c)^5 + 24360*C*a^2*b
^2*tan(1/2*d*x + 1/2*c)^5 + 1890*A*a*b^3*tan(1/2*d*x + 1/2*c)^5 + 4200*C*a*b^3*tan(1/2*d*x + 1/2*c)^5 + 4060*A
*b^4*tan(1/2*d*x + 1/2*c)^5 + 6300*C*b^4*tan(1/2*d*x + 1/2*c)^5 - 840*A*a^4*tan(1/2*d*x + 1/2*c)^3 - 1400*C*a^
4*tan(1/2*d*x + 1/2*c)^3 - 980*A*a^3*b*tan(1/2*d*x + 1/2*c)^3 - 2520*C*a^3*b*tan(1/2*d*x + 1/2*c)^3 - 8400*A*a
^2*b^2*tan(1/2*d*x + 1/2*c)^3 - 11760*C*a^2*b^2*tan(1/2*d*x + 1/2*c)^3 - 2520*A*a*b^3*tan(1/2*d*x + 1/2*c)^3 -
3360*C*a*b^3*tan(1/2*d*x + 1/2*c)^3 - 1960*A*b^4*tan(1/2*d*x + 1/2*c)^3 - 2520*C*b^4*tan(1/2*d*x + 1/2*c)^3 +
420*A*a^4*tan(1/2*d*x + 1/2*c) + 420*C*a^4*tan(1/2*d*x + 1/2*c) + 1155*A*a^3*b*tan(1/2*d*x + 1/2*c) + 1050*C*
a^3*b*tan(1/2*d*x + 1/2*c) + 2520*A*a^2*b^2*tan(1/2*d*x + 1/2*c) + 2520*C*a^2*b^2*tan(1/2*d*x + 1/2*c) + 1050*
A*a*b^3*tan(1/2*d*x + 1/2*c) + 840*C*a*b^3*tan(1/2*d*x + 1/2*c) + 420*A*b^4*tan(1/2*d*x + 1/2*c) + 420*C*b^4*t
an(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^7)/d

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maple [A]  time = 0.54, size = 591, normalized size = 1.66 \[ \frac {2 A \,b^{4} \tan \left (d x +c \right )}{3 d}+\frac {8 a^{4} C \tan \left (d x +c \right )}{15 d}+\frac {5 A \,a^{3} b \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )}{6 d}+\frac {8 A \,a^{2} b^{2} \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{5 d}+\frac {6 A \,a^{4} \tan \left (d x +c \right ) \left (\sec ^{4}\left (d x +c \right )\right )}{35 d}+\frac {C \,b^{4} \tan \left (d x +c \right )}{d}+\frac {16 A \,a^{4} \tan \left (d x +c \right )}{35 d}+\frac {2 C a \,b^{3} \tan \left (d x +c \right ) \sec \left (d x +c \right )}{d}+\frac {3 a A \,b^{3} \tan \left (d x +c \right ) \sec \left (d x +c \right )}{2 d}+\frac {3 a^{3} b C \tan \left (d x +c \right ) \sec \left (d x +c \right )}{2 d}+\frac {5 A \,a^{3} b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{4 d}+\frac {8 A \,a^{4} \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{35 d}+\frac {5 A \,a^{3} b \sec \left (d x +c \right ) \tan \left (d x +c \right )}{4 d}+\frac {6 A \,a^{2} b^{2} \tan \left (d x +c \right ) \left (\sec ^{4}\left (d x +c \right )\right )}{5 d}+\frac {2 C \,a^{2} b^{2} \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{d}+\frac {2 A \,a^{3} b \tan \left (d x +c \right ) \left (\sec ^{5}\left (d x +c \right )\right )}{3 d}+\frac {a^{3} b C \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )}{d}+\frac {a A \,b^{3} \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )}{d}+\frac {A \,a^{4} \tan \left (d x +c \right ) \left (\sec ^{6}\left (d x +c \right )\right )}{7 d}+\frac {a^{4} C \tan \left (d x +c \right ) \left (\sec ^{4}\left (d x +c \right )\right )}{5 d}+\frac {3 a^{3} b C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {16 A \,a^{2} b^{2} \tan \left (d x +c \right )}{5 d}+\frac {3 a A \,b^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {4 a^{4} C \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{15 d}+\frac {4 C \,a^{2} b^{2} \tan \left (d x +c \right )}{d}+\frac {2 C a \,b^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {A \,b^{4} \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+b*cos(d*x+c))^4*(A+C*cos(d*x+c)^2)*sec(d*x+c)^8,x)

[Out]

2/3/d*A*b^4*tan(d*x+c)+8/15/d*a^4*C*tan(d*x+c)+8/35/d*A*a^4*tan(d*x+c)*sec(d*x+c)^2+1/d*C*b^4*tan(d*x+c)+16/35
/d*A*a^4*tan(d*x+c)+2/3/d*A*a^3*b*tan(d*x+c)*sec(d*x+c)^5+1/d*a^3*b*C*tan(d*x+c)*sec(d*x+c)^3+1/d*a*A*b^3*tan(
d*x+c)*sec(d*x+c)^3+2/d*C*a*b^3*tan(d*x+c)*sec(d*x+c)+6/5/d*A*a^2*b^2*tan(d*x+c)*sec(d*x+c)^4+2/d*C*a^2*b^2*ta
n(d*x+c)*sec(d*x+c)^2+8/5/d*A*a^2*b^2*tan(d*x+c)*sec(d*x+c)^2+3/2/d*a*A*b^3*tan(d*x+c)*sec(d*x+c)+5/6/d*A*a^3*
b*tan(d*x+c)*sec(d*x+c)^3+3/2/d*a^3*b*C*tan(d*x+c)*sec(d*x+c)+5/4/d*A*a^3*b*ln(sec(d*x+c)+tan(d*x+c))+5/4/d*A*
a^3*b*sec(d*x+c)*tan(d*x+c)+6/35/d*A*a^4*tan(d*x+c)*sec(d*x+c)^4+4/15/d*a^4*C*tan(d*x+c)*sec(d*x+c)^2+1/5/d*a^
4*C*tan(d*x+c)*sec(d*x+c)^4+1/3/d*A*b^4*tan(d*x+c)*sec(d*x+c)^2+1/7/d*A*a^4*tan(d*x+c)*sec(d*x+c)^6+3/2/d*a^3*
b*C*ln(sec(d*x+c)+tan(d*x+c))+16/5/d*A*a^2*b^2*tan(d*x+c)+3/2/d*a*A*b^3*ln(sec(d*x+c)+tan(d*x+c))+4/d*C*a^2*b^
2*tan(d*x+c)+2/d*C*a*b^3*ln(sec(d*x+c)+tan(d*x+c))

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maxima [A]  time = 0.38, size = 472, normalized size = 1.33 \[ \frac {24 \, {\left (5 \, \tan \left (d x + c\right )^{7} + 21 \, \tan \left (d x + c\right )^{5} + 35 \, \tan \left (d x + c\right )^{3} + 35 \, \tan \left (d x + c\right )\right )} A a^{4} + 56 \, {\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 )} C a^{4} + 336 \, {\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^{2} b^{2} + 1680 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a^{2} b^{2} + 280 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A b^{4} - 35 \, A a^{3} b {\left (\frac {2 \, {\left (15 \, \sin \left (d x + c\right )^{5} - 40 \, \sin \left (d x + c\right )^{3} + 33 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{6} - 3 \, \sin \left (d x + c\right )^{4} + 3 \, \sin \left (d x + c\right )^{2} - 1} - 15 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 15 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 210 \, C a^{3} b {\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 )} - 210 \, A a b^{3} {\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 )} - 840 \, C a b^{3} {\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 )} + 840 \, C b^{4} \tan \left (d x + c\right )}{840 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^4*(A+C*cos(d*x+c)^2)*sec(d*x+c)^8,x, algorithm="maxima")

[Out]

1/840*(24*(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 + 56*(3*tan(d*x +
c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c))*C*a^4 + 336*(3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x +
c))*A*a^2*b^2 + 1680*(tan(d*x + c)^3 + 3*tan(d*x + c))*C*a^2*b^2 + 280*(tan(d*x + c)^3 + 3*tan(d*x + c))*A*b^4
- 35*A*a^3*b*(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)) - 210*C*a^3*b*(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)) - 210*A*a*b^3*(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*b^3*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c)
+ 1) + log(sin(d*x + c) - 1)) + 840*C*b^4*tan(d*x + c))/d

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mupad [B]  time = 4.94, size = 755, normalized size = 2.13 \[ \frac {a\,b\,\mathrm {atanh}\left (\frac {a\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (5\,A\,a^2+6\,A\,b^2+6\,C\,a^2+8\,C\,b^2\right )}{6\,A\,a\,b^3+5\,A\,a^3\,b+8\,C\,a\,b^3+6\,C\,a^3\,b}\right )\,\left (5\,A\,a^2+6\,A\,b^2+6\,C\,a^2+8\,C\,b^2\right )}{2\,d}-\frac {\left (2\,A\,a^4+2\,A\,b^4+2\,C\,a^4+2\,C\,b^4+12\,A\,a^2\,b^2+12\,C\,a^2\,b^2-5\,A\,a\,b^3-\frac {11\,A\,a^3\,b}{2}-4\,C\,a\,b^3-5\,C\,a^3\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}+\left (12\,A\,a\,b^3-\frac {28\,A\,b^4}{3}-\frac {20\,C\,a^4}{3}-12\,C\,b^4-40\,A\,a^2\,b^2-56\,C\,a^2\,b^2-4\,A\,a^4+\frac {14\,A\,a^3\,b}{3}+16\,C\,a\,b^3+12\,C\,a^3\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}+\left (\frac {86\,A\,a^4}{5}+\frac {58\,A\,b^4}{3}+\frac {226\,C\,a^4}{15}+30\,C\,b^4+\frac {452\,A\,a^2\,b^2}{5}+116\,C\,a^2\,b^2-9\,A\,a\,b^3-\frac {85\,A\,a^3\,b}{6}-20\,C\,a\,b^3-9\,C\,a^3\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (-\frac {424\,A\,a^4}{35}-24\,A\,b^4-\frac {104\,C\,a^4}{5}-40\,C\,b^4-\frac {624\,A\,a^2\,b^2}{5}-144\,C\,a^2\,b^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {86\,A\,a^4}{5}+\frac {58\,A\,b^4}{3}+\frac {226\,C\,a^4}{15}+30\,C\,b^4+\frac {452\,A\,a^2\,b^2}{5}+116\,C\,a^2\,b^2+9\,A\,a\,b^3+\frac {85\,A\,a^3\,b}{6}+20\,C\,a\,b^3+9\,C\,a^3\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-4\,A\,a^4-\frac {28\,A\,b^4}{3}-\frac {20\,C\,a^4}{3}-12\,C\,b^4-40\,A\,a^2\,b^2-56\,C\,a^2\,b^2-12\,A\,a\,b^3-\frac {14\,A\,a^3\,b}{3}-16\,C\,a\,b^3-12\,C\,a^3\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (2\,A\,a^4+2\,A\,b^4+2\,C\,a^4+2\,C\,b^4+12\,A\,a^2\,b^2+12\,C\,a^2\,b^2+5\,A\,a\,b^3+\frac {11\,A\,a^3\,b}{2}+4\,C\,a\,b^3+5\,C\,a^3\,b\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{14}-7\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}+21\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-35\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+35\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-21\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+7\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + C*cos(c + d*x)^2)*(a + b*cos(c + d*x))^4)/cos(c + d*x)^8,x)

[Out]

(a*b*atanh((a*b*tan(c/2 + (d*x)/2)*(5*A*a^2 + 6*A*b^2 + 6*C*a^2 + 8*C*b^2))/(6*A*a*b^3 + 5*A*a^3*b + 8*C*a*b^3
+ 6*C*a^3*b))*(5*A*a^2 + 6*A*b^2 + 6*C*a^2 + 8*C*b^2))/(2*d) - (tan(c/2 + (d*x)/2)*(2*A*a^4 + 2*A*b^4 + 2*C*a
^4 + 2*C*b^4 + 12*A*a^2*b^2 + 12*C*a^2*b^2 + 5*A*a*b^3 + (11*A*a^3*b)/2 + 4*C*a*b^3 + 5*C*a^3*b) - tan(c/2 + (
d*x)/2)^7*((424*A*a^4)/35 + 24*A*b^4 + (104*C*a^4)/5 + 40*C*b^4 + (624*A*a^2*b^2)/5 + 144*C*a^2*b^2) + tan(c/2
+ (d*x)/2)^13*(2*A*a^4 + 2*A*b^4 + 2*C*a^4 + 2*C*b^4 + 12*A*a^2*b^2 + 12*C*a^2*b^2 - 5*A*a*b^3 - (11*A*a^3*b)
/2 - 4*C*a*b^3 - 5*C*a^3*b) - tan(c/2 + (d*x)/2)^3*(4*A*a^4 + (28*A*b^4)/3 + (20*C*a^4)/3 + 12*C*b^4 + 40*A*a^
2*b^2 + 56*C*a^2*b^2 + 12*A*a*b^3 + (14*A*a^3*b)/3 + 16*C*a*b^3 + 12*C*a^3*b) - tan(c/2 + (d*x)/2)^11*(4*A*a^4
+ (28*A*b^4)/3 + (20*C*a^4)/3 + 12*C*b^4 + 40*A*a^2*b^2 + 56*C*a^2*b^2 - 12*A*a*b^3 - (14*A*a^3*b)/3 - 16*C*a
*b^3 - 12*C*a^3*b) + tan(c/2 + (d*x)/2)^5*((86*A*a^4)/5 + (58*A*b^4)/3 + (226*C*a^4)/15 + 30*C*b^4 + (452*A*a^
2*b^2)/5 + 116*C*a^2*b^2 + 9*A*a*b^3 + (85*A*a^3*b)/6 + 20*C*a*b^3 + 9*C*a^3*b) + tan(c/2 + (d*x)/2)^9*((86*A*
a^4)/5 + (58*A*b^4)/3 + (226*C*a^4)/15 + 30*C*b^4 + (452*A*a^2*b^2)/5 + 116*C*a^2*b^2 - 9*A*a*b^3 - (85*A*a^3*
b)/6 - 20*C*a*b^3 - 9*C*a^3*b))/(d*(7*tan(c/2 + (d*x)/2)^2 - 21*tan(c/2 + (d*x)/2)^4 + 35*tan(c/2 + (d*x)/2)^6
- 35*tan(c/2 + (d*x)/2)^8 + 21*tan(c/2 + (d*x)/2)^10 - 7*tan(c/2 + (d*x)/2)^12 + tan(c/2 + (d*x)/2)^14 - 1))

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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+b*cos(d*x+c))**4*(A+C*cos(d*x+c)**2)*sec(d*x+c)**8,x)

[Out]

Timed out

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