3.963 \(\int (a+b \cos (c+d x))^3 (A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^7(c+d x) \, dx\)

Optimal. Leaf size=336 \[ \frac {a \tan (c+d x) \sec ^3(c+d x) \left (5 a^2 (5 A+6 C)+42 a b B+6 A b^2\right )}{120 d}+\frac {\tan (c+d x) \left (8 a^3 B+6 a^2 b (4 A+5 C)+30 a b^2 B+5 b^3 (2 A+3 C)\right )}{15 d}+\frac {\left (a^3 (5 A+6 C)+18 a^2 b B+6 a b^2 (3 A+4 C)+8 b^3 B\right ) \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac {\tan (c+d x) \sec ^2(c+d x) \left (4 a^3 B+3 a^2 b (4 A+5 C)+12 a b^2 B+A b^3\right )}{15 d}+\frac {\tan (c+d x) \sec (c+d x) \left (a^3 (5 A+6 C)+18 a^2 b B+6 a b^2 (3 A+4 C)+8 b^3 B\right )}{16 d}+\frac {(2 a B+A b) \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^2}{10 d}+\frac {A \tan (c+d x) \sec ^5(c+d x) (a+b \cos (c+d x))^3}{6 d} \]

[Out]

1/16*(18*a^2*b*B+8*b^3*B+6*a*b^2*(3*A+4*C)+a^3*(5*A+6*C))*arctanh(sin(d*x+c))/d+1/15*(8*a^3*B+30*a*b^2*B+5*b^3
*(2*A+3*C)+6*a^2*b*(4*A+5*C))*tan(d*x+c)/d+1/16*(18*a^2*b*B+8*b^3*B+6*a*b^2*(3*A+4*C)+a^3*(5*A+6*C))*sec(d*x+c
)*tan(d*x+c)/d+1/15*(A*b^3+4*a^3*B+12*a*b^2*B+3*a^2*b*(4*A+5*C))*sec(d*x+c)^2*tan(d*x+c)/d+1/120*a*(6*A*b^2+42
*a*b*B+5*a^2*(5*A+6*C))*sec(d*x+c)^3*tan(d*x+c)/d+1/10*(A*b+2*B*a)*(a+b*cos(d*x+c))^2*sec(d*x+c)^4*tan(d*x+c)/
d+1/6*A*(a+b*cos(d*x+c))^3*sec(d*x+c)^5*tan(d*x+c)/d

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Rubi [A]  time = 0.91, antiderivative size = 336, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.195, Rules used = {3047, 3031, 3021, 2748, 3768, 3770, 3767, 8} \[ \frac {\tan (c+d x) \left (6 a^2 b (4 A+5 C)+8 a^3 B+30 a b^2 B+5 b^3 (2 A+3 C)\right )}{15 d}+\frac {\left (a^3 (5 A+6 C)+18 a^2 b B+6 a b^2 (3 A+4 C)+8 b^3 B\right ) \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac {a \tan (c+d x) \sec ^3(c+d x) \left (5 a^2 (5 A+6 C)+42 a b B+6 A b^2\right )}{120 d}+\frac {\tan (c+d x) \sec ^2(c+d x) \left (3 a^2 b (4 A+5 C)+4 a^3 B+12 a b^2 B+A b^3\right )}{15 d}+\frac {\tan (c+d x) \sec (c+d x) \left (a^3 (5 A+6 C)+18 a^2 b B+6 a b^2 (3 A+4 C)+8 b^3 B\right )}{16 d}+\frac {(2 a B+A b) \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^2}{10 d}+\frac {A \tan (c+d x) \sec ^5(c+d x) (a+b \cos (c+d x))^3}{6 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Cos[c + d*x])^3*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^7,x]

[Out]

((18*a^2*b*B + 8*b^3*B + 6*a*b^2*(3*A + 4*C) + a^3*(5*A + 6*C))*ArcTanh[Sin[c + d*x]])/(16*d) + ((8*a^3*B + 30
*a*b^2*B + 5*b^3*(2*A + 3*C) + 6*a^2*b*(4*A + 5*C))*Tan[c + d*x])/(15*d) + ((18*a^2*b*B + 8*b^3*B + 6*a*b^2*(3
*A + 4*C) + a^3*(5*A + 6*C))*Sec[c + d*x]*Tan[c + d*x])/(16*d) + ((A*b^3 + 4*a^3*B + 12*a*b^2*B + 3*a^2*b*(4*A
+ 5*C))*Sec[c + d*x]^2*Tan[c + d*x])/(15*d) + (a*(6*A*b^2 + 42*a*b*B + 5*a^2*(5*A + 6*C))*Sec[c + d*x]^3*Tan[
c + d*x])/(120*d) + ((A*b + 2*a*B)*(a + b*Cos[c + d*x])^2*Sec[c + d*x]^4*Tan[c + d*x])/(10*d) + (A*(a + b*Cos[
c + d*x])^3*Sec[c + d*x]^5*Tan[c + d*x])/(6*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 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))^3 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^7(c+d x) \, dx &=\frac {A (a+b \cos (c+d x))^3 \sec ^5(c+d x) \tan (c+d x)}{6 d}+\frac {1}{6} \int (a+b \cos (c+d x))^2 \left (3 (A b+2 a B)+(5 a A+6 b B+6 a C) \cos (c+d x)+2 b (A+3 C) \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx\\ &=\frac {(A b+2 a B) (a+b \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{10 d}+\frac {A (a+b \cos (c+d x))^3 \sec ^5(c+d x) \tan (c+d x)}{6 d}+\frac {1}{30} \int (a+b \cos (c+d x)) \left (6 A b^2+42 a b B+5 a^2 (5 A+6 C)+\left (24 a^2 B+30 b^2 B+a b (47 A+60 C)\right ) \cos (c+d x)+2 b (8 A b+6 a B+15 b C) \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx\\ &=\frac {a \left (6 A b^2+42 a b B+5 a^2 (5 A+6 C)\right ) \sec ^3(c+d x) \tan (c+d x)}{120 d}+\frac {(A b+2 a B) (a+b \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{10 d}+\frac {A (a+b \cos (c+d x))^3 \sec ^5(c+d x) \tan (c+d x)}{6 d}-\frac {1}{120} \int \left (-24 \left (A b^3+4 a^3 B+12 a b^2 B+3 a^2 b (4 A+5 C)\right )-15 \left (18 a^2 b B+8 b^3 B+6 a b^2 (3 A+4 C)+a^3 (5 A+6 C)\right ) \cos (c+d x)-8 b^2 (8 A b+6 a B+15 b C) \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx\\ &=\frac {\left (A b^3+4 a^3 B+12 a b^2 B+3 a^2 b (4 A+5 C)\right ) \sec ^2(c+d x) \tan (c+d x)}{15 d}+\frac {a \left (6 A b^2+42 a b B+5 a^2 (5 A+6 C)\right ) \sec ^3(c+d x) \tan (c+d x)}{120 d}+\frac {(A b+2 a B) (a+b \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{10 d}+\frac {A (a+b \cos (c+d x))^3 \sec ^5(c+d x) \tan (c+d x)}{6 d}-\frac {1}{360} \int \left (-45 \left (18 a^2 b B+8 b^3 B+6 a b^2 (3 A+4 C)+a^3 (5 A+6 C)\right )-24 \left (8 a^3 B+30 a b^2 B+5 b^3 (2 A+3 C)+6 a^2 b (4 A+5 C)\right ) \cos (c+d x)\right ) \sec ^3(c+d x) \, dx\\ &=\frac {\left (A b^3+4 a^3 B+12 a b^2 B+3 a^2 b (4 A+5 C)\right ) \sec ^2(c+d x) \tan (c+d x)}{15 d}+\frac {a \left (6 A b^2+42 a b B+5 a^2 (5 A+6 C)\right ) \sec ^3(c+d x) \tan (c+d x)}{120 d}+\frac {(A b+2 a B) (a+b \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{10 d}+\frac {A (a+b \cos (c+d x))^3 \sec ^5(c+d x) \tan (c+d x)}{6 d}-\frac {1}{15} \left (-8 a^3 B-30 a b^2 B-5 b^3 (2 A+3 C)-6 a^2 b (4 A+5 C)\right ) \int \sec ^2(c+d x) \, dx-\frac {1}{8} \left (-18 a^2 b B-8 b^3 B-6 a b^2 (3 A+4 C)-a^3 (5 A+6 C)\right ) \int \sec ^3(c+d x) \, dx\\ &=\frac {\left (18 a^2 b B+8 b^3 B+6 a b^2 (3 A+4 C)+a^3 (5 A+6 C)\right ) \sec (c+d x) \tan (c+d x)}{16 d}+\frac {\left (A b^3+4 a^3 B+12 a b^2 B+3 a^2 b (4 A+5 C)\right ) \sec ^2(c+d x) \tan (c+d x)}{15 d}+\frac {a \left (6 A b^2+42 a b B+5 a^2 (5 A+6 C)\right ) \sec ^3(c+d x) \tan (c+d x)}{120 d}+\frac {(A b+2 a B) (a+b \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{10 d}+\frac {A (a+b \cos (c+d x))^3 \sec ^5(c+d x) \tan (c+d x)}{6 d}-\frac {1}{16} \left (-18 a^2 b B-8 b^3 B-6 a b^2 (3 A+4 C)-a^3 (5 A+6 C)\right ) \int \sec (c+d x) \, dx-\frac {\left (8 a^3 B+30 a b^2 B+5 b^3 (2 A+3 C)+6 a^2 b (4 A+5 C)\right ) \operatorname {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{15 d}\\ &=\frac {\left (18 a^2 b B+8 b^3 B+6 a b^2 (3 A+4 C)+a^3 (5 A+6 C)\right ) \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac {\left (8 a^3 B+30 a b^2 B+5 b^3 (2 A+3 C)+6 a^2 b (4 A+5 C)\right ) \tan (c+d x)}{15 d}+\frac {\left (18 a^2 b B+8 b^3 B+6 a b^2 (3 A+4 C)+a^3 (5 A+6 C)\right ) \sec (c+d x) \tan (c+d x)}{16 d}+\frac {\left (A b^3+4 a^3 B+12 a b^2 B+3 a^2 b (4 A+5 C)\right ) \sec ^2(c+d x) \tan (c+d x)}{15 d}+\frac {a \left (6 A b^2+42 a b B+5 a^2 (5 A+6 C)\right ) \sec ^3(c+d x) \tan (c+d x)}{120 d}+\frac {(A b+2 a B) (a+b \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{10 d}+\frac {A (a+b \cos (c+d x))^3 \sec ^5(c+d x) \tan (c+d x)}{6 d}\\ \end {align*}

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Mathematica [A]  time = 2.91, size = 252, normalized size = 0.75 \[ \frac {15 \left (a^3 (5 A+6 C)+18 a^2 b B+6 a b^2 (3 A+4 C)+8 b^3 B\right ) \tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \left (40 a^3 A \sec ^5(c+d x)+10 a \sec ^3(c+d x) \left (a^2 (5 A+6 C)+18 a b B+18 A b^2\right )+16 \left (3 a^2 (a B+3 A b) \tan ^4(c+d x)+5 \tan ^2(c+d x) \left (2 a^3 B+3 a^2 b (2 A+C)+3 a b^2 B+A b^3\right )+15 \left (a^3 B+3 a^2 b (A+C)+3 a b^2 B+b^3 (A+C)\right )\right )+15 \sec (c+d x) \left (a^3 (5 A+6 C)+18 a^2 b B+6 a b^2 (3 A+4 C)+8 b^3 B\right )\right )}{240 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Cos[c + d*x])^3*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^7,x]

[Out]

(15*(18*a^2*b*B + 8*b^3*B + 6*a*b^2*(3*A + 4*C) + a^3*(5*A + 6*C))*ArcTanh[Sin[c + d*x]] + Tan[c + d*x]*(15*(1
8*a^2*b*B + 8*b^3*B + 6*a*b^2*(3*A + 4*C) + a^3*(5*A + 6*C))*Sec[c + d*x] + 10*a*(18*A*b^2 + 18*a*b*B + a^2*(5
*A + 6*C))*Sec[c + d*x]^3 + 40*a^3*A*Sec[c + d*x]^5 + 16*(15*(a^3*B + 3*a*b^2*B + 3*a^2*b*(A + C) + b^3*(A + C
)) + 5*(A*b^3 + 2*a^3*B + 3*a*b^2*B + 3*a^2*b*(2*A + C))*Tan[c + d*x]^2 + 3*a^2*(3*A*b + a*B)*Tan[c + d*x]^4))
)/(240*d)

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fricas [A]  time = 0.48, size = 342, normalized size = 1.02 \[ \frac {15 \, {\left ({\left (5 \, A + 6 \, C\right )} a^{3} + 18 \, B a^{2} b + 6 \, {\left (3 \, A + 4 \, C\right )} a b^{2} + 8 \, B b^{3}\right )} \cos \left (d x + c\right )^{6} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left ({\left (5 \, A + 6 \, C\right )} a^{3} + 18 \, B a^{2} b + 6 \, {\left (3 \, A + 4 \, C\right )} a b^{2} + 8 \, B b^{3}\right )} \cos \left (d x + c\right )^{6} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (16 \, {\left (8 \, B a^{3} + 6 \, {\left (4 \, A + 5 \, C\right )} a^{2} b + 30 \, B a b^{2} + 5 \, {\left (2 \, A + 3 \, C\right )} b^{3}\right )} \cos \left (d x + c\right )^{5} + 15 \, {\left ({\left (5 \, A + 6 \, C\right )} a^{3} + 18 \, B a^{2} b + 6 \, {\left (3 \, A + 4 \, C\right )} a b^{2} + 8 \, B b^{3}\right )} \cos \left (d x + c\right )^{4} + 40 \, A a^{3} + 16 \, {\left (4 \, B a^{3} + 3 \, {\left (4 \, A + 5 \, C\right )} a^{2} b + 15 \, B a b^{2} + 5 \, A b^{3}\right )} \cos \left (d x + c\right )^{3} + 10 \, {\left ({\left (5 \, A + 6 \, C\right )} a^{3} + 18 \, B a^{2} b + 18 \, A a b^{2}\right )} \cos \left (d x + c\right )^{2} + 48 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{480 \, d \cos \left (d x + c\right )^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^3*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^7,x, algorithm="fricas")

[Out]

1/480*(15*((5*A + 6*C)*a^3 + 18*B*a^2*b + 6*(3*A + 4*C)*a*b^2 + 8*B*b^3)*cos(d*x + c)^6*log(sin(d*x + c) + 1)
- 15*((5*A + 6*C)*a^3 + 18*B*a^2*b + 6*(3*A + 4*C)*a*b^2 + 8*B*b^3)*cos(d*x + c)^6*log(-sin(d*x + c) + 1) + 2*
(16*(8*B*a^3 + 6*(4*A + 5*C)*a^2*b + 30*B*a*b^2 + 5*(2*A + 3*C)*b^3)*cos(d*x + c)^5 + 15*((5*A + 6*C)*a^3 + 18
*B*a^2*b + 6*(3*A + 4*C)*a*b^2 + 8*B*b^3)*cos(d*x + c)^4 + 40*A*a^3 + 16*(4*B*a^3 + 3*(4*A + 5*C)*a^2*b + 15*B
*a*b^2 + 5*A*b^3)*cos(d*x + c)^3 + 10*((5*A + 6*C)*a^3 + 18*B*a^2*b + 18*A*a*b^2)*cos(d*x + c)^2 + 48*(B*a^3 +
3*A*a^2*b)*cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^6)

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giac [B]  time = 0.44, size = 1370, normalized size = 4.08 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^3*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^7,x, algorithm="giac")

[Out]

1/240*(15*(5*A*a^3 + 6*C*a^3 + 18*B*a^2*b + 18*A*a*b^2 + 24*C*a*b^2 + 8*B*b^3)*log(abs(tan(1/2*d*x + 1/2*c) +
1)) - 15*(5*A*a^3 + 6*C*a^3 + 18*B*a^2*b + 18*A*a*b^2 + 24*C*a*b^2 + 8*B*b^3)*log(abs(tan(1/2*d*x + 1/2*c) - 1
)) + 2*(165*A*a^3*tan(1/2*d*x + 1/2*c)^11 - 240*B*a^3*tan(1/2*d*x + 1/2*c)^11 + 150*C*a^3*tan(1/2*d*x + 1/2*c)
^11 - 720*A*a^2*b*tan(1/2*d*x + 1/2*c)^11 + 450*B*a^2*b*tan(1/2*d*x + 1/2*c)^11 - 720*C*a^2*b*tan(1/2*d*x + 1/
2*c)^11 + 450*A*a*b^2*tan(1/2*d*x + 1/2*c)^11 - 720*B*a*b^2*tan(1/2*d*x + 1/2*c)^11 + 360*C*a*b^2*tan(1/2*d*x
+ 1/2*c)^11 - 240*A*b^3*tan(1/2*d*x + 1/2*c)^11 + 120*B*b^3*tan(1/2*d*x + 1/2*c)^11 - 240*C*b^3*tan(1/2*d*x +
1/2*c)^11 + 25*A*a^3*tan(1/2*d*x + 1/2*c)^9 + 560*B*a^3*tan(1/2*d*x + 1/2*c)^9 - 210*C*a^3*tan(1/2*d*x + 1/2*c
)^9 + 1680*A*a^2*b*tan(1/2*d*x + 1/2*c)^9 - 630*B*a^2*b*tan(1/2*d*x + 1/2*c)^9 + 2640*C*a^2*b*tan(1/2*d*x + 1/
2*c)^9 - 630*A*a*b^2*tan(1/2*d*x + 1/2*c)^9 + 2640*B*a*b^2*tan(1/2*d*x + 1/2*c)^9 - 1080*C*a*b^2*tan(1/2*d*x +
1/2*c)^9 + 880*A*b^3*tan(1/2*d*x + 1/2*c)^9 - 360*B*b^3*tan(1/2*d*x + 1/2*c)^9 + 1200*C*b^3*tan(1/2*d*x + 1/2
*c)^9 + 450*A*a^3*tan(1/2*d*x + 1/2*c)^7 - 1248*B*a^3*tan(1/2*d*x + 1/2*c)^7 + 60*C*a^3*tan(1/2*d*x + 1/2*c)^7
- 3744*A*a^2*b*tan(1/2*d*x + 1/2*c)^7 + 180*B*a^2*b*tan(1/2*d*x + 1/2*c)^7 - 4320*C*a^2*b*tan(1/2*d*x + 1/2*c
)^7 + 180*A*a*b^2*tan(1/2*d*x + 1/2*c)^7 - 4320*B*a*b^2*tan(1/2*d*x + 1/2*c)^7 + 720*C*a*b^2*tan(1/2*d*x + 1/2
*c)^7 - 1440*A*b^3*tan(1/2*d*x + 1/2*c)^7 + 240*B*b^3*tan(1/2*d*x + 1/2*c)^7 - 2400*C*b^3*tan(1/2*d*x + 1/2*c)
^7 + 450*A*a^3*tan(1/2*d*x + 1/2*c)^5 + 1248*B*a^3*tan(1/2*d*x + 1/2*c)^5 + 60*C*a^3*tan(1/2*d*x + 1/2*c)^5 +
3744*A*a^2*b*tan(1/2*d*x + 1/2*c)^5 + 180*B*a^2*b*tan(1/2*d*x + 1/2*c)^5 + 4320*C*a^2*b*tan(1/2*d*x + 1/2*c)^5
+ 180*A*a*b^2*tan(1/2*d*x + 1/2*c)^5 + 4320*B*a*b^2*tan(1/2*d*x + 1/2*c)^5 + 720*C*a*b^2*tan(1/2*d*x + 1/2*c)
^5 + 1440*A*b^3*tan(1/2*d*x + 1/2*c)^5 + 240*B*b^3*tan(1/2*d*x + 1/2*c)^5 + 2400*C*b^3*tan(1/2*d*x + 1/2*c)^5
+ 25*A*a^3*tan(1/2*d*x + 1/2*c)^3 - 560*B*a^3*tan(1/2*d*x + 1/2*c)^3 - 210*C*a^3*tan(1/2*d*x + 1/2*c)^3 - 1680
*A*a^2*b*tan(1/2*d*x + 1/2*c)^3 - 630*B*a^2*b*tan(1/2*d*x + 1/2*c)^3 - 2640*C*a^2*b*tan(1/2*d*x + 1/2*c)^3 - 6
30*A*a*b^2*tan(1/2*d*x + 1/2*c)^3 - 2640*B*a*b^2*tan(1/2*d*x + 1/2*c)^3 - 1080*C*a*b^2*tan(1/2*d*x + 1/2*c)^3
- 880*A*b^3*tan(1/2*d*x + 1/2*c)^3 - 360*B*b^3*tan(1/2*d*x + 1/2*c)^3 - 1200*C*b^3*tan(1/2*d*x + 1/2*c)^3 + 16
5*A*a^3*tan(1/2*d*x + 1/2*c) + 240*B*a^3*tan(1/2*d*x + 1/2*c) + 150*C*a^3*tan(1/2*d*x + 1/2*c) + 720*A*a^2*b*t
an(1/2*d*x + 1/2*c) + 450*B*a^2*b*tan(1/2*d*x + 1/2*c) + 720*C*a^2*b*tan(1/2*d*x + 1/2*c) + 450*A*a*b^2*tan(1/
2*d*x + 1/2*c) + 720*B*a*b^2*tan(1/2*d*x + 1/2*c) + 360*C*a*b^2*tan(1/2*d*x + 1/2*c) + 240*A*b^3*tan(1/2*d*x +
1/2*c) + 120*B*b^3*tan(1/2*d*x + 1/2*c) + 240*C*b^3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^6)/d

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maple [A]  time = 0.62, size = 644, normalized size = 1.92 \[ \frac {3 C a \,b^{2} \tan \left (d x +c \right ) \sec \left (d x +c \right )}{2 d}+\frac {C \,a^{2} b \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{d}+\frac {B a \,b^{2} \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{d}+\frac {5 A \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16 d}+\frac {2 A \,b^{3} \tan \left (d x +c \right )}{3 d}+\frac {b^{3} B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {3 C \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d}+\frac {b^{3} B \tan \left (d x +c \right ) \sec \left (d x +c \right )}{2 d}+\frac {8 a^{3} B \tan \left (d x +c \right )}{15 d}+\frac {b^{3} C \tan \left (d x +c \right )}{d}+\frac {5 A \,a^{3} \sec \left (d x +c \right ) \tan \left (d x +c \right )}{16 d}+\frac {3 A \,a^{2} b \tan \left (d x +c \right ) \left (\sec ^{4}\left (d x +c \right )\right )}{5 d}+\frac {9 a^{2} b B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d}+\frac {5 A \,a^{3} \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )}{24 d}+\frac {3 a^{2} b B \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )}{4 d}+\frac {3 A a \,b^{2} \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )}{4 d}+\frac {4 A \,a^{2} b \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{5 d}+\frac {3 C \,a^{3} \tan \left (d x +c \right ) \sec \left (d x +c \right )}{8 d}+\frac {2 C \,a^{2} b \tan \left (d x +c \right )}{d}+\frac {2 B a \,b^{2} \tan \left (d x +c \right )}{d}+\frac {3 C a \,b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {a^{3} B \tan \left (d x +c \right ) \left (\sec ^{4}\left (d x +c \right )\right )}{5 d}+\frac {A \,b^{3} \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{3 d}+\frac {A \,a^{3} \tan \left (d x +c \right ) \left (\sec ^{5}\left (d x +c \right )\right )}{6 d}+\frac {C \,a^{3} \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )}{4 d}+\frac {9 a^{2} b B \tan \left (d x +c \right ) \sec \left (d x +c \right )}{8 d}+\frac {4 a^{3} B \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{15 d}+\frac {8 A \,a^{2} b \tan \left (d x +c \right )}{5 d}+\frac {9 A a \,b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d}+\frac {9 A a \,b^{2} \tan \left (d x +c \right ) \sec \left (d x +c \right )}{8 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*cos(d*x+c))^3*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^7,x)

[Out]

3/2/d*C*a*b^2*tan(d*x+c)*sec(d*x+c)+3/5/d*A*a^2*b*tan(d*x+c)*sec(d*x+c)^4+1/d*C*a^2*b*tan(d*x+c)*sec(d*x+c)^2+
1/d*B*a*b^2*tan(d*x+c)*sec(d*x+c)^2+3/4/d*a^2*b*B*tan(d*x+c)*sec(d*x+c)^3+5/16/d*A*a^3*ln(sec(d*x+c)+tan(d*x+c
))+2/3/d*A*b^3*tan(d*x+c)+1/2/d*b^3*B*ln(sec(d*x+c)+tan(d*x+c))+3/8/d*C*a^3*ln(sec(d*x+c)+tan(d*x+c))+1/2/d*b^
3*B*tan(d*x+c)*sec(d*x+c)+8/15/d*a^3*B*tan(d*x+c)+1/d*b^3*C*tan(d*x+c)+5/24/d*A*a^3*tan(d*x+c)*sec(d*x+c)^3+4/
15/d*a^3*B*tan(d*x+c)*sec(d*x+c)^2+5/16/d*A*a^3*sec(d*x+c)*tan(d*x+c)+9/8/d*a^2*b*B*ln(sec(d*x+c)+tan(d*x+c))+
3/4/d*A*a*b^2*tan(d*x+c)*sec(d*x+c)^3+1/5/d*a^3*B*tan(d*x+c)*sec(d*x+c)^4+1/3/d*A*b^3*tan(d*x+c)*sec(d*x+c)^2+
1/6/d*A*a^3*tan(d*x+c)*sec(d*x+c)^5+1/4/d*C*a^3*tan(d*x+c)*sec(d*x+c)^3+3/8/d*C*a^3*tan(d*x+c)*sec(d*x+c)+2/d*
C*a^2*b*tan(d*x+c)+2/d*B*a*b^2*tan(d*x+c)+3/2/d*C*a*b^2*ln(sec(d*x+c)+tan(d*x+c))+9/8/d*a^2*b*B*tan(d*x+c)*sec
(d*x+c)+8/5/d*A*a^2*b*tan(d*x+c)+9/8/d*A*a*b^2*ln(sec(d*x+c)+tan(d*x+c))+9/8/d*A*a*b^2*tan(d*x+c)*sec(d*x+c)+4
/5/d*A*a^2*b*tan(d*x+c)*sec(d*x+c)^2

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maxima [A]  time = 0.38, size = 565, normalized size = 1.68 \[ \frac {32 \, {\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 )} B a^{3} + 96 \, {\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 + 480 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a^{2} b + 480 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a b^{2} + 160 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A b^{3} - 5 \, A a^{3} {\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 )} - 30 \, C a^{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 )} - 90 \, B a^{2} 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 )} - 90 \, A a b^{2} {\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 )} - 360 \, C a b^{2} {\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 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 )} + 480 \, C b^{3} \tan \left (d x + c\right )}{480 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^3*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^7,x, algorithm="maxima")

[Out]

1/480*(32*(3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c))*B*a^3 + 96*(3*tan(d*x + c)^5 + 10*tan(d*x +
c)^3 + 15*tan(d*x + c))*A*a^2*b + 480*(tan(d*x + c)^3 + 3*tan(d*x + c))*C*a^2*b + 480*(tan(d*x + c)^3 + 3*tan
(d*x + c))*B*a*b^2 + 160*(tan(d*x + c)^3 + 3*tan(d*x + c))*A*b^3 - 5*A*a^3*(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)) - 30*C*a^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)) - 90*B*a^2*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)) - 90*A*a*b^2
*(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)) - 360*C*a*b^2*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*
x + c) - 1)) - 120*B*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))
+ 480*C*b^3*tan(d*x + c))/d

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mupad [B]  time = 3.98, size = 766, normalized size = 2.28 \[ \frac {\mathrm {atanh}\left (\frac {4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {5\,A\,a^3}{16}+\frac {B\,b^3}{2}+\frac {3\,C\,a^3}{8}+\frac {9\,A\,a\,b^2}{8}+\frac {9\,B\,a^2\,b}{8}+\frac {3\,C\,a\,b^2}{2}\right )}{\frac {5\,A\,a^3}{4}+2\,B\,b^3+\frac {3\,C\,a^3}{2}+\frac {9\,A\,a\,b^2}{2}+\frac {9\,B\,a^2\,b}{2}+6\,C\,a\,b^2}\right )\,\left (\frac {5\,A\,a^3}{8}+B\,b^3+\frac {3\,C\,a^3}{4}+\frac {9\,A\,a\,b^2}{4}+\frac {9\,B\,a^2\,b}{4}+3\,C\,a\,b^2\right )}{d}+\frac {\left (\frac {11\,A\,a^3}{8}-2\,A\,b^3-2\,B\,a^3+B\,b^3+\frac {5\,C\,a^3}{4}-2\,C\,b^3+\frac {15\,A\,a\,b^2}{4}-6\,A\,a^2\,b-6\,B\,a\,b^2+\frac {15\,B\,a^2\,b}{4}+3\,C\,a\,b^2-6\,C\,a^2\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}+\left (\frac {5\,A\,a^3}{24}+\frac {22\,A\,b^3}{3}+\frac {14\,B\,a^3}{3}-3\,B\,b^3-\frac {7\,C\,a^3}{4}+10\,C\,b^3-\frac {21\,A\,a\,b^2}{4}+14\,A\,a^2\,b+22\,B\,a\,b^2-\frac {21\,B\,a^2\,b}{4}-9\,C\,a\,b^2+22\,C\,a^2\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (\frac {15\,A\,a^3}{4}-12\,A\,b^3-\frac {52\,B\,a^3}{5}+2\,B\,b^3+\frac {C\,a^3}{2}-20\,C\,b^3+\frac {3\,A\,a\,b^2}{2}-\frac {156\,A\,a^2\,b}{5}-36\,B\,a\,b^2+\frac {3\,B\,a^2\,b}{2}+6\,C\,a\,b^2-36\,C\,a^2\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {15\,A\,a^3}{4}+12\,A\,b^3+\frac {52\,B\,a^3}{5}+2\,B\,b^3+\frac {C\,a^3}{2}+20\,C\,b^3+\frac {3\,A\,a\,b^2}{2}+\frac {156\,A\,a^2\,b}{5}+36\,B\,a\,b^2+\frac {3\,B\,a^2\,b}{2}+6\,C\,a\,b^2+36\,C\,a^2\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {5\,A\,a^3}{24}-\frac {22\,A\,b^3}{3}-\frac {14\,B\,a^3}{3}-3\,B\,b^3-\frac {7\,C\,a^3}{4}-10\,C\,b^3-\frac {21\,A\,a\,b^2}{4}-14\,A\,a^2\,b-22\,B\,a\,b^2-\frac {21\,B\,a^2\,b}{4}-9\,C\,a\,b^2-22\,C\,a^2\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (\frac {11\,A\,a^3}{8}+2\,A\,b^3+2\,B\,a^3+B\,b^3+\frac {5\,C\,a^3}{4}+2\,C\,b^3+\frac {15\,A\,a\,b^2}{4}+6\,A\,a^2\,b+6\,B\,a\,b^2+\frac {15\,B\,a^2\,b}{4}+3\,C\,a\,b^2+6\,C\,a^2\,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 )}^{12}-6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-20\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-6\,{\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 + b*cos(c + d*x))^3*(A + B*cos(c + d*x) + C*cos(c + d*x)^2))/cos(c + d*x)^7,x)

[Out]

(atanh((4*tan(c/2 + (d*x)/2)*((5*A*a^3)/16 + (B*b^3)/2 + (3*C*a^3)/8 + (9*A*a*b^2)/8 + (9*B*a^2*b)/8 + (3*C*a*
b^2)/2))/((5*A*a^3)/4 + 2*B*b^3 + (3*C*a^3)/2 + (9*A*a*b^2)/2 + (9*B*a^2*b)/2 + 6*C*a*b^2))*((5*A*a^3)/8 + B*b
^3 + (3*C*a^3)/4 + (9*A*a*b^2)/4 + (9*B*a^2*b)/4 + 3*C*a*b^2))/d + (tan(c/2 + (d*x)/2)*((11*A*a^3)/8 + 2*A*b^3
+ 2*B*a^3 + B*b^3 + (5*C*a^3)/4 + 2*C*b^3 + (15*A*a*b^2)/4 + 6*A*a^2*b + 6*B*a*b^2 + (15*B*a^2*b)/4 + 3*C*a*b
^2 + 6*C*a^2*b) + tan(c/2 + (d*x)/2)^11*((11*A*a^3)/8 - 2*A*b^3 - 2*B*a^3 + B*b^3 + (5*C*a^3)/4 - 2*C*b^3 + (1
5*A*a*b^2)/4 - 6*A*a^2*b - 6*B*a*b^2 + (15*B*a^2*b)/4 + 3*C*a*b^2 - 6*C*a^2*b) - tan(c/2 + (d*x)/2)^3*((22*A*b
^3)/3 - (5*A*a^3)/24 + (14*B*a^3)/3 + 3*B*b^3 + (7*C*a^3)/4 + 10*C*b^3 + (21*A*a*b^2)/4 + 14*A*a^2*b + 22*B*a*
b^2 + (21*B*a^2*b)/4 + 9*C*a*b^2 + 22*C*a^2*b) + tan(c/2 + (d*x)/2)^9*((5*A*a^3)/24 + (22*A*b^3)/3 + (14*B*a^3
)/3 - 3*B*b^3 - (7*C*a^3)/4 + 10*C*b^3 - (21*A*a*b^2)/4 + 14*A*a^2*b + 22*B*a*b^2 - (21*B*a^2*b)/4 - 9*C*a*b^2
+ 22*C*a^2*b) + tan(c/2 + (d*x)/2)^5*((15*A*a^3)/4 + 12*A*b^3 + (52*B*a^3)/5 + 2*B*b^3 + (C*a^3)/2 + 20*C*b^3
+ (3*A*a*b^2)/2 + (156*A*a^2*b)/5 + 36*B*a*b^2 + (3*B*a^2*b)/2 + 6*C*a*b^2 + 36*C*a^2*b) + tan(c/2 + (d*x)/2)
^7*((15*A*a^3)/4 - 12*A*b^3 - (52*B*a^3)/5 + 2*B*b^3 + (C*a^3)/2 - 20*C*b^3 + (3*A*a*b^2)/2 - (156*A*a^2*b)/5
- 36*B*a*b^2 + (3*B*a^2*b)/2 + 6*C*a*b^2 - 36*C*a^2*b))/(d*(15*tan(c/2 + (d*x)/2)^4 - 6*tan(c/2 + (d*x)/2)^2 -
20*tan(c/2 + (d*x)/2)^6 + 15*tan(c/2 + (d*x)/2)^8 - 6*tan(c/2 + (d*x)/2)^10 + tan(c/2 + (d*x)/2)^12 + 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))**3*(A+B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)**7,x)

[Out]

Timed out

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