3.557 \(\int (a+b \cos (c+d x))^4 (A+C \cos ^2(c+d x)) \sec ^7(c+d x) \, dx\)

Optimal. Leaf size=307 \[ \frac{4 a b \left (2 a^2 (4 A+5 C)+5 b^2 (2 A+3 C)\right ) \tan (c+d x)}{15 d}+\frac{\left (12 a^2 b^2 (3 A+4 C)+a^4 (5 A+6 C)+8 b^4 (A+2 C)\right ) \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac{a b \left (a^2 (39 A+50 C)+4 A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{60 d}+\frac{\left (10 a^2 b^2 (49 A+66 C)+15 a^4 (5 A+6 C)+24 A b^4\right ) \tan (c+d x) \sec (c+d x)}{240 d}+\frac{\left (5 a^2 (5 A+6 C)+12 A b^2\right ) \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))^2}{120 d}+\frac{A \tan (c+d x) \sec ^5(c+d x) (a+b \cos (c+d x))^4}{6 d}+\frac{2 A b \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^3}{15 d} \]

[Out]

((8*b^4*(A + 2*C) + 12*a^2*b^2*(3*A + 4*C) + a^4*(5*A + 6*C))*ArcTanh[Sin[c + d*x]])/(16*d) + (4*a*b*(5*b^2*(2
*A + 3*C) + 2*a^2*(4*A + 5*C))*Tan[c + d*x])/(15*d) + ((24*A*b^4 + 15*a^4*(5*A + 6*C) + 10*a^2*b^2*(49*A + 66*
C))*Sec[c + d*x]*Tan[c + d*x])/(240*d) + (a*b*(4*A*b^2 + a^2*(39*A + 50*C))*Sec[c + d*x]^2*Tan[c + d*x])/(60*d
) + ((12*A*b^2 + 5*a^2*(5*A + 6*C))*(a + b*Cos[c + d*x])^2*Sec[c + d*x]^3*Tan[c + d*x])/(120*d) + (2*A*b*(a +
b*Cos[c + d*x])^3*Sec[c + d*x]^4*Tan[c + d*x])/(15*d) + (A*(a + b*Cos[c + d*x])^4*Sec[c + d*x]^5*Tan[c + d*x])
/(6*d)

________________________________________________________________________________________

Rubi [A]  time = 1.12453, antiderivative size = 307, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 8, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.242, Rules used = {3048, 3047, 3031, 3021, 2748, 3767, 8, 3770} \[ \frac{4 a b \left (2 a^2 (4 A+5 C)+5 b^2 (2 A+3 C)\right ) \tan (c+d x)}{15 d}+\frac{\left (12 a^2 b^2 (3 A+4 C)+a^4 (5 A+6 C)+8 b^4 (A+2 C)\right ) \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac{a b \left (a^2 (39 A+50 C)+4 A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{60 d}+\frac{\left (10 a^2 b^2 (49 A+66 C)+15 a^4 (5 A+6 C)+24 A b^4\right ) \tan (c+d x) \sec (c+d x)}{240 d}+\frac{\left (5 a^2 (5 A+6 C)+12 A b^2\right ) \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))^2}{120 d}+\frac{A \tan (c+d x) \sec ^5(c+d x) (a+b \cos (c+d x))^4}{6 d}+\frac{2 A b \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^3}{15 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Cos[c + d*x])^4*(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^7,x]

[Out]

((8*b^4*(A + 2*C) + 12*a^2*b^2*(3*A + 4*C) + a^4*(5*A + 6*C))*ArcTanh[Sin[c + d*x]])/(16*d) + (4*a*b*(5*b^2*(2
*A + 3*C) + 2*a^2*(4*A + 5*C))*Tan[c + d*x])/(15*d) + ((24*A*b^4 + 15*a^4*(5*A + 6*C) + 10*a^2*b^2*(49*A + 66*
C))*Sec[c + d*x]*Tan[c + d*x])/(240*d) + (a*b*(4*A*b^2 + a^2*(39*A + 50*C))*Sec[c + d*x]^2*Tan[c + d*x])/(60*d
) + ((12*A*b^2 + 5*a^2*(5*A + 6*C))*(a + b*Cos[c + d*x])^2*Sec[c + d*x]^3*Tan[c + d*x])/(120*d) + (2*A*b*(a +
b*Cos[c + d*x])^3*Sec[c + d*x]^4*Tan[c + d*x])/(15*d) + (A*(a + b*Cos[c + d*x])^4*Sec[c + d*x]^5*Tan[c + d*x])
/(6*d)

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 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 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 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 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]

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 ^7(c+d x) \, dx &=\frac{A (a+b \cos (c+d x))^4 \sec ^5(c+d x) \tan (c+d x)}{6 d}+\frac{1}{6} \int (a+b \cos (c+d x))^3 \left (4 A b+a (5 A+6 C) \cos (c+d x)+b (A+6 C) \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx\\ &=\frac{2 A b (a+b \cos (c+d x))^3 \sec ^4(c+d x) \tan (c+d x)}{15 d}+\frac{A (a+b \cos (c+d x))^4 \sec ^5(c+d x) \tan (c+d x)}{6 d}+\frac{1}{30} \int (a+b \cos (c+d x))^2 \left (12 A b^2+5 a^2 (5 A+6 C)+2 a b (23 A+30 C) \cos (c+d x)+3 b^2 (3 A+10 C) \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx\\ &=\frac{\left (12 A b^2+5 a^2 (5 A+6 C)\right ) (a+b \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{120 d}+\frac{2 A b (a+b \cos (c+d x))^3 \sec ^4(c+d x) \tan (c+d x)}{15 d}+\frac{A (a+b \cos (c+d x))^4 \sec ^5(c+d x) \tan (c+d x)}{6 d}+\frac{1}{120} \int (a+b \cos (c+d x)) \left (6 b \left (4 A b^2+a^2 (39 A+50 C)\right )+a \left (15 a^2 (5 A+6 C)+8 b^2 (32 A+45 C)\right ) \cos (c+d x)+b \left (24 b^2 (2 A+5 C)+5 a^2 (5 A+6 C)\right ) \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx\\ &=\frac{a b \left (4 A b^2+a^2 (39 A+50 C)\right ) \sec ^2(c+d x) \tan (c+d x)}{60 d}+\frac{\left (12 A b^2+5 a^2 (5 A+6 C)\right ) (a+b \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{120 d}+\frac{2 A b (a+b \cos (c+d x))^3 \sec ^4(c+d x) \tan (c+d x)}{15 d}+\frac{A (a+b \cos (c+d x))^4 \sec ^5(c+d x) \tan (c+d x)}{6 d}-\frac{1}{360} \int \left (-3 \left (24 A b^4+15 a^4 (5 A+6 C)+10 a^2 b^2 (49 A+66 C)\right )-96 a b \left (5 b^2 (2 A+3 C)+2 a^2 (4 A+5 C)\right ) \cos (c+d x)-3 b^2 \left (24 b^2 (2 A+5 C)+5 a^2 (5 A+6 C)\right ) \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx\\ &=\frac{\left (24 A b^4+15 a^4 (5 A+6 C)+10 a^2 b^2 (49 A+66 C)\right ) \sec (c+d x) \tan (c+d x)}{240 d}+\frac{a b \left (4 A b^2+a^2 (39 A+50 C)\right ) \sec ^2(c+d x) \tan (c+d x)}{60 d}+\frac{\left (12 A b^2+5 a^2 (5 A+6 C)\right ) (a+b \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{120 d}+\frac{2 A b (a+b \cos (c+d x))^3 \sec ^4(c+d x) \tan (c+d x)}{15 d}+\frac{A (a+b \cos (c+d x))^4 \sec ^5(c+d x) \tan (c+d x)}{6 d}-\frac{1}{720} \int \left (-192 a b \left (5 b^2 (2 A+3 C)+2 a^2 (4 A+5 C)\right )-45 \left (8 b^4 (A+2 C)+12 a^2 b^2 (3 A+4 C)+a^4 (5 A+6 C)\right ) \cos (c+d x)\right ) \sec ^2(c+d x) \, dx\\ &=\frac{\left (24 A b^4+15 a^4 (5 A+6 C)+10 a^2 b^2 (49 A+66 C)\right ) \sec (c+d x) \tan (c+d x)}{240 d}+\frac{a b \left (4 A b^2+a^2 (39 A+50 C)\right ) \sec ^2(c+d x) \tan (c+d x)}{60 d}+\frac{\left (12 A b^2+5 a^2 (5 A+6 C)\right ) (a+b \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{120 d}+\frac{2 A b (a+b \cos (c+d x))^3 \sec ^4(c+d x) \tan (c+d x)}{15 d}+\frac{A (a+b \cos (c+d x))^4 \sec ^5(c+d x) \tan (c+d x)}{6 d}+\frac{1}{15} \left (4 a b \left (5 b^2 (2 A+3 C)+2 a^2 (4 A+5 C)\right )\right ) \int \sec ^2(c+d x) \, dx-\frac{1}{16} \left (-8 b^4 (A+2 C)-12 a^2 b^2 (3 A+4 C)-a^4 (5 A+6 C)\right ) \int \sec (c+d x) \, dx\\ &=\frac{\left (8 b^4 (A+2 C)+12 a^2 b^2 (3 A+4 C)+a^4 (5 A+6 C)\right ) \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac{\left (24 A b^4+15 a^4 (5 A+6 C)+10 a^2 b^2 (49 A+66 C)\right ) \sec (c+d x) \tan (c+d x)}{240 d}+\frac{a b \left (4 A b^2+a^2 (39 A+50 C)\right ) \sec ^2(c+d x) \tan (c+d x)}{60 d}+\frac{\left (12 A b^2+5 a^2 (5 A+6 C)\right ) (a+b \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{120 d}+\frac{2 A b (a+b \cos (c+d x))^3 \sec ^4(c+d x) \tan (c+d x)}{15 d}+\frac{A (a+b \cos (c+d x))^4 \sec ^5(c+d x) \tan (c+d x)}{6 d}-\frac{\left (4 a b \left (5 b^2 (2 A+3 C)+2 a^2 (4 A+5 C)\right )\right ) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{15 d}\\ &=\frac{\left (8 b^4 (A+2 C)+12 a^2 b^2 (3 A+4 C)+a^4 (5 A+6 C)\right ) \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac{4 a b \left (5 b^2 (2 A+3 C)+2 a^2 (4 A+5 C)\right ) \tan (c+d x)}{15 d}+\frac{\left (24 A b^4+15 a^4 (5 A+6 C)+10 a^2 b^2 (49 A+66 C)\right ) \sec (c+d x) \tan (c+d x)}{240 d}+\frac{a b \left (4 A b^2+a^2 (39 A+50 C)\right ) \sec ^2(c+d x) \tan (c+d x)}{60 d}+\frac{\left (12 A b^2+5 a^2 (5 A+6 C)\right ) (a+b \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{120 d}+\frac{2 A b (a+b \cos (c+d x))^3 \sec ^4(c+d x) \tan (c+d x)}{15 d}+\frac{A (a+b \cos (c+d x))^4 \sec ^5(c+d x) \tan (c+d x)}{6 d}\\ \end{align*}

Mathematica [A]  time = 4.90567, size = 204, normalized size = 0.66 \[ \frac{15 \left (12 a^2 b^2 (3 A+4 C)+a^4 (5 A+6 C)+8 b^4 (A+2 C)\right ) \tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \left (64 a b \left (5 \left (a^2 (2 A+C)+A b^2\right ) \tan ^2(c+d x)+15 \left (a^2+b^2\right ) (A+C)+3 a^2 A \tan ^4(c+d x)\right )+10 a^2 \left (a^2 (5 A+6 C)+36 A b^2\right ) \sec ^3(c+d x)+15 \left (12 a^2 b^2 (3 A+4 C)+a^4 (5 A+6 C)+8 A b^4\right ) \sec (c+d x)+40 a^4 A \sec ^5(c+d x)\right )}{240 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Cos[c + d*x])^4*(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^7,x]

[Out]

(15*(8*b^4*(A + 2*C) + 12*a^2*b^2*(3*A + 4*C) + a^4*(5*A + 6*C))*ArcTanh[Sin[c + d*x]] + Tan[c + d*x]*(15*(8*A
*b^4 + 12*a^2*b^2*(3*A + 4*C) + a^4*(5*A + 6*C))*Sec[c + d*x] + 10*a^2*(36*A*b^2 + a^2*(5*A + 6*C))*Sec[c + d*
x]^3 + 40*a^4*A*Sec[c + d*x]^5 + 64*a*b*(15*(a^2 + b^2)*(A + C) + 5*(A*b^2 + a^2*(2*A + C))*Tan[c + d*x]^2 + 3
*a^2*A*Tan[c + d*x]^4)))/(240*d)

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Maple [A]  time = 0.073, size = 511, normalized size = 1.7 \begin{align*} \text{result too large to display} \end{align*}

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)^7,x)

[Out]

1/2/d*A*b^4*sec(d*x+c)*tan(d*x+c)+1/2/d*A*b^4*ln(sec(d*x+c)+tan(d*x+c))+1/d*C*b^4*ln(sec(d*x+c)+tan(d*x+c))+8/
3/d*a*A*b^3*tan(d*x+c)+4/3/d*a*A*b^3*tan(d*x+c)*sec(d*x+c)^2+4/d*C*a*b^3*tan(d*x+c)+3/2/d*a^2*A*b^2*tan(d*x+c)
*sec(d*x+c)^3+9/4/d*a^2*A*b^2*sec(d*x+c)*tan(d*x+c)+9/4/d*a^2*A*b^2*ln(sec(d*x+c)+tan(d*x+c))+3/d*a^2*b^2*C*se
c(d*x+c)*tan(d*x+c)+3/d*a^2*b^2*C*ln(sec(d*x+c)+tan(d*x+c))+32/15/d*A*a^3*b*tan(d*x+c)+4/5/d*A*a^3*b*tan(d*x+c
)*sec(d*x+c)^4+16/15/d*A*a^3*b*tan(d*x+c)*sec(d*x+c)^2+8/3/d*a^3*b*C*tan(d*x+c)+4/3/d*a^3*b*C*tan(d*x+c)*sec(d
*x+c)^2+1/6/d*A*a^4*tan(d*x+c)*sec(d*x+c)^5+5/24/d*A*a^4*tan(d*x+c)*sec(d*x+c)^3+5/16/d*A*a^4*sec(d*x+c)*tan(d
*x+c)+5/16/d*A*a^4*ln(sec(d*x+c)+tan(d*x+c))+1/4/d*a^4*C*tan(d*x+c)*sec(d*x+c)^3+3/8/d*a^4*C*sec(d*x+c)*tan(d*
x+c)+3/8/d*a^4*C*ln(sec(d*x+c)+tan(d*x+c))

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Maxima [A]  time = 1.0382, size = 629, normalized size = 2.05 \begin{align*} \frac{128 \,{\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^{3} b + 640 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a^{3} b + 640 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a b^{3} - 5 \, A a^{4}{\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^{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 )} - 180 \, A a^{2} 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 )} - 720 \, C a^{2} 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 \, A b^{4}{\left (\frac{2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 240 \, C b^{4}{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 1920 \, C a b^{3} \tan \left (d x + c\right )}{480 \, d} \end{align*}

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)^7,x, algorithm="maxima")

[Out]

1/480*(128*(3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c))*A*a^3*b + 640*(tan(d*x + c)^3 + 3*tan(d*x
+ c))*C*a^3*b + 640*(tan(d*x + c)^3 + 3*tan(d*x + c))*A*a*b^3 - 5*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)) - 30*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)) - 180*A*a^2*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)) - 720*C*a^2*
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*A*b^4*(2*sin(d
*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) + 240*C*b^4*(log(sin(d*x + c) +
1) - log(sin(d*x + c) - 1)) + 1920*C*a*b^3*tan(d*x + c))/d

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Fricas [A]  time = 1.72725, size = 716, normalized size = 2.33 \begin{align*} \frac{15 \,{\left ({\left (5 \, A + 6 \, C\right )} a^{4} + 12 \,{\left (3 \, A + 4 \, C\right )} a^{2} b^{2} + 8 \,{\left (A + 2 \, C\right )} b^{4}\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^{4} + 12 \,{\left (3 \, A + 4 \, C\right )} a^{2} b^{2} + 8 \,{\left (A + 2 \, C\right )} b^{4}\right )} \cos \left (d x + c\right )^{6} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (192 \, A a^{3} b \cos \left (d x + c\right ) + 64 \,{\left (2 \,{\left (4 \, A + 5 \, C\right )} a^{3} b + 5 \,{\left (2 \, A + 3 \, C\right )} a b^{3}\right )} \cos \left (d x + c\right )^{5} + 40 \, A a^{4} + 15 \,{\left ({\left (5 \, A + 6 \, C\right )} a^{4} + 12 \,{\left (3 \, A + 4 \, C\right )} a^{2} b^{2} + 8 \, A b^{4}\right )} \cos \left (d x + c\right )^{4} + 64 \,{\left ({\left (4 \, A + 5 \, C\right )} a^{3} b + 5 \, A a b^{3}\right )} \cos \left (d x + c\right )^{3} + 10 \,{\left ({\left (5 \, A + 6 \, C\right )} a^{4} + 36 \, A a^{2} b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{480 \, d \cos \left (d x + c\right )^{6}} \end{align*}

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)^7,x, algorithm="fricas")

[Out]

1/480*(15*((5*A + 6*C)*a^4 + 12*(3*A + 4*C)*a^2*b^2 + 8*(A + 2*C)*b^4)*cos(d*x + c)^6*log(sin(d*x + c) + 1) -
15*((5*A + 6*C)*a^4 + 12*(3*A + 4*C)*a^2*b^2 + 8*(A + 2*C)*b^4)*cos(d*x + c)^6*log(-sin(d*x + c) + 1) + 2*(192
*A*a^3*b*cos(d*x + c) + 64*(2*(4*A + 5*C)*a^3*b + 5*(2*A + 3*C)*a*b^3)*cos(d*x + c)^5 + 40*A*a^4 + 15*((5*A +
6*C)*a^4 + 12*(3*A + 4*C)*a^2*b^2 + 8*A*b^4)*cos(d*x + c)^4 + 64*((4*A + 5*C)*a^3*b + 5*A*a*b^3)*cos(d*x + c)^
3 + 10*((5*A + 6*C)*a^4 + 36*A*a^2*b^2)*cos(d*x + c)^2)*sin(d*x + c))/(d*cos(d*x + c)^6)

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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+b*cos(d*x+c))**4*(A+C*cos(d*x+c)**2)*sec(d*x+c)**7,x)

[Out]

Timed out

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Giac [B]  time = 1.55314, size = 1485, normalized size = 4.84 \begin{align*} \text{result too large to display} \end{align*}

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)^7,x, algorithm="giac")

[Out]

1/240*(15*(5*A*a^4 + 6*C*a^4 + 36*A*a^2*b^2 + 48*C*a^2*b^2 + 8*A*b^4 + 16*C*b^4)*log(abs(tan(1/2*d*x + 1/2*c)
+ 1)) - 15*(5*A*a^4 + 6*C*a^4 + 36*A*a^2*b^2 + 48*C*a^2*b^2 + 8*A*b^4 + 16*C*b^4)*log(abs(tan(1/2*d*x + 1/2*c)
- 1)) + 2*(165*A*a^4*tan(1/2*d*x + 1/2*c)^11 + 150*C*a^4*tan(1/2*d*x + 1/2*c)^11 - 960*A*a^3*b*tan(1/2*d*x +
1/2*c)^11 - 960*C*a^3*b*tan(1/2*d*x + 1/2*c)^11 + 900*A*a^2*b^2*tan(1/2*d*x + 1/2*c)^11 + 720*C*a^2*b^2*tan(1/
2*d*x + 1/2*c)^11 - 960*A*a*b^3*tan(1/2*d*x + 1/2*c)^11 - 960*C*a*b^3*tan(1/2*d*x + 1/2*c)^11 + 120*A*b^4*tan(
1/2*d*x + 1/2*c)^11 + 25*A*a^4*tan(1/2*d*x + 1/2*c)^9 - 210*C*a^4*tan(1/2*d*x + 1/2*c)^9 + 2240*A*a^3*b*tan(1/
2*d*x + 1/2*c)^9 + 3520*C*a^3*b*tan(1/2*d*x + 1/2*c)^9 - 1260*A*a^2*b^2*tan(1/2*d*x + 1/2*c)^9 - 2160*C*a^2*b^
2*tan(1/2*d*x + 1/2*c)^9 + 3520*A*a*b^3*tan(1/2*d*x + 1/2*c)^9 + 4800*C*a*b^3*tan(1/2*d*x + 1/2*c)^9 - 360*A*b
^4*tan(1/2*d*x + 1/2*c)^9 + 450*A*a^4*tan(1/2*d*x + 1/2*c)^7 + 60*C*a^4*tan(1/2*d*x + 1/2*c)^7 - 4992*A*a^3*b*
tan(1/2*d*x + 1/2*c)^7 - 5760*C*a^3*b*tan(1/2*d*x + 1/2*c)^7 + 360*A*a^2*b^2*tan(1/2*d*x + 1/2*c)^7 + 1440*C*a
^2*b^2*tan(1/2*d*x + 1/2*c)^7 - 5760*A*a*b^3*tan(1/2*d*x + 1/2*c)^7 - 9600*C*a*b^3*tan(1/2*d*x + 1/2*c)^7 + 24
0*A*b^4*tan(1/2*d*x + 1/2*c)^7 + 450*A*a^4*tan(1/2*d*x + 1/2*c)^5 + 60*C*a^4*tan(1/2*d*x + 1/2*c)^5 + 4992*A*a
^3*b*tan(1/2*d*x + 1/2*c)^5 + 5760*C*a^3*b*tan(1/2*d*x + 1/2*c)^5 + 360*A*a^2*b^2*tan(1/2*d*x + 1/2*c)^5 + 144
0*C*a^2*b^2*tan(1/2*d*x + 1/2*c)^5 + 5760*A*a*b^3*tan(1/2*d*x + 1/2*c)^5 + 9600*C*a*b^3*tan(1/2*d*x + 1/2*c)^5
+ 240*A*b^4*tan(1/2*d*x + 1/2*c)^5 + 25*A*a^4*tan(1/2*d*x + 1/2*c)^3 - 210*C*a^4*tan(1/2*d*x + 1/2*c)^3 - 224
0*A*a^3*b*tan(1/2*d*x + 1/2*c)^3 - 3520*C*a^3*b*tan(1/2*d*x + 1/2*c)^3 - 1260*A*a^2*b^2*tan(1/2*d*x + 1/2*c)^3
- 2160*C*a^2*b^2*tan(1/2*d*x + 1/2*c)^3 - 3520*A*a*b^3*tan(1/2*d*x + 1/2*c)^3 - 4800*C*a*b^3*tan(1/2*d*x + 1/
2*c)^3 - 360*A*b^4*tan(1/2*d*x + 1/2*c)^3 + 165*A*a^4*tan(1/2*d*x + 1/2*c) + 150*C*a^4*tan(1/2*d*x + 1/2*c) +
960*A*a^3*b*tan(1/2*d*x + 1/2*c) + 960*C*a^3*b*tan(1/2*d*x + 1/2*c) + 900*A*a^2*b^2*tan(1/2*d*x + 1/2*c) + 720
*C*a^2*b^2*tan(1/2*d*x + 1/2*c) + 960*A*a*b^3*tan(1/2*d*x + 1/2*c) + 960*C*a*b^3*tan(1/2*d*x + 1/2*c) + 120*A*
b^4*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^6)/d