3.249 \(\int (a+b \cos (c+d x))^4 (A+B \cos (c+d x)) \sec ^7(c+d x) \, dx\)

Optimal. Leaf size=324 \[ \frac{\left (32 a^3 A b+60 a^2 b^2 B+8 a^4 B+40 a A b^3+15 b^4 B\right ) \tan (c+d x)}{15 d}+\frac{\left (36 a^2 A b^2+5 a^4 A+24 a^3 b B+32 a b^3 B+8 A b^4\right ) \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac{a^2 \left (25 a^2 A+72 a b B+48 A b^2\right ) \tan (c+d x) \sec ^3(c+d x)}{120 d}+\frac{a \left (16 a^2 A b+4 a^3 B+27 a b^2 B+13 A b^3\right ) \tan (c+d x) \sec ^2(c+d x)}{15 d}+\frac{\left (36 a^2 A b^2+5 a^4 A+24 a^3 b B+32 a b^3 B+8 A b^4\right ) \tan (c+d x) \sec (c+d x)}{16 d}+\frac{a (2 a B+3 A b) \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^2}{10 d}+\frac{a A \tan (c+d x) \sec ^5(c+d x) (a+b \cos (c+d x))^3}{6 d} \]

[Out]

((5*a^4*A + 36*a^2*A*b^2 + 8*A*b^4 + 24*a^3*b*B + 32*a*b^3*B)*ArcTanh[Sin[c + d*x]])/(16*d) + ((32*a^3*A*b + 4
0*a*A*b^3 + 8*a^4*B + 60*a^2*b^2*B + 15*b^4*B)*Tan[c + d*x])/(15*d) + ((5*a^4*A + 36*a^2*A*b^2 + 8*A*b^4 + 24*
a^3*b*B + 32*a*b^3*B)*Sec[c + d*x]*Tan[c + d*x])/(16*d) + (a*(16*a^2*A*b + 13*A*b^3 + 4*a^3*B + 27*a*b^2*B)*Se
c[c + d*x]^2*Tan[c + d*x])/(15*d) + (a^2*(25*a^2*A + 48*A*b^2 + 72*a*b*B)*Sec[c + d*x]^3*Tan[c + d*x])/(120*d)
+ (a*(3*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*(a + b*Cos[c + d*x])^3
*Sec[c + d*x]^5*Tan[c + d*x])/(6*d)

________________________________________________________________________________________

Rubi [A]  time = 0.802034, antiderivative size = 324, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 9, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.29, Rules used = {2989, 3047, 3031, 3021, 2748, 3768, 3770, 3767, 8} \[ \frac{\left (32 a^3 A b+60 a^2 b^2 B+8 a^4 B+40 a A b^3+15 b^4 B\right ) \tan (c+d x)}{15 d}+\frac{\left (36 a^2 A b^2+5 a^4 A+24 a^3 b B+32 a b^3 B+8 A b^4\right ) \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac{a^2 \left (25 a^2 A+72 a b B+48 A b^2\right ) \tan (c+d x) \sec ^3(c+d x)}{120 d}+\frac{a \left (16 a^2 A b+4 a^3 B+27 a b^2 B+13 A b^3\right ) \tan (c+d x) \sec ^2(c+d x)}{15 d}+\frac{\left (36 a^2 A b^2+5 a^4 A+24 a^3 b B+32 a b^3 B+8 A b^4\right ) \tan (c+d x) \sec (c+d x)}{16 d}+\frac{a (2 a B+3 A b) \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^2}{10 d}+\frac{a A \tan (c+d x) \sec ^5(c+d x) (a+b \cos (c+d x))^3}{6 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((5*a^4*A + 36*a^2*A*b^2 + 8*A*b^4 + 24*a^3*b*B + 32*a*b^3*B)*ArcTanh[Sin[c + d*x]])/(16*d) + ((32*a^3*A*b + 4
0*a*A*b^3 + 8*a^4*B + 60*a^2*b^2*B + 15*b^4*B)*Tan[c + d*x])/(15*d) + ((5*a^4*A + 36*a^2*A*b^2 + 8*A*b^4 + 24*
a^3*b*B + 32*a*b^3*B)*Sec[c + d*x]*Tan[c + d*x])/(16*d) + (a*(16*a^2*A*b + 13*A*b^3 + 4*a^3*B + 27*a*b^2*B)*Se
c[c + d*x]^2*Tan[c + d*x])/(15*d) + (a^2*(25*a^2*A + 48*A*b^2 + 72*a*b*B)*Sec[c + d*x]^3*Tan[c + d*x])/(120*d)
+ (a*(3*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*(a + b*Cos[c + d*x])^3
*Sec[c + d*x]^5*Tan[c + d*x])/(6*d)

Rule 2989

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*c - a*d)*(B*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)
*(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 - 2)*(c + d*Sin[e + f*x])^(n + 1)*Simp[b*(b*c - a*d)*(B*c - A*d)*(m - 1) + a*d*(a*A*c + b*B*c - (
A*b + a*B)*d)*(n + 1) + (b*(b*d*(B*c - A*d) + a*(A*c*d + B*(c^2 - 2*d^2)))*(n + 1) - a*(b*c - a*d)*(B*c - A*d)
*(n + 2))*Sin[e + f*x] + b*(d*(A*b*c + a*B*c - a*A*d)*(m + n + 1) - b*B*(c^2*m + d^2*(n + 1)))*Sin[e + f*x]^2,
x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2,
0] && GtQ[m, 1] && 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 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+b \cos (c+d x))^4 (A+B \cos (c+d x)) \sec ^7(c+d x) \, dx &=\frac{a 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 (3 A b+2 a B)+\left (5 a^2 A+6 A b^2+12 a b B\right ) \cos (c+d x)+2 b (a A+3 b B) \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx\\ &=\frac{a (3 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 (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 (a \left (25 a^2 A+48 A b^2+72 a b B\right )+\left (71 a^2 A b+30 A b^3+24 a^3 B+90 a b^2 B\right ) \cos (c+d x)+2 b \left (14 a A b+6 a^2 B+15 b^2 B\right ) \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx\\ &=\frac{a^2 \left (25 a^2 A+48 A b^2+72 a b B\right ) \sec ^3(c+d x) \tan (c+d x)}{120 d}+\frac{a (3 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 (a+b \cos (c+d x))^3 \sec ^5(c+d x) \tan (c+d x)}{6 d}-\frac{1}{120} \int \left (-24 a \left (16 a^2 A b+13 A b^3+4 a^3 B+27 a b^2 B\right )-15 \left (5 a^4 A+36 a^2 A b^2+8 A b^4+24 a^3 b B+32 a b^3 B\right ) \cos (c+d x)-8 b^2 \left (14 a A b+6 a^2 B+15 b^2 B\right ) \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx\\ &=\frac{a \left (16 a^2 A b+13 A b^3+4 a^3 B+27 a b^2 B\right ) \sec ^2(c+d x) \tan (c+d x)}{15 d}+\frac{a^2 \left (25 a^2 A+48 A b^2+72 a b B\right ) \sec ^3(c+d x) \tan (c+d x)}{120 d}+\frac{a (3 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 (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 (5 a^4 A+36 a^2 A b^2+8 A b^4+24 a^3 b B+32 a b^3 B\right )-24 \left (32 a^3 A b+40 a A b^3+8 a^4 B+60 a^2 b^2 B+15 b^4 B\right ) \cos (c+d x)\right ) \sec ^3(c+d x) \, dx\\ &=\frac{a \left (16 a^2 A b+13 A b^3+4 a^3 B+27 a b^2 B\right ) \sec ^2(c+d x) \tan (c+d x)}{15 d}+\frac{a^2 \left (25 a^2 A+48 A b^2+72 a b B\right ) \sec ^3(c+d x) \tan (c+d x)}{120 d}+\frac{a (3 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 (a+b \cos (c+d x))^3 \sec ^5(c+d x) \tan (c+d x)}{6 d}-\frac{1}{8} \left (-5 a^4 A-36 a^2 A b^2-8 A b^4-24 a^3 b B-32 a b^3 B\right ) \int \sec ^3(c+d x) \, dx-\frac{1}{15} \left (-32 a^3 A b-40 a A b^3-8 a^4 B-60 a^2 b^2 B-15 b^4 B\right ) \int \sec ^2(c+d x) \, dx\\ &=\frac{\left (5 a^4 A+36 a^2 A b^2+8 A b^4+24 a^3 b B+32 a b^3 B\right ) \sec (c+d x) \tan (c+d x)}{16 d}+\frac{a \left (16 a^2 A b+13 A b^3+4 a^3 B+27 a b^2 B\right ) \sec ^2(c+d x) \tan (c+d x)}{15 d}+\frac{a^2 \left (25 a^2 A+48 A b^2+72 a b B\right ) \sec ^3(c+d x) \tan (c+d x)}{120 d}+\frac{a (3 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 (a+b \cos (c+d x))^3 \sec ^5(c+d x) \tan (c+d x)}{6 d}-\frac{1}{16} \left (-5 a^4 A-36 a^2 A b^2-8 A b^4-24 a^3 b B-32 a b^3 B\right ) \int \sec (c+d x) \, dx-\frac{\left (32 a^3 A b+40 a A b^3+8 a^4 B+60 a^2 b^2 B+15 b^4 B\right ) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{15 d}\\ &=\frac{\left (5 a^4 A+36 a^2 A b^2+8 A b^4+24 a^3 b B+32 a b^3 B\right ) \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac{\left (32 a^3 A b+40 a A b^3+8 a^4 B+60 a^2 b^2 B+15 b^4 B\right ) \tan (c+d x)}{15 d}+\frac{\left (5 a^4 A+36 a^2 A b^2+8 A b^4+24 a^3 b B+32 a b^3 B\right ) \sec (c+d x) \tan (c+d x)}{16 d}+\frac{a \left (16 a^2 A b+13 A b^3+4 a^3 B+27 a b^2 B\right ) \sec ^2(c+d x) \tan (c+d x)}{15 d}+\frac{a^2 \left (25 a^2 A+48 A b^2+72 a b B\right ) \sec ^3(c+d x) \tan (c+d x)}{120 d}+\frac{a (3 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 (a+b \cos (c+d x))^3 \sec ^5(c+d x) \tan (c+d x)}{6 d}\\ \end{align*}

Mathematica [A]  time = 2.64182, size = 244, normalized size = 0.75 \[ \frac{15 \left (36 a^2 A b^2+5 a^4 A+24 a^3 b B+32 a b^3 B+8 A b^4\right ) \tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \left (160 a \left (4 a^2 A b+a^3 B+3 a b^2 B+2 A b^3\right ) \tan ^2(c+d x)+10 a^2 \left (5 a^2 A+24 a b B+36 A b^2\right ) \sec ^3(c+d x)+15 \left (36 a^2 A b^2+5 a^4 A+24 a^3 b B+32 a b^3 B+8 A b^4\right ) \sec (c+d x)+240 \left (4 a^3 A b+6 a^2 b^2 B+a^4 B+4 a A b^3+b^4 B\right )+48 a^3 (a B+4 A b) \tan ^4(c+d x)+40 a^4 A \sec ^5(c+d x)\right )}{240 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(15*(5*a^4*A + 36*a^2*A*b^2 + 8*A*b^4 + 24*a^3*b*B + 32*a*b^3*B)*ArcTanh[Sin[c + d*x]] + Tan[c + d*x]*(240*(4*
a^3*A*b + 4*a*A*b^3 + a^4*B + 6*a^2*b^2*B + b^4*B) + 15*(5*a^4*A + 36*a^2*A*b^2 + 8*A*b^4 + 24*a^3*b*B + 32*a*
b^3*B)*Sec[c + d*x] + 10*a^2*(5*a^2*A + 36*A*b^2 + 24*a*b*B)*Sec[c + d*x]^3 + 40*a^4*A*Sec[c + d*x]^5 + 160*a*
(4*a^2*A*b + 2*A*b^3 + a^3*B + 3*a*b^2*B)*Tan[c + d*x]^2 + 48*a^3*(4*A*b + a*B)*Tan[c + d*x]^4))/(240*d)

________________________________________________________________________________________

Maple [A]  time = 0.098, size = 550, 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+B*cos(d*x+c))*sec(d*x+c)^7,x)

[Out]

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

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Maxima [A]  time = 1.02618, size = 640, normalized size = 1.98 \begin{align*} \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^{4} + 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 + 960 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a^{2} b^{2} + 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 )} - 120 \, B 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 )} - 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 )} - 480 \, B 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 )} - 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 )} + 480 \, B b^{4} \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+B*cos(d*x+c))*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^4 + 128*(3*tan(d*x + c)^5 + 10*tan(d*x
+ c)^3 + 15*tan(d*x + c))*A*a^3*b + 960*(tan(d*x + c)^3 + 3*tan(d*x + c))*B*a^2*b^2 + 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)) - 120*B*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)) - 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)) - 480*B*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)) - 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)) + 480*B*b^4*tan(d*x + c))/d

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Fricas [A]  time = 1.67112, size = 797, normalized size = 2.46 \begin{align*} \frac{15 \,{\left (5 \, A a^{4} + 24 \, B a^{3} b + 36 \, A a^{2} b^{2} + 32 \, B a b^{3} + 8 \, A b^{4}\right )} \cos \left (d x + c\right )^{6} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \,{\left (5 \, A a^{4} + 24 \, B a^{3} b + 36 \, A a^{2} b^{2} + 32 \, B a b^{3} + 8 \, A b^{4}\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^{4} + 32 \, A a^{3} b + 60 \, B a^{2} b^{2} + 40 \, A a b^{3} + 15 \, B b^{4}\right )} \cos \left (d x + c\right )^{5} + 40 \, A a^{4} + 15 \,{\left (5 \, A a^{4} + 24 \, B a^{3} b + 36 \, A a^{2} b^{2} + 32 \, B a b^{3} + 8 \, A b^{4}\right )} \cos \left (d x + c\right )^{4} + 32 \,{\left (2 \, B a^{4} + 8 \, A a^{3} b + 15 \, B a^{2} b^{2} + 10 \, A a b^{3}\right )} \cos \left (d x + c\right )^{3} + 10 \,{\left (5 \, A a^{4} + 24 \, B a^{3} b + 36 \, A a^{2} b^{2}\right )} \cos \left (d x + c\right )^{2} + 48 \,{\left (B a^{4} + 4 \, A a^{3} b\right )} \cos \left (d x + c\right )\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+B*cos(d*x+c))*sec(d*x+c)^7,x, algorithm="fricas")

[Out]

1/480*(15*(5*A*a^4 + 24*B*a^3*b + 36*A*a^2*b^2 + 32*B*a*b^3 + 8*A*b^4)*cos(d*x + c)^6*log(sin(d*x + c) + 1) -
15*(5*A*a^4 + 24*B*a^3*b + 36*A*a^2*b^2 + 32*B*a*b^3 + 8*A*b^4)*cos(d*x + c)^6*log(-sin(d*x + c) + 1) + 2*(16*
(8*B*a^4 + 32*A*a^3*b + 60*B*a^2*b^2 + 40*A*a*b^3 + 15*B*b^4)*cos(d*x + c)^5 + 40*A*a^4 + 15*(5*A*a^4 + 24*B*a
^3*b + 36*A*a^2*b^2 + 32*B*a*b^3 + 8*A*b^4)*cos(d*x + c)^4 + 32*(2*B*a^4 + 8*A*a^3*b + 15*B*a^2*b^2 + 10*A*a*b
^3)*cos(d*x + c)^3 + 10*(5*A*a^4 + 24*B*a^3*b + 36*A*a^2*b^2)*cos(d*x + c)^2 + 48*(B*a^4 + 4*A*a^3*b)*cos(d*x
+ c))*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+B*cos(d*x+c))*sec(d*x+c)**7,x)

[Out]

Timed out

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Giac [B]  time = 1.70022, size = 1601, normalized size = 4.94 \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+B*cos(d*x+c))*sec(d*x+c)^7,x, algorithm="giac")

[Out]

1/240*(15*(5*A*a^4 + 24*B*a^3*b + 36*A*a^2*b^2 + 32*B*a*b^3 + 8*A*b^4)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 15
*(5*A*a^4 + 24*B*a^3*b + 36*A*a^2*b^2 + 32*B*a*b^3 + 8*A*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 - 240*B*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 + 600*B*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 - 1440*B*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 + 480*B*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
- 240*B*b^4*tan(1/2*d*x + 1/2*c)^11 + 25*A*a^4*tan(1/2*d*x + 1/2*c)^9 + 560*B*a^4*tan(1/2*d*x + 1/2*c)^9 + 22
40*A*a^3*b*tan(1/2*d*x + 1/2*c)^9 - 840*B*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
+ 5280*B*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 - 1440*B*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 + 1200*B*b^4*tan(1/2*d*x + 1/2*c)^9 + 450*A*a^4*tan(1/2*d*x + 1/2*c)
^7 - 1248*B*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 + 240*B*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 - 8640*B*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 + 960*B*a*b^3*tan(1/2*d*x + 1/2*c)^7 + 240*A*b^4*tan(1/2*d*x + 1/2*c)^7 - 2400*B*b^4*tan(1/2*d*x +
1/2*c)^7 + 450*A*a^4*tan(1/2*d*x + 1/2*c)^5 + 1248*B*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 + 240*B*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 + 8640*B*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 + 960*B*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 + 2400*B*b^4*tan(1/2*d*x + 1/2*c)^5 + 25*A*a^4*tan(1/2*d*x + 1/2*c)^3 - 560*B*a^4*tan(1/2*d*x +
1/2*c)^3 - 2240*A*a^3*b*tan(1/2*d*x + 1/2*c)^3 - 840*B*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 - 5280*B*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 - 1440*B*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 - 1200*B*b^4*tan(1/2*d*x + 1/2*c)^3 + 165*A*a^4*tan(1/2
*d*x + 1/2*c) + 240*B*a^4*tan(1/2*d*x + 1/2*c) + 960*A*a^3*b*tan(1/2*d*x + 1/2*c) + 600*B*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) + 1440*B*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) + 480*B*a*b^3*tan(1/2*d*x + 1/2*c) + 120*A*b^4*tan(1/2*d*x + 1/2*c) + 240*B*b^4*tan(1/2*d*x + 1/2*c))/(
tan(1/2*d*x + 1/2*c)^2 - 1)^6)/d