3.962 \(\int (a+b \cos (c+d x))^3 (A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^6(c+d x) \, dx\)

Optimal. Leaf size=278 \[ \frac{\tan (c+d x) \left (2 a^3 (4 A+5 C)+30 a^2 b B+15 a b^2 (2 A+3 C)+15 b^3 B\right )}{15 d}+\frac{\left (3 a^2 b (3 A+4 C)+3 a^3 B+12 a b^2 B+4 b^3 (A+2 C)\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{a \tan (c+d x) \sec ^2(c+d x) \left (2 a^2 (4 A+5 C)+15 a b B+3 A b^2\right )}{30 d}+\frac{\tan (c+d x) \sec (c+d x) \left (15 a^2 b (3 A+4 C)+15 a^3 B+50 a b^2 B+6 A b^3\right )}{40 d}+\frac{(5 a B+3 A b) \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))^2}{20 d}+\frac{A \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^3}{5 d} \]

[Out]

((3*a^3*B + 12*a*b^2*B + 4*b^3*(A + 2*C) + 3*a^2*b*(3*A + 4*C))*ArcTanh[Sin[c + d*x]])/(8*d) + ((30*a^2*b*B +
15*b^3*B + 15*a*b^2*(2*A + 3*C) + 2*a^3*(4*A + 5*C))*Tan[c + d*x])/(15*d) + ((6*A*b^3 + 15*a^3*B + 50*a*b^2*B
+ 15*a^2*b*(3*A + 4*C))*Sec[c + d*x]*Tan[c + d*x])/(40*d) + (a*(3*A*b^2 + 15*a*b*B + 2*a^2*(4*A + 5*C))*Sec[c
+ d*x]^2*Tan[c + d*x])/(30*d) + ((3*A*b + 5*a*B)*(a + b*Cos[c + d*x])^2*Sec[c + d*x]^3*Tan[c + d*x])/(20*d) +
(A*(a + b*Cos[c + d*x])^3*Sec[c + d*x]^4*Tan[c + d*x])/(5*d)

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Rubi [A]  time = 0.917961, antiderivative size = 278, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.171, Rules used = {3047, 3031, 3021, 2748, 3767, 8, 3770} \[ \frac{\tan (c+d x) \left (2 a^3 (4 A+5 C)+30 a^2 b B+15 a b^2 (2 A+3 C)+15 b^3 B\right )}{15 d}+\frac{\left (3 a^2 b (3 A+4 C)+3 a^3 B+12 a b^2 B+4 b^3 (A+2 C)\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{a \tan (c+d x) \sec ^2(c+d x) \left (2 a^2 (4 A+5 C)+15 a b B+3 A b^2\right )}{30 d}+\frac{\tan (c+d x) \sec (c+d x) \left (15 a^2 b (3 A+4 C)+15 a^3 B+50 a b^2 B+6 A b^3\right )}{40 d}+\frac{(5 a B+3 A b) \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))^2}{20 d}+\frac{A \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^3}{5 d} \]

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]^6,x]

[Out]

((3*a^3*B + 12*a*b^2*B + 4*b^3*(A + 2*C) + 3*a^2*b*(3*A + 4*C))*ArcTanh[Sin[c + d*x]])/(8*d) + ((30*a^2*b*B +
15*b^3*B + 15*a*b^2*(2*A + 3*C) + 2*a^3*(4*A + 5*C))*Tan[c + d*x])/(15*d) + ((6*A*b^3 + 15*a^3*B + 50*a*b^2*B
+ 15*a^2*b*(3*A + 4*C))*Sec[c + d*x]*Tan[c + d*x])/(40*d) + (a*(3*A*b^2 + 15*a*b*B + 2*a^2*(4*A + 5*C))*Sec[c
+ d*x]^2*Tan[c + d*x])/(30*d) + ((3*A*b + 5*a*B)*(a + b*Cos[c + d*x])^2*Sec[c + d*x]^3*Tan[c + d*x])/(20*d) +
(A*(a + b*Cos[c + d*x])^3*Sec[c + d*x]^4*Tan[c + d*x])/(5*d)

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))^3 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx &=\frac{A (a+b \cos (c+d x))^3 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac{1}{5} \int (a+b \cos (c+d x))^2 \left (3 A b+5 a B+(4 a A+5 b B+5 a C) \cos (c+d x)+b (A+5 C) \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx\\ &=\frac{(3 A b+5 a B) (a+b \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac{A (a+b \cos (c+d x))^3 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac{1}{20} \int (a+b \cos (c+d x)) \left (2 \left (3 A b^2+15 a b B+2 a^2 (4 A+5 C)\right )+\left (15 a^2 B+20 b^2 B+a b (29 A+40 C)\right ) \cos (c+d x)+b (7 A b+5 a B+20 b C) \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx\\ &=\frac{a \left (3 A b^2+15 a b B+2 a^2 (4 A+5 C)\right ) \sec ^2(c+d x) \tan (c+d x)}{30 d}+\frac{(3 A b+5 a B) (a+b \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac{A (a+b \cos (c+d x))^3 \sec ^4(c+d x) \tan (c+d x)}{5 d}-\frac{1}{60} \int \left (-3 \left (6 A b^3+15 a^3 B+50 a b^2 B+15 a^2 b (3 A+4 C)\right )-4 \left (30 a^2 b B+15 b^3 B+15 a b^2 (2 A+3 C)+2 a^3 (4 A+5 C)\right ) \cos (c+d x)-3 b^2 (7 A b+5 a B+20 b C) \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx\\ &=\frac{\left (6 A b^3+15 a^3 B+50 a b^2 B+15 a^2 b (3 A+4 C)\right ) \sec (c+d x) \tan (c+d x)}{40 d}+\frac{a \left (3 A b^2+15 a b B+2 a^2 (4 A+5 C)\right ) \sec ^2(c+d x) \tan (c+d x)}{30 d}+\frac{(3 A b+5 a B) (a+b \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac{A (a+b \cos (c+d x))^3 \sec ^4(c+d x) \tan (c+d x)}{5 d}-\frac{1}{120} \int \left (-8 \left (30 a^2 b B+15 b^3 B+15 a b^2 (2 A+3 C)+2 a^3 (4 A+5 C)\right )-15 \left (3 a^3 B+12 a b^2 B+4 b^3 (A+2 C)+3 a^2 b (3 A+4 C)\right ) \cos (c+d x)\right ) \sec ^2(c+d x) \, dx\\ &=\frac{\left (6 A b^3+15 a^3 B+50 a b^2 B+15 a^2 b (3 A+4 C)\right ) \sec (c+d x) \tan (c+d x)}{40 d}+\frac{a \left (3 A b^2+15 a b B+2 a^2 (4 A+5 C)\right ) \sec ^2(c+d x) \tan (c+d x)}{30 d}+\frac{(3 A b+5 a B) (a+b \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac{A (a+b \cos (c+d x))^3 \sec ^4(c+d x) \tan (c+d x)}{5 d}-\frac{1}{8} \left (-3 a^3 B-12 a b^2 B-4 b^3 (A+2 C)-3 a^2 b (3 A+4 C)\right ) \int \sec (c+d x) \, dx-\frac{1}{15} \left (-30 a^2 b B-15 b^3 B-15 a b^2 (2 A+3 C)-2 a^3 (4 A+5 C)\right ) \int \sec ^2(c+d x) \, dx\\ &=\frac{\left (3 a^3 B+12 a b^2 B+4 b^3 (A+2 C)+3 a^2 b (3 A+4 C)\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{\left (6 A b^3+15 a^3 B+50 a b^2 B+15 a^2 b (3 A+4 C)\right ) \sec (c+d x) \tan (c+d x)}{40 d}+\frac{a \left (3 A b^2+15 a b B+2 a^2 (4 A+5 C)\right ) \sec ^2(c+d x) \tan (c+d x)}{30 d}+\frac{(3 A b+5 a B) (a+b \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac{A (a+b \cos (c+d x))^3 \sec ^4(c+d x) \tan (c+d x)}{5 d}-\frac{\left (30 a^2 b B+15 b^3 B+15 a b^2 (2 A+3 C)+2 a^3 (4 A+5 C)\right ) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{15 d}\\ &=\frac{\left (3 a^3 B+12 a b^2 B+4 b^3 (A+2 C)+3 a^2 b (3 A+4 C)\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{\left (30 a^2 b B+15 b^3 B+15 a b^2 (2 A+3 C)+2 a^3 (4 A+5 C)\right ) \tan (c+d x)}{15 d}+\frac{\left (6 A b^3+15 a^3 B+50 a b^2 B+15 a^2 b (3 A+4 C)\right ) \sec (c+d x) \tan (c+d x)}{40 d}+\frac{a \left (3 A b^2+15 a b B+2 a^2 (4 A+5 C)\right ) \sec ^2(c+d x) \tan (c+d x)}{30 d}+\frac{(3 A b+5 a B) (a+b \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac{A (a+b \cos (c+d x))^3 \sec ^4(c+d x) \tan (c+d x)}{5 d}\\ \end{align*}

Mathematica [A]  time = 4.6066, size = 204, normalized size = 0.73 \[ \frac{15 \left (3 a^2 b (3 A+4 C)+3 a^3 B+12 a b^2 B+4 b^3 (A+2 C)\right ) \tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \left (8 \left (5 a \tan ^2(c+d x) \left (a^2 (2 A+C)+3 a b B+3 A b^2\right )+15 \left (a^3 (A+C)+3 a^2 b B+3 a b^2 (A+C)+b^3 B\right )+3 a^3 A \tan ^4(c+d x)\right )+15 \sec (c+d x) \left (3 a^2 b (3 A+4 C)+3 a^3 B+12 a b^2 B+4 A b^3\right )+30 a^2 (a B+3 A b) \sec ^3(c+d x)\right )}{120 d} \]

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]^6,x]

[Out]

(15*(3*a^3*B + 12*a*b^2*B + 4*b^3*(A + 2*C) + 3*a^2*b*(3*A + 4*C))*ArcTanh[Sin[c + d*x]] + Tan[c + d*x]*(15*(4
*A*b^3 + 3*a^3*B + 12*a*b^2*B + 3*a^2*b*(3*A + 4*C))*Sec[c + d*x] + 30*a^2*(3*A*b + a*B)*Sec[c + d*x]^3 + 8*(1
5*(3*a^2*b*B + b^3*B + a^3*(A + C) + 3*a*b^2*(A + C)) + 5*a*(3*A*b^2 + 3*a*b*B + a^2*(2*A + C))*Tan[c + d*x]^2
+ 3*a^3*A*Tan[c + d*x]^4)))/(120*d)

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Maple [A]  time = 0.079, size = 504, normalized size = 1.8 \begin{align*}{\frac{A{b}^{3}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{A{b}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+{\frac{{b}^{3}B\tan \left ( dx+c \right ) }{d}}+{\frac{C{b}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+2\,{\frac{aA{b}^{2}\tan \left ( dx+c \right ) }{d}}+{\frac{aA{b}^{2}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{d}}+{\frac{3\,a{b}^{2}B\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{3\,a{b}^{2}B\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+3\,{\frac{Ca{b}^{2}\tan \left ( dx+c \right ) }{d}}+{\frac{3\,A{a}^{2}b\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{4\,d}}+{\frac{9\,A{a}^{2}b\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{8\,d}}+{\frac{9\,A{a}^{2}b\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}}+2\,{\frac{{a}^{2}bB\tan \left ( dx+c \right ) }{d}}+{\frac{{a}^{2}bB\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{d}}+{\frac{3\,{a}^{2}bC\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{3\,{a}^{2}bC\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+{\frac{8\,A{a}^{3}\tan \left ( dx+c \right ) }{15\,d}}+{\frac{A{a}^{3}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{4}}{5\,d}}+{\frac{4\,A{a}^{3}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{15\,d}}+{\frac{{a}^{3}B\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{4\,d}}+{\frac{3\,{a}^{3}B\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{8\,d}}+{\frac{3\,{a}^{3}B\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}}+{\frac{2\,{a}^{3}C\tan \left ( dx+c \right ) }{3\,d}}+{\frac{{a}^{3}C\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3\,d}} \end{align*}

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)^6,x)

[Out]

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

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Maxima [A]  time = 1.01116, size = 610, normalized size = 2.19 \begin{align*} \frac{16 \,{\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} + 80 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a^{3} + 240 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a^{2} b + 240 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a b^{2} - 15 \, B 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 )} - 45 \, A 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 )} - 180 \, C a^{2} b{\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 )} - 180 \, B 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 )} - 60 \, 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 \, C b^{3}{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 720 \, C a b^{2} \tan \left (d x + c\right ) + 240 \, B b^{3} \tan \left (d x + c\right )}{240 \, d} \end{align*}

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)^6,x, algorithm="maxima")

[Out]

1/240*(16*(3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c))*A*a^3 + 80*(tan(d*x + c)^3 + 3*tan(d*x + c)
)*C*a^3 + 240*(tan(d*x + c)^3 + 3*tan(d*x + c))*B*a^2*b + 240*(tan(d*x + c)^3 + 3*tan(d*x + c))*A*a*b^2 - 15*B
*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)) - 45*A*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)) - 180*C*a^2*b*(2*sin(d*x + c)/(sin(d*x + c)^2 -
1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) - 180*B*a*b^2*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(s
in(d*x + c) + 1) + log(sin(d*x + c) - 1)) - 60*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*C*b^3*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 720*C*a*b^2*tan(d*x
+ c) + 240*B*b^3*tan(d*x + c))/d

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Fricas [A]  time = 2.30307, size = 711, normalized size = 2.56 \begin{align*} \frac{15 \,{\left (3 \, B a^{3} + 3 \,{\left (3 \, A + 4 \, C\right )} a^{2} b + 12 \, B a b^{2} + 4 \,{\left (A + 2 \, C\right )} b^{3}\right )} \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \,{\left (3 \, B a^{3} + 3 \,{\left (3 \, A + 4 \, C\right )} a^{2} b + 12 \, B a b^{2} + 4 \,{\left (A + 2 \, C\right )} b^{3}\right )} \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (8 \,{\left (2 \,{\left (4 \, A + 5 \, C\right )} a^{3} + 30 \, B a^{2} b + 15 \,{\left (2 \, A + 3 \, C\right )} a b^{2} + 15 \, B b^{3}\right )} \cos \left (d x + c\right )^{4} + 24 \, A a^{3} + 15 \,{\left (3 \, B a^{3} + 3 \,{\left (3 \, A + 4 \, C\right )} a^{2} b + 12 \, B a b^{2} + 4 \, A b^{3}\right )} \cos \left (d x + c\right )^{3} + 8 \,{\left ({\left (4 \, A + 5 \, C\right )} a^{3} + 15 \, B a^{2} b + 15 \, A a b^{2}\right )} \cos \left (d x + c\right )^{2} + 30 \,{\left (B a^{3} + 3 \, A a^{2} b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{240 \, d \cos \left (d x + c\right )^{5}} \end{align*}

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)^6,x, algorithm="fricas")

[Out]

1/240*(15*(3*B*a^3 + 3*(3*A + 4*C)*a^2*b + 12*B*a*b^2 + 4*(A + 2*C)*b^3)*cos(d*x + c)^5*log(sin(d*x + c) + 1)
- 15*(3*B*a^3 + 3*(3*A + 4*C)*a^2*b + 12*B*a*b^2 + 4*(A + 2*C)*b^3)*cos(d*x + c)^5*log(-sin(d*x + c) + 1) + 2*
(8*(2*(4*A + 5*C)*a^3 + 30*B*a^2*b + 15*(2*A + 3*C)*a*b^2 + 15*B*b^3)*cos(d*x + c)^4 + 24*A*a^3 + 15*(3*B*a^3
+ 3*(3*A + 4*C)*a^2*b + 12*B*a*b^2 + 4*A*b^3)*cos(d*x + c)^3 + 8*((4*A + 5*C)*a^3 + 15*B*a^2*b + 15*A*a*b^2)*c
os(d*x + c)^2 + 30*(B*a^3 + 3*A*a^2*b)*cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^5)

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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))**3*(A+B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)**6,x)

[Out]

Timed out

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Giac [B]  time = 1.27234, size = 1335, normalized size = 4.8 \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))^3*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^6,x, algorithm="giac")

[Out]

1/120*(15*(3*B*a^3 + 9*A*a^2*b + 12*C*a^2*b + 12*B*a*b^2 + 4*A*b^3 + 8*C*b^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1
)) - 15*(3*B*a^3 + 9*A*a^2*b + 12*C*a^2*b + 12*B*a*b^2 + 4*A*b^3 + 8*C*b^3)*log(abs(tan(1/2*d*x + 1/2*c) - 1))
- 2*(120*A*a^3*tan(1/2*d*x + 1/2*c)^9 - 75*B*a^3*tan(1/2*d*x + 1/2*c)^9 + 120*C*a^3*tan(1/2*d*x + 1/2*c)^9 -
225*A*a^2*b*tan(1/2*d*x + 1/2*c)^9 + 360*B*a^2*b*tan(1/2*d*x + 1/2*c)^9 - 180*C*a^2*b*tan(1/2*d*x + 1/2*c)^9 +
360*A*a*b^2*tan(1/2*d*x + 1/2*c)^9 - 180*B*a*b^2*tan(1/2*d*x + 1/2*c)^9 + 360*C*a*b^2*tan(1/2*d*x + 1/2*c)^9
- 60*A*b^3*tan(1/2*d*x + 1/2*c)^9 + 120*B*b^3*tan(1/2*d*x + 1/2*c)^9 - 160*A*a^3*tan(1/2*d*x + 1/2*c)^7 + 30*B
*a^3*tan(1/2*d*x + 1/2*c)^7 - 320*C*a^3*tan(1/2*d*x + 1/2*c)^7 + 90*A*a^2*b*tan(1/2*d*x + 1/2*c)^7 - 960*B*a^2
*b*tan(1/2*d*x + 1/2*c)^7 + 360*C*a^2*b*tan(1/2*d*x + 1/2*c)^7 - 960*A*a*b^2*tan(1/2*d*x + 1/2*c)^7 + 360*B*a*
b^2*tan(1/2*d*x + 1/2*c)^7 - 1440*C*a*b^2*tan(1/2*d*x + 1/2*c)^7 + 120*A*b^3*tan(1/2*d*x + 1/2*c)^7 - 480*B*b^
3*tan(1/2*d*x + 1/2*c)^7 + 464*A*a^3*tan(1/2*d*x + 1/2*c)^5 + 400*C*a^3*tan(1/2*d*x + 1/2*c)^5 + 1200*B*a^2*b*
tan(1/2*d*x + 1/2*c)^5 + 1200*A*a*b^2*tan(1/2*d*x + 1/2*c)^5 + 2160*C*a*b^2*tan(1/2*d*x + 1/2*c)^5 + 720*B*b^3
*tan(1/2*d*x + 1/2*c)^5 - 160*A*a^3*tan(1/2*d*x + 1/2*c)^3 - 30*B*a^3*tan(1/2*d*x + 1/2*c)^3 - 320*C*a^3*tan(1
/2*d*x + 1/2*c)^3 - 90*A*a^2*b*tan(1/2*d*x + 1/2*c)^3 - 960*B*a^2*b*tan(1/2*d*x + 1/2*c)^3 - 360*C*a^2*b*tan(1
/2*d*x + 1/2*c)^3 - 960*A*a*b^2*tan(1/2*d*x + 1/2*c)^3 - 360*B*a*b^2*tan(1/2*d*x + 1/2*c)^3 - 1440*C*a*b^2*tan
(1/2*d*x + 1/2*c)^3 - 120*A*b^3*tan(1/2*d*x + 1/2*c)^3 - 480*B*b^3*tan(1/2*d*x + 1/2*c)^3 + 120*A*a^3*tan(1/2*
d*x + 1/2*c) + 75*B*a^3*tan(1/2*d*x + 1/2*c) + 120*C*a^3*tan(1/2*d*x + 1/2*c) + 225*A*a^2*b*tan(1/2*d*x + 1/2*
c) + 360*B*a^2*b*tan(1/2*d*x + 1/2*c) + 180*C*a^2*b*tan(1/2*d*x + 1/2*c) + 360*A*a*b^2*tan(1/2*d*x + 1/2*c) +
180*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) + 60*A*b^3*tan(1/2*d*x + 1/2*c) + 120*B*b^
3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^5)/d