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

Optimal. Leaf size=232 \[ \frac{\tan (c+d x) \left (2 a^2 (4 A+5 C)+20 a b B+5 b^2 (2 A+3 C)\right )}{15 d}+\frac{\left (3 a^2 B+6 a A b+8 a b C+4 b^2 B\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{\tan (c+d x) \sec ^2(c+d x) \left (a^2 (4 A+5 C)+10 a b B+2 A b^2\right )}{15 d}+\frac{\tan (c+d x) \sec (c+d x) \left (3 a^2 B+6 a A b+8 a b C+4 b^2 B\right )}{8 d}+\frac{a (5 a B+2 A b) \tan (c+d x) \sec ^3(c+d x)}{20 d}+\frac{A \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^2}{5 d} \]

[Out]

((6*a*A*b + 3*a^2*B + 4*b^2*B + 8*a*b*C)*ArcTanh[Sin[c + d*x]])/(8*d) + ((20*a*b*B + 5*b^2*(2*A + 3*C) + 2*a^2
*(4*A + 5*C))*Tan[c + d*x])/(15*d) + ((6*a*A*b + 3*a^2*B + 4*b^2*B + 8*a*b*C)*Sec[c + d*x]*Tan[c + d*x])/(8*d)
+ ((2*A*b^2 + 10*a*b*B + a^2*(4*A + 5*C))*Sec[c + d*x]^2*Tan[c + d*x])/(15*d) + (a*(2*A*b + 5*a*B)*Sec[c + d*
x]^3*Tan[c + d*x])/(20*d) + (A*(a + b*Cos[c + d*x])^2*Sec[c + d*x]^4*Tan[c + d*x])/(5*d)

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Rubi [A]  time = 0.507059, antiderivative size = 232, normalized size of antiderivative = 1., number of steps used = 8, 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} \[ \frac{\tan (c+d x) \left (2 a^2 (4 A+5 C)+20 a b B+5 b^2 (2 A+3 C)\right )}{15 d}+\frac{\left (3 a^2 B+6 a A b+8 a b C+4 b^2 B\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{\tan (c+d x) \sec ^2(c+d x) \left (a^2 (4 A+5 C)+10 a b B+2 A b^2\right )}{15 d}+\frac{\tan (c+d x) \sec (c+d x) \left (3 a^2 B+6 a A b+8 a b C+4 b^2 B\right )}{8 d}+\frac{a (5 a B+2 A b) \tan (c+d x) \sec ^3(c+d x)}{20 d}+\frac{A \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^2}{5 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Cos[c + d*x])^2*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^6,x]

[Out]

((6*a*A*b + 3*a^2*B + 4*b^2*B + 8*a*b*C)*ArcTanh[Sin[c + d*x]])/(8*d) + ((20*a*b*B + 5*b^2*(2*A + 3*C) + 2*a^2
*(4*A + 5*C))*Tan[c + d*x])/(15*d) + ((6*a*A*b + 3*a^2*B + 4*b^2*B + 8*a*b*C)*Sec[c + d*x]*Tan[c + d*x])/(8*d)
+ ((2*A*b^2 + 10*a*b*B + a^2*(4*A + 5*C))*Sec[c + d*x]^2*Tan[c + d*x])/(15*d) + (a*(2*A*b + 5*a*B)*Sec[c + d*
x]^3*Tan[c + d*x])/(20*d) + (A*(a + b*Cos[c + d*x])^2*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 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))^2 \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))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac{1}{5} \int (a+b \cos (c+d x)) \left (2 A b+5 a B+(4 a A+5 b B+5 a C) \cos (c+d x)+b (2 A+5 C) \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx\\ &=\frac{a (2 A b+5 a B) \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac{A (a+b \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}-\frac{1}{20} \int \left (-4 \left (2 A b^2+10 a b B+a^2 (4 A+5 C)\right )-5 \left (6 a A b+3 a^2 B+4 b^2 B+8 a b C\right ) \cos (c+d x)-4 b^2 (2 A+5 C) \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx\\ &=\frac{\left (2 A b^2+10 a b B+a^2 (4 A+5 C)\right ) \sec ^2(c+d x) \tan (c+d x)}{15 d}+\frac{a (2 A b+5 a B) \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac{A (a+b \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}-\frac{1}{60} \int \left (-15 \left (6 a A b+3 a^2 B+4 b^2 B+8 a b C\right )-4 \left (20 a b B+5 b^2 (2 A+3 C)+2 a^2 (4 A+5 C)\right ) \cos (c+d x)\right ) \sec ^3(c+d x) \, dx\\ &=\frac{\left (2 A b^2+10 a b B+a^2 (4 A+5 C)\right ) \sec ^2(c+d x) \tan (c+d x)}{15 d}+\frac{a (2 A b+5 a B) \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac{A (a+b \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}-\frac{1}{4} \left (-6 a A b-3 a^2 B-4 b^2 B-8 a b C\right ) \int \sec ^3(c+d x) \, dx-\frac{1}{15} \left (-20 a b B-5 b^2 (2 A+3 C)-2 a^2 (4 A+5 C)\right ) \int \sec ^2(c+d x) \, dx\\ &=\frac{\left (6 a A b+3 a^2 B+4 b^2 B+8 a b C\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac{\left (2 A b^2+10 a b B+a^2 (4 A+5 C)\right ) \sec ^2(c+d x) \tan (c+d x)}{15 d}+\frac{a (2 A b+5 a B) \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac{A (a+b \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}-\frac{1}{8} \left (-6 a A b-3 a^2 B-4 b^2 B-8 a b C\right ) \int \sec (c+d x) \, dx-\frac{\left (20 a b B+5 b^2 (2 A+3 C)+2 a^2 (4 A+5 C)\right ) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{15 d}\\ &=\frac{\left (6 a A b+3 a^2 B+4 b^2 B+8 a b C\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{\left (20 a b B+5 b^2 (2 A+3 C)+2 a^2 (4 A+5 C)\right ) \tan (c+d x)}{15 d}+\frac{\left (6 a A b+3 a^2 B+4 b^2 B+8 a b C\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac{\left (2 A b^2+10 a b B+a^2 (4 A+5 C)\right ) \sec ^2(c+d x) \tan (c+d x)}{15 d}+\frac{a (2 A b+5 a B) \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac{A (a+b \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}\\ \end{align*}

Mathematica [A]  time = 2.47562, size = 167, normalized size = 0.72 \[ \frac{15 \left (3 a^2 B+6 a A b+8 a b C+4 b^2 B\right ) \tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \left (8 \left (5 \tan ^2(c+d x) \left (a^2 (2 A+C)+2 a b B+A b^2\right )+15 \left (a^2 (A+C)+2 a b B+b^2 (A+C)\right )+3 a^2 A \tan ^4(c+d x)\right )+15 \sec (c+d x) \left (3 a^2 B+6 a A b+8 a b C+4 b^2 B\right )+30 a (a B+2 A b) \sec ^3(c+d x)\right )}{120 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Cos[c + d*x])^2*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^6,x]

[Out]

(15*(6*a*A*b + 3*a^2*B + 4*b^2*B + 8*a*b*C)*ArcTanh[Sin[c + d*x]] + Tan[c + d*x]*(15*(6*a*A*b + 3*a^2*B + 4*b^
2*B + 8*a*b*C)*Sec[c + d*x] + 30*a*(2*A*b + a*B)*Sec[c + d*x]^3 + 8*(15*(2*a*b*B + a^2*(A + C) + b^2*(A + C))
+ 5*(A*b^2 + 2*a*b*B + a^2*(2*A + C))*Tan[c + d*x]^2 + 3*a^2*A*Tan[c + d*x]^4)))/(120*d)

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Maple [A]  time = 0.069, size = 404, normalized size = 1.7 \begin{align*}{\frac{2\,A{b}^{2}\tan \left ( dx+c \right ) }{3\,d}}+{\frac{A{b}^{2}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3\,d}}+{\frac{{b}^{2}B\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{{b}^{2}B\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+{\frac{{b}^{2}C\tan \left ( dx+c \right ) }{d}}+{\frac{aAb \left ( \sec \left ( dx+c \right ) \right ) ^{3}\tan \left ( dx+c \right ) }{2\,d}}+{\frac{3\,aAb\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{4\,d}}+{\frac{3\,aAb\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{4\,d}}+{\frac{4\,abB\tan \left ( dx+c \right ) }{3\,d}}+{\frac{2\,abB\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3\,d}}+{\frac{abC\tan \left ( dx+c \right ) \sec \left ( dx+c \right ) }{d}}+{\frac{abC\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{8\,A{a}^{2}\tan \left ( dx+c \right ) }{15\,d}}+{\frac{A{a}^{2}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{4}}{5\,d}}+{\frac{4\,A{a}^{2}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{15\,d}}+{\frac{{a}^{2}B \left ( \sec \left ( dx+c \right ) \right ) ^{3}\tan \left ( dx+c \right ) }{4\,d}}+{\frac{3\,{a}^{2}B\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{8\,d}}+{\frac{3\,{a}^{2}B\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}}+{\frac{2\,{a}^{2}C\tan \left ( dx+c \right ) }{3\,d}}+{\frac{{a}^{2}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))^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^6,x)

[Out]

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

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Maxima [A]  time = 1.033, size = 482, normalized size = 2.08 \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^{2} + 80 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a^{2} + 160 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a b + 80 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A b^{2} - 15 \, B a^{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 )} - 30 \, A a 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 )} - 120 \, C a 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 )} - 60 \, B 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 )} + 240 \, C b^{2} \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))^2*(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^2 + 80*(tan(d*x + c)^3 + 3*tan(d*x + c)
)*C*a^2 + 160*(tan(d*x + c)^3 + 3*tan(d*x + c))*B*a*b + 80*(tan(d*x + c)^3 + 3*tan(d*x + c))*A*b^2 - 15*B*a^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*l
og(sin(d*x + c) - 1)) - 30*A*a*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)) - 120*C*a*b*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(
sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) - 60*B*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)) + 240*C*b^2*tan(d*x + c))/d

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Fricas [A]  time = 1.87832, size = 598, normalized size = 2.58 \begin{align*} \frac{15 \,{\left (3 \, B a^{2} + 2 \,{\left (3 \, A + 4 \, C\right )} a b + 4 \, B b^{2}\right )} \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \,{\left (3 \, B a^{2} + 2 \,{\left (3 \, A + 4 \, C\right )} a b + 4 \, B b^{2}\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^{2} + 20 \, B a b + 5 \,{\left (2 \, A + 3 \, C\right )} b^{2}\right )} \cos \left (d x + c\right )^{4} + 15 \,{\left (3 \, B a^{2} + 2 \,{\left (3 \, A + 4 \, C\right )} a b + 4 \, B b^{2}\right )} \cos \left (d x + c\right )^{3} + 24 \, A a^{2} + 8 \,{\left ({\left (4 \, A + 5 \, C\right )} a^{2} + 10 \, B a b + 5 \, A b^{2}\right )} \cos \left (d x + c\right )^{2} + 30 \,{\left (B a^{2} + 2 \, A a 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))^2*(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^2 + 2*(3*A + 4*C)*a*b + 4*B*b^2)*cos(d*x + c)^5*log(sin(d*x + c) + 1) - 15*(3*B*a^2 + 2*(3*A
+ 4*C)*a*b + 4*B*b^2)*cos(d*x + c)^5*log(-sin(d*x + c) + 1) + 2*(8*(2*(4*A + 5*C)*a^2 + 20*B*a*b + 5*(2*A + 3*
C)*b^2)*cos(d*x + c)^4 + 15*(3*B*a^2 + 2*(3*A + 4*C)*a*b + 4*B*b^2)*cos(d*x + c)^3 + 24*A*a^2 + 8*((4*A + 5*C)
*a^2 + 10*B*a*b + 5*A*b^2)*cos(d*x + c)^2 + 30*(B*a^2 + 2*A*a*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))**2*(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 = 1.20769, size = 1034, normalized size = 4.46 \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))^2*(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^2 + 6*A*a*b + 8*C*a*b + 4*B*b^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 15*(3*B*a^2 + 6*A*a*b +
8*C*a*b + 4*B*b^2)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(120*A*a^2*tan(1/2*d*x + 1/2*c)^9 - 75*B*a^2*tan(1/
2*d*x + 1/2*c)^9 + 120*C*a^2*tan(1/2*d*x + 1/2*c)^9 - 150*A*a*b*tan(1/2*d*x + 1/2*c)^9 + 240*B*a*b*tan(1/2*d*x
+ 1/2*c)^9 - 120*C*a*b*tan(1/2*d*x + 1/2*c)^9 + 120*A*b^2*tan(1/2*d*x + 1/2*c)^9 - 60*B*b^2*tan(1/2*d*x + 1/2
*c)^9 + 120*C*b^2*tan(1/2*d*x + 1/2*c)^9 - 160*A*a^2*tan(1/2*d*x + 1/2*c)^7 + 30*B*a^2*tan(1/2*d*x + 1/2*c)^7
- 320*C*a^2*tan(1/2*d*x + 1/2*c)^7 + 60*A*a*b*tan(1/2*d*x + 1/2*c)^7 - 640*B*a*b*tan(1/2*d*x + 1/2*c)^7 + 240*
C*a*b*tan(1/2*d*x + 1/2*c)^7 - 320*A*b^2*tan(1/2*d*x + 1/2*c)^7 + 120*B*b^2*tan(1/2*d*x + 1/2*c)^7 - 480*C*b^2
*tan(1/2*d*x + 1/2*c)^7 + 464*A*a^2*tan(1/2*d*x + 1/2*c)^5 + 400*C*a^2*tan(1/2*d*x + 1/2*c)^5 + 800*B*a*b*tan(
1/2*d*x + 1/2*c)^5 + 400*A*b^2*tan(1/2*d*x + 1/2*c)^5 + 720*C*b^2*tan(1/2*d*x + 1/2*c)^5 - 160*A*a^2*tan(1/2*d
*x + 1/2*c)^3 - 30*B*a^2*tan(1/2*d*x + 1/2*c)^3 - 320*C*a^2*tan(1/2*d*x + 1/2*c)^3 - 60*A*a*b*tan(1/2*d*x + 1/
2*c)^3 - 640*B*a*b*tan(1/2*d*x + 1/2*c)^3 - 240*C*a*b*tan(1/2*d*x + 1/2*c)^3 - 320*A*b^2*tan(1/2*d*x + 1/2*c)^
3 - 120*B*b^2*tan(1/2*d*x + 1/2*c)^3 - 480*C*b^2*tan(1/2*d*x + 1/2*c)^3 + 120*A*a^2*tan(1/2*d*x + 1/2*c) + 75*
B*a^2*tan(1/2*d*x + 1/2*c) + 120*C*a^2*tan(1/2*d*x + 1/2*c) + 150*A*a*b*tan(1/2*d*x + 1/2*c) + 240*B*a*b*tan(1
/2*d*x + 1/2*c) + 120*C*a*b*tan(1/2*d*x + 1/2*c) + 120*A*b^2*tan(1/2*d*x + 1/2*c) + 60*B*b^2*tan(1/2*d*x + 1/2
*c) + 120*C*b^2*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^5)/d