3.970 \(\int (a+b \cos (c+d x))^4 (A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^5(c+d x) \, dx\)

Optimal. Leaf size=293 \[ -\frac{b^2 \sin (c+d x) \left (3 a^2 (3 A+4 C)+32 a b B+2 b^2 (13 A-12 C)\right )}{24 d}+\frac{a \tan (c+d x) \left (a^2 b (23 A+36 C)+8 a^3 B+36 a b^2 B+12 A b^3\right )}{12 d}+\frac{\left (24 a^2 b^2 (A+2 C)+a^4 (3 A+4 C)+16 a^3 b B+32 a b^3 B+8 A b^4\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{\tan (c+d x) \sec (c+d x) \left (a^2 (3 A+4 C)+8 a b B+4 A b^2\right ) (a+b \cos (c+d x))^2}{8 d}+\frac{(a B+A b) \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))^3}{3 d}+\frac{A \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))^4}{4 d}+b^3 x (4 a C+b B) \]

[Out]

b^3*(b*B + 4*a*C)*x + ((8*A*b^4 + 16*a^3*b*B + 32*a*b^3*B + 24*a^2*b^2*(A + 2*C) + a^4*(3*A + 4*C))*ArcTanh[Si
n[c + d*x]])/(8*d) - (b^2*(32*a*b*B + 2*b^2*(13*A - 12*C) + 3*a^2*(3*A + 4*C))*Sin[c + d*x])/(24*d) + (a*(12*A
*b^3 + 8*a^3*B + 36*a*b^2*B + a^2*b*(23*A + 36*C))*Tan[c + d*x])/(12*d) + ((4*A*b^2 + 8*a*b*B + a^2*(3*A + 4*C
))*(a + b*Cos[c + d*x])^2*Sec[c + d*x]*Tan[c + d*x])/(8*d) + ((A*b + a*B)*(a + b*Cos[c + d*x])^3*Sec[c + d*x]^
2*Tan[c + d*x])/(3*d) + (A*(a + b*Cos[c + d*x])^4*Sec[c + d*x]^3*Tan[c + d*x])/(4*d)

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Rubi [A]  time = 1.06484, antiderivative size = 293, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.122, Rules used = {3047, 3031, 3023, 2735, 3770} \[ -\frac{b^2 \sin (c+d x) \left (3 a^2 (3 A+4 C)+32 a b B+2 b^2 (13 A-12 C)\right )}{24 d}+\frac{a \tan (c+d x) \left (a^2 b (23 A+36 C)+8 a^3 B+36 a b^2 B+12 A b^3\right )}{12 d}+\frac{\left (24 a^2 b^2 (A+2 C)+a^4 (3 A+4 C)+16 a^3 b B+32 a b^3 B+8 A b^4\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{\tan (c+d x) \sec (c+d x) \left (a^2 (3 A+4 C)+8 a b B+4 A b^2\right ) (a+b \cos (c+d x))^2}{8 d}+\frac{(a B+A b) \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))^3}{3 d}+\frac{A \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))^4}{4 d}+b^3 x (4 a C+b B) \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Cos[c + d*x])^4*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^5,x]

[Out]

b^3*(b*B + 4*a*C)*x + ((8*A*b^4 + 16*a^3*b*B + 32*a*b^3*B + 24*a^2*b^2*(A + 2*C) + a^4*(3*A + 4*C))*ArcTanh[Si
n[c + d*x]])/(8*d) - (b^2*(32*a*b*B + 2*b^2*(13*A - 12*C) + 3*a^2*(3*A + 4*C))*Sin[c + d*x])/(24*d) + (a*(12*A
*b^3 + 8*a^3*B + 36*a*b^2*B + a^2*b*(23*A + 36*C))*Tan[c + d*x])/(12*d) + ((4*A*b^2 + 8*a*b*B + a^2*(3*A + 4*C
))*(a + b*Cos[c + d*x])^2*Sec[c + d*x]*Tan[c + d*x])/(8*d) + ((A*b + a*B)*(a + b*Cos[c + d*x])^3*Sec[c + d*x]^
2*Tan[c + d*x])/(3*d) + (A*(a + b*Cos[c + d*x])^4*Sec[c + d*x]^3*Tan[c + d*x])/(4*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 3023

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*
(m + 2)), Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x]
, x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

Rule 2735

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*x)/d
, x] - Dist[(b*c - a*d)/d, Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d
, 0]

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+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx &=\frac{A (a+b \cos (c+d x))^4 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac{1}{4} \int (a+b \cos (c+d x))^3 \left (4 (A b+a B)+(3 a A+4 b B+4 a C) \cos (c+d x)-b (A-4 C) \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx\\ &=\frac{(A b+a B) (a+b \cos (c+d x))^3 \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac{A (a+b \cos (c+d x))^4 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac{1}{12} \int (a+b \cos (c+d x))^2 \left (3 \left (4 A b^2+8 a b B+a^2 (3 A+4 C)\right )+2 \left (4 a^2 B+6 b^2 B+a b (7 A+12 C)\right ) \cos (c+d x)-b (7 A b+4 a B-12 b C) \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx\\ &=\frac{\left (4 A b^2+8 a b B+a^2 (3 A+4 C)\right ) (a+b \cos (c+d x))^2 \sec (c+d x) \tan (c+d x)}{8 d}+\frac{(A b+a B) (a+b \cos (c+d x))^3 \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac{A (a+b \cos (c+d x))^4 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac{1}{24} \int (a+b \cos (c+d x)) \left (2 \left (12 A b^3+8 a^3 B+36 a b^2 B+\frac{1}{2} a^2 (46 A b+72 b C)\right )+\left (32 a^2 b B+24 b^3 B+3 a^3 (3 A+4 C)+2 a b^2 (13 A+36 C)\right ) \cos (c+d x)-b \left (32 a b B+2 b^2 (13 A-12 C)+3 a^2 (3 A+4 C)\right ) \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx\\ &=\frac{a \left (12 A b^3+8 a^3 B+36 a b^2 B+a^2 b (23 A+36 C)\right ) \tan (c+d x)}{12 d}+\frac{\left (4 A b^2+8 a b B+a^2 (3 A+4 C)\right ) (a+b \cos (c+d x))^2 \sec (c+d x) \tan (c+d x)}{8 d}+\frac{(A b+a B) (a+b \cos (c+d x))^3 \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac{A (a+b \cos (c+d x))^4 \sec ^3(c+d x) \tan (c+d x)}{4 d}-\frac{1}{24} \int \left (-3 \left (8 A b^4+16 a^3 b B+32 a b^3 B+24 a^2 b^2 (A+2 C)+a^4 (3 A+4 C)\right )-24 b^3 (b B+4 a C) \cos (c+d x)+b^2 \left (32 a b B+2 b^2 (13 A-12 C)+3 a^2 (3 A+4 C)\right ) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx\\ &=-\frac{b^2 \left (32 a b B+2 b^2 (13 A-12 C)+3 a^2 (3 A+4 C)\right ) \sin (c+d x)}{24 d}+\frac{a \left (12 A b^3+8 a^3 B+36 a b^2 B+a^2 b (23 A+36 C)\right ) \tan (c+d x)}{12 d}+\frac{\left (4 A b^2+8 a b B+a^2 (3 A+4 C)\right ) (a+b \cos (c+d x))^2 \sec (c+d x) \tan (c+d x)}{8 d}+\frac{(A b+a B) (a+b \cos (c+d x))^3 \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac{A (a+b \cos (c+d x))^4 \sec ^3(c+d x) \tan (c+d x)}{4 d}-\frac{1}{24} \int \left (-3 \left (8 A b^4+16 a^3 b B+32 a b^3 B+24 a^2 b^2 (A+2 C)+a^4 (3 A+4 C)\right )-24 b^3 (b B+4 a C) \cos (c+d x)\right ) \sec (c+d x) \, dx\\ &=b^3 (b B+4 a C) x-\frac{b^2 \left (32 a b B+2 b^2 (13 A-12 C)+3 a^2 (3 A+4 C)\right ) \sin (c+d x)}{24 d}+\frac{a \left (12 A b^3+8 a^3 B+36 a b^2 B+a^2 b (23 A+36 C)\right ) \tan (c+d x)}{12 d}+\frac{\left (4 A b^2+8 a b B+a^2 (3 A+4 C)\right ) (a+b \cos (c+d x))^2 \sec (c+d x) \tan (c+d x)}{8 d}+\frac{(A b+a B) (a+b \cos (c+d x))^3 \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac{A (a+b \cos (c+d x))^4 \sec ^3(c+d x) \tan (c+d x)}{4 d}-\frac{1}{8} \left (-8 A b^4-16 a^3 b B-32 a b^3 B-24 a^2 b^2 (A+2 C)-a^4 (3 A+4 C)\right ) \int \sec (c+d x) \, dx\\ &=b^3 (b B+4 a C) x+\frac{\left (8 A b^4+16 a^3 b B+32 a b^3 B+24 a^2 b^2 (A+2 C)+a^4 (3 A+4 C)\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}-\frac{b^2 \left (32 a b B+2 b^2 (13 A-12 C)+3 a^2 (3 A+4 C)\right ) \sin (c+d x)}{24 d}+\frac{a \left (12 A b^3+8 a^3 B+36 a b^2 B+a^2 b (23 A+36 C)\right ) \tan (c+d x)}{12 d}+\frac{\left (4 A b^2+8 a b B+a^2 (3 A+4 C)\right ) (a+b \cos (c+d x))^2 \sec (c+d x) \tan (c+d x)}{8 d}+\frac{(A b+a B) (a+b \cos (c+d x))^3 \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac{A (a+b \cos (c+d x))^4 \sec ^3(c+d x) \tan (c+d x)}{4 d}\\ \end{align*}

Mathematica [A]  time = 2.45643, size = 462, normalized size = 1.58 \[ \frac{3 \tan (c+d x) \sec ^3(c+d x) \left (24 a^2 A b^2+a^4 (11 A+4 C)+16 a^3 b B+4 b^4 C\right )+32 a \tan (c+d x) \sec ^2(c+d x) \left (a^2 (8 A b+6 b C)+2 a^3 B+9 a b^2 B+6 A b^3\right )-12 \left (24 a^2 b^2 (A+2 C)+a^4 (3 A+4 C)+16 a^3 b B+32 a b^3 B+8 A b^4\right ) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )+\sec ^4(c+d x) \left (72 a^2 A b^2 \sin (3 (c+d x))+32 a^3 A b \sin (4 (c+d x))+9 a^4 A \sin (3 (c+d x))+72 a^2 b^2 B \sin (4 (c+d x))+48 a^3 b B \sin (3 (c+d x))+48 a^3 b C \sin (4 (c+d x))+8 a^4 B \sin (4 (c+d x))+12 a^4 C \sin (3 (c+d x))+48 a A b^3 \sin (4 (c+d x))+48 b^3 (c+d x) (4 a C+b B) \cos (2 (c+d x))+12 b^3 (c+d x) (4 a C+b B) \cos (4 (c+d x))+144 a b^3 c C+144 a b^3 C d x+36 b^4 B c+36 b^4 B d x+18 b^4 C \sin (3 (c+d x))+6 b^4 C \sin (5 (c+d x))\right )}{96 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Cos[c + d*x])^4*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^5,x]

[Out]

(-12*(8*A*b^4 + 16*a^3*b*B + 32*a*b^3*B + 24*a^2*b^2*(A + 2*C) + a^4*(3*A + 4*C))*(Log[Cos[(c + d*x)/2] - Sin[
(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) + Sec[c + d*x]^4*(36*b^4*B*c + 144*a*b^3*c*C + 36*b^
4*B*d*x + 144*a*b^3*C*d*x + 48*b^3*(b*B + 4*a*C)*(c + d*x)*Cos[2*(c + d*x)] + 12*b^3*(b*B + 4*a*C)*(c + d*x)*C
os[4*(c + d*x)] + 9*a^4*A*Sin[3*(c + d*x)] + 72*a^2*A*b^2*Sin[3*(c + d*x)] + 48*a^3*b*B*Sin[3*(c + d*x)] + 12*
a^4*C*Sin[3*(c + d*x)] + 18*b^4*C*Sin[3*(c + d*x)] + 32*a^3*A*b*Sin[4*(c + d*x)] + 48*a*A*b^3*Sin[4*(c + d*x)]
+ 8*a^4*B*Sin[4*(c + d*x)] + 72*a^2*b^2*B*Sin[4*(c + d*x)] + 48*a^3*b*C*Sin[4*(c + d*x)] + 6*b^4*C*Sin[5*(c +
d*x)]) + 32*a*(6*A*b^3 + 2*a^3*B + 9*a*b^2*B + a^2*(8*A*b + 6*b*C))*Sec[c + d*x]^2*Tan[c + d*x] + 3*(24*a^2*A
*b^2 + 16*a^3*b*B + 4*b^4*C + a^4*(11*A + 4*C))*Sec[c + d*x]^3*Tan[c + d*x])/(96*d)

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Maple [A]  time = 0.082, size = 457, normalized size = 1.6 \begin{align*}{\frac{A{b}^{4}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{b}^{4}Bx+{\frac{B{b}^{4}c}{d}}+{\frac{C{b}^{4}\sin \left ( dx+c \right ) }{d}}+4\,{\frac{aA{b}^{3}\tan \left ( dx+c \right ) }{d}}+4\,{\frac{a{b}^{3}B\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+4\,a{b}^{3}Cx+4\,{\frac{Ca{b}^{3}c}{d}}+3\,{\frac{{a}^{2}A{b}^{2}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{d}}+3\,{\frac{{a}^{2}A{b}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+6\,{\frac{{a}^{2}{b}^{2}B\tan \left ( dx+c \right ) }{d}}+6\,{\frac{{a}^{2}{b}^{2}C\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{8\,A{a}^{3}b\tan \left ( dx+c \right ) }{3\,d}}+{\frac{4\,A{a}^{3}b\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3\,d}}+2\,{\frac{{a}^{3}bB\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{d}}+2\,{\frac{{a}^{3}bB\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+4\,{\frac{{a}^{3}bC\tan \left ( dx+c \right ) }{d}}+{\frac{A{a}^{4}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{4\,d}}+{\frac{3\,A{a}^{4}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{8\,d}}+{\frac{3\,A{a}^{4}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}}+{\frac{2\,{a}^{4}B\tan \left ( dx+c \right ) }{3\,d}}+{\frac{{a}^{4}B\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3\,d}}+{\frac{{a}^{4}C\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{{a}^{4}C\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}} \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)+C*cos(d*x+c)^2)*sec(d*x+c)^5,x)

[Out]

1/d*A*b^4*ln(sec(d*x+c)+tan(d*x+c))+b^4*B*x+1/d*b^4*B*c+1/d*C*b^4*sin(d*x+c)+4/d*a*A*b^3*tan(d*x+c)+4/d*a*b^3*
B*ln(sec(d*x+c)+tan(d*x+c))+4*a*b^3*C*x+4/d*C*a*b^3*c+3/d*a^2*A*b^2*sec(d*x+c)*tan(d*x+c)+3/d*a^2*A*b^2*ln(sec
(d*x+c)+tan(d*x+c))+6/d*a^2*b^2*B*tan(d*x+c)+6/d*a^2*b^2*C*ln(sec(d*x+c)+tan(d*x+c))+8/3/d*A*a^3*b*tan(d*x+c)+
4/3/d*A*a^3*b*tan(d*x+c)*sec(d*x+c)^2+2/d*a^3*b*B*sec(d*x+c)*tan(d*x+c)+2/d*a^3*b*B*ln(sec(d*x+c)+tan(d*x+c))+
4/d*a^3*b*C*tan(d*x+c)+1/4/d*A*a^4*tan(d*x+c)*sec(d*x+c)^3+3/8/d*A*a^4*sec(d*x+c)*tan(d*x+c)+3/8/d*A*a^4*ln(se
c(d*x+c)+tan(d*x+c))+2/3/d*a^4*B*tan(d*x+c)+1/3/d*a^4*B*tan(d*x+c)*sec(d*x+c)^2+1/2/d*a^4*C*sec(d*x+c)*tan(d*x
+c)+1/2/d*a^4*C*ln(sec(d*x+c)+tan(d*x+c))

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Maxima [A]  time = 1.00991, size = 582, normalized size = 1.99 \begin{align*} \frac{16 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a^{4} + 64 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a^{3} b + 192 \,{\left (d x + c\right )} C a b^{3} + 48 \,{\left (d x + c\right )} B b^{4} - 3 \, A 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 )} - 12 \, C a^{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 )} - 48 \, B a^{3} 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 )} - 72 \, A 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 )} + 144 \, C a^{2} b^{2}{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 96 \, B a b^{3}{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 24 \, A b^{4}{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 48 \, C b^{4} \sin \left (d x + c\right ) + 192 \, C a^{3} b \tan \left (d x + c\right ) + 288 \, B a^{2} b^{2} \tan \left (d x + c\right ) + 192 \, A a b^{3} \tan \left (d x + c\right )}{48 \, 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)+C*cos(d*x+c)^2)*sec(d*x+c)^5,x, algorithm="maxima")

[Out]

1/48*(16*(tan(d*x + c)^3 + 3*tan(d*x + c))*B*a^4 + 64*(tan(d*x + c)^3 + 3*tan(d*x + c))*A*a^3*b + 192*(d*x + c
)*C*a*b^3 + 48*(d*x + c)*B*b^4 - 3*A*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)) - 12*C*a^4*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1)
- log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) - 48*B*a^3*b*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(
d*x + c) + 1) + log(sin(d*x + c) - 1)) - 72*A*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)) + 144*C*a^2*b^2*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 96*B*a*b^3*(lo
g(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 24*A*b^4*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 48*C
*b^4*sin(d*x + c) + 192*C*a^3*b*tan(d*x + c) + 288*B*a^2*b^2*tan(d*x + c) + 192*A*a*b^3*tan(d*x + c))/d

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Fricas [A]  time = 2.0737, size = 725, normalized size = 2.47 \begin{align*} \frac{48 \,{\left (4 \, C a b^{3} + B b^{4}\right )} d x \cos \left (d x + c\right )^{4} + 3 \,{\left ({\left (3 \, A + 4 \, C\right )} a^{4} + 16 \, B a^{3} b + 24 \,{\left (A + 2 \, C\right )} a^{2} b^{2} + 32 \, B a b^{3} + 8 \, A b^{4}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left ({\left (3 \, A + 4 \, C\right )} a^{4} + 16 \, B a^{3} b + 24 \,{\left (A + 2 \, C\right )} a^{2} b^{2} + 32 \, B a b^{3} + 8 \, A b^{4}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (24 \, C b^{4} \cos \left (d x + c\right )^{4} + 6 \, A a^{4} + 16 \,{\left (B a^{4} + 2 \,{\left (2 \, A + 3 \, C\right )} a^{3} b + 9 \, B a^{2} b^{2} + 6 \, A a b^{3}\right )} \cos \left (d x + c\right )^{3} + 3 \,{\left ({\left (3 \, A + 4 \, C\right )} a^{4} + 16 \, B a^{3} b + 24 \, A a^{2} b^{2}\right )} \cos \left (d x + c\right )^{2} + 8 \,{\left (B a^{4} + 4 \, A a^{3} b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )^{4}} \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)+C*cos(d*x+c)^2)*sec(d*x+c)^5,x, algorithm="fricas")

[Out]

1/48*(48*(4*C*a*b^3 + B*b^4)*d*x*cos(d*x + c)^4 + 3*((3*A + 4*C)*a^4 + 16*B*a^3*b + 24*(A + 2*C)*a^2*b^2 + 32*
B*a*b^3 + 8*A*b^4)*cos(d*x + c)^4*log(sin(d*x + c) + 1) - 3*((3*A + 4*C)*a^4 + 16*B*a^3*b + 24*(A + 2*C)*a^2*b
^2 + 32*B*a*b^3 + 8*A*b^4)*cos(d*x + c)^4*log(-sin(d*x + c) + 1) + 2*(24*C*b^4*cos(d*x + c)^4 + 6*A*a^4 + 16*(
B*a^4 + 2*(2*A + 3*C)*a^3*b + 9*B*a^2*b^2 + 6*A*a*b^3)*cos(d*x + c)^3 + 3*((3*A + 4*C)*a^4 + 16*B*a^3*b + 24*A
*a^2*b^2)*cos(d*x + c)^2 + 8*(B*a^4 + 4*A*a^3*b)*cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^4)

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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)+C*cos(d*x+c)**2)*sec(d*x+c)**5,x)

[Out]

Timed out

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Giac [B]  time = 1.43607, size = 1134, normalized size = 3.87 \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)+C*cos(d*x+c)^2)*sec(d*x+c)^5,x, algorithm="giac")

[Out]

1/24*(48*C*b^4*tan(1/2*d*x + 1/2*c)/(tan(1/2*d*x + 1/2*c)^2 + 1) + 24*(4*C*a*b^3 + B*b^4)*(d*x + c) + 3*(3*A*a
^4 + 4*C*a^4 + 16*B*a^3*b + 24*A*a^2*b^2 + 48*C*a^2*b^2 + 32*B*a*b^3 + 8*A*b^4)*log(abs(tan(1/2*d*x + 1/2*c) +
1)) - 3*(3*A*a^4 + 4*C*a^4 + 16*B*a^3*b + 24*A*a^2*b^2 + 48*C*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*(15*A*a^4*tan(1/2*d*x + 1/2*c)^7 - 24*B*a^4*tan(1/2*d*x + 1/2*c)^7 + 12*C*a^4*tan(1/2*
d*x + 1/2*c)^7 - 96*A*a^3*b*tan(1/2*d*x + 1/2*c)^7 + 48*B*a^3*b*tan(1/2*d*x + 1/2*c)^7 - 96*C*a^3*b*tan(1/2*d*
x + 1/2*c)^7 + 72*A*a^2*b^2*tan(1/2*d*x + 1/2*c)^7 - 144*B*a^2*b^2*tan(1/2*d*x + 1/2*c)^7 - 96*A*a*b^3*tan(1/2
*d*x + 1/2*c)^7 + 9*A*a^4*tan(1/2*d*x + 1/2*c)^5 + 40*B*a^4*tan(1/2*d*x + 1/2*c)^5 - 12*C*a^4*tan(1/2*d*x + 1/
2*c)^5 + 160*A*a^3*b*tan(1/2*d*x + 1/2*c)^5 - 48*B*a^3*b*tan(1/2*d*x + 1/2*c)^5 + 288*C*a^3*b*tan(1/2*d*x + 1/
2*c)^5 - 72*A*a^2*b^2*tan(1/2*d*x + 1/2*c)^5 + 432*B*a^2*b^2*tan(1/2*d*x + 1/2*c)^5 + 288*A*a*b^3*tan(1/2*d*x
+ 1/2*c)^5 + 9*A*a^4*tan(1/2*d*x + 1/2*c)^3 - 40*B*a^4*tan(1/2*d*x + 1/2*c)^3 - 12*C*a^4*tan(1/2*d*x + 1/2*c)^
3 - 160*A*a^3*b*tan(1/2*d*x + 1/2*c)^3 - 48*B*a^3*b*tan(1/2*d*x + 1/2*c)^3 - 288*C*a^3*b*tan(1/2*d*x + 1/2*c)^
3 - 72*A*a^2*b^2*tan(1/2*d*x + 1/2*c)^3 - 432*B*a^2*b^2*tan(1/2*d*x + 1/2*c)^3 - 288*A*a*b^3*tan(1/2*d*x + 1/2
*c)^3 + 15*A*a^4*tan(1/2*d*x + 1/2*c) + 24*B*a^4*tan(1/2*d*x + 1/2*c) + 12*C*a^4*tan(1/2*d*x + 1/2*c) + 96*A*a
^3*b*tan(1/2*d*x + 1/2*c) + 48*B*a^3*b*tan(1/2*d*x + 1/2*c) + 96*C*a^3*b*tan(1/2*d*x + 1/2*c) + 72*A*a^2*b^2*t
an(1/2*d*x + 1/2*c) + 144*B*a^2*b^2*tan(1/2*d*x + 1/2*c) + 96*A*a*b^3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2
*c)^2 - 1)^4)/d