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

Optimal. Leaf size=303 \[ -\frac{b \sin (c+d x) \left (4 a^3 (2 A+3 C)+39 a^2 b B+4 a b^2 (11 A-6 C)-6 b^3 B\right )}{6 d}+\frac{a \left (4 a^2 b (A+2 C)+a^3 B+12 a b^2 B+8 A b^3\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac{b^2 \sin (c+d x) \cos (c+d x) \left (a^2 (4 A+6 C)+18 a b B+3 b^2 (6 A-C)\right )}{6 d}+\frac{\tan (c+d x) \left (a^2 (4 A+6 C)+15 a b B+12 A b^2\right ) (a+b \cos (c+d x))^2}{6 d}+\frac{1}{2} b^2 x \left (12 a^2 C+8 a b B+2 A b^2+b^2 C\right )+\frac{(3 a B+4 A b) \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^3}{6 d}+\frac{A \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))^4}{3 d} \]

[Out]

(b^2*(2*A*b^2 + 8*a*b*B + 12*a^2*C + b^2*C)*x)/2 + (a*(8*A*b^3 + a^3*B + 12*a*b^2*B + 4*a^2*b*(A + 2*C))*ArcTa
nh[Sin[c + d*x]])/(2*d) - (b*(39*a^2*b*B - 6*b^3*B + 4*a*b^2*(11*A - 6*C) + 4*a^3*(2*A + 3*C))*Sin[c + d*x])/(
6*d) - (b^2*(18*a*b*B + 3*b^2*(6*A - C) + a^2*(4*A + 6*C))*Cos[c + d*x]*Sin[c + d*x])/(6*d) + ((12*A*b^2 + 15*
a*b*B + a^2*(4*A + 6*C))*(a + b*Cos[c + d*x])^2*Tan[c + d*x])/(6*d) + ((4*A*b + 3*a*B)*(a + b*Cos[c + d*x])^3*
Sec[c + d*x]*Tan[c + d*x])/(6*d) + (A*(a + b*Cos[c + d*x])^4*Sec[c + d*x]^2*Tan[c + d*x])/(3*d)

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Rubi [A]  time = 1.07708, antiderivative size = 303, 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, 3033, 3023, 2735, 3770} \[ -\frac{b \sin (c+d x) \left (4 a^3 (2 A+3 C)+39 a^2 b B+4 a b^2 (11 A-6 C)-6 b^3 B\right )}{6 d}+\frac{a \left (4 a^2 b (A+2 C)+a^3 B+12 a b^2 B+8 A b^3\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac{b^2 \sin (c+d x) \cos (c+d x) \left (a^2 (4 A+6 C)+18 a b B+3 b^2 (6 A-C)\right )}{6 d}+\frac{\tan (c+d x) \left (a^2 (4 A+6 C)+15 a b B+12 A b^2\right ) (a+b \cos (c+d x))^2}{6 d}+\frac{1}{2} b^2 x \left (12 a^2 C+8 a b B+2 A b^2+b^2 C\right )+\frac{(3 a B+4 A b) \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^3}{6 d}+\frac{A \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))^4}{3 d} \]

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]^4,x]

[Out]

(b^2*(2*A*b^2 + 8*a*b*B + 12*a^2*C + b^2*C)*x)/2 + (a*(8*A*b^3 + a^3*B + 12*a*b^2*B + 4*a^2*b*(A + 2*C))*ArcTa
nh[Sin[c + d*x]])/(2*d) - (b*(39*a^2*b*B - 6*b^3*B + 4*a*b^2*(11*A - 6*C) + 4*a^3*(2*A + 3*C))*Sin[c + d*x])/(
6*d) - (b^2*(18*a*b*B + 3*b^2*(6*A - C) + a^2*(4*A + 6*C))*Cos[c + d*x]*Sin[c + d*x])/(6*d) + ((12*A*b^2 + 15*
a*b*B + a^2*(4*A + 6*C))*(a + b*Cos[c + d*x])^2*Tan[c + d*x])/(6*d) + ((4*A*b + 3*a*B)*(a + b*Cos[c + d*x])^3*
Sec[c + d*x]*Tan[c + d*x])/(6*d) + (A*(a + b*Cos[c + d*x])^4*Sec[c + d*x]^2*Tan[c + d*x])/(3*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 3033

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[(C*d*Cos[e + f*x]*Sin[e + f*x]*(a + b
*Sin[e + f*x])^(m + 1))/(b*f*(m + 3)), x] + Dist[1/(b*(m + 3)), Int[(a + b*Sin[e + f*x])^m*Simp[a*C*d + A*b*c*
(m + 3) + b*(B*c*(m + 3) + d*(C*(m + 2) + A*(m + 3)))*Sin[e + f*x] - (2*a*C*d - b*(c*C + B*d)*(m + 3))*Sin[e +
f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m}, x] && NeQ[b*c - a*d, 0] && NeQ[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 ^4(c+d x) \, dx &=\frac{A (a+b \cos (c+d x))^4 \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac{1}{3} \int (a+b \cos (c+d x))^3 \left (4 A b+3 a B+(2 a A+3 b B+3 a C) \cos (c+d x)-b (2 A-3 C) \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx\\ &=\frac{(4 A b+3 a B) (a+b \cos (c+d x))^3 \sec (c+d x) \tan (c+d x)}{6 d}+\frac{A (a+b \cos (c+d x))^4 \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac{1}{6} \int (a+b \cos (c+d x))^2 \left (12 A b^2+15 a b B+a^2 (4 A+6 C)+\left (3 a^2 B+6 b^2 B+4 a b (A+3 C)\right ) \cos (c+d x)-6 b (2 A b+a B-b C) \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx\\ &=\frac{\left (12 A b^2+15 a b B+a^2 (4 A+6 C)\right ) (a+b \cos (c+d x))^2 \tan (c+d x)}{6 d}+\frac{(4 A b+3 a B) (a+b \cos (c+d x))^3 \sec (c+d x) \tan (c+d x)}{6 d}+\frac{A (a+b \cos (c+d x))^4 \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac{1}{6} \int (a+b \cos (c+d x)) \left (3 \left (8 A b^3+a^3 B+12 a b^2 B+4 a^2 b (A+2 C)\right )-b \left (8 a A b+3 a^2 B-6 b^2 B-18 a b C\right ) \cos (c+d x)-2 b \left (18 a b B+3 b^2 (6 A-C)+a^2 (4 A+6 C)\right ) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx\\ &=-\frac{b^2 \left (18 a b B+3 b^2 (6 A-C)+a^2 (4 A+6 C)\right ) \cos (c+d x) \sin (c+d x)}{6 d}+\frac{\left (12 A b^2+15 a b B+a^2 (4 A+6 C)\right ) (a+b \cos (c+d x))^2 \tan (c+d x)}{6 d}+\frac{(4 A b+3 a B) (a+b \cos (c+d x))^3 \sec (c+d x) \tan (c+d x)}{6 d}+\frac{A (a+b \cos (c+d x))^4 \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac{1}{12} \int \left (6 a \left (8 A b^3+a^3 B+12 a b^2 B+4 a^2 b (A+2 C)\right )+6 b^2 \left (2 A b^2+8 a b B+12 a^2 C+b^2 C\right ) \cos (c+d x)-2 b \left (39 a^2 b B-6 b^3 B+4 a b^2 (11 A-6 C)+4 a^3 (2 A+3 C)\right ) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx\\ &=-\frac{b \left (39 a^2 b B-6 b^3 B+4 a b^2 (11 A-6 C)+4 a^3 (2 A+3 C)\right ) \sin (c+d x)}{6 d}-\frac{b^2 \left (18 a b B+3 b^2 (6 A-C)+a^2 (4 A+6 C)\right ) \cos (c+d x) \sin (c+d x)}{6 d}+\frac{\left (12 A b^2+15 a b B+a^2 (4 A+6 C)\right ) (a+b \cos (c+d x))^2 \tan (c+d x)}{6 d}+\frac{(4 A b+3 a B) (a+b \cos (c+d x))^3 \sec (c+d x) \tan (c+d x)}{6 d}+\frac{A (a+b \cos (c+d x))^4 \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac{1}{12} \int \left (6 a \left (8 A b^3+a^3 B+12 a b^2 B+4 a^2 b (A+2 C)\right )+6 b^2 \left (2 A b^2+8 a b B+12 a^2 C+b^2 C\right ) \cos (c+d x)\right ) \sec (c+d x) \, dx\\ &=\frac{1}{2} b^2 \left (2 A b^2+8 a b B+12 a^2 C+b^2 C\right ) x-\frac{b \left (39 a^2 b B-6 b^3 B+4 a b^2 (11 A-6 C)+4 a^3 (2 A+3 C)\right ) \sin (c+d x)}{6 d}-\frac{b^2 \left (18 a b B+3 b^2 (6 A-C)+a^2 (4 A+6 C)\right ) \cos (c+d x) \sin (c+d x)}{6 d}+\frac{\left (12 A b^2+15 a b B+a^2 (4 A+6 C)\right ) (a+b \cos (c+d x))^2 \tan (c+d x)}{6 d}+\frac{(4 A b+3 a B) (a+b \cos (c+d x))^3 \sec (c+d x) \tan (c+d x)}{6 d}+\frac{A (a+b \cos (c+d x))^4 \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac{1}{2} \left (a \left (8 A b^3+a^3 B+12 a b^2 B+4 a^2 b (A+2 C)\right )\right ) \int \sec (c+d x) \, dx\\ &=\frac{1}{2} b^2 \left (2 A b^2+8 a b B+12 a^2 C+b^2 C\right ) x+\frac{a \left (8 A b^3+a^3 B+12 a b^2 B+4 a^2 b (A+2 C)\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac{b \left (39 a^2 b B-6 b^3 B+4 a b^2 (11 A-6 C)+4 a^3 (2 A+3 C)\right ) \sin (c+d x)}{6 d}-\frac{b^2 \left (18 a b B+3 b^2 (6 A-C)+a^2 (4 A+6 C)\right ) \cos (c+d x) \sin (c+d x)}{6 d}+\frac{\left (12 A b^2+15 a b B+a^2 (4 A+6 C)\right ) (a+b \cos (c+d x))^2 \tan (c+d x)}{6 d}+\frac{(4 A b+3 a B) (a+b \cos (c+d x))^3 \sec (c+d x) \tan (c+d x)}{6 d}+\frac{A (a+b \cos (c+d x))^4 \sec ^2(c+d x) \tan (c+d x)}{3 d}\\ \end{align*}

Mathematica [A]  time = 2.25235, size = 351, normalized size = 1.16 \[ \frac{\sec ^3(c+d x) \left (36 b^2 (c+d x) \cos (c+d x) \left (12 a^2 C+8 a b B+2 A b^2+b^2 C\right )+12 b^2 (c+d x) \cos (3 (c+d x)) \left (12 a^2 C+8 a b B+2 A b^2+b^2 C\right )+2 \sin (c+d x) \left (12 \cos (c+d x) \left (8 a^3 A b+2 a^4 B+12 a b^3 C+3 b^4 B\right )+4 \cos (2 (c+d x)) \left (36 a^2 A b^2+a^4 (4 A+6 C)+24 a^3 b B+3 b^4 C\right )+144 a^2 A b^2+32 a^4 A+96 a^3 b B+24 a^4 C+48 a b^3 C \cos (3 (c+d x))+12 b^4 B \cos (3 (c+d x))+3 b^4 C \cos (4 (c+d x))+9 b^4 C\right )-48 a \cos ^3(c+d x) \left (4 a^2 b (A+2 C)+a^3 B+12 a b^2 B+8 A b^3\right ) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )\right )}{96 d} \]

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]^4,x]

[Out]

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

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Maple [A]  time = 0.081, size = 377, normalized size = 1.2 \begin{align*} A{b}^{4}x+{\frac{A{b}^{4}c}{d}}+{\frac{{b}^{4}B\sin \left ( dx+c \right ) }{d}}+{\frac{C{b}^{4}\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2\,d}}+{\frac{{b}^{4}Cx}{2}}+{\frac{C{b}^{4}c}{2\,d}}+4\,{\frac{aA{b}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+4\,a{b}^{3}Bx+4\,{\frac{Ba{b}^{3}c}{d}}+4\,{\frac{Ca{b}^{3}\sin \left ( dx+c \right ) }{d}}+6\,{\frac{{a}^{2}A{b}^{2}\tan \left ( dx+c \right ) }{d}}+6\,{\frac{{a}^{2}{b}^{2}B\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+6\,{a}^{2}{b}^{2}Cx+6\,{\frac{C{a}^{2}{b}^{2}c}{d}}+2\,{\frac{A{a}^{3}b\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{d}}+2\,{\frac{A{a}^{3}b\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+4\,{\frac{{a}^{3}bB\tan \left ( dx+c \right ) }{d}}+4\,{\frac{{a}^{3}bC\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{2\,A{a}^{4}\tan \left ( dx+c \right ) }{3\,d}}+{\frac{A{a}^{4}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3\,d}}+{\frac{{a}^{4}B\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{{a}^{4}B\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+{\frac{{a}^{4}C\tan \left ( dx+c \right ) }{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)^4,x)

[Out]

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

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Maxima [A]  time = 1.04641, size = 452, normalized size = 1.49 \begin{align*} \frac{4 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a^{4} + 72 \,{\left (d x + c\right )} C a^{2} b^{2} + 48 \,{\left (d x + c\right )} B a b^{3} + 12 \,{\left (d x + c\right )} A b^{4} + 3 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C b^{4} - 3 \, B 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 )} - 12 \, A 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 )} + 24 \, C a^{3} b{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 36 \, B 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 )} + 24 \, A 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 )} + 48 \, C a b^{3} \sin \left (d x + c\right ) + 12 \, B b^{4} \sin \left (d x + c\right ) + 12 \, C a^{4} \tan \left (d x + c\right ) + 48 \, B a^{3} b \tan \left (d x + c\right ) + 72 \, A a^{2} b^{2} \tan \left (d x + c\right )}{12 \, 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)^4,x, algorithm="maxima")

[Out]

1/12*(4*(tan(d*x + c)^3 + 3*tan(d*x + c))*A*a^4 + 72*(d*x + c)*C*a^2*b^2 + 48*(d*x + c)*B*a*b^3 + 12*(d*x + c)
*A*b^4 + 3*(2*d*x + 2*c + sin(2*d*x + 2*c))*C*b^4 - 3*B*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)) - 12*A*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)) + 24*C*a^3*b*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 36*B*a^2*b^2*(log(sin(
d*x + c) + 1) - log(sin(d*x + c) - 1)) + 24*A*a*b^3*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 48*C*a*b
^3*sin(d*x + c) + 12*B*b^4*sin(d*x + c) + 12*C*a^4*tan(d*x + c) + 48*B*a^3*b*tan(d*x + c) + 72*A*a^2*b^2*tan(d
*x + c))/d

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Fricas [A]  time = 2.07981, size = 644, normalized size = 2.13 \begin{align*} \frac{6 \,{\left (12 \, C a^{2} b^{2} + 8 \, B a b^{3} +{\left (2 \, A + C\right )} b^{4}\right )} d x \cos \left (d x + c\right )^{3} + 3 \,{\left (B a^{4} + 4 \,{\left (A + 2 \, C\right )} a^{3} b + 12 \, B a^{2} b^{2} + 8 \, A a b^{3}\right )} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left (B a^{4} + 4 \,{\left (A + 2 \, C\right )} a^{3} b + 12 \, B a^{2} b^{2} + 8 \, A a b^{3}\right )} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (3 \, C b^{4} \cos \left (d x + c\right )^{4} + 2 \, A a^{4} + 6 \,{\left (4 \, C a b^{3} + B b^{4}\right )} \cos \left (d x + c\right )^{3} + 2 \,{\left ({\left (2 \, A + 3 \, C\right )} a^{4} + 12 \, B a^{3} b + 18 \, A a^{2} b^{2}\right )} \cos \left (d x + c\right )^{2} + 3 \,{\left (B a^{4} + 4 \, A a^{3} b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{12 \, d \cos \left (d x + c\right )^{3}} \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)^4,x, algorithm="fricas")

[Out]

1/12*(6*(12*C*a^2*b^2 + 8*B*a*b^3 + (2*A + C)*b^4)*d*x*cos(d*x + c)^3 + 3*(B*a^4 + 4*(A + 2*C)*a^3*b + 12*B*a^
2*b^2 + 8*A*a*b^3)*cos(d*x + c)^3*log(sin(d*x + c) + 1) - 3*(B*a^4 + 4*(A + 2*C)*a^3*b + 12*B*a^2*b^2 + 8*A*a*
b^3)*cos(d*x + c)^3*log(-sin(d*x + c) + 1) + 2*(3*C*b^4*cos(d*x + c)^4 + 2*A*a^4 + 6*(4*C*a*b^3 + B*b^4)*cos(d
*x + c)^3 + 2*((2*A + 3*C)*a^4 + 12*B*a^3*b + 18*A*a^2*b^2)*cos(d*x + c)^2 + 3*(B*a^4 + 4*A*a^3*b)*cos(d*x + c
))*sin(d*x + c))/(d*cos(d*x + c)^3)

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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)**4,x)

[Out]

Timed out

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Giac [A]  time = 1.38283, size = 743, normalized size = 2.45 \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)^4,x, algorithm="giac")

[Out]

1/6*(3*(12*C*a^2*b^2 + 8*B*a*b^3 + 2*A*b^4 + C*b^4)*(d*x + c) + 3*(B*a^4 + 4*A*a^3*b + 8*C*a^3*b + 12*B*a^2*b^
2 + 8*A*a*b^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(B*a^4 + 4*A*a^3*b + 8*C*a^3*b + 12*B*a^2*b^2 + 8*A*a*b^
3)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 6*(8*C*a*b^3*tan(1/2*d*x + 1/2*c)^3 + 2*B*b^4*tan(1/2*d*x + 1/2*c)^3 -
C*b^4*tan(1/2*d*x + 1/2*c)^3 + 8*C*a*b^3*tan(1/2*d*x + 1/2*c) + 2*B*b^4*tan(1/2*d*x + 1/2*c) + C*b^4*tan(1/2*
d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^2 - 2*(6*A*a^4*tan(1/2*d*x + 1/2*c)^5 - 3*B*a^4*tan(1/2*d*x + 1/2*c
)^5 + 6*C*a^4*tan(1/2*d*x + 1/2*c)^5 - 12*A*a^3*b*tan(1/2*d*x + 1/2*c)^5 + 24*B*a^3*b*tan(1/2*d*x + 1/2*c)^5 +
36*A*a^2*b^2*tan(1/2*d*x + 1/2*c)^5 - 4*A*a^4*tan(1/2*d*x + 1/2*c)^3 - 12*C*a^4*tan(1/2*d*x + 1/2*c)^3 - 48*B
*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 + 6*A*a^4*tan(1/2*d*x + 1/2*c) + 3*B*a^4*t
an(1/2*d*x + 1/2*c) + 6*C*a^4*tan(1/2*d*x + 1/2*c) + 12*A*a^3*b*tan(1/2*d*x + 1/2*c) + 24*B*a^3*b*tan(1/2*d*x
+ 1/2*c) + 36*A*a^2*b^2*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^3)/d