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

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

[Out]

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

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Rubi [A]  time = 0.97243, antiderivative size = 293, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.146, Rules used = {4094, 4076, 4047, 8, 4045, 3770} \[ \frac{a \sin (c+d x) \left (a^2 b (23 A+36 C)+8 a^3 B+36 a b^2 B+12 A b^3\right )}{12 d}-\frac{b^2 \tan (c+d x) \left (3 a^2 (3 A+4 C)+32 a b B+2 b^2 (13 A-12 C)\right )}{24 d}+\frac{\sin (c+d x) \cos (c+d x) \left (a^2 (3 A+4 C)+8 a b B+4 A b^2\right ) (a+b \sec (c+d x))^2}{8 d}+\frac{1}{8} x \left (24 a^2 b^2 (A+2 C)+a^4 (3 A+4 C)+16 a^3 b B+32 a b^3 B+8 A b^4\right )+\frac{(a B+A b) \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^3}{3 d}+\frac{A \sin (c+d x) \cos ^3(c+d x) (a+b \sec (c+d x))^4}{4 d}+\frac{b^3 (4 a C+b B) \tanh ^{-1}(\sin (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4094

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^
(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(A*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(d*
Csc[e + f*x])^n)/(f*n), x] - Dist[1/(d*n), Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^(n + 1)*Simp[A*b*
m - a*B*n - (b*B*n + a*(C*n + A*(n + 1)))*Csc[e + f*x] - b*(C*n + A*(m + n + 1))*Csc[e + f*x]^2, x], x], x] /;
FreeQ[{a, b, d, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 0] && LeQ[n, -1]

Rule 4076

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^
(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> -Simp[(b*C*Csc[e + f*x]*Cot[e + f*x]*(d*Csc[e + f*x
])^n)/(f*(n + 2)), x] + Dist[1/(n + 2), Int[(d*Csc[e + f*x])^n*Simp[A*a*(n + 2) + (B*a*(n + 2) + b*(C*(n + 1)
+ A*(n + 2)))*Csc[e + f*x] + (a*C + B*b)*(n + 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C,
n}, x] &&  !LtQ[n, -1]

Rule 4047

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(
C_.)), x_Symbol] :> Dist[B/b, Int[(b*Csc[e + f*x])^(m + 1), x], x] + Int[(b*Csc[e + f*x])^m*(A + C*Csc[e + f*x
]^2), x] /; FreeQ[{b, e, f, A, B, C, m}, x]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 4045

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Simp[(A*Cot[e
+ f*x]*(b*Csc[e + f*x])^m)/(f*m), x] + Dist[(C*m + A*(m + 1))/(b^2*m), Int[(b*Csc[e + f*x])^(m + 2), x], x] /
; FreeQ[{b, e, f, A, C}, x] && NeQ[C*m + A*(m + 1), 0] && LeQ[m, -1]

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

Mathematica [A]  time = 3.97338, size = 382, normalized size = 1.3 \[ \frac{32 a \sin (c+d x) \left (4 a^2 b (5 A+6 C)+5 a^3 B+36 a b^2 B+24 A b^3\right )+a^2 \sec (c+d x) \left (3 \sin (3 (c+d x)) \left (a^2 (9 A+8 C)+32 a b B+48 A b^2\right )+a (8 (a B+4 A b) \sin (4 (c+d x))+3 a A \sin (5 (c+d x)))\right )+24 \left (\tan (c+d x) \left (6 a^2 A b^2+a^4 (A+C)+4 a^3 b B+8 b^4 C\right )+24 a^2 A b^2 c+24 a^2 A b^2 d x+3 a^4 A c+3 a^4 A d x+48 a^2 b^2 c C+48 a^2 b^2 C d x+16 a^3 b B c+16 a^3 b B d x+4 a^4 c C+4 a^4 C d x-8 b^3 (4 a C+b B) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+32 a b^3 B c+32 a b^3 B d x+32 a b^3 C \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+8 A b^4 c+8 A b^4 d x+8 b^4 B \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )}{192 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(32*a*(24*A*b^3 + 5*a^3*B + 36*a*b^2*B + 4*a^2*b*(5*A + 6*C))*Sin[c + d*x] + a^2*Sec[c + d*x]*(3*(48*A*b^2 + 3
2*a*b*B + a^2*(9*A + 8*C))*Sin[3*(c + d*x)] + a*(8*(4*A*b + a*B)*Sin[4*(c + d*x)] + 3*a*A*Sin[5*(c + d*x)])) +
24*(3*a^4*A*c + 24*a^2*A*b^2*c + 8*A*b^4*c + 16*a^3*b*B*c + 32*a*b^3*B*c + 4*a^4*c*C + 48*a^2*b^2*c*C + 3*a^4
*A*d*x + 24*a^2*A*b^2*d*x + 8*A*b^4*d*x + 16*a^3*b*B*d*x + 32*a*b^3*B*d*x + 4*a^4*C*d*x + 48*a^2*b^2*C*d*x - 8
*b^3*(b*B + 4*a*C)*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 8*b^4*B*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]
+ 32*a*b^3*C*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + (6*a^2*A*b^2 + 4*a^3*b*B + 8*b^4*C + a^4*(A + C))*Tan
[c + d*x]))/(192*d)

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Maple [A]  time = 0.082, size = 434, normalized size = 1.5 \begin{align*}{\frac{3\,{a}^{4}Ax}{8}}+6\,{\frac{C{a}^{2}{b}^{2}c}{d}}+4\,{\frac{Ba{b}^{3}c}{d}}+{\frac{3\,A{a}^{4}c}{8\,d}}+{\frac{8\,A{a}^{3}b\sin \left ( dx+c \right ) }{3\,d}}+{\frac{{a}^{4}Cc}{2\,d}}+{\frac{2\,B{a}^{4}\sin \left ( dx+c \right ) }{3\,d}}+2\,B{a}^{3}bx+3\,A{a}^{2}{b}^{2}x+4\,{\frac{Aa{b}^{3}\sin \left ( dx+c \right ) }{d}}+A{b}^{4}x+6\,{\frac{{a}^{2}{b}^{2}B\sin \left ( dx+c \right ) }{d}}+4\,{\frac{Ca{b}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{C{b}^{4}\tan \left ( dx+c \right ) }{d}}+{\frac{B{b}^{4}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{3\,A{a}^{4}\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{8\,d}}+2\,{\frac{B{a}^{3}bc}{d}}+3\,{\frac{A{a}^{2}{b}^{2}c}{d}}+{\frac{B\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}{a}^{4}}{3\,d}}+{\frac{A{a}^{4}\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{4\,d}}+{\frac{{a}^{4}C\sin \left ( dx+c \right ) \cos \left ( dx+c \right ) }{2\,d}}+4\,{\frac{{a}^{3}bC\sin \left ( dx+c \right ) }{d}}+{\frac{4\,A\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}{a}^{3}b}{3\,d}}+2\,{\frac{B{a}^{3}b\sin \left ( dx+c \right ) \cos \left ( dx+c \right ) }{d}}+3\,{\frac{A{a}^{2}{b}^{2}\sin \left ( dx+c \right ) \cos \left ( dx+c \right ) }{d}}+{\frac{A{b}^{4}c}{d}}+4\,a{b}^{3}Bx+6\,C{a}^{2}{b}^{2}x+{\frac{{a}^{4}Cx}{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.044, size = 412, normalized size = 1.41 \begin{align*} \frac{3 \,{\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{4} - 32 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a^{4} + 24 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{4} - 128 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a^{3} b + 96 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{3} b + 144 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{2} b^{2} + 576 \,{\left (d x + c\right )} C a^{2} b^{2} + 384 \,{\left (d x + c\right )} B a b^{3} + 96 \,{\left (d x + c\right )} A b^{4} + 192 \, C 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 \, B b^{4}{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 384 \, C a^{3} b \sin \left (d x + c\right ) + 576 \, B a^{2} b^{2} \sin \left (d x + c\right ) + 384 \, A a b^{3} \sin \left (d x + c\right ) + 96 \, C b^{4} \tan \left (d x + c\right )}{96 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(a+b*sec(d*x+c))^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

1/96*(3*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*A*a^4 - 32*(sin(d*x + c)^3 - 3*sin(d*x + c))*B
*a^4 + 24*(2*d*x + 2*c + sin(2*d*x + 2*c))*C*a^4 - 128*(sin(d*x + c)^3 - 3*sin(d*x + c))*A*a^3*b + 96*(2*d*x +
2*c + sin(2*d*x + 2*c))*B*a^3*b + 144*(2*d*x + 2*c + sin(2*d*x + 2*c))*A*a^2*b^2 + 576*(d*x + c)*C*a^2*b^2 +
384*(d*x + c)*B*a*b^3 + 96*(d*x + c)*A*b^4 + 192*C*a*b^3*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 48*
B*b^4*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 384*C*a^3*b*sin(d*x + c) + 576*B*a^2*b^2*sin(d*x + c)
+ 384*A*a*b^3*sin(d*x + c) + 96*C*b^4*tan(d*x + c))/d

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Fricas [A]  time = 0.615794, size = 636, normalized size = 2.17 \begin{align*} \frac{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 )} d x \cos \left (d x + c\right ) + 12 \,{\left (4 \, C a b^{3} + B b^{4}\right )} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) - 12 \,{\left (4 \, C a b^{3} + B b^{4}\right )} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) +{\left (6 \, A a^{4} \cos \left (d x + c\right )^{4} + 24 \, C b^{4} + 8 \,{\left (B a^{4} + 4 \, A a^{3} b\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} + 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 )\right )} \sin \left (d x + c\right )}{24 \, d \cos \left (d x + c\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(a+b*sec(d*x+c))^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

1/24*(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)*d*x*cos(d*x + c) + 12*(4*
C*a*b^3 + B*b^4)*cos(d*x + c)*log(sin(d*x + c) + 1) - 12*(4*C*a*b^3 + B*b^4)*cos(d*x + c)*log(-sin(d*x + c) +
1) + (6*A*a^4*cos(d*x + c)^4 + 24*C*b^4 + 8*(B*a^4 + 4*A*a^3*b)*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 + 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))*s
in(d*x + c))/(d*cos(d*x + c))

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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(cos(d*x+c)**4*(a+b*sec(d*x+c))**4*(A+B*sec(d*x+c)+C*sec(d*x+c)**2),x)

[Out]

Timed out

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Giac [B]  time = 1.35701, size = 1083, normalized size = 3.7 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(a+b*sec(d*x+c))^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),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) - 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)*(d*x + c) - 24*(4*C*a*b^3 + B*b^4)*log(abs(tan(1/2*d*x + 1/2*c)
+ 1)) + 24*(4*C*a*b^3 + B*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*tan(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