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

Optimal. Leaf size=246 \[ \frac {a b \left (a^2 (23 A+36 C)+12 A b^2\right ) \sin (c+d x)}{12 d}-\frac {b^2 \left (3 a^2 (3 A+4 C)+2 b^2 (13 A-12 C)\right ) \tan (c+d x)}{24 d}+\frac {\left (a^2 (3 A+4 C)+4 A b^2\right ) \sin (c+d x) \cos (c+d x) (a+b \sec (c+d x))^2}{8 d}+\frac {1}{8} x \left (a^4 (3 A+4 C)+24 a^2 b^2 (A+2 C)+8 A b^4\right )+\frac {A \sin (c+d x) \cos ^3(c+d x) (a+b \sec (c+d x))^4}{4 d}+\frac {A b \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^3}{3 d}+\frac {4 a b^3 C \tanh ^{-1}(\sin (c+d x))}{d} \]

[Out]

1/8*(8*A*b^4+24*a^2*b^2*(A+2*C)+a^4*(3*A+4*C))*x+4*a*b^3*C*arctanh(sin(d*x+c))/d+1/12*a*b*(12*A*b^2+a^2*(23*A+
36*C))*sin(d*x+c)/d+1/8*(4*A*b^2+a^2*(3*A+4*C))*cos(d*x+c)*(a+b*sec(d*x+c))^2*sin(d*x+c)/d+1/3*A*b*cos(d*x+c)^
2*(a+b*sec(d*x+c))^3*sin(d*x+c)/d+1/4*A*cos(d*x+c)^3*(a+b*sec(d*x+c))^4*sin(d*x+c)/d-1/24*b^2*(2*b^2*(13*A-12*
C)+3*a^2*(3*A+4*C))*tan(d*x+c)/d

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Rubi [A]  time = 0.85, antiderivative size = 246, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.212, Rules used = {4095, 4094, 4076, 4047, 8, 4045, 3770} \[ \frac {a b \left (a^2 (23 A+36 C)+12 A b^2\right ) \sin (c+d x)}{12 d}-\frac {b^2 \left (3 a^2 (3 A+4 C)+2 b^2 (13 A-12 C)\right ) \tan (c+d x)}{24 d}+\frac {\left (a^2 (3 A+4 C)+4 A b^2\right ) \sin (c+d x) \cos (c+d x) (a+b \sec (c+d x))^2}{8 d}+\frac {1}{8} x \left (24 a^2 b^2 (A+2 C)+a^4 (3 A+4 C)+8 A b^4\right )+\frac {A \sin (c+d x) \cos ^3(c+d x) (a+b \sec (c+d x))^4}{4 d}+\frac {A b \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^3}{3 d}+\frac {4 a b^3 C \tanh ^{-1}(\sin (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

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 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 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 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 4095

Int[((A_.) + 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] - Dis
t[1/(d*n), Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^(n + 1)*Simp[A*b*m - 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, C}, x] && NeQ[a^2 - b^2,
0] && GtQ[m, 0] && LeQ[n, -1]

Rubi steps

\begin {align*} \int \cos ^4(c+d x) (a+b \sec (c+d x))^4 \left (A+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 (3 A+4 C) \sec (c+d x)-b (A-4 C) \sec ^2(c+d x)\right ) \, dx\\ &=\frac {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+a^2 (3 A+4 C)\right )+2 a b (7 A+12 C) \sec (c+d x)-b^2 (7 A-12 C) \sec ^2(c+d x)\right ) \, dx\\ &=\frac {\left (4 A b^2+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 \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+\frac {1}{2} a^2 (46 A b+72 b C)\right )+a \left (3 a^2 (3 A+4 C)+2 b^2 (13 A+36 C)\right ) \sec (c+d x)-b \left (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+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 \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 (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+\frac {1}{2} a^2 (46 A b+72 b C)\right )+3 \left (8 A b^4+24 a^2 b^2 (A+2 C)+a^4 (3 A+4 C)\right ) \sec (c+d x)+96 a b^3 C \sec ^2(c+d x)\right ) \, dx\\ &=\frac {\left (4 A b^2+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 \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 (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+\frac {1}{2} a^2 (46 A b+72 b C)\right )+96 a b^3 C \sec ^2(c+d x)\right ) \, dx+\frac {1}{8} \left (8 A b^4+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+24 a^2 b^2 (A+2 C)+a^4 (3 A+4 C)\right ) x+\frac {a b \left (12 A b^2+a^2 (23 A+36 C)\right ) \sin (c+d x)}{12 d}+\frac {\left (4 A b^2+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 \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 (2 b^2 (13 A-12 C)+3 a^2 (3 A+4 C)\right ) \tan (c+d x)}{24 d}+\left (4 a b^3 C\right ) \int \sec (c+d x) \, dx\\ &=\frac {1}{8} \left (8 A b^4+24 a^2 b^2 (A+2 C)+a^4 (3 A+4 C)\right ) x+\frac {4 a b^3 C \tanh ^{-1}(\sin (c+d x))}{d}+\frac {a b \left (12 A b^2+a^2 (23 A+36 C)\right ) \sin (c+d x)}{12 d}+\frac {\left (4 A b^2+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 \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 (2 b^2 (13 A-12 C)+3 a^2 (3 A+4 C)\right ) \tan (c+d x)}{24 d}\\ \end {align*}

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Mathematica [A]  time = 1.70, size = 270, normalized size = 1.10 \[ \frac {3 a^4 A \sin (4 (c+d x))+32 a^3 A b \sin (3 (c+d x))+24 a^2 \left (a^2 (A+C)+6 A b^2\right ) \sin (2 (c+d x))+96 a b \left (a^2 (3 A+4 C)+4 A b^2\right ) \sin (c+d x)+12 (c+d x) \left (a^4 (3 A+4 C)+24 a^2 b^2 (A+2 C)+8 A b^4\right )-384 a b^3 C \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+384 a b^3 C \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )+\frac {96 b^4 C \sin \left (\frac {1}{2} (c+d x)\right )}{\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )}+\frac {96 b^4 C \sin \left (\frac {1}{2} (c+d x)\right )}{\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )}}{96 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(12*(8*A*b^4 + 24*a^2*b^2*(A + 2*C) + a^4*(3*A + 4*C))*(c + d*x) - 384*a*b^3*C*Log[Cos[(c + d*x)/2] - Sin[(c +
d*x)/2]] + 384*a*b^3*C*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + (96*b^4*C*Sin[(c + d*x)/2])/(Cos[(c + d*x)/
2] - Sin[(c + d*x)/2]) + (96*b^4*C*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]) + 96*a*b*(4*A*b^2 +
a^2*(3*A + 4*C))*Sin[c + d*x] + 24*a^2*(6*A*b^2 + a^2*(A + C))*Sin[2*(c + d*x)] + 32*a^3*A*b*Sin[3*(c + d*x)]
+ 3*a^4*A*Sin[4*(c + d*x)])/(96*d)

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fricas [A]  time = 0.50, size = 203, normalized size = 0.83 \[ \frac {48 \, C a b^{3} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) - 48 \, C a b^{3} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) + 3 \, {\left ({\left (3 \, A + 4 \, C\right )} a^{4} + 24 \, {\left (A + 2 \, C\right )} a^{2} b^{2} + 8 \, A b^{4}\right )} d x \cos \left (d x + c\right ) + {\left (6 \, A a^{4} \cos \left (d x + c\right )^{4} + 32 \, A a^{3} b \cos \left (d x + c\right )^{3} + 24 \, C b^{4} + 3 \, {\left ({\left (3 \, A + 4 \, C\right )} a^{4} + 24 \, A a^{2} b^{2}\right )} \cos \left (d x + c\right )^{2} + 32 \, {\left ({\left (2 \, A + 3 \, C\right )} a^{3} b + 3 \, A a b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{24 \, d \cos \left (d x + c\right )} \]

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+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.33, size = 558, normalized size = 2.27 \[ \frac {96 \, C a b^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 96 \, C a b^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {48 \, C b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1} + 3 \, {\left (3 \, A a^{4} + 4 \, C a^{4} + 24 \, A a^{2} b^{2} + 48 \, C a^{2} b^{2} + 8 \, A b^{4}\right )} {\left (d x + c\right )} - \frac {2 \, {\left (15 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 12 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 96 \, A a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 96 \, C a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 72 \, A a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 96 \, A a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 9 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 160 \, A a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 288 \, C a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 72 \, A a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 288 \, A a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 9 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 12 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 160 \, A a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 288 \, C a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 72 \, A a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 288 \, A a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 15 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 12 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 96 \, A a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 96 \, C a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 72 \, A a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 96 \, A a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{4}}}{24 \, d} \]

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+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

1/24*(96*C*a*b^3*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 96*C*a*b^3*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 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 + 24*A*a^2*b^2 + 48*C*a^2*b^2 + 8*A*
b^4)*(d*x + c) - 2*(15*A*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 - 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 - 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 + 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 - 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 - 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 - 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 - 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 - 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) - 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) - 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) - 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

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maple [A]  time = 1.35, size = 296, normalized size = 1.20 \[ \frac {A \,a^{4} \sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )}{4 d}+\frac {3 A \,a^{4} \cos \left (d x +c \right ) \sin \left (d x +c \right )}{8 d}+\frac {3 A \,a^{4} x}{8}+\frac {3 A \,a^{4} c}{8 d}+\frac {a^{4} C \sin \left (d x +c \right ) \cos \left (d x +c \right )}{2 d}+\frac {a^{4} C x}{2}+\frac {C \,a^{4} c}{2 d}+\frac {4 A \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right ) a^{3} b}{3 d}+\frac {8 A \,a^{3} b \sin \left (d x +c \right )}{3 d}+\frac {4 a^{3} b C \sin \left (d x +c \right )}{d}+\frac {3 A \,a^{2} b^{2} \sin \left (d x +c \right ) \cos \left (d x +c \right )}{d}+3 A x \,a^{2} b^{2}+\frac {3 A \,a^{2} b^{2} c}{d}+6 C \,a^{2} b^{2} x +\frac {6 C \,a^{2} b^{2} c}{d}+\frac {4 a A \,b^{3} \sin \left (d x +c \right )}{d}+\frac {4 C a \,b^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+A x \,b^{4}+\frac {A \,b^{4} c}{d}+\frac {C \,b^{4} \tan \left (d x +c \right )}{d} \]

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+C*sec(d*x+c)^2),x)

[Out]

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

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maxima [A]  time = 0.36, size = 204, normalized size = 0.83 \[ \frac {3 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{4} + 24 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{4} - 128 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a^{3} b + 144 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{2} b^{2} + 576 \, {\left (d x + c\right )} C a^{2} b^{2} + 96 \, {\left (d x + c\right )} A b^{4} + 192 \, C a b^{3} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 384 \, C a^{3} b \sin \left (d x + c\right ) + 384 \, A a b^{3} \sin \left (d x + c\right ) + 96 \, C b^{4} \tan \left (d x + c\right )}{96 \, d} \]

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+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 + 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 + 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 + 96*(d*x + c)*A*b^4 + 192*C*a*b^3*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 384*C*a^
3*b*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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mupad [B]  time = 4.39, size = 395, normalized size = 1.61 \[ \frac {A\,a^4\,{\cos \left (c+d\,x\right )}^3\,\sin \left (c+d\,x\right )}{4\,d}+\frac {C\,b^4\,\sin \left (c+d\,x\right )}{d\,\cos \left (c+d\,x\right )}+\frac {3\,A\,a^4\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )}{8\,d}+\frac {4\,A\,a\,b^3\,\sin \left (c+d\,x\right )}{d}+\frac {8\,A\,a^3\,b\,\sin \left (c+d\,x\right )}{3\,d}+\frac {C\,a^4\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )}{2\,d}+\frac {4\,C\,a^3\,b\,\sin \left (c+d\,x\right )}{d}+\frac {3\,A\,a^2\,b^2\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )}{d}+\frac {4\,A\,a^3\,b\,{\cos \left (c+d\,x\right )}^2\,\sin \left (c+d\,x\right )}{3\,d}-\frac {A\,a^4\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,3{}\mathrm {i}}{4\,d}-\frac {A\,b^4\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,2{}\mathrm {i}}{d}-\frac {C\,a^4\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,1{}\mathrm {i}}{d}-\frac {C\,a\,b^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,8{}\mathrm {i}}{d}-\frac {A\,a^2\,b^2\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,6{}\mathrm {i}}{d}-\frac {C\,a^2\,b^2\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,12{}\mathrm {i}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(A*a^4*cos(c + d*x)^3*sin(c + d*x))/(4*d) - (A*b^4*atanh((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*2i)/d - (
C*a^4*atanh((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*1i)/d - (C*a*b^3*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2
+ (d*x)/2))*8i)/d - (A*a^4*atanh((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*3i)/(4*d) + (C*b^4*sin(c + d*x))/
(d*cos(c + d*x)) - (A*a^2*b^2*atanh((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*6i)/d - (C*a^2*b^2*atanh((sin(
c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*12i)/d + (3*A*a^4*cos(c + d*x)*sin(c + d*x))/(8*d) + (4*A*a*b^3*sin(c +
d*x))/d + (8*A*a^3*b*sin(c + d*x))/(3*d) + (C*a^4*cos(c + d*x)*sin(c + d*x))/(2*d) + (4*C*a^3*b*sin(c + d*x))
/d + (3*A*a^2*b^2*cos(c + d*x)*sin(c + d*x))/d + (4*A*a^3*b*cos(c + d*x)^2*sin(c + d*x))/(3*d)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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

[Out]

Timed out

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