3.299 \(\int \cos ^3(c+d x) (a+b \sec (c+d x))^3 (A+B \sec (c+d x)) \, dx\)

Optimal. Leaf size=145 \[ \frac {a \left (2 a^2 A+9 a b B+8 A b^2\right ) \sin (c+d x)}{3 d}+\frac {a^2 (3 a B+5 A b) \sin (c+d x) \cos (c+d x)}{6 d}+\frac {1}{2} x \left (a^3 B+3 a^2 A b+6 a b^2 B+2 A b^3\right )+\frac {a A \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^2}{3 d}+\frac {b^3 B \tanh ^{-1}(\sin (c+d x))}{d} \]

[Out]

1/2*(3*A*a^2*b+2*A*b^3+B*a^3+6*B*a*b^2)*x+b^3*B*arctanh(sin(d*x+c))/d+1/3*a*(2*A*a^2+8*A*b^2+9*B*a*b)*sin(d*x+
c)/d+1/6*a^2*(5*A*b+3*B*a)*cos(d*x+c)*sin(d*x+c)/d+1/3*a*A*cos(d*x+c)^2*(a+b*sec(d*x+c))^2*sin(d*x+c)/d

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Rubi [A]  time = 0.35, antiderivative size = 145, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {4025, 4074, 4047, 8, 4045, 3770} \[ \frac {a \left (2 a^2 A+9 a b B+8 A b^2\right ) \sin (c+d x)}{3 d}+\frac {1}{2} x \left (3 a^2 A b+a^3 B+6 a b^2 B+2 A b^3\right )+\frac {a^2 (3 a B+5 A b) \sin (c+d x) \cos (c+d x)}{6 d}+\frac {a A \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^2}{3 d}+\frac {b^3 B \tanh ^{-1}(\sin (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^3*(a + b*Sec[c + d*x])^3*(A + B*Sec[c + d*x]),x]

[Out]

((3*a^2*A*b + 2*A*b^3 + a^3*B + 6*a*b^2*B)*x)/2 + (b^3*B*ArcTanh[Sin[c + d*x]])/d + (a*(2*a^2*A + 8*A*b^2 + 9*
a*b*B)*Sin[c + d*x])/(3*d) + (a^2*(5*A*b + 3*a*B)*Cos[c + d*x]*Sin[c + d*x])/(6*d) + (a*A*Cos[c + d*x]^2*(a +
b*Sec[c + d*x])^2*Sin[c + d*x])/(3*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 4025

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

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 4074

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[(A*a*Cot[e + f*x]*(d*Csc[e + f*x])^n)/(f*n), x]
+ Dist[1/(d*n), Int[(d*Csc[e + f*x])^(n + 1)*Simp[n*(B*a + A*b) + (n*(a*C + B*b) + A*a*(n + 1))*Csc[e + f*x]
+ b*C*n*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C}, x] && LtQ[n, -1]

Rubi steps

\begin {align*} \int \cos ^3(c+d x) (a+b \sec (c+d x))^3 (A+B \sec (c+d x)) \, dx &=\frac {a A \cos ^2(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{3 d}-\frac {1}{3} \int \cos ^2(c+d x) (a+b \sec (c+d x)) \left (-a (5 A b+3 a B)-\left (2 a^2 A+3 A b^2+6 a b B\right ) \sec (c+d x)-3 b^2 B \sec ^2(c+d x)\right ) \, dx\\ &=\frac {a^2 (5 A b+3 a B) \cos (c+d x) \sin (c+d x)}{6 d}+\frac {a A \cos ^2(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{3 d}+\frac {1}{6} \int \cos (c+d x) \left (2 a \left (2 a^2 A+8 A b^2+9 a b B\right )+3 \left (3 a^2 A b+2 A b^3+a^3 B+6 a b^2 B\right ) \sec (c+d x)+6 b^3 B \sec ^2(c+d x)\right ) \, dx\\ &=\frac {a^2 (5 A b+3 a B) \cos (c+d x) \sin (c+d x)}{6 d}+\frac {a A \cos ^2(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{3 d}+\frac {1}{6} \int \cos (c+d x) \left (2 a \left (2 a^2 A+8 A b^2+9 a b B\right )+6 b^3 B \sec ^2(c+d x)\right ) \, dx+\frac {1}{2} \left (3 a^2 A b+2 A b^3+a^3 B+6 a b^2 B\right ) \int 1 \, dx\\ &=\frac {1}{2} \left (3 a^2 A b+2 A b^3+a^3 B+6 a b^2 B\right ) x+\frac {a \left (2 a^2 A+8 A b^2+9 a b B\right ) \sin (c+d x)}{3 d}+\frac {a^2 (5 A b+3 a B) \cos (c+d x) \sin (c+d x)}{6 d}+\frac {a A \cos ^2(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{3 d}+\left (b^3 B\right ) \int \sec (c+d x) \, dx\\ &=\frac {1}{2} \left (3 a^2 A b+2 A b^3+a^3 B+6 a b^2 B\right ) x+\frac {b^3 B \tanh ^{-1}(\sin (c+d x))}{d}+\frac {a \left (2 a^2 A+8 A b^2+9 a b B\right ) \sin (c+d x)}{3 d}+\frac {a^2 (5 A b+3 a B) \cos (c+d x) \sin (c+d x)}{6 d}+\frac {a A \cos ^2(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{3 d}\\ \end {align*}

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Mathematica [A]  time = 0.38, size = 159, normalized size = 1.10 \[ \frac {a^3 A \sin (3 (c+d x))+9 a \left (a^2 A+4 a b B+4 A b^2\right ) \sin (c+d x)+3 a^2 (a B+3 A b) \sin (2 (c+d x))+6 (c+d x) \left (a^3 B+3 a^2 A b+6 a b^2 B+2 A b^3\right )-12 b^3 B \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+12 b^3 B \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )}{12 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^3*(a + b*Sec[c + d*x])^3*(A + B*Sec[c + d*x]),x]

[Out]

(6*(3*a^2*A*b + 2*A*b^3 + a^3*B + 6*a*b^2*B)*(c + d*x) - 12*b^3*B*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 1
2*b^3*B*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + 9*a*(a^2*A + 4*A*b^2 + 4*a*b*B)*Sin[c + d*x] + 3*a^2*(3*A*b
+ a*B)*Sin[2*(c + d*x)] + a^3*A*Sin[3*(c + d*x)])/(12*d)

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fricas [A]  time = 0.51, size = 131, normalized size = 0.90 \[ \frac {3 \, B b^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, B b^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 3 \, {\left (B a^{3} + 3 \, A a^{2} b + 6 \, B a b^{2} + 2 \, A b^{3}\right )} d x + {\left (2 \, A a^{3} \cos \left (d x + c\right )^{2} + 4 \, A a^{3} + 18 \, B a^{2} b + 18 \, A a b^{2} + 3 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(a+b*sec(d*x+c))^3*(A+B*sec(d*x+c)),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.67, size = 314, normalized size = 2.17 \[ \frac {6 \, B b^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 6 \, B b^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + 3 \, {\left (B a^{3} + 3 \, A a^{2} b + 6 \, B a b^{2} + 2 \, A b^{3}\right )} {\left (d x + c\right )} + \frac {2 \, {\left (6 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 9 \, A a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 18 \, B a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 18 \, A a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 4 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 36 \, B a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 36 \, A a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 9 \, A a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 18 \, B a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 18 \, A a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(a+b*sec(d*x+c))^3*(A+B*sec(d*x+c)),x, algorithm="giac")

[Out]

1/6*(6*B*b^3*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 6*B*b^3*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 3*(B*a^3 + 3*A*
a^2*b + 6*B*a*b^2 + 2*A*b^3)*(d*x + c) + 2*(6*A*a^3*tan(1/2*d*x + 1/2*c)^5 - 3*B*a^3*tan(1/2*d*x + 1/2*c)^5 -
9*A*a^2*b*tan(1/2*d*x + 1/2*c)^5 + 18*B*a^2*b*tan(1/2*d*x + 1/2*c)^5 + 18*A*a*b^2*tan(1/2*d*x + 1/2*c)^5 + 4*A
*a^3*tan(1/2*d*x + 1/2*c)^3 + 36*B*a^2*b*tan(1/2*d*x + 1/2*c)^3 + 36*A*a*b^2*tan(1/2*d*x + 1/2*c)^3 + 6*A*a^3*
tan(1/2*d*x + 1/2*c) + 3*B*a^3*tan(1/2*d*x + 1/2*c) + 9*A*a^2*b*tan(1/2*d*x + 1/2*c) + 18*B*a^2*b*tan(1/2*d*x
+ 1/2*c) + 18*A*a*b^2*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^3)/d

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maple [A]  time = 2.00, size = 207, normalized size = 1.43 \[ \frac {A \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right ) a^{3}}{3 d}+\frac {2 a^{3} A \sin \left (d x +c \right )}{3 d}+\frac {a^{3} B \cos \left (d x +c \right ) \sin \left (d x +c \right )}{2 d}+\frac {a^{3} B x}{2}+\frac {a^{3} B c}{2 d}+\frac {3 A \,a^{2} b \cos \left (d x +c \right ) \sin \left (d x +c \right )}{2 d}+\frac {3 A x \,a^{2} b}{2}+\frac {3 A \,a^{2} b c}{2 d}+\frac {3 a^{2} b B \sin \left (d x +c \right )}{d}+\frac {3 A a \,b^{2} \sin \left (d x +c \right )}{d}+3 B x a \,b^{2}+\frac {3 B a \,b^{2} c}{d}+A x \,b^{3}+\frac {A \,b^{3} c}{d}+\frac {b^{3} B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^3*(a+b*sec(d*x+c))^3*(A+B*sec(d*x+c)),x)

[Out]

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

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maxima [A]  time = 0.70, size = 152, normalized size = 1.05 \[ -\frac {4 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a^{3} - 3 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{3} - 9 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{2} b - 36 \, {\left (d x + c\right )} B a b^{2} - 12 \, {\left (d x + c\right )} A b^{3} - 6 \, B b^{3} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 36 \, B a^{2} b \sin \left (d x + c\right ) - 36 \, A a b^{2} \sin \left (d x + c\right )}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(a+b*sec(d*x+c))^3*(A+B*sec(d*x+c)),x, algorithm="maxima")

[Out]

-1/12*(4*(sin(d*x + c)^3 - 3*sin(d*x + c))*A*a^3 - 3*(2*d*x + 2*c + sin(2*d*x + 2*c))*B*a^3 - 9*(2*d*x + 2*c +
sin(2*d*x + 2*c))*A*a^2*b - 36*(d*x + c)*B*a*b^2 - 12*(d*x + c)*A*b^3 - 6*B*b^3*(log(sin(d*x + c) + 1) - log(
sin(d*x + c) - 1)) - 36*B*a^2*b*sin(d*x + c) - 36*A*a*b^2*sin(d*x + c))/d

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mupad [B]  time = 3.90, size = 1924, normalized size = 13.27 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^3*(A + B/cos(c + d*x))*(a + b/cos(c + d*x))^3,x)

[Out]

(tan(c/2 + (d*x)/2)*(2*A*a^3 + B*a^3 + 6*A*a*b^2 + 3*A*a^2*b + 6*B*a^2*b) + tan(c/2 + (d*x)/2)^3*((4*A*a^3)/3
+ 12*A*a*b^2 + 12*B*a^2*b) + tan(c/2 + (d*x)/2)^5*(2*A*a^3 - B*a^3 + 6*A*a*b^2 - 3*A*a^2*b + 6*B*a^2*b))/(d*(3
*tan(c/2 + (d*x)/2)^2 + 3*tan(c/2 + (d*x)/2)^4 + tan(c/2 + (d*x)/2)^6 + 1)) + (atan((((A*b^3*1i + (B*a^3*1i)/2
+ (A*a^2*b*3i)/2 + B*a*b^2*3i)*(32*A*b^3 + 16*B*a^3 + 32*B*b^3 + 48*A*a^2*b + 96*B*a*b^2) + tan(c/2 + (d*x)/2
)*(32*A^2*b^6 + 8*B^2*a^6 + 32*B^2*b^6 + 96*A^2*a^2*b^4 + 72*A^2*a^4*b^2 + 288*B^2*a^2*b^4 + 96*B^2*a^4*b^2 +
192*A*B*a*b^5 + 48*A*B*a^5*b + 320*A*B*a^3*b^3))*(A*b^3*1i + (B*a^3*1i)/2 + (A*a^2*b*3i)/2 + B*a*b^2*3i)*1i -
((A*b^3*1i + (B*a^3*1i)/2 + (A*a^2*b*3i)/2 + B*a*b^2*3i)*(32*A*b^3 + 16*B*a^3 + 32*B*b^3 + 48*A*a^2*b + 96*B*a
*b^2) - tan(c/2 + (d*x)/2)*(32*A^2*b^6 + 8*B^2*a^6 + 32*B^2*b^6 + 96*A^2*a^2*b^4 + 72*A^2*a^4*b^2 + 288*B^2*a^
2*b^4 + 96*B^2*a^4*b^2 + 192*A*B*a*b^5 + 48*A*B*a^5*b + 320*A*B*a^3*b^3))*(A*b^3*1i + (B*a^3*1i)/2 + (A*a^2*b*
3i)/2 + B*a*b^2*3i)*1i)/(((A*b^3*1i + (B*a^3*1i)/2 + (A*a^2*b*3i)/2 + B*a*b^2*3i)*(32*A*b^3 + 16*B*a^3 + 32*B*
b^3 + 48*A*a^2*b + 96*B*a*b^2) + tan(c/2 + (d*x)/2)*(32*A^2*b^6 + 8*B^2*a^6 + 32*B^2*b^6 + 96*A^2*a^2*b^4 + 72
*A^2*a^4*b^2 + 288*B^2*a^2*b^4 + 96*B^2*a^4*b^2 + 192*A*B*a*b^5 + 48*A*B*a^5*b + 320*A*B*a^3*b^3))*(A*b^3*1i +
(B*a^3*1i)/2 + (A*a^2*b*3i)/2 + B*a*b^2*3i) + ((A*b^3*1i + (B*a^3*1i)/2 + (A*a^2*b*3i)/2 + B*a*b^2*3i)*(32*A*
b^3 + 16*B*a^3 + 32*B*b^3 + 48*A*a^2*b + 96*B*a*b^2) - tan(c/2 + (d*x)/2)*(32*A^2*b^6 + 8*B^2*a^6 + 32*B^2*b^6
+ 96*A^2*a^2*b^4 + 72*A^2*a^4*b^2 + 288*B^2*a^2*b^4 + 96*B^2*a^4*b^2 + 192*A*B*a*b^5 + 48*A*B*a^5*b + 320*A*B
*a^3*b^3))*(A*b^3*1i + (B*a^3*1i)/2 + (A*a^2*b*3i)/2 + B*a*b^2*3i) - 64*A*B^2*b^9 + 64*A^2*B*b^9 - 192*B^3*a*b
^8 + 576*B^3*a^2*b^7 - 32*B^3*a^3*b^6 + 192*B^3*a^4*b^5 + 16*B^3*a^6*b^3 + 384*A*B^2*a*b^8 - 96*A*B^2*a^2*b^7
+ 640*A*B^2*a^3*b^6 + 96*A*B^2*a^5*b^4 + 192*A^2*B*a^2*b^7 + 144*A^2*B*a^4*b^5))*(2*A*b^3 + B*a^3 + 3*A*a^2*b
+ 6*B*a*b^2))/d - (B*b^3*atan((B*b^3*(tan(c/2 + (d*x)/2)*(32*A^2*b^6 + 8*B^2*a^6 + 32*B^2*b^6 + 96*A^2*a^2*b^4
+ 72*A^2*a^4*b^2 + 288*B^2*a^2*b^4 + 96*B^2*a^4*b^2 + 192*A*B*a*b^5 + 48*A*B*a^5*b + 320*A*B*a^3*b^3) + B*b^3
*(32*A*b^3 + 16*B*a^3 + 32*B*b^3 + 48*A*a^2*b + 96*B*a*b^2))*1i + B*b^3*(tan(c/2 + (d*x)/2)*(32*A^2*b^6 + 8*B^
2*a^6 + 32*B^2*b^6 + 96*A^2*a^2*b^4 + 72*A^2*a^4*b^2 + 288*B^2*a^2*b^4 + 96*B^2*a^4*b^2 + 192*A*B*a*b^5 + 48*A
*B*a^5*b + 320*A*B*a^3*b^3) - B*b^3*(32*A*b^3 + 16*B*a^3 + 32*B*b^3 + 48*A*a^2*b + 96*B*a*b^2))*1i)/(64*A^2*B*
b^9 - 64*A*B^2*b^9 - 192*B^3*a*b^8 + B*b^3*(tan(c/2 + (d*x)/2)*(32*A^2*b^6 + 8*B^2*a^6 + 32*B^2*b^6 + 96*A^2*a
^2*b^4 + 72*A^2*a^4*b^2 + 288*B^2*a^2*b^4 + 96*B^2*a^4*b^2 + 192*A*B*a*b^5 + 48*A*B*a^5*b + 320*A*B*a^3*b^3) +
B*b^3*(32*A*b^3 + 16*B*a^3 + 32*B*b^3 + 48*A*a^2*b + 96*B*a*b^2)) - B*b^3*(tan(c/2 + (d*x)/2)*(32*A^2*b^6 + 8
*B^2*a^6 + 32*B^2*b^6 + 96*A^2*a^2*b^4 + 72*A^2*a^4*b^2 + 288*B^2*a^2*b^4 + 96*B^2*a^4*b^2 + 192*A*B*a*b^5 + 4
8*A*B*a^5*b + 320*A*B*a^3*b^3) - B*b^3*(32*A*b^3 + 16*B*a^3 + 32*B*b^3 + 48*A*a^2*b + 96*B*a*b^2)) + 576*B^3*a
^2*b^7 - 32*B^3*a^3*b^6 + 192*B^3*a^4*b^5 + 16*B^3*a^6*b^3 + 384*A*B^2*a*b^8 - 96*A*B^2*a^2*b^7 + 640*A*B^2*a^
3*b^6 + 96*A*B^2*a^5*b^4 + 192*A^2*B*a^2*b^7 + 144*A^2*B*a^4*b^5))*2i)/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)**3*(a+b*sec(d*x+c))**3*(A+B*sec(d*x+c)),x)

[Out]

Timed out

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