3.659 \(\int \cos ^3(c+d x) (a+b \sec (c+d x))^3 (A+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=163 \[ \frac {a \left (a^2 (2 A+3 C)+3 A b^2\right ) \sin (c+d x)}{3 d}+\frac {1}{2} b x \left (3 a^2 (A+2 C)+2 A b^2\right )+\frac {A \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^3}{3 d}+\frac {A b \sin (c+d x) \cos (c+d x) (a+b \sec (c+d x))^2}{2 d}+\frac {3 a b^2 C \tanh ^{-1}(\sin (c+d x))}{d}-\frac {b^3 (5 A-6 C) \tan (c+d x)}{6 d} \]

[Out]

1/2*b*(2*A*b^2+3*a^2*(A+2*C))*x+3*a*b^2*C*arctanh(sin(d*x+c))/d+1/3*a*(3*A*b^2+a^2*(2*A+3*C))*sin(d*x+c)/d+1/2
*A*b*cos(d*x+c)*(a+b*sec(d*x+c))^2*sin(d*x+c)/d+1/3*A*cos(d*x+c)^2*(a+b*sec(d*x+c))^3*sin(d*x+c)/d-1/6*b^3*(5*
A-6*C)*tan(d*x+c)/d

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Rubi [A]  time = 0.52, antiderivative size = 163, normalized size of antiderivative = 1.00, number of steps used = 7, 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} \[ \frac {a \left (a^2 (2 A+3 C)+3 A b^2\right ) \sin (c+d x)}{3 d}+\frac {1}{2} b x \left (3 a^2 (A+2 C)+2 A b^2\right )+\frac {A \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^3}{3 d}+\frac {A b \sin (c+d x) \cos (c+d x) (a+b \sec (c+d x))^2}{2 d}+\frac {3 a b^2 C \tanh ^{-1}(\sin (c+d x))}{d}-\frac {b^3 (5 A-6 C) \tan (c+d x)}{6 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^3*(a + b*Sec[c + d*x])^3*(A + C*Sec[c + d*x]^2),x]

[Out]

(b*(2*A*b^2 + 3*a^2*(A + 2*C))*x)/2 + (3*a*b^2*C*ArcTanh[Sin[c + d*x]])/d + (a*(3*A*b^2 + a^2*(2*A + 3*C))*Sin
[c + d*x])/(3*d) + (A*b*Cos[c + d*x]*(a + b*Sec[c + d*x])^2*Sin[c + d*x])/(2*d) + (A*Cos[c + d*x]^2*(a + b*Sec
[c + d*x])^3*Sin[c + d*x])/(3*d) - (b^3*(5*A - 6*C)*Tan[c + d*x])/(6*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 ^3(c+d x) (a+b \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right ) \, dx &=\frac {A \cos ^2(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{3 d}+\frac {1}{3} \int \cos ^2(c+d x) (a+b \sec (c+d x))^2 \left (3 A b+a (2 A+3 C) \sec (c+d x)-b (A-3 C) \sec ^2(c+d x)\right ) \, dx\\ &=\frac {A b \cos (c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{2 d}+\frac {A \cos ^2(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{3 d}+\frac {1}{6} \int \cos (c+d x) (a+b \sec (c+d x)) \left (2 \left (3 A b^2+\frac {1}{2} a^2 (4 A+6 C)\right )+a b (5 A+12 C) \sec (c+d x)-b^2 (5 A-6 C) \sec ^2(c+d x)\right ) \, dx\\ &=\frac {A b \cos (c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{2 d}+\frac {A \cos ^2(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{3 d}-\frac {b^3 (5 A-6 C) \tan (c+d x)}{6 d}+\frac {1}{6} \int \cos (c+d x) \left (2 a \left (3 A b^2+\frac {1}{2} a^2 (4 A+6 C)\right )+3 b \left (2 A b^2+3 a^2 (A+2 C)\right ) \sec (c+d x)+18 a b^2 C \sec ^2(c+d x)\right ) \, dx\\ &=\frac {A b \cos (c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{2 d}+\frac {A \cos ^2(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{3 d}-\frac {b^3 (5 A-6 C) \tan (c+d x)}{6 d}+\frac {1}{6} \int \cos (c+d x) \left (2 a \left (3 A b^2+\frac {1}{2} a^2 (4 A+6 C)\right )+18 a b^2 C \sec ^2(c+d x)\right ) \, dx+\frac {1}{2} \left (b \left (2 A b^2+3 a^2 (A+2 C)\right )\right ) \int 1 \, dx\\ &=\frac {1}{2} b \left (2 A b^2+3 a^2 (A+2 C)\right ) x+\frac {a \left (3 A b^2+a^2 (2 A+3 C)\right ) \sin (c+d x)}{3 d}+\frac {A b \cos (c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{2 d}+\frac {A \cos ^2(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{3 d}-\frac {b^3 (5 A-6 C) \tan (c+d x)}{6 d}+\left (3 a b^2 C\right ) \int \sec (c+d x) \, dx\\ &=\frac {1}{2} b \left (2 A b^2+3 a^2 (A+2 C)\right ) x+\frac {3 a b^2 C \tanh ^{-1}(\sin (c+d x))}{d}+\frac {a \left (3 A b^2+a^2 (2 A+3 C)\right ) \sin (c+d x)}{3 d}+\frac {A b \cos (c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{2 d}+\frac {A \cos ^2(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{3 d}-\frac {b^3 (5 A-6 C) \tan (c+d x)}{6 d}\\ \end {align*}

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Mathematica [A]  time = 1.01, size = 184, normalized size = 1.13 \[ \frac {a^3 A \sin (3 (c+d x))+3 a \left (a^2 (3 A+4 C)+12 A b^2\right ) \sin (c+d x)+9 a^2 A b \sin (2 (c+d x))+18 a^2 A b c+18 a^2 A b d x+36 a^2 b c C+36 a^2 b C d x-36 a b^2 C \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+36 a b^2 C \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )+12 A b^3 c+12 A b^3 d x+12 b^3 C \tan (c+d x)}{12 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^3*(a + b*Sec[c + d*x])^3*(A + C*Sec[c + d*x]^2),x]

[Out]

(18*a^2*A*b*c + 12*A*b^3*c + 36*a^2*b*c*C + 18*a^2*A*b*d*x + 12*A*b^3*d*x + 36*a^2*b*C*d*x - 36*a*b^2*C*Log[Co
s[(c + d*x)/2] - Sin[(c + d*x)/2]] + 36*a*b^2*C*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + 3*a*(12*A*b^2 + a^2
*(3*A + 4*C))*Sin[c + d*x] + 9*a^2*A*b*Sin[2*(c + d*x)] + a^3*A*Sin[3*(c + d*x)] + 12*b^3*C*Tan[c + d*x])/(12*
d)

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fricas [A]  time = 0.50, size = 158, normalized size = 0.97 \[ \frac {9 \, C a b^{2} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) - 9 \, C a b^{2} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) + 3 \, {\left (3 \, {\left (A + 2 \, C\right )} a^{2} b + 2 \, A b^{3}\right )} d x \cos \left (d x + c\right ) + {\left (2 \, A a^{3} \cos \left (d x + c\right )^{3} + 9 \, A a^{2} b \cos \left (d x + c\right )^{2} + 6 \, C b^{3} + 2 \, {\left ({\left (2 \, A + 3 \, C\right )} a^{3} + 9 \, A a b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \, d \cos \left (d x + c\right )} \]

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

[Out]

1/6*(9*C*a*b^2*cos(d*x + c)*log(sin(d*x + c) + 1) - 9*C*a*b^2*cos(d*x + c)*log(-sin(d*x + c) + 1) + 3*(3*(A +
2*C)*a^2*b + 2*A*b^3)*d*x*cos(d*x + c) + (2*A*a^3*cos(d*x + c)^3 + 9*A*a^2*b*cos(d*x + c)^2 + 6*C*b^3 + 2*((2*
A + 3*C)*a^3 + 9*A*a*b^2)*cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c))

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giac [A]  time = 0.29, size = 306, normalized size = 1.88 \[ \frac {18 \, C a b^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 18 \, C a b^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {12 \, C b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1} + 3 \, {\left (3 \, A a^{2} b + 6 \, C a^{2} b + 2 \, A b^{3}\right )} {\left (d x + c\right )} + \frac {2 \, {\left (6 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 9 \, A a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 18 \, A a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 4 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 12 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 36 \, A a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 9 \, A a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 18 \, A a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3}}}{6 \, d} \]

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+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

1/6*(18*C*a*b^2*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 18*C*a*b^2*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 12*C*b^3*
tan(1/2*d*x + 1/2*c)/(tan(1/2*d*x + 1/2*c)^2 - 1) + 3*(3*A*a^2*b + 6*C*a^2*b + 2*A*b^3)*(d*x + c) + 2*(6*A*a^3
*tan(1/2*d*x + 1/2*c)^5 + 6*C*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*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 + 12*C*a^3*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) + 6*C*a^3*tan(1/2*d*x + 1/2*c) + 9*A*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 = 1.10, size = 183, normalized size = 1.12 \[ \frac {A \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right ) a^{3}}{3 d}+\frac {2 a^{3} A \sin \left (d x +c \right )}{3 d}+\frac {a^{3} C \sin \left (d x +c \right )}{d}+\frac {3 A \,a^{2} b \sin \left (d x +c \right ) \cos \left (d x +c \right )}{2 d}+\frac {3 a^{2} A b x}{2}+\frac {3 A \,a^{2} b c}{2 d}+3 C x \,a^{2} b +\frac {3 C \,a^{2} b c}{d}+\frac {3 A a \,b^{2} \sin \left (d x +c \right )}{d}+\frac {3 C a \,b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+A x \,b^{3}+\frac {A \,b^{3} c}{d}+\frac {b^{3} C \tan \left (d x +c \right )}{d} \]

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

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maxima [A]  time = 0.37, size = 141, normalized size = 0.87 \[ -\frac {4 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a^{3} - 9 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{2} b - 36 \, {\left (d x + c\right )} C a^{2} b - 12 \, {\left (d x + c\right )} A b^{3} - 18 \, C a b^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 12 \, C a^{3} \sin \left (d x + c\right ) - 36 \, A a b^{2} \sin \left (d x + c\right ) - 12 \, C b^{3} \tan \left (d x + c\right )}{12 \, d} \]

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

[Out]

-1/12*(4*(sin(d*x + c)^3 - 3*sin(d*x + c))*A*a^3 - 9*(2*d*x + 2*c + sin(2*d*x + 2*c))*A*a^2*b - 36*(d*x + c)*C
*a^2*b - 12*(d*x + c)*A*b^3 - 18*C*a*b^2*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) - 12*C*a^3*sin(d*x +
c) - 36*A*a*b^2*sin(d*x + c) - 12*C*b^3*tan(d*x + c))/d

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mupad [B]  time = 5.07, size = 238, normalized size = 1.46 \[ \frac {2\,A\,b^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+3\,A\,a^2\,b\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+6\,C\,a^2\,b\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )-C\,a\,b^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,6{}\mathrm {i}}{d}+\frac {\frac {5\,A\,a^3\,\sin \left (2\,c+2\,d\,x\right )}{12}+\frac {A\,a^3\,\sin \left (4\,c+4\,d\,x\right )}{24}+\frac {C\,a^3\,\sin \left (2\,c+2\,d\,x\right )}{2}+C\,b^3\,\sin \left (c+d\,x\right )+\frac {3\,A\,a^2\,b\,\sin \left (c+d\,x\right )}{8}+\frac {3\,A\,a\,b^2\,\sin \left (2\,c+2\,d\,x\right )}{2}+\frac {3\,A\,a^2\,b\,\sin \left (3\,c+3\,d\,x\right )}{8}}{d\,\cos \left (c+d\,x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^3*(A + C/cos(c + d*x)^2)*(a + b/cos(c + d*x))^3,x)

[Out]

(2*A*b^3*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)) + 3*A*a^2*b*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)) +
6*C*a^2*b*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)) - C*a*b^2*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/
2))*6i)/d + ((5*A*a^3*sin(2*c + 2*d*x))/12 + (A*a^3*sin(4*c + 4*d*x))/24 + (C*a^3*sin(2*c + 2*d*x))/2 + C*b^3*
sin(c + d*x) + (3*A*a^2*b*sin(c + d*x))/8 + (3*A*a*b^2*sin(2*c + 2*d*x))/2 + (3*A*a^2*b*sin(3*c + 3*d*x))/8)/(
d*cos(c + d*x))

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

[Out]

Timed out

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