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

Optimal. Leaf size=156 \[ \frac{a^3 (6 A+7 B+5 C) \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac{(6 A-3 B-5 C) \tan (c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{6 d}+a^3 x (3 A+B)-\frac{(3 A-C) \tan (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{3 a d}+\frac{5 a^3 (B+C) \tan (c+d x)}{2 d}+\frac{A \sin (c+d x) (a \sec (c+d x)+a)^3}{d} \]

[Out]

a^3*(3*A + B)*x + (a^3*(6*A + 7*B + 5*C)*ArcTanh[Sin[c + d*x]])/(2*d) + (A*(a + a*Sec[c + d*x])^3*Sin[c + d*x]
)/d + (5*a^3*(B + C)*Tan[c + d*x])/(2*d) - ((3*A - C)*(a^2 + a^2*Sec[c + d*x])^2*Tan[c + d*x])/(3*a*d) - ((6*A
- 3*B - 5*C)*(a^3 + a^3*Sec[c + d*x])*Tan[c + d*x])/(6*d)

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Rubi [A]  time = 0.284327, antiderivative size = 156, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 39, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {4086, 3917, 3914, 3767, 8, 3770} \[ \frac{a^3 (6 A+7 B+5 C) \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac{(6 A-3 B-5 C) \tan (c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{6 d}+a^3 x (3 A+B)-\frac{(3 A-C) \tan (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{3 a d}+\frac{5 a^3 (B+C) \tan (c+d x)}{2 d}+\frac{A \sin (c+d x) (a \sec (c+d x)+a)^3}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

a^3*(3*A + B)*x + (a^3*(6*A + 7*B + 5*C)*ArcTanh[Sin[c + d*x]])/(2*d) + (A*(a + a*Sec[c + d*x])^3*Sin[c + d*x]
)/d + (5*a^3*(B + C)*Tan[c + d*x])/(2*d) - ((3*A - C)*(a^2 + a^2*Sec[c + d*x])^2*Tan[c + d*x])/(3*a*d) - ((6*A
- 3*B - 5*C)*(a^3 + a^3*Sec[c + d*x])*Tan[c + d*x])/(6*d)

Rule 4086

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/(b*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[a*A*m -
b*B*n - b*(A*(m + n + 1) + C*n)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, m}, x] && EqQ[a^2 -
b^2, 0] &&  !LtQ[m, -2^(-1)] && (LtQ[n, -2^(-1)] || EqQ[m + n + 1, 0])

Rule 3917

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_)), x_Symbol] :> -Simp[(b*
d*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 1))/(f*m), x] + Dist[1/m, Int[(a + b*Csc[e + f*x])^(m - 1)*Simp[a*c*m
+ (b*c*m + a*d*(2*m - 1))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && Gt
Q[m, 1] && EqQ[a^2 - b^2, 0] && IntegerQ[2*m]

Rule 3914

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_)), x_Symbol] :> Simp[a*c*x, x]
+ (Dist[b*d, Int[Csc[e + f*x]^2, x], x] + Dist[b*c + a*d, Int[Csc[e + f*x], x], x]) /; FreeQ[{a, b, c, d, e,
f}, x] && NeQ[b*c - a*d, 0] && NeQ[b*c + a*d, 0]

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

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]

Rubi steps

\begin{align*} \int \cos (c+d x) (a+a \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\frac{A (a+a \sec (c+d x))^3 \sin (c+d x)}{d}+\frac{\int (a+a \sec (c+d x))^3 (a (3 A+B)-a (3 A-C) \sec (c+d x)) \, dx}{a}\\ &=\frac{A (a+a \sec (c+d x))^3 \sin (c+d x)}{d}-\frac{(3 A-C) \left (a^2+a^2 \sec (c+d x)\right )^2 \tan (c+d x)}{3 a d}+\frac{\int (a+a \sec (c+d x))^2 \left (3 a^2 (3 A+B)-a^2 (6 A-3 B-5 C) \sec (c+d x)\right ) \, dx}{3 a}\\ &=\frac{A (a+a \sec (c+d x))^3 \sin (c+d x)}{d}-\frac{(3 A-C) \left (a^2+a^2 \sec (c+d x)\right )^2 \tan (c+d x)}{3 a d}-\frac{(6 A-3 B-5 C) \left (a^3+a^3 \sec (c+d x)\right ) \tan (c+d x)}{6 d}+\frac{\int (a+a \sec (c+d x)) \left (6 a^3 (3 A+B)+15 a^3 (B+C) \sec (c+d x)\right ) \, dx}{6 a}\\ &=a^3 (3 A+B) x+\frac{A (a+a \sec (c+d x))^3 \sin (c+d x)}{d}-\frac{(3 A-C) \left (a^2+a^2 \sec (c+d x)\right )^2 \tan (c+d x)}{3 a d}-\frac{(6 A-3 B-5 C) \left (a^3+a^3 \sec (c+d x)\right ) \tan (c+d x)}{6 d}+\frac{1}{2} \left (5 a^3 (B+C)\right ) \int \sec ^2(c+d x) \, dx+\frac{1}{2} \left (a^3 (6 A+7 B+5 C)\right ) \int \sec (c+d x) \, dx\\ &=a^3 (3 A+B) x+\frac{a^3 (6 A+7 B+5 C) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{A (a+a \sec (c+d x))^3 \sin (c+d x)}{d}-\frac{(3 A-C) \left (a^2+a^2 \sec (c+d x)\right )^2 \tan (c+d x)}{3 a d}-\frac{(6 A-3 B-5 C) \left (a^3+a^3 \sec (c+d x)\right ) \tan (c+d x)}{6 d}-\frac{\left (5 a^3 (B+C)\right ) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{2 d}\\ &=a^3 (3 A+B) x+\frac{a^3 (6 A+7 B+5 C) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{A (a+a \sec (c+d x))^3 \sin (c+d x)}{d}+\frac{5 a^3 (B+C) \tan (c+d x)}{2 d}-\frac{(3 A-C) \left (a^2+a^2 \sec (c+d x)\right )^2 \tan (c+d x)}{3 a d}-\frac{(6 A-3 B-5 C) \left (a^3+a^3 \sec (c+d x)\right ) \tan (c+d x)}{6 d}\\ \end{align*}

Mathematica [B]  time = 6.45498, size = 1503, normalized size = 9.63 \[ \text{result too large to display} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((3*A + B)*x*Cos[c + d*x]^5*Sec[c/2 + (d*x)/2]^6*(a + a*Sec[c + d*x])^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2
))/(4*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])) + ((-6*A - 7*B - 5*C)*Cos[c + d*x]^5*Log[Cos[c/2 + (d
*x)/2] - Sin[c/2 + (d*x)/2]]*Sec[c/2 + (d*x)/2]^6*(a + a*Sec[c + d*x])^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^
2))/(8*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])) + ((6*A + 7*B + 5*C)*Cos[c + d*x]^5*Log[Cos[c/2 +
(d*x)/2] + Sin[c/2 + (d*x)/2]]*Sec[c/2 + (d*x)/2]^6*(a + a*Sec[c + d*x])^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x
]^2))/(8*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])) + (A*Cos[d*x]*Cos[c + d*x]^5*Sec[c/2 + (d*x)/2]^
6*(a + a*Sec[c + d*x])^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*Sin[c])/(4*d*(A + 2*C + 2*B*Cos[c + d*x] + A*
Cos[2*c + 2*d*x])) + (A*Cos[c]*Cos[c + d*x]^5*Sec[c/2 + (d*x)/2]^6*(a + a*Sec[c + d*x])^3*(A + B*Sec[c + d*x]
+ C*Sec[c + d*x]^2)*Sin[d*x])/(4*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])) + (C*Cos[c + d*x]^5*Sec[
c/2 + (d*x)/2]^6*(a + a*Sec[c + d*x])^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*Sin[(d*x)/2])/(24*d*(A + 2*C +
2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(Cos[c/2] - Sin[c/2])*(Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2])^3) + (
Cos[c + d*x]^5*Sec[c/2 + (d*x)/2]^6*(a + a*Sec[c + d*x])^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*(3*B*Cos[c/
2] + 10*C*Cos[c/2] - 3*B*Sin[c/2] - 8*C*Sin[c/2]))/(48*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(Co
s[c/2] - Sin[c/2])*(Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2])^2) + (Cos[c + d*x]^5*Sec[c/2 + (d*x)/2]^6*(a + a*
Sec[c + d*x])^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*(3*A*Sin[(d*x)/2] + 9*B*Sin[(d*x)/2] + 11*C*Sin[(d*x)/
2]))/(12*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(Cos[c/2] - Sin[c/2])*(Cos[c/2 + (d*x)/2] - Sin[c
/2 + (d*x)/2])) + (C*Cos[c + d*x]^5*Sec[c/2 + (d*x)/2]^6*(a + a*Sec[c + d*x])^3*(A + B*Sec[c + d*x] + C*Sec[c
+ d*x]^2)*Sin[(d*x)/2])/(24*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(Cos[c/2] + Sin[c/2])*(Cos[c/2
+ (d*x)/2] + Sin[c/2 + (d*x)/2])^3) + (Cos[c + d*x]^5*Sec[c/2 + (d*x)/2]^6*(a + a*Sec[c + d*x])^3*(A + B*Sec[
c + d*x] + C*Sec[c + d*x]^2)*(-3*B*Cos[c/2] - 10*C*Cos[c/2] - 3*B*Sin[c/2] - 8*C*Sin[c/2]))/(48*d*(A + 2*C + 2
*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(Cos[c/2] + Sin[c/2])*(Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2])^2) + (Co
s[c + d*x]^5*Sec[c/2 + (d*x)/2]^6*(a + a*Sec[c + d*x])^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*(3*A*Sin[(d*x
)/2] + 9*B*Sin[(d*x)/2] + 11*C*Sin[(d*x)/2]))/(12*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(Cos[c/2
] + Sin[c/2])*(Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2]))

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Maple [A]  time = 0.107, size = 226, normalized size = 1.5 \begin{align*}{\frac{A{a}^{3}\sin \left ( dx+c \right ) }{d}}+{a}^{3}Bx+{\frac{B{a}^{3}c}{d}}+{\frac{5\,{a}^{3}C\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+3\,{a}^{3}Ax+3\,{\frac{A{a}^{3}c}{d}}+{\frac{7\,B{a}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+{\frac{11\,{a}^{3}C\tan \left ( dx+c \right ) }{3\,d}}+3\,{\frac{A{a}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+3\,{\frac{B{a}^{3}\tan \left ( dx+c \right ) }{d}}+{\frac{3\,{a}^{3}C\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{A{a}^{3}\tan \left ( dx+c \right ) }{d}}+{\frac{B{a}^{3}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{{a}^{3}C\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^3*A*sin(d*x+c)/d+a^3*B*x+1/d*B*a^3*c+5/2/d*a^3*C*ln(sec(d*x+c)+tan(d*x+c))+3*a^3*A*x+3/d*A*a^3*c+7/2/d*B*a^3
*ln(sec(d*x+c)+tan(d*x+c))+11/3*a^3*C*tan(d*x+c)/d+3/d*A*a^3*ln(sec(d*x+c)+tan(d*x+c))+3/d*B*a^3*tan(d*x+c)+3/
2/d*a^3*C*sec(d*x+c)*tan(d*x+c)+1/d*A*a^3*tan(d*x+c)+1/2/d*B*a^3*sec(d*x+c)*tan(d*x+c)+1/3/d*a^3*C*tan(d*x+c)*
sec(d*x+c)^2

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Maxima [A]  time = 0.965363, size = 370, normalized size = 2.37 \begin{align*} \frac{36 \,{\left (d x + c\right )} A a^{3} + 12 \,{\left (d x + c\right )} B a^{3} + 4 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a^{3} - 3 \, B a^{3}{\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 )} - 9 \, C a^{3}{\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 )} + 18 \, A a^{3}{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 18 \, B a^{3}{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 6 \, C a^{3}{\left (\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} \sin \left (d x + c\right ) + 12 \, A a^{3} \tan \left (d x + c\right ) + 36 \, B a^{3} \tan \left (d x + c\right ) + 36 \, C a^{3} \tan \left (d x + c\right )}{12 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*(a+a*sec(d*x+c))^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

1/12*(36*(d*x + c)*A*a^3 + 12*(d*x + c)*B*a^3 + 4*(tan(d*x + c)^3 + 3*tan(d*x + c))*C*a^3 - 3*B*a^3*(2*sin(d*x
+ c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) - 9*C*a^3*(2*sin(d*x + c)/(sin(d*x
+ c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) + 18*A*a^3*(log(sin(d*x + c) + 1) - log(sin(d*x
+ c) - 1)) + 18*B*a^3*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 6*C*a^3*(log(sin(d*x + c) + 1) - log(s
in(d*x + c) - 1)) + 12*A*a^3*sin(d*x + c) + 12*A*a^3*tan(d*x + c) + 36*B*a^3*tan(d*x + c) + 36*C*a^3*tan(d*x +
c))/d

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Fricas [A]  time = 0.557569, size = 425, normalized size = 2.72 \begin{align*} \frac{12 \,{\left (3 \, A + B\right )} a^{3} d x \cos \left (d x + c\right )^{3} + 3 \,{\left (6 \, A + 7 \, B + 5 \, C\right )} a^{3} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left (6 \, A + 7 \, B + 5 \, C\right )} a^{3} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (6 \, A a^{3} \cos \left (d x + c\right )^{3} + 2 \,{\left (3 \, A + 9 \, B + 11 \, C\right )} a^{3} \cos \left (d x + c\right )^{2} + 3 \,{\left (B + 3 \, C\right )} a^{3} \cos \left (d x + c\right ) + 2 \, C a^{3}\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(cos(d*x+c)*(a+a*sec(d*x+c))^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

1/12*(12*(3*A + B)*a^3*d*x*cos(d*x + c)^3 + 3*(6*A + 7*B + 5*C)*a^3*cos(d*x + c)^3*log(sin(d*x + c) + 1) - 3*(
6*A + 7*B + 5*C)*a^3*cos(d*x + c)^3*log(-sin(d*x + c) + 1) + 2*(6*A*a^3*cos(d*x + c)^3 + 2*(3*A + 9*B + 11*C)*
a^3*cos(d*x + c)^2 + 3*(B + 3*C)*a^3*cos(d*x + c) + 2*C*a^3)*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(cos(d*x+c)*(a+a*sec(d*x+c))**3*(A+B*sec(d*x+c)+C*sec(d*x+c)**2),x)

[Out]

Timed out

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Giac [A]  time = 1.3687, size = 389, normalized size = 2.49 \begin{align*} \frac{\frac{12 \, A a^{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} + 6 \,{\left (3 \, A a^{3} + B a^{3}\right )}{\left (d x + c\right )} + 3 \,{\left (6 \, A a^{3} + 7 \, B a^{3} + 5 \, C a^{3}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 3 \,{\left (6 \, A a^{3} + 7 \, B a^{3} + 5 \, C a^{3}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) - \frac{2 \,{\left (6 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 15 \, B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 15 \, C a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 12 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 36 \, B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 40 \, C a^{3} \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 ) + 21 \, B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 33 \, C a^{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 )}^{3}}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*(a+a*sec(d*x+c))^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

1/6*(12*A*a^3*tan(1/2*d*x + 1/2*c)/(tan(1/2*d*x + 1/2*c)^2 + 1) + 6*(3*A*a^3 + B*a^3)*(d*x + c) + 3*(6*A*a^3 +
7*B*a^3 + 5*C*a^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(6*A*a^3 + 7*B*a^3 + 5*C*a^3)*log(abs(tan(1/2*d*x +
1/2*c) - 1)) - 2*(6*A*a^3*tan(1/2*d*x + 1/2*c)^5 + 15*B*a^3*tan(1/2*d*x + 1/2*c)^5 + 15*C*a^3*tan(1/2*d*x + 1
/2*c)^5 - 12*A*a^3*tan(1/2*d*x + 1/2*c)^3 - 36*B*a^3*tan(1/2*d*x + 1/2*c)^3 - 40*C*a^3*tan(1/2*d*x + 1/2*c)^3
+ 6*A*a^3*tan(1/2*d*x + 1/2*c) + 21*B*a^3*tan(1/2*d*x + 1/2*c) + 33*C*a^3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x +
1/2*c)^2 - 1)^3)/d