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

Optimal. Leaf size=192 \[ \frac{b \tan (c+d x) \left (a^2 (-(6 A-8 C))+9 a b B+b^2 (3 A+2 C)\right )}{3 d}+\frac{\left (6 a^2 b B+2 a^3 C+3 a b^2 (2 A+C)+b^3 B\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+a^2 x (a B+3 A b)-\frac{b^2 \tan (c+d x) \sec (c+d x) (6 a A-5 a C-3 b B)}{6 d}-\frac{b (3 A-C) \tan (c+d x) (a+b \sec (c+d x))^2}{3 d}+\frac{A \sin (c+d x) (a+b \sec (c+d x))^3}{d} \]

[Out]

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

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Rubi [A]  time = 0.371263, antiderivative size = 192, 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 = {4094, 4056, 4048, 3770, 3767, 8} \[ \frac{b \tan (c+d x) \left (a^2 (-(6 A-8 C))+9 a b B+b^2 (3 A+2 C)\right )}{3 d}+\frac{\left (6 a^2 b B+2 a^3 C+3 a b^2 (2 A+C)+b^3 B\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+a^2 x (a B+3 A b)-\frac{b^2 \tan (c+d x) \sec (c+d x) (6 a A-5 a C-3 b B)}{6 d}-\frac{b (3 A-C) \tan (c+d x) (a+b \sec (c+d x))^2}{3 d}+\frac{A \sin (c+d x) (a+b \sec (c+d x))^3}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

a^2*(3*A*b + a*B)*x + ((6*a^2*b*B + b^3*B + 2*a^3*C + 3*a*b^2*(2*A + C))*ArcTanh[Sin[c + d*x]])/(2*d) + (A*(a
+ b*Sec[c + d*x])^3*Sin[c + d*x])/d + (b*(9*a*b*B - a^2*(6*A - 8*C) + b^2*(3*A + 2*C))*Tan[c + d*x])/(3*d) - (
b^2*(6*a*A - 3*b*B - 5*a*C)*Sec[c + d*x]*Tan[c + d*x])/(6*d) - (b*(3*A - C)*(a + b*Sec[c + d*x])^2*Tan[c + d*x
])/(3*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 4056

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

Rule 4048

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

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

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]

Rubi steps

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

Mathematica [B]  time = 6.58619, size = 1335, normalized size = 6.95 \[ \text{result too large to display} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^2*(3*A*b + a*B)*(c + d*x)*Cos[c + d*x]^5*(a + b*Sec[c + d*x])^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/
(d*(b + a*Cos[c + d*x])^3*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])) + ((-6*a*A*b^2 - 6*a^2*b*B - b^3*
B - 2*a^3*C - 3*a*b^2*C)*Cos[c + d*x]^5*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]*(a + b*Sec[c + d*x])^3*(A + B
*Sec[c + d*x] + C*Sec[c + d*x]^2))/(d*(b + a*Cos[c + d*x])^3*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])
) + ((6*a*A*b^2 + 6*a^2*b*B + b^3*B + 2*a^3*C + 3*a*b^2*C)*Cos[c + d*x]^5*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)
/2]]*(a + b*Sec[c + d*x])^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(d*(b + a*Cos[c + d*x])^3*(A + 2*C + 2*B*
Cos[c + d*x] + A*Cos[2*c + 2*d*x])) + ((3*b^3*B + 9*a*b^2*C + b^3*C)*Cos[c + d*x]^5*(a + b*Sec[c + d*x])^3*(A
+ B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(6*d*(b + a*Cos[c + d*x])^3*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*
d*x])*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2) + (b^3*C*Cos[c + d*x]^5*(a + b*Sec[c + d*x])^3*(A + B*Sec[c + d
*x] + C*Sec[c + d*x]^2)*Sin[(c + d*x)/2])/(3*d*(b + a*Cos[c + d*x])^3*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c
+ 2*d*x])*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^3) + (b^3*C*Cos[c + d*x]^5*(a + b*Sec[c + d*x])^3*(A + B*Sec[c
+ d*x] + C*Sec[c + d*x]^2)*Sin[(c + d*x)/2])/(3*d*(b + a*Cos[c + d*x])^3*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[
2*c + 2*d*x])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3) + ((-3*b^3*B - 9*a*b^2*C - b^3*C)*Cos[c + d*x]^5*(a + b
*Sec[c + d*x])^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(6*d*(b + a*Cos[c + d*x])^3*(A + 2*C + 2*B*Cos[c + d
*x] + A*Cos[2*c + 2*d*x])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2) + (2*Cos[c + d*x]^5*(a + b*Sec[c + d*x])^3*
(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*(3*A*b^3*Sin[(c + d*x)/2] + 9*a*b^2*B*Sin[(c + d*x)/2] + 9*a^2*b*C*Sin
[(c + d*x)/2] + 2*b^3*C*Sin[(c + d*x)/2]))/(3*d*(b + a*Cos[c + d*x])^3*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c
+ 2*d*x])*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])) + (2*Cos[c + d*x]^5*(a + b*Sec[c + d*x])^3*(A + B*Sec[c + d*
x] + C*Sec[c + d*x]^2)*(3*A*b^3*Sin[(c + d*x)/2] + 9*a*b^2*B*Sin[(c + d*x)/2] + 9*a^2*b*C*Sin[(c + d*x)/2] + 2
*b^3*C*Sin[(c + d*x)/2]))/(3*d*(b + a*Cos[c + d*x])^3*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(Cos[(
c + d*x)/2] + Sin[(c + d*x)/2])) + (2*a^3*A*Cos[c + d*x]^5*(a + b*Sec[c + d*x])^3*(A + B*Sec[c + d*x] + C*Sec[
c + d*x]^2)*Sin[c + d*x])/(d*(b + a*Cos[c + d*x])^3*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x]))

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Maple [A]  time = 0.078, size = 294, 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{{a}^{3}C\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+3\,{a}^{2}Abx+3\,{\frac{A{a}^{2}bc}{d}}+3\,{\frac{B{a}^{2}b\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+3\,{\frac{{a}^{2}bC\tan \left ( dx+c \right ) }{d}}+3\,{\frac{Aa{b}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+3\,{\frac{Ba{b}^{2}\tan \left ( dx+c \right ) }{d}}+{\frac{3\,Ca{b}^{2}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{3\,Ca{b}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+{\frac{A{b}^{3}\tan \left ( dx+c \right ) }{d}}+{\frac{B{b}^{3}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{B{b}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+{\frac{2\,C{b}^{3}\tan \left ( dx+c \right ) }{3\,d}}+{\frac{C{b}^{3}\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+b*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+1/d*a^3*C*ln(sec(d*x+c)+tan(d*x+c))+3*a^2*A*b*x+3/d*A*a^2*b*c+3/d*B*a^2
*b*ln(sec(d*x+c)+tan(d*x+c))+3/d*a^2*b*C*tan(d*x+c)+3/d*A*a*b^2*ln(sec(d*x+c)+tan(d*x+c))+3/d*B*a*b^2*tan(d*x+
c)+3/2/d*C*a*b^2*sec(d*x+c)*tan(d*x+c)+3/2/d*C*a*b^2*ln(sec(d*x+c)+tan(d*x+c))+1/d*A*b^3*tan(d*x+c)+1/2/d*B*b^
3*sec(d*x+c)*tan(d*x+c)+1/2/d*B*b^3*ln(sec(d*x+c)+tan(d*x+c))+2/3/d*C*b^3*tan(d*x+c)+1/3/d*C*b^3*tan(d*x+c)*se
c(d*x+c)^2

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Maxima [A]  time = 1.03056, size = 378, normalized size = 1.97 \begin{align*} \frac{12 \,{\left (d x + c\right )} B a^{3} + 36 \,{\left (d x + c\right )} A a^{2} b + 4 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C b^{3} - 9 \, C a b^{2}{\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 )} - 3 \, B b^{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 )} + 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 )} + 18 \, B a^{2} b{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 18 \, A 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 \, A a^{3} \sin \left (d x + c\right ) + 36 \, C a^{2} b \tan \left (d x + c\right ) + 36 \, B a b^{2} \tan \left (d x + c\right ) + 12 \, A b^{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+b*sec(d*x+c))^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 0.578599, size = 543, normalized size = 2.83 \begin{align*} \frac{12 \,{\left (B a^{3} + 3 \, A a^{2} b\right )} d x \cos \left (d x + c\right )^{3} + 3 \,{\left (2 \, C a^{3} + 6 \, B a^{2} b + 3 \,{\left (2 \, A + C\right )} a b^{2} + B b^{3}\right )} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left (2 \, C a^{3} + 6 \, B a^{2} b + 3 \,{\left (2 \, A + C\right )} a b^{2} + B b^{3}\right )} \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 \, C b^{3} + 2 \,{\left (9 \, C a^{2} b + 9 \, B a b^{2} +{\left (3 \, A + 2 \, C\right )} b^{3}\right )} \cos \left (d x + c\right )^{2} + 3 \,{\left (3 \, C a b^{2} + B b^{3}\right )} \cos \left (d x + c\right )\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+b*sec(d*x+c))^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

1/12*(12*(B*a^3 + 3*A*a^2*b)*d*x*cos(d*x + c)^3 + 3*(2*C*a^3 + 6*B*a^2*b + 3*(2*A + C)*a*b^2 + B*b^3)*cos(d*x
+ c)^3*log(sin(d*x + c) + 1) - 3*(2*C*a^3 + 6*B*a^2*b + 3*(2*A + C)*a*b^2 + B*b^3)*cos(d*x + c)^3*log(-sin(d*x
+ c) + 1) + 2*(6*A*a^3*cos(d*x + c)^3 + 2*C*b^3 + 2*(9*C*a^2*b + 9*B*a*b^2 + (3*A + 2*C)*b^3)*cos(d*x + c)^2
+ 3*(3*C*a*b^2 + B*b^3)*cos(d*x + c))*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+b*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 [B]  time = 1.37315, size = 591, normalized size = 3.08 \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 (B a^{3} + 3 \, A a^{2} b\right )}{\left (d x + c\right )} + 3 \,{\left (2 \, C a^{3} + 6 \, B a^{2} b + 6 \, A a b^{2} + 3 \, C a b^{2} + B b^{3}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 3 \,{\left (2 \, C a^{3} + 6 \, B a^{2} b + 6 \, A a b^{2} + 3 \, C a b^{2} + B b^{3}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) - \frac{2 \,{\left (18 \, C a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 18 \, B a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 9 \, C a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 6 \, A b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 3 \, B b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 6 \, C b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 36 \, C a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 36 \, B a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 12 \, A b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 4 \, C b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 18 \, C a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 18 \, B a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 9 \, C a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 6 \, A b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 3 \, B b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 6 \, C 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 )}^{3}}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*(a+b*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*(B*a^3 + 3*A*a^2*b)*(d*x + c) + 3*(2*C*a^3
+ 6*B*a^2*b + 6*A*a*b^2 + 3*C*a*b^2 + B*b^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(2*C*a^3 + 6*B*a^2*b + 6*
A*a*b^2 + 3*C*a*b^2 + B*b^3)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(18*C*a^2*b*tan(1/2*d*x + 1/2*c)^5 + 18*B*
a*b^2*tan(1/2*d*x + 1/2*c)^5 - 9*C*a*b^2*tan(1/2*d*x + 1/2*c)^5 + 6*A*b^3*tan(1/2*d*x + 1/2*c)^5 - 3*B*b^3*tan
(1/2*d*x + 1/2*c)^5 + 6*C*b^3*tan(1/2*d*x + 1/2*c)^5 - 36*C*a^2*b*tan(1/2*d*x + 1/2*c)^3 - 36*B*a*b^2*tan(1/2*
d*x + 1/2*c)^3 - 12*A*b^3*tan(1/2*d*x + 1/2*c)^3 - 4*C*b^3*tan(1/2*d*x + 1/2*c)^3 + 18*C*a^2*b*tan(1/2*d*x + 1
/2*c) + 18*B*a*b^2*tan(1/2*d*x + 1/2*c) + 9*C*a*b^2*tan(1/2*d*x + 1/2*c) + 6*A*b^3*tan(1/2*d*x + 1/2*c) + 3*B*
b^3*tan(1/2*d*x + 1/2*c) + 6*C*b^3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^3)/d