3.892 \(\int \cos ^3(c+d x) (a+b \sec (c+d x))^4 (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=303 \[ -\frac{b \tan (c+d x) \left (4 a^3 (2 A+3 C)+39 a^2 b B+4 a b^2 (11 A-6 C)-6 b^3 B\right )}{6 d}+\frac{b^2 \left (12 a^2 C+8 a b B+2 A b^2+b^2 C\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{\sin (c+d x) \left (a^2 (4 A+6 C)+15 a b B+12 A b^2\right ) (a+b \sec (c+d x))^2}{6 d}-\frac{b^2 \tan (c+d x) \sec (c+d x) \left (a^2 (4 A+6 C)+18 a b B+3 b^2 (6 A-C)\right )}{6 d}+\frac{1}{2} a x \left (4 a^2 b (A+2 C)+a^3 B+12 a b^2 B+8 A b^3\right )+\frac{(3 a B+4 A b) \sin (c+d x) \cos (c+d x) (a+b \sec (c+d x))^3}{6 d}+\frac{A \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^4}{3 d} \]

[Out]

(a*(8*A*b^3 + a^3*B + 12*a*b^2*B + 4*a^2*b*(A + 2*C))*x)/2 + (b^2*(2*A*b^2 + 8*a*b*B + 12*a^2*C + b^2*C)*ArcTa
nh[Sin[c + d*x]])/(2*d) + ((12*A*b^2 + 15*a*b*B + a^2*(4*A + 6*C))*(a + b*Sec[c + d*x])^2*Sin[c + d*x])/(6*d)
+ ((4*A*b + 3*a*B)*Cos[c + d*x]*(a + b*Sec[c + d*x])^3*Sin[c + d*x])/(6*d) + (A*Cos[c + d*x]^2*(a + b*Sec[c +
d*x])^4*Sin[c + d*x])/(3*d) - (b*(39*a^2*b*B - 6*b^3*B + 4*a*b^2*(11*A - 6*C) + 4*a^3*(2*A + 3*C))*Tan[c + d*x
])/(6*d) - (b^2*(18*a*b*B + 3*b^2*(6*A - C) + a^2*(4*A + 6*C))*Sec[c + d*x]*Tan[c + d*x])/(6*d)

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Rubi [A]  time = 0.850054, antiderivative size = 303, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 5, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.122, Rules used = {4094, 4048, 3770, 3767, 8} \[ -\frac{b \tan (c+d x) \left (4 a^3 (2 A+3 C)+39 a^2 b B+4 a b^2 (11 A-6 C)-6 b^3 B\right )}{6 d}+\frac{b^2 \left (12 a^2 C+8 a b B+2 A b^2+b^2 C\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{\sin (c+d x) \left (a^2 (4 A+6 C)+15 a b B+12 A b^2\right ) (a+b \sec (c+d x))^2}{6 d}-\frac{b^2 \tan (c+d x) \sec (c+d x) \left (a^2 (4 A+6 C)+18 a b B+3 b^2 (6 A-C)\right )}{6 d}+\frac{1}{2} a x \left (4 a^2 b (A+2 C)+a^3 B+12 a b^2 B+8 A b^3\right )+\frac{(3 a B+4 A b) \sin (c+d x) \cos (c+d x) (a+b \sec (c+d x))^3}{6 d}+\frac{A \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^4}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^3*(a + b*Sec[c + d*x])^4*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

(a*(8*A*b^3 + a^3*B + 12*a*b^2*B + 4*a^2*b*(A + 2*C))*x)/2 + (b^2*(2*A*b^2 + 8*a*b*B + 12*a^2*C + b^2*C)*ArcTa
nh[Sin[c + d*x]])/(2*d) + ((12*A*b^2 + 15*a*b*B + a^2*(4*A + 6*C))*(a + b*Sec[c + d*x])^2*Sin[c + d*x])/(6*d)
+ ((4*A*b + 3*a*B)*Cos[c + d*x]*(a + b*Sec[c + d*x])^3*Sin[c + d*x])/(6*d) + (A*Cos[c + d*x]^2*(a + b*Sec[c +
d*x])^4*Sin[c + d*x])/(3*d) - (b*(39*a^2*b*B - 6*b^3*B + 4*a*b^2*(11*A - 6*C) + 4*a^3*(2*A + 3*C))*Tan[c + d*x
])/(6*d) - (b^2*(18*a*b*B + 3*b^2*(6*A - C) + a^2*(4*A + 6*C))*Sec[c + d*x]*Tan[c + d*x])/(6*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 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 ^3(c+d x) (a+b \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\frac{A \cos ^2(c+d x) (a+b \sec (c+d x))^4 \sin (c+d x)}{3 d}+\frac{1}{3} \int \cos ^2(c+d x) (a+b \sec (c+d x))^3 \left (4 A b+3 a B+(2 a A+3 b B+3 a C) \sec (c+d x)-b (2 A-3 C) \sec ^2(c+d x)\right ) \, dx\\ &=\frac{(4 A b+3 a B) \cos (c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{6 d}+\frac{A \cos ^2(c+d x) (a+b \sec (c+d x))^4 \sin (c+d x)}{3 d}+\frac{1}{6} \int \cos (c+d x) (a+b \sec (c+d x))^2 \left (12 A b^2+15 a b B+a^2 (4 A+6 C)+\left (3 a^2 B+6 b^2 B+4 a b (A+3 C)\right ) \sec (c+d x)-6 b (2 A b+a B-b C) \sec ^2(c+d x)\right ) \, dx\\ &=\frac{\left (12 A b^2+15 a b B+a^2 (4 A+6 C)\right ) (a+b \sec (c+d x))^2 \sin (c+d x)}{6 d}+\frac{(4 A b+3 a B) \cos (c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{6 d}+\frac{A \cos ^2(c+d x) (a+b \sec (c+d x))^4 \sin (c+d x)}{3 d}+\frac{1}{6} \int (a+b \sec (c+d x)) \left (3 \left (8 A b^3+a^3 B+12 a b^2 B+4 a^2 b (A+2 C)\right )-b \left (8 a A b+3 a^2 B-6 b^2 B-18 a b C\right ) \sec (c+d x)-2 b \left (18 a b B+3 b^2 (6 A-C)+a^2 (4 A+6 C)\right ) \sec ^2(c+d x)\right ) \, dx\\ &=\frac{\left (12 A b^2+15 a b B+a^2 (4 A+6 C)\right ) (a+b \sec (c+d x))^2 \sin (c+d x)}{6 d}+\frac{(4 A b+3 a B) \cos (c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{6 d}+\frac{A \cos ^2(c+d x) (a+b \sec (c+d x))^4 \sin (c+d x)}{3 d}-\frac{b^2 \left (18 a b B+3 b^2 (6 A-C)+a^2 (4 A+6 C)\right ) \sec (c+d x) \tan (c+d x)}{6 d}+\frac{1}{12} \int \left (6 a \left (8 A b^3+a^3 B+12 a b^2 B+4 a^2 b (A+2 C)\right )+6 b^2 \left (2 A b^2+8 a b B+12 a^2 C+b^2 C\right ) \sec (c+d x)-2 b \left (39 a^2 b B-6 b^3 B+4 a b^2 (11 A-6 C)+4 a^3 (2 A+3 C)\right ) \sec ^2(c+d x)\right ) \, dx\\ &=\frac{1}{2} a \left (8 A b^3+a^3 B+12 a b^2 B+4 a^2 b (A+2 C)\right ) x+\frac{\left (12 A b^2+15 a b B+a^2 (4 A+6 C)\right ) (a+b \sec (c+d x))^2 \sin (c+d x)}{6 d}+\frac{(4 A b+3 a B) \cos (c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{6 d}+\frac{A \cos ^2(c+d x) (a+b \sec (c+d x))^4 \sin (c+d x)}{3 d}-\frac{b^2 \left (18 a b B+3 b^2 (6 A-C)+a^2 (4 A+6 C)\right ) \sec (c+d x) \tan (c+d x)}{6 d}+\frac{1}{2} \left (b^2 \left (2 A b^2+8 a b B+12 a^2 C+b^2 C\right )\right ) \int \sec (c+d x) \, dx-\frac{1}{6} \left (b \left (39 a^2 b B-6 b^3 B+4 a b^2 (11 A-6 C)+4 a^3 (2 A+3 C)\right )\right ) \int \sec ^2(c+d x) \, dx\\ &=\frac{1}{2} a \left (8 A b^3+a^3 B+12 a b^2 B+4 a^2 b (A+2 C)\right ) x+\frac{b^2 \left (2 A b^2+8 a b B+12 a^2 C+b^2 C\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{\left (12 A b^2+15 a b B+a^2 (4 A+6 C)\right ) (a+b \sec (c+d x))^2 \sin (c+d x)}{6 d}+\frac{(4 A b+3 a B) \cos (c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{6 d}+\frac{A \cos ^2(c+d x) (a+b \sec (c+d x))^4 \sin (c+d x)}{3 d}-\frac{b^2 \left (18 a b B+3 b^2 (6 A-C)+a^2 (4 A+6 C)\right ) \sec (c+d x) \tan (c+d x)}{6 d}+\frac{\left (b \left (39 a^2 b B-6 b^3 B+4 a b^2 (11 A-6 C)+4 a^3 (2 A+3 C)\right )\right ) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{6 d}\\ &=\frac{1}{2} a \left (8 A b^3+a^3 B+12 a b^2 B+4 a^2 b (A+2 C)\right ) x+\frac{b^2 \left (2 A b^2+8 a b B+12 a^2 C+b^2 C\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{\left (12 A b^2+15 a b B+a^2 (4 A+6 C)\right ) (a+b \sec (c+d x))^2 \sin (c+d x)}{6 d}+\frac{(4 A b+3 a B) \cos (c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{6 d}+\frac{A \cos ^2(c+d x) (a+b \sec (c+d x))^4 \sin (c+d x)}{3 d}-\frac{b \left (39 a^2 b B-6 b^3 B+4 a b^2 (11 A-6 C)+4 a^3 (2 A+3 C)\right ) \tan (c+d x)}{6 d}-\frac{b^2 \left (18 a b B+3 b^2 (6 A-C)+a^2 (4 A+6 C)\right ) \sec (c+d x) \tan (c+d x)}{6 d}\\ \end{align*}

Mathematica [A]  time = 5.2081, size = 370, normalized size = 1.22 \[ \frac{6 a (c+d x) \left (4 a^2 b (A+2 C)+a^3 B+12 a b^2 B+8 A b^3\right )+3 a^2 \sin (c+d x) \left (a^2 (3 A+4 C)+16 a b B+24 A b^2\right )-6 b^2 \left (12 a^2 C+8 a b B+2 A b^2+b^2 C\right ) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+6 b^2 \left (12 a^2 C+8 a b B+2 A b^2+b^2 C\right ) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+3 a^3 (a B+4 A b) \sin (2 (c+d x))+a^4 A \sin (3 (c+d x))+\frac{12 b^3 (4 a C+b B) \sin \left (\frac{1}{2} (c+d x)\right )}{\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )}+\frac{12 b^3 (4 a C+b B) \sin \left (\frac{1}{2} (c+d x)\right )}{\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )}+\frac{3 b^4 C}{\left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}-\frac{3 b^4 C}{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2}}{12 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^3*(a + b*Sec[c + d*x])^4*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

(6*a*(8*A*b^3 + a^3*B + 12*a*b^2*B + 4*a^2*b*(A + 2*C))*(c + d*x) - 6*b^2*(2*A*b^2 + 8*a*b*B + 12*a^2*C + b^2*
C)*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 6*b^2*(2*A*b^2 + 8*a*b*B + 12*a^2*C + b^2*C)*Log[Cos[(c + d*x)/2
] + Sin[(c + d*x)/2]] + (3*b^4*C)/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2 + (12*b^3*(b*B + 4*a*C)*Sin[(c + d*x
)/2])/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2]) - (3*b^4*C)/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2 + (12*b^3*(b*B
+ 4*a*C)*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]) + 3*a^2*(24*A*b^2 + 16*a*b*B + a^2*(3*A + 4*
C))*Sin[c + d*x] + 3*a^3*(4*A*b + a*B)*Sin[2*(c + d*x)] + a^4*A*Sin[3*(c + d*x)])/(12*d)

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Maple [A]  time = 0.09, size = 374, normalized size = 1.2 \begin{align*}{\frac{A\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}{a}^{4}}{3\,d}}+{\frac{2\,A{a}^{4}\sin \left ( dx+c \right ) }{3\,d}}+{\frac{B{a}^{4}\sin \left ( dx+c \right ) \cos \left ( dx+c \right ) }{2\,d}}+{\frac{B{a}^{4}x}{2}}+{\frac{B{a}^{4}c}{2\,d}}+{\frac{{a}^{4}C\sin \left ( dx+c \right ) }{d}}+2\,{\frac{A{a}^{3}b\sin \left ( dx+c \right ) \cos \left ( dx+c \right ) }{d}}+2\,{a}^{3}Abx+2\,{\frac{A{a}^{3}bc}{d}}+4\,{\frac{B{a}^{3}b\sin \left ( dx+c \right ) }{d}}+4\,{a}^{3}bCx+4\,{\frac{C{a}^{3}bc}{d}}+6\,{\frac{A{a}^{2}{b}^{2}\sin \left ( dx+c \right ) }{d}}+6\,{a}^{2}{b}^{2}Bx+6\,{\frac{B{a}^{2}{b}^{2}c}{d}}+6\,{\frac{C{a}^{2}{b}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+4\,Aa{b}^{3}x+4\,{\frac{Aa{b}^{3}c}{d}}+4\,{\frac{a{b}^{3}B\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+4\,{\frac{Ca{b}^{3}\tan \left ( dx+c \right ) }{d}}+{\frac{A{b}^{4}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{B{b}^{4}\tan \left ( dx+c \right ) }{d}}+{\frac{C{b}^{4}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{C{b}^{4}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^3*(a+b*sec(d*x+c))^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x)

[Out]

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

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Maxima [A]  time = 1.03411, size = 420, normalized size = 1.39 \begin{align*} -\frac{4 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a^{4} - 3 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{4} - 12 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{3} b - 48 \,{\left (d x + c\right )} C a^{3} b - 72 \,{\left (d x + c\right )} B a^{2} b^{2} - 48 \,{\left (d x + c\right )} A a b^{3} + 3 \, C b^{4}{\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 )} - 36 \, C a^{2} b^{2}{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 24 \, B 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 )} - 6 \, A b^{4}{\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^{4} \sin \left (d x + c\right ) - 48 \, B a^{3} b \sin \left (d x + c\right ) - 72 \, A a^{2} b^{2} \sin \left (d x + c\right ) - 48 \, C a b^{3} \tan \left (d x + c\right ) - 12 \, B b^{4} \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)^3*(a+b*sec(d*x+c))^4*(A+B*sec(d*x+c)+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^4 - 3*(2*d*x + 2*c + sin(2*d*x + 2*c))*B*a^4 - 12*(2*d*x + 2*c
+ sin(2*d*x + 2*c))*A*a^3*b - 48*(d*x + c)*C*a^3*b - 72*(d*x + c)*B*a^2*b^2 - 48*(d*x + c)*A*a*b^3 + 3*C*b^4*(
2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) - 36*C*a^2*b^2*(log(sin(d
*x + c) + 1) - log(sin(d*x + c) - 1)) - 24*B*a*b^3*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) - 6*A*b^4*(
log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) - 12*C*a^4*sin(d*x + c) - 48*B*a^3*b*sin(d*x + c) - 72*A*a^2*b^
2*sin(d*x + c) - 48*C*a*b^3*tan(d*x + c) - 12*B*b^4*tan(d*x + c))/d

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Fricas [A]  time = 0.617009, size = 628, normalized size = 2.07 \begin{align*} \frac{6 \,{\left (B a^{4} + 4 \,{\left (A + 2 \, C\right )} a^{3} b + 12 \, B a^{2} b^{2} + 8 \, A a b^{3}\right )} d x \cos \left (d x + c\right )^{2} + 3 \,{\left (12 \, C a^{2} b^{2} + 8 \, B a b^{3} +{\left (2 \, A + C\right )} b^{4}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left (12 \, C a^{2} b^{2} + 8 \, B a b^{3} +{\left (2 \, A + C\right )} b^{4}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (2 \, A a^{4} \cos \left (d x + c\right )^{4} + 3 \, C b^{4} + 3 \,{\left (B a^{4} + 4 \, A a^{3} b\right )} \cos \left (d x + c\right )^{3} + 2 \,{\left ({\left (2 \, A + 3 \, C\right )} a^{4} + 12 \, B a^{3} b + 18 \, A a^{2} b^{2}\right )} \cos \left (d x + c\right )^{2} + 6 \,{\left (4 \, C a b^{3} + B b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{12 \, d \cos \left (d x + c\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(a+b*sec(d*x+c))^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

1/12*(6*(B*a^4 + 4*(A + 2*C)*a^3*b + 12*B*a^2*b^2 + 8*A*a*b^3)*d*x*cos(d*x + c)^2 + 3*(12*C*a^2*b^2 + 8*B*a*b^
3 + (2*A + C)*b^4)*cos(d*x + c)^2*log(sin(d*x + c) + 1) - 3*(12*C*a^2*b^2 + 8*B*a*b^3 + (2*A + C)*b^4)*cos(d*x
+ c)^2*log(-sin(d*x + c) + 1) + 2*(2*A*a^4*cos(d*x + c)^4 + 3*C*b^4 + 3*(B*a^4 + 4*A*a^3*b)*cos(d*x + c)^3 +
2*((2*A + 3*C)*a^4 + 12*B*a^3*b + 18*A*a^2*b^2)*cos(d*x + c)^2 + 6*(4*C*a*b^3 + B*b^4)*cos(d*x + c))*sin(d*x +
c))/(d*cos(d*x + c)^2)

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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)**3*(a+b*sec(d*x+c))**4*(A+B*sec(d*x+c)+C*sec(d*x+c)**2),x)

[Out]

Timed out

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Giac [A]  time = 1.39076, size = 733, normalized size = 2.42 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(a+b*sec(d*x+c))^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

1/6*(3*(B*a^4 + 4*A*a^3*b + 8*C*a^3*b + 12*B*a^2*b^2 + 8*A*a*b^3)*(d*x + c) + 3*(12*C*a^2*b^2 + 8*B*a*b^3 + 2*
A*b^4 + C*b^4)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(12*C*a^2*b^2 + 8*B*a*b^3 + 2*A*b^4 + C*b^4)*log(abs(tan
(1/2*d*x + 1/2*c) - 1)) - 6*(8*C*a*b^3*tan(1/2*d*x + 1/2*c)^3 + 2*B*b^4*tan(1/2*d*x + 1/2*c)^3 - C*b^4*tan(1/2
*d*x + 1/2*c)^3 - 8*C*a*b^3*tan(1/2*d*x + 1/2*c) - 2*B*b^4*tan(1/2*d*x + 1/2*c) - C*b^4*tan(1/2*d*x + 1/2*c))/
(tan(1/2*d*x + 1/2*c)^2 - 1)^2 + 2*(6*A*a^4*tan(1/2*d*x + 1/2*c)^5 - 3*B*a^4*tan(1/2*d*x + 1/2*c)^5 + 6*C*a^4*
tan(1/2*d*x + 1/2*c)^5 - 12*A*a^3*b*tan(1/2*d*x + 1/2*c)^5 + 24*B*a^3*b*tan(1/2*d*x + 1/2*c)^5 + 36*A*a^2*b^2*
tan(1/2*d*x + 1/2*c)^5 + 4*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 + 48*B*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 + 6*A*a^4*tan(1/2*d*x + 1/2*c) + 3*B*a^4*tan(1/2*d*x + 1
/2*c) + 6*C*a^4*tan(1/2*d*x + 1/2*c) + 12*A*a^3*b*tan(1/2*d*x + 1/2*c) + 24*B*a^3*b*tan(1/2*d*x + 1/2*c) + 36*
A*a^2*b^2*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^3)/d