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

Optimal. Leaf size=274 \[ -\frac{b \tan (c+d x) \left (a^2 b (39 A-34 C)+12 a^3 B-24 a b^2 B-2 b^3 (3 A+2 C)\right )}{6 d}+\frac{b \left (12 a^2 b B+8 a^3 C+4 a b^2 (2 A+C)+b^3 B\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac{b^2 \tan (c+d x) \sec (c+d x) \left (6 a^2 B+2 a b (9 A-4 C)-3 b^2 B\right )}{6 d}+\frac{1}{2} a^2 x \left (a^2 (A+2 C)+8 a b B+12 A b^2\right )-\frac{b \tan (c+d x) (6 a B+15 A b-2 b C) (a+b \sec (c+d x))^2}{6 d}+\frac{(a B+2 A b) \sin (c+d x) (a+b \sec (c+d x))^3}{d}+\frac{A \sin (c+d x) \cos (c+d x) (a+b \sec (c+d x))^4}{2 d} \]

[Out]

(a^2*(12*A*b^2 + 8*a*b*B + a^2*(A + 2*C))*x)/2 + (b*(12*a^2*b*B + b^3*B + 8*a^3*C + 4*a*b^2*(2*A + C))*ArcTanh
[Sin[c + d*x]])/(2*d) + ((2*A*b + a*B)*(a + b*Sec[c + d*x])^3*Sin[c + d*x])/d + (A*Cos[c + d*x]*(a + b*Sec[c +
d*x])^4*Sin[c + d*x])/(2*d) - (b*(12*a^3*B - 24*a*b^2*B + a^2*b*(39*A - 34*C) - 2*b^3*(3*A + 2*C))*Tan[c + d*
x])/(6*d) - (b^2*(6*a^2*B - 3*b^2*B + 2*a*b*(9*A - 4*C))*Sec[c + d*x]*Tan[c + d*x])/(6*d) - (b*(15*A*b + 6*a*B
- 2*b*C)*(a + b*Sec[c + d*x])^2*Tan[c + d*x])/(6*d)

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Rubi [A]  time = 0.677395, antiderivative size = 274, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.146, Rules used = {4094, 4056, 4048, 3770, 3767, 8} \[ -\frac{b \tan (c+d x) \left (a^2 b (39 A-34 C)+12 a^3 B-24 a b^2 B-2 b^3 (3 A+2 C)\right )}{6 d}+\frac{b \left (12 a^2 b B+8 a^3 C+4 a b^2 (2 A+C)+b^3 B\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac{b^2 \tan (c+d x) \sec (c+d x) \left (6 a^2 B+2 a b (9 A-4 C)-3 b^2 B\right )}{6 d}+\frac{1}{2} a^2 x \left (a^2 (A+2 C)+8 a b B+12 A b^2\right )-\frac{b \tan (c+d x) (6 a B+15 A b-2 b C) (a+b \sec (c+d x))^2}{6 d}+\frac{(a B+2 A b) \sin (c+d x) (a+b \sec (c+d x))^3}{d}+\frac{A \sin (c+d x) \cos (c+d x) (a+b \sec (c+d x))^4}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(12*A*b^2 + 8*a*b*B + a^2*(A + 2*C))*x)/2 + (b*(12*a^2*b*B + b^3*B + 8*a^3*C + 4*a*b^2*(2*A + C))*ArcTanh
[Sin[c + d*x]])/(2*d) + ((2*A*b + a*B)*(a + b*Sec[c + d*x])^3*Sin[c + d*x])/d + (A*Cos[c + d*x]*(a + b*Sec[c +
d*x])^4*Sin[c + d*x])/(2*d) - (b*(12*a^3*B - 24*a*b^2*B + a^2*b*(39*A - 34*C) - 2*b^3*(3*A + 2*C))*Tan[c + d*
x])/(6*d) - (b^2*(6*a^2*B - 3*b^2*B + 2*a*b*(9*A - 4*C))*Sec[c + d*x]*Tan[c + d*x])/(6*d) - (b*(15*A*b + 6*a*B
- 2*b*C)*(a + b*Sec[c + d*x])^2*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 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 ^2(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 (c+d x) (a+b \sec (c+d x))^4 \sin (c+d x)}{2 d}+\frac{1}{2} \int \cos (c+d x) (a+b \sec (c+d x))^3 \left (2 (2 A b+a B)+(2 b B+a (A+2 C)) \sec (c+d x)-b (3 A-2 C) \sec ^2(c+d x)\right ) \, dx\\ &=\frac{(2 A b+a B) (a+b \sec (c+d x))^3 \sin (c+d x)}{d}+\frac{A \cos (c+d x) (a+b \sec (c+d x))^4 \sin (c+d x)}{2 d}+\frac{1}{2} \int (a+b \sec (c+d x))^2 \left (12 A b^2+8 a b B+a^2 (A+2 C)-2 b (a A-b B-2 a C) \sec (c+d x)-b (15 A b+6 a B-2 b C) \sec ^2(c+d x)\right ) \, dx\\ &=\frac{(2 A b+a B) (a+b \sec (c+d x))^3 \sin (c+d x)}{d}+\frac{A \cos (c+d x) (a+b \sec (c+d x))^4 \sin (c+d x)}{2 d}-\frac{b (15 A b+6 a B-2 b C) (a+b \sec (c+d x))^2 \tan (c+d x)}{6 d}+\frac{1}{6} \int (a+b \sec (c+d x)) \left (3 a \left (12 A b^2+8 a b B+a^2 (A+2 C)\right )+b \left (18 a b B-3 a^2 (A-6 C)+2 b^2 (3 A+2 C)\right ) \sec (c+d x)-2 b \left (18 a A b+6 a^2 B-3 b^2 B-8 a b C\right ) \sec ^2(c+d x)\right ) \, dx\\ &=\frac{(2 A b+a B) (a+b \sec (c+d x))^3 \sin (c+d x)}{d}+\frac{A \cos (c+d x) (a+b \sec (c+d x))^4 \sin (c+d x)}{2 d}-\frac{b^2 \left (6 a^2 B-3 b^2 B+2 a b (9 A-4 C)\right ) \sec (c+d x) \tan (c+d x)}{6 d}-\frac{b (15 A b+6 a B-2 b C) (a+b \sec (c+d x))^2 \tan (c+d x)}{6 d}+\frac{1}{12} \int \left (6 a^2 \left (12 A b^2+8 a b B+a^2 (A+2 C)\right )+6 b \left (12 a^2 b B+b^3 B+8 a^3 C+4 a b^2 (2 A+C)\right ) \sec (c+d x)-2 b \left (12 a^3 B-24 a b^2 B-2 b^3 (3 A+2 C)+a^2 (39 A b-34 b C)\right ) \sec ^2(c+d x)\right ) \, dx\\ &=\frac{1}{2} a^2 \left (12 A b^2+8 a b B+a^2 (A+2 C)\right ) x+\frac{(2 A b+a B) (a+b \sec (c+d x))^3 \sin (c+d x)}{d}+\frac{A \cos (c+d x) (a+b \sec (c+d x))^4 \sin (c+d x)}{2 d}-\frac{b^2 \left (6 a^2 B-3 b^2 B+2 a b (9 A-4 C)\right ) \sec (c+d x) \tan (c+d x)}{6 d}-\frac{b (15 A b+6 a B-2 b C) (a+b \sec (c+d x))^2 \tan (c+d x)}{6 d}+\frac{1}{2} \left (b \left (12 a^2 b B+b^3 B+8 a^3 C+4 a b^2 (2 A+C)\right )\right ) \int \sec (c+d x) \, dx-\frac{1}{6} \left (b \left (12 a^3 B-24 a b^2 B+a^2 b (39 A-34 C)-2 b^3 (3 A+2 C)\right )\right ) \int \sec ^2(c+d x) \, dx\\ &=\frac{1}{2} a^2 \left (12 A b^2+8 a b B+a^2 (A+2 C)\right ) x+\frac{b \left (12 a^2 b B+b^3 B+8 a^3 C+4 a b^2 (2 A+C)\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{(2 A b+a B) (a+b \sec (c+d x))^3 \sin (c+d x)}{d}+\frac{A \cos (c+d x) (a+b \sec (c+d x))^4 \sin (c+d x)}{2 d}-\frac{b^2 \left (6 a^2 B-3 b^2 B+2 a b (9 A-4 C)\right ) \sec (c+d x) \tan (c+d x)}{6 d}-\frac{b (15 A b+6 a B-2 b C) (a+b \sec (c+d x))^2 \tan (c+d x)}{6 d}+\frac{\left (b \left (12 a^3 B-24 a b^2 B+a^2 b (39 A-34 C)-2 b^3 (3 A+2 C)\right )\right ) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{6 d}\\ &=\frac{1}{2} a^2 \left (12 A b^2+8 a b B+a^2 (A+2 C)\right ) x+\frac{b \left (12 a^2 b B+b^3 B+8 a^3 C+4 a b^2 (2 A+C)\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{(2 A b+a B) (a+b \sec (c+d x))^3 \sin (c+d x)}{d}+\frac{A \cos (c+d x) (a+b \sec (c+d x))^4 \sin (c+d x)}{2 d}-\frac{b \left (12 a^3 B-24 a b^2 B+a^2 b (39 A-34 C)-2 b^3 (3 A+2 C)\right ) \tan (c+d x)}{6 d}-\frac{b^2 \left (6 a^2 B-3 b^2 B+2 a b (9 A-4 C)\right ) \sec (c+d x) \tan (c+d x)}{6 d}-\frac{b (15 A b+6 a B-2 b C) (a+b \sec (c+d x))^2 \tan (c+d x)}{6 d}\\ \end{align*}

Mathematica [A]  time = 2.45191, size = 348, normalized size = 1.27 \[ \frac{\sec ^3(c+d x) \left (36 a^2 (c+d x) \cos (c+d x) \left (a^2 (A+2 C)+8 a b B+12 A b^2\right )+12 a^2 (c+d x) \cos (3 (c+d x)) \left (a^2 (A+2 C)+8 a b B+12 A b^2\right )+2 \sin (c+d x) \left (12 \cos (c+d x) \left (12 a^3 A b+3 a^4 B+8 a b^3 C+2 b^4 B\right )+4 \cos (2 (c+d x)) \left (3 a^4 A+36 a^2 b^2 C+24 a b^3 B+6 A b^4+4 b^4 C\right )+48 a^3 A b \cos (3 (c+d x))+3 a^4 A \cos (4 (c+d x))+9 a^4 A+144 a^2 b^2 C+12 a^4 B \cos (3 (c+d x))+96 a b^3 B+24 A b^4+32 b^4 C\right )-48 b \cos ^3(c+d x) \left (12 a^2 b B+8 a^3 C+4 a b^2 (2 A+C)+b^3 B\right ) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )\right )}{96 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[c + d*x]^3*(36*a^2*(12*A*b^2 + 8*a*b*B + a^2*(A + 2*C))*(c + d*x)*Cos[c + d*x] + 12*a^2*(12*A*b^2 + 8*a*b
*B + a^2*(A + 2*C))*(c + d*x)*Cos[3*(c + d*x)] - 48*b*(12*a^2*b*B + b^3*B + 8*a^3*C + 4*a*b^2*(2*A + C))*Cos[c
+ d*x]^3*(Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) + 2*(9*a^4*A +
24*A*b^4 + 96*a*b^3*B + 144*a^2*b^2*C + 32*b^4*C + 12*(12*a^3*A*b + 3*a^4*B + 2*b^4*B + 8*a*b^3*C)*Cos[c + d*
x] + 4*(3*a^4*A + 6*A*b^4 + 24*a*b^3*B + 36*a^2*b^2*C + 4*b^4*C)*Cos[2*(c + d*x)] + 48*a^3*A*b*Cos[3*(c + d*x)
] + 12*a^4*B*Cos[3*(c + d*x)] + 3*a^4*A*Cos[4*(c + d*x)])*Sin[c + d*x]))/(96*d)

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Maple [A]  time = 0.091, size = 377, normalized size = 1.4 \begin{align*}{\frac{A{a}^{4}\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2\,d}}+{\frac{{a}^{4}Ax}{2}}+{\frac{A{a}^{4}c}{2\,d}}+{\frac{B{a}^{4}\sin \left ( dx+c \right ) }{d}}+{a}^{4}Cx+{\frac{C{a}^{4}c}{d}}+4\,{\frac{A{a}^{3}b\sin \left ( dx+c \right ) }{d}}+4\,B{a}^{3}bx+4\,{\frac{B{a}^{3}bc}{d}}+4\,{\frac{{a}^{3}bC\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+6\,A{a}^{2}{b}^{2}x+6\,{\frac{A{a}^{2}{b}^{2}c}{d}}+6\,{\frac{{a}^{2}{b}^{2}B\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+6\,{\frac{C{a}^{2}{b}^{2}\tan \left ( dx+c \right ) }{d}}+4\,{\frac{Aa{b}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+4\,{\frac{a{b}^{3}B\tan \left ( dx+c \right ) }{d}}+2\,{\frac{Ca{b}^{3}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{d}}+2\,{\frac{Ca{b}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{A{b}^{4}\tan \left ( dx+c \right ) }{d}}+{\frac{B{b}^{4}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{B{b}^{4}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+{\frac{2\,C{b}^{4}\tan \left ( dx+c \right ) }{3\,d}}+{\frac{C{b}^{4}\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)^2*(a+b*sec(d*x+c))^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x)

[Out]

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

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Maxima [A]  time = 1.0535, size = 452, normalized size = 1.65 \begin{align*} \frac{3 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{4} + 12 \,{\left (d x + c\right )} C a^{4} + 48 \,{\left (d x + c\right )} B a^{3} b + 72 \,{\left (d x + c\right )} A a^{2} b^{2} + 4 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C b^{4} - 12 \, C a 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 )} - 3 \, B 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 )} + 24 \, C a^{3} b{\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^{2}{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 24 \, A 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 )} + 12 \, B a^{4} \sin \left (d x + c\right ) + 48 \, A a^{3} b \sin \left (d x + c\right ) + 72 \, C a^{2} b^{2} \tan \left (d x + c\right ) + 48 \, B a b^{3} \tan \left (d x + c\right ) + 12 \, A 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)^2*(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*(3*(2*d*x + 2*c + sin(2*d*x + 2*c))*A*a^4 + 12*(d*x + c)*C*a^4 + 48*(d*x + c)*B*a^3*b + 72*(d*x + c)*A*a^
2*b^2 + 4*(tan(d*x + c)^3 + 3*tan(d*x + c))*C*b^4 - 12*C*a*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)) - 3*B*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)) + 24*C*a^3*b*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 36*B*a^2*b^2*(log(sin(
d*x + c) + 1) - log(sin(d*x + c) - 1)) + 24*A*a*b^3*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 12*B*a^4
*sin(d*x + c) + 48*A*a^3*b*sin(d*x + c) + 72*C*a^2*b^2*tan(d*x + c) + 48*B*a*b^3*tan(d*x + c) + 12*A*b^4*tan(d
*x + c))/d

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Fricas [A]  time = 0.615278, size = 644, normalized size = 2.35 \begin{align*} \frac{6 \,{\left ({\left (A + 2 \, C\right )} a^{4} + 8 \, B a^{3} b + 12 \, A a^{2} b^{2}\right )} d x \cos \left (d x + c\right )^{3} + 3 \,{\left (8 \, C a^{3} b + 12 \, B a^{2} b^{2} + 4 \,{\left (2 \, A + C\right )} a b^{3} + B b^{4}\right )} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left (8 \, C a^{3} b + 12 \, B a^{2} b^{2} + 4 \,{\left (2 \, A + C\right )} a b^{3} + B b^{4}\right )} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (3 \, A a^{4} \cos \left (d x + c\right )^{4} + 2 \, C b^{4} + 6 \,{\left (B a^{4} + 4 \, A a^{3} b\right )} \cos \left (d x + c\right )^{3} + 2 \,{\left (18 \, C a^{2} b^{2} + 12 \, B a b^{3} +{\left (3 \, A + 2 \, C\right )} b^{4}\right )} \cos \left (d x + c\right )^{2} + 3 \,{\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 )^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*(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*((A + 2*C)*a^4 + 8*B*a^3*b + 12*A*a^2*b^2)*d*x*cos(d*x + c)^3 + 3*(8*C*a^3*b + 12*B*a^2*b^2 + 4*(2*A +
C)*a*b^3 + B*b^4)*cos(d*x + c)^3*log(sin(d*x + c) + 1) - 3*(8*C*a^3*b + 12*B*a^2*b^2 + 4*(2*A + C)*a*b^3 + B*
b^4)*cos(d*x + c)^3*log(-sin(d*x + c) + 1) + 2*(3*A*a^4*cos(d*x + c)^4 + 2*C*b^4 + 6*(B*a^4 + 4*A*a^3*b)*cos(d
*x + c)^3 + 2*(18*C*a^2*b^2 + 12*B*a*b^3 + (3*A + 2*C)*b^4)*cos(d*x + c)^2 + 3*(4*C*a*b^3 + B*b^4)*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)**2*(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 [B]  time = 1.40329, size = 743, normalized size = 2.71 \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)^2*(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*(A*a^4 + 2*C*a^4 + 8*B*a^3*b + 12*A*a^2*b^2)*(d*x + c) + 3*(8*C*a^3*b + 12*B*a^2*b^2 + 8*A*a*b^3 + 4*C*
a*b^3 + B*b^4)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(8*C*a^3*b + 12*B*a^2*b^2 + 8*A*a*b^3 + 4*C*a*b^3 + B*b^
4)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 6*(A*a^4*tan(1/2*d*x + 1/2*c)^3 - 2*B*a^4*tan(1/2*d*x + 1/2*c)^3 - 8*A
*a^3*b*tan(1/2*d*x + 1/2*c)^3 - A*a^4*tan(1/2*d*x + 1/2*c) - 2*B*a^4*tan(1/2*d*x + 1/2*c) - 8*A*a^3*b*tan(1/2*
d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^2 - 2*(36*C*a^2*b^2*tan(1/2*d*x + 1/2*c)^5 + 24*B*a*b^3*tan(1/2*d*x
+ 1/2*c)^5 - 12*C*a*b^3*tan(1/2*d*x + 1/2*c)^5 + 6*A*b^4*tan(1/2*d*x + 1/2*c)^5 - 3*B*b^4*tan(1/2*d*x + 1/2*c
)^5 + 6*C*b^4*tan(1/2*d*x + 1/2*c)^5 - 72*C*a^2*b^2*tan(1/2*d*x + 1/2*c)^3 - 48*B*a*b^3*tan(1/2*d*x + 1/2*c)^3
- 12*A*b^4*tan(1/2*d*x + 1/2*c)^3 - 4*C*b^4*tan(1/2*d*x + 1/2*c)^3 + 36*C*a^2*b^2*tan(1/2*d*x + 1/2*c) + 24*B
*a*b^3*tan(1/2*d*x + 1/2*c) + 12*C*a*b^3*tan(1/2*d*x + 1/2*c) + 6*A*b^4*tan(1/2*d*x + 1/2*c) + 3*B*b^4*tan(1/2
*d*x + 1/2*c) + 6*C*b^4*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^3)/d