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

Optimal. Leaf size=273 \[ \frac{b \tan (c+d x) \left (a^3 (-(12 A-19 C))+34 a^2 b B+8 a b^2 (3 A+2 C)+4 b^3 B\right )}{6 d}+\frac{\left (24 a^2 b^2 (2 A+C)+32 a^3 b B+8 a^4 C+16 a b^3 B+b^4 (4 A+3 C)\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{b^2 \tan (c+d x) \sec (c+d x) \left (a^2 (-(24 A-26 C))+32 a b B+3 b^2 (4 A+3 C)\right )}{24 d}+a^3 x (a B+4 A b)-\frac{b \tan (c+d x) (12 a A-7 a C-4 b B) (a+b \sec (c+d x))^2}{12 d}-\frac{b (4 A-C) \tan (c+d x) (a+b \sec (c+d x))^3}{4 d}+\frac{A \sin (c+d x) (a+b \sec (c+d x))^4}{d} \]

[Out]

a^3*(4*A*b + a*B)*x + ((32*a^3*b*B + 16*a*b^3*B + 8*a^4*C + 24*a^2*b^2*(2*A + C) + b^4*(4*A + 3*C))*ArcTanh[Si
n[c + d*x]])/(8*d) + (A*(a + b*Sec[c + d*x])^4*Sin[c + d*x])/d + (b*(34*a^2*b*B + 4*b^3*B - a^3*(12*A - 19*C)
+ 8*a*b^2*(3*A + 2*C))*Tan[c + d*x])/(6*d) + (b^2*(32*a*b*B - a^2*(24*A - 26*C) + 3*b^2*(4*A + 3*C))*Sec[c + d
*x]*Tan[c + d*x])/(24*d) - (b*(12*a*A - 4*b*B - 7*a*C)*(a + b*Sec[c + d*x])^2*Tan[c + d*x])/(12*d) - (b*(4*A -
C)*(a + b*Sec[c + d*x])^3*Tan[c + d*x])/(4*d)

________________________________________________________________________________________

Rubi [A]  time = 0.582644, antiderivative size = 273, normalized size of antiderivative = 1., number of steps used = 8, 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} \[ \frac{b \tan (c+d x) \left (a^3 (-(12 A-19 C))+34 a^2 b B+8 a b^2 (3 A+2 C)+4 b^3 B\right )}{6 d}+\frac{\left (24 a^2 b^2 (2 A+C)+32 a^3 b B+8 a^4 C+16 a b^3 B+b^4 (4 A+3 C)\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{b^2 \tan (c+d x) \sec (c+d x) \left (a^2 (-(24 A-26 C))+32 a b B+3 b^2 (4 A+3 C)\right )}{24 d}+a^3 x (a B+4 A b)-\frac{b \tan (c+d x) (12 a A-7 a C-4 b B) (a+b \sec (c+d x))^2}{12 d}-\frac{b (4 A-C) \tan (c+d x) (a+b \sec (c+d x))^3}{4 d}+\frac{A \sin (c+d x) (a+b \sec (c+d x))^4}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

a^3*(4*A*b + a*B)*x + ((32*a^3*b*B + 16*a*b^3*B + 8*a^4*C + 24*a^2*b^2*(2*A + C) + b^4*(4*A + 3*C))*ArcTanh[Si
n[c + d*x]])/(8*d) + (A*(a + b*Sec[c + d*x])^4*Sin[c + d*x])/d + (b*(34*a^2*b*B + 4*b^3*B - a^3*(12*A - 19*C)
+ 8*a*b^2*(3*A + 2*C))*Tan[c + d*x])/(6*d) + (b^2*(32*a*b*B - a^2*(24*A - 26*C) + 3*b^2*(4*A + 3*C))*Sec[c + d
*x]*Tan[c + d*x])/(24*d) - (b*(12*a*A - 4*b*B - 7*a*C)*(a + b*Sec[c + d*x])^2*Tan[c + d*x])/(12*d) - (b*(4*A -
C)*(a + b*Sec[c + d*x])^3*Tan[c + d*x])/(4*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))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\frac{A (a+b \sec (c+d x))^4 \sin (c+d x)}{d}+\int (a+b \sec (c+d x))^3 \left (4 A b+a B+(b B+a C) \sec (c+d x)-b (4 A-C) \sec ^2(c+d x)\right ) \, dx\\ &=\frac{A (a+b \sec (c+d x))^4 \sin (c+d x)}{d}-\frac{b (4 A-C) (a+b \sec (c+d x))^3 \tan (c+d x)}{4 d}+\frac{1}{4} \int (a+b \sec (c+d x))^2 \left (4 a (4 A b+a B)+\left (4 A b^2+8 a b B+4 a^2 C+3 b^2 C\right ) \sec (c+d x)-b (12 a A-4 b B-7 a C) \sec ^2(c+d x)\right ) \, dx\\ &=\frac{A (a+b \sec (c+d x))^4 \sin (c+d x)}{d}-\frac{b (12 a A-4 b B-7 a C) (a+b \sec (c+d x))^2 \tan (c+d x)}{12 d}-\frac{b (4 A-C) (a+b \sec (c+d x))^3 \tan (c+d x)}{4 d}+\frac{1}{12} \int (a+b \sec (c+d x)) \left (12 a^2 (4 A b+a B)+\left (36 a^2 b B+8 b^3 B+12 a^3 C+a b^2 (36 A+23 C)\right ) \sec (c+d x)+b \left (32 a b B-a^2 (24 A-26 C)+3 b^2 (4 A+3 C)\right ) \sec ^2(c+d x)\right ) \, dx\\ &=\frac{A (a+b \sec (c+d x))^4 \sin (c+d x)}{d}+\frac{b^2 \left (32 a b B-a^2 (24 A-26 C)+3 b^2 (4 A+3 C)\right ) \sec (c+d x) \tan (c+d x)}{24 d}-\frac{b (12 a A-4 b B-7 a C) (a+b \sec (c+d x))^2 \tan (c+d x)}{12 d}-\frac{b (4 A-C) (a+b \sec (c+d x))^3 \tan (c+d x)}{4 d}+\frac{1}{24} \int \left (24 a^3 (4 A b+a B)+3 \left (32 a^3 b B+16 a b^3 B+8 a^4 C+24 a^2 b^2 (2 A+C)+b^4 (4 A+3 C)\right ) \sec (c+d x)+4 b \left (34 a^2 b B+4 b^3 B-a^3 (12 A-19 C)+8 a b^2 (3 A+2 C)\right ) \sec ^2(c+d x)\right ) \, dx\\ &=a^3 (4 A b+a B) x+\frac{A (a+b \sec (c+d x))^4 \sin (c+d x)}{d}+\frac{b^2 \left (32 a b B-a^2 (24 A-26 C)+3 b^2 (4 A+3 C)\right ) \sec (c+d x) \tan (c+d x)}{24 d}-\frac{b (12 a A-4 b B-7 a C) (a+b \sec (c+d x))^2 \tan (c+d x)}{12 d}-\frac{b (4 A-C) (a+b \sec (c+d x))^3 \tan (c+d x)}{4 d}+\frac{1}{6} \left (b \left (34 a^2 b B+4 b^3 B-a^3 (12 A-19 C)+8 a b^2 (3 A+2 C)\right )\right ) \int \sec ^2(c+d x) \, dx+\frac{1}{8} \left (32 a^3 b B+16 a b^3 B+8 a^4 C+24 a^2 b^2 (2 A+C)+b^4 (4 A+3 C)\right ) \int \sec (c+d x) \, dx\\ &=a^3 (4 A b+a B) x+\frac{\left (32 a^3 b B+16 a b^3 B+8 a^4 C+24 a^2 b^2 (2 A+C)+b^4 (4 A+3 C)\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{A (a+b \sec (c+d x))^4 \sin (c+d x)}{d}+\frac{b^2 \left (32 a b B-a^2 (24 A-26 C)+3 b^2 (4 A+3 C)\right ) \sec (c+d x) \tan (c+d x)}{24 d}-\frac{b (12 a A-4 b B-7 a C) (a+b \sec (c+d x))^2 \tan (c+d x)}{12 d}-\frac{b (4 A-C) (a+b \sec (c+d x))^3 \tan (c+d x)}{4 d}-\frac{\left (b \left (34 a^2 b B+4 b^3 B-a^3 (12 A-19 C)+8 a b^2 (3 A+2 C)\right )\right ) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{6 d}\\ &=a^3 (4 A b+a B) x+\frac{\left (32 a^3 b B+16 a b^3 B+8 a^4 C+24 a^2 b^2 (2 A+C)+b^4 (4 A+3 C)\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{A (a+b \sec (c+d x))^4 \sin (c+d x)}{d}+\frac{b \left (34 a^2 b B+4 b^3 B-a^3 (12 A-19 C)+8 a b^2 (3 A+2 C)\right ) \tan (c+d x)}{6 d}+\frac{b^2 \left (32 a b B-a^2 (24 A-26 C)+3 b^2 (4 A+3 C)\right ) \sec (c+d x) \tan (c+d x)}{24 d}-\frac{b (12 a A-4 b B-7 a C) (a+b \sec (c+d x))^2 \tan (c+d x)}{12 d}-\frac{b (4 A-C) (a+b \sec (c+d x))^3 \tan (c+d x)}{4 d}\\ \end{align*}

Mathematica [B]  time = 6.89277, size = 813, normalized size = 2.98 \[ \frac{\left (-8 C a^4-32 b B a^3-48 A b^2 a^2-24 b^2 C a^2-16 b^3 B a-4 A b^4-3 b^4 C\right ) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right ) (a+b \sec (c+d x))^4 \left (C \sec ^2(c+d x)+B \sec (c+d x)+A\right ) \cos ^6(c+d x)}{4 d (b+a \cos (c+d x))^4 (\cos (2 c+2 d x) A+A+2 C+2 B \cos (c+d x))}+\frac{\left (8 C a^4+32 b B a^3+48 A b^2 a^2+24 b^2 C a^2+16 b^3 B a+4 A b^4+3 b^4 C\right ) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )+\sin \left (\frac{1}{2} (c+d x)\right )\right ) (a+b \sec (c+d x))^4 \left (C \sec ^2(c+d x)+B \sec (c+d x)+A\right ) \cos ^6(c+d x)}{4 d (b+a \cos (c+d x))^4 (\cos (2 c+2 d x) A+A+2 C+2 B \cos (c+d x))}+\frac{(a+b \sec (c+d x))^4 \left (C \sec ^2(c+d x)+B \sec (c+d x)+A\right ) \left (36 B (c+d x) a^4+48 B (c+d x) \cos (2 (c+d x)) a^4+12 B (c+d x) \cos (4 (c+d x)) a^4+12 A \sin (c+d x) a^4+18 A \sin (3 (c+d x)) a^4+6 A \sin (5 (c+d x)) a^4+144 A b (c+d x) a^3+192 A b (c+d x) \cos (2 (c+d x)) a^3+48 A b (c+d x) \cos (4 (c+d x)) a^3+96 b C \sin (2 (c+d x)) a^3+48 b C \sin (4 (c+d x)) a^3+72 b^2 C \sin (c+d x) a^2+144 b^2 B \sin (2 (c+d x)) a^2+72 b^2 C \sin (3 (c+d x)) a^2+72 b^2 B \sin (4 (c+d x)) a^2+48 b^3 B \sin (c+d x) a+96 A b^3 \sin (2 (c+d x)) a+128 b^3 C \sin (2 (c+d x)) a+48 b^3 B \sin (3 (c+d x)) a+48 A b^3 \sin (4 (c+d x)) a+32 b^3 C \sin (4 (c+d x)) a+12 A b^4 \sin (c+d x)+33 b^4 C \sin (c+d x)+32 b^4 B \sin (2 (c+d x))+12 A b^4 \sin (3 (c+d x))+9 b^4 C \sin (3 (c+d x))+8 b^4 B \sin (4 (c+d x))\right ) \cos ^2(c+d x)}{48 d (b+a \cos (c+d x))^4 (\cos (2 c+2 d x) A+A+2 C+2 B \cos (c+d x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-48*a^2*A*b^2 - 4*A*b^4 - 32*a^3*b*B - 16*a*b^3*B - 8*a^4*C - 24*a^2*b^2*C - 3*b^4*C)*Cos[c + d*x]^6*Log[Cos
[(c + d*x)/2] - Sin[(c + d*x)/2]]*(a + b*Sec[c + d*x])^4*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(4*d*(b + a*
Cos[c + d*x])^4*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])) + ((48*a^2*A*b^2 + 4*A*b^4 + 32*a^3*b*B + 1
6*a*b^3*B + 8*a^4*C + 24*a^2*b^2*C + 3*b^4*C)*Cos[c + d*x]^6*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]*(a + b*S
ec[c + d*x])^4*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(4*d*(b + a*Cos[c + d*x])^4*(A + 2*C + 2*B*Cos[c + d*x
] + A*Cos[2*c + 2*d*x])) + (Cos[c + d*x]^2*(a + b*Sec[c + d*x])^4*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*(144
*a^3*A*b*(c + d*x) + 36*a^4*B*(c + d*x) + 192*a^3*A*b*(c + d*x)*Cos[2*(c + d*x)] + 48*a^4*B*(c + d*x)*Cos[2*(c
+ d*x)] + 48*a^3*A*b*(c + d*x)*Cos[4*(c + d*x)] + 12*a^4*B*(c + d*x)*Cos[4*(c + d*x)] + 12*a^4*A*Sin[c + d*x]
+ 12*A*b^4*Sin[c + d*x] + 48*a*b^3*B*Sin[c + d*x] + 72*a^2*b^2*C*Sin[c + d*x] + 33*b^4*C*Sin[c + d*x] + 96*a*
A*b^3*Sin[2*(c + d*x)] + 144*a^2*b^2*B*Sin[2*(c + d*x)] + 32*b^4*B*Sin[2*(c + d*x)] + 96*a^3*b*C*Sin[2*(c + d*
x)] + 128*a*b^3*C*Sin[2*(c + d*x)] + 18*a^4*A*Sin[3*(c + d*x)] + 12*A*b^4*Sin[3*(c + d*x)] + 48*a*b^3*B*Sin[3*
(c + d*x)] + 72*a^2*b^2*C*Sin[3*(c + d*x)] + 9*b^4*C*Sin[3*(c + d*x)] + 48*a*A*b^3*Sin[4*(c + d*x)] + 72*a^2*b
^2*B*Sin[4*(c + d*x)] + 8*b^4*B*Sin[4*(c + d*x)] + 48*a^3*b*C*Sin[4*(c + d*x)] + 32*a*b^3*C*Sin[4*(c + d*x)] +
6*a^4*A*Sin[5*(c + d*x)]))/(48*d*(b + a*Cos[c + d*x])^4*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x]))

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Maple [A]  time = 0.09, size = 457, normalized size = 1.7 \begin{align*}{\frac{A{a}^{4}\sin \left ( dx+c \right ) }{d}}+B{a}^{4}x+{\frac{B{a}^{4}c}{d}}+{\frac{{a}^{4}C\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+4\,{a}^{3}Abx+4\,{\frac{A{a}^{3}bc}{d}}+4\,{\frac{B{a}^{3}b\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+4\,{\frac{{a}^{3}bC\tan \left ( dx+c \right ) }{d}}+6\,{\frac{A{a}^{2}{b}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+6\,{\frac{{a}^{2}{b}^{2}B\tan \left ( dx+c \right ) }{d}}+3\,{\frac{C{a}^{2}{b}^{2}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{d}}+3\,{\frac{C{a}^{2}{b}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+4\,{\frac{Aa{b}^{3}\tan \left ( dx+c \right ) }{d}}+2\,{\frac{a{b}^{3}B\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{d}}+2\,{\frac{a{b}^{3}B\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{8\,Ca{b}^{3}\tan \left ( dx+c \right ) }{3\,d}}+{\frac{4\,Ca{b}^{3}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3\,d}}+{\frac{A{b}^{4}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{A{b}^{4}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+{\frac{2\,B{b}^{4}\tan \left ( dx+c \right ) }{3\,d}}+{\frac{B{b}^{4}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3\,d}}+{\frac{C{b}^{4}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{4\,d}}+{\frac{3\,C{b}^{4}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{8\,d}}+{\frac{3\,C{b}^{4}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,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))^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x)

[Out]

1/d*A*a^4*sin(d*x+c)+B*a^4*x+1/d*B*a^4*c+1/d*a^4*C*ln(sec(d*x+c)+tan(d*x+c))+4*a^3*A*b*x+4/d*A*a^3*b*c+4/d*B*a
^3*b*ln(sec(d*x+c)+tan(d*x+c))+4/d*a^3*b*C*tan(d*x+c)+6/d*A*a^2*b^2*ln(sec(d*x+c)+tan(d*x+c))+6/d*a^2*b^2*B*ta
n(d*x+c)+3/d*C*a^2*b^2*sec(d*x+c)*tan(d*x+c)+3/d*C*a^2*b^2*ln(sec(d*x+c)+tan(d*x+c))+4/d*A*a*b^3*tan(d*x+c)+2/
d*a*b^3*B*sec(d*x+c)*tan(d*x+c)+2/d*a*b^3*B*ln(sec(d*x+c)+tan(d*x+c))+8/3/d*C*a*b^3*tan(d*x+c)+4/3/d*C*a*b^3*t
an(d*x+c)*sec(d*x+c)^2+1/2/d*A*b^4*sec(d*x+c)*tan(d*x+c)+1/2/d*A*b^4*ln(sec(d*x+c)+tan(d*x+c))+2/3/d*B*b^4*tan
(d*x+c)+1/3/d*B*b^4*tan(d*x+c)*sec(d*x+c)^2+1/4/d*C*b^4*tan(d*x+c)*sec(d*x+c)^3+3/8/d*C*b^4*sec(d*x+c)*tan(d*x
+c)+3/8/d*C*b^4*ln(sec(d*x+c)+tan(d*x+c))

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Maxima [A]  time = 1.08799, size = 582, normalized size = 2.13 \begin{align*} \frac{48 \,{\left (d x + c\right )} B a^{4} + 192 \,{\left (d x + c\right )} A a^{3} b + 64 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a b^{3} + 16 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B b^{4} - 3 \, C b^{4}{\left (\frac{2 \,{\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 72 \, C a^{2} 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 )} - 48 \, B 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 )} - 12 \, A 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^{4}{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 96 \, B 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 )} + 144 \, A 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 )} + 48 \, A a^{4} \sin \left (d x + c\right ) + 192 \, C a^{3} b \tan \left (d x + c\right ) + 288 \, B a^{2} b^{2} \tan \left (d x + c\right ) + 192 \, A a b^{3} \tan \left (d x + c\right )}{48 \, 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))^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

1/48*(48*(d*x + c)*B*a^4 + 192*(d*x + c)*A*a^3*b + 64*(tan(d*x + c)^3 + 3*tan(d*x + c))*C*a*b^3 + 16*(tan(d*x
+ c)^3 + 3*tan(d*x + c))*B*b^4 - 3*C*b^4*(2*(3*sin(d*x + c)^3 - 5*sin(d*x + c))/(sin(d*x + c)^4 - 2*sin(d*x +
c)^2 + 1) - 3*log(sin(d*x + c) + 1) + 3*log(sin(d*x + c) - 1)) - 72*C*a^2*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)) - 48*B*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)) - 12*A*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^4*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 96*B*a^3*b*(log(sin
(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 144*A*a^2*b^2*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 48*A
*a^4*sin(d*x + c) + 192*C*a^3*b*tan(d*x + c) + 288*B*a^2*b^2*tan(d*x + c) + 192*A*a*b^3*tan(d*x + c))/d

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Fricas [A]  time = 0.629816, size = 725, normalized size = 2.66 \begin{align*} \frac{48 \,{\left (B a^{4} + 4 \, A a^{3} b\right )} d x \cos \left (d x + c\right )^{4} + 3 \,{\left (8 \, C a^{4} + 32 \, B a^{3} b + 24 \,{\left (2 \, A + C\right )} a^{2} b^{2} + 16 \, B a b^{3} +{\left (4 \, A + 3 \, C\right )} b^{4}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left (8 \, C a^{4} + 32 \, B a^{3} b + 24 \,{\left (2 \, A + C\right )} a^{2} b^{2} + 16 \, B a b^{3} +{\left (4 \, A + 3 \, C\right )} b^{4}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (24 \, A a^{4} \cos \left (d x + c\right )^{4} + 6 \, C b^{4} + 16 \,{\left (6 \, C a^{3} b + 9 \, B a^{2} b^{2} + 2 \,{\left (3 \, A + 2 \, C\right )} a b^{3} + B b^{4}\right )} \cos \left (d x + c\right )^{3} + 3 \,{\left (24 \, C a^{2} b^{2} + 16 \, B a b^{3} +{\left (4 \, A + 3 \, C\right )} b^{4}\right )} \cos \left (d x + c\right )^{2} + 8 \,{\left (4 \, C a b^{3} + B b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(48*(B*a^4 + 4*A*a^3*b)*d*x*cos(d*x + c)^4 + 3*(8*C*a^4 + 32*B*a^3*b + 24*(2*A + C)*a^2*b^2 + 16*B*a*b^3
+ (4*A + 3*C)*b^4)*cos(d*x + c)^4*log(sin(d*x + c) + 1) - 3*(8*C*a^4 + 32*B*a^3*b + 24*(2*A + C)*a^2*b^2 + 16*
B*a*b^3 + (4*A + 3*C)*b^4)*cos(d*x + c)^4*log(-sin(d*x + c) + 1) + 2*(24*A*a^4*cos(d*x + c)^4 + 6*C*b^4 + 16*(
6*C*a^3*b + 9*B*a^2*b^2 + 2*(3*A + 2*C)*a*b^3 + B*b^4)*cos(d*x + c)^3 + 3*(24*C*a^2*b^2 + 16*B*a*b^3 + (4*A +
3*C)*b^4)*cos(d*x + c)^2 + 8*(4*C*a*b^3 + B*b^4)*cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^4)

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

[Out]

Timed out

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Giac [B]  time = 1.45465, size = 1134, normalized size = 4.15 \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)*(a+b*sec(d*x+c))^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

1/24*(48*A*a^4*tan(1/2*d*x + 1/2*c)/(tan(1/2*d*x + 1/2*c)^2 + 1) + 24*(B*a^4 + 4*A*a^3*b)*(d*x + c) + 3*(8*C*a
^4 + 32*B*a^3*b + 48*A*a^2*b^2 + 24*C*a^2*b^2 + 16*B*a*b^3 + 4*A*b^4 + 3*C*b^4)*log(abs(tan(1/2*d*x + 1/2*c) +
1)) - 3*(8*C*a^4 + 32*B*a^3*b + 48*A*a^2*b^2 + 24*C*a^2*b^2 + 16*B*a*b^3 + 4*A*b^4 + 3*C*b^4)*log(abs(tan(1/2
*d*x + 1/2*c) - 1)) - 2*(96*C*a^3*b*tan(1/2*d*x + 1/2*c)^7 + 144*B*a^2*b^2*tan(1/2*d*x + 1/2*c)^7 - 72*C*a^2*b
^2*tan(1/2*d*x + 1/2*c)^7 + 96*A*a*b^3*tan(1/2*d*x + 1/2*c)^7 - 48*B*a*b^3*tan(1/2*d*x + 1/2*c)^7 + 96*C*a*b^3
*tan(1/2*d*x + 1/2*c)^7 - 12*A*b^4*tan(1/2*d*x + 1/2*c)^7 + 24*B*b^4*tan(1/2*d*x + 1/2*c)^7 - 15*C*b^4*tan(1/2
*d*x + 1/2*c)^7 - 288*C*a^3*b*tan(1/2*d*x + 1/2*c)^5 - 432*B*a^2*b^2*tan(1/2*d*x + 1/2*c)^5 + 72*C*a^2*b^2*tan
(1/2*d*x + 1/2*c)^5 - 288*A*a*b^3*tan(1/2*d*x + 1/2*c)^5 + 48*B*a*b^3*tan(1/2*d*x + 1/2*c)^5 - 160*C*a*b^3*tan
(1/2*d*x + 1/2*c)^5 + 12*A*b^4*tan(1/2*d*x + 1/2*c)^5 - 40*B*b^4*tan(1/2*d*x + 1/2*c)^5 - 9*C*b^4*tan(1/2*d*x
+ 1/2*c)^5 + 288*C*a^3*b*tan(1/2*d*x + 1/2*c)^3 + 432*B*a^2*b^2*tan(1/2*d*x + 1/2*c)^3 + 72*C*a^2*b^2*tan(1/2*
d*x + 1/2*c)^3 + 288*A*a*b^3*tan(1/2*d*x + 1/2*c)^3 + 48*B*a*b^3*tan(1/2*d*x + 1/2*c)^3 + 160*C*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 + 40*B*b^4*tan(1/2*d*x + 1/2*c)^3 - 9*C*b^4*tan(1/2*d*x + 1/2
*c)^3 - 96*C*a^3*b*tan(1/2*d*x + 1/2*c) - 144*B*a^2*b^2*tan(1/2*d*x + 1/2*c) - 72*C*a^2*b^2*tan(1/2*d*x + 1/2*
c) - 96*A*a*b^3*tan(1/2*d*x + 1/2*c) - 48*B*a*b^3*tan(1/2*d*x + 1/2*c) - 96*C*a*b^3*tan(1/2*d*x + 1/2*c) - 12*
A*b^4*tan(1/2*d*x + 1/2*c) - 24*B*b^4*tan(1/2*d*x + 1/2*c) - 15*C*b^4*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2
*c)^2 - 1)^4)/d