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

Optimal. Leaf size=384 \[ \frac {\tan (c+d x) \left (-4 a^3 C+24 a^2 b B+a b^2 (70 A+53 C)+32 b^3 B\right ) (a+b \sec (c+d x))^2}{120 b d}+\frac {\left (8 a^4 (2 A+C)+32 a^3 b B+12 a^2 b^2 (4 A+3 C)+24 a b^3 B+b^4 (6 A+5 C)\right ) \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac {\tan (c+d x) \sec (c+d x) \left (-8 a^4 C+48 a^3 b B+2 a^2 b^2 (130 A+89 C)+232 a b^3 B+15 b^4 (6 A+5 C)\right )}{240 d}+\frac {\tan (c+d x) \left (-4 a^5 C+24 a^4 b B+a^3 b^2 (190 A+121 C)+224 a^2 b^3 B+32 a b^4 (5 A+4 C)+32 b^5 B\right )}{60 b d}+\frac {\tan (c+d x) \left (4 a (6 b B-a C)+5 b^2 (6 A+5 C)\right ) (a+b \sec (c+d x))^3}{120 b d}+\frac {(6 b B-a C) \tan (c+d x) (a+b \sec (c+d x))^4}{30 b d}+\frac {C \tan (c+d x) (a+b \sec (c+d x))^5}{6 b d} \]

[Out]

1/16*(32*a^3*b*B+24*a*b^3*B+8*a^4*(2*A+C)+12*a^2*b^2*(4*A+3*C)+b^4*(6*A+5*C))*arctanh(sin(d*x+c))/d+1/60*(24*a
^4*b*B+224*a^2*b^3*B+32*b^5*B-4*a^5*C+32*a*b^4*(5*A+4*C)+a^3*b^2*(190*A+121*C))*tan(d*x+c)/b/d+1/240*(48*a^3*b
*B+232*a*b^3*B-8*a^4*C+15*b^4*(6*A+5*C)+2*a^2*b^2*(130*A+89*C))*sec(d*x+c)*tan(d*x+c)/d+1/120*(24*a^2*b*B+32*b
^3*B-4*a^3*C+a*b^2*(70*A+53*C))*(a+b*sec(d*x+c))^2*tan(d*x+c)/b/d+1/120*(5*b^2*(6*A+5*C)+4*a*(6*B*b-C*a))*(a+b
*sec(d*x+c))^3*tan(d*x+c)/b/d+1/30*(6*B*b-C*a)*(a+b*sec(d*x+c))^4*tan(d*x+c)/b/d+1/6*C*(a+b*sec(d*x+c))^5*tan(
d*x+c)/b/d

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Rubi [A]  time = 0.88, antiderivative size = 384, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.180, Rules used = {4082, 4002, 3997, 3787, 3770, 3767, 8} \[ \frac {\tan (c+d x) \left (a^3 b^2 (190 A+121 C)+224 a^2 b^3 B+24 a^4 b B-4 a^5 C+32 a b^4 (5 A+4 C)+32 b^5 B\right )}{60 b d}+\frac {\left (12 a^2 b^2 (4 A+3 C)+8 a^4 (2 A+C)+32 a^3 b B+24 a b^3 B+b^4 (6 A+5 C)\right ) \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac {\tan (c+d x) \left (24 a^2 b B-4 a^3 C+a b^2 (70 A+53 C)+32 b^3 B\right ) (a+b \sec (c+d x))^2}{120 b d}+\frac {\tan (c+d x) \sec (c+d x) \left (2 a^2 b^2 (130 A+89 C)+48 a^3 b B-8 a^4 C+232 a b^3 B+15 b^4 (6 A+5 C)\right )}{240 d}+\frac {\tan (c+d x) \left (4 a (6 b B-a C)+5 b^2 (6 A+5 C)\right ) (a+b \sec (c+d x))^3}{120 b d}+\frac {(6 b B-a C) \tan (c+d x) (a+b \sec (c+d x))^4}{30 b d}+\frac {C \tan (c+d x) (a+b \sec (c+d x))^5}{6 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((32*a^3*b*B + 24*a*b^3*B + 8*a^4*(2*A + C) + 12*a^2*b^2*(4*A + 3*C) + b^4*(6*A + 5*C))*ArcTanh[Sin[c + d*x]])
/(16*d) + ((24*a^4*b*B + 224*a^2*b^3*B + 32*b^5*B - 4*a^5*C + 32*a*b^4*(5*A + 4*C) + a^3*b^2*(190*A + 121*C))*
Tan[c + d*x])/(60*b*d) + ((48*a^3*b*B + 232*a*b^3*B - 8*a^4*C + 15*b^4*(6*A + 5*C) + 2*a^2*b^2*(130*A + 89*C))
*Sec[c + d*x]*Tan[c + d*x])/(240*d) + ((24*a^2*b*B + 32*b^3*B - 4*a^3*C + a*b^2*(70*A + 53*C))*(a + b*Sec[c +
d*x])^2*Tan[c + d*x])/(120*b*d) + ((5*b^2*(6*A + 5*C) + 4*a*(6*b*B - a*C))*(a + b*Sec[c + d*x])^3*Tan[c + d*x]
)/(120*b*d) + ((6*b*B - a*C)*(a + b*Sec[c + d*x])^4*Tan[c + d*x])/(30*b*d) + (C*(a + b*Sec[c + d*x])^5*Tan[c +
d*x])/(6*b*d)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, 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 3770

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

Rule 3787

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(d*
Csc[e + f*x])^n, x], x] + Dist[b/d, Int[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]

Rule 3997

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))*(csc[(e_.) + (f_.)*(x_)]*(B_.
) + (A_)), x_Symbol] :> -Simp[(b*B*Cot[e + f*x]*(d*Csc[e + f*x])^n)/(f*(n + 1)), x] + Dist[1/(n + 1), Int[(d*C
sc[e + f*x])^n*Simp[A*a*(n + 1) + B*b*n + (A*b + B*a)*(n + 1)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f
, A, B}, x] && NeQ[A*b - a*B, 0] &&  !LeQ[n, -1]

Rule 4002

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

Rule 4082

Int[csc[(e_.) + (f_.)*(x_)]*((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 + 1))/(b*f*(m
+ 2)), x] + Dist[1/(b*(m + 2)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*Simp[b*A*(m + 2) + b*C*(m + 1) + (b*B*
(m + 2) - a*C)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

Rubi steps

\begin {align*} \int \sec (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 {C (a+b \sec (c+d x))^5 \tan (c+d x)}{6 b d}+\frac {\int \sec (c+d x) (a+b \sec (c+d x))^4 (b (6 A+5 C)+(6 b B-a C) \sec (c+d x)) \, dx}{6 b}\\ &=\frac {(6 b B-a C) (a+b \sec (c+d x))^4 \tan (c+d x)}{30 b d}+\frac {C (a+b \sec (c+d x))^5 \tan (c+d x)}{6 b d}+\frac {\int \sec (c+d x) (a+b \sec (c+d x))^3 \left (3 b (10 a A+8 b B+7 a C)+\left (5 b^2 (6 A+5 C)+4 a (6 b B-a C)\right ) \sec (c+d x)\right ) \, dx}{30 b}\\ &=\frac {\left (5 b^2 (6 A+5 C)+4 a (6 b B-a C)\right ) (a+b \sec (c+d x))^3 \tan (c+d x)}{120 b d}+\frac {(6 b B-a C) (a+b \sec (c+d x))^4 \tan (c+d x)}{30 b d}+\frac {C (a+b \sec (c+d x))^5 \tan (c+d x)}{6 b d}+\frac {\int \sec (c+d x) (a+b \sec (c+d x))^2 \left (3 b \left (56 a b B+8 a^2 (5 A+3 C)+5 b^2 (6 A+5 C)\right )+3 \left (24 a^2 b B+32 b^3 B-4 a^3 C+a b^2 (70 A+53 C)\right ) \sec (c+d x)\right ) \, dx}{120 b}\\ &=\frac {\left (24 a^2 b B+32 b^3 B-4 a^3 C+a b^2 (70 A+53 C)\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{120 b d}+\frac {\left (5 b^2 (6 A+5 C)+4 a (6 b B-a C)\right ) (a+b \sec (c+d x))^3 \tan (c+d x)}{120 b d}+\frac {(6 b B-a C) (a+b \sec (c+d x))^4 \tan (c+d x)}{30 b d}+\frac {C (a+b \sec (c+d x))^5 \tan (c+d x)}{6 b d}+\frac {\int \sec (c+d x) (a+b \sec (c+d x)) \left (3 b \left (216 a^2 b B+64 b^3 B+8 a^3 (15 A+8 C)+a b^2 (230 A+181 C)\right )+3 \left (48 a^3 b B+232 a b^3 B-8 a^4 C+15 b^4 (6 A+5 C)+2 a^2 b^2 (130 A+89 C)\right ) \sec (c+d x)\right ) \, dx}{360 b}\\ &=\frac {\left (48 a^3 b B+232 a b^3 B-8 a^4 C+15 b^4 (6 A+5 C)+2 a^2 b^2 (130 A+89 C)\right ) \sec (c+d x) \tan (c+d x)}{240 d}+\frac {\left (24 a^2 b B+32 b^3 B-4 a^3 C+a b^2 (70 A+53 C)\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{120 b d}+\frac {\left (5 b^2 (6 A+5 C)+4 a (6 b B-a C)\right ) (a+b \sec (c+d x))^3 \tan (c+d x)}{120 b d}+\frac {(6 b B-a C) (a+b \sec (c+d x))^4 \tan (c+d x)}{30 b d}+\frac {C (a+b \sec (c+d x))^5 \tan (c+d x)}{6 b d}+\frac {\int \sec (c+d x) \left (45 b \left (32 a^3 b B+24 a b^3 B+8 a^4 (2 A+C)+12 a^2 b^2 (4 A+3 C)+b^4 (6 A+5 C)\right )+12 \left (24 a^4 b B+224 a^2 b^3 B+32 b^5 B-4 a^5 C+32 a b^4 (5 A+4 C)+a^3 b^2 (190 A+121 C)\right ) \sec (c+d x)\right ) \, dx}{720 b}\\ &=\frac {\left (48 a^3 b B+232 a b^3 B-8 a^4 C+15 b^4 (6 A+5 C)+2 a^2 b^2 (130 A+89 C)\right ) \sec (c+d x) \tan (c+d x)}{240 d}+\frac {\left (24 a^2 b B+32 b^3 B-4 a^3 C+a b^2 (70 A+53 C)\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{120 b d}+\frac {\left (5 b^2 (6 A+5 C)+4 a (6 b B-a C)\right ) (a+b \sec (c+d x))^3 \tan (c+d x)}{120 b d}+\frac {(6 b B-a C) (a+b \sec (c+d x))^4 \tan (c+d x)}{30 b d}+\frac {C (a+b \sec (c+d x))^5 \tan (c+d x)}{6 b d}+\frac {1}{16} \left (32 a^3 b B+24 a b^3 B+8 a^4 (2 A+C)+12 a^2 b^2 (4 A+3 C)+b^4 (6 A+5 C)\right ) \int \sec (c+d x) \, dx+\frac {\left (24 a^4 b B+224 a^2 b^3 B+32 b^5 B-4 a^5 C+32 a b^4 (5 A+4 C)+a^3 b^2 (190 A+121 C)\right ) \int \sec ^2(c+d x) \, dx}{60 b}\\ &=\frac {\left (32 a^3 b B+24 a b^3 B+8 a^4 (2 A+C)+12 a^2 b^2 (4 A+3 C)+b^4 (6 A+5 C)\right ) \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac {\left (48 a^3 b B+232 a b^3 B-8 a^4 C+15 b^4 (6 A+5 C)+2 a^2 b^2 (130 A+89 C)\right ) \sec (c+d x) \tan (c+d x)}{240 d}+\frac {\left (24 a^2 b B+32 b^3 B-4 a^3 C+a b^2 (70 A+53 C)\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{120 b d}+\frac {\left (5 b^2 (6 A+5 C)+4 a (6 b B-a C)\right ) (a+b \sec (c+d x))^3 \tan (c+d x)}{120 b d}+\frac {(6 b B-a C) (a+b \sec (c+d x))^4 \tan (c+d x)}{30 b d}+\frac {C (a+b \sec (c+d x))^5 \tan (c+d x)}{6 b d}-\frac {\left (24 a^4 b B+224 a^2 b^3 B+32 b^5 B-4 a^5 C+32 a b^4 (5 A+4 C)+a^3 b^2 (190 A+121 C)\right ) \operatorname {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{60 b d}\\ &=\frac {\left (32 a^3 b B+24 a b^3 B+8 a^4 (2 A+C)+12 a^2 b^2 (4 A+3 C)+b^4 (6 A+5 C)\right ) \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac {\left (24 a^4 b B+224 a^2 b^3 B+32 b^5 B-4 a^5 C+32 a b^4 (5 A+4 C)+a^3 b^2 (190 A+121 C)\right ) \tan (c+d x)}{60 b d}+\frac {\left (48 a^3 b B+232 a b^3 B-8 a^4 C+15 b^4 (6 A+5 C)+2 a^2 b^2 (130 A+89 C)\right ) \sec (c+d x) \tan (c+d x)}{240 d}+\frac {\left (24 a^2 b B+32 b^3 B-4 a^3 C+a b^2 (70 A+53 C)\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{120 b d}+\frac {\left (5 b^2 (6 A+5 C)+4 a (6 b B-a C)\right ) (a+b \sec (c+d x))^3 \tan (c+d x)}{120 b d}+\frac {(6 b B-a C) (a+b \sec (c+d x))^4 \tan (c+d x)}{30 b d}+\frac {C (a+b \sec (c+d x))^5 \tan (c+d x)}{6 b d}\\ \end {align*}

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Mathematica [A]  time = 3.54, size = 424, normalized size = 1.10 \[ -\frac {\sec ^6(c+d x) \left (A \cos ^2(c+d x)+B \cos (c+d x)+C\right ) \left (-10 b^2 \sin (c+d x) \cos ^2(c+d x) \left (36 a^2 C+24 a b B+6 A b^2+5 b^2 C\right )-32 b \sin (c+d x) \cos ^3(c+d x) \left (10 a^3 C+15 a^2 b B+2 a b^2 (5 A+4 C)+2 b^3 B\right )-16 \sin (c+d x) \cos ^5(c+d x) \left (15 a^4 B+20 a^3 b (3 A+2 C)+60 a^2 b^2 B+8 a b^3 (5 A+4 C)+8 b^4 B\right )-15 \sin (c+d x) \cos ^4(c+d x) \left (8 a^4 C+32 a^3 b B+12 a^2 b^2 (4 A+3 C)+24 a b^3 B+b^4 (6 A+5 C)\right )+15 \cos ^6(c+d x) \left (8 a^4 (2 A+C)+32 a^3 b B+12 a^2 b^2 (4 A+3 C)+24 a b^3 B+b^4 (6 A+5 C)\right ) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )-8 b^3 \sin (c+d x) (6 (4 a C+b B) \cos (c+d x)+5 b C)\right )}{120 d (A \cos (2 (c+d x))+A+2 B \cos (c+d x)+2 C)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/120*((C + B*Cos[c + d*x] + A*Cos[c + d*x]^2)*Sec[c + d*x]^6*(15*(32*a^3*b*B + 24*a*b^3*B + 8*a^4*(2*A + C)
+ 12*a^2*b^2*(4*A + 3*C) + b^4*(6*A + 5*C))*Cos[c + d*x]^6*(Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos
[(c + d*x)/2] + Sin[(c + d*x)/2]]) - 10*b^2*(6*A*b^2 + 24*a*b*B + 36*a^2*C + 5*b^2*C)*Cos[c + d*x]^2*Sin[c + d
*x] - 32*b*(15*a^2*b*B + 2*b^3*B + 10*a^3*C + 2*a*b^2*(5*A + 4*C))*Cos[c + d*x]^3*Sin[c + d*x] - 15*(32*a^3*b*
B + 24*a*b^3*B + 8*a^4*C + 12*a^2*b^2*(4*A + 3*C) + b^4*(6*A + 5*C))*Cos[c + d*x]^4*Sin[c + d*x] - 16*(15*a^4*
B + 60*a^2*b^2*B + 8*b^4*B + 20*a^3*b*(3*A + 2*C) + 8*a*b^3*(5*A + 4*C))*Cos[c + d*x]^5*Sin[c + d*x] - 8*b^3*(
5*b*C + 6*(b*B + 4*a*C)*Cos[c + d*x])*Sin[c + d*x]))/(d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*(c + d*x)]))

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fricas [A]  time = 0.73, size = 391, normalized size = 1.02 \[ \frac {15 \, {\left (8 \, {\left (2 \, A + C\right )} a^{4} + 32 \, B a^{3} b + 12 \, {\left (4 \, A + 3 \, C\right )} a^{2} b^{2} + 24 \, B a b^{3} + {\left (6 \, A + 5 \, C\right )} b^{4}\right )} \cos \left (d x + c\right )^{6} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (8 \, {\left (2 \, A + C\right )} a^{4} + 32 \, B a^{3} b + 12 \, {\left (4 \, A + 3 \, C\right )} a^{2} b^{2} + 24 \, B a b^{3} + {\left (6 \, A + 5 \, C\right )} b^{4}\right )} \cos \left (d x + c\right )^{6} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (16 \, {\left (15 \, B a^{4} + 20 \, {\left (3 \, A + 2 \, C\right )} a^{3} b + 60 \, B a^{2} b^{2} + 8 \, {\left (5 \, A + 4 \, C\right )} a b^{3} + 8 \, B b^{4}\right )} \cos \left (d x + c\right )^{5} + 40 \, C b^{4} + 15 \, {\left (8 \, C a^{4} + 32 \, B a^{3} b + 12 \, {\left (4 \, A + 3 \, C\right )} a^{2} b^{2} + 24 \, B a b^{3} + {\left (6 \, A + 5 \, C\right )} b^{4}\right )} \cos \left (d x + c\right )^{4} + 32 \, {\left (10 \, C a^{3} b + 15 \, B a^{2} b^{2} + 2 \, {\left (5 \, A + 4 \, C\right )} a b^{3} + 2 \, B b^{4}\right )} \cos \left (d x + c\right )^{3} + 10 \, {\left (36 \, C a^{2} b^{2} + 24 \, B a b^{3} + {\left (6 \, A + 5 \, C\right )} b^{4}\right )} \cos \left (d x + c\right )^{2} + 48 \, {\left (4 \, C a b^{3} + B b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{480 \, d \cos \left (d x + c\right )^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(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]

1/480*(15*(8*(2*A + C)*a^4 + 32*B*a^3*b + 12*(4*A + 3*C)*a^2*b^2 + 24*B*a*b^3 + (6*A + 5*C)*b^4)*cos(d*x + c)^
6*log(sin(d*x + c) + 1) - 15*(8*(2*A + C)*a^4 + 32*B*a^3*b + 12*(4*A + 3*C)*a^2*b^2 + 24*B*a*b^3 + (6*A + 5*C)
*b^4)*cos(d*x + c)^6*log(-sin(d*x + c) + 1) + 2*(16*(15*B*a^4 + 20*(3*A + 2*C)*a^3*b + 60*B*a^2*b^2 + 8*(5*A +
4*C)*a*b^3 + 8*B*b^4)*cos(d*x + c)^5 + 40*C*b^4 + 15*(8*C*a^4 + 32*B*a^3*b + 12*(4*A + 3*C)*a^2*b^2 + 24*B*a*
b^3 + (6*A + 5*C)*b^4)*cos(d*x + c)^4 + 32*(10*C*a^3*b + 15*B*a^2*b^2 + 2*(5*A + 4*C)*a*b^3 + 2*B*b^4)*cos(d*x
+ c)^3 + 10*(36*C*a^2*b^2 + 24*B*a*b^3 + (6*A + 5*C)*b^4)*cos(d*x + c)^2 + 48*(4*C*a*b^3 + B*b^4)*cos(d*x + c
))*sin(d*x + c))/(d*cos(d*x + c)^6)

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giac [B]  time = 0.47, size = 1658, normalized size = 4.32 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(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]

1/240*(15*(16*A*a^4 + 8*C*a^4 + 32*B*a^3*b + 48*A*a^2*b^2 + 36*C*a^2*b^2 + 24*B*a*b^3 + 6*A*b^4 + 5*C*b^4)*log
(abs(tan(1/2*d*x + 1/2*c) + 1)) - 15*(16*A*a^4 + 8*C*a^4 + 32*B*a^3*b + 48*A*a^2*b^2 + 36*C*a^2*b^2 + 24*B*a*b
^3 + 6*A*b^4 + 5*C*b^4)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(240*B*a^4*tan(1/2*d*x + 1/2*c)^11 - 120*C*a^4*
tan(1/2*d*x + 1/2*c)^11 + 960*A*a^3*b*tan(1/2*d*x + 1/2*c)^11 - 480*B*a^3*b*tan(1/2*d*x + 1/2*c)^11 + 960*C*a^
3*b*tan(1/2*d*x + 1/2*c)^11 - 720*A*a^2*b^2*tan(1/2*d*x + 1/2*c)^11 + 1440*B*a^2*b^2*tan(1/2*d*x + 1/2*c)^11 -
900*C*a^2*b^2*tan(1/2*d*x + 1/2*c)^11 + 960*A*a*b^3*tan(1/2*d*x + 1/2*c)^11 - 600*B*a*b^3*tan(1/2*d*x + 1/2*c
)^11 + 960*C*a*b^3*tan(1/2*d*x + 1/2*c)^11 - 150*A*b^4*tan(1/2*d*x + 1/2*c)^11 + 240*B*b^4*tan(1/2*d*x + 1/2*c
)^11 - 165*C*b^4*tan(1/2*d*x + 1/2*c)^11 - 1200*B*a^4*tan(1/2*d*x + 1/2*c)^9 + 360*C*a^4*tan(1/2*d*x + 1/2*c)^
9 - 4800*A*a^3*b*tan(1/2*d*x + 1/2*c)^9 + 1440*B*a^3*b*tan(1/2*d*x + 1/2*c)^9 - 3520*C*a^3*b*tan(1/2*d*x + 1/2
*c)^9 + 2160*A*a^2*b^2*tan(1/2*d*x + 1/2*c)^9 - 5280*B*a^2*b^2*tan(1/2*d*x + 1/2*c)^9 + 1260*C*a^2*b^2*tan(1/2
*d*x + 1/2*c)^9 - 3520*A*a*b^3*tan(1/2*d*x + 1/2*c)^9 + 840*B*a*b^3*tan(1/2*d*x + 1/2*c)^9 - 2240*C*a*b^3*tan(
1/2*d*x + 1/2*c)^9 + 210*A*b^4*tan(1/2*d*x + 1/2*c)^9 - 560*B*b^4*tan(1/2*d*x + 1/2*c)^9 - 25*C*b^4*tan(1/2*d*
x + 1/2*c)^9 + 2400*B*a^4*tan(1/2*d*x + 1/2*c)^7 - 240*C*a^4*tan(1/2*d*x + 1/2*c)^7 + 9600*A*a^3*b*tan(1/2*d*x
+ 1/2*c)^7 - 960*B*a^3*b*tan(1/2*d*x + 1/2*c)^7 + 5760*C*a^3*b*tan(1/2*d*x + 1/2*c)^7 - 1440*A*a^2*b^2*tan(1/
2*d*x + 1/2*c)^7 + 8640*B*a^2*b^2*tan(1/2*d*x + 1/2*c)^7 - 360*C*a^2*b^2*tan(1/2*d*x + 1/2*c)^7 + 5760*A*a*b^3
*tan(1/2*d*x + 1/2*c)^7 - 240*B*a*b^3*tan(1/2*d*x + 1/2*c)^7 + 4992*C*a*b^3*tan(1/2*d*x + 1/2*c)^7 - 60*A*b^4*
tan(1/2*d*x + 1/2*c)^7 + 1248*B*b^4*tan(1/2*d*x + 1/2*c)^7 - 450*C*b^4*tan(1/2*d*x + 1/2*c)^7 - 2400*B*a^4*tan
(1/2*d*x + 1/2*c)^5 - 240*C*a^4*tan(1/2*d*x + 1/2*c)^5 - 9600*A*a^3*b*tan(1/2*d*x + 1/2*c)^5 - 960*B*a^3*b*tan
(1/2*d*x + 1/2*c)^5 - 5760*C*a^3*b*tan(1/2*d*x + 1/2*c)^5 - 1440*A*a^2*b^2*tan(1/2*d*x + 1/2*c)^5 - 8640*B*a^2
*b^2*tan(1/2*d*x + 1/2*c)^5 - 360*C*a^2*b^2*tan(1/2*d*x + 1/2*c)^5 - 5760*A*a*b^3*tan(1/2*d*x + 1/2*c)^5 - 240
*B*a*b^3*tan(1/2*d*x + 1/2*c)^5 - 4992*C*a*b^3*tan(1/2*d*x + 1/2*c)^5 - 60*A*b^4*tan(1/2*d*x + 1/2*c)^5 - 1248
*B*b^4*tan(1/2*d*x + 1/2*c)^5 - 450*C*b^4*tan(1/2*d*x + 1/2*c)^5 + 1200*B*a^4*tan(1/2*d*x + 1/2*c)^3 + 360*C*a
^4*tan(1/2*d*x + 1/2*c)^3 + 4800*A*a^3*b*tan(1/2*d*x + 1/2*c)^3 + 1440*B*a^3*b*tan(1/2*d*x + 1/2*c)^3 + 3520*C
*a^3*b*tan(1/2*d*x + 1/2*c)^3 + 2160*A*a^2*b^2*tan(1/2*d*x + 1/2*c)^3 + 5280*B*a^2*b^2*tan(1/2*d*x + 1/2*c)^3
+ 1260*C*a^2*b^2*tan(1/2*d*x + 1/2*c)^3 + 3520*A*a*b^3*tan(1/2*d*x + 1/2*c)^3 + 840*B*a*b^3*tan(1/2*d*x + 1/2*
c)^3 + 2240*C*a*b^3*tan(1/2*d*x + 1/2*c)^3 + 210*A*b^4*tan(1/2*d*x + 1/2*c)^3 + 560*B*b^4*tan(1/2*d*x + 1/2*c)
^3 - 25*C*b^4*tan(1/2*d*x + 1/2*c)^3 - 240*B*a^4*tan(1/2*d*x + 1/2*c) - 120*C*a^4*tan(1/2*d*x + 1/2*c) - 960*A
*a^3*b*tan(1/2*d*x + 1/2*c) - 480*B*a^3*b*tan(1/2*d*x + 1/2*c) - 960*C*a^3*b*tan(1/2*d*x + 1/2*c) - 720*A*a^2*
b^2*tan(1/2*d*x + 1/2*c) - 1440*B*a^2*b^2*tan(1/2*d*x + 1/2*c) - 900*C*a^2*b^2*tan(1/2*d*x + 1/2*c) - 960*A*a*
b^3*tan(1/2*d*x + 1/2*c) - 600*B*a*b^3*tan(1/2*d*x + 1/2*c) - 960*C*a*b^3*tan(1/2*d*x + 1/2*c) - 150*A*b^4*tan
(1/2*d*x + 1/2*c) - 240*B*b^4*tan(1/2*d*x + 1/2*c) - 165*C*b^4*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 -
1)^6)/d

________________________________________________________________________________________

maple [B]  time = 1.77, size = 745, normalized size = 1.94 \[ \frac {a^{4} B \tan \left (d x +c \right )}{d}+\frac {8 B \,b^{4} \tan \left (d x +c \right )}{15 d}+\frac {3 A \,b^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d}+\frac {5 C \,b^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16 d}+\frac {a^{4} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {A \,a^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {A \,b^{4} \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )}{4 d}+\frac {C \,b^{4} \tan \left (d x +c \right ) \left (\sec ^{5}\left (d x +c \right )\right )}{6 d}+\frac {5 C \,b^{4} \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )}{24 d}+\frac {a^{4} C \sec \left (d x +c \right ) \tan \left (d x +c \right )}{2 d}+\frac {3 A \,a^{2} b^{2} \sec \left (d x +c \right ) \tan \left (d x +c \right )}{d}+\frac {4 A \,a^{3} b \tan \left (d x +c \right )}{d}+\frac {2 B \,a^{3} b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {3 A \,a^{2} b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {4 a^{2} b^{2} B \tan \left (d x +c \right )}{d}+\frac {32 C a \,b^{3} \tan \left (d x +c \right )}{15 d}+\frac {8 a A \,b^{3} \tan \left (d x +c \right )}{3 d}+\frac {9 C \,a^{2} b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{4 d}+\frac {3 B a \,b^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {3 A \,b^{4} \sec \left (d x +c \right ) \tan \left (d x +c \right )}{8 d}+\frac {5 C \,b^{4} \sec \left (d x +c \right ) \tan \left (d x +c \right )}{16 d}+\frac {8 a^{3} b C \tan \left (d x +c \right )}{3 d}+\frac {B \,b^{4} \tan \left (d x +c \right ) \left (\sec ^{4}\left (d x +c \right )\right )}{5 d}+\frac {9 C \,a^{2} b^{2} \sec \left (d x +c \right ) \tan \left (d x +c \right )}{4 d}+\frac {3 B a \,b^{3} \sec \left (d x +c \right ) \tan \left (d x +c \right )}{2 d}+\frac {2 B \,a^{3} b \sec \left (d x +c \right ) \tan \left (d x +c \right )}{d}+\frac {4 C a \,b^{3} \tan \left (d x +c \right ) \left (\sec ^{4}\left (d x +c \right )\right )}{5 d}+\frac {16 C a \,b^{3} \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{15 d}+\frac {B a \,b^{3} \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )}{d}+\frac {4 a^{3} b C \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{3 d}+\frac {4 a A \,b^{3} \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{3 d}+\frac {3 C \,a^{2} b^{2} \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )}{2 d}+\frac {2 a^{2} b^{2} B \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{d}+\frac {4 B \,b^{4} \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{15 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(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^4*B*tan(d*x+c)+8/15/d*B*b^4*tan(d*x+c)+3/8/d*A*b^4*ln(sec(d*x+c)+tan(d*x+c))+5/16/d*C*b^4*ln(sec(d*x+c)+
tan(d*x+c))+1/2/d*a^4*C*ln(sec(d*x+c)+tan(d*x+c))+1/d*A*a^4*ln(sec(d*x+c)+tan(d*x+c))+1/2/d*a^4*C*sec(d*x+c)*t
an(d*x+c)+3/d*A*a^2*b^2*sec(d*x+c)*tan(d*x+c)+4/5/d*C*a*b^3*tan(d*x+c)*sec(d*x+c)^4+4/d*A*a^3*b*tan(d*x+c)+1/5
/d*B*b^4*tan(d*x+c)*sec(d*x+c)^4+4/15/d*B*b^4*tan(d*x+c)*sec(d*x+c)^2+2/d*B*a^3*b*ln(sec(d*x+c)+tan(d*x+c))+3/
d*A*a^2*b^2*ln(sec(d*x+c)+tan(d*x+c))+4/d*a^2*b^2*B*tan(d*x+c)+32/15/d*C*a*b^3*tan(d*x+c)+8/3/d*a*A*b^3*tan(d*
x+c)+9/4/d*C*a^2*b^2*ln(sec(d*x+c)+tan(d*x+c))+3/2/d*B*a*b^3*ln(sec(d*x+c)+tan(d*x+c))+1/4/d*A*b^4*tan(d*x+c)*
sec(d*x+c)^3+3/8/d*A*b^4*sec(d*x+c)*tan(d*x+c)+1/6/d*C*b^4*tan(d*x+c)*sec(d*x+c)^5+5/24/d*C*b^4*tan(d*x+c)*sec
(d*x+c)^3+5/16/d*C*b^4*sec(d*x+c)*tan(d*x+c)+8/3/d*a^3*b*C*tan(d*x+c)+16/15/d*C*a*b^3*tan(d*x+c)*sec(d*x+c)^2+
9/4/d*C*a^2*b^2*sec(d*x+c)*tan(d*x+c)+1/d*B*a*b^3*tan(d*x+c)*sec(d*x+c)^3+3/2/d*B*a*b^3*sec(d*x+c)*tan(d*x+c)+
2/d*B*a^3*b*sec(d*x+c)*tan(d*x+c)+4/3/d*a^3*b*C*tan(d*x+c)*sec(d*x+c)^2+4/3/d*a*A*b^3*tan(d*x+c)*sec(d*x+c)^2+
3/2/d*C*a^2*b^2*tan(d*x+c)*sec(d*x+c)^3+2/d*a^2*b^2*B*tan(d*x+c)*sec(d*x+c)^2

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maxima [A]  time = 0.38, size = 653, normalized size = 1.70 \[ \frac {640 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a^{3} b + 960 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a^{2} b^{2} + 640 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a b^{3} + 128 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} C a b^{3} + 32 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} B b^{4} - 5 \, C b^{4} {\left (\frac {2 \, {\left (15 \, \sin \left (d x + c\right )^{5} - 40 \, \sin \left (d x + c\right )^{3} + 33 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{6} - 3 \, \sin \left (d x + c\right )^{4} + 3 \, \sin \left (d x + c\right )^{2} - 1} - 15 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 15 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 180 \, C a^{2} b^{2} {\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 )} - 120 \, B a b^{3} {\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 )} - 30 \, A 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 )} - 120 \, C a^{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 )} - 480 \, B a^{3} b {\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 )} - 720 \, A 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 )} + 480 \, A a^{4} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 480 \, B a^{4} \tan \left (d x + c\right ) + 1920 \, A a^{3} b \tan \left (d x + c\right )}{480 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(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]

1/480*(640*(tan(d*x + c)^3 + 3*tan(d*x + c))*C*a^3*b + 960*(tan(d*x + c)^3 + 3*tan(d*x + c))*B*a^2*b^2 + 640*(
tan(d*x + c)^3 + 3*tan(d*x + c))*A*a*b^3 + 128*(3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c))*C*a*b^
3 + 32*(3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c))*B*b^4 - 5*C*b^4*(2*(15*sin(d*x + c)^5 - 40*sin
(d*x + c)^3 + 33*sin(d*x + c))/(sin(d*x + c)^6 - 3*sin(d*x + c)^4 + 3*sin(d*x + c)^2 - 1) - 15*log(sin(d*x + c
) + 1) + 15*log(sin(d*x + c) - 1)) - 180*C*a^2*b^2*(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)) - 120*B*a*b^3*(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)) -
30*A*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)) - 120*C*a^4*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log
(sin(d*x + c) - 1)) - 480*B*a^3*b*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x +
c) - 1)) - 720*A*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)
) + 480*A*a^4*log(sec(d*x + c) + tan(d*x + c)) + 480*B*a^4*tan(d*x + c) + 1920*A*a^3*b*tan(d*x + c))/d

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mupad [B]  time = 6.32, size = 942, normalized size = 2.45 \[ \frac {\left (\frac {5\,A\,b^4}{4}-2\,B\,a^4-2\,B\,b^4+C\,a^4+\frac {11\,C\,b^4}{8}+6\,A\,a^2\,b^2-12\,B\,a^2\,b^2+\frac {15\,C\,a^2\,b^2}{2}-8\,A\,a\,b^3-8\,A\,a^3\,b+5\,B\,a\,b^3+4\,B\,a^3\,b-8\,C\,a\,b^3-8\,C\,a^3\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}+\left (10\,B\,a^4-\frac {7\,A\,b^4}{4}+\frac {14\,B\,b^4}{3}-3\,C\,a^4+\frac {5\,C\,b^4}{24}-18\,A\,a^2\,b^2+44\,B\,a^2\,b^2-\frac {21\,C\,a^2\,b^2}{2}+\frac {88\,A\,a\,b^3}{3}+40\,A\,a^3\,b-7\,B\,a\,b^3-12\,B\,a^3\,b+\frac {56\,C\,a\,b^3}{3}+\frac {88\,C\,a^3\,b}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (\frac {A\,b^4}{2}-20\,B\,a^4-\frac {52\,B\,b^4}{5}+2\,C\,a^4+\frac {15\,C\,b^4}{4}+12\,A\,a^2\,b^2-72\,B\,a^2\,b^2+3\,C\,a^2\,b^2-48\,A\,a\,b^3-80\,A\,a^3\,b+2\,B\,a\,b^3+8\,B\,a^3\,b-\frac {208\,C\,a\,b^3}{5}-48\,C\,a^3\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {A\,b^4}{2}+20\,B\,a^4+\frac {52\,B\,b^4}{5}+2\,C\,a^4+\frac {15\,C\,b^4}{4}+12\,A\,a^2\,b^2+72\,B\,a^2\,b^2+3\,C\,a^2\,b^2+48\,A\,a\,b^3+80\,A\,a^3\,b+2\,B\,a\,b^3+8\,B\,a^3\,b+\frac {208\,C\,a\,b^3}{5}+48\,C\,a^3\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {5\,C\,b^4}{24}-10\,B\,a^4-\frac {14\,B\,b^4}{3}-3\,C\,a^4-\frac {7\,A\,b^4}{4}-18\,A\,a^2\,b^2-44\,B\,a^2\,b^2-\frac {21\,C\,a^2\,b^2}{2}-\frac {88\,A\,a\,b^3}{3}-40\,A\,a^3\,b-7\,B\,a\,b^3-12\,B\,a^3\,b-\frac {56\,C\,a\,b^3}{3}-\frac {88\,C\,a^3\,b}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (\frac {5\,A\,b^4}{4}+2\,B\,a^4+2\,B\,b^4+C\,a^4+\frac {11\,C\,b^4}{8}+6\,A\,a^2\,b^2+12\,B\,a^2\,b^2+\frac {15\,C\,a^2\,b^2}{2}+8\,A\,a\,b^3+8\,A\,a^3\,b+5\,B\,a\,b^3+4\,B\,a^3\,b+8\,C\,a\,b^3+8\,C\,a^3\,b\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}-6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-20\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {\mathrm {atanh}\left (\frac {4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (A\,a^4+\frac {3\,A\,b^4}{8}+\frac {C\,a^4}{2}+\frac {5\,C\,b^4}{16}+3\,A\,a^2\,b^2+\frac {9\,C\,a^2\,b^2}{4}+\frac {3\,B\,a\,b^3}{2}+2\,B\,a^3\,b\right )}{4\,A\,a^4+\frac {3\,A\,b^4}{2}+2\,C\,a^4+\frac {5\,C\,b^4}{4}+12\,A\,a^2\,b^2+9\,C\,a^2\,b^2+6\,B\,a\,b^3+8\,B\,a^3\,b}\right )\,\left (2\,A\,a^4+\frac {3\,A\,b^4}{4}+C\,a^4+\frac {5\,C\,b^4}{8}+6\,A\,a^2\,b^2+\frac {9\,C\,a^2\,b^2}{2}+3\,B\,a\,b^3+4\,B\,a^3\,b\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(tan(c/2 + (d*x)/2)*((5*A*b^4)/4 + 2*B*a^4 + 2*B*b^4 + C*a^4 + (11*C*b^4)/8 + 6*A*a^2*b^2 + 12*B*a^2*b^2 + (15
*C*a^2*b^2)/2 + 8*A*a*b^3 + 8*A*a^3*b + 5*B*a*b^3 + 4*B*a^3*b + 8*C*a*b^3 + 8*C*a^3*b) + tan(c/2 + (d*x)/2)^11
*((5*A*b^4)/4 - 2*B*a^4 - 2*B*b^4 + C*a^4 + (11*C*b^4)/8 + 6*A*a^2*b^2 - 12*B*a^2*b^2 + (15*C*a^2*b^2)/2 - 8*A
*a*b^3 - 8*A*a^3*b + 5*B*a*b^3 + 4*B*a^3*b - 8*C*a*b^3 - 8*C*a^3*b) - tan(c/2 + (d*x)/2)^3*((7*A*b^4)/4 + 10*B
*a^4 + (14*B*b^4)/3 + 3*C*a^4 - (5*C*b^4)/24 + 18*A*a^2*b^2 + 44*B*a^2*b^2 + (21*C*a^2*b^2)/2 + (88*A*a*b^3)/3
+ 40*A*a^3*b + 7*B*a*b^3 + 12*B*a^3*b + (56*C*a*b^3)/3 + (88*C*a^3*b)/3) + tan(c/2 + (d*x)/2)^9*(10*B*a^4 - (
7*A*b^4)/4 + (14*B*b^4)/3 - 3*C*a^4 + (5*C*b^4)/24 - 18*A*a^2*b^2 + 44*B*a^2*b^2 - (21*C*a^2*b^2)/2 + (88*A*a*
b^3)/3 + 40*A*a^3*b - 7*B*a*b^3 - 12*B*a^3*b + (56*C*a*b^3)/3 + (88*C*a^3*b)/3) + tan(c/2 + (d*x)/2)^5*((A*b^4
)/2 + 20*B*a^4 + (52*B*b^4)/5 + 2*C*a^4 + (15*C*b^4)/4 + 12*A*a^2*b^2 + 72*B*a^2*b^2 + 3*C*a^2*b^2 + 48*A*a*b^
3 + 80*A*a^3*b + 2*B*a*b^3 + 8*B*a^3*b + (208*C*a*b^3)/5 + 48*C*a^3*b) + tan(c/2 + (d*x)/2)^7*((A*b^4)/2 - 20*
B*a^4 - (52*B*b^4)/5 + 2*C*a^4 + (15*C*b^4)/4 + 12*A*a^2*b^2 - 72*B*a^2*b^2 + 3*C*a^2*b^2 - 48*A*a*b^3 - 80*A*
a^3*b + 2*B*a*b^3 + 8*B*a^3*b - (208*C*a*b^3)/5 - 48*C*a^3*b))/(d*(15*tan(c/2 + (d*x)/2)^4 - 6*tan(c/2 + (d*x)
/2)^2 - 20*tan(c/2 + (d*x)/2)^6 + 15*tan(c/2 + (d*x)/2)^8 - 6*tan(c/2 + (d*x)/2)^10 + tan(c/2 + (d*x)/2)^12 +
1)) + (atanh((4*tan(c/2 + (d*x)/2)*(A*a^4 + (3*A*b^4)/8 + (C*a^4)/2 + (5*C*b^4)/16 + 3*A*a^2*b^2 + (9*C*a^2*b^
2)/4 + (3*B*a*b^3)/2 + 2*B*a^3*b))/(4*A*a^4 + (3*A*b^4)/2 + 2*C*a^4 + (5*C*b^4)/4 + 12*A*a^2*b^2 + 9*C*a^2*b^2
+ 6*B*a*b^3 + 8*B*a^3*b))*(2*A*a^4 + (3*A*b^4)/4 + C*a^4 + (5*C*b^4)/8 + 6*A*a^2*b^2 + (9*C*a^2*b^2)/2 + 3*B*
a*b^3 + 4*B*a^3*b))/d

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \sec {\left (c + d x \right )}\right )^{4} \left (A + B \sec {\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right ) \sec {\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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