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

Optimal. Leaf size=233 \[ \frac {\tan (c+d x) \left (5 a^2 (3 A+2 C)+20 a b B+2 b^2 (5 A+4 C)\right )}{15 d}+\frac {\left (4 a^2 B+8 a A b+6 a b C+3 b^2 B\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {\tan (c+d x) \sec ^2(c+d x) \left (2 a^2 C+10 a b B+5 A b^2+4 b^2 C\right )}{15 d}+\frac {\tan (c+d x) \sec (c+d x) \left (4 a^2 B+8 a A b+6 a b C+3 b^2 B\right )}{8 d}+\frac {b (2 a C+5 b B) \tan (c+d x) \sec ^3(c+d x)}{20 d}+\frac {C \tan (c+d x) \sec ^2(c+d x) (a+b \sec (c+d x))^2}{5 d} \]

[Out]

1/8*(8*A*a*b+4*B*a^2+3*B*b^2+6*C*a*b)*arctanh(sin(d*x+c))/d+1/15*(20*a*b*B+5*a^2*(3*A+2*C)+2*b^2*(5*A+4*C))*ta
n(d*x+c)/d+1/8*(8*A*a*b+4*B*a^2+3*B*b^2+6*C*a*b)*sec(d*x+c)*tan(d*x+c)/d+1/15*(5*A*b^2+10*B*a*b+2*C*a^2+4*C*b^
2)*sec(d*x+c)^2*tan(d*x+c)/d+1/20*b*(5*B*b+2*C*a)*sec(d*x+c)^3*tan(d*x+c)/d+1/5*C*sec(d*x+c)^2*(a+b*sec(d*x+c)
)^2*tan(d*x+c)/d

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Rubi [A]  time = 0.59, antiderivative size = 281, normalized size of antiderivative = 1.21, number of steps used = 8, number of rules used = 8, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.195, Rules used = {4092, 4082, 4002, 3997, 3787, 3770, 3767, 8} \[ -\frac {\tan (c+d x) \left (-4 a^2 b^2 (5 A+3 C)+5 a^3 b B-2 a^4 C-40 a b^3 B-4 b^4 (5 A+4 C)\right )}{30 b^2 d}+\frac {\left (4 a^2 B+8 a A b+6 a b C+3 b^2 B\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {\tan (c+d x) \left (2 a^2 C-5 a b B+20 A b^2+16 b^2 C\right ) (a+b \sec (c+d x))^2}{60 b^2 d}-\frac {\tan (c+d x) \sec (c+d x) \left (10 a^2 b B-4 a^3 C-2 a b^2 (20 A+13 C)-45 b^3 B\right )}{120 b d}+\frac {(5 b B-2 a C) \tan (c+d x) (a+b \sec (c+d x))^3}{20 b^2 d}+\frac {C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^3}{5 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((8*a*A*b + 4*a^2*B + 3*b^2*B + 6*a*b*C)*ArcTanh[Sin[c + d*x]])/(8*d) - ((5*a^3*b*B - 40*a*b^3*B - 2*a^4*C - 4
*a^2*b^2*(5*A + 3*C) - 4*b^4*(5*A + 4*C))*Tan[c + d*x])/(30*b^2*d) - ((10*a^2*b*B - 45*b^3*B - 4*a^3*C - 2*a*b
^2*(20*A + 13*C))*Sec[c + d*x]*Tan[c + d*x])/(120*b*d) + ((20*A*b^2 - 5*a*b*B + 2*a^2*C + 16*b^2*C)*(a + b*Sec
[c + d*x])^2*Tan[c + d*x])/(60*b^2*d) + ((5*b*B - 2*a*C)*(a + b*Sec[c + d*x])^3*Tan[c + d*x])/(20*b^2*d) + (C*
Sec[c + d*x]*(a + b*Sec[c + d*x])^3*Tan[c + d*x])/(5*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]

Rule 4092

Int[csc[(e_.) + (f_.)*(x_)]^2*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(
e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Simp[(C*Csc[e + f*x]*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m
+ 1))/(b*f*(m + 3)), x] + Dist[1/(b*(m + 3)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*Simp[a*C + b*(C*(m + 2)
+ A*(m + 3))*Csc[e + f*x] - (2*a*C - b*B*(m + 3))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m
}, x] && NeQ[a^2 - b^2, 0] &&  !LtQ[m, -1]

Rubi steps

\begin {align*} \int \sec ^2(c+d x) (a+b \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\frac {C \sec (c+d x) (a+b \sec (c+d x))^3 \tan (c+d x)}{5 b d}+\frac {\int \sec (c+d x) (a+b \sec (c+d x))^2 \left (a C+b (5 A+4 C) \sec (c+d x)+(5 b B-2 a C) \sec ^2(c+d x)\right ) \, dx}{5 b}\\ &=\frac {(5 b B-2 a C) (a+b \sec (c+d x))^3 \tan (c+d x)}{20 b^2 d}+\frac {C \sec (c+d x) (a+b \sec (c+d x))^3 \tan (c+d x)}{5 b d}+\frac {\int \sec (c+d x) (a+b \sec (c+d x))^2 \left (b (15 b B-2 a C)+\left (20 A b^2-5 a b B+2 a^2 C+16 b^2 C\right ) \sec (c+d x)\right ) \, dx}{20 b^2}\\ &=\frac {\left (20 A b^2-5 a b B+2 a^2 C+16 b^2 C\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{60 b^2 d}+\frac {(5 b B-2 a C) (a+b \sec (c+d x))^3 \tan (c+d x)}{20 b^2 d}+\frac {C \sec (c+d x) (a+b \sec (c+d x))^3 \tan (c+d x)}{5 b d}+\frac {\int \sec (c+d x) (a+b \sec (c+d x)) \left (b \left (40 A b^2+35 a b B-2 a^2 C+32 b^2 C\right )-\left (10 a^2 b B-45 b^3 B-4 a^3 C-2 a b^2 (20 A+13 C)\right ) \sec (c+d x)\right ) \, dx}{60 b^2}\\ &=-\frac {\left (10 a^2 b B-45 b^3 B-4 a^3 C-2 a b^2 (20 A+13 C)\right ) \sec (c+d x) \tan (c+d x)}{120 b d}+\frac {\left (20 A b^2-5 a b B+2 a^2 C+16 b^2 C\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{60 b^2 d}+\frac {(5 b B-2 a C) (a+b \sec (c+d x))^3 \tan (c+d x)}{20 b^2 d}+\frac {C \sec (c+d x) (a+b \sec (c+d x))^3 \tan (c+d x)}{5 b d}+\frac {\int \sec (c+d x) \left (15 b^2 \left (8 a A b+4 a^2 B+3 b^2 B+6 a b C\right )-4 \left (5 a^3 b B-40 a b^3 B-2 a^4 C-4 a^2 b^2 (5 A+3 C)-4 b^4 (5 A+4 C)\right ) \sec (c+d x)\right ) \, dx}{120 b^2}\\ &=-\frac {\left (10 a^2 b B-45 b^3 B-4 a^3 C-2 a b^2 (20 A+13 C)\right ) \sec (c+d x) \tan (c+d x)}{120 b d}+\frac {\left (20 A b^2-5 a b B+2 a^2 C+16 b^2 C\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{60 b^2 d}+\frac {(5 b B-2 a C) (a+b \sec (c+d x))^3 \tan (c+d x)}{20 b^2 d}+\frac {C \sec (c+d x) (a+b \sec (c+d x))^3 \tan (c+d x)}{5 b d}+\frac {1}{8} \left (8 a A b+4 a^2 B+3 b^2 B+6 a b C\right ) \int \sec (c+d x) \, dx-\frac {\left (5 a^3 b B-40 a b^3 B-2 a^4 C-4 a^2 b^2 (5 A+3 C)-4 b^4 (5 A+4 C)\right ) \int \sec ^2(c+d x) \, dx}{30 b^2}\\ &=\frac {\left (8 a A b+4 a^2 B+3 b^2 B+6 a b C\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}-\frac {\left (10 a^2 b B-45 b^3 B-4 a^3 C-2 a b^2 (20 A+13 C)\right ) \sec (c+d x) \tan (c+d x)}{120 b d}+\frac {\left (20 A b^2-5 a b B+2 a^2 C+16 b^2 C\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{60 b^2 d}+\frac {(5 b B-2 a C) (a+b \sec (c+d x))^3 \tan (c+d x)}{20 b^2 d}+\frac {C \sec (c+d x) (a+b \sec (c+d x))^3 \tan (c+d x)}{5 b d}+\frac {\left (5 a^3 b B-40 a b^3 B-2 a^4 C-4 a^2 b^2 (5 A+3 C)-4 b^4 (5 A+4 C)\right ) \operatorname {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{30 b^2 d}\\ &=\frac {\left (8 a A b+4 a^2 B+3 b^2 B+6 a b C\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}-\frac {\left (5 a^3 b B-40 a b^3 B-2 a^4 C-4 a^2 b^2 (5 A+3 C)-4 b^4 (5 A+4 C)\right ) \tan (c+d x)}{30 b^2 d}-\frac {\left (10 a^2 b B-45 b^3 B-4 a^3 C-2 a b^2 (20 A+13 C)\right ) \sec (c+d x) \tan (c+d x)}{120 b d}+\frac {\left (20 A b^2-5 a b B+2 a^2 C+16 b^2 C\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{60 b^2 d}+\frac {(5 b B-2 a C) (a+b \sec (c+d x))^3 \tan (c+d x)}{20 b^2 d}+\frac {C \sec (c+d x) (a+b \sec (c+d x))^3 \tan (c+d x)}{5 b d}\\ \end {align*}

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Mathematica [A]  time = 2.15, size = 371, normalized size = 1.59 \[ \frac {\sec ^5(c+d x) \left (A \cos ^2(c+d x)+B \cos (c+d x)+C\right ) \left (2 \sin (c+d x) \left (15 \cos (c+d x) \left (12 a^2 B+24 a A b+34 a b C+17 b^2 B\right )+48 \cos (2 (c+d x)) \left (5 a^2 (A+C)+10 a b B+b^2 (5 A+4 C)\right )+60 a^2 A \cos (4 (c+d x))+180 a^2 A+60 a^2 B \cos (3 (c+d x))+40 a^2 C \cos (4 (c+d x))+200 a^2 C+120 a A b \cos (3 (c+d x))+80 a b B \cos (4 (c+d x))+400 a b B+90 a b C \cos (3 (c+d x))+40 A b^2 \cos (4 (c+d x))+200 A b^2+45 b^2 B \cos (3 (c+d x))+32 b^2 C \cos (4 (c+d x))+256 b^2 C\right )-120 \cos ^5(c+d x) \left (4 a^2 B+8 a A b+6 a b C+3 b^2 B\right ) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )\right )}{480 d (A \cos (2 (c+d x))+A+2 B \cos (c+d x)+2 C)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((C + B*Cos[c + d*x] + A*Cos[c + d*x]^2)*Sec[c + d*x]^5*(-120*(8*a*A*b + 4*a^2*B + 3*b^2*B + 6*a*b*C)*Cos[c +
d*x]^5*(Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) + 2*(180*a^2*A +
200*A*b^2 + 400*a*b*B + 200*a^2*C + 256*b^2*C + 15*(24*a*A*b + 12*a^2*B + 17*b^2*B + 34*a*b*C)*Cos[c + d*x] +
48*(10*a*b*B + 5*a^2*(A + C) + b^2*(5*A + 4*C))*Cos[2*(c + d*x)] + 120*a*A*b*Cos[3*(c + d*x)] + 60*a^2*B*Cos[3
*(c + d*x)] + 45*b^2*B*Cos[3*(c + d*x)] + 90*a*b*C*Cos[3*(c + d*x)] + 60*a^2*A*Cos[4*(c + d*x)] + 40*A*b^2*Cos
[4*(c + d*x)] + 80*a*b*B*Cos[4*(c + d*x)] + 40*a^2*C*Cos[4*(c + d*x)] + 32*b^2*C*Cos[4*(c + d*x)])*Sin[c + d*x
]))/(480*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*(c + d*x)]))

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fricas [A]  time = 0.93, size = 243, normalized size = 1.04 \[ \frac {15 \, {\left (4 \, B a^{2} + 2 \, {\left (4 \, A + 3 \, C\right )} a b + 3 \, B b^{2}\right )} \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (4 \, B a^{2} + 2 \, {\left (4 \, A + 3 \, C\right )} a b + 3 \, B b^{2}\right )} \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (8 \, {\left (5 \, {\left (3 \, A + 2 \, C\right )} a^{2} + 20 \, B a b + 2 \, {\left (5 \, A + 4 \, C\right )} b^{2}\right )} \cos \left (d x + c\right )^{4} + 15 \, {\left (4 \, B a^{2} + 2 \, {\left (4 \, A + 3 \, C\right )} a b + 3 \, B b^{2}\right )} \cos \left (d x + c\right )^{3} + 24 \, C b^{2} + 8 \, {\left (5 \, C a^{2} + 10 \, B a b + {\left (5 \, A + 4 \, C\right )} b^{2}\right )} \cos \left (d x + c\right )^{2} + 30 \, {\left (2 \, C a b + B b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{240 \, d \cos \left (d x + c\right )^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/240*(15*(4*B*a^2 + 2*(4*A + 3*C)*a*b + 3*B*b^2)*cos(d*x + c)^5*log(sin(d*x + c) + 1) - 15*(4*B*a^2 + 2*(4*A
+ 3*C)*a*b + 3*B*b^2)*cos(d*x + c)^5*log(-sin(d*x + c) + 1) + 2*(8*(5*(3*A + 2*C)*a^2 + 20*B*a*b + 2*(5*A + 4*
C)*b^2)*cos(d*x + c)^4 + 15*(4*B*a^2 + 2*(4*A + 3*C)*a*b + 3*B*b^2)*cos(d*x + c)^3 + 24*C*b^2 + 8*(5*C*a^2 + 1
0*B*a*b + (5*A + 4*C)*b^2)*cos(d*x + c)^2 + 30*(2*C*a*b + B*b^2)*cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^5
)

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giac [B]  time = 0.32, size = 766, normalized size = 3.29 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/120*(15*(4*B*a^2 + 8*A*a*b + 6*C*a*b + 3*B*b^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 15*(4*B*a^2 + 8*A*a*b +
6*C*a*b + 3*B*b^2)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(120*A*a^2*tan(1/2*d*x + 1/2*c)^9 - 60*B*a^2*tan(1/
2*d*x + 1/2*c)^9 + 120*C*a^2*tan(1/2*d*x + 1/2*c)^9 - 120*A*a*b*tan(1/2*d*x + 1/2*c)^9 + 240*B*a*b*tan(1/2*d*x
+ 1/2*c)^9 - 150*C*a*b*tan(1/2*d*x + 1/2*c)^9 + 120*A*b^2*tan(1/2*d*x + 1/2*c)^9 - 75*B*b^2*tan(1/2*d*x + 1/2
*c)^9 + 120*C*b^2*tan(1/2*d*x + 1/2*c)^9 - 480*A*a^2*tan(1/2*d*x + 1/2*c)^7 + 120*B*a^2*tan(1/2*d*x + 1/2*c)^7
- 320*C*a^2*tan(1/2*d*x + 1/2*c)^7 + 240*A*a*b*tan(1/2*d*x + 1/2*c)^7 - 640*B*a*b*tan(1/2*d*x + 1/2*c)^7 + 60
*C*a*b*tan(1/2*d*x + 1/2*c)^7 - 320*A*b^2*tan(1/2*d*x + 1/2*c)^7 + 30*B*b^2*tan(1/2*d*x + 1/2*c)^7 - 160*C*b^2
*tan(1/2*d*x + 1/2*c)^7 + 720*A*a^2*tan(1/2*d*x + 1/2*c)^5 + 400*C*a^2*tan(1/2*d*x + 1/2*c)^5 + 800*B*a*b*tan(
1/2*d*x + 1/2*c)^5 + 400*A*b^2*tan(1/2*d*x + 1/2*c)^5 + 464*C*b^2*tan(1/2*d*x + 1/2*c)^5 - 480*A*a^2*tan(1/2*d
*x + 1/2*c)^3 - 120*B*a^2*tan(1/2*d*x + 1/2*c)^3 - 320*C*a^2*tan(1/2*d*x + 1/2*c)^3 - 240*A*a*b*tan(1/2*d*x +
1/2*c)^3 - 640*B*a*b*tan(1/2*d*x + 1/2*c)^3 - 60*C*a*b*tan(1/2*d*x + 1/2*c)^3 - 320*A*b^2*tan(1/2*d*x + 1/2*c)
^3 - 30*B*b^2*tan(1/2*d*x + 1/2*c)^3 - 160*C*b^2*tan(1/2*d*x + 1/2*c)^3 + 120*A*a^2*tan(1/2*d*x + 1/2*c) + 60*
B*a^2*tan(1/2*d*x + 1/2*c) + 120*C*a^2*tan(1/2*d*x + 1/2*c) + 120*A*a*b*tan(1/2*d*x + 1/2*c) + 240*B*a*b*tan(1
/2*d*x + 1/2*c) + 150*C*a*b*tan(1/2*d*x + 1/2*c) + 120*A*b^2*tan(1/2*d*x + 1/2*c) + 75*B*b^2*tan(1/2*d*x + 1/2
*c) + 120*C*b^2*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^5)/d

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maple [A]  time = 1.55, size = 404, normalized size = 1.73 \[ \frac {a^{2} A \tan \left (d x +c \right )}{d}+\frac {a^{2} B \sec \left (d x +c \right ) \tan \left (d x +c \right )}{2 d}+\frac {B \,a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {2 a^{2} C \tan \left (d x +c \right )}{3 d}+\frac {a^{2} C \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{3 d}+\frac {a A b \sec \left (d x +c \right ) \tan \left (d x +c \right )}{d}+\frac {A a b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {4 B a b \tan \left (d x +c \right )}{3 d}+\frac {2 B a b \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{3 d}+\frac {C a b \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )}{2 d}+\frac {3 a b C \sec \left (d x +c \right ) \tan \left (d x +c \right )}{4 d}+\frac {3 C a b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{4 d}+\frac {2 A \,b^{2} \tan \left (d x +c \right )}{3 d}+\frac {A \,b^{2} \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{3 d}+\frac {b^{2} B \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )}{4 d}+\frac {3 b^{2} B \sec \left (d x +c \right ) \tan \left (d x +c \right )}{8 d}+\frac {3 b^{2} B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d}+\frac {8 b^{2} C \tan \left (d x +c \right )}{15 d}+\frac {b^{2} C \tan \left (d x +c \right ) \left (\sec ^{4}\left (d x +c \right )\right )}{5 d}+\frac {4 b^{2} C \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{15 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^2*A*tan(d*x+c)/d+1/2*a^2*B*sec(d*x+c)*tan(d*x+c)/d+1/2/d*B*a^2*ln(sec(d*x+c)+tan(d*x+c))+2/3/d*a^2*C*tan(d*x
+c)+1/3/d*a^2*C*tan(d*x+c)*sec(d*x+c)^2+a*A*b*sec(d*x+c)*tan(d*x+c)/d+1/d*A*a*b*ln(sec(d*x+c)+tan(d*x+c))+4/3/
d*B*a*b*tan(d*x+c)+2/3/d*B*a*b*tan(d*x+c)*sec(d*x+c)^2+1/2/d*C*a*b*tan(d*x+c)*sec(d*x+c)^3+3/4*a*b*C*sec(d*x+c
)*tan(d*x+c)/d+3/4/d*C*a*b*ln(sec(d*x+c)+tan(d*x+c))+2/3/d*A*b^2*tan(d*x+c)+1/3/d*A*b^2*tan(d*x+c)*sec(d*x+c)^
2+1/4/d*b^2*B*tan(d*x+c)*sec(d*x+c)^3+3/8/d*b^2*B*sec(d*x+c)*tan(d*x+c)+3/8/d*b^2*B*ln(sec(d*x+c)+tan(d*x+c))+
8/15*b^2*C*tan(d*x+c)/d+1/5/d*b^2*C*tan(d*x+c)*sec(d*x+c)^4+4/15/d*b^2*C*tan(d*x+c)*sec(d*x+c)^2

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maxima [A]  time = 0.36, size = 357, normalized size = 1.53 \[ \frac {80 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a^{2} + 160 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a b + 80 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A b^{2} + 16 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} C b^{2} - 30 \, C a b {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 15 \, B b^{2} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 60 \, B a^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 120 \, A a b {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 240 \, A a^{2} \tan \left (d x + c\right )}{240 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/240*(80*(tan(d*x + c)^3 + 3*tan(d*x + c))*C*a^2 + 160*(tan(d*x + c)^3 + 3*tan(d*x + c))*B*a*b + 80*(tan(d*x
+ c)^3 + 3*tan(d*x + c))*A*b^2 + 16*(3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c))*C*b^2 - 30*C*a*b*
(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*l
og(sin(d*x + c) - 1)) - 15*B*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)) - 60*B*a^2*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(s
in(d*x + c) + 1) + log(sin(d*x + c) - 1)) - 120*A*a*b*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c)
+ 1) + log(sin(d*x + c) - 1)) + 240*A*a^2*tan(d*x + c))/d

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mupad [B]  time = 7.48, size = 455, normalized size = 1.95 \[ \frac {\mathrm {atanh}\left (\frac {4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {B\,a^2}{2}+\frac {3\,B\,b^2}{8}+A\,a\,b+\frac {3\,C\,a\,b}{4}\right )}{2\,B\,a^2+\frac {3\,B\,b^2}{2}+4\,A\,a\,b+3\,C\,a\,b}\right )\,\left (B\,a^2+\frac {3\,B\,b^2}{4}+2\,A\,a\,b+\frac {3\,C\,a\,b}{2}\right )}{d}-\frac {\left (2\,A\,a^2+2\,A\,b^2-B\,a^2-\frac {5\,B\,b^2}{4}+2\,C\,a^2+2\,C\,b^2-2\,A\,a\,b+4\,B\,a\,b-\frac {5\,C\,a\,b}{2}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (2\,B\,a^2-\frac {16\,A\,b^2}{3}-8\,A\,a^2+\frac {B\,b^2}{2}-\frac {16\,C\,a^2}{3}-\frac {8\,C\,b^2}{3}+4\,A\,a\,b-\frac {32\,B\,a\,b}{3}+C\,a\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (12\,A\,a^2+\frac {20\,A\,b^2}{3}+\frac {20\,C\,a^2}{3}+\frac {116\,C\,b^2}{15}+\frac {40\,B\,a\,b}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-8\,A\,a^2-\frac {16\,A\,b^2}{3}-2\,B\,a^2-\frac {B\,b^2}{2}-\frac {16\,C\,a^2}{3}-\frac {8\,C\,b^2}{3}-4\,A\,a\,b-\frac {32\,B\,a\,b}{3}-C\,a\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (2\,A\,a^2+2\,A\,b^2+B\,a^2+\frac {5\,B\,b^2}{4}+2\,C\,a^2+2\,C\,b^2+2\,A\,a\,b+4\,B\,a\,b+\frac {5\,C\,a\,b}{2}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(atanh((4*tan(c/2 + (d*x)/2)*((B*a^2)/2 + (3*B*b^2)/8 + A*a*b + (3*C*a*b)/4))/(2*B*a^2 + (3*B*b^2)/2 + 4*A*a*b
+ 3*C*a*b))*(B*a^2 + (3*B*b^2)/4 + 2*A*a*b + (3*C*a*b)/2))/d - (tan(c/2 + (d*x)/2)*(2*A*a^2 + 2*A*b^2 + B*a^2
+ (5*B*b^2)/4 + 2*C*a^2 + 2*C*b^2 + 2*A*a*b + 4*B*a*b + (5*C*a*b)/2) + tan(c/2 + (d*x)/2)^9*(2*A*a^2 + 2*A*b^
2 - B*a^2 - (5*B*b^2)/4 + 2*C*a^2 + 2*C*b^2 - 2*A*a*b + 4*B*a*b - (5*C*a*b)/2) - tan(c/2 + (d*x)/2)^3*(8*A*a^2
+ (16*A*b^2)/3 + 2*B*a^2 + (B*b^2)/2 + (16*C*a^2)/3 + (8*C*b^2)/3 + 4*A*a*b + (32*B*a*b)/3 + C*a*b) - tan(c/2
+ (d*x)/2)^7*(8*A*a^2 + (16*A*b^2)/3 - 2*B*a^2 - (B*b^2)/2 + (16*C*a^2)/3 + (8*C*b^2)/3 - 4*A*a*b + (32*B*a*b
)/3 - C*a*b) + tan(c/2 + (d*x)/2)^5*(12*A*a^2 + (20*A*b^2)/3 + (20*C*a^2)/3 + (116*C*b^2)/15 + (40*B*a*b)/3))/
(d*(5*tan(c/2 + (d*x)/2)^2 - 10*tan(c/2 + (d*x)/2)^4 + 10*tan(c/2 + (d*x)/2)^6 - 5*tan(c/2 + (d*x)/2)^8 + tan(
c/2 + (d*x)/2)^10 - 1))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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