\(\int \sec ^2(c+d x) (a+b \sec (c+d x))^2 (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [870]

Optimal result

Integrand size = 41, antiderivative size = 233 \[ \int \sec ^2(c+d x) (a+b \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {\left (8 a A b+4 a^2 B+3 b^2 B+6 a b C\right ) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {\left (20 a b B+5 a^2 (3 A+2 C)+2 b^2 (5 A+4 C)\right ) \tan (c+d x)}{15 d}+\frac {\left (8 a A b+4 a^2 B+3 b^2 B+6 a b C\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {\left (5 A b^2+10 a b B+2 a^2 C+4 b^2 C\right ) \sec ^2(c+d x) \tan (c+d x)}{15 d}+\frac {b (5 b B+2 a C) \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac {C \sec ^2(c+d x) (a+b \sec (c+d x))^2 \tan (c+d x)}{5 d} \]

[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*B*a*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

Rubi [A] (verified)

Time = 0.66 (sec) , antiderivative size = 281, normalized size of antiderivative = 1.21, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.195, Rules used = {4177, 4167, 4087, 4082, 3872, 3855, 3852, 8} \[ \int \sec ^2(c+d x) (a+b \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {\left (4 a^2 B+8 a A b+6 a b C+3 b^2 B\right ) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {\tan (c+d x) \left (2 a^2 C-5 a b B+20 A b^2+16 b^2 C\right ) (a+b \sec (c+d x))^2}{60 b^2 d}-\frac {\tan (c+d x) \sec (c+d x) \left (-4 a^3 C+10 a^2 b B-2 a b^2 (20 A+13 C)-45 b^3 B\right )}{120 b d}-\frac {\tan (c+d x) \left (-2 a^4 C+5 a^3 b B-4 a^2 b^2 (5 A+3 C)-40 a b^3 B-4 b^4 (5 A+4 C)\right )}{30 b^2 d}+\frac {(5 b B-2 a C) \tan (c+d x) (a+b \sec (c+d x))^3}{20 b^2 d}+\frac {C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^3}{5 b d} \]

[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 3852

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 3855

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

Rule 3872

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 4082

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
*Csc[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 4087

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] /;
FreeQ[{a, b, A, B, e, f}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0] && GtQ[m, 0]

Rule 4167

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 4177

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*} \text {integral}& = \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 ) \text {arctanh}(\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 ) \text {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 ) \text {arctanh}(\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*}

Mathematica [A] (verified)

Time = 3.64 (sec) , antiderivative size = 178, normalized size of antiderivative = 0.76 \[ \int \sec ^2(c+d x) (a+b \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {15 \left (8 a A b+4 a^2 B+3 b^2 B+6 a b C\right ) \text {arctanh}(\sin (c+d x))+\left (8 \left (20 a b B+5 a^2 (3 A+2 C)+2 b^2 (5 A+4 C)\right )+15 \left (8 a A b+4 a^2 B+3 b^2 B+6 a b C\right ) \sec (c+d x)+8 \left (5 A b^2+10 a b B+5 a^2 C+4 b^2 C\right ) \sec ^2(c+d x)+30 b (b B+2 a C) \sec ^3(c+d x)+24 b^2 C \sec ^4(c+d x)\right ) \tan (c+d x)}{120 d} \]

[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]

(15*(8*a*A*b + 4*a^2*B + 3*b^2*B + 6*a*b*C)*ArcTanh[Sin[c + d*x]] + (8*(20*a*b*B + 5*a^2*(3*A + 2*C) + 2*b^2*(
5*A + 4*C)) + 15*(8*a*A*b + 4*a^2*B + 3*b^2*B + 6*a*b*C)*Sec[c + d*x] + 8*(5*A*b^2 + 10*a*b*B + 5*a^2*C + 4*b^
2*C)*Sec[c + d*x]^2 + 30*b*(b*B + 2*a*C)*Sec[c + d*x]^3 + 24*b^2*C*Sec[c + d*x]^4)*Tan[c + d*x])/(120*d)

Maple [A] (verified)

Time = 1.38 (sec) , antiderivative size = 197, normalized size of antiderivative = 0.85

method	result	size
parts	\(\frac {\left (2 a A b +B \,a^{2}\right ) \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}+\frac {\left (B \,b^{2}+2 C a b \right ) \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}-\frac {\left (A \,b^{2}+2 B a b +C \,a^{2}\right ) \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}-\frac {C \,b^{2} \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )}{d}+\frac {a^{2} A \tan \left (d x +c \right )}{d}\)	\(197\)
derivativedivides	\(\frac {a^{2} A \tan \left (d x +c \right )+B \,a^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-C \,a^{2} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+2 a A b \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-2 B a b \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+2 C a b \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-A \,b^{2} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+B \,b^{2} \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-C \,b^{2} \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )}{d}\)	\(292\)
default	\(\frac {a^{2} A \tan \left (d x +c \right )+B \,a^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-C \,a^{2} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+2 a A b \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-2 B a b \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+2 C a b \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-A \,b^{2} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+B \,b^{2} \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-C \,b^{2} \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )}{d}\)	\(292\)
parallelrisch	\(\frac {-600 \left (\frac {B \,a^{2}}{2}+b \left (A +\frac {3 C}{4}\right ) a +\frac {3 B \,b^{2}}{8}\right ) \left (\frac {\cos \left (5 d x +5 c \right )}{5}+\cos \left (3 d x +3 c \right )+2 \cos \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+600 \left (\frac {B \,a^{2}}{2}+b \left (A +\frac {3 C}{4}\right ) a +\frac {3 B \,b^{2}}{8}\right ) \left (\frac {\cos \left (5 d x +5 c \right )}{5}+\cos \left (3 d x +3 c \right )+2 \cos \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\left (\left (360 A +400 C \right ) a^{2}+800 B a b +400 b^{2} \left (A +\frac {4 C}{5}\right )\right ) \sin \left (3 d x +3 c \right )+\left (\left (120 A +80 C \right ) a^{2}+160 B a b +80 b^{2} \left (A +\frac {4 C}{5}\right )\right ) \sin \left (5 d x +5 c \right )+\left (240 B \,a^{2}+480 b \left (A +\frac {7 C}{4}\right ) a +420 B \,b^{2}\right ) \sin \left (2 d x +2 c \right )+\left (120 B \,a^{2}+240 b \left (A +\frac {3 C}{4}\right ) a +90 B \,b^{2}\right ) \sin \left (4 d x +4 c \right )+240 \left (\left (A +\frac {4 C}{3}\right ) a^{2}+\frac {8 B a b}{3}+\frac {4 b^{2} \left (A +2 C \right )}{3}\right ) \sin \left (d x +c \right )}{600 d \left (\frac {\cos \left (5 d x +5 c \right )}{5}+\cos \left (3 d x +3 c \right )+2 \cos \left (d x +c \right )\right )}\)	\(337\)
norman	\(\frac {-\frac {4 \left (45 a^{2} A +25 A \,b^{2}+50 B a b +25 C \,a^{2}+29 C \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{15 d}-\frac {\left (8 a^{2} A -8 a A b +8 A \,b^{2}-4 B \,a^{2}+16 B a b -5 B \,b^{2}+8 C \,a^{2}-10 C a b +8 C \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{4 d}-\frac {\left (8 a^{2} A +8 a A b +8 A \,b^{2}+4 B \,a^{2}+16 B a b +5 B \,b^{2}+8 C \,a^{2}+10 C a b +8 C \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {\left (48 a^{2} A -24 a A b +32 A \,b^{2}-12 B \,a^{2}+64 B a b -3 B \,b^{2}+32 C \,a^{2}-6 C a b +16 C \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{6 d}+\frac {\left (48 a^{2} A +24 a A b +32 A \,b^{2}+12 B \,a^{2}+64 B a b +3 B \,b^{2}+32 C \,a^{2}+6 C a b +16 C \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{6 d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{5}}-\frac {\left (8 a A b +4 B \,a^{2}+3 B \,b^{2}+6 C a b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}+\frac {\left (8 a A b +4 B \,a^{2}+3 B \,b^{2}+6 C a b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}\)	\(415\)
risch	\(-\frac {i \left (-45 B \,b^{2} {\mathrm e}^{i \left (d x +c \right )}-120 a^{2} A -160 B a b -80 A \,b^{2}-80 C \,a^{2}-64 C \,b^{2}-90 C b a \,{\mathrm e}^{i \left (d x +c \right )}-480 a^{2} A \,{\mathrm e}^{2 i \left (d x +c \right )}-240 A \,b^{2} {\mathrm e}^{6 i \left (d x +c \right )}-240 C \,a^{2} {\mathrm e}^{6 i \left (d x +c \right )}-720 A \,a^{2} {\mathrm e}^{4 i \left (d x +c \right )}+45 B \,b^{2} {\mathrm e}^{9 i \left (d x +c \right )}-480 A \,a^{2} {\mathrm e}^{6 i \left (d x +c \right )}+120 B \,a^{2} {\mathrm e}^{7 i \left (d x +c \right )}+210 B \,b^{2} {\mathrm e}^{7 i \left (d x +c \right )}-120 B \,a^{2} {\mathrm e}^{3 i \left (d x +c \right )}-60 a^{2} B \,{\mathrm e}^{i \left (d x +c \right )}-210 B \,b^{2} {\mathrm e}^{3 i \left (d x +c \right )}-480 B a b \,{\mathrm e}^{6 i \left (d x +c \right )}+240 A a b \,{\mathrm e}^{7 i \left (d x +c \right )}+420 C a b \,{\mathrm e}^{7 i \left (d x +c \right )}-240 a A b \,{\mathrm e}^{3 i \left (d x +c \right )}-420 C a b \,{\mathrm e}^{3 i \left (d x +c \right )}+120 a A b \,{\mathrm e}^{9 i \left (d x +c \right )}+90 C a b \,{\mathrm e}^{9 i \left (d x +c \right )}-120 a A b \,{\mathrm e}^{i \left (d x +c \right )}-1120 B a b \,{\mathrm e}^{4 i \left (d x +c \right )}-800 B a b \,{\mathrm e}^{2 i \left (d x +c \right )}+60 B \,a^{2} {\mathrm e}^{9 i \left (d x +c \right )}-120 A \,a^{2} {\mathrm e}^{8 i \left (d x +c \right )}-640 C \,b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-560 A \,b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-560 C \,a^{2} {\mathrm e}^{4 i \left (d x +c \right )}-400 A \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-400 C \,a^{2} {\mathrm e}^{2 i \left (d x +c \right )}-320 C \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}\right )}{60 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{5}}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a A b}{d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B \,a^{2}}{2 d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B \,b^{2}}{8 d}-\frac {3 a b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{4 d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a A b}{d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B \,a^{2}}{2 d}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B \,b^{2}}{8 d}+\frac {3 a b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{4 d}\)	\(676\)

[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,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 243, normalized size of antiderivative = 1.04 \[ \int \sec ^2(c+d x) (a+b \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {15 \, {\left (4 \, B a^{2} + 2 \, {\left (4 \, A + 3 \, C\right )} a b + 3 \, B b^{2}\right )} \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (4 \, B a^{2} + 2 \, {\left (4 \, A + 3 \, C\right )} a b + 3 \, B b^{2}\right )} \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (8 \, {\left (5 \, {\left (3 \, A + 2 \, C\right )} a^{2} + 20 \, B a b + 2 \, {\left (5 \, A + 4 \, C\right )} b^{2}\right )} \cos \left (d x + c\right )^{4} + 15 \, {\left (4 \, B a^{2} + 2 \, {\left (4 \, A + 3 \, C\right )} a b + 3 \, B b^{2}\right )} \cos \left (d x + c\right )^{3} + 24 \, C b^{2} + 8 \, {\left (5 \, C a^{2} + 10 \, B a b + {\left (5 \, A + 4 \, C\right )} b^{2}\right )} \cos \left (d x + c\right )^{2} + 30 \, {\left (2 \, C a b + B b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{240 \, d \cos \left (d x + c\right )^{5}} \]

[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
)

Sympy [F]

\[ \int \sec ^2(c+d x) (a+b \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int \left (a + b \sec {\left (c + d x \right )}\right )^{2} \left (A + B \sec {\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right ) \sec ^{2}{\left (c + d x \right )}\, dx \]

[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)

Maxima [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 357, normalized size of antiderivative = 1.53 \[ \int \sec ^2(c+d x) (a+b \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {80 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a^{2} + 160 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a b + 80 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A b^{2} + 16 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} C b^{2} - 30 \, C a b {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 15 \, B b^{2} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 60 \, B a^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 120 \, A a b {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 240 \, A a^{2} \tan \left (d x + c\right )}{240 \, d} \]

[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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 766 vs. \(2 (221) = 442\).

Time = 0.34 (sec) , antiderivative size = 766, normalized size of antiderivative = 3.29 \[ \int \sec ^2(c+d x) (a+b \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Too large to display} \]

[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

Mupad [B] (verification not implemented)

Time = 19.83 (sec) , antiderivative size = 455, normalized size of antiderivative = 1.95 \[ \int \sec ^2(c+d x) (a+b \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {\mathrm {atanh}\left (\frac {4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {B\,a^2}{2}+\frac {3\,B\,b^2}{8}+A\,a\,b+\frac {3\,C\,a\,b}{4}\right )}{2\,B\,a^2+\frac {3\,B\,b^2}{2}+4\,A\,a\,b+3\,C\,a\,b}\right )\,\left (B\,a^2+\frac {3\,B\,b^2}{4}+2\,A\,a\,b+\frac {3\,C\,a\,b}{2}\right )}{d}-\frac {\left (2\,A\,a^2+2\,A\,b^2-B\,a^2-\frac {5\,B\,b^2}{4}+2\,C\,a^2+2\,C\,b^2-2\,A\,a\,b+4\,B\,a\,b-\frac {5\,C\,a\,b}{2}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (2\,B\,a^2-\frac {16\,A\,b^2}{3}-8\,A\,a^2+\frac {B\,b^2}{2}-\frac {16\,C\,a^2}{3}-\frac {8\,C\,b^2}{3}+4\,A\,a\,b-\frac {32\,B\,a\,b}{3}+C\,a\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (12\,A\,a^2+\frac {20\,A\,b^2}{3}+\frac {20\,C\,a^2}{3}+\frac {116\,C\,b^2}{15}+\frac {40\,B\,a\,b}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-8\,A\,a^2-\frac {16\,A\,b^2}{3}-2\,B\,a^2-\frac {B\,b^2}{2}-\frac {16\,C\,a^2}{3}-\frac {8\,C\,b^2}{3}-4\,A\,a\,b-\frac {32\,B\,a\,b}{3}-C\,a\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (2\,A\,a^2+2\,A\,b^2+B\,a^2+\frac {5\,B\,b^2}{4}+2\,C\,a^2+2\,C\,b^2+2\,A\,a\,b+4\,B\,a\,b+\frac {5\,C\,a\,b}{2}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]

[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))