\(\int \sec ^2(c+d x) (a+b \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx\) [286]

Optimal result

Integrand size = 31, antiderivative size = 179 \[ \int \sec ^2(c+d x) (a+b \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx=\frac {\left (8 a A b+4 a^2 B+3 b^2 B\right ) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {\left (4 a^2 A b+4 A b^3-a^3 B+8 a b^2 B\right ) \tan (c+d x)}{6 b d}+\frac {\left (8 a A b-2 a^2 B+9 b^2 B\right ) \sec (c+d x) \tan (c+d x)}{24 d}+\frac {(4 A b-a B) (a+b \sec (c+d x))^2 \tan (c+d x)}{12 b d}+\frac {B (a+b \sec (c+d x))^3 \tan (c+d x)}{4 b d} \] Output:

1/8*(8*A*a*b+4*B*a^2+3*B*b^2)*arctanh(sin(d*x+c))/d+1/6*(4*A*a^2*b+4*A*b^3 
-B*a^3+8*B*a*b^2)*tan(d*x+c)/b/d+1/24*(8*A*a*b-2*B*a^2+9*B*b^2)*sec(d*x+c) 
*tan(d*x+c)/d+1/12*(4*A*b-B*a)*(a+b*sec(d*x+c))^2*tan(d*x+c)/b/d+1/4*B*(a+ 
b*sec(d*x+c))^3*tan(d*x+c)/b/d
 

Mathematica [A] (verified)

Time = 0.50 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.67 \[ \int \sec ^2(c+d x) (a+b \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx=\frac {3 \left (8 a A b+4 a^2 B+3 b^2 B\right ) \text {arctanh}(\sin (c+d x))+\tan (c+d x) \left (24 \left (a^2 A+A b^2+2 a b B\right )+3 \left (8 a A b+4 a^2 B+3 b^2 B\right ) \sec (c+d x)+6 b^2 B \sec ^3(c+d x)+8 b (A b+2 a B) \tan ^2(c+d x)\right )}{24 d} \] Input:

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

Output:

(3*(8*a*A*b + 4*a^2*B + 3*b^2*B)*ArcTanh[Sin[c + d*x]] + Tan[c + d*x]*(24* 
(a^2*A + A*b^2 + 2*a*b*B) + 3*(8*a*A*b + 4*a^2*B + 3*b^2*B)*Sec[c + d*x] + 
 6*b^2*B*Sec[c + d*x]^3 + 8*b*(A*b + 2*a*B)*Tan[c + d*x]^2))/(24*d)
 

Rubi [A] (verified)

Time = 1.04 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.06, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.387, Rules used = {3042, 4498, 3042, 4490, 3042, 4485, 3042, 4274, 3042, 4254, 24, 4257}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^2 \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx\)

\(\Big \downarrow \) 4498

\(\displaystyle \frac {\int \sec (c+d x) (a+b \sec (c+d x))^2 (3 b B+(4 A b-a B) \sec (c+d x))dx}{4 b}+\frac {B \tan (c+d x) (a+b \sec (c+d x))^3}{4 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^2 \left (3 b B+(4 A b-a B) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx}{4 b}+\frac {B \tan (c+d x) (a+b \sec (c+d x))^3}{4 b d}\)

\(\Big \downarrow \) 4490

\(\displaystyle \frac {\frac {1}{3} \int \sec (c+d x) (a+b \sec (c+d x)) \left (b (8 A b+7 a B)+\left (-2 B a^2+8 A b a+9 b^2 B\right ) \sec (c+d x)\right )dx+\frac {(4 A b-a B) \tan (c+d x) (a+b \sec (c+d x))^2}{3 d}}{4 b}+\frac {B \tan (c+d x) (a+b \sec (c+d x))^3}{4 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{3} \int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right ) \left (b (8 A b+7 a B)+\left (-2 B a^2+8 A b a+9 b^2 B\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx+\frac {(4 A b-a B) \tan (c+d x) (a+b \sec (c+d x))^2}{3 d}}{4 b}+\frac {B \tan (c+d x) (a+b \sec (c+d x))^3}{4 b d}\)

\(\Big \downarrow \) 4485

\(\displaystyle \frac {\frac {1}{3} \left (\frac {1}{2} \int \sec (c+d x) \left (3 b \left (4 B a^2+8 A b a+3 b^2 B\right )+4 \left (-B a^3+4 A b a^2+8 b^2 B a+4 A b^3\right ) \sec (c+d x)\right )dx+\frac {b \left (-2 a^2 B+8 a A b+9 b^2 B\right ) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {(4 A b-a B) \tan (c+d x) (a+b \sec (c+d x))^2}{3 d}}{4 b}+\frac {B \tan (c+d x) (a+b \sec (c+d x))^3}{4 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{3} \left (\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (3 b \left (4 B a^2+8 A b a+3 b^2 B\right )+4 \left (-B a^3+4 A b a^2+8 b^2 B a+4 A b^3\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx+\frac {b \left (-2 a^2 B+8 a A b+9 b^2 B\right ) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {(4 A b-a B) \tan (c+d x) (a+b \sec (c+d x))^2}{3 d}}{4 b}+\frac {B \tan (c+d x) (a+b \sec (c+d x))^3}{4 b d}\)

\(\Big \downarrow \) 4274

\(\displaystyle \frac {\frac {1}{3} \left (\frac {1}{2} \left (3 b \left (4 a^2 B+8 a A b+3 b^2 B\right ) \int \sec (c+d x)dx+4 \left (a^3 (-B)+4 a^2 A b+8 a b^2 B+4 A b^3\right ) \int \sec ^2(c+d x)dx\right )+\frac {b \left (-2 a^2 B+8 a A b+9 b^2 B\right ) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {(4 A b-a B) \tan (c+d x) (a+b \sec (c+d x))^2}{3 d}}{4 b}+\frac {B \tan (c+d x) (a+b \sec (c+d x))^3}{4 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{3} \left (\frac {1}{2} \left (3 b \left (4 a^2 B+8 a A b+3 b^2 B\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+4 \left (a^3 (-B)+4 a^2 A b+8 a b^2 B+4 A b^3\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )^2dx\right )+\frac {b \left (-2 a^2 B+8 a A b+9 b^2 B\right ) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {(4 A b-a B) \tan (c+d x) (a+b \sec (c+d x))^2}{3 d}}{4 b}+\frac {B \tan (c+d x) (a+b \sec (c+d x))^3}{4 b d}\)

\(\Big \downarrow \) 4254

\(\displaystyle \frac {\frac {1}{3} \left (\frac {1}{2} \left (3 b \left (4 a^2 B+8 a A b+3 b^2 B\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx-\frac {4 \left (a^3 (-B)+4 a^2 A b+8 a b^2 B+4 A b^3\right ) \int 1d(-\tan (c+d x))}{d}\right )+\frac {b \left (-2 a^2 B+8 a A b+9 b^2 B\right ) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {(4 A b-a B) \tan (c+d x) (a+b \sec (c+d x))^2}{3 d}}{4 b}+\frac {B \tan (c+d x) (a+b \sec (c+d x))^3}{4 b d}\)

\(\Big \downarrow \) 24

\(\displaystyle \frac {\frac {1}{3} \left (\frac {1}{2} \left (3 b \left (4 a^2 B+8 a A b+3 b^2 B\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {4 \left (a^3 (-B)+4 a^2 A b+8 a b^2 B+4 A b^3\right ) \tan (c+d x)}{d}\right )+\frac {b \left (-2 a^2 B+8 a A b+9 b^2 B\right ) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {(4 A b-a B) \tan (c+d x) (a+b \sec (c+d x))^2}{3 d}}{4 b}+\frac {B \tan (c+d x) (a+b \sec (c+d x))^3}{4 b d}\)

\(\Big \downarrow \) 4257

\(\displaystyle \frac {\frac {1}{3} \left (\frac {b \left (-2 a^2 B+8 a A b+9 b^2 B\right ) \tan (c+d x) \sec (c+d x)}{2 d}+\frac {1}{2} \left (\frac {3 b \left (4 a^2 B+8 a A b+3 b^2 B\right ) \text {arctanh}(\sin (c+d x))}{d}+\frac {4 \left (a^3 (-B)+4 a^2 A b+8 a b^2 B+4 A b^3\right ) \tan (c+d x)}{d}\right )\right )+\frac {(4 A b-a B) \tan (c+d x) (a+b \sec (c+d x))^2}{3 d}}{4 b}+\frac {B \tan (c+d x) (a+b \sec (c+d x))^3}{4 b d}\)

Input:

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

Output:

(B*(a + b*Sec[c + d*x])^3*Tan[c + d*x])/(4*b*d) + (((4*A*b - a*B)*(a + b*S 
ec[c + d*x])^2*Tan[c + d*x])/(3*d) + ((b*(8*a*A*b - 2*a^2*B + 9*b^2*B)*Sec 
[c + d*x]*Tan[c + d*x])/(2*d) + ((3*b*(8*a*A*b + 4*a^2*B + 3*b^2*B)*ArcTan 
h[Sin[c + d*x]])/d + (4*(4*a^2*A*b + 4*A*b^3 - a^3*B + 8*a*b^2*B)*Tan[c + 
d*x])/d)/2)/3)/(4*b)
 

Defintions of rubi rules used

rule 24

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

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4254

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Simp[-d^(-1)   Subst[Int[Exp 
andIntegrand[(1 + x^2)^(n/2 - 1), x], x], x, Cot[c + d*x]], x] /; FreeQ[{c, 
 d}, x] && IGtQ[n/2, 0]
 

rule 4257

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

rule 4274

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

rule 4485

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] + Simp[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 4490

Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(cs 
c[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-B)*Cot[e + f*x]*(( 
a + b*Csc[e + f*x])^m/(f*(m + 1))), x] + Simp[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 4498

Int[csc[(e_.) + (f_.)*(x_)]^2*(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 + 1)/(b*f*(m + 2))), x] + Simp[1/(b*(m + 2))   Int 
[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*Simp[b*B*(m + 1) + (A*b*(m + 2) - a*B) 
*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, m}, x] && NeQ[A*b - a 
*B, 0] &&  !LtQ[m, -1]
 

Maple [A] (verified)

Time = 1.37 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.83

method	result	size
parts	\(-\frac {\left (A \,b^{2}+2 B a b \right ) \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}+\frac {\left (2 A a b +B \,a^{2}\right ) \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}+\frac {A \,a^{2} \tan \left (d x +c \right )}{d}+\frac {b^{2} B \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\)	\(148\)
derivativedivides	\(\frac {A \,a^{2} \tan \left (d x +c \right )+B \,a^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+2 A a b \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-2 B a b \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )-A \,b^{2} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+b^{2} B \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\)	\(185\)
default	\(\frac {A \,a^{2} \tan \left (d x +c \right )+B \,a^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+2 A a b \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-2 B a b \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )-A \,b^{2} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+b^{2} B \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\)	\(185\)
parallelrisch	\(\frac {-96 \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (A a b +\frac {1}{2} B \,a^{2}+\frac {3}{8} b^{2} B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+96 \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (A a b +\frac {1}{2} B \,a^{2}+\frac {3}{8} b^{2} B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\left (48 A \,a^{2}+64 A \,b^{2}+128 B a b \right ) \sin \left (2 d x +2 c \right )+\left (48 A a b +24 B \,a^{2}+18 b^{2} B \right ) \sin \left (3 d x +3 c \right )+\left (24 A \,a^{2}+16 A \,b^{2}+32 B a b \right ) \sin \left (4 d x +4 c \right )+48 \sin \left (d x +c \right ) \left (A a b +\frac {1}{2} B \,a^{2}+\frac {11}{8} b^{2} B \right )}{24 d \left (\cos \left (4 d x +4 c \right )+4 \cos \left (2 d x +2 c \right )+3\right )}\)	\(246\)
norman	\(\frac {-\frac {\left (8 A \,a^{2}-8 A a b +8 A \,b^{2}-4 B \,a^{2}+16 B a b -5 b^{2} B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{4 d}+\frac {\left (8 A \,a^{2}+8 A a b +8 A \,b^{2}+4 B \,a^{2}+16 B a b +5 b^{2} B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {\left (72 A \,a^{2}-24 A a b +40 A \,b^{2}-12 B \,a^{2}+80 B a b +9 b^{2} B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{12 d}-\frac {\left (72 A \,a^{2}+24 A a b +40 A \,b^{2}+12 B \,a^{2}+80 B a b -9 b^{2} B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{12 d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{4}}-\frac {\left (8 A a b +4 B \,a^{2}+3 b^{2} B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}+\frac {\left (8 A a b +4 B \,a^{2}+3 b^{2} B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}\)	\(291\)
risch	\(-\frac {i \left (24 A a b \,{\mathrm e}^{7 i \left (d x +c \right )}+12 B \,a^{2} {\mathrm e}^{7 i \left (d x +c \right )}+9 B \,b^{2} {\mathrm e}^{7 i \left (d x +c \right )}-24 A \,a^{2} {\mathrm e}^{6 i \left (d x +c \right )}+24 A a b \,{\mathrm e}^{5 i \left (d x +c \right )}+12 B \,a^{2} {\mathrm e}^{5 i \left (d x +c \right )}+33 B \,b^{2} {\mathrm e}^{5 i \left (d x +c \right )}-72 A \,a^{2} {\mathrm e}^{4 i \left (d x +c \right )}-48 A \,b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-96 B a b \,{\mathrm e}^{4 i \left (d x +c \right )}-24 A a b \,{\mathrm e}^{3 i \left (d x +c \right )}-12 B \,a^{2} {\mathrm e}^{3 i \left (d x +c \right )}-33 B \,b^{2} {\mathrm e}^{3 i \left (d x +c \right )}-72 A \,a^{2} {\mathrm e}^{2 i \left (d x +c \right )}-64 A \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-128 B a b \,{\mathrm e}^{2 i \left (d x +c \right )}-24 A a b \,{\mathrm e}^{i \left (d x +c \right )}-12 B \,a^{2} {\mathrm e}^{i \left (d x +c \right )}-9 b^{2} B \,{\mathrm e}^{i \left (d x +c \right )}-24 A \,a^{2}-16 A \,b^{2}-32 B a b \right )}{12 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A a b}{d}-\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{2 d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) b^{2} B}{8 d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A a b}{d}+\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{2 d}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) b^{2} B}{8 d}\)	\(447\)

Input:

int(sec(d*x+c)^2*(a+b*sec(d*x+c))^2*(A+B*sec(d*x+c)),x,method=_RETURNVERBO 
SE)
 

Output:

-(A*b^2+2*B*a*b)/d*(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+c)+(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)))+A*a^2/d*tan(d*x+c) 
+b^2*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)))
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.01 \[ \int \sec ^2(c+d x) (a+b \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx=\frac {3 \, {\left (4 \, B a^{2} + 8 \, A a b + 3 \, B b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (4 \, B a^{2} + 8 \, A a b + 3 \, B b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (8 \, {\left (3 \, A a^{2} + 4 \, B a b + 2 \, A b^{2}\right )} \cos \left (d x + c\right )^{3} + 6 \, B b^{2} + 3 \, {\left (4 \, B a^{2} + 8 \, A a b + 3 \, B b^{2}\right )} \cos \left (d x + c\right )^{2} + 8 \, {\left (2 \, B a b + A b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )^{4}} \] Input:

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

Output:

1/48*(3*(4*B*a^2 + 8*A*a*b + 3*B*b^2)*cos(d*x + c)^4*log(sin(d*x + c) + 1) 
 - 3*(4*B*a^2 + 8*A*a*b + 3*B*b^2)*cos(d*x + c)^4*log(-sin(d*x + c) + 1) + 
 2*(8*(3*A*a^2 + 4*B*a*b + 2*A*b^2)*cos(d*x + c)^3 + 6*B*b^2 + 3*(4*B*a^2 
+ 8*A*a*b + 3*B*b^2)*cos(d*x + c)^2 + 8*(2*B*a*b + A*b^2)*cos(d*x + c))*si 
n(d*x + c))/(d*cos(d*x + c)^4)
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 228, normalized size of antiderivative = 1.27 \[ \int \sec ^2(c+d x) (a+b \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx=\frac {32 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a b + 16 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A b^{2} - 3 \, B b^{2} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 12 \, B a^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 24 \, A a b {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 48 \, A a^{2} \tan \left (d x + c\right )}{48 \, d} \] Input:

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

Output:

1/48*(32*(tan(d*x + c)^3 + 3*tan(d*x + c))*B*a*b + 16*(tan(d*x + c)^3 + 3* 
tan(d*x + c))*A*b^2 - 3*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)) - 12*B*a^2*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d* 
x + c) + 1) + log(sin(d*x + c) - 1)) - 24*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)) + 48*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. 478 vs. \(2 (167) = 334\).

Time = 0.16 (sec) , antiderivative size = 478, normalized size of antiderivative = 2.67 \[ \int \sec ^2(c+d x) (a+b \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx =\text {Too large to display} \] Input:

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

Output:

1/24*(3*(4*B*a^2 + 8*A*a*b + 3*B*b^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 
 3*(4*B*a^2 + 8*A*a*b + 3*B*b^2)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(2 
4*A*a^2*tan(1/2*d*x + 1/2*c)^7 - 12*B*a^2*tan(1/2*d*x + 1/2*c)^7 - 24*A*a* 
b*tan(1/2*d*x + 1/2*c)^7 + 48*B*a*b*tan(1/2*d*x + 1/2*c)^7 + 24*A*b^2*tan( 
1/2*d*x + 1/2*c)^7 - 15*B*b^2*tan(1/2*d*x + 1/2*c)^7 - 72*A*a^2*tan(1/2*d* 
x + 1/2*c)^5 + 12*B*a^2*tan(1/2*d*x + 1/2*c)^5 + 24*A*a*b*tan(1/2*d*x + 1/ 
2*c)^5 - 80*B*a*b*tan(1/2*d*x + 1/2*c)^5 - 40*A*b^2*tan(1/2*d*x + 1/2*c)^5 
 - 9*B*b^2*tan(1/2*d*x + 1/2*c)^5 + 72*A*a^2*tan(1/2*d*x + 1/2*c)^3 + 12*B 
*a^2*tan(1/2*d*x + 1/2*c)^3 + 24*A*a*b*tan(1/2*d*x + 1/2*c)^3 + 80*B*a*b*t 
an(1/2*d*x + 1/2*c)^3 + 40*A*b^2*tan(1/2*d*x + 1/2*c)^3 - 9*B*b^2*tan(1/2* 
d*x + 1/2*c)^3 - 24*A*a^2*tan(1/2*d*x + 1/2*c) - 12*B*a^2*tan(1/2*d*x + 1/ 
2*c) - 24*A*a*b*tan(1/2*d*x + 1/2*c) - 48*B*a*b*tan(1/2*d*x + 1/2*c) - 24* 
A*b^2*tan(1/2*d*x + 1/2*c) - 15*B*b^2*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 
 1/2*c)^2 - 1)^4)/d
 

Mupad [B] (verification not implemented)

Time = 15.11 (sec) , antiderivative size = 317, normalized size of antiderivative = 1.77 \[ \int \sec ^2(c+d x) (a+b \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx=\frac {\mathrm {atanh}\left (\frac {4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {B\,a^2}{2}+A\,a\,b+\frac {3\,B\,b^2}{8}\right )}{2\,B\,a^2+4\,A\,a\,b+\frac {3\,B\,b^2}{2}}\right )\,\left (B\,a^2+2\,A\,a\,b+\frac {3\,B\,b^2}{4}\right )}{d}-\frac {\left (2\,A\,a^2+2\,A\,b^2-B\,a^2-\frac {5\,B\,b^2}{4}-2\,A\,a\,b+4\,B\,a\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (B\,a^2-\frac {10\,A\,b^2}{3}-6\,A\,a^2-\frac {3\,B\,b^2}{4}+2\,A\,a\,b-\frac {20\,B\,a\,b}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (6\,A\,a^2+\frac {10\,A\,b^2}{3}+B\,a^2-\frac {3\,B\,b^2}{4}+2\,A\,a\,b+\frac {20\,B\,a\,b}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (-2\,A\,a^2-2\,A\,b^2-B\,a^2-\frac {5\,B\,b^2}{4}-2\,A\,a\,b-4\,B\,a\,b\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )} \] Input:

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

Output:

(atanh((4*tan(c/2 + (d*x)/2)*((B*a^2)/2 + (3*B*b^2)/8 + A*a*b))/(2*B*a^2 + 
 (3*B*b^2)/2 + 4*A*a*b))*(B*a^2 + (3*B*b^2)/4 + 2*A*a*b))/d - (tan(c/2 + ( 
d*x)/2)^7*(2*A*a^2 + 2*A*b^2 - B*a^2 - (5*B*b^2)/4 - 2*A*a*b + 4*B*a*b) + 
tan(c/2 + (d*x)/2)^3*(6*A*a^2 + (10*A*b^2)/3 + B*a^2 - (3*B*b^2)/4 + 2*A*a 
*b + (20*B*a*b)/3) - tan(c/2 + (d*x)/2)^5*(6*A*a^2 + (10*A*b^2)/3 - B*a^2 
+ (3*B*b^2)/4 - 2*A*a*b + (20*B*a*b)/3) - tan(c/2 + (d*x)/2)*(2*A*a^2 + 2* 
A*b^2 + B*a^2 + (5*B*b^2)/4 + 2*A*a*b + 4*B*a*b))/(d*(6*tan(c/2 + (d*x)/2) 
^4 - 4*tan(c/2 + (d*x)/2)^2 - 4*tan(c/2 + (d*x)/2)^6 + tan(c/2 + (d*x)/2)^ 
8 + 1))
 

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 426, normalized size of antiderivative = 2.38 \[ \int \sec ^2(c+d x) (a+b \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx=\frac {-8 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} a^{3}-16 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} a \,b^{2}+8 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a^{3}+24 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a \,b^{2}-12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{4} a^{2} b -3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{4} b^{3}+24 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} a^{2} b +6 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} b^{3}-12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a^{2} b -3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) b^{3}+12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{4} a^{2} b +3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{4} b^{3}-24 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} a^{2} b -6 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} b^{3}+12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a^{2} b +3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) b^{3}-12 \sin \left (d x +c \right )^{3} a^{2} b -3 \sin \left (d x +c \right )^{3} b^{3}+12 \sin \left (d x +c \right ) a^{2} b +5 \sin \left (d x +c \right ) b^{3}}{8 d \left (\sin \left (d x +c \right )^{4}-2 \sin \left (d x +c \right )^{2}+1\right )} \] Input:

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

Output:

( - 8*cos(c + d*x)*sin(c + d*x)**3*a**3 - 16*cos(c + d*x)*sin(c + d*x)**3* 
a*b**2 + 8*cos(c + d*x)*sin(c + d*x)*a**3 + 24*cos(c + d*x)*sin(c + d*x)*a 
*b**2 - 12*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*a**2*b - 3*log(tan((c 
 + d*x)/2) - 1)*sin(c + d*x)**4*b**3 + 24*log(tan((c + d*x)/2) - 1)*sin(c 
+ d*x)**2*a**2*b + 6*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*b**3 - 12*l 
og(tan((c + d*x)/2) - 1)*a**2*b - 3*log(tan((c + d*x)/2) - 1)*b**3 + 12*lo 
g(tan((c + d*x)/2) + 1)*sin(c + d*x)**4*a**2*b + 3*log(tan((c + d*x)/2) + 
1)*sin(c + d*x)**4*b**3 - 24*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a** 
2*b - 6*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*b**3 + 12*log(tan((c + d 
*x)/2) + 1)*a**2*b + 3*log(tan((c + d*x)/2) + 1)*b**3 - 12*sin(c + d*x)**3 
*a**2*b - 3*sin(c + d*x)**3*b**3 + 12*sin(c + d*x)*a**2*b + 5*sin(c + d*x) 
*b**3)/(8*d*(sin(c + d*x)**4 - 2*sin(c + d*x)**2 + 1))