\(\int \frac {\sec ^2(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^3} \, dx\) [331]

Optimal result

Integrand size = 31, antiderivative size = 180 \[ \int \frac {\sec ^2(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^3} \, dx=-\frac {\left (3 a A b-a^2 B-2 b^2 B\right ) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{5/2} (a+b)^{5/2} d}+\frac {a (A b-a B) \tan (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}+\frac {\left (a^2 A b+2 A b^3+a^3 B-4 a b^2 B\right ) \tan (c+d x)}{2 b \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))} \] Output:

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

Mathematica [A] (verified)

Time = 0.64 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.87 \[ \int \frac {\sec ^2(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^3} \, dx=\frac {-\frac {2 \left (-3 a A b+a^2 B+2 b^2 B\right ) \text {arctanh}\left (\frac {(-a+b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}}+\frac {(-A b+a B) \sin (c+d x)}{(a-b) (a+b) (b+a \cos (c+d x))^2}+\frac {\left (2 a^2 A+A b^2-3 a b B\right ) \sin (c+d x)}{(a-b)^2 (a+b)^2 (b+a \cos (c+d x))}}{2 d} \] Input:

Integrate[(Sec[c + d*x]^2*(A + B*Sec[c + d*x]))/(a + b*Sec[c + d*x])^3,x]
 

Output:

((-2*(-3*a*A*b + a^2*B + 2*b^2*B)*ArcTanh[((-a + b)*Tan[(c + d*x)/2])/Sqrt 
[a^2 - b^2]])/(a^2 - b^2)^(5/2) + ((-(A*b) + a*B)*Sin[c + d*x])/((a - b)*( 
a + b)*(b + a*Cos[c + d*x])^2) + ((2*a^2*A + A*b^2 - 3*a*b*B)*Sin[c + d*x] 
)/((a - b)^2*(a + b)^2*(b + a*Cos[c + d*x])))/(2*d)
 

Rubi [A] (verified)

Time = 0.88 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.14, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.387, Rules used = {3042, 4497, 25, 3042, 4491, 25, 27, 3042, 4318, 3042, 3138, 221}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4497

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4491

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4318

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3138

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

\(\Big \downarrow \) 221

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

Input:

Int[(Sec[c + d*x]^2*(A + B*Sec[c + d*x]))/(a + b*Sec[c + d*x])^3,x]
 

Output:

(a*(A*b - a*B)*Tan[c + d*x])/(2*b*(a^2 - b^2)*d*(a + b*Sec[c + d*x])^2) - 
((2*b*(3*a*A*b - a^2*B - 2*b^2*B)*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/S 
qrt[a + b]])/(Sqrt[a - b]*Sqrt[a + b]*(a^2 - b^2)*d) - ((a^2*A*b + 2*A*b^3 
 + a^3*B - 4*a*b^2*B)*Tan[c + d*x])/((a^2 - b^2)*d*(a + b*Sec[c + d*x])))/ 
(2*b*(a^2 - b^2))
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 221

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x 
/Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
 

rule 3042

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

rule 3138

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{ 
e = FreeFactors[Tan[(c + d*x)/2], x]}, Simp[2*(e/d)   Subst[Int[1/(a + b + 
(a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] 
 && NeQ[a^2 - b^2, 0]
 

rule 4318

Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbo 
l] :> Simp[1/b   Int[1/(1 + (a/b)*Sin[e + f*x]), x], x] /; FreeQ[{a, b, e, 
f}, x] && NeQ[a^2 - b^2, 0]
 

rule 4491

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

rule 4497

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

Maple [A] (verified)

Time = 0.42 (sec) , antiderivative size = 238, normalized size of antiderivative = 1.32

method	result	size
derivativedivides	\(\frac {\frac {-\frac {\left (2 A \,a^{2}+A a b +2 A \,b^{2}-B \,a^{2}-4 B a b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{\left (a -b \right ) \left (a^{2}+2 a b +b^{2}\right )}+\frac {\left (2 A \,a^{2}-A a b +2 A \,b^{2}+B \,a^{2}-4 B a b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a +b \right ) \left (a^{2}-2 a b +b^{2}\right )}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a -b \right )^{2}}-\frac {\left (3 A a b -B \,a^{2}-2 b^{2} B \right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}}{d}\)	\(238\)
default	\(\frac {\frac {-\frac {\left (2 A \,a^{2}+A a b +2 A \,b^{2}-B \,a^{2}-4 B a b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{\left (a -b \right ) \left (a^{2}+2 a b +b^{2}\right )}+\frac {\left (2 A \,a^{2}-A a b +2 A \,b^{2}+B \,a^{2}-4 B a b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a +b \right ) \left (a^{2}-2 a b +b^{2}\right )}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a -b \right )^{2}}-\frac {\left (3 A a b -B \,a^{2}-2 b^{2} B \right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}}{d}\)	\(238\)
risch	\(\frac {i \left (3 A \,a^{3} b \,{\mathrm e}^{3 i \left (d x +c \right )}-B \,a^{4} {\mathrm e}^{3 i \left (d x +c \right )}-2 B \,a^{2} b^{2} {\mathrm e}^{3 i \left (d x +c \right )}+2 a^{4} A \,{\mathrm e}^{2 i \left (d x +c \right )}+5 A \,a^{2} b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+2 A \,b^{4} {\mathrm e}^{2 i \left (d x +c \right )}-3 B \,a^{3} b \,{\mathrm e}^{2 i \left (d x +c \right )}-6 B a \,b^{3} {\mathrm e}^{2 i \left (d x +c \right )}+5 A \,a^{3} b \,{\mathrm e}^{i \left (d x +c \right )}+4 A a \,b^{3} {\mathrm e}^{i \left (d x +c \right )}+B \,a^{4} {\mathrm e}^{i \left (d x +c \right )}-10 B \,a^{2} b^{2} {\mathrm e}^{i \left (d x +c \right )}+2 a^{4} A +A \,a^{2} b^{2}-3 a^{3} b B \right )}{a \left (-a^{2}+b^{2}\right )^{2} d \left (a \,{\mathrm e}^{2 i \left (d x +c \right )}+2 b \,{\mathrm e}^{i \left (d x +c \right )}+a \right )^{2}}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-\sqrt {a^{2}-b^{2}}\, b}{\sqrt {a^{2}-b^{2}}\, a}\right ) A a b}{2 \sqrt {a^{2}-b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-\sqrt {a^{2}-b^{2}}\, b}{\sqrt {a^{2}-b^{2}}\, a}\right ) B \,a^{2}}{2 \sqrt {a^{2}-b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-\sqrt {a^{2}-b^{2}}\, b}{\sqrt {a^{2}-b^{2}}\, a}\right ) b^{2} B}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{\sqrt {a^{2}-b^{2}}\, a}\right ) A a b}{2 \sqrt {a^{2}-b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{\sqrt {a^{2}-b^{2}}\, a}\right ) B \,a^{2}}{2 \sqrt {a^{2}-b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{\sqrt {a^{2}-b^{2}}\, a}\right ) b^{2} B}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d}\)	\(771\)

Input:

int(sec(d*x+c)^2*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))^3,x,method=_RETURNVERBO 
SE)
 

Output:

1/d*(2*(-1/2*(2*A*a^2+A*a*b+2*A*b^2-B*a^2-4*B*a*b)/(a-b)/(a^2+2*a*b+b^2)*t 
an(1/2*d*x+1/2*c)^3+1/2*(2*A*a^2-A*a*b+2*A*b^2+B*a^2-4*B*a*b)/(a+b)/(a^2-2 
*a*b+b^2)*tan(1/2*d*x+1/2*c))/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2 
*b-a-b)^2-(3*A*a*b-B*a^2-2*B*b^2)/(a^4-2*a^2*b^2+b^4)/((a+b)*(a-b))^(1/2)* 
arctanh((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2)))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 346 vs. \(2 (165) = 330\).

Time = 0.12 (sec) , antiderivative size = 750, normalized size of antiderivative = 4.17 \[ \int \frac {\sec ^2(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^3} \, dx =\text {Too large to display} \] Input:

integrate(sec(d*x+c)^2*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))^3,x, algorithm="f 
ricas")
 

Output:

[1/4*((B*a^2*b^2 - 3*A*a*b^3 + 2*B*b^4 + (B*a^4 - 3*A*a^3*b + 2*B*a^2*b^2) 
*cos(d*x + c)^2 + 2*(B*a^3*b - 3*A*a^2*b^2 + 2*B*a*b^3)*cos(d*x + c))*sqrt 
(a^2 - b^2)*log((2*a*b*cos(d*x + c) - (a^2 - 2*b^2)*cos(d*x + c)^2 + 2*sqr 
t(a^2 - b^2)*(b*cos(d*x + c) + a)*sin(d*x + c) + 2*a^2 - b^2)/(a^2*cos(d*x 
 + c)^2 + 2*a*b*cos(d*x + c) + b^2)) + 2*(B*a^5 + A*a^4*b - 5*B*a^3*b^2 + 
A*a^2*b^3 + 4*B*a*b^4 - 2*A*b^5 + (2*A*a^5 - 3*B*a^4*b - A*a^3*b^2 + 3*B*a 
^2*b^3 - A*a*b^4)*cos(d*x + c))*sin(d*x + c))/((a^8 - 3*a^6*b^2 + 3*a^4*b^ 
4 - a^2*b^6)*d*cos(d*x + c)^2 + 2*(a^7*b - 3*a^5*b^3 + 3*a^3*b^5 - a*b^7)* 
d*cos(d*x + c) + (a^6*b^2 - 3*a^4*b^4 + 3*a^2*b^6 - b^8)*d), 1/2*((B*a^2*b 
^2 - 3*A*a*b^3 + 2*B*b^4 + (B*a^4 - 3*A*a^3*b + 2*B*a^2*b^2)*cos(d*x + c)^ 
2 + 2*(B*a^3*b - 3*A*a^2*b^2 + 2*B*a*b^3)*cos(d*x + c))*sqrt(-a^2 + b^2)*a 
rctan(-sqrt(-a^2 + b^2)*(b*cos(d*x + c) + a)/((a^2 - b^2)*sin(d*x + c))) + 
 (B*a^5 + A*a^4*b - 5*B*a^3*b^2 + A*a^2*b^3 + 4*B*a*b^4 - 2*A*b^5 + (2*A*a 
^5 - 3*B*a^4*b - A*a^3*b^2 + 3*B*a^2*b^3 - A*a*b^4)*cos(d*x + c))*sin(d*x 
+ c))/((a^8 - 3*a^6*b^2 + 3*a^4*b^4 - a^2*b^6)*d*cos(d*x + c)^2 + 2*(a^7*b 
 - 3*a^5*b^3 + 3*a^3*b^5 - a*b^7)*d*cos(d*x + c) + (a^6*b^2 - 3*a^4*b^4 + 
3*a^2*b^6 - b^8)*d)]
 

Sympy [F]

\[ \int \frac {\sec ^2(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^3} \, dx=\int \frac {\left (A + B \sec {\left (c + d x \right )}\right ) \sec ^{2}{\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{3}}\, dx \] Input:

integrate(sec(d*x+c)**2*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))**3,x)
 

Output:

Integral((A + B*sec(c + d*x))*sec(c + d*x)**2/(a + b*sec(c + d*x))**3, x)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {\sec ^2(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^3} \, dx=\text {Exception raised: ValueError} \] Input:

integrate(sec(d*x+c)^2*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))^3,x, algorithm="m 
axima")
 

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(4*a^2-4*b^2>0)', see `assume?` f 
or more de
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 400 vs. \(2 (165) = 330\).

Time = 0.23 (sec) , antiderivative size = 400, normalized size of antiderivative = 2.22 \[ \int \frac {\sec ^2(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^3} \, dx=\frac {\frac {{\left (B a^{2} - 3 \, A a b + 2 \, B b^{2}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {-a^{2} + b^{2}}}\right )\right )}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {-a^{2} + b^{2}}} - \frac {2 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - A a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3 \, B a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + A a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 4 \, B a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, A b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - A a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, B a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - A a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 4 \, B a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, A b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a - b\right )}^{2}}}{d} \] Input:

integrate(sec(d*x+c)^2*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))^3,x, algorithm="g 
iac")
 

Output:

((B*a^2 - 3*A*a*b + 2*B*b^2)*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(-2*a + 
2*b) + arctan(-(a*tan(1/2*d*x + 1/2*c) - b*tan(1/2*d*x + 1/2*c))/sqrt(-a^2 
 + b^2)))/((a^4 - 2*a^2*b^2 + b^4)*sqrt(-a^2 + b^2)) - (2*A*a^3*tan(1/2*d* 
x + 1/2*c)^3 - B*a^3*tan(1/2*d*x + 1/2*c)^3 - A*a^2*b*tan(1/2*d*x + 1/2*c) 
^3 - 3*B*a^2*b*tan(1/2*d*x + 1/2*c)^3 + A*a*b^2*tan(1/2*d*x + 1/2*c)^3 + 4 
*B*a*b^2*tan(1/2*d*x + 1/2*c)^3 - 2*A*b^3*tan(1/2*d*x + 1/2*c)^3 - 2*A*a^3 
*tan(1/2*d*x + 1/2*c) - B*a^3*tan(1/2*d*x + 1/2*c) - A*a^2*b*tan(1/2*d*x + 
 1/2*c) + 3*B*a^2*b*tan(1/2*d*x + 1/2*c) - A*a*b^2*tan(1/2*d*x + 1/2*c) + 
4*B*a*b^2*tan(1/2*d*x + 1/2*c) - 2*A*b^3*tan(1/2*d*x + 1/2*c))/((a^4 - 2*a 
^2*b^2 + b^4)*(a*tan(1/2*d*x + 1/2*c)^2 - b*tan(1/2*d*x + 1/2*c)^2 - a - b 
)^2))/d
 

Mupad [B] (verification not implemented)

Time = 14.81 (sec) , antiderivative size = 251, normalized size of antiderivative = 1.39 \[ \int \frac {\sec ^2(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^3} \, dx=\frac {\mathrm {atanh}\left (\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,a-2\,b\right )\,\left (a^2-2\,a\,b+b^2\right )}{2\,\sqrt {a+b}\,{\left (a-b\right )}^{5/2}}\right )\,\left (B\,a^2-3\,A\,a\,b+2\,B\,b^2\right )}{d\,{\left (a+b\right )}^{5/2}\,{\left (a-b\right )}^{5/2}}-\frac {\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (2\,A\,a^2+2\,A\,b^2-B\,a^2+A\,a\,b-4\,B\,a\,b\right )}{{\left (a+b\right )}^2\,\left (a-b\right )}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,A\,a^2+2\,A\,b^2+B\,a^2-A\,a\,b-4\,B\,a\,b\right )}{\left (a+b\right )\,\left (a^2-2\,a\,b+b^2\right )}}{d\,\left (2\,a\,b-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (2\,a^2-2\,b^2\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (a^2-2\,a\,b+b^2\right )+a^2+b^2\right )} \] Input:

int((A + B/cos(c + d*x))/(cos(c + d*x)^2*(a + b/cos(c + d*x))^3),x)
 

Output:

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

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.07 \[ \int \frac {\sec ^2(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^3} \, dx=\frac {-2 \sqrt {-a^{2}+b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b}{\sqrt {-a^{2}+b^{2}}}\right ) \cos \left (d x +c \right ) a b -2 \sqrt {-a^{2}+b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b}{\sqrt {-a^{2}+b^{2}}}\right ) b^{2}+\sin \left (d x +c \right ) a^{3}-\sin \left (d x +c \right ) a \,b^{2}}{d \left (\cos \left (d x +c \right ) a^{5}-2 \cos \left (d x +c \right ) a^{3} b^{2}+\cos \left (d x +c \right ) a \,b^{4}+a^{4} b -2 a^{2} b^{3}+b^{5}\right )} \] Input:

int(sec(d*x+c)^2*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))^3,x)
 

Output:

( - 2*sqrt( - a**2 + b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/ 
sqrt( - a**2 + b**2))*cos(c + d*x)*a*b - 2*sqrt( - a**2 + b**2)*atan((tan( 
(c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt( - a**2 + b**2))*b**2 + sin(c + 
d*x)*a**3 - sin(c + d*x)*a*b**2)/(d*(cos(c + d*x)*a**5 - 2*cos(c + d*x)*a* 
*3*b**2 + cos(c + d*x)*a*b**4 + a**4*b - 2*a**2*b**3 + b**5))