\(\int \frac {A+B \sec (c+d x)}{(a+b \sec (c+d x))^3} \, dx\) [333]

Optimal result

Integrand size = 23, antiderivative size = 205 \[ \int \frac {A+B \sec (c+d x)}{(a+b \sec (c+d x))^3} \, dx=\frac {A x}{a^3}-\frac {\left (6 a^4 A b-5 a^2 A b^3+2 A b^5-2 a^5 B-a^3 b^2 B\right ) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^3 (a-b)^{5/2} (a+b)^{5/2} d}+\frac {b (A b-a B) \tan (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}+\frac {b \left (5 a^2 A b-2 A b^3-3 a^3 B\right ) \tan (c+d x)}{2 a^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))} \] Output:

A*x/a^3-(6*A*a^4*b-5*A*a^2*b^3+2*A*b^5-2*B*a^5-B*a^3*b^2)*arctanh((a-b)^(1 
/2)*tan(1/2*d*x+1/2*c)/(a+b)^(1/2))/a^3/(a-b)^(5/2)/(a+b)^(5/2)/d+1/2*b*(A 
*b-B*a)*tan(d*x+c)/a/(a^2-b^2)/d/(a+b*sec(d*x+c))^2+1/2*b*(5*A*a^2*b-2*A*b 
^3-3*B*a^3)*tan(d*x+c)/a^2/(a^2-b^2)^2/d/(a+b*sec(d*x+c))
 

Mathematica [A] (verified)

Time = 1.75 (sec) , antiderivative size = 267, normalized size of antiderivative = 1.30 \[ \int \frac {A+B \sec (c+d x)}{(a+b \sec (c+d x))^3} \, dx=\frac {(b+a \cos (c+d x)) \sec ^2(c+d x) (A+B \sec (c+d x)) \left (2 A (c+d x) (b+a \cos (c+d x))^2-\frac {2 \left (-6 a^4 A b+5 a^2 A b^3-2 A b^5+2 a^5 B+a^3 b^2 B\right ) \text {arctanh}\left (\frac {(-a+b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right ) (b+a \cos (c+d x))^2}{\left (a^2-b^2\right )^{5/2}}+\frac {a b^2 (-A b+a B) \sin (c+d x)}{(a-b) (a+b)}-\frac {a b \left (-6 a^2 A b+3 A b^3+4 a^3 B-a b^2 B\right ) (b+a \cos (c+d x)) \sin (c+d x)}{(a-b)^2 (a+b)^2}\right )}{2 a^3 d (B+A \cos (c+d x)) (a+b \sec (c+d x))^3} \] Input:

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

Output:

((b + a*Cos[c + d*x])*Sec[c + d*x]^2*(A + B*Sec[c + d*x])*(2*A*(c + d*x)*( 
b + a*Cos[c + d*x])^2 - (2*(-6*a^4*A*b + 5*a^2*A*b^3 - 2*A*b^5 + 2*a^5*B + 
 a^3*b^2*B)*ArcTanh[((-a + b)*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]]*(b + a*Co 
s[c + d*x])^2)/(a^2 - b^2)^(5/2) + (a*b^2*(-(A*b) + a*B)*Sin[c + d*x])/((a 
 - b)*(a + b)) - (a*b*(-6*a^2*A*b + 3*A*b^3 + 4*a^3*B - a*b^2*B)*(b + a*Co 
s[c + d*x])*Sin[c + d*x])/((a - b)^2*(a + b)^2)))/(2*a^3*d*(B + A*Cos[c + 
d*x])*(a + b*Sec[c + d*x])^3)
 

Rubi [A] (verified)

Time = 1.13 (sec) , antiderivative size = 249, normalized size of antiderivative = 1.21, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.565, Rules used = {3042, 4411, 25, 3042, 4548, 25, 3042, 4407, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4411

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4548

\(\displaystyle \frac {\frac {b \left (-3 a^3 B+5 a^2 A b-2 A b^3\right ) \tan (c+d x)}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}-\frac {\int -\frac {2 A \left (a^2-b^2\right )^2-a \left (-2 B a^3+4 A b a^2-b^2 B a-A b^3\right ) \sec (c+d x)}{a+b \sec (c+d x)}dx}{a \left (a^2-b^2\right )}}{2 a \left (a^2-b^2\right )}+\frac {b (A b-a B) \tan (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {\int \frac {2 A \left (a^2-b^2\right )^2-a \left (-2 B a^3+4 A b a^2-b^2 B a-A b^3\right ) \sec (c+d x)}{a+b \sec (c+d x)}dx}{a \left (a^2-b^2\right )}+\frac {b \left (-3 a^3 B+5 a^2 A b-2 A b^3\right ) \tan (c+d x)}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{2 a \left (a^2-b^2\right )}+\frac {b (A b-a B) \tan (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {2 A \left (a^2-b^2\right )^2-a \left (-2 B a^3+4 A b a^2-b^2 B a-A b^3\right ) \csc \left (c+d x+\frac {\pi }{2}\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{a \left (a^2-b^2\right )}+\frac {b \left (-3 a^3 B+5 a^2 A b-2 A b^3\right ) \tan (c+d x)}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{2 a \left (a^2-b^2\right )}+\frac {b (A b-a B) \tan (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\)

\(\Big \downarrow \) 4407

\(\displaystyle \frac {\frac {\frac {2 A x \left (a^2-b^2\right )^2}{a}-\frac {\left (-2 a^5 B+6 a^4 A b-a^3 b^2 B-5 a^2 A b^3+2 A b^5\right ) \int \frac {\sec (c+d x)}{a+b \sec (c+d x)}dx}{a}}{a \left (a^2-b^2\right )}+\frac {b \left (-3 a^3 B+5 a^2 A b-2 A b^3\right ) \tan (c+d x)}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{2 a \left (a^2-b^2\right )}+\frac {b (A b-a B) \tan (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {2 A x \left (a^2-b^2\right )^2}{a}-\frac {\left (-2 a^5 B+6 a^4 A b-a^3 b^2 B-5 a^2 A b^3+2 A b^5\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{a}}{a \left (a^2-b^2\right )}+\frac {b \left (-3 a^3 B+5 a^2 A b-2 A b^3\right ) \tan (c+d x)}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{2 a \left (a^2-b^2\right )}+\frac {b (A b-a B) \tan (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\)

\(\Big \downarrow \) 4318

\(\displaystyle \frac {\frac {\frac {2 A x \left (a^2-b^2\right )^2}{a}-\frac {\left (-2 a^5 B+6 a^4 A b-a^3 b^2 B-5 a^2 A b^3+2 A b^5\right ) \int \frac {1}{\frac {a \cos (c+d x)}{b}+1}dx}{a b}}{a \left (a^2-b^2\right )}+\frac {b \left (-3 a^3 B+5 a^2 A b-2 A b^3\right ) \tan (c+d x)}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{2 a \left (a^2-b^2\right )}+\frac {b (A b-a B) \tan (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {2 A x \left (a^2-b^2\right )^2}{a}-\frac {\left (-2 a^5 B+6 a^4 A b-a^3 b^2 B-5 a^2 A b^3+2 A b^5\right ) \int \frac {1}{\frac {a \sin \left (c+d x+\frac {\pi }{2}\right )}{b}+1}dx}{a b}}{a \left (a^2-b^2\right )}+\frac {b \left (-3 a^3 B+5 a^2 A b-2 A b^3\right ) \tan (c+d x)}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{2 a \left (a^2-b^2\right )}+\frac {b (A b-a B) \tan (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\)

\(\Big \downarrow \) 3138

\(\displaystyle \frac {\frac {\frac {2 A x \left (a^2-b^2\right )^2}{a}-\frac {2 \left (-2 a^5 B+6 a^4 A b-a^3 b^2 B-5 a^2 A b^3+2 A b^5\right ) \int \frac {1}{\left (1-\frac {a}{b}\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right )+\frac {a+b}{b}}d\tan \left (\frac {1}{2} (c+d x)\right )}{a b d}}{a \left (a^2-b^2\right )}+\frac {b \left (-3 a^3 B+5 a^2 A b-2 A b^3\right ) \tan (c+d x)}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{2 a \left (a^2-b^2\right )}+\frac {b (A b-a B) \tan (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {b (A b-a B) \tan (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}+\frac {\frac {b \left (-3 a^3 B+5 a^2 A b-2 A b^3\right ) \tan (c+d x)}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}+\frac {\frac {2 A x \left (a^2-b^2\right )^2}{a}-\frac {2 \left (-2 a^5 B+6 a^4 A b-a^3 b^2 B-5 a^2 A b^3+2 A b^5\right ) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a d \sqrt {a-b} \sqrt {a+b}}}{a \left (a^2-b^2\right )}}{2 a \left (a^2-b^2\right )}\)

Input:

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

Output:

(b*(A*b - a*B)*Tan[c + d*x])/(2*a*(a^2 - b^2)*d*(a + b*Sec[c + d*x])^2) + 
(((2*A*(a^2 - b^2)^2*x)/a - (2*(6*a^4*A*b - 5*a^2*A*b^3 + 2*A*b^5 - 2*a^5* 
B - a^3*b^2*B)*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(a*Sqr 
t[a - b]*Sqrt[a + b]*d))/(a*(a^2 - b^2)) + (b*(5*a^2*A*b - 2*A*b^3 - 3*a^3 
*B)*Tan[c + d*x])/(a*(a^2 - b^2)*d*(a + b*Sec[c + d*x])))/(2*a*(a^2 - b^2) 
)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], 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 4407

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + 
 (a_)), x_Symbol] :> Simp[c*(x/a), x] - Simp[(b*c - a*d)/a   Int[Csc[e + f* 
x]/(a + b*Csc[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c 
- a*d, 0]
 

rule 4411

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(d 
_.) + (c_)), x_Symbol] :> Simp[b*(b*c - a*d)*Cot[e + f*x]*((a + b*Csc[e + f 
*x])^(m + 1)/(a*f*(m + 1)*(a^2 - b^2))), x] + Simp[1/(a*(m + 1)*(a^2 - b^2) 
)   Int[(a + b*Csc[e + f*x])^(m + 1)*Simp[c*(a^2 - b^2)*(m + 1) - (a*(b*c - 
 a*d)*(m + 1))*Csc[e + f*x] + b*(b*c - a*d)*(m + 2)*Csc[e + f*x]^2, x], x], 
 x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] && N 
eQ[a^2 - b^2, 0] && IntegerQ[2*m]
 

rule 4548

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

Maple [A] (verified)

Time = 0.40 (sec) , antiderivative size = 287, normalized size of antiderivative = 1.40

method	result	size
derivativedivides	\(\frac {\frac {\frac {2 \left (-\frac {\left (6 A \,a^{2} b +A a \,b^{2}-2 A \,b^{3}-4 B \,a^{3}-B \,a^{2} b \right ) a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{2 \left (a -b \right ) \left (a^{2}+2 a b +b^{2}\right )}+\frac {b a \left (6 A \,a^{2} b -A a \,b^{2}-2 A \,b^{3}-4 B \,a^{3}+B \,a^{2} b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a +b \right ) \left (a -b \right )^{2}}\right )}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a -b \right )^{2}}-\frac {\left (6 A \,a^{4} b -5 A \,a^{2} b^{3}+2 A \,b^{5}-2 B \,a^{5}-B \,a^{3} b^{2}\right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}}{a^{3}}+\frac {2 A \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{3}}}{d}\)	\(287\)
default	\(\frac {\frac {\frac {2 \left (-\frac {\left (6 A \,a^{2} b +A a \,b^{2}-2 A \,b^{3}-4 B \,a^{3}-B \,a^{2} b \right ) a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{2 \left (a -b \right ) \left (a^{2}+2 a b +b^{2}\right )}+\frac {b a \left (6 A \,a^{2} b -A a \,b^{2}-2 A \,b^{3}-4 B \,a^{3}+B \,a^{2} b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a +b \right ) \left (a -b \right )^{2}}\right )}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a -b \right )^{2}}-\frac {\left (6 A \,a^{4} b -5 A \,a^{2} b^{3}+2 A \,b^{5}-2 B \,a^{5}-B \,a^{3} b^{2}\right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}}{a^{3}}+\frac {2 A \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{3}}}{d}\)	\(287\)
risch	\(\text {Expression too large to display}\)	\(1170\)

Input:

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

Output:

1/d*(2/a^3*((-1/2*(6*A*a^2*b+A*a*b^2-2*A*b^3-4*B*a^3-B*a^2*b)*a*b/(a-b)/(a 
^2+2*a*b+b^2)*tan(1/2*d*x+1/2*c)^3+1/2*b*a*(6*A*a^2*b-A*a*b^2-2*A*b^3-4*B* 
a^3+B*a^2*b)/(a+b)/(a-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)^2-1/2*(6*A*a^4*b-5*A*a^2*b^3+2*A*b^5-2*B*a^5-B*a^ 
3*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)))+2*A/a^3*arctan(tan(1/2*d*x+1/2*c)))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 547 vs. \(2 (190) = 380\).

Time = 0.16 (sec) , antiderivative size = 1152, normalized size of antiderivative = 5.62 \[ \int \frac {A+B \sec (c+d x)}{(a+b \sec (c+d x))^3} \, dx=\text {Too large to display} \] Input:

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

Output:

[1/4*(4*(A*a^8 - 3*A*a^6*b^2 + 3*A*a^4*b^4 - A*a^2*b^6)*d*x*cos(d*x + c)^2 
 + 8*(A*a^7*b - 3*A*a^5*b^3 + 3*A*a^3*b^5 - A*a*b^7)*d*x*cos(d*x + c) + 4* 
(A*a^6*b^2 - 3*A*a^4*b^4 + 3*A*a^2*b^6 - A*b^8)*d*x - (2*B*a^5*b^2 - 6*A*a 
^4*b^3 + B*a^3*b^4 + 5*A*a^2*b^5 - 2*A*b^7 + (2*B*a^7 - 6*A*a^6*b + B*a^5* 
b^2 + 5*A*a^4*b^3 - 2*A*a^2*b^5)*cos(d*x + c)^2 + 2*(2*B*a^6*b - 6*A*a^5*b 
^2 + B*a^4*b^3 + 5*A*a^3*b^4 - 2*A*a*b^6)*cos(d*x + c))*sqrt(a^2 - b^2)*lo 
g((2*a*b*cos(d*x + c) - (a^2 - 2*b^2)*cos(d*x + c)^2 - 2*sqrt(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*(3*B*a^6*b^2 - 5*A*a^5*b^3 - 3*B*a^4*b^4 + 7*A* 
a^3*b^5 - 2*A*a*b^7 + (4*B*a^7*b - 6*A*a^6*b^2 - 5*B*a^5*b^3 + 9*A*a^4*b^4 
 + B*a^3*b^5 - 3*A*a^2*b^6)*cos(d*x + c))*sin(d*x + c))/((a^11 - 3*a^9*b^2 
 + 3*a^7*b^4 - a^5*b^6)*d*cos(d*x + c)^2 + 2*(a^10*b - 3*a^8*b^3 + 3*a^6*b 
^5 - a^4*b^7)*d*cos(d*x + c) + (a^9*b^2 - 3*a^7*b^4 + 3*a^5*b^6 - a^3*b^8) 
*d), 1/2*(2*(A*a^8 - 3*A*a^6*b^2 + 3*A*a^4*b^4 - A*a^2*b^6)*d*x*cos(d*x + 
c)^2 + 4*(A*a^7*b - 3*A*a^5*b^3 + 3*A*a^3*b^5 - A*a*b^7)*d*x*cos(d*x + c) 
+ 2*(A*a^6*b^2 - 3*A*a^4*b^4 + 3*A*a^2*b^6 - A*b^8)*d*x + (2*B*a^5*b^2 - 6 
*A*a^4*b^3 + B*a^3*b^4 + 5*A*a^2*b^5 - 2*A*b^7 + (2*B*a^7 - 6*A*a^6*b + B* 
a^5*b^2 + 5*A*a^4*b^3 - 2*A*a^2*b^5)*cos(d*x + c)^2 + 2*(2*B*a^6*b - 6*A*a 
^5*b^2 + B*a^4*b^3 + 5*A*a^3*b^4 - 2*A*a*b^6)*cos(d*x + c))*sqrt(-a^2 + b^ 
2)*arctan(-sqrt(-a^2 + b^2)*(b*cos(d*x + c) + a)/((a^2 - b^2)*sin(d*x +...
 

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

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

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. 457 vs. \(2 (190) = 380\).

Time = 0.25 (sec) , antiderivative size = 457, normalized size of antiderivative = 2.23 \[ \int \frac {A+B \sec (c+d x)}{(a+b \sec (c+d x))^3} \, dx=\frac {\frac {{\left (2 \, B a^{5} - 6 \, A a^{4} b + B a^{3} b^{2} + 5 \, A a^{2} b^{3} - 2 \, A b^{5}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {-a^{2} + b^{2}}}\right )\right )}}{{\left (a^{7} - 2 \, a^{5} b^{2} + a^{3} b^{4}\right )} \sqrt {-a^{2} + b^{2}}} + \frac {{\left (d x + c\right )} A}{a^{3}} + \frac {4 \, B a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 6 \, A a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3 \, B a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 5 \, A a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - B a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3 \, A a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, A b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 4 \, B a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, A a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, B a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 5 \, A a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + B a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, A a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, A b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a - b\right )}^{2}}}{d} \] Input:

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

Output:

((2*B*a^5 - 6*A*a^4*b + B*a^3*b^2 + 5*A*a^2*b^3 - 2*A*b^5)*(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*t 
an(1/2*d*x + 1/2*c))/sqrt(-a^2 + b^2)))/((a^7 - 2*a^5*b^2 + a^3*b^4)*sqrt( 
-a^2 + b^2)) + (d*x + c)*A/a^3 + (4*B*a^4*b*tan(1/2*d*x + 1/2*c)^3 - 6*A*a 
^3*b^2*tan(1/2*d*x + 1/2*c)^3 - 3*B*a^3*b^2*tan(1/2*d*x + 1/2*c)^3 + 5*A*a 
^2*b^3*tan(1/2*d*x + 1/2*c)^3 - B*a^2*b^3*tan(1/2*d*x + 1/2*c)^3 + 3*A*a*b 
^4*tan(1/2*d*x + 1/2*c)^3 - 2*A*b^5*tan(1/2*d*x + 1/2*c)^3 - 4*B*a^4*b*tan 
(1/2*d*x + 1/2*c) + 6*A*a^3*b^2*tan(1/2*d*x + 1/2*c) - 3*B*a^3*b^2*tan(1/2 
*d*x + 1/2*c) + 5*A*a^2*b^3*tan(1/2*d*x + 1/2*c) + B*a^2*b^3*tan(1/2*d*x + 
 1/2*c) - 3*A*a*b^4*tan(1/2*d*x + 1/2*c) - 2*A*b^5*tan(1/2*d*x + 1/2*c))/( 
(a^6 - 2*a^4*b^2 + a^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 = 21.33 (sec) , antiderivative size = 6909, normalized size of antiderivative = 33.70 \[ \int \frac {A+B \sec (c+d x)}{(a+b \sec (c+d x))^3} \, dx=\text {Too large to display} \] Input:

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

Output:

(2*A*atan(((A*((8*tan(c/2 + (d*x)/2)*(4*A^2*a^10 + 8*A^2*b^10 + 4*B^2*a^10 
 - 8*A^2*a*b^9 - 8*A^2*a^9*b - 32*A^2*a^2*b^8 + 32*A^2*a^3*b^7 + 57*A^2*a^ 
4*b^6 - 48*A^2*a^5*b^5 - 52*A^2*a^6*b^4 + 32*A^2*a^7*b^3 + 24*A^2*a^8*b^2 
+ B^2*a^6*b^4 + 4*B^2*a^8*b^2 - 24*A*B*a^9*b - 4*A*B*a^3*b^7 + 2*A*B*a^5*b 
^5 + 8*A*B*a^7*b^3))/(a^10*b + a^11 - a^4*b^7 - a^5*b^6 + 3*a^6*b^5 + 3*a^ 
7*b^4 - 3*a^8*b^3 - 3*a^9*b^2) + (A*((8*(4*A*a^15 + 4*B*a^15 - 4*A*a^6*b^9 
 + 2*A*a^7*b^8 + 18*A*a^8*b^7 - 4*A*a^9*b^6 - 36*A*a^10*b^5 + 6*A*a^11*b^4 
 + 34*A*a^12*b^3 - 8*A*a^13*b^2 - 2*B*a^8*b^7 + 2*B*a^9*b^6 + 6*B*a^12*b^3 
 - 6*B*a^13*b^2 - 12*A*a^14*b - 4*B*a^14*b))/(a^12*b + a^13 - a^6*b^7 - a^ 
7*b^6 + 3*a^8*b^5 + 3*a^9*b^4 - 3*a^10*b^3 - 3*a^11*b^2) - (A*tan(c/2 + (d 
*x)/2)*(8*a^15*b - 8*a^6*b^10 + 8*a^7*b^9 + 32*a^8*b^8 - 32*a^9*b^7 - 48*a 
^10*b^6 + 48*a^11*b^5 + 32*a^12*b^4 - 32*a^13*b^3 - 8*a^14*b^2)*8i)/(a^3*( 
a^10*b + a^11 - a^4*b^7 - a^5*b^6 + 3*a^6*b^5 + 3*a^7*b^4 - 3*a^8*b^3 - 3* 
a^9*b^2)))*1i)/a^3))/a^3 + (A*((8*tan(c/2 + (d*x)/2)*(4*A^2*a^10 + 8*A^2*b 
^10 + 4*B^2*a^10 - 8*A^2*a*b^9 - 8*A^2*a^9*b - 32*A^2*a^2*b^8 + 32*A^2*a^3 
*b^7 + 57*A^2*a^4*b^6 - 48*A^2*a^5*b^5 - 52*A^2*a^6*b^4 + 32*A^2*a^7*b^3 + 
 24*A^2*a^8*b^2 + B^2*a^6*b^4 + 4*B^2*a^8*b^2 - 24*A*B*a^9*b - 4*A*B*a^3*b 
^7 + 2*A*B*a^5*b^5 + 8*A*B*a^7*b^3))/(a^10*b + a^11 - a^4*b^7 - a^5*b^6 + 
3*a^6*b^5 + 3*a^7*b^4 - 3*a^8*b^3 - 3*a^9*b^2) - (A*((8*(4*A*a^15 + 4*B*a^ 
15 - 4*A*a^6*b^9 + 2*A*a^7*b^8 + 18*A*a^8*b^7 - 4*A*a^9*b^6 - 36*A*a^10...
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 380, normalized size of antiderivative = 1.85 \[ \int \frac {A+B \sec (c+d x)}{(a+b \sec (c+d x))^3} \, dx=\frac {-4 \sqrt {-a^{2}+b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b}{\sqrt {-a^{2}+b^{2}}}\right ) \cos \left (d x +c \right ) a^{3} b +2 \sqrt {-a^{2}+b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b}{\sqrt {-a^{2}+b^{2}}}\right ) \cos \left (d x +c \right ) a \,b^{3}-4 \sqrt {-a^{2}+b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b}{\sqrt {-a^{2}+b^{2}}}\right ) a^{2} b^{2}+2 \sqrt {-a^{2}+b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b}{\sqrt {-a^{2}+b^{2}}}\right ) b^{4}+\cos \left (d x +c \right ) a^{5} d x -2 \cos \left (d x +c \right ) a^{3} b^{2} d x +\cos \left (d x +c \right ) a \,b^{4} d x +\sin \left (d x +c \right ) a^{3} b^{2}-\sin \left (d x +c \right ) a \,b^{4}+a^{4} b d x -2 a^{2} b^{3} d x +b^{5} d x}{a^{2} d \left (\cos \left (d x +c \right ) a^{5}-2 \cos \left (d x +c \right ) a^{3} b^{2}+\cos \left (d x +c \right ) a \,b^{4}+a^{4} b -2 a^{2} b^{3}+b^{5}\right )} \] Input:

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

Output:

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