\(\int \frac {\sec (c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^2} \, dx\) [323]

Optimal result

Integrand size = 29, antiderivative size = 100 \[ \int \frac {\sec (c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^2} \, dx=\frac {2 (a A-b B) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{3/2} (a+b)^{3/2} d}-\frac {(A b-a B) \tan (c+d x)}{\left (a^2-b^2\right ) d (a+b \sec (c+d x))} \] Output:

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

Mathematica [A] (verified)

Time = 0.39 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.97 \[ \int \frac {\sec (c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^2} \, dx=\frac {-\frac {2 (a A-b B) \text {arctanh}\left (\frac {(-a+b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2}}+\frac {(-A b+a B) \sin (c+d x)}{(a-b) (a+b) (b+a \cos (c+d x))}}{d} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.11, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.310, Rules used = {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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4491

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4318

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3138

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

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

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]
 

Maple [A] (verified)

Time = 0.23 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.32

method	result	size
derivativedivides	\(\frac {\frac {2 \left (A b -B a \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a -b \right )}+\frac {2 \left (A a -B b \right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a +b \right ) \left (a -b \right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}}{d}\)	\(132\)
default	\(\frac {\frac {2 \left (A b -B a \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a -b \right )}+\frac {2 \left (A a -B b \right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a +b \right ) \left (a -b \right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}}{d}\)	\(132\)
risch	\(\frac {2 i \left (-A b +B a \right ) \left (b \,{\mathrm e}^{i \left (d x +c \right )}+a \right )}{a \left (a^{2}-b^{2}\right ) d \left (a \,{\mathrm e}^{2 i \left (d x +c \right )}+2 b \,{\mathrm e}^{i \left (d x +c \right )}+a \right )}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{\sqrt {a^{2}-b^{2}}\, a}\right ) A}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{\sqrt {a^{2}-b^{2}}\, a}\right ) B b}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-\sqrt {a^{2}-b^{2}}\, b}{\sqrt {a^{2}-b^{2}}\, a}\right ) A a}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-\sqrt {a^{2}-b^{2}}\, b}{\sqrt {a^{2}-b^{2}}\, a}\right ) B b}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d}\)	\(400\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 389, normalized size of antiderivative = 3.89 \[ \int \frac {\sec (c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^2} \, dx=\left [\frac {{\left (A a b - B b^{2} + {\left (A a^{2} - B a b\right )} \cos \left (d x + c\right )\right )} \sqrt {a^{2} - b^{2}} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) - {\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {a^{2} - b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right ) + 2 \, a^{2} - b^{2}}{a^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + b^{2}}\right ) + 2 \, {\left (B a^{3} - A a^{2} b - B a b^{2} + A b^{3}\right )} \sin \left (d x + c\right )}{2 \, {\left ({\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} d \cos \left (d x + c\right ) + {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} d\right )}}, \frac {{\left (A a b - B b^{2} + {\left (A a^{2} - B a b\right )} \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}} \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right )}\right ) + {\left (B a^{3} - A a^{2} b - B a b^{2} + A b^{3}\right )} \sin \left (d x + c\right )}{{\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} d \cos \left (d x + c\right ) + {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} d}\right ] \] Input:

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

Output:

[1/2*((A*a*b - B*b^2 + (A*a^2 - B*a*b)*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*sqrt(a^2 - b^2)*(b*c 
os(d*x + c) + a)*sin(d*x + c) + 2*a^2 - b^2)/(a^2*cos(d*x + c)^2 + 2*a*b*c 
os(d*x + c) + b^2)) + 2*(B*a^3 - A*a^2*b - B*a*b^2 + A*b^3)*sin(d*x + c))/ 
((a^5 - 2*a^3*b^2 + a*b^4)*d*cos(d*x + c) + (a^4*b - 2*a^2*b^3 + b^5)*d), 
((A*a*b - B*b^2 + (A*a^2 - B*a*b)*cos(d*x + c))*sqrt(-a^2 + b^2)*arctan(-s 
qrt(-a^2 + b^2)*(b*cos(d*x + c) + a)/((a^2 - b^2)*sin(d*x + c))) + (B*a^3 
- A*a^2*b - B*a*b^2 + A*b^3)*sin(d*x + c))/((a^5 - 2*a^3*b^2 + a*b^4)*d*co 
s(d*x + c) + (a^4*b - 2*a^2*b^3 + b^5)*d)]
 

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

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

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 [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.72 \[ \int \frac {\sec (c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^2} \, dx=-\frac {2 \, {\left (\frac {{\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (2 \, a - 2 \, b\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {-a^{2} + b^{2}}}\right )\right )} {\left (A a - B b\right )}}{{\left (a^{2} - b^{2}\right )} \sqrt {-a^{2} + b^{2}}} + \frac {B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a - b\right )} {\left (a^{2} - b^{2}\right )}}\right )}}{d} \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 11.94 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.06 \[ \int \frac {\sec (c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^2} \, dx=\frac {2\,\mathrm {atanh}\left (\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a-b}}{\sqrt {a+b}}\right )\,\left (A\,a-B\,b\right )}{d\,{\left (a+b\right )}^{3/2}\,{\left (a-b\right )}^{3/2}}-\frac {2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (A\,b-B\,a\right )}{d\,\left (a+b\right )\,\left (a-b\right )\,\left (\left (b-a\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a+b\right )} \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.64 \[ \int \frac {\sec (c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^2} \, dx=\frac {2 \sqrt {-a^{2}+b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b}{\sqrt {-a^{2}+b^{2}}}\right )}{d \left (a^{2}-b^{2}\right )} \] Input:

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

Output:

(2*sqrt( - a**2 + b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqr 
t( - a**2 + b**2)))/(d*(a**2 - b**2))