\(\int \frac {\sec (c+d x) (A+B \sec (c+d x))}{a+b \sec (c+d x)} \, dx\) [314]

Optimal result

Integrand size = 29, antiderivative size = 76 \[ \int \frac {\sec (c+d x) (A+B \sec (c+d x))}{a+b \sec (c+d x)} \, dx=\frac {B \text {arctanh}(\sin (c+d x))}{b d}+\frac {2 (A b-a B) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{\sqrt {a-b} b \sqrt {a+b} d} \] Output:

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

Mathematica [A] (verified)

Time = 0.38 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.47 \[ \int \frac {\sec (c+d x) (A+B \sec (c+d x))}{a+b \sec (c+d x)} \, dx=\frac {\frac {2 (-A b+a B) \text {arctanh}\left (\frac {(-a+b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}+B \left (-\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+\log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )}{b d} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.276, Rules used = {3042, 4486, 3042, 4257, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4486

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4257

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

\(\Big \downarrow \) 4318

\(\displaystyle \frac {(A b-a B) \int \frac {1}{\frac {a \cos (c+d x)}{b}+1}dx}{b^2}+\frac {B \text {arctanh}(\sin (c+d x))}{b d}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3138

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

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

Defintions of rubi rules used

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 4257

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

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 4486

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

Maple [A] (verified)

Time = 0.29 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.21

method	result	size
derivativedivides	\(\frac {-\frac {2 \left (-A b +B a \right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{b \sqrt {\left (a +b \right ) \left (a -b \right )}}-\frac {B \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{b}+\frac {B \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{b}}{d}\)	\(92\)
default	\(\frac {-\frac {2 \left (-A b +B a \right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{b \sqrt {\left (a +b \right ) \left (a -b \right )}}-\frac {B \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{b}+\frac {B \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{b}}{d}\)	\(92\)
risch	\(\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{\sqrt {a^{2}-b^{2}}\, a}\right ) A}{\sqrt {a^{2}-b^{2}}\, d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{\sqrt {a^{2}-b^{2}}\, a}\right ) B a}{\sqrt {a^{2}-b^{2}}\, d b}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {-i a^{2}+i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{\sqrt {a^{2}-b^{2}}\, a}\right ) A}{\sqrt {a^{2}-b^{2}}\, d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {-i a^{2}+i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{\sqrt {a^{2}-b^{2}}\, a}\right ) B a}{\sqrt {a^{2}-b^{2}}\, d b}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{d b}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{d b}\)	\(327\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.43 (sec) , antiderivative size = 316, normalized size of antiderivative = 4.16 \[ \int \frac {\sec (c+d x) (A+B \sec (c+d x))}{a+b \sec (c+d x)} \, dx=\left [-\frac {{\left (B a - A b\right )} \sqrt {a^{2} - b^{2}} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) - {\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {a^{2} - b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right ) + 2 \, a^{2} - b^{2}}{a^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + b^{2}}\right ) - {\left (B a^{2} - B b^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left (B a^{2} - B b^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right )}{2 \, {\left (a^{2} b - b^{3}\right )} d}, -\frac {2 \, {\left (B a - A b\right )} \sqrt {-a^{2} + b^{2}} \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right )}\right ) - {\left (B a^{2} - B b^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left (B a^{2} - B b^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right )}{2 \, {\left (a^{2} b - b^{3}\right )} d}\right ] \] Input:

integrate(sec(d*x+c)*(A+B*sec(d*x+c))/(a+b*sec(d*x+c)),x, algorithm="frica 
s")
 

Output:

[-1/2*((B*a - A*b)*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*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)) - (B*a^2 - B*b 
^2)*log(sin(d*x + c) + 1) + (B*a^2 - B*b^2)*log(-sin(d*x + c) + 1))/((a^2* 
b - b^3)*d), -1/2*(2*(B*a - A*b)*sqrt(-a^2 + b^2)*arctan(-sqrt(-a^2 + b^2) 
*(b*cos(d*x + c) + a)/((a^2 - b^2)*sin(d*x + c))) - (B*a^2 - B*b^2)*log(si 
n(d*x + c) + 1) + (B*a^2 - B*b^2)*log(-sin(d*x + c) + 1))/((a^2*b - b^3)*d 
)]
 

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

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

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 = 0.16 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.67 \[ \int \frac {\sec (c+d x) (A+B \sec (c+d x))}{a+b \sec (c+d x)} \, dx=\frac {\frac {B \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{b} - \frac {B \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{b} + \frac {2 \, {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (2 \, a - 2 \, b\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {-a^{2} + b^{2}}}\right )\right )} {\left (B a - A b\right )}}{\sqrt {-a^{2} + b^{2}} b}}{d} \] Input:

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

Output:

(B*log(abs(tan(1/2*d*x + 1/2*c) + 1))/b - B*log(abs(tan(1/2*d*x + 1/2*c) - 
 1))/b + 2*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(2*a - 2*b) + arctan((a*ta 
n(1/2*d*x + 1/2*c) - b*tan(1/2*d*x + 1/2*c))/sqrt(-a^2 + b^2)))*(B*a - A*b 
)/(sqrt(-a^2 + b^2)*b))/d
 

Mupad [B] (verification not implemented)

Time = 12.58 (sec) , antiderivative size = 573, normalized size of antiderivative = 7.54 \[ \int \frac {\sec (c+d x) (A+B \sec (c+d x))}{a+b \sec (c+d x)} \, dx =\text {Too large to display} \] Input:

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

Output:

(A*a^2*log((a*sin(c/2 + (d*x)/2) - b*sin(c/2 + (d*x)/2) + cos(c/2 + (d*x)/ 
2)*(a^2 - b^2)^(1/2))/cos(c/2 + (d*x)/2)))/(d*(a^2 - b^2)^(3/2)) - (A*log( 
(a*cos(c/2 + (d*x)/2) + b*cos(c/2 + (d*x)/2) - sin(c/2 + (d*x)/2)*(a^2 - b 
^2)^(1/2))/cos(c/2 + (d*x)/2))*((a + b)*(a - b))^(1/2))/(d*(a^2 - b^2)) - 
(2*B*b*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/(d*(a^2 - b^2)) - (A* 
b^2*log((a*sin(c/2 + (d*x)/2) - b*sin(c/2 + (d*x)/2) + cos(c/2 + (d*x)/2)* 
(a^2 - b^2)^(1/2))/cos(c/2 + (d*x)/2)))/(d*(a^2 - b^2)^(3/2)) + (2*B*a^2*a 
tanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/(b*d*(a^2 - b^2)) - (B*a^3*lo 
g((a*sin(c/2 + (d*x)/2) - b*sin(c/2 + (d*x)/2) + cos(c/2 + (d*x)/2)*(a^2 - 
 b^2)^(1/2))/cos(c/2 + (d*x)/2)))/(b*d*(a^2 - b^2)^(3/2)) + (B*a*b*log((a* 
sin(c/2 + (d*x)/2) - b*sin(c/2 + (d*x)/2) + cos(c/2 + (d*x)/2)*(a^2 - b^2) 
^(1/2))/cos(c/2 + (d*x)/2)))/(d*(a^2 - b^2)^(3/2)) + (B*a*log((a*cos(c/2 + 
 (d*x)/2) + b*cos(c/2 + (d*x)/2) - sin(c/2 + (d*x)/2)*(a^2 - b^2)^(1/2))/c 
os(c/2 + (d*x)/2))*((a + b)*(a - b))^(1/2))/(b*d*(a^2 - b^2))
                                                                                    
                                                                                    
 

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.41 \[ \int \frac {\sec (c+d x) (A+B \sec (c+d x))}{a+b \sec (c+d x)} \, dx=\frac {-\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d} \] Input:

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

Output:

( - log(tan((c + d*x)/2) - 1) + log(tan((c + d*x)/2) + 1))/d