\(\int \frac {\sec ^2(c+d x) (A+B \sec (c+d x))}{a+b \sec (c+d x)} \, dx\) [313]

Optimal result

Integrand size = 31, antiderivative size = 98 \[ \int \frac {\sec ^2(c+d x) (A+B \sec (c+d x))}{a+b \sec (c+d x)} \, dx=\frac {(A b-a B) \text {arctanh}(\sin (c+d x))}{b^2 d}-\frac {2 a (A b-a B) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{\sqrt {a-b} b^2 \sqrt {a+b} d}+\frac {B \tan (c+d x)}{b d} \] Output:

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

Mathematica [A] (verified)

Time = 0.80 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.33 \[ \int \frac {\sec ^2(c+d x) (A+B \sec (c+d x))}{a+b \sec (c+d x)} \, dx=\frac {-\frac {2 a (-A b+a B) \text {arctanh}\left (\frac {(-a+b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}-(A b-a B) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )+b B \tan (c+d x)}{b^2 d} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.63 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.97, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.355, Rules used = {3042, 4498, 27, 3042, 4276, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4498

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4276

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4257

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

\(\Big \downarrow \) 4318

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3138

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

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

Defintions of rubi rules used

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 4257

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

rule 4276

Int[csc[(e_.) + (f_.)*(x_)]^2/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Sym 
bol] :> Simp[1/b   Int[Csc[e + f*x], x], x] - Simp[a/b   Int[Csc[e + f*x]/( 
a + b*Csc[e + f*x]), x], x] /; FreeQ[{a, b, e, f}, 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 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 = 0.36 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.47

method	result	size
derivativedivides	\(\frac {-\frac {B}{b \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\left (-A b +B a \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{b^{2}}-\frac {2 a \left (A b -B a \right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{b^{2} \sqrt {\left (a +b \right ) \left (a -b \right )}}-\frac {B}{b \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {\left (A b -B a \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{b^{2}}}{d}\)	\(144\)
default	\(\frac {-\frac {B}{b \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\left (-A b +B a \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{b^{2}}-\frac {2 a \left (A b -B a \right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{b^{2} \sqrt {\left (a +b \right ) \left (a -b \right )}}-\frac {B}{b \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {\left (A b -B a \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{b^{2}}}{d}\)	\(144\)
risch	\(\frac {2 i B}{d b \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{b d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B a}{b^{2} d}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-\sqrt {a^{2}-b^{2}}\, b}{\sqrt {a^{2}-b^{2}}\, a}\right ) A}{\sqrt {a^{2}-b^{2}}\, d b}-\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-\sqrt {a^{2}-b^{2}}\, b}{\sqrt {a^{2}-b^{2}}\, a}\right ) B}{\sqrt {a^{2}-b^{2}}\, d \,b^{2}}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{\sqrt {a^{2}-b^{2}}\, a}\right ) A}{\sqrt {a^{2}-b^{2}}\, d b}+\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{\sqrt {a^{2}-b^{2}}\, a}\right ) B}{\sqrt {a^{2}-b^{2}}\, d \,b^{2}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{b d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B a}{b^{2} d}\)	\(411\)

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 208 vs. \(2 (90) = 180\).

Time = 0.19 (sec) , antiderivative size = 472, normalized size of antiderivative = 4.82 \[ \int \frac {\sec ^2(c+d x) (A+B \sec (c+d x))}{a+b \sec (c+d x)} \, dx=\left [-\frac {{\left (B a^{2} - A a b\right )} \sqrt {a^{2} - b^{2}} \cos \left (d x + c\right ) \log \left (\frac {2 \, a b \cos \left (d x + c\right ) - {\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {a^{2} - b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right ) + 2 \, a^{2} - b^{2}}{a^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + b^{2}}\right ) + {\left (B a^{3} - A a^{2} b - B a b^{2} + A b^{3}\right )} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (B a^{3} - A a^{2} b - B a b^{2} + A b^{3}\right )} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (B a^{2} b - B b^{3}\right )} \sin \left (d x + c\right )}{2 \, {\left (a^{2} b^{2} - b^{4}\right )} d \cos \left (d x + c\right )}, \frac {2 \, {\left (B a^{2} - A a b\right )} \sqrt {-a^{2} + b^{2}} \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right )}\right ) \cos \left (d x + c\right ) - {\left (B a^{3} - A a^{2} b - B a b^{2} + A b^{3}\right )} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left (B a^{3} - A a^{2} b - B a b^{2} + A b^{3}\right )} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (B a^{2} b - B b^{3}\right )} \sin \left (d x + c\right )}{2 \, {\left (a^{2} b^{2} - b^{4}\right )} d \cos \left (d x + c\right )}\right ] \] Input:

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

Output:

[-1/2*((B*a^2 - A*a*b)*sqrt(a^2 - b^2)*cos(d*x + c)*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^3 - A*a^2*b - B*a*b^2 + A*b^3)*cos(d*x + c)*log(sin(d*x + c) + 1 
) - (B*a^3 - A*a^2*b - B*a*b^2 + A*b^3)*cos(d*x + c)*log(-sin(d*x + c) + 1 
) - 2*(B*a^2*b - B*b^3)*sin(d*x + c))/((a^2*b^2 - b^4)*d*cos(d*x + c)), 1/ 
2*(2*(B*a^2 - 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)))*cos(d*x + c) - (B*a^3 - A*a^2*b - B* 
a*b^2 + A*b^3)*cos(d*x + c)*log(sin(d*x + c) + 1) + (B*a^3 - A*a^2*b - B*a 
*b^2 + A*b^3)*cos(d*x + c)*log(-sin(d*x + c) + 1) + 2*(B*a^2*b - B*b^3)*si 
n(d*x + c))/((a^2*b^2 - b^4)*d*cos(d*x + c))]
 

Sympy [F]

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

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

Maxima [F(-2)]

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

integrate(sec(d*x+c)^2*(A+B*sec(d*x+c))/(a+b*sec(d*x+c)),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 = 176, normalized size of antiderivative = 1.80 \[ \int \frac {\sec ^2(c+d x) (A+B \sec (c+d x))}{a+b \sec (c+d x)} \, dx=-\frac {\frac {{\left (B a - A b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{b^{2}} - \frac {{\left (B a - A b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{b^{2}} + \frac {2 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} b} - \frac {2 \, {\left (B a^{2} - A a b\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {-a^{2} + b^{2}}}\right )\right )}}{\sqrt {-a^{2} + b^{2}} b^{2}}}{d} \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 12.42 (sec) , antiderivative size = 719, normalized size of antiderivative = 7.34 \[ \int \frac {\sec ^2(c+d x) (A+B \sec (c+d x))}{a+b \sec (c+d x)} \, dx =\text {Too large to display} \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.18 \[ \int \frac {\sec ^2(c+d x) (A+B \sec (c+d x))}{a+b \sec (c+d x)} \, dx=\frac {\sin \left (d x +c \right )}{\cos \left (d x +c \right ) d} \] Input:

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

Output:

sin(c + d*x)/(cos(c + d*x)*d)