\(\int \frac {\sec ^2(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^2} \, dx\) [322]

Optimal result

Integrand size = 31, antiderivative size = 131 \[ \int \frac {\sec ^2(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^2} \, dx=\frac {B \text {arctanh}(\sin (c+d x))}{b^2 d}-\frac {2 \left (A b^3+a^3 B-2 a b^2 B\right ) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{3/2} b^2 (a+b)^{3/2} d}+\frac {a (A b-a B) \tan (c+d x)}{b \left (a^2-b^2\right ) d (a+b \sec (c+d x))} \] Output:

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

Mathematica [A] (verified)

Time = 0.80 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.46 \[ \int \frac {\sec ^2(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^2} \, dx=\frac {\cos (c+d x) (A+B \sec (c+d x)) \left (\frac {2 \left (A b^3+a \left (a^2-2 b^2\right ) B\right ) \text {arctanh}\left (\frac {(-a+b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2}}-B \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+B \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+\frac {a b (-A b+a B) \sin (c+d x)}{(-a+b) (a+b) (b+a \cos (c+d x))}\right )}{b^2 d (B+A \cos (c+d x))} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.85 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.21, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.355, Rules used = {3042, 4497, 25, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4497

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4486

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4257

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

\(\Big \downarrow \) 4318

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3138

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

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

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 = 0.44 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.41

method	result	size
derivativedivides	\(\frac {-\frac {B \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{b^{2}}+\frac {B \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{b^{2}}+\frac {-\frac {2 a b \left (A b -B a \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a -b \right )}-\frac {2 \left (A \,b^{3}+B \,a^{3}-2 B a \,b^{2}\right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a +b \right ) \left (a -b \right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}}{b^{2}}}{d}\)	\(185\)
default	\(\frac {-\frac {B \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{b^{2}}+\frac {B \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{b^{2}}+\frac {-\frac {2 a b \left (A b -B a \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a -b \right )}-\frac {2 \left (A \,b^{3}+B \,a^{3}-2 B a \,b^{2}\right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a +b \right ) \left (a -b \right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}}{b^{2}}}{d}\)	\(185\)
risch	\(-\frac {2 i \left (-A b +B a \right ) \left (b \,{\mathrm e}^{i \left (d x +c \right )}+a \right )}{\left (a^{2}-b^{2}\right ) d b \left (a \,{\mathrm e}^{2 i \left (d x +c \right )}+2 b \,{\mathrm e}^{i \left (d x +c \right )}+a \right )}+\frac {b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-\sqrt {a^{2}-b^{2}}\, b}{\sqrt {a^{2}-b^{2}}\, a}\right ) A}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d}+\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-\sqrt {a^{2}-b^{2}}\, b}{\sqrt {a^{2}-b^{2}}\, a}\right ) B}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,b^{2}}-\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-\sqrt {a^{2}-b^{2}}\, b}{\sqrt {a^{2}-b^{2}}\, a}\right ) B a}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d}-\frac {b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{\sqrt {a^{2}-b^{2}}\, a}\right ) A}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d}-\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{\sqrt {a^{2}-b^{2}}\, a}\right ) B}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,b^{2}}+\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{\sqrt {a^{2}-b^{2}}\, a}\right ) B a}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{b^{2} d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{b^{2} d}\)	\(619\)

Input:

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

Output:

1/d*(-B/b^2*ln(tan(1/2*d*x+1/2*c)-1)+B/b^2*ln(tan(1/2*d*x+1/2*c)+1)+2/b^2* 
(-a*b*(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)-(A*b^3+B*a^3-2*B*a*b^2)/(a+b)/(a-b)/((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. 319 vs. \(2 (123) = 246\).

Time = 2.06 (sec) , antiderivative size = 694, normalized size of antiderivative = 5.30 \[ \int \frac {\sec ^2(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^2} \, dx =\text {Too large to display} \] Input:

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

Output:

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

Sympy [F]

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

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

Maxima [F(-2)]

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

integrate(sec(d*x+c)^2*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))^2,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 [A] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 231, normalized size of antiderivative = 1.76 \[ \int \frac {\sec ^2(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^2} \, dx=-\frac {\frac {2 \, {\left (B a^{3} - 2 \, B a b^{2} + A b^{3}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {-a^{2} + b^{2}}}\right )\right )}}{{\left (a^{2} b^{2} - b^{4}\right )} \sqrt {-a^{2} + b^{2}}} - \frac {B \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{b^{2}} + \frac {B \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{b^{2}} - \frac {2 \, {\left (B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (a^{2} b - b^{3}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a - b\right )}}}{d} \] Input:

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

Output:

-(2*(B*a^3 - 2*B*a*b^2 + A*b^3)*(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^2*b^2 - b^4)*sqrt(-a^2 + b^2)) - B*log(abs(tan(1/2*d*x + 
1/2*c) + 1))/b^2 + B*log(abs(tan(1/2*d*x + 1/2*c) - 1))/b^2 - 2*(B*a^2*tan 
(1/2*d*x + 1/2*c) - A*a*b*tan(1/2*d*x + 1/2*c))/((a^2*b - b^3)*(a*tan(1/2* 
d*x + 1/2*c)^2 - b*tan(1/2*d*x + 1/2*c)^2 - a - b)))/d
 

Mupad [B] (verification not implemented)

Time = 18.55 (sec) , antiderivative size = 3751, normalized size of antiderivative = 28.63 \[ \int \frac {\sec ^2(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^2} \, dx=\text {Too large to display} \] Input:

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

Output:

- (B*atan(((B*((B*((32*(A*a^2*b^7 - B*b^9 - A*b^9 - A*a^3*b^6 + B*a^2*b^7 
- 3*B*a^3*b^6 + B*a^5*b^4 + A*a*b^8 + 2*B*a*b^8))/(a*b^5 + b^6 - a^2*b^4 - 
 a^3*b^3) - (32*B*tan(c/2 + (d*x)/2)*(2*a*b^9 - 2*a^2*b^8 - 4*a^3*b^7 + 4* 
a^4*b^6 + 2*a^5*b^5 - 2*a^6*b^4))/(b^2*(a*b^4 + b^5 - a^2*b^3 - a^3*b^2))) 
)/b^2 - (32*tan(c/2 + (d*x)/2)*(A^2*b^6 + 2*B^2*a^6 + B^2*b^6 - 2*B^2*a*b^ 
5 - 2*B^2*a^5*b + 3*B^2*a^2*b^4 + 4*B^2*a^3*b^3 - 5*B^2*a^4*b^2 - 4*A*B*a* 
b^5 + 2*A*B*a^3*b^3))/(a*b^4 + b^5 - a^2*b^3 - a^3*b^2))*1i)/b^2 - (B*((B* 
((32*(A*a^2*b^7 - B*b^9 - A*b^9 - A*a^3*b^6 + B*a^2*b^7 - 3*B*a^3*b^6 + B* 
a^5*b^4 + A*a*b^8 + 2*B*a*b^8))/(a*b^5 + b^6 - a^2*b^4 - a^3*b^3) + (32*B* 
tan(c/2 + (d*x)/2)*(2*a*b^9 - 2*a^2*b^8 - 4*a^3*b^7 + 4*a^4*b^6 + 2*a^5*b^ 
5 - 2*a^6*b^4))/(b^2*(a*b^4 + b^5 - a^2*b^3 - a^3*b^2))))/b^2 + (32*tan(c/ 
2 + (d*x)/2)*(A^2*b^6 + 2*B^2*a^6 + B^2*b^6 - 2*B^2*a*b^5 - 2*B^2*a^5*b + 
3*B^2*a^2*b^4 + 4*B^2*a^3*b^3 - 5*B^2*a^4*b^2 - 4*A*B*a*b^5 + 2*A*B*a^3*b^ 
3))/(a*b^4 + b^5 - a^2*b^3 - a^3*b^2))*1i)/b^2)/((B*((B*((32*(A*a^2*b^7 - 
B*b^9 - A*b^9 - A*a^3*b^6 + B*a^2*b^7 - 3*B*a^3*b^6 + B*a^5*b^4 + A*a*b^8 
+ 2*B*a*b^8))/(a*b^5 + b^6 - a^2*b^4 - a^3*b^3) - (32*B*tan(c/2 + (d*x)/2) 
*(2*a*b^9 - 2*a^2*b^8 - 4*a^3*b^7 + 4*a^4*b^6 + 2*a^5*b^5 - 2*a^6*b^4))/(b 
^2*(a*b^4 + b^5 - a^2*b^3 - a^3*b^2))))/b^2 - (32*tan(c/2 + (d*x)/2)*(A^2* 
b^6 + 2*B^2*a^6 + B^2*b^6 - 2*B^2*a*b^5 - 2*B^2*a^5*b + 3*B^2*a^2*b^4 + 4* 
B^2*a^3*b^3 - 5*B^2*a^4*b^2 - 4*A*B*a*b^5 + 2*A*B*a^3*b^3))/(a*b^4 + b^...
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.04 \[ \int \frac {\sec ^2(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^2} \, dx=\frac {-2 \sqrt {-a^{2}+b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b}{\sqrt {-a^{2}+b^{2}}}\right ) a -\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a^{2}+\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) b^{2}+\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a^{2}-\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) b^{2}}{b d \left (a^{2}-b^{2}\right )} \] Input:

int(sec(d*x+c)^2*(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)/ 
sqrt( - a**2 + b**2))*a - log(tan((c + d*x)/2) - 1)*a**2 + log(tan((c + d* 
x)/2) - 1)*b**2 + log(tan((c + d*x)/2) + 1)*a**2 - log(tan((c + d*x)/2) + 
1)*b**2)/(b*d*(a**2 - b**2))