\(\int \frac {A+C \sec ^2(c+d x)}{a+b \sec (c+d x)} \, dx\) [679]

Optimal result

Integrand size = 25, antiderivative size = 88 \[ \int \frac {A+C \sec ^2(c+d x)}{a+b \sec (c+d x)} \, dx=\frac {A x}{a}+\frac {C \text {arctanh}(\sin (c+d x))}{b d}-\frac {2 \left (A b^2+a^2 C\right ) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a \sqrt {a-b} b \sqrt {a+b} d} \] Output:

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

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 0.59 (sec) , antiderivative size = 239, normalized size of antiderivative = 2.72 \[ \int \frac {A+C \sec ^2(c+d x)}{a+b \sec (c+d x)} \, dx=\frac {2 \left (C+A \cos ^2(c+d x)\right ) \left (\sqrt {a^2-b^2} \left (A b d x-a C \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+a C \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right ) \sqrt {(\cos (c)-i \sin (c))^2}+2 \left (A b^2+a^2 C\right ) \arctan \left (\frac {(i \cos (c)+\sin (c)) \left (a \sin (c)+(-b+a \cos (c)) \tan \left (\frac {d x}{2}\right )\right )}{\sqrt {a^2-b^2} \sqrt {(\cos (c)-i \sin (c))^2}}\right ) (i \cos (c)+\sin (c))\right )}{a b \sqrt {a^2-b^2} d (A+2 C+A \cos (2 (c+d x))) \sqrt {(\cos (c)-i \sin (c))^2}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.59 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.03, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 4539, 3042, 4257, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4539

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4257

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

\(\Big \downarrow \) 4407

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4318

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3138

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

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

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 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 4539

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))/(csc[(e_.) + (f_.)*(x_)]*(b_. 
) + (a_)), x_Symbol] :> Simp[C/b   Int[Csc[e + f*x], x], x] + Simp[1/b   In 
t[(A*b - a*C*Csc[e + f*x])/(a + b*Csc[e + f*x]), x], x] /; FreeQ[{a, b, e, 
f, A, C}, x]
 

Maple [A] (verified)

Time = 0.40 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.30

method	result	size
derivativedivides	\(\frac {\frac {2 A \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}+\frac {C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{b}-\frac {C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{b}-\frac {2 \left (A \,b^{2}+C \,a^{2}\right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{b a \sqrt {\left (a +b \right ) \left (a -b \right )}}}{d}\)	\(114\)
default	\(\frac {\frac {2 A \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}+\frac {C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{b}-\frac {C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{b}-\frac {2 \left (A \,b^{2}+C \,a^{2}\right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{b a \sqrt {\left (a +b \right ) \left (a -b \right )}}}{d}\)	\(114\)
risch	\(\frac {A x}{a}+\frac {b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) A}{\sqrt {a^{2}-b^{2}}\, d a}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) C}{\sqrt {a^{2}-b^{2}}\, d b}-\frac {b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) A}{\sqrt {a^{2}-b^{2}}\, d a}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) C}{\sqrt {a^{2}-b^{2}}\, d b}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{d b}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{d b}\)	\(345\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.41 (sec) , antiderivative size = 362, normalized size of antiderivative = 4.11 \[ \int \frac {A+C \sec ^2(c+d x)}{a+b \sec (c+d x)} \, dx=\left [\frac {2 \, {\left (A a^{2} b - A b^{3}\right )} d x + {\left (C a^{2} + A b^{2}\right )} \sqrt {a^{2} - b^{2}} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) - {\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {a^{2} - b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right ) + 2 \, a^{2} - b^{2}}{a^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + b^{2}}\right ) + {\left (C a^{3} - C a b^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (C a^{3} - C a b^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right )}{2 \, {\left (a^{3} b - a b^{3}\right )} d}, \frac {2 \, {\left (A a^{2} b - A b^{3}\right )} d x - 2 \, {\left (C a^{2} + A b^{2}\right )} \sqrt {-a^{2} + b^{2}} \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right )}\right ) + {\left (C a^{3} - C a b^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (C a^{3} - C a b^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right )}{2 \, {\left (a^{3} b - a b^{3}\right )} d}\right ] \] Input:

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

Output:

[1/2*(2*(A*a^2*b - A*b^3)*d*x + (C*a^2 + A*b^2)*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)) + (C*a^3 - C*a*b^2)*log(sin(d*x + c) + 1) - (C*a^3 - C*a*b^ 
2)*log(-sin(d*x + c) + 1))/((a^3*b - a*b^3)*d), 1/2*(2*(A*a^2*b - A*b^3)*d 
*x - 2*(C*a^2 + A*b^2)*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))) + (C*a^3 - C*a*b^2)*log(sin(d*x + 
c) + 1) - (C*a^3 - C*a*b^2)*log(-sin(d*x + c) + 1))/((a^3*b - a*b^3)*d)]
 

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

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

Time = 0.34 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.64 \[ \int \frac {A+C \sec ^2(c+d x)}{a+b \sec (c+d x)} \, dx=\frac {\frac {{\left (d x + c\right )} A}{a} + \frac {C \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{b} - \frac {C \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{b} - \frac {2 \, {\left (C a^{2} + A b^{2}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {-a^{2} + b^{2}}}\right )\right )}}{\sqrt {-a^{2} + b^{2}} a b}}{d} \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 17.99 (sec) , antiderivative size = 3656, normalized size of antiderivative = 41.55 \[ \int \frac {A+C \sec ^2(c+d x)}{a+b \sec (c+d x)} \, dx=\text {Too large to display} \] Input:

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

Output:

(2*A*atan((16384*A^5*b^4*tan(c/2 + (d*x)/2))/(16384*A^5*b^4 + 16384*A*C^4* 
a^4 + 32768*A^4*C*b^4 - (16384*A^5*b^5)/a + 32768*A^2*C^3*a^2*b^2 + 32768* 
A^3*C^2*a^2*b^2 - 16384*A*C^4*a^3*b - 32768*A^2*C^3*a*b^3 - 32768*A^3*C^2* 
a*b^3 - (32768*A^4*C*b^5)/a) + (16384*A^5*b^5*tan(c/2 + (d*x)/2))/(16384*A 
^5*b^5 - 16384*A*C^4*a^5 + 32768*A^4*C*b^5 - 16384*A^5*a*b^4 + 32768*A^2*C 
^3*a^2*b^3 - 32768*A^2*C^3*a^3*b^2 + 32768*A^3*C^2*a^2*b^3 - 32768*A^3*C^2 
*a^3*b^2 + 16384*A*C^4*a^4*b - 32768*A^4*C*a*b^4) + (16384*A*C^4*a^4*tan(c 
/2 + (d*x)/2))/(16384*A^5*b^4 + 16384*A*C^4*a^4 + 32768*A^4*C*b^4 - (16384 
*A^5*b^5)/a + 32768*A^2*C^3*a^2*b^2 + 32768*A^3*C^2*a^2*b^2 - 16384*A*C^4* 
a^3*b - 32768*A^2*C^3*a*b^3 - 32768*A^3*C^2*a*b^3 - (32768*A^4*C*b^5)/a) + 
 (32768*A^4*C*b^4*tan(c/2 + (d*x)/2))/(16384*A^5*b^4 + 16384*A*C^4*a^4 + 3 
2768*A^4*C*b^4 - (16384*A^5*b^5)/a + 32768*A^2*C^3*a^2*b^2 + 32768*A^3*C^2 
*a^2*b^2 - 16384*A*C^4*a^3*b - 32768*A^2*C^3*a*b^3 - 32768*A^3*C^2*a*b^3 - 
 (32768*A^4*C*b^5)/a) + (32768*A^4*C*b^5*tan(c/2 + (d*x)/2))/(16384*A^5*b^ 
5 - 16384*A*C^4*a^5 + 32768*A^4*C*b^5 - 16384*A^5*a*b^4 + 32768*A^2*C^3*a^ 
2*b^3 - 32768*A^2*C^3*a^3*b^2 + 32768*A^3*C^2*a^2*b^3 - 32768*A^3*C^2*a^3* 
b^2 + 16384*A*C^4*a^4*b - 32768*A^4*C*a*b^4) - (16384*A*C^4*a^3*b*tan(c/2 
+ (d*x)/2))/(16384*A^5*b^4 + 16384*A*C^4*a^4 + 32768*A^4*C*b^4 - (16384*A^ 
5*b^5)/a + 32768*A^2*C^3*a^2*b^2 + 32768*A^3*C^2*a^2*b^2 - 16384*A*C^4*a^3 
*b - 32768*A^2*C^3*a*b^3 - 32768*A^3*C^2*a*b^3 - (32768*A^4*C*b^5)/a) -...
 

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 208, normalized size of antiderivative = 2.36 \[ \int \frac {A+C \sec ^2(c+d x)}{a+b \sec (c+d x)} \, dx=\frac {-2 \sqrt {-a^{2}+b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b}{\sqrt {-a^{2}+b^{2}}}\right ) a c -2 \sqrt {-a^{2}+b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b}{\sqrt {-a^{2}+b^{2}}}\right ) b^{2}-\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a^{2} c +\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) b^{2} c +\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a^{2} c -\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) b^{2} c +a^{2} b d x -b^{3} d x}{b d \left (a^{2}-b^{2}\right )} \] Input:

int((A+C*sec(d*x+c)^2)/(a+b*sec(d*x+c)),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*c - 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**2 - log(tan((c + d*x)/2) 
- 1)*a**2*c + log(tan((c + d*x)/2) - 1)*b**2*c + log(tan((c + d*x)/2) + 1) 
*a**2*c - log(tan((c + d*x)/2) + 1)*b**2*c + a**2*b*d*x - b**3*d*x)/(b*d*( 
a**2 - b**2))