\(\int \frac {\sec ^2(c+d x)}{a+b \cos (c+d x)} \, dx\) [456]

Optimal result

Integrand size = 21, antiderivative size = 85 \[ \int \frac {\sec ^2(c+d x)}{a+b \cos (c+d x)} \, dx=\frac {2 b^2 \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^2 \sqrt {a-b} \sqrt {a+b} d}-\frac {b \text {arctanh}(\sin (c+d x))}{a^2 d}+\frac {\tan (c+d x)}{a d} \] Output:

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

Mathematica [A] (verified)

Time = 0.70 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.35 \[ \int \frac {\sec ^2(c+d x)}{a+b \cos (c+d x)} \, dx=\frac {-\frac {2 b^2 \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2+b^2}}\right )}{\sqrt {-a^2+b^2}}+b \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )+a \tan (c+d x)}{a^2 d} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.04, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.476, Rules used = {3042, 3281, 25, 27, 3042, 3226, 3042, 3138, 218, 4257}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3281

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3226

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3138

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

\(\Big \downarrow \) 218

\(\displaystyle \frac {\tan (c+d x)}{a d}-\frac {b \left (\frac {\int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a}-\frac {2 b \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a d \sqrt {a-b} \sqrt {a+b}}\right )}{a}\)

\(\Big \downarrow \) 4257

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

Input:

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

Output:

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

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 218

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R 
t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[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 3226

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + ( 
f_.)*(x_)])), x_Symbol] :> Simp[b/(b*c - a*d)   Int[1/(a + b*Sin[e + f*x]), 
 x], x] - Simp[d/(b*c - a*d)   Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[ 
{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]
 

rule 3281

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
 (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b^2)*Cos[e + f*x]*(a + b*Sin[e + f* 
x])^(m + 1)*((c + d*Sin[e + f*x])^(n + 1)/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2 
))), x] + Simp[1/((m + 1)*(b*c - a*d)*(a^2 - b^2))   Int[(a + b*Sin[e + f*x 
])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[a*(b*c - a*d)*(m + 1) + b^2*d*(m + n 
 + 2) - (b^2*c + b*(b*c - a*d)*(m + 1))*Sin[e + f*x] - b^2*d*(m + n + 3)*Si 
n[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a* 
d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -1] && IntegersQ[ 
2*m, 2*n] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(IntegerQ[2* 
n] && LtQ[n, -1] && ((IntegerQ[n] &&  !IntegerQ[m]) || EqQ[a, 0])))
 

rule 4257

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

Maple [A] (verified)

Time = 1.09 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.45

method	result	size
derivativedivides	\(\frac {-\frac {1}{a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{2}}+\frac {2 b^{2} \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{a^{2} \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {1}{a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{2}}}{d}\)	\(123\)
default	\(\frac {-\frac {1}{a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{2}}+\frac {2 b^{2} \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{a^{2} \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {1}{a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{2}}}{d}\)	\(123\)
risch	\(\frac {2 i}{d a \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}-\frac {b^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, d \,a^{2}}+\frac {b^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, d \,a^{2}}+\frac {b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{a^{2} d}-\frac {b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{a^{2} d}\)	\(216\)

Input:

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

Output:

1/d*(-1/a/(tan(1/2*d*x+1/2*c)-1)+b/a^2*ln(tan(1/2*d*x+1/2*c)-1)+2*b^2/a^2/ 
((a-b)*(a+b))^(1/2)*arctan((a-b)*tan(1/2*d*x+1/2*c)/((a-b)*(a+b))^(1/2))-1 
/a/(tan(1/2*d*x+1/2*c)+1)-b/a^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. 157 vs. \(2 (76) = 152\).

Time = 0.13 (sec) , antiderivative size = 382, normalized size of antiderivative = 4.49 \[ \int \frac {\sec ^2(c+d x)}{a+b \cos (c+d x)} \, dx=\left [-\frac {\sqrt {-a^{2} + b^{2}} b^{2} \cos \left (d x + c\right ) \log \left (\frac {2 \, a b \cos \left (d x + c\right ) + {\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {-a^{2} + b^{2}} {\left (a \cos \left (d x + c\right ) + b\right )} \sin \left (d x + c\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}}\right ) + {\left (a^{2} b - b^{3}\right )} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (a^{2} b - b^{3}\right )} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (a^{3} - a b^{2}\right )} \sin \left (d x + c\right )}{2 \, {\left (a^{4} - a^{2} b^{2}\right )} d \cos \left (d x + c\right )}, \frac {2 \, \sqrt {a^{2} - b^{2}} b^{2} \arctan \left (-\frac {a \cos \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \sin \left (d x + c\right )}\right ) \cos \left (d x + c\right ) - {\left (a^{2} b - b^{3}\right )} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left (a^{2} b - b^{3}\right )} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (a^{3} - a b^{2}\right )} \sin \left (d x + c\right )}{2 \, {\left (a^{4} - a^{2} b^{2}\right )} d \cos \left (d x + c\right )}\right ] \] Input:

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

Output:

[-1/2*(sqrt(-a^2 + b^2)*b^2*cos(d*x + c)*log((2*a*b*cos(d*x + c) + (2*a^2 
- b^2)*cos(d*x + c)^2 + 2*sqrt(-a^2 + b^2)*(a*cos(d*x + c) + b)*sin(d*x + 
c) - a^2 + 2*b^2)/(b^2*cos(d*x + c)^2 + 2*a*b*cos(d*x + c) + a^2)) + (a^2* 
b - b^3)*cos(d*x + c)*log(sin(d*x + c) + 1) - (a^2*b - b^3)*cos(d*x + c)*l 
og(-sin(d*x + c) + 1) - 2*(a^3 - a*b^2)*sin(d*x + c))/((a^4 - a^2*b^2)*d*c 
os(d*x + c)), 1/2*(2*sqrt(a^2 - b^2)*b^2*arctan(-(a*cos(d*x + c) + b)/(sqr 
t(a^2 - b^2)*sin(d*x + c)))*cos(d*x + c) - (a^2*b - b^3)*cos(d*x + c)*log( 
sin(d*x + c) + 1) + (a^2*b - b^3)*cos(d*x + c)*log(-sin(d*x + c) + 1) + 2* 
(a^3 - a*b^2)*sin(d*x + c))/((a^4 - a^2*b^2)*d*cos(d*x + c))]
 

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

integrate(sec(d*x+c)^2/(a+b*cos(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*b^2-4*a^2>0)', see `assume?` f 
or more de
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 153 vs. \(2 (76) = 152\).

Time = 0.56 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.80 \[ \int \frac {\sec ^2(c+d x)}{a+b \cos (c+d x)} \, dx=-\frac {\frac {2 \, {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )} b^{2}}{\sqrt {a^{2} - b^{2}} a^{2}} + \frac {b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{2}} - \frac {b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{2}} + \frac {2 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} a}}{d} \] Input:

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

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

Mupad [B] (verification not implemented)

Time = 40.85 (sec) , antiderivative size = 324, normalized size of antiderivative = 3.81 \[ \int \frac {\sec ^2(c+d x)}{a+b \cos (c+d x)} \, dx=\frac {a^3\,\sin \left (c+d\,x\right )-a\,b^2\,\sin \left (c+d\,x\right )}{a^2\,d\,\cos \left (c+d\,x\right )\,\left (a^2-b^2\right )}-\frac {2\,a^2\,b\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )-2\,b^3\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+2\,b^2\,\mathrm {atanh}\left (\frac {a^5\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}+2\,b^3\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,{\left (b^2-a^2\right )}^{3/2}-2\,b^5\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}+3\,a^2\,b^3\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}-a^3\,b^2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}-a^4\,b\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,{\left (a\,b^2-a^3\right )}^2}\right )\,\sqrt {b^2-a^2}}{a^2\,d\,\left (a^2-b^2\right )} \] Input:

int(1/(cos(c + d*x)^2*(a + b*cos(c + d*x))),x)
 

Output:

(a^3*sin(c + d*x) - a*b^2*sin(c + d*x))/(a^2*d*cos(c + d*x)*(a^2 - b^2)) - 
 (2*a^2*b*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)) - 2*b^3*atanh(sin(c 
/2 + (d*x)/2)/cos(c/2 + (d*x)/2)) + 2*b^2*atanh((a^5*sin(c/2 + (d*x)/2)*(b 
^2 - a^2)^(1/2) + 2*b^3*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(3/2) - 2*b^5*sin(c 
/2 + (d*x)/2)*(b^2 - a^2)^(1/2) + 3*a^2*b^3*sin(c/2 + (d*x)/2)*(b^2 - a^2) 
^(1/2) - a^3*b^2*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2) - a^4*b*sin(c/2 + (d 
*x)/2)*(b^2 - a^2)^(1/2))/(cos(c/2 + (d*x)/2)*(a*b^2 - a^3)^2))*(b^2 - a^2 
)^(1/2))/(a^2*d*(a^2 - b^2))
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 200, normalized size of antiderivative = 2.35 \[ \int \frac {\sec ^2(c+d x)}{a+b \cos (c+d x)} \, dx=\frac {2 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b}{\sqrt {a^{2}-b^{2}}}\right ) \cos \left (d x +c \right ) b^{2}+\cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a^{2} b -\cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) b^{3}-\cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a^{2} b +\cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) b^{3}+\sin \left (d x +c \right ) a^{3}-\sin \left (d x +c \right ) a \,b^{2}}{\cos \left (d x +c \right ) a^{2} d \left (a^{2}-b^{2}\right )} \] Input:

int(sec(d*x+c)^2/(a+b*cos(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))*cos(c + d*x)*b**2 + cos(c + d*x)*log(tan((c + d*x)/2) - 1)*a* 
*2*b - cos(c + d*x)*log(tan((c + d*x)/2) - 1)*b**3 - cos(c + d*x)*log(tan( 
(c + d*x)/2) + 1)*a**2*b + cos(c + d*x)*log(tan((c + d*x)/2) + 1)*b**3 + s 
in(c + d*x)*a**3 - sin(c + d*x)*a*b**2)/(cos(c + d*x)*a**2*d*(a**2 - b**2) 
)