\(\int \frac {\cos ^2(e+f x)}{a+b \sec ^2(e+f x)} \, dx\) [190]

Optimal result

Integrand size = 23, antiderivative size = 75 \[ \int \frac {\cos ^2(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\frac {(a-2 b) x}{2 a^2}+\frac {b^{3/2} \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{a^2 \sqrt {a+b} f}+\frac {\cos (e+f x) \sin (e+f x)}{2 a f} \] Output:

1/2*(a-2*b)*x/a^2+b^(3/2)*arctan(b^(1/2)*tan(f*x+e)/(a+b)^(1/2))/a^2/(a+b) 
^(1/2)/f+1/2*cos(f*x+e)*sin(f*x+e)/a/f
 

Mathematica [A] (verified)

Time = 0.38 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.89 \[ \int \frac {\cos ^2(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\frac {2 (a-2 b) (e+f x)+\frac {4 b^{3/2} \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{\sqrt {a+b}}+a \sin (2 (e+f x))}{4 a^2 f} \] Input:

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

Output:

(2*(a - 2*b)*(e + f*x) + (4*b^(3/2)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + 
 b]])/Sqrt[a + b] + a*Sin[2*(e + f*x)])/(4*a^2*f)
 

Rubi [A] (verified)

Time = 0.28 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.21, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3042, 4634, 316, 25, 397, 216, 218}

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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\sec (e+f x)^2 \left (a+b \sec (e+f x)^2\right )}dx\)

\(\Big \downarrow \) 4634

\(\displaystyle \frac {\int \frac {1}{\left (\tan ^2(e+f x)+1\right )^2 \left (b \tan ^2(e+f x)+a+b\right )}d\tan (e+f x)}{f}\)

\(\Big \downarrow \) 316

\(\displaystyle \frac {\frac {\tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right )}-\frac {\int -\frac {b \tan ^2(e+f x)+a-b}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a+b\right )}d\tan (e+f x)}{2 a}}{f}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {\int \frac {b \tan ^2(e+f x)+a-b}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a+b\right )}d\tan (e+f x)}{2 a}+\frac {\tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right )}}{f}\)

\(\Big \downarrow \) 397

\(\displaystyle \frac {\frac {\frac {2 b^2 \int \frac {1}{b \tan ^2(e+f x)+a+b}d\tan (e+f x)}{a}+\frac {(a-2 b) \int \frac {1}{\tan ^2(e+f x)+1}d\tan (e+f x)}{a}}{2 a}+\frac {\tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right )}}{f}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {\frac {\frac {2 b^2 \int \frac {1}{b \tan ^2(e+f x)+a+b}d\tan (e+f x)}{a}+\frac {(a-2 b) \arctan (\tan (e+f x))}{a}}{2 a}+\frac {\tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right )}}{f}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {\frac {\frac {2 b^{3/2} \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{a \sqrt {a+b}}+\frac {(a-2 b) \arctan (\tan (e+f x))}{a}}{2 a}+\frac {\tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right )}}{f}\)

Input:

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

Output:

((((a - 2*b)*ArcTan[Tan[e + f*x]])/a + (2*b^(3/2)*ArcTan[(Sqrt[b]*Tan[e + 
f*x])/Sqrt[a + b]])/(a*Sqrt[a + b]))/(2*a) + Tan[e + f*x]/(2*a*(1 + Tan[e 
+ f*x]^2)))/f
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 216

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[b, 0])
 

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 316

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim 
p[(-b)*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(2*a*(p + 1)*(b*c - a*d)) 
), x] + Simp[1/(2*a*(p + 1)*(b*c - a*d))   Int[(a + b*x^2)^(p + 1)*(c + d*x 
^2)^q*Simp[b*c + 2*(p + 1)*(b*c - a*d) + d*b*(2*(p + q + 2) + 1)*x^2, x], x 
], x] /; FreeQ[{a, b, c, d, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &&  ! 
( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomialQ[a, b, c, d, 2, 
 p, q, x]
 

rule 397

Int[((e_) + (f_.)*(x_)^2)/(((a_) + (b_.)*(x_)^2)*((c_) + (d_.)*(x_)^2)), x_ 
Symbol] :> Simp[(b*e - a*f)/(b*c - a*d)   Int[1/(a + b*x^2), x], x] - Simp[ 
(d*e - c*f)/(b*c - a*d)   Int[1/(c + d*x^2), x], x] /; FreeQ[{a, b, c, d, e 
, f}, x]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4634

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_) 
)^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[ff/f 
Subst[Int[(1 + ff^2*x^2)^(m/2 - 1)*ExpandToSum[a + b*(1 + ff^2*x^2)^(n/2), 
x]^p, x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ 
[m/2] && IntegerQ[n/2]
 

Maple [A] (verified)

Time = 0.70 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.01

method	result	size
derivativedivides	\(\frac {\frac {b^{2} \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {\left (a +b \right ) b}}\right )}{a^{2} \sqrt {\left (a +b \right ) b}}+\frac {\frac {a \tan \left (f x +e \right )}{2+2 \tan \left (f x +e \right )^{2}}+\frac {\left (a -2 b \right ) \arctan \left (\tan \left (f x +e \right )\right )}{2}}{a^{2}}}{f}\)	\(76\)
default	\(\frac {\frac {b^{2} \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {\left (a +b \right ) b}}\right )}{a^{2} \sqrt {\left (a +b \right ) b}}+\frac {\frac {a \tan \left (f x +e \right )}{2+2 \tan \left (f x +e \right )^{2}}+\frac {\left (a -2 b \right ) \arctan \left (\tan \left (f x +e \right )\right )}{2}}{a^{2}}}{f}\)	\(76\)
risch	\(\frac {x}{2 a}-\frac {x b}{a^{2}}-\frac {i {\mathrm e}^{2 i \left (f x +e \right )}}{8 a f}+\frac {i {\mathrm e}^{-2 i \left (f x +e \right )}}{8 a f}+\frac {\sqrt {-\left (a +b \right ) b}\, b \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {-\left (a +b \right ) b}-a -2 b}{a}\right )}{2 \left (a +b \right ) f \,a^{2}}-\frac {\sqrt {-\left (a +b \right ) b}\, b \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {-\left (a +b \right ) b}+a +2 b}{a}\right )}{2 \left (a +b \right ) f \,a^{2}}\)	\(160\)

Input:

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

Output:

1/f*(b^2/a^2/((a+b)*b)^(1/2)*arctan(b*tan(f*x+e)/((a+b)*b)^(1/2))+1/a^2*(1 
/2*a*tan(f*x+e)/(1+tan(f*x+e)^2)+1/2*(a-2*b)*arctan(tan(f*x+e))))
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 272, normalized size of antiderivative = 3.63 \[ \int \frac {\cos ^2(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\left [\frac {2 \, {\left (a - 2 \, b\right )} f x + 2 \, a \cos \left (f x + e\right ) \sin \left (f x + e\right ) + b \sqrt {-\frac {b}{a + b}} \log \left (\frac {{\left (a^{2} + 8 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (3 \, a b + 4 \, b^{2}\right )} \cos \left (f x + e\right )^{2} - 4 \, {\left ({\left (a^{2} + 3 \, a b + 2 \, b^{2}\right )} \cos \left (f x + e\right )^{3} - {\left (a b + b^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {-\frac {b}{a + b}} \sin \left (f x + e\right ) + b^{2}}{a^{2} \cos \left (f x + e\right )^{4} + 2 \, a b \cos \left (f x + e\right )^{2} + b^{2}}\right )}{4 \, a^{2} f}, \frac {{\left (a - 2 \, b\right )} f x + a \cos \left (f x + e\right ) \sin \left (f x + e\right ) - b \sqrt {\frac {b}{a + b}} \arctan \left (\frac {{\left ({\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{2} - b\right )} \sqrt {\frac {b}{a + b}}}{2 \, b \cos \left (f x + e\right ) \sin \left (f x + e\right )}\right )}{2 \, a^{2} f}\right ] \] Input:

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

Output:

[1/4*(2*(a - 2*b)*f*x + 2*a*cos(f*x + e)*sin(f*x + e) + b*sqrt(-b/(a + b)) 
*log(((a^2 + 8*a*b + 8*b^2)*cos(f*x + e)^4 - 2*(3*a*b + 4*b^2)*cos(f*x + e 
)^2 - 4*((a^2 + 3*a*b + 2*b^2)*cos(f*x + e)^3 - (a*b + b^2)*cos(f*x + e))* 
sqrt(-b/(a + b))*sin(f*x + e) + b^2)/(a^2*cos(f*x + e)^4 + 2*a*b*cos(f*x + 
 e)^2 + b^2)))/(a^2*f), 1/2*((a - 2*b)*f*x + a*cos(f*x + e)*sin(f*x + e) - 
 b*sqrt(b/(a + b))*arctan(1/2*((a + 2*b)*cos(f*x + e)^2 - b)*sqrt(b/(a + b 
))/(b*cos(f*x + e)*sin(f*x + e))))/(a^2*f)]
 

Sympy [F]

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

integrate(cos(f*x+e)**2/(a+b*sec(f*x+e)**2),x)
 

Output:

Integral(cos(e + f*x)**2/(a + b*sec(e + f*x)**2), x)
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.96 \[ \int \frac {\cos ^2(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\frac {\frac {2 \, b^{2} \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{\sqrt {{\left (a + b\right )} b} a^{2}} + \frac {{\left (f x + e\right )} {\left (a - 2 \, b\right )}}{a^{2}} + \frac {\tan \left (f x + e\right )}{a \tan \left (f x + e\right )^{2} + a}}{2 \, f} \] Input:

integrate(cos(f*x+e)^2/(a+b*sec(f*x+e)^2),x, algorithm="maxima")
 

Output:

1/2*(2*b^2*arctan(b*tan(f*x + e)/sqrt((a + b)*b))/(sqrt((a + b)*b)*a^2) + 
(f*x + e)*(a - 2*b)/a^2 + tan(f*x + e)/(a*tan(f*x + e)^2 + a))/f
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.25 \[ \int \frac {\cos ^2(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\frac {\frac {2 \, {\left (\pi \left \lfloor \frac {f x + e}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (b\right ) + \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b + b^{2}}}\right )\right )} b^{2}}{\sqrt {a b + b^{2}} a^{2}} + \frac {{\left (f x + e\right )} {\left (a - 2 \, b\right )}}{a^{2}} + \frac {\tan \left (f x + e\right )}{{\left (\tan \left (f x + e\right )^{2} + 1\right )} a}}{2 \, f} \] Input:

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

Output:

1/2*(2*(pi*floor((f*x + e)/pi + 1/2)*sgn(b) + arctan(b*tan(f*x + e)/sqrt(a 
*b + b^2)))*b^2/(sqrt(a*b + b^2)*a^2) + (f*x + e)*(a - 2*b)/a^2 + tan(f*x 
+ e)/((tan(f*x + e)^2 + 1)*a))/f
 

Mupad [B] (verification not implemented)

Time = 17.43 (sec) , antiderivative size = 373, normalized size of antiderivative = 4.97 \[ \int \frac {\cos ^2(e+f x)}{a+b \sec ^2(e+f x)} \, dx=-\frac {2\,b^2\,\mathrm {atan}\left (\frac {\sin \left (e+f\,x\right )}{\cos \left (e+f\,x\right )}\right )-a\,\left (\frac {b\,\sin \left (2\,e+2\,f\,x\right )}{2}-b\,\mathrm {atan}\left (\frac {\sin \left (e+f\,x\right )}{\cos \left (e+f\,x\right )}\right )\right )-a^2\,\left (\frac {\sin \left (2\,e+2\,f\,x\right )}{2}+\mathrm {atan}\left (\frac {\sin \left (e+f\,x\right )}{\cos \left (e+f\,x\right )}\right )\right )+\mathrm {atan}\left (\frac {a\,\sin \left (e+f\,x\right )\,{\left (-b^4-a\,b^3\right )}^{3/2}\,4{}\mathrm {i}+b\,\sin \left (e+f\,x\right )\,{\left (-b^4-a\,b^3\right )}^{3/2}\,8{}\mathrm {i}+b^5\,\sin \left (e+f\,x\right )\,\sqrt {-b^4-a\,b^3}\,8{}\mathrm {i}+a\,b^4\,\sin \left (e+f\,x\right )\,\sqrt {-b^4-a\,b^3}\,12{}\mathrm {i}+a^4\,b\,\sin \left (e+f\,x\right )\,\sqrt {-b^4-a\,b^3}\,1{}\mathrm {i}+a^2\,b^3\,\sin \left (e+f\,x\right )\,\sqrt {-b^4-a\,b^3}\,1{}\mathrm {i}-a^3\,b^2\,\sin \left (e+f\,x\right )\,\sqrt {-b^4-a\,b^3}\,2{}\mathrm {i}}{-\cos \left (e+f\,x\right )\,a^5\,b^2+\cos \left (e+f\,x\right )\,a^4\,b^3+5\,\cos \left (e+f\,x\right )\,a^3\,b^4+3\,\cos \left (e+f\,x\right )\,a^2\,b^5}\right )\,\sqrt {-b^4-a\,b^3}\,2{}\mathrm {i}}{f\,\left (2\,a^3+2\,b\,a^2\right )} \] Input:

int(cos(e + f*x)^2/(a + b/cos(e + f*x)^2),x)
                                                                                    
                                                                                    
 

Output:

-(atan((a*sin(e + f*x)*(- a*b^3 - b^4)^(3/2)*4i + b*sin(e + f*x)*(- a*b^3 
- b^4)^(3/2)*8i + b^5*sin(e + f*x)*(- a*b^3 - b^4)^(1/2)*8i + a*b^4*sin(e 
+ f*x)*(- a*b^3 - b^4)^(1/2)*12i + a^4*b*sin(e + f*x)*(- a*b^3 - b^4)^(1/2 
)*1i + a^2*b^3*sin(e + f*x)*(- a*b^3 - b^4)^(1/2)*1i - a^3*b^2*sin(e + f*x 
)*(- a*b^3 - b^4)^(1/2)*2i)/(3*a^2*b^5*cos(e + f*x) + 5*a^3*b^4*cos(e + f* 
x) + a^4*b^3*cos(e + f*x) - a^5*b^2*cos(e + f*x)))*(- a*b^3 - b^4)^(1/2)*2 
i + 2*b^2*atan(sin(e + f*x)/cos(e + f*x)) - a*((b*sin(2*e + 2*f*x))/2 - b* 
atan(sin(e + f*x)/cos(e + f*x))) - a^2*(sin(2*e + 2*f*x)/2 + atan(sin(e + 
f*x)/cos(e + f*x))))/(f*(2*a^2*b + 2*a^3))
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.95 \[ \int \frac {\cos ^2(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\frac {2 \sqrt {b}\, \sqrt {a +b}\, \mathit {atan} \left (\frac {\sqrt {a +b}\, \tan \left (\frac {f x}{2}+\frac {e}{2}\right )-\sqrt {a}}{\sqrt {b}}\right ) b +2 \sqrt {b}\, \sqrt {a +b}\, \mathit {atan} \left (\frac {\sqrt {a +b}\, \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+\sqrt {a}}{\sqrt {b}}\right ) b +\cos \left (f x +e \right ) \sin \left (f x +e \right ) a^{2}+\cos \left (f x +e \right ) \sin \left (f x +e \right ) a b +a^{2} e +a^{2} f x -a b e -a b f x -2 b^{2} e -2 b^{2} f x}{2 a^{2} f \left (a +b \right )} \] Input:

int(cos(f*x+e)^2/(a+b*sec(f*x+e)^2),x)
 

Output:

(2*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*tan((e + f*x)/2) - sqrt(a))/sqrt( 
b))*b + 2*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*tan((e + f*x)/2) + sqrt(a) 
)/sqrt(b))*b + cos(e + f*x)*sin(e + f*x)*a**2 + cos(e + f*x)*sin(e + f*x)* 
a*b + a**2*e + a**2*f*x - a*b*e - a*b*f*x - 2*b**2*e - 2*b**2*f*x)/(2*a**2 
*f*(a + b))