\(\int \frac {\cos ^2(x)}{a+b \csc (x)} \, dx\) [11]

Optimal result

Integrand size = 13, antiderivative size = 80 \[ \int \frac {\cos ^2(x)}{a+b \csc (x)} \, dx=\frac {\left (a^2-2 b^2\right ) x}{2 a^3}+\frac {2 b \sqrt {a^2-b^2} \text {arctanh}\left (\frac {a+b \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{a^3}-\frac {\cos (x) (2 b-a \sin (x))}{2 a^2} \] Output:

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

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.94 \[ \int \frac {\cos ^2(x)}{a+b \csc (x)} \, dx=\frac {2 a^2 x-4 b^2 x+8 b \sqrt {-a^2+b^2} \arctan \left (\frac {a+b \tan \left (\frac {x}{2}\right )}{\sqrt {-a^2+b^2}}\right )-4 a b \cos (x)+a^2 \sin (2 x)}{4 a^3} \] Input:

Integrate[Cos[x]^2/(a + b*Csc[x]),x]
 

Output:

(2*a^2*x - 4*b^2*x + 8*b*Sqrt[-a^2 + b^2]*ArcTan[(a + b*Tan[x/2])/Sqrt[-a^ 
2 + b^2]] - 4*a*b*Cos[x] + a^2*Sin[2*x])/(4*a^3)
 

Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.15, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.846, Rules used = {3042, 4360, 3042, 3344, 25, 3042, 3214, 3042, 3139, 1083, 219}

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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\cos (x)^2}{a+b \csc (x)}dx\)

\(\Big \downarrow \) 4360

\(\displaystyle \int \frac {\sin (x) \cos ^2(x)}{a \sin (x)+b}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sin (x) \cos (x)^2}{a \sin (x)+b}dx\)

\(\Big \downarrow \) 3344

\(\displaystyle \frac {\int -\frac {a b-\left (a^2-2 b^2\right ) \sin (x)}{b+a \sin (x)}dx}{2 a^2}-\frac {\cos (x) (2 b-a \sin (x))}{2 a^2}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {\int \frac {a b-\left (a^2-2 b^2\right ) \sin (x)}{b+a \sin (x)}dx}{2 a^2}-\frac {\cos (x) (2 b-a \sin (x))}{2 a^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\int \frac {a b-\left (a^2-2 b^2\right ) \sin (x)}{b+a \sin (x)}dx}{2 a^2}-\frac {\cos (x) (2 b-a \sin (x))}{2 a^2}\)

\(\Big \downarrow \) 3214

\(\displaystyle -\frac {\frac {2 b \left (a^2-b^2\right ) \int \frac {1}{b+a \sin (x)}dx}{a}-\frac {x \left (a^2-2 b^2\right )}{a}}{2 a^2}-\frac {\cos (x) (2 b-a \sin (x))}{2 a^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\frac {2 b \left (a^2-b^2\right ) \int \frac {1}{b+a \sin (x)}dx}{a}-\frac {x \left (a^2-2 b^2\right )}{a}}{2 a^2}-\frac {\cos (x) (2 b-a \sin (x))}{2 a^2}\)

\(\Big \downarrow \) 3139

\(\displaystyle -\frac {\frac {4 b \left (a^2-b^2\right ) \int \frac {1}{b \tan ^2\left (\frac {x}{2}\right )+2 a \tan \left (\frac {x}{2}\right )+b}d\tan \left (\frac {x}{2}\right )}{a}-\frac {x \left (a^2-2 b^2\right )}{a}}{2 a^2}-\frac {\cos (x) (2 b-a \sin (x))}{2 a^2}\)

\(\Big \downarrow \) 1083

\(\displaystyle -\frac {-\frac {8 b \left (a^2-b^2\right ) \int \frac {1}{4 \left (a^2-b^2\right )-\left (2 a+2 b \tan \left (\frac {x}{2}\right )\right )^2}d\left (2 a+2 b \tan \left (\frac {x}{2}\right )\right )}{a}-\frac {x \left (a^2-2 b^2\right )}{a}}{2 a^2}-\frac {\cos (x) (2 b-a \sin (x))}{2 a^2}\)

\(\Big \downarrow \) 219

\(\displaystyle -\frac {-\frac {4 b \sqrt {a^2-b^2} \text {arctanh}\left (\frac {2 a+2 b \tan \left (\frac {x}{2}\right )}{2 \sqrt {a^2-b^2}}\right )}{a}-\frac {x \left (a^2-2 b^2\right )}{a}}{2 a^2}-\frac {\cos (x) (2 b-a \sin (x))}{2 a^2}\)

Input:

Int[Cos[x]^2/(a + b*Csc[x]),x]
 

Output:

-1/2*(-(((a^2 - 2*b^2)*x)/a) - (4*b*Sqrt[a^2 - b^2]*ArcTanh[(2*a + 2*b*Tan 
[x/2])/(2*Sqrt[a^2 - b^2])])/a)/a^2 - (Cos[x]*(2*b - a*Sin[x]))/(2*a^2)
 

Defintions of rubi rules used

rule 25

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

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 1083

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[-2   Subst[I 
nt[1/Simp[b^2 - 4*a*c - x^2, x], x], x, b + 2*c*x], x] /; FreeQ[{a, b, c}, 
x]
 

rule 3042

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

rule 3139

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = Fre 
eFactors[Tan[(c + d*x)/2], x]}, Simp[2*(e/d)   Subst[Int[1/(a + 2*b*e*x + a 
*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 3214

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_. 
)*(x_)]), x_Symbol] :> Simp[b*(x/d), x] - Simp[(b*c - a*d)/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 3344

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[g*(g* 
Cos[e + f*x])^(p - 1)*(a + b*Sin[e + f*x])^(m + 1)*((b*c*(m + p + 1) - a*d* 
p + b*d*(m + p)*Sin[e + f*x])/(b^2*f*(m + p)*(m + p + 1))), x] + Simp[g^2*( 
(p - 1)/(b^2*(m + p)*(m + p + 1)))   Int[(g*Cos[e + f*x])^(p - 2)*(a + b*Si 
n[e + f*x])^m*Simp[b*(a*d*m + b*c*(m + p + 1)) + (a*b*c*(m + p + 1) - d*(a^ 
2*p - b^2*(m + p)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, g, 
 m}, x] && NeQ[a^2 - b^2, 0] && GtQ[p, 1] && NeQ[m + p, 0] && NeQ[m + p + 1 
, 0] && IntegerQ[2*m]
 

rule 4360

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + 
(a_))^(m_.), x_Symbol] :> Int[(g*Cos[e + f*x])^p*((b + a*Sin[e + f*x])^m/Si 
n[e + f*x]^m), x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]
 

Maple [A] (verified)

Time = 0.23 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.51

method	result	size
default	\(\frac {\frac {2 \left (-\frac {\tan \left (\frac {x}{2}\right )^{3} a^{2}}{2}-b a \tan \left (\frac {x}{2}\right )^{2}+\frac {a^{2} \tan \left (\frac {x}{2}\right )}{2}-b a \right )}{\left (1+\tan \left (\frac {x}{2}\right )^{2}\right )^{2}}+\left (a^{2}-2 b^{2}\right ) \arctan \left (\tan \left (\frac {x}{2}\right )\right )}{a^{3}}-\frac {2 b \left (a^{2}-b^{2}\right ) \arctan \left (\frac {2 b \tan \left (\frac {x}{2}\right )+2 a}{2 \sqrt {-a^{2}+b^{2}}}\right )}{a^{3} \sqrt {-a^{2}+b^{2}}}\)	\(121\)
risch	\(\frac {x}{2 a}-\frac {x \,b^{2}}{a^{3}}-\frac {b \,{\mathrm e}^{i x}}{2 a^{2}}-\frac {b \,{\mathrm e}^{-i x}}{2 a^{2}}+\frac {\sqrt {a^{2}-b^{2}}\, b \ln \left ({\mathrm e}^{i x}+\frac {i b +\sqrt {a^{2}-b^{2}}}{a}\right )}{a^{3}}-\frac {\sqrt {a^{2}-b^{2}}\, b \ln \left ({\mathrm e}^{i x}+\frac {i b -\sqrt {a^{2}-b^{2}}}{a}\right )}{a^{3}}+\frac {\sin \left (2 x \right )}{4 a}\)	\(137\)

Input:

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

Output:

2/a^3*((-1/2*tan(1/2*x)^3*a^2-b*a*tan(1/2*x)^2+1/2*a^2*tan(1/2*x)-b*a)/(1+ 
tan(1/2*x)^2)^2+1/2*(a^2-2*b^2)*arctan(tan(1/2*x)))-2*b*(a^2-b^2)/a^3/(-a^ 
2+b^2)^(1/2)*arctan(1/2*(2*b*tan(1/2*x)+2*a)/(-a^2+b^2)^(1/2))
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 205, normalized size of antiderivative = 2.56 \[ \int \frac {\cos ^2(x)}{a+b \csc (x)} \, dx=\left [\frac {a^{2} \cos \left (x\right ) \sin \left (x\right ) - 2 \, a b \cos \left (x\right ) + \sqrt {a^{2} - b^{2}} b \log \left (\frac {{\left (a^{2} - 2 \, b^{2}\right )} \cos \left (x\right )^{2} + 2 \, a b \sin \left (x\right ) + a^{2} + b^{2} + 2 \, {\left (b \cos \left (x\right ) \sin \left (x\right ) + a \cos \left (x\right )\right )} \sqrt {a^{2} - b^{2}}}{a^{2} \cos \left (x\right )^{2} - 2 \, a b \sin \left (x\right ) - a^{2} - b^{2}}\right ) + {\left (a^{2} - 2 \, b^{2}\right )} x}{2 \, a^{3}}, \frac {a^{2} \cos \left (x\right ) \sin \left (x\right ) - 2 \, a b \cos \left (x\right ) + 2 \, \sqrt {-a^{2} + b^{2}} b \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \sin \left (x\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \cos \left (x\right )}\right ) + {\left (a^{2} - 2 \, b^{2}\right )} x}{2 \, a^{3}}\right ] \] Input:

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

Output:

[1/2*(a^2*cos(x)*sin(x) - 2*a*b*cos(x) + sqrt(a^2 - b^2)*b*log(((a^2 - 2*b 
^2)*cos(x)^2 + 2*a*b*sin(x) + a^2 + b^2 + 2*(b*cos(x)*sin(x) + a*cos(x))*s 
qrt(a^2 - b^2))/(a^2*cos(x)^2 - 2*a*b*sin(x) - a^2 - b^2)) + (a^2 - 2*b^2) 
*x)/a^3, 1/2*(a^2*cos(x)*sin(x) - 2*a*b*cos(x) + 2*sqrt(-a^2 + b^2)*b*arct 
an(-sqrt(-a^2 + b^2)*(b*sin(x) + a)/((a^2 - b^2)*cos(x))) + (a^2 - 2*b^2)* 
x)/a^3]
 

Sympy [F]

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

integrate(cos(x)**2/(a+b*csc(x)),x)
 

Output:

Integral(cos(x)**2/(a + b*csc(x)), x)
 

Maxima [F(-2)]

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

integrate(cos(x)^2/(a+b*csc(x)),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 = 0.17 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.51 \[ \int \frac {\cos ^2(x)}{a+b \csc (x)} \, dx=\frac {{\left (a^{2} - 2 \, b^{2}\right )} x}{2 \, a^{3}} - \frac {2 \, {\left (a^{2} b - b^{3}\right )} {\left (\pi \left \lfloor \frac {x}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (b\right ) + \arctan \left (\frac {b \tan \left (\frac {1}{2} \, x\right ) + a}{\sqrt {-a^{2} + b^{2}}}\right )\right )}}{\sqrt {-a^{2} + b^{2}} a^{3}} - \frac {a \tan \left (\frac {1}{2} \, x\right )^{3} + 2 \, b \tan \left (\frac {1}{2} \, x\right )^{2} - a \tan \left (\frac {1}{2} \, x\right ) + 2 \, b}{{\left (\tan \left (\frac {1}{2} \, x\right )^{2} + 1\right )}^{2} a^{2}} \] Input:

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

Output:

1/2*(a^2 - 2*b^2)*x/a^3 - 2*(a^2*b - b^3)*(pi*floor(1/2*x/pi + 1/2)*sgn(b) 
 + arctan((b*tan(1/2*x) + a)/sqrt(-a^2 + b^2)))/(sqrt(-a^2 + b^2)*a^3) - ( 
a*tan(1/2*x)^3 + 2*b*tan(1/2*x)^2 - a*tan(1/2*x) + 2*b)/((tan(1/2*x)^2 + 1 
)^2*a^2)
 

Mupad [B] (verification not implemented)

Time = 15.74 (sec) , antiderivative size = 362, normalized size of antiderivative = 4.52 \[ \int \frac {\cos ^2(x)}{a+b \csc (x)} \, dx=-\frac {\frac {2\,b}{a^2}-\frac {\mathrm {tan}\left (\frac {x}{2}\right )}{a}+\frac {{\mathrm {tan}\left (\frac {x}{2}\right )}^3}{a}+\frac {2\,b\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2}{a^2}}{{\mathrm {tan}\left (\frac {x}{2}\right )}^4+2\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2+1}-\frac {2\,b\,\mathrm {atanh}\left (\frac {32\,b^2\,\mathrm {tan}\left (\frac {x}{2}\right )\,\sqrt {a^2-b^2}}{16\,b^3-\frac {16\,b^5}{a^2}-\frac {32\,b^4\,\mathrm {tan}\left (\frac {x}{2}\right )}{a}+32\,a\,b^2\,\mathrm {tan}\left (\frac {x}{2}\right )}-\frac {16\,b^3\,\sqrt {a^2-b^2}}{32\,b^4\,\mathrm {tan}\left (\frac {x}{2}\right )-16\,a\,b^3+\frac {16\,b^5}{a}-32\,a^2\,b^2\,\mathrm {tan}\left (\frac {x}{2}\right )}+\frac {16\,b^4\,\mathrm {tan}\left (\frac {x}{2}\right )\,\sqrt {a^2-b^2}}{-32\,\mathrm {tan}\left (\frac {x}{2}\right )\,a^3\,b^2-16\,a^2\,b^3+32\,\mathrm {tan}\left (\frac {x}{2}\right )\,a\,b^4+16\,b^5}\right )\,\sqrt {a^2-b^2}}{a^3}-\frac {\mathrm {atan}\left (\frac {16\,b^5\,\mathrm {tan}\left (\frac {x}{2}\right )}{8\,a^4\,b-24\,a^2\,b^3+16\,b^5}-\frac {24\,b^3\,\mathrm {tan}\left (\frac {x}{2}\right )}{8\,a^2\,b-24\,b^3+\frac {16\,b^5}{a^2}}+\frac {8\,a\,b\,\mathrm {tan}\left (\frac {x}{2}\right )}{8\,a\,b-\frac {24\,b^3}{a}+\frac {16\,b^5}{a^3}}\right )\,\left (a^2\,1{}\mathrm {i}-b^2\,2{}\mathrm {i}\right )\,1{}\mathrm {i}}{a^3} \] Input:

int(cos(x)^2/(a + b/sin(x)),x)
 

Output:

- ((2*b)/a^2 - tan(x/2)/a + tan(x/2)^3/a + (2*b*tan(x/2)^2)/a^2)/(2*tan(x/ 
2)^2 + tan(x/2)^4 + 1) - (atan((16*b^5*tan(x/2))/(8*a^4*b + 16*b^5 - 24*a^ 
2*b^3) - (24*b^3*tan(x/2))/(8*a^2*b - 24*b^3 + (16*b^5)/a^2) + (8*a*b*tan( 
x/2))/(8*a*b - (24*b^3)/a + (16*b^5)/a^3))*(a^2*1i - b^2*2i)*1i)/a^3 - (2* 
b*atanh((32*b^2*tan(x/2)*(a^2 - b^2)^(1/2))/(16*b^3 - (16*b^5)/a^2 - (32*b 
^4*tan(x/2))/a + 32*a*b^2*tan(x/2)) - (16*b^3*(a^2 - b^2)^(1/2))/(32*b^4*t 
an(x/2) - 16*a*b^3 + (16*b^5)/a - 32*a^2*b^2*tan(x/2)) + (16*b^4*tan(x/2)* 
(a^2 - b^2)^(1/2))/(16*b^5 - 16*a^2*b^3 - 32*a^3*b^2*tan(x/2) + 32*a*b^4*t 
an(x/2)))*(a^2 - b^2)^(1/2))/a^3
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.82 \[ \int \frac {\cos ^2(x)}{a+b \csc (x)} \, dx=\frac {4 \sqrt {-a^{2}+b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {x}{2}\right ) b +a}{\sqrt {-a^{2}+b^{2}}}\right ) b +\cos \left (x \right ) \sin \left (x \right ) a^{2}-2 \cos \left (x \right ) a b +a^{2} x -2 b^{2} x}{2 a^{3}} \] Input:

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

Output:

(4*sqrt( - a**2 + b**2)*atan((tan(x/2)*b + a)/sqrt( - a**2 + b**2))*b + co 
s(x)*sin(x)*a**2 - 2*cos(x)*a*b + a**2*x - 2*b**2*x)/(2*a**3)