Integrand size = 13, antiderivative size = 67 \[ \int \frac {\csc ^2(x)}{a+b \cos (x)} \, dx=-\frac {2 b^2 \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{(a-b)^{3/2} (a+b)^{3/2}}+\frac {(b-a \cos (x)) \csc (x)}{a^2-b^2} \] Output:
-2*b^2*arctan((a-b)^(1/2)*tan(1/2*x)/(a+b)^(1/2))/(a-b)^(3/2)/(a+b)^(3/2)+ (b-a*cos(x))*csc(x)/(a^2-b^2)
Time = 0.40 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.99 \[ \int \frac {\csc ^2(x)}{a+b \cos (x)} \, dx=-\frac {2 b^2 \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {x}{2}\right )}{\sqrt {-a^2+b^2}}\right )}{\left (-a^2+b^2\right )^{3/2}}+\frac {(b-a \cos (x)) \csc (x)}{a^2-b^2} \] Input:
Integrate[Csc[x]^2/(a + b*Cos[x]),x]
Output:
(-2*b^2*ArcTanh[((a - b)*Tan[x/2])/Sqrt[-a^2 + b^2]])/(-a^2 + b^2)^(3/2) + ((b - a*Cos[x])*Csc[x])/(a^2 - b^2)
Time = 0.32 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.16, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {3042, 3175, 27, 3042, 3138, 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.
\(\displaystyle \int \frac {\csc ^2(x)}{a+b \cos (x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\cos \left (x-\frac {\pi }{2}\right )^2 \left (a-b \sin \left (x-\frac {\pi }{2}\right )\right )}dx\) |
\(\Big \downarrow \) 3175 |
\(\displaystyle \frac {\csc (x) (b-a \cos (x))}{a^2-b^2}-\frac {\int \frac {b^2}{a+b \cos (x)}dx}{a^2-b^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\csc (x) (b-a \cos (x))}{a^2-b^2}-\frac {b^2 \int \frac {1}{a+b \cos (x)}dx}{a^2-b^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\csc (x) (b-a \cos (x))}{a^2-b^2}-\frac {b^2 \int \frac {1}{a+b \sin \left (x+\frac {\pi }{2}\right )}dx}{a^2-b^2}\) |
\(\Big \downarrow \) 3138 |
\(\displaystyle \frac {\csc (x) (b-a \cos (x))}{a^2-b^2}-\frac {2 b^2 \int \frac {1}{(a-b) \tan ^2\left (\frac {x}{2}\right )+a+b}d\tan \left (\frac {x}{2}\right )}{a^2-b^2}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {\csc (x) (b-a \cos (x))}{a^2-b^2}-\frac {2 b^2 \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{\sqrt {a-b} \sqrt {a+b} \left (a^2-b^2\right )}\) |
Input:
Int[Csc[x]^2/(a + b*Cos[x]),x]
Output:
(-2*b^2*ArcTan[(Sqrt[a - b]*Tan[x/2])/Sqrt[a + b]])/(Sqrt[a - b]*Sqrt[a + b]*(a^2 - b^2)) + ((b - a*Cos[x])*Csc[x])/(a^2 - b^2)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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]
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]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_), x_Symbol] :> Simp[(g*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^ (m + 1)*((b - a*Sin[e + f*x])/(f*g*(a^2 - b^2)*(p + 1))), x] + Simp[1/(g^2* (a^2 - b^2)*(p + 1)) Int[(g*Cos[e + f*x])^(p + 2)*(a + b*Sin[e + f*x])^m* (a^2*(p + 2) - b^2*(m + p + 2) + a*b*(m + p + 3)*Sin[e + f*x]), x], x] /; F reeQ[{a, b, e, f, g, m}, x] && NeQ[a^2 - b^2, 0] && LtQ[p, -1] && IntegersQ [2*m, 2*p]
Time = 0.52 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.16
method | result | size |
default | \(\frac {\tan \left (\frac {x}{2}\right )}{2 a -2 b}-\frac {1}{2 \left (a +b \right ) \tan \left (\frac {x}{2}\right )}-\frac {2 b^{2} \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {x}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a -b \right ) \left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}\) | \(78\) |
risch | \(-\frac {2 i \left (-{\mathrm e}^{i x} b +a \right )}{\left ({\mathrm e}^{2 i x}-1\right ) \left (a^{2}-b^{2}\right )}+\frac {b^{2} \ln \left ({\mathrm e}^{i x}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right )}-\frac {b^{2} \ln \left ({\mathrm e}^{i x}+\frac {-i a^{2}+i b^{2}+a \sqrt {-a^{2}+b^{2}}}{\sqrt {-a^{2}+b^{2}}\, b}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right )}\) | \(186\) |
Input:
int(csc(x)^2/(a+b*cos(x)),x,method=_RETURNVERBOSE)
Output:
1/2/(a-b)*tan(1/2*x)-1/2/(a+b)/tan(1/2*x)-2/(a-b)*b^2/(a+b)/((a-b)*(a+b))^ (1/2)*arctan((a-b)*tan(1/2*x)/((a-b)*(a+b))^(1/2))
Time = 0.10 (sec) , antiderivative size = 230, normalized size of antiderivative = 3.43 \[ \int \frac {\csc ^2(x)}{a+b \cos (x)} \, dx=\left [\frac {\sqrt {-a^{2} + b^{2}} b^{2} \log \left (\frac {2 \, a b \cos \left (x\right ) + {\left (2 \, a^{2} - b^{2}\right )} \cos \left (x\right )^{2} + 2 \, \sqrt {-a^{2} + b^{2}} {\left (a \cos \left (x\right ) + b\right )} \sin \left (x\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (x\right )^{2} + 2 \, a b \cos \left (x\right ) + a^{2}}\right ) \sin \left (x\right ) + 2 \, a^{2} b - 2 \, b^{3} - 2 \, {\left (a^{3} - a b^{2}\right )} \cos \left (x\right )}{2 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sin \left (x\right )}, -\frac {\sqrt {a^{2} - b^{2}} b^{2} \arctan \left (-\frac {a \cos \left (x\right ) + b}{\sqrt {a^{2} - b^{2}} \sin \left (x\right )}\right ) \sin \left (x\right ) - a^{2} b + b^{3} + {\left (a^{3} - a b^{2}\right )} \cos \left (x\right )}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sin \left (x\right )}\right ] \] Input:
integrate(csc(x)^2/(a+b*cos(x)),x, algorithm="fricas")
Output:
[1/2*(sqrt(-a^2 + b^2)*b^2*log((2*a*b*cos(x) + (2*a^2 - b^2)*cos(x)^2 + 2* sqrt(-a^2 + b^2)*(a*cos(x) + b)*sin(x) - a^2 + 2*b^2)/(b^2*cos(x)^2 + 2*a* b*cos(x) + a^2))*sin(x) + 2*a^2*b - 2*b^3 - 2*(a^3 - a*b^2)*cos(x))/((a^4 - 2*a^2*b^2 + b^4)*sin(x)), -(sqrt(a^2 - b^2)*b^2*arctan(-(a*cos(x) + b)/( sqrt(a^2 - b^2)*sin(x)))*sin(x) - a^2*b + b^3 + (a^3 - a*b^2)*cos(x))/((a^ 4 - 2*a^2*b^2 + b^4)*sin(x))]
\[ \int \frac {\csc ^2(x)}{a+b \cos (x)} \, dx=\int \frac {\csc ^{2}{\left (x \right )}}{a + b \cos {\left (x \right )}}\, dx \] Input:
integrate(csc(x)**2/(a+b*cos(x)),x)
Output:
Integral(csc(x)**2/(a + b*cos(x)), x)
Exception generated. \[ \int \frac {\csc ^2(x)}{a+b \cos (x)} \, dx=\text {Exception raised: ValueError} \] Input:
integrate(csc(x)^2/(a+b*cos(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*b^2-4*a^2>0)', see `assume?` f or more de
Time = 0.15 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.36 \[ \int \frac {\csc ^2(x)}{a+b \cos (x)} \, dx=\frac {2 \, {\left (\pi \left \lfloor \frac {x}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, x\right ) - b \tan \left (\frac {1}{2} \, x\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )} b^{2}}{{\left (a^{2} - b^{2}\right )}^{\frac {3}{2}}} + \frac {\tan \left (\frac {1}{2} \, x\right )}{2 \, {\left (a - b\right )}} - \frac {1}{2 \, {\left (a + b\right )} \tan \left (\frac {1}{2} \, x\right )} \] Input:
integrate(csc(x)^2/(a+b*cos(x)),x, algorithm="giac")
Output:
2*(pi*floor(1/2*x/pi + 1/2)*sgn(-2*a + 2*b) + arctan(-(a*tan(1/2*x) - b*ta n(1/2*x))/sqrt(a^2 - b^2)))*b^2/(a^2 - b^2)^(3/2) + 1/2*tan(1/2*x)/(a - b) - 1/2/((a + b)*tan(1/2*x))
Time = 44.81 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.28 \[ \int \frac {\csc ^2(x)}{a+b \cos (x)} \, dx=\frac {\mathrm {tan}\left (\frac {x}{2}\right )}{2\,a-2\,b}-\frac {2\,b^2\,\mathrm {atan}\left (\frac {\mathrm {tan}\left (\frac {x}{2}\right )\,\left (a^2-b^2\right )}{{\left (a+b\right )}^{3/2}\,\sqrt {a-b}}\right )}{{\left (a+b\right )}^{3/2}\,{\left (a-b\right )}^{3/2}}-\frac {a-b}{\mathrm {tan}\left (\frac {x}{2}\right )\,\left (a+b\right )\,\left (2\,a-2\,b\right )} \] Input:
int(1/(sin(x)^2*(a + b*cos(x))),x)
Output:
tan(x/2)/(2*a - 2*b) - (2*b^2*atan((tan(x/2)*(a^2 - b^2))/((a + b)^(3/2)*( a - b)^(1/2))))/((a + b)^(3/2)*(a - b)^(3/2)) - (a - b)/(tan(x/2)*(a + b)* (2*a - 2*b))
Time = 0.16 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.37 \[ \int \frac {\csc ^2(x)}{a+b \cos (x)} \, dx=\frac {-2 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {x}{2}\right ) a -\tan \left (\frac {x}{2}\right ) b}{\sqrt {a^{2}-b^{2}}}\right ) \sin \left (x \right ) b^{2}-\cos \left (x \right ) a^{3}+\cos \left (x \right ) a \,b^{2}+a^{2} b -b^{3}}{\sin \left (x \right ) \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )} \] Input:
int(csc(x)^2/(a+b*cos(x)),x)
Output:
( - 2*sqrt(a**2 - b**2)*atan((tan(x/2)*a - tan(x/2)*b)/sqrt(a**2 - b**2))* sin(x)*b**2 - cos(x)*a**3 + cos(x)*a*b**2 + a**2*b - b**3)/(sin(x)*(a**4 - 2*a**2*b**2 + b**4))