Integrand size = 11, antiderivative size = 67 \[ \int \frac {1}{(a \cot (x)+b \csc (x))^2} \, dx=-\frac {x}{a^2}+\frac {2 b \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{a^2 \sqrt {a-b} \sqrt {a+b}}+\frac {\sin (x)}{a (b+a \cos (x))} \]
-x/a^2+sin(x)/a/(b+a*cos(x))+2*b*arctanh((a-b)^(1/2)*tan(1/2*x)/(a+b)^(1/2 ))/a^2/(a-b)^(1/2)/(a+b)^(1/2)
Time = 0.24 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.06 \[ \int \frac {1}{(a \cot (x)+b \csc (x))^2} \, dx=-\frac {\frac {2 b \text {arctanh}\left (\frac {(-a+b) \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}+\frac {b x+a x \cos (x)-a \sin (x)}{b+a \cos (x)}}{a^2} \]
-(((2*b*ArcTanh[((-a + b)*Tan[x/2])/Sqrt[a^2 - b^2]])/Sqrt[a^2 - b^2] + (b *x + a*x*Cos[x] - a*Sin[x])/(b + a*Cos[x]))/a^2)
Time = 0.41 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.07, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.909, Rules used = {3042, 4892, 3042, 3172, 25, 3042, 3214, 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.
| \(\displaystyle \int \frac {1}{(a \cot (x)+b \csc (x))^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{(a \cot (x)+b \csc (x))^2}dx\) |
| \(\Big \downarrow \) 4892 |
| \(\displaystyle \int \frac {\sin ^2(x)}{(a \cos (x)+b)^2}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\cos \left (x-\frac {\pi }{2}\right )^2}{\left (b-a \sin \left (x-\frac {\pi }{2}\right )\right )^2}dx\) |
| \(\Big \downarrow \) 3172 |
| \(\displaystyle \frac {\int -\frac {\cos (x)}{b+a \cos (x)}dx}{a}+\frac {\sin (x)}{a (a \cos (x)+b)}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\sin (x)}{a (a \cos (x)+b)}-\frac {\int \frac {\cos (x)}{b+a \cos (x)}dx}{a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\sin (x)}{a (a \cos (x)+b)}-\frac {\int \frac {\sin \left (x+\frac {\pi }{2}\right )}{b+a \sin \left (x+\frac {\pi }{2}\right )}dx}{a}\) |
| \(\Big \downarrow \) 3214 |
| \(\displaystyle \frac {\sin (x)}{a (a \cos (x)+b)}-\frac {\frac {x}{a}-\frac {b \int \frac {1}{b+a \cos (x)}dx}{a}}{a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\sin (x)}{a (a \cos (x)+b)}-\frac {\frac {x}{a}-\frac {b \int \frac {1}{b+a \sin \left (x+\frac {\pi }{2}\right )}dx}{a}}{a}\) |
| \(\Big \downarrow \) 3138 |
| \(\displaystyle \frac {\sin (x)}{a (a \cos (x)+b)}-\frac {\frac {x}{a}-\frac {2 b \int \frac {1}{-\left ((a-b) \tan ^2\left (\frac {x}{2}\right )\right )+a+b}d\tan \left (\frac {x}{2}\right )}{a}}{a}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {\sin (x)}{a (a \cos (x)+b)}-\frac {\frac {x}{a}-\frac {2 b \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{a \sqrt {a-b} \sqrt {a+b}}}{a}\) |
-((x/a - (2*b*ArcTanh[(Sqrt[a - b]*Tan[x/2])/Sqrt[a + b]])/(a*Sqrt[a - b]* Sqrt[a + b]))/a) + Sin[x]/(a*(b + a*Cos[x]))
3.3.89.3.1 Defintions of rubi rules used
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]
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*(g*Cos[e + f*x])^(p - 1)*((a + b*Sin[e + f*x ])^(m + 1)/(b*f*(m + 1))), x] + Simp[g^2*((p - 1)/(b*(m + 1))) Int[(g*Cos [e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 1)*Sin[e + f*x], x], x] /; Fre eQ[{a, b, e, f, g}, x] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && GtQ[p, 1] && I ntegersQ[2*m, 2*p]
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]
Int[(cot[(c_.) + (d_.)*(x_)]^(n_.)*(a_.) + csc[(c_.) + (d_.)*(x_)]^(n_.)*(b _.))^(p_)*(u_.), x_Symbol] :> Int[ActivateTrig[u]*Csc[c + d*x]^(n*p)*(b + a *Cos[c + d*x]^n)^p, x] /; FreeQ[{a, b, c, d}, x] && IntegersQ[n, p]
Time = 1.44 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.28
| method | result | size |
| default | \(\frac {-\frac {2 a \tan \left (\frac {x}{2}\right )}{\tan \left (\frac {x}{2}\right )^{2} a -\tan \left (\frac {x}{2}\right )^{2} b -a -b}+\frac {2 b \,\operatorname {arctanh}\left (\frac {\tan \left (\frac {x}{2}\right ) \left (a -b \right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}}{a^{2}}-\frac {2 \arctan \left (\tan \left (\frac {x}{2}\right )\right )}{a^{2}}\) | \(86\) |
| risch | \(-\frac {x}{a^{2}}+\frac {2 i \left (b \,{\mathrm e}^{i x}+a \right )}{a^{2} \left (a \,{\mathrm e}^{2 i x}+2 b \,{\mathrm e}^{i x}+a \right )}+\frac {b \ln \left ({\mathrm e}^{i x}+\frac {i a^{2}-i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, a^{2}}-\frac {b \ln \left ({\mathrm e}^{i x}+\frac {-i a^{2}+i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, a^{2}}\) | \(171\) |
2/a^2*(-a*tan(1/2*x)/(tan(1/2*x)^2*a-tan(1/2*x)^2*b-a-b)+b/((a+b)*(a-b))^( 1/2)*arctanh(tan(1/2*x)*(a-b)/((a+b)*(a-b))^(1/2)))-2/a^2*arctan(tan(1/2*x ))
Leaf count of result is larger than twice the leaf count of optimal. 132 vs. \(2 (57) = 114\).
Time = 0.28 (sec) , antiderivative size = 307, normalized size of antiderivative = 4.58 \[ \int \frac {1}{(a \cot (x)+b \csc (x))^2} \, dx=\left [-\frac {2 \, {\left (a^{3} - a b^{2}\right )} x \cos \left (x\right ) - {\left (a b \cos \left (x\right ) + b^{2}\right )} \sqrt {a^{2} - b^{2}} \log \left (\frac {2 \, a b \cos \left (x\right ) - {\left (a^{2} - 2 \, b^{2}\right )} \cos \left (x\right )^{2} + 2 \, \sqrt {a^{2} - b^{2}} {\left (b \cos \left (x\right ) + a\right )} \sin \left (x\right ) + 2 \, a^{2} - b^{2}}{a^{2} \cos \left (x\right )^{2} + 2 \, a b \cos \left (x\right ) + b^{2}}\right ) + 2 \, {\left (a^{2} b - b^{3}\right )} x - 2 \, {\left (a^{3} - a b^{2}\right )} \sin \left (x\right )}{2 \, {\left (a^{4} b - a^{2} b^{3} + {\left (a^{5} - a^{3} b^{2}\right )} \cos \left (x\right )\right )}}, -\frac {{\left (a^{3} - a b^{2}\right )} x \cos \left (x\right ) - {\left (a b \cos \left (x\right ) + b^{2}\right )} \sqrt {-a^{2} + b^{2}} \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \cos \left (x\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (x\right )}\right ) + {\left (a^{2} b - b^{3}\right )} x - {\left (a^{3} - a b^{2}\right )} \sin \left (x\right )}{a^{4} b - a^{2} b^{3} + {\left (a^{5} - a^{3} b^{2}\right )} \cos \left (x\right )}\right ] \]
[-1/2*(2*(a^3 - a*b^2)*x*cos(x) - (a*b*cos(x) + b^2)*sqrt(a^2 - b^2)*log(( 2*a*b*cos(x) - (a^2 - 2*b^2)*cos(x)^2 + 2*sqrt(a^2 - b^2)*(b*cos(x) + a)*s in(x) + 2*a^2 - b^2)/(a^2*cos(x)^2 + 2*a*b*cos(x) + b^2)) + 2*(a^2*b - b^3 )*x - 2*(a^3 - a*b^2)*sin(x))/(a^4*b - a^2*b^3 + (a^5 - a^3*b^2)*cos(x)), -((a^3 - a*b^2)*x*cos(x) - (a*b*cos(x) + b^2)*sqrt(-a^2 + b^2)*arctan(-sqr t(-a^2 + b^2)*(b*cos(x) + a)/((a^2 - b^2)*sin(x))) + (a^2*b - b^3)*x - (a^ 3 - a*b^2)*sin(x))/(a^4*b - a^2*b^3 + (a^5 - a^3*b^2)*cos(x))]
\[ \int \frac {1}{(a \cot (x)+b \csc (x))^2} \, dx=\int \frac {1}{\left (a \cot {\left (x \right )} + b \csc {\left (x \right )}\right )^{2}}\, dx \]
Exception generated. \[ \int \frac {1}{(a \cot (x)+b \csc (x))^2} \, dx=\text {Exception raised: ValueError} \]
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
Time = 0.26 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.60 \[ \int \frac {1}{(a \cot (x)+b \csc (x))^2} \, dx=\frac {2 \, {\left (\pi \left \lfloor \frac {x}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, x\right ) - b \tan \left (\frac {1}{2} \, x\right )}{\sqrt {-a^{2} + b^{2}}}\right )\right )} b}{\sqrt {-a^{2} + b^{2}} a^{2}} - \frac {x}{a^{2}} - \frac {2 \, \tan \left (\frac {1}{2} \, x\right )}{{\left (a \tan \left (\frac {1}{2} \, x\right )^{2} - b \tan \left (\frac {1}{2} \, x\right )^{2} - a - b\right )} a} \]
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/(sqrt(-a^2 + b^2)*a^2) - x/a^2 - 2*tan(1/2* x)/((a*tan(1/2*x)^2 - b*tan(1/2*x)^2 - a - b)*a)
Time = 29.27 (sec) , antiderivative size = 440, normalized size of antiderivative = 6.57 \[ \int \frac {1}{(a \cot (x)+b \csc (x))^2} \, dx=\frac {a^3\,\sin \left (x\right )+b^2\,\left (-a\,\sin \left (x\right )+\mathrm {atan}\left (\frac {-a^5\,\sin \left (\frac {x}{2}\right )\,\sqrt {a^2-b^2}\,1{}\mathrm {i}+b^3\,\sin \left (\frac {x}{2}\right )\,{\left (a^2-b^2\right )}^{3/2}\,2{}\mathrm {i}+b^5\,\sin \left (\frac {x}{2}\right )\,\sqrt {a^2-b^2}\,2{}\mathrm {i}+a^4\,b\,\sin \left (\frac {x}{2}\right )\,\sqrt {a^2-b^2}\,1{}\mathrm {i}-a^2\,b^3\,\sin \left (\frac {x}{2}\right )\,\sqrt {a^2-b^2}\,3{}\mathrm {i}+a^3\,b^2\,\sin \left (\frac {x}{2}\right )\,\sqrt {a^2-b^2}\,1{}\mathrm {i}}{\cos \left (\frac {x}{2}\right )\,a^6-2\,\cos \left (\frac {x}{2}\right )\,a^4\,b^2+\cos \left (\frac {x}{2}\right )\,a^2\,b^4}\right )\,\sqrt {a^2-b^2}\,2{}\mathrm {i}\right )+a\,b\,\mathrm {atan}\left (\frac {-a^5\,\sin \left (\frac {x}{2}\right )\,\sqrt {a^2-b^2}\,1{}\mathrm {i}+b^3\,\sin \left (\frac {x}{2}\right )\,{\left (a^2-b^2\right )}^{3/2}\,2{}\mathrm {i}+b^5\,\sin \left (\frac {x}{2}\right )\,\sqrt {a^2-b^2}\,2{}\mathrm {i}+a^4\,b\,\sin \left (\frac {x}{2}\right )\,\sqrt {a^2-b^2}\,1{}\mathrm {i}-a^2\,b^3\,\sin \left (\frac {x}{2}\right )\,\sqrt {a^2-b^2}\,3{}\mathrm {i}+a^3\,b^2\,\sin \left (\frac {x}{2}\right )\,\sqrt {a^2-b^2}\,1{}\mathrm {i}}{\cos \left (\frac {x}{2}\right )\,a^6-2\,\cos \left (\frac {x}{2}\right )\,a^4\,b^2+\cos \left (\frac {x}{2}\right )\,a^2\,b^4}\right )\,\cos \left (x\right )\,\sqrt {a^2-b^2}\,2{}\mathrm {i}}{\cos \left (x\right )\,a^5+a^4\,b-\cos \left (x\right )\,a^3\,b^2-a^2\,b^3}-\frac {2\,\mathrm {atan}\left (\frac {\sin \left (\frac {x}{2}\right )}{\cos \left (\frac {x}{2}\right )}\right )}{a^2} \]
(a^3*sin(x) + b^2*(atan((b^3*sin(x/2)*(a^2 - b^2)^(3/2)*2i - a^5*sin(x/2)* (a^2 - b^2)^(1/2)*1i + b^5*sin(x/2)*(a^2 - b^2)^(1/2)*2i + a^4*b*sin(x/2)* (a^2 - b^2)^(1/2)*1i - a^2*b^3*sin(x/2)*(a^2 - b^2)^(1/2)*3i + a^3*b^2*sin (x/2)*(a^2 - b^2)^(1/2)*1i)/(a^6*cos(x/2) + a^2*b^4*cos(x/2) - 2*a^4*b^2*c os(x/2)))*(a^2 - b^2)^(1/2)*2i - a*sin(x)) + a*b*atan((b^3*sin(x/2)*(a^2 - b^2)^(3/2)*2i - a^5*sin(x/2)*(a^2 - b^2)^(1/2)*1i + b^5*sin(x/2)*(a^2 - b ^2)^(1/2)*2i + a^4*b*sin(x/2)*(a^2 - b^2)^(1/2)*1i - a^2*b^3*sin(x/2)*(a^2 - b^2)^(1/2)*3i + a^3*b^2*sin(x/2)*(a^2 - b^2)^(1/2)*1i)/(a^6*cos(x/2) + a^2*b^4*cos(x/2) - 2*a^4*b^2*cos(x/2)))*cos(x)*(a^2 - b^2)^(1/2)*2i)/(a^4* b - a^2*b^3 + a^5*cos(x) - a^3*b^2*cos(x)) - (2*atan(sin(x/2)/cos(x/2)))/a ^2