Integrand size = 13, antiderivative size = 61 \[ \int \frac {\cot ^2(x)}{a+b \csc (x)} \, dx=-\frac {x}{a}-\frac {\text {arctanh}(\cos (x))}{b}+\frac {2 \sqrt {a^2-b^2} \text {arctanh}\left (\frac {a+b \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{a b} \]
Time = 0.10 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.16 \[ \int \frac {\cot ^2(x)}{a+b \csc (x)} \, dx=\frac {-b x+2 \sqrt {-a^2+b^2} \arctan \left (\frac {a+b \tan \left (\frac {x}{2}\right )}{\sqrt {-a^2+b^2}}\right )-a \log \left (\cos \left (\frac {x}{2}\right )\right )+a \log \left (\sin \left (\frac {x}{2}\right )\right )}{a b} \]
(-(b*x) + 2*Sqrt[-a^2 + b^2]*ArcTan[(a + b*Tan[x/2])/Sqrt[-a^2 + b^2]] - a *Log[Cos[x/2]] + a*Log[Sin[x/2]])/(a*b)
Time = 0.59 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.20, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.077, Rules used = {3042, 4382, 3042, 4539, 25, 3042, 4257, 4407, 3042, 4318, 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.
| \(\displaystyle \int \frac {\cot ^2(x)}{a+b \csc (x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\cot (x)^2}{a+b \csc (x)}dx\) |
| \(\Big \downarrow \) 4382 |
| \(\displaystyle \int \frac {\csc ^2(x)-1}{a+b \csc (x)}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\csc (x)^2-1}{a+b \csc (x)}dx\) |
| \(\Big \downarrow \) 4539 |
| \(\displaystyle \frac {\int -\frac {b+a \csc (x)}{a+b \csc (x)}dx}{b}+\frac {\int \csc (x)dx}{b}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \csc (x)dx}{b}-\frac {\int \frac {b+a \csc (x)}{a+b \csc (x)}dx}{b}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \csc (x)dx}{b}-\frac {\int \frac {b+a \csc (x)}{a+b \csc (x)}dx}{b}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle -\frac {\int \frac {b+a \csc (x)}{a+b \csc (x)}dx}{b}-\frac {\text {arctanh}(\cos (x))}{b}\) |
| \(\Big \downarrow \) 4407 |
| \(\displaystyle -\frac {\frac {\left (a^2-b^2\right ) \int \frac {\csc (x)}{a+b \csc (x)}dx}{a}+\frac {b x}{a}}{b}-\frac {\text {arctanh}(\cos (x))}{b}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\frac {\left (a^2-b^2\right ) \int \frac {\csc (x)}{a+b \csc (x)}dx}{a}+\frac {b x}{a}}{b}-\frac {\text {arctanh}(\cos (x))}{b}\) |
| \(\Big \downarrow \) 4318 |
| \(\displaystyle -\frac {\frac {\left (a^2-b^2\right ) \int \frac {1}{\frac {a \sin (x)}{b}+1}dx}{a b}+\frac {b x}{a}}{b}-\frac {\text {arctanh}(\cos (x))}{b}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\frac {\left (a^2-b^2\right ) \int \frac {1}{\frac {a \sin (x)}{b}+1}dx}{a b}+\frac {b x}{a}}{b}-\frac {\text {arctanh}(\cos (x))}{b}\) |
| \(\Big \downarrow \) 3139 |
| \(\displaystyle -\frac {\frac {2 \left (a^2-b^2\right ) \int \frac {1}{\tan ^2\left (\frac {x}{2}\right )+\frac {2 a \tan \left (\frac {x}{2}\right )}{b}+1}d\tan \left (\frac {x}{2}\right )}{a b}+\frac {b x}{a}}{b}-\frac {\text {arctanh}(\cos (x))}{b}\) |
| \(\Big \downarrow \) 1083 |
| \(\displaystyle -\frac {\frac {b x}{a}-\frac {4 \left (a^2-b^2\right ) \int \frac {1}{-\left (\frac {2 a}{b}+2 \tan \left (\frac {x}{2}\right )\right )^2-4 \left (1-\frac {a^2}{b^2}\right )}d\left (\frac {2 a}{b}+2 \tan \left (\frac {x}{2}\right )\right )}{a b}}{b}-\frac {\text {arctanh}(\cos (x))}{b}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle -\frac {\frac {b x}{a}-\frac {2 \sqrt {a^2-b^2} \text {arctanh}\left (\frac {b \left (\frac {2 a}{b}+2 \tan \left (\frac {x}{2}\right )\right )}{2 \sqrt {a^2-b^2}}\right )}{a}}{b}-\frac {\text {arctanh}(\cos (x))}{b}\) |
-(ArcTanh[Cos[x]]/b) - ((b*x)/a - (2*Sqrt[a^2 - b^2]*ArcTanh[(b*((2*a)/b + 2*Tan[x/2]))/(2*Sqrt[a^2 - b^2])])/a)/b
3.1.21.3.1 Defintions of rubi rules used
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])
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]
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]
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbo l] :> Simp[1/b Int[1/(1 + (a/b)*Sin[e + f*x]), x], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]
Int[cot[(c_.) + (d_.)*(x_)]^2*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Int[(-1 + Csc[c + d*x]^2)*(a + b*Csc[c + d*x])^n, x] /; FreeQ[ {a, b, c, d, n}, x] && NeQ[a^2 - b^2, 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[c*(x/a), x] - Simp[(b*c - a*d)/a Int[Csc[e + f* x]/(a + b*Csc[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))/(csc[(e_.) + (f_.)*(x_)]*(b_. ) + (a_)), x_Symbol] :> Simp[C/b Int[Csc[e + f*x], x], x] + Simp[1/b In t[(A*b - a*C*Csc[e + f*x])/(a + b*Csc[e + f*x]), x], x] /; FreeQ[{a, b, e, f, A, C}, x]
Time = 0.68 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.23
| method | result | size |
| default | \(\frac {\ln \left (\tan \left (\frac {x}{2}\right )\right )}{b}+\frac {\left (-2 a^{2}+2 b^{2}\right ) \arctan \left (\frac {2 b \tan \left (\frac {x}{2}\right )+2 a}{2 \sqrt {-a^{2}+b^{2}}}\right )}{a b \sqrt {-a^{2}+b^{2}}}-\frac {2 \arctan \left (\tan \left (\frac {x}{2}\right )\right )}{a}\) | \(75\) |
| risch | \(-\frac {x}{a}-\frac {\sqrt {a^{2}-b^{2}}\, \ln \left ({\mathrm e}^{i x}-\frac {-i b +\sqrt {a^{2}-b^{2}}}{a}\right )}{b a}+\frac {\sqrt {a^{2}-b^{2}}\, \ln \left ({\mathrm e}^{i x}+\frac {i b +\sqrt {a^{2}-b^{2}}}{a}\right )}{b a}-\frac {\ln \left ({\mathrm e}^{i x}+1\right )}{b}+\frac {\ln \left ({\mathrm e}^{i x}-1\right )}{b}\) | \(125\) |
1/b*ln(tan(1/2*x))+(-2*a^2+2*b^2)/a/b/(-a^2+b^2)^(1/2)*arctan(1/2*(2*b*tan (1/2*x)+2*a)/(-a^2+b^2)^(1/2))-2/a*arctan(tan(1/2*x))
Time = 0.28 (sec) , antiderivative size = 206, normalized size of antiderivative = 3.38 \[ \int \frac {\cot ^2(x)}{a+b \csc (x)} \, dx=\left [-\frac {2 \, b x + a \log \left (\frac {1}{2} \, \cos \left (x\right ) + \frac {1}{2}\right ) - a \log \left (-\frac {1}{2} \, \cos \left (x\right ) + \frac {1}{2}\right ) - \sqrt {a^{2} - b^{2}} \log \left (\frac {{\left (a^{2} - 2 \, b^{2}\right )} \cos \left (x\right )^{2} + 2 \, a b \sin \left (x\right ) + a^{2} + b^{2} + 2 \, {\left (b \cos \left (x\right ) \sin \left (x\right ) + a \cos \left (x\right )\right )} \sqrt {a^{2} - b^{2}}}{a^{2} \cos \left (x\right )^{2} - 2 \, a b \sin \left (x\right ) - a^{2} - b^{2}}\right )}{2 \, a b}, -\frac {2 \, b x + a \log \left (\frac {1}{2} \, \cos \left (x\right ) + \frac {1}{2}\right ) - a \log \left (-\frac {1}{2} \, \cos \left (x\right ) + \frac {1}{2}\right ) - 2 \, \sqrt {-a^{2} + b^{2}} \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \sin \left (x\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \cos \left (x\right )}\right )}{2 \, a b}\right ] \]
[-1/2*(2*b*x + a*log(1/2*cos(x) + 1/2) - a*log(-1/2*cos(x) + 1/2) - sqrt(a ^2 - b^2)*log(((a^2 - 2*b^2)*cos(x)^2 + 2*a*b*sin(x) + a^2 + b^2 + 2*(b*co s(x)*sin(x) + a*cos(x))*sqrt(a^2 - b^2))/(a^2*cos(x)^2 - 2*a*b*sin(x) - a^ 2 - b^2)))/(a*b), -1/2*(2*b*x + a*log(1/2*cos(x) + 1/2) - a*log(-1/2*cos(x ) + 1/2) - 2*sqrt(-a^2 + b^2)*arctan(-sqrt(-a^2 + b^2)*(b*sin(x) + a)/((a^ 2 - b^2)*cos(x))))/(a*b)]
\[ \int \frac {\cot ^2(x)}{a+b \csc (x)} \, dx=\int \frac {\cot ^{2}{\left (x \right )}}{a + b \csc {\left (x \right )}}\, dx \]
Exception generated. \[ \int \frac {\cot ^2(x)}{a+b \csc (x)} \, dx=\text {Exception raised: ValueError} \]
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 = 0.29 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.31 \[ \int \frac {\cot ^2(x)}{a+b \csc (x)} \, dx=-\frac {x}{a} + \frac {\log \left ({\left | \tan \left (\frac {1}{2} \, x\right ) \right |}\right )}{b} - \frac {2 \, {\left (\pi \left \lfloor \frac {x}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (b\right ) + \arctan \left (\frac {b \tan \left (\frac {1}{2} \, x\right ) + a}{\sqrt {-a^{2} + b^{2}}}\right )\right )} {\left (a^{2} - b^{2}\right )}}{\sqrt {-a^{2} + b^{2}} a b} \]
-x/a + log(abs(tan(1/2*x)))/b - 2*(pi*floor(1/2*x/pi + 1/2)*sgn(b) + arcta n((b*tan(1/2*x) + a)/sqrt(-a^2 + b^2)))*(a^2 - b^2)/(sqrt(-a^2 + b^2)*a*b)
Time = 19.06 (sec) , antiderivative size = 697, normalized size of antiderivative = 11.43 \[ \int \frac {\cot ^2(x)}{a+b \csc (x)} \, dx=\frac {2\,\mathrm {atan}\left (\frac {64\,b^3}{-64\,a^3-64\,\mathrm {tan}\left (\frac {x}{2}\right )\,a^2\,b+64\,a\,b^2+64\,\mathrm {tan}\left (\frac {x}{2}\right )\,b^3}+\frac {64\,a^3\,\mathrm {tan}\left (\frac {x}{2}\right )}{-64\,a^3-64\,\mathrm {tan}\left (\frac {x}{2}\right )\,a^2\,b+64\,a\,b^2+64\,\mathrm {tan}\left (\frac {x}{2}\right )\,b^3}-\frac {64\,a^2\,b}{-64\,a^3-64\,\mathrm {tan}\left (\frac {x}{2}\right )\,a^2\,b+64\,a\,b^2+64\,\mathrm {tan}\left (\frac {x}{2}\right )\,b^3}-\frac {64\,a\,b^2\,\mathrm {tan}\left (\frac {x}{2}\right )}{-64\,a^3-64\,\mathrm {tan}\left (\frac {x}{2}\right )\,a^2\,b+64\,a\,b^2+64\,\mathrm {tan}\left (\frac {x}{2}\right )\,b^3}\right )}{a}+\frac {\ln \left (\mathrm {tan}\left (\frac {x}{2}\right )\right )}{b}-\frac {2\,\mathrm {atanh}\left (\frac {512\,a^4\,\sqrt {a^2-b^2}}{256\,a\,b^4-64\,b^5\,\mathrm {tan}\left (\frac {x}{2}\right )+512\,a^5-768\,a^3\,b^2+832\,a^2\,b^3\,\mathrm {tan}\left (\frac {x}{2}\right )+\frac {1024\,a^6\,\mathrm {tan}\left (\frac {x}{2}\right )}{b}-1792\,a^4\,b\,\mathrm {tan}\left (\frac {x}{2}\right )}-\frac {512\,a^2\,\sqrt {a^2-b^2}}{256\,a\,b^2-64\,b^3\,\mathrm {tan}\left (\frac {x}{2}\right )-768\,a^3+\frac {512\,a^5}{b^2}-\frac {1792\,a^4\,\mathrm {tan}\left (\frac {x}{2}\right )}{b}+\frac {1024\,a^6\,\mathrm {tan}\left (\frac {x}{2}\right )}{b^3}+832\,a^2\,b\,\mathrm {tan}\left (\frac {x}{2}\right )}+\frac {64\,b^2\,\sqrt {a^2-b^2}}{256\,a\,b^2-64\,b^3\,\mathrm {tan}\left (\frac {x}{2}\right )-768\,a^3+\frac {512\,a^5}{b^2}-\frac {1792\,a^4\,\mathrm {tan}\left (\frac {x}{2}\right )}{b}+\frac {1024\,a^6\,\mathrm {tan}\left (\frac {x}{2}\right )}{b^3}+832\,a^2\,b\,\mathrm {tan}\left (\frac {x}{2}\right )}-\frac {1280\,a^3\,\mathrm {tan}\left (\frac {x}{2}\right )\,\sqrt {a^2-b^2}}{256\,a\,b^3-64\,b^4\,\mathrm {tan}\left (\frac {x}{2}\right )-1792\,a^4\,\mathrm {tan}\left (\frac {x}{2}\right )-768\,a^3\,b+\frac {512\,a^5}{b}+832\,a^2\,b^2\,\mathrm {tan}\left (\frac {x}{2}\right )+\frac {1024\,a^6\,\mathrm {tan}\left (\frac {x}{2}\right )}{b^2}}+\frac {1024\,a^5\,\mathrm {tan}\left (\frac {x}{2}\right )\,\sqrt {a^2-b^2}}{1024\,\mathrm {tan}\left (\frac {x}{2}\right )\,a^6+512\,a^5\,b-1792\,\mathrm {tan}\left (\frac {x}{2}\right )\,a^4\,b^2-768\,a^3\,b^3+832\,\mathrm {tan}\left (\frac {x}{2}\right )\,a^2\,b^4+256\,a\,b^5-64\,\mathrm {tan}\left (\frac {x}{2}\right )\,b^6}+\frac {320\,a\,b\,\mathrm {tan}\left (\frac {x}{2}\right )\,\sqrt {a^2-b^2}}{256\,a\,b^2-64\,b^3\,\mathrm {tan}\left (\frac {x}{2}\right )-768\,a^3+\frac {512\,a^5}{b^2}-\frac {1792\,a^4\,\mathrm {tan}\left (\frac {x}{2}\right )}{b}+\frac {1024\,a^6\,\mathrm {tan}\left (\frac {x}{2}\right )}{b^3}+832\,a^2\,b\,\mathrm {tan}\left (\frac {x}{2}\right )}\right )\,\sqrt {a^2-b^2}}{a\,b} \]
(2*atan((64*b^3)/(64*b^3*tan(x/2) + 64*a*b^2 - 64*a^3 - 64*a^2*b*tan(x/2)) + (64*a^3*tan(x/2))/(64*b^3*tan(x/2) + 64*a*b^2 - 64*a^3 - 64*a^2*b*tan(x /2)) - (64*a^2*b)/(64*b^3*tan(x/2) + 64*a*b^2 - 64*a^3 - 64*a^2*b*tan(x/2) ) - (64*a*b^2*tan(x/2))/(64*b^3*tan(x/2) + 64*a*b^2 - 64*a^3 - 64*a^2*b*ta n(x/2))))/a + log(tan(x/2))/b - (2*atanh((512*a^4*(a^2 - b^2)^(1/2))/(256* a*b^4 - 64*b^5*tan(x/2) + 512*a^5 - 768*a^3*b^2 + 832*a^2*b^3*tan(x/2) + ( 1024*a^6*tan(x/2))/b - 1792*a^4*b*tan(x/2)) - (512*a^2*(a^2 - b^2)^(1/2))/ (256*a*b^2 - 64*b^3*tan(x/2) - 768*a^3 + (512*a^5)/b^2 - (1792*a^4*tan(x/2 ))/b + (1024*a^6*tan(x/2))/b^3 + 832*a^2*b*tan(x/2)) + (64*b^2*(a^2 - b^2) ^(1/2))/(256*a*b^2 - 64*b^3*tan(x/2) - 768*a^3 + (512*a^5)/b^2 - (1792*a^4 *tan(x/2))/b + (1024*a^6*tan(x/2))/b^3 + 832*a^2*b*tan(x/2)) - (1280*a^3*t an(x/2)*(a^2 - b^2)^(1/2))/(256*a*b^3 - 64*b^4*tan(x/2) - 1792*a^4*tan(x/2 ) - 768*a^3*b + (512*a^5)/b + 832*a^2*b^2*tan(x/2) + (1024*a^6*tan(x/2))/b ^2) + (1024*a^5*tan(x/2)*(a^2 - b^2)^(1/2))/(1024*a^6*tan(x/2) - 64*b^6*ta n(x/2) + 256*a*b^5 + 512*a^5*b - 768*a^3*b^3 + 832*a^2*b^4*tan(x/2) - 1792 *a^4*b^2*tan(x/2)) + (320*a*b*tan(x/2)*(a^2 - b^2)^(1/2))/(256*a*b^2 - 64* b^3*tan(x/2) - 768*a^3 + (512*a^5)/b^2 - (1792*a^4*tan(x/2))/b + (1024*a^6 *tan(x/2))/b^3 + 832*a^2*b*tan(x/2)))*(a^2 - b^2)^(1/2))/(a*b)