Integrand size = 11, antiderivative size = 93 \[ \int \frac {\csc (x)}{(a+b \sin (x))^2} \, dx=-\frac {2 b \left (2 a^2-b^2\right ) \arctan \left (\frac {b+a \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{a^2 \left (a^2-b^2\right )^{3/2}}-\frac {\text {arctanh}(\cos (x))}{a^2}-\frac {b^2 \cos (x)}{a \left (a^2-b^2\right ) (a+b \sin (x))} \]
-2*b*(2*a^2-b^2)*arctan((b+a*tan(1/2*x))/(a^2-b^2)^(1/2))/a^2/(a^2-b^2)^(3 /2)-arctanh(cos(x))/a^2-b^2*cos(x)/a/(a^2-b^2)/(a+b*sin(x))
Time = 0.27 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.06 \[ \int \frac {\csc (x)}{(a+b \sin (x))^2} \, dx=\frac {\frac {2 b \left (-2 a^2+b^2\right ) \arctan \left (\frac {b+a \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2}}-\log \left (\cos \left (\frac {x}{2}\right )\right )+\log \left (\sin \left (\frac {x}{2}\right )\right )-\frac {a b^2 \cos (x)}{(a-b) (a+b) (a+b \sin (x))}}{a^2} \]
((2*b*(-2*a^2 + b^2)*ArcTan[(b + a*Tan[x/2])/Sqrt[a^2 - b^2]])/(a^2 - b^2) ^(3/2) - Log[Cos[x/2]] + Log[Sin[x/2]] - (a*b^2*Cos[x])/((a - b)*(a + b)*( a + b*Sin[x])))/a^2
Time = 0.57 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.33, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.818, Rules used = {3042, 3281, 3042, 3480, 3042, 3139, 1083, 217, 4257}
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.
| \(\displaystyle \int \frac {\csc (x)}{(a+b \sin (x))^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sin (x) (a+b \sin (x))^2}dx\) |
| \(\Big \downarrow \) 3281 |
| \(\displaystyle \frac {\int \frac {\csc (x) \left (a^2-b \sin (x) a-b^2\right )}{a+b \sin (x)}dx}{a \left (a^2-b^2\right )}-\frac {b^2 \cos (x)}{a \left (a^2-b^2\right ) (a+b \sin (x))}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {a^2-b \sin (x) a-b^2}{\sin (x) (a+b \sin (x))}dx}{a \left (a^2-b^2\right )}-\frac {b^2 \cos (x)}{a \left (a^2-b^2\right ) (a+b \sin (x))}\) |
| \(\Big \downarrow \) 3480 |
| \(\displaystyle \frac {\frac {\left (a^2-b^2\right ) \int \csc (x)dx}{a}-\frac {b \left (2 a^2-b^2\right ) \int \frac {1}{a+b \sin (x)}dx}{a}}{a \left (a^2-b^2\right )}-\frac {b^2 \cos (x)}{a \left (a^2-b^2\right ) (a+b \sin (x))}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\left (a^2-b^2\right ) \int \csc (x)dx}{a}-\frac {b \left (2 a^2-b^2\right ) \int \frac {1}{a+b \sin (x)}dx}{a}}{a \left (a^2-b^2\right )}-\frac {b^2 \cos (x)}{a \left (a^2-b^2\right ) (a+b \sin (x))}\) |
| \(\Big \downarrow \) 3139 |
| \(\displaystyle \frac {\frac {\left (a^2-b^2\right ) \int \csc (x)dx}{a}-\frac {2 b \left (2 a^2-b^2\right ) \int \frac {1}{a \tan ^2\left (\frac {x}{2}\right )+2 b \tan \left (\frac {x}{2}\right )+a}d\tan \left (\frac {x}{2}\right )}{a}}{a \left (a^2-b^2\right )}-\frac {b^2 \cos (x)}{a \left (a^2-b^2\right ) (a+b \sin (x))}\) |
| \(\Big \downarrow \) 1083 |
| \(\displaystyle \frac {\frac {4 b \left (2 a^2-b^2\right ) \int \frac {1}{-\left (2 b+2 a \tan \left (\frac {x}{2}\right )\right )^2-4 \left (a^2-b^2\right )}d\left (2 b+2 a \tan \left (\frac {x}{2}\right )\right )}{a}+\frac {\left (a^2-b^2\right ) \int \csc (x)dx}{a}}{a \left (a^2-b^2\right )}-\frac {b^2 \cos (x)}{a \left (a^2-b^2\right ) (a+b \sin (x))}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {\frac {\left (a^2-b^2\right ) \int \csc (x)dx}{a}-\frac {2 b \left (2 a^2-b^2\right ) \arctan \left (\frac {2 a \tan \left (\frac {x}{2}\right )+2 b}{2 \sqrt {a^2-b^2}}\right )}{a \sqrt {a^2-b^2}}}{a \left (a^2-b^2\right )}-\frac {b^2 \cos (x)}{a \left (a^2-b^2\right ) (a+b \sin (x))}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {-\frac {2 b \left (2 a^2-b^2\right ) \arctan \left (\frac {2 a \tan \left (\frac {x}{2}\right )+2 b}{2 \sqrt {a^2-b^2}}\right )}{a \sqrt {a^2-b^2}}-\frac {\left (a^2-b^2\right ) \text {arctanh}(\cos (x))}{a}}{a \left (a^2-b^2\right )}-\frac {b^2 \cos (x)}{a \left (a^2-b^2\right ) (a+b \sin (x))}\) |
((-2*b*(2*a^2 - b^2)*ArcTan[(2*b + 2*a*Tan[x/2])/(2*Sqrt[a^2 - b^2])])/(a* Sqrt[a^2 - b^2]) - ((a^2 - b^2)*ArcTanh[Cos[x]])/a)/(a*(a^2 - b^2)) - (b^2 *Cos[x])/(a*(a^2 - b^2)*(a + b*Sin[x]))
3.2.90.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( -1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & & (LtQ[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[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b^2)*Cos[e + f*x]*(a + b*Sin[e + f* x])^(m + 1)*((c + d*Sin[e + f*x])^(n + 1)/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2 ))), x] + Simp[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)) Int[(a + b*Sin[e + f*x ])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[a*(b*c - a*d)*(m + 1) + b^2*d*(m + n + 2) - (b^2*c + b*(b*c - a*d)*(m + 1))*Sin[e + f*x] - b^2*d*(m + n + 3)*Si n[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a* d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -1] && IntegersQ[ 2*m, 2*n] && ((EqQ[a, 0] && IntegerQ[m] && !IntegerQ[n]) || !(IntegerQ[2* n] && LtQ[n, -1] && ((IntegerQ[n] && !IntegerQ[m]) || EqQ[a, 0])))
Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((a_.) + (b_.)*sin[(e_.) + (f_ .)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Simp[(A*b - a*B)/(b*c - a*d) Int[1/(a + b*Sin[e + f*x]), x], x] + Simp[(B*c - A*d)/ (b*c - a*d) Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f , A, B}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 0.67 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.32
| method | result | size |
| default | \(-\frac {4 b \left (\frac {\frac {b^{2} \tan \left (\frac {x}{2}\right )}{2 a^{2}-2 b^{2}}+\frac {a b}{2 a^{2}-2 b^{2}}}{a \left (\tan ^{2}\left (\frac {x}{2}\right )\right )+2 b \tan \left (\frac {x}{2}\right )+a}+\frac {\left (2 a^{2}-b^{2}\right ) \arctan \left (\frac {2 a \tan \left (\frac {x}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{2 \left (a^{2}-b^{2}\right )^{\frac {3}{2}}}\right )}{a^{2}}+\frac {\ln \left (\tan \left (\frac {x}{2}\right )\right )}{a^{2}}\) | \(123\) |
| risch | \(\frac {2 i b \left (-i a \,{\mathrm e}^{i x}+b \right )}{a \left (-a^{2}+b^{2}\right ) \left (b \,{\mathrm e}^{2 i x}-b +2 i a \,{\mathrm e}^{i x}\right )}+\frac {2 i b \ln \left ({\mathrm e}^{i x}+\frac {i \left (\sqrt {a^{2}-b^{2}}\, a -a^{2}+b^{2}\right )}{\sqrt {a^{2}-b^{2}}\, b}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right )}-\frac {i b^{3} \ln \left ({\mathrm e}^{i x}+\frac {i \left (\sqrt {a^{2}-b^{2}}\, a -a^{2}+b^{2}\right )}{\sqrt {a^{2}-b^{2}}\, b}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) a^{2}}-\frac {2 i b \ln \left ({\mathrm e}^{i x}+\frac {i \left (\sqrt {a^{2}-b^{2}}\, a +a^{2}-b^{2}\right )}{\sqrt {a^{2}-b^{2}}\, b}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right )}+\frac {i b^{3} \ln \left ({\mathrm e}^{i x}+\frac {i \left (\sqrt {a^{2}-b^{2}}\, a +a^{2}-b^{2}\right )}{\sqrt {a^{2}-b^{2}}\, b}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) a^{2}}-\frac {\ln \left ({\mathrm e}^{i x}+1\right )}{a^{2}}+\frac {\ln \left ({\mathrm e}^{i x}-1\right )}{a^{2}}\) | \(380\) |
-4/a^2*b*((1/2*b^2/(a^2-b^2)*tan(1/2*x)+1/2*a*b/(a^2-b^2))/(a*tan(1/2*x)^2 +2*b*tan(1/2*x)+a)+1/2*(2*a^2-b^2)/(a^2-b^2)^(3/2)*arctan(1/2*(2*a*tan(1/2 *x)+2*b)/(a^2-b^2)^(1/2)))+1/a^2*ln(tan(1/2*x))
Leaf count of result is larger than twice the leaf count of optimal. 224 vs. \(2 (87) = 174\).
Time = 0.51 (sec) , antiderivative size = 511, normalized size of antiderivative = 5.49 \[ \int \frac {\csc (x)}{(a+b \sin (x))^2} \, dx=\left [-\frac {{\left (2 \, a^{3} b - a b^{3} + {\left (2 \, a^{2} b^{2} - b^{4}\right )} \sin \left (x\right )\right )} \sqrt {-a^{2} + b^{2}} \log \left (-\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (x\right )^{2} - 2 \, a b \sin \left (x\right ) - a^{2} - b^{2} - 2 \, {\left (a \cos \left (x\right ) \sin \left (x\right ) + b \cos \left (x\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (x\right )^{2} - 2 \, a b \sin \left (x\right ) - a^{2} - b^{2}}\right ) + 2 \, {\left (a^{3} b^{2} - a b^{4}\right )} \cos \left (x\right ) + {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4} + {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} \sin \left (x\right )\right )} \log \left (\frac {1}{2} \, \cos \left (x\right ) + \frac {1}{2}\right ) - {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4} + {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} \sin \left (x\right )\right )} \log \left (-\frac {1}{2} \, \cos \left (x\right ) + \frac {1}{2}\right )}{2 \, {\left (a^{7} - 2 \, a^{5} b^{2} + a^{3} b^{4} + {\left (a^{6} b - 2 \, a^{4} b^{3} + a^{2} b^{5}\right )} \sin \left (x\right )\right )}}, \frac {2 \, {\left (2 \, a^{3} b - a b^{3} + {\left (2 \, a^{2} b^{2} - b^{4}\right )} \sin \left (x\right )\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \sin \left (x\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (x\right )}\right ) - 2 \, {\left (a^{3} b^{2} - a b^{4}\right )} \cos \left (x\right ) - {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4} + {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} \sin \left (x\right )\right )} \log \left (\frac {1}{2} \, \cos \left (x\right ) + \frac {1}{2}\right ) + {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4} + {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} \sin \left (x\right )\right )} \log \left (-\frac {1}{2} \, \cos \left (x\right ) + \frac {1}{2}\right )}{2 \, {\left (a^{7} - 2 \, a^{5} b^{2} + a^{3} b^{4} + {\left (a^{6} b - 2 \, a^{4} b^{3} + a^{2} b^{5}\right )} \sin \left (x\right )\right )}}\right ] \]
[-1/2*((2*a^3*b - a*b^3 + (2*a^2*b^2 - b^4)*sin(x))*sqrt(-a^2 + b^2)*log(- ((2*a^2 - b^2)*cos(x)^2 - 2*a*b*sin(x) - a^2 - b^2 - 2*(a*cos(x)*sin(x) + b*cos(x))*sqrt(-a^2 + b^2))/(b^2*cos(x)^2 - 2*a*b*sin(x) - a^2 - b^2)) + 2 *(a^3*b^2 - a*b^4)*cos(x) + (a^5 - 2*a^3*b^2 + a*b^4 + (a^4*b - 2*a^2*b^3 + b^5)*sin(x))*log(1/2*cos(x) + 1/2) - (a^5 - 2*a^3*b^2 + a*b^4 + (a^4*b - 2*a^2*b^3 + b^5)*sin(x))*log(-1/2*cos(x) + 1/2))/(a^7 - 2*a^5*b^2 + a^3*b ^4 + (a^6*b - 2*a^4*b^3 + a^2*b^5)*sin(x)), 1/2*(2*(2*a^3*b - a*b^3 + (2*a ^2*b^2 - b^4)*sin(x))*sqrt(a^2 - b^2)*arctan(-(a*sin(x) + b)/(sqrt(a^2 - b ^2)*cos(x))) - 2*(a^3*b^2 - a*b^4)*cos(x) - (a^5 - 2*a^3*b^2 + a*b^4 + (a^ 4*b - 2*a^2*b^3 + b^5)*sin(x))*log(1/2*cos(x) + 1/2) + (a^5 - 2*a^3*b^2 + a*b^4 + (a^4*b - 2*a^2*b^3 + b^5)*sin(x))*log(-1/2*cos(x) + 1/2))/(a^7 - 2 *a^5*b^2 + a^3*b^4 + (a^6*b - 2*a^4*b^3 + a^2*b^5)*sin(x))]
\[ \int \frac {\csc (x)}{(a+b \sin (x))^2} \, dx=\int \frac {\csc {\left (x \right )}}{\left (a + b \sin {\left (x \right )}\right )^{2}}\, dx \]
Exception generated. \[ \int \frac {\csc (x)}{(a+b \sin (x))^2} \, dx=\text {Exception raised: ValueError} \]
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*b^2-4*a^2>0)', see `assume?` f or more de
Time = 0.33 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.44 \[ \int \frac {\csc (x)}{(a+b \sin (x))^2} \, dx=-\frac {2 \, {\left (2 \, a^{2} b - b^{3}\right )} {\left (\pi \left \lfloor \frac {x}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, x\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{{\left (a^{4} - a^{2} b^{2}\right )} \sqrt {a^{2} - b^{2}}} - \frac {2 \, {\left (b^{3} \tan \left (\frac {1}{2} \, x\right ) + a b^{2}\right )}}{{\left (a^{4} - a^{2} b^{2}\right )} {\left (a \tan \left (\frac {1}{2} \, x\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, x\right ) + a\right )}} + \frac {\log \left ({\left | \tan \left (\frac {1}{2} \, x\right ) \right |}\right )}{a^{2}} \]
-2*(2*a^2*b - b^3)*(pi*floor(1/2*x/pi + 1/2)*sgn(a) + arctan((a*tan(1/2*x) + b)/sqrt(a^2 - b^2)))/((a^4 - a^2*b^2)*sqrt(a^2 - b^2)) - 2*(b^3*tan(1/2 *x) + a*b^2)/((a^4 - a^2*b^2)*(a*tan(1/2*x)^2 + 2*b*tan(1/2*x) + a)) + log (abs(tan(1/2*x)))/a^2
Time = 7.61 (sec) , antiderivative size = 1356, normalized size of antiderivative = 14.58 \[ \int \frac {\csc (x)}{(a+b \sin (x))^2} \, dx=\text {Too large to display} \]
log(tan(x/2))/a^2 - ((2*b^2)/(a*(a^2 - b^2)) + (2*b^3*tan(x/2))/(a^2*(a^2 - b^2)))/(a + 2*b*tan(x/2) + a*tan(x/2)^2) - (b*atan(((b*(2*a^2 - b^2)*(-( a + b)^3*(a - b)^3)^(1/2)*((2*tan(x/2)*(a^6 - 4*b^6 + 11*a^2*b^4 - 8*a^4*b ^2))/(a*b^4 + a^5 - 2*a^3*b^2) - (2*(3*a^4*b - 2*a^2*b^3))/(a^4 - a^2*b^2) + (b*((2*(a^6*b - a^4*b^3))/(a^4 - a^2*b^2) - (2*tan(x/2)*(3*a^8 - 4*a^2* b^6 + 11*a^4*b^4 - 10*a^6*b^2))/(a*b^4 + a^5 - 2*a^3*b^2))*(2*a^2 - b^2)*( -(a + b)^3*(a - b)^3)^(1/2))/(a^8 - a^2*b^6 + 3*a^4*b^4 - 3*a^6*b^2))*1i)/ (a^8 - a^2*b^6 + 3*a^4*b^4 - 3*a^6*b^2) - (b*(2*a^2 - b^2)*(-(a + b)^3*(a - b)^3)^(1/2)*((2*(3*a^4*b - 2*a^2*b^3))/(a^4 - a^2*b^2) - (2*tan(x/2)*(a^ 6 - 4*b^6 + 11*a^2*b^4 - 8*a^4*b^2))/(a*b^4 + a^5 - 2*a^3*b^2) + (b*((2*(a ^6*b - a^4*b^3))/(a^4 - a^2*b^2) - (2*tan(x/2)*(3*a^8 - 4*a^2*b^6 + 11*a^4 *b^4 - 10*a^6*b^2))/(a*b^4 + a^5 - 2*a^3*b^2))*(2*a^2 - b^2)*(-(a + b)^3*( a - b)^3)^(1/2))/(a^8 - a^2*b^6 + 3*a^4*b^4 - 3*a^6*b^2))*1i)/(a^8 - a^2*b ^6 + 3*a^4*b^4 - 3*a^6*b^2))/((4*(2*a^2*b - b^3))/(a^4 - a^2*b^2) + (4*tan (x/2)*(2*b^4 - 4*a^2*b^2))/(a*b^4 + a^5 - 2*a^3*b^2) + (b*(2*a^2 - b^2)*(- (a + b)^3*(a - b)^3)^(1/2)*((2*tan(x/2)*(a^6 - 4*b^6 + 11*a^2*b^4 - 8*a^4* b^2))/(a*b^4 + a^5 - 2*a^3*b^2) - (2*(3*a^4*b - 2*a^2*b^3))/(a^4 - a^2*b^2 ) + (b*((2*(a^6*b - a^4*b^3))/(a^4 - a^2*b^2) - (2*tan(x/2)*(3*a^8 - 4*a^2 *b^6 + 11*a^4*b^4 - 10*a^6*b^2))/(a*b^4 + a^5 - 2*a^3*b^2))*(2*a^2 - b^2)* (-(a + b)^3*(a - b)^3)^(1/2))/(a^8 - a^2*b^6 + 3*a^4*b^4 - 3*a^6*b^2)))...