Integrand size = 13, antiderivative size = 65 \[ \int \frac {\sec ^2(x)}{a+b \csc (x)} \, dx=\frac {2 a b \text {arctanh}\left (\frac {a+b \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2}}-\frac {\sec (x) (b-a \sin (x))}{a^2-b^2} \]
2*a*b*arctanh((a+b*tan(1/2*x))/(a^2-b^2)^(1/2))/(a^2-b^2)^(3/2)-sec(x)*(b- a*sin(x))/(a^2-b^2)
Time = 0.42 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.49 \[ \int \frac {\sec ^2(x)}{a+b \csc (x)} \, dx=\frac {2 a b \arctan \left (\frac {a+b \tan \left (\frac {x}{2}\right )}{\sqrt {-a^2+b^2}}\right )}{\left (-a^2+b^2\right )^{3/2}}+\sin \left (\frac {x}{2}\right ) \left (\frac {1}{(a+b) \left (\cos \left (\frac {x}{2}\right )-\sin \left (\frac {x}{2}\right )\right )}+\frac {1}{(a-b) \left (\cos \left (\frac {x}{2}\right )+\sin \left (\frac {x}{2}\right )\right )}\right ) \]
(2*a*b*ArcTan[(a + b*Tan[x/2])/Sqrt[-a^2 + b^2]])/(-a^2 + b^2)^(3/2) + Sin [x/2]*(1/((a + b)*(Cos[x/2] - Sin[x/2])) + 1/((a - b)*(Cos[x/2] + Sin[x/2] )))
Time = 0.38 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.09, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.769, Rules used = {3042, 4360, 3042, 3345, 25, 27, 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.
\(\displaystyle \int \frac {\sec ^2(x)}{a+b \csc (x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\cos (x)^2 (a+b \csc (x))}dx\) |
\(\Big \downarrow \) 4360 |
\(\displaystyle \int \frac {\tan (x) \sec (x)}{a \sin (x)+b}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin (x)}{\cos (x)^2 (a \sin (x)+b)}dx\) |
\(\Big \downarrow \) 3345 |
\(\displaystyle \frac {\int -\frac {a b}{b+a \sin (x)}dx}{a^2-b^2}-\frac {\sec (x) (b-a \sin (x))}{a^2-b^2}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\int \frac {a b}{b+a \sin (x)}dx}{a^2-b^2}-\frac {\sec (x) (b-a \sin (x))}{a^2-b^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {a b \int \frac {1}{b+a \sin (x)}dx}{a^2-b^2}-\frac {\sec (x) (b-a \sin (x))}{a^2-b^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {a b \int \frac {1}{b+a \sin (x)}dx}{a^2-b^2}-\frac {\sec (x) (b-a \sin (x))}{a^2-b^2}\) |
\(\Big \downarrow \) 3139 |
\(\displaystyle -\frac {2 a b \int \frac {1}{b \tan ^2\left (\frac {x}{2}\right )+2 a \tan \left (\frac {x}{2}\right )+b}d\tan \left (\frac {x}{2}\right )}{a^2-b^2}-\frac {\sec (x) (b-a \sin (x))}{a^2-b^2}\) |
\(\Big \downarrow \) 1083 |
\(\displaystyle \frac {4 a b \int \frac {1}{4 \left (a^2-b^2\right )-\left (2 a+2 b \tan \left (\frac {x}{2}\right )\right )^2}d\left (2 a+2 b \tan \left (\frac {x}{2}\right )\right )}{a^2-b^2}-\frac {\sec (x) (b-a \sin (x))}{a^2-b^2}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {2 a b \text {arctanh}\left (\frac {2 a+2 b \tan \left (\frac {x}{2}\right )}{2 \sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2}}-\frac {\sec (x) (b-a \sin (x))}{a^2-b^2}\) |
(2*a*b*ArcTanh[(2*a + 2*b*Tan[x/2])/(2*Sqrt[a^2 - b^2])])/(a^2 - b^2)^(3/2 ) - (Sec[x]*(b - a*Sin[x]))/(a^2 - b^2)
3.1.14.3.1 Defintions of rubi rules used
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[(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[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(g*Co s[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m + 1)*((b*c - a*d - (a*c - b*d)* 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*Simp[c*(a^2*(p + 2) - b^2*(m + p + 2)) + a*b*d*m + b*(a*c - b*d)*(m + p + 3)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && NeQ[a^2 - b^2, 0] && Lt Q[p, -1] && IntegerQ[2*m]
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]
Time = 0.54 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.42
method | result | size |
default | \(-\frac {4}{\left (4 a +4 b \right ) \left (\tan \left (\frac {x}{2}\right )-1\right )}-\frac {4}{\left (4 a -4 b \right ) \left (\tan \left (\frac {x}{2}\right )+1\right )}-\frac {2 a b \arctan \left (\frac {2 b \tan \left (\frac {x}{2}\right )+2 a}{2 \sqrt {-a^{2}+b^{2}}}\right )}{\left (a +b \right ) \left (a -b \right ) \sqrt {-a^{2}+b^{2}}}\) | \(92\) |
risch | \(\frac {-2 i a +2 b \,{\mathrm e}^{i x}}{\left ({\mathrm e}^{2 i x}+1\right ) \left (-a^{2}+b^{2}\right )}+\frac {b a \ln \left ({\mathrm e}^{i x}+\frac {i b \sqrt {a^{2}-b^{2}}+a^{2}-b^{2}}{\sqrt {a^{2}-b^{2}}\, a}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right )}-\frac {b a \ln \left ({\mathrm e}^{i x}+\frac {i b \sqrt {a^{2}-b^{2}}-a^{2}+b^{2}}{\sqrt {a^{2}-b^{2}}\, a}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right )}\) | \(181\) |
-4/(4*a+4*b)/(tan(1/2*x)-1)-4/(4*a-4*b)/(tan(1/2*x)+1)-2*a*b/(a+b)/(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))
Time = 0.27 (sec) , antiderivative size = 243, normalized size of antiderivative = 3.74 \[ \int \frac {\sec ^2(x)}{a+b \csc (x)} \, dx=\left [-\frac {\sqrt {a^{2} - b^{2}} a b \cos \left (x\right ) \log \left (-\frac {{\left (a^{2} - 2 \, b^{2}\right )} \cos \left (x\right )^{2} + 2 \, a b \sin \left (x\right ) + a^{2} + b^{2} - 2 \, {\left (b \cos \left (x\right ) \sin \left (x\right ) + a \cos \left (x\right )\right )} \sqrt {a^{2} - b^{2}}}{a^{2} \cos \left (x\right )^{2} - 2 \, a b \sin \left (x\right ) - a^{2} - b^{2}}\right ) + 2 \, a^{2} b - 2 \, b^{3} - 2 \, {\left (a^{3} - a b^{2}\right )} \sin \left (x\right )}{2 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \cos \left (x\right )}, \frac {\sqrt {-a^{2} + b^{2}} a b \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \sin \left (x\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \cos \left (x\right )}\right ) \cos \left (x\right ) - a^{2} b + b^{3} + {\left (a^{3} - a b^{2}\right )} \sin \left (x\right )}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \cos \left (x\right )}\right ] \]
[-1/2*(sqrt(a^2 - b^2)*a*b*cos(x)*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))*sqrt(a^2 - b^2))/(a^2*cos (x)^2 - 2*a*b*sin(x) - a^2 - b^2)) + 2*a^2*b - 2*b^3 - 2*(a^3 - a*b^2)*sin (x))/((a^4 - 2*a^2*b^2 + b^4)*cos(x)), (sqrt(-a^2 + b^2)*a*b*arctan(-sqrt( -a^2 + b^2)*(b*sin(x) + a)/((a^2 - b^2)*cos(x)))*cos(x) - a^2*b + b^3 + (a ^3 - a*b^2)*sin(x))/((a^4 - 2*a^2*b^2 + b^4)*cos(x))]
\[ \int \frac {\sec ^2(x)}{a+b \csc (x)} \, dx=\int \frac {\sec ^{2}{\left (x \right )}}{a + b \csc {\left (x \right )}}\, dx \]
Exception generated. \[ \int \frac {\sec ^2(x)}{a+b \csc (x)} \, dx=\text {Exception raised: ValueError} \]
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 = 0.27 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.46 \[ \int \frac {\sec ^2(x)}{a+b \csc (x)} \, dx=-\frac {2 \, {\left (\pi \left \lfloor \frac {x}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (b\right ) + \arctan \left (\frac {b \tan \left (\frac {1}{2} \, x\right ) + a}{\sqrt {-a^{2} + b^{2}}}\right )\right )} a b}{{\left (a^{2} - b^{2}\right )} \sqrt {-a^{2} + b^{2}}} - \frac {2 \, {\left (a \tan \left (\frac {1}{2} \, x\right ) - b\right )}}{{\left (a^{2} - b^{2}\right )} {\left (\tan \left (\frac {1}{2} \, x\right )^{2} - 1\right )}} \]
-2*(pi*floor(1/2*x/pi + 1/2)*sgn(b) + arctan((b*tan(1/2*x) + a)/sqrt(-a^2 + b^2)))*a*b/((a^2 - b^2)*sqrt(-a^2 + b^2)) - 2*(a*tan(1/2*x) - b)/((a^2 - b^2)*(tan(1/2*x)^2 - 1))
Time = 18.98 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.60 \[ \int \frac {\sec ^2(x)}{a+b \csc (x)} \, dx=\frac {\frac {2\,b}{a^2-b^2}-\frac {2\,a\,\mathrm {tan}\left (\frac {x}{2}\right )}{a^2-b^2}}{{\mathrm {tan}\left (\frac {x}{2}\right )}^2-1}+\frac {2\,a\,b\,\mathrm {atanh}\left (\frac {2\,a^3-2\,a\,b^2+2\,b\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (a^2-b^2\right )}{2\,{\left (a+b\right )}^{3/2}\,{\left (a-b\right )}^{3/2}}\right )}{{\left (a+b\right )}^{3/2}\,{\left (a-b\right )}^{3/2}} \]