\(\int \frac {\csc ^2(x)}{a+b \csc (x)} \, dx\) [42]

Optimal result

Integrand size = 13, antiderivative size = 53 \[ \int \frac {\csc ^2(x)}{a+b \csc (x)} \, dx=-\frac {\text {arctanh}(\cos (x))}{b}+\frac {2 a \text {arctanh}\left (\frac {a+b \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{b \sqrt {a^2-b^2}} \]

[Out]

-arctanh(cos(x))/b+2*a*arctanh((a+b*tan(1/2*x))/(a^2-b^2)^(1/2))/b/(a^2-b^2)^(1/2)

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {3874, 3855, 3916, 2739, 632, 212} \[ \int \frac {\csc ^2(x)}{a+b \csc (x)} \, dx=\frac {2 a \text {arctanh}\left (\frac {a+b \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{b \sqrt {a^2-b^2}}-\frac {\text {arctanh}(\cos (x))}{b} \]

[In]

Int[Csc[x]^2/(a + b*Csc[x]),x]

[Out]

-(ArcTanh[Cos[x]]/b) + (2*a*ArcTanh[(a + b*Tan[x/2])/Sqrt[a^2 - b^2]])/(b*Sqrt[a^2 - b^2])

Rule 212

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] && (GtQ[a, 0] || LtQ[b, 0])

Rule 632

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 2739

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[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]

Rule 3855

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3874

Int[csc[(e_.) + (f_.)*(x_)]^2/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[1/b, Int[Csc[e + f*x],
x], x] - Dist[a/b, Int[Csc[e + f*x]/(a + b*Csc[e + f*x]), x], x] /; FreeQ[{a, b, e, f}, x]

Rule 3916

Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[1/b, Int[1/(1 + (a/b)*Si
n[e + f*x]), x], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {\int \csc (x) \, dx}{b}-\frac {a \int \frac {\csc (x)}{a+b \csc (x)} \, dx}{b} \\ & = -\frac {\text {arctanh}(\cos (x))}{b}-\frac {a \int \frac {1}{1+\frac {a \sin (x)}{b}} \, dx}{b^2} \\ & = -\frac {\text {arctanh}(\cos (x))}{b}-\frac {(2 a) \text {Subst}\left (\int \frac {1}{1+\frac {2 a x}{b}+x^2} \, dx,x,\tan \left (\frac {x}{2}\right )\right )}{b^2} \\ & = -\frac {\text {arctanh}(\cos (x))}{b}+\frac {(4 a) \text {Subst}\left (\int \frac {1}{-4 \left (1-\frac {a^2}{b^2}\right )-x^2} \, dx,x,\frac {2 a}{b}+2 \tan \left (\frac {x}{2}\right )\right )}{b^2} \\ & = -\frac {\text {arctanh}(\cos (x))}{b}+\frac {2 a \text {arctanh}\left (\frac {b \left (\frac {a}{b}+\tan \left (\frac {x}{2}\right )\right )}{\sqrt {a^2-b^2}}\right )}{b \sqrt {a^2-b^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.17 \[ \int \frac {\csc ^2(x)}{a+b \csc (x)} \, dx=\frac {-\frac {2 a \arctan \left (\frac {a+b \tan \left (\frac {x}{2}\right )}{\sqrt {-a^2+b^2}}\right )}{\sqrt {-a^2+b^2}}-\log \left (\cos \left (\frac {x}{2}\right )\right )+\log \left (\sin \left (\frac {x}{2}\right )\right )}{b} \]

[In]

Integrate[Csc[x]^2/(a + b*Csc[x]),x]

[Out]

((-2*a*ArcTan[(a + b*Tan[x/2])/Sqrt[-a^2 + b^2]])/Sqrt[-a^2 + b^2] - Log[Cos[x/2]] + Log[Sin[x/2]])/b

Maple [A] (verified)

Time = 0.35 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.00

method	result	size
default	\(-\frac {2 a \arctan \left (\frac {2 b \tan \left (\frac {x}{2}\right )+2 a}{2 \sqrt {-a^{2}+b^{2}}}\right )}{b \sqrt {-a^{2}+b^{2}}}+\frac {\ln \left (\tan \left (\frac {x}{2}\right )\right )}{b}\)	\(53\)
risch	\(-\frac {a \ln \left ({\mathrm e}^{i x}+\frac {i b \sqrt {a^{2}-b^{2}}-a^{2}+b^{2}}{\sqrt {a^{2}-b^{2}}\, a}\right )}{\sqrt {a^{2}-b^{2}}\, b}+\frac {a \ln \left ({\mathrm e}^{i x}+\frac {i b \sqrt {a^{2}-b^{2}}+a^{2}-b^{2}}{\sqrt {a^{2}-b^{2}}\, a}\right )}{\sqrt {a^{2}-b^{2}}\, b}-\frac {\ln \left ({\mathrm e}^{i x}+1\right )}{b}+\frac {\ln \left ({\mathrm e}^{i x}-1\right )}{b}\)	\(152\)

[In]

int(csc(x)^2/(a+b*csc(x)),x,method=_RETURNVERBOSE)

[Out]

-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))+1/b*ln(tan(1/2*x))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 100 vs. \(2 (47) = 94\).

Time = 0.29 (sec) , antiderivative size = 245, normalized size of antiderivative = 4.62 \[ \int \frac {\csc ^2(x)}{a+b \csc (x)} \, dx=\left [\frac {\sqrt {a^{2} - b^{2}} a \log \left (\frac {{\left (a^{2} - 2 \, b^{2}\right )} \cos \left (x\right )^{2} + 2 \, a b \sin \left (x\right ) + a^{2} + b^{2} + 2 \, {\left (b \cos \left (x\right ) \sin \left (x\right ) + a \cos \left (x\right )\right )} \sqrt {a^{2} - b^{2}}}{a^{2} \cos \left (x\right )^{2} - 2 \, a b \sin \left (x\right ) - a^{2} - b^{2}}\right ) - {\left (a^{2} - b^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (x\right ) + \frac {1}{2}\right ) + {\left (a^{2} - b^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (x\right ) + \frac {1}{2}\right )}{2 \, {\left (a^{2} b - b^{3}\right )}}, \frac {2 \, \sqrt {-a^{2} + b^{2}} a \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \sin \left (x\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \cos \left (x\right )}\right ) - {\left (a^{2} - b^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (x\right ) + \frac {1}{2}\right ) + {\left (a^{2} - b^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (x\right ) + \frac {1}{2}\right )}{2 \, {\left (a^{2} b - b^{3}\right )}}\right ] \]

[In]

integrate(csc(x)^2/(a+b*csc(x)),x, algorithm="fricas")

[Out]

[1/2*(sqrt(a^2 - b^2)*a*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)) - (a^2 - b^2)*log(1/2*cos(x) + 1/2) + (a^2 - b^2
)*log(-1/2*cos(x) + 1/2))/(a^2*b - b^3), 1/2*(2*sqrt(-a^2 + b^2)*a*arctan(-sqrt(-a^2 + b^2)*(b*sin(x) + a)/((a
^2 - b^2)*cos(x))) - (a^2 - b^2)*log(1/2*cos(x) + 1/2) + (a^2 - b^2)*log(-1/2*cos(x) + 1/2))/(a^2*b - b^3)]

Sympy [F]

\[ \int \frac {\csc ^2(x)}{a+b \csc (x)} \, dx=\int \frac {\csc ^{2}{\left (x \right )}}{a + b \csc {\left (x \right )}}\, dx \]

[In]

integrate(csc(x)**2/(a+b*csc(x)),x)

[Out]

Integral(csc(x)**2/(a + b*csc(x)), x)

Maxima [F(-2)]

Exception generated. \[ \int \frac {\csc ^2(x)}{a+b \csc (x)} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate(csc(x)^2/(a+b*csc(x)),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a^2-4*b^2>0)', see `assume?`
for more de

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.19 \[ \int \frac {\csc ^2(x)}{a+b \csc (x)} \, dx=-\frac {2 \, {\left (\pi \left \lfloor \frac {x}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (b\right ) + \arctan \left (\frac {b \tan \left (\frac {1}{2} \, x\right ) + a}{\sqrt {-a^{2} + b^{2}}}\right )\right )} a}{\sqrt {-a^{2} + b^{2}} b} + \frac {\log \left ({\left | \tan \left (\frac {1}{2} \, x\right ) \right |}\right )}{b} \]

[In]

integrate(csc(x)^2/(a+b*csc(x)),x, algorithm="giac")

[Out]

-2*(pi*floor(1/2*x/pi + 1/2)*sgn(b) + arctan((b*tan(1/2*x) + a)/sqrt(-a^2 + b^2)))*a/(sqrt(-a^2 + b^2)*b) + lo
g(abs(tan(1/2*x)))/b

Mupad [B] (verification not implemented)

Time = 18.29 (sec) , antiderivative size = 129, normalized size of antiderivative = 2.43 \[ \int \frac {\csc ^2(x)}{a+b \csc (x)} \, dx=\frac {\ln \left (\frac {\sin \left (\frac {x}{2}\right )}{\cos \left (\frac {x}{2}\right )}\right )}{b}-\frac {2\,a\,\mathrm {atanh}\left (\frac {\sqrt {a^2-b^2}\,\left (4{}\mathrm {i}\,\sin \left (\frac {x}{2}\right )\,a^2+2{}\mathrm {i}\,\cos \left (\frac {x}{2}\right )\,a\,b-1{}\mathrm {i}\,\sin \left (\frac {x}{2}\right )\,b^2\right )}{a^3\,\sin \left (\frac {x}{2}\right )\,4{}\mathrm {i}+b\,\cos \left (\frac {x}{2}\right )\,\left (a^2-b^2\right )\,1{}\mathrm {i}+a^2\,b\,\cos \left (\frac {x}{2}\right )\,1{}\mathrm {i}-a\,b^2\,\sin \left (\frac {x}{2}\right )\,3{}\mathrm {i}}\right )}{b\,\sqrt {a^2-b^2}} \]

[In]

int(1/(sin(x)^2*(a + b/sin(x))),x)

[Out]

log(sin(x/2)/cos(x/2))/b - (2*a*atanh(((a^2 - b^2)^(1/2)*(a^2*sin(x/2)*4i - b^2*sin(x/2)*1i + a*b*cos(x/2)*2i)
)/(a^3*sin(x/2)*4i + b*cos(x/2)*(a^2 - b^2)*1i + a^2*b*cos(x/2)*1i - a*b^2*sin(x/2)*3i)))/(b*(a^2 - b^2)^(1/2)
)