3.43 \(\int \frac {\csc (x)}{a+b \csc (x)} \, dx\)

Optimal. Leaf size=40 \[ -\frac {2 \tanh ^{-1}\left (\frac {a+b \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 40, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {3831, 2660, 618, 206} \[ -\frac {2 \tanh ^{-1}\left (\frac {a+b \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}} \]

Antiderivative was successfully verified.

[In]

Int[Csc[x]/(a + b*Csc[x]),x]

[Out]

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

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 618

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 2660

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 3831

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

Rubi steps

\begin {align*} \int \frac {\csc (x)}{a+b \csc (x)} \, dx &=\frac {\int \frac {1}{1+\frac {a \sin (x)}{b}} \, dx}{b}\\ &=\frac {2 \operatorname {Subst}\left (\int \frac {1}{1+\frac {2 a x}{b}+x^2} \, dx,x,\tan \left (\frac {x}{2}\right )\right )}{b}\\ &=-\frac {4 \operatorname {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}\\ &=-\frac {2 \tanh ^{-1}\left (\frac {b \left (\frac {a}{b}+\tan \left (\frac {x}{2}\right )\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 40, normalized size = 1.00 \[ \frac {2 \tan ^{-1}\left (\frac {a+b \tan \left (\frac {x}{2}\right )}{\sqrt {b^2-a^2}}\right )}{\sqrt {b^2-a^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[x]/(a + b*Csc[x]),x]

[Out]

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

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fricas [A]  time = 0.53, size = 154, normalized size = 3.85 \[ \left [\frac {\log \left (-\frac {{\left (a^{2} - 2 \, b^{2}\right )} \cos \relax (x)^{2} + 2 \, a b \sin \relax (x) + a^{2} + b^{2} - 2 \, {\left (b \cos \relax (x) \sin \relax (x) + a \cos \relax (x)\right )} \sqrt {a^{2} - b^{2}}}{a^{2} \cos \relax (x)^{2} - 2 \, a b \sin \relax (x) - a^{2} - b^{2}}\right )}{2 \, \sqrt {a^{2} - b^{2}}}, -\frac {\sqrt {-a^{2} + b^{2}} \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \sin \relax (x) + a\right )}}{{\left (a^{2} - b^{2}\right )} \cos \relax (x)}\right )}{a^{2} - b^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.45, size = 48, normalized size = 1.20 \[ \frac {2 \, {\left (\pi \left \lfloor \frac {x}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (b) + \arctan \left (\frac {b \tan \left (\frac {1}{2} \, x\right ) + a}{\sqrt {-a^{2} + b^{2}}}\right )\right )}}{\sqrt {-a^{2} + b^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(x)/(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)))/sqrt(-a^2 + b^2)

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maple [A]  time = 0.18, size = 39, normalized size = 0.98 \[ \frac {2 \arctan \left (\frac {2 \tan \left (\frac {x}{2}\right ) b +2 a}{2 \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(x)/(a+b*csc(x)),x)

[Out]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(x)/(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 details)Is 4*a^2-4*b^2 positive or negative?

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mupad [B]  time = 0.29, size = 36, normalized size = 0.90 \[ -\frac {2\,\mathrm {atanh}\left (\frac {a+b\,\mathrm {tan}\left (\frac {x}{2}\right )}{\sqrt {a+b}\,\sqrt {a-b}}\right )}{\sqrt {a+b}\,\sqrt {a-b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sin(x)*(a + b/sin(x))),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\csc {\relax (x )}}{a + b \csc {\relax (x )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(csc(x)/(a + b*csc(x)), x)

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