3.80 \(\int \frac {\text {csch}(x)}{a+b \text {csch}(x)} \, dx\)

Optimal. Leaf size=37 \[ -\frac {2 \tanh ^{-1}\left (\frac {a-b \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2+b^2}}\right )}{\sqrt {a^2+b^2}} \]

[Out]

-2*arctanh((a-b*tanh(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 = 37, 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} \[ -\frac {2 \tanh ^{-1}\left (\frac {a-b \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2+b^2}}\right )}{\sqrt {a^2+b^2}} \]

Antiderivative was successfully verified.

[In]

Int[Csch[x]/(a + b*Csch[x]),x]

[Out]

(-2*ArcTanh[(a - b*Tanh[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 {\text {csch}(x)}{a+b \text {csch}(x)} \, dx &=\frac {\int \frac {1}{1+\frac {a \sinh (x)}{b}} \, dx}{b}\\ &=\frac {2 \operatorname {Subst}\left (\int \frac {1}{1+\frac {2 a x}{b}-x^2} \, dx,x,\tanh \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 \tanh \left (\frac {x}{2}\right )\right )}{b}\\ &=-\frac {2 \tanh ^{-1}\left (\frac {b \left (\frac {a}{b}-\tanh \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 = 0.03, size = 45, normalized size = 1.22 \[ \frac {2 \tan ^{-1}\left (\frac {a-b \tanh \left (\frac {x}{2}\right )}{\sqrt {-a^2-b^2}}\right )}{\sqrt {-a^2-b^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Csch[x]/(a + b*Csch[x]),x]

[Out]

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

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fricas [B]  time = 0.40, size = 111, normalized size = 3.00 \[ \frac {\log \left (\frac {a^{2} \cosh \relax (x)^{2} + a^{2} \sinh \relax (x)^{2} + 2 \, a b \cosh \relax (x) + a^{2} + 2 \, b^{2} + 2 \, {\left (a^{2} \cosh \relax (x) + a b\right )} \sinh \relax (x) - 2 \, \sqrt {a^{2} + b^{2}} {\left (a \cosh \relax (x) + a \sinh \relax (x) + b\right )}}{a \cosh \relax (x)^{2} + a \sinh \relax (x)^{2} + 2 \, b \cosh \relax (x) + 2 \, {\left (a \cosh \relax (x) + b\right )} \sinh \relax (x) - a}\right )}{\sqrt {a^{2} + b^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log((a^2*cosh(x)^2 + a^2*sinh(x)^2 + 2*a*b*cosh(x) + a^2 + 2*b^2 + 2*(a^2*cosh(x) + a*b)*sinh(x) - 2*sqrt(a^2
+ b^2)*(a*cosh(x) + a*sinh(x) + b))/(a*cosh(x)^2 + a*sinh(x)^2 + 2*b*cosh(x) + 2*(a*cosh(x) + b)*sinh(x) - a))
/sqrt(a^2 + b^2)

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giac [A]  time = 0.14, size = 56, normalized size = 1.51 \[ \frac {\log \left (\frac {{\left | 2 \, a e^{x} + 2 \, b - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, a e^{x} + 2 \, b + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{\sqrt {a^{2} + b^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(x)/(a+b*csch(x)),x, algorithm="giac")

[Out]

log(abs(2*a*e^x + 2*b - 2*sqrt(a^2 + b^2))/abs(2*a*e^x + 2*b + 2*sqrt(a^2 + b^2)))/sqrt(a^2 + b^2)

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maple [A]  time = 0.08, size = 35, normalized size = 0.95 \[ \frac {2 \arctanh \left (\frac {2 \tanh \left (\frac {x}{2}\right ) b -2 a}{2 \sqrt {a^{2}+b^{2}}}\right )}{\sqrt {a^{2}+b^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.40, size = 54, normalized size = 1.46 \[ \frac {\log \left (\frac {a e^{\left (-x\right )} - b - \sqrt {a^{2} + b^{2}}}{a e^{\left (-x\right )} - b + \sqrt {a^{2} + b^{2}}}\right )}{\sqrt {a^{2} + b^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(x)/(a+b*csch(x)),x, algorithm="maxima")

[Out]

log((a*e^(-x) - b - sqrt(a^2 + b^2))/(a*e^(-x) - b + sqrt(a^2 + b^2)))/sqrt(a^2 + b^2)

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mupad [B]  time = 1.52, size = 49, normalized size = 1.32 \[ \frac {2\,\mathrm {atan}\left (\frac {b}{\sqrt {-a^2-b^2}}+\frac {a\,{\mathrm {e}}^x}{\sqrt {-a^2-b^2}}\right )}{\sqrt {-a^2-b^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sinh(x)*(a + b/sinh(x))),x)

[Out]

(2*atan(b/(- a^2 - b^2)^(1/2) + (a*exp(x))/(- a^2 - b^2)^(1/2)))/(- a^2 - b^2)^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(csch(x)/(a + b*csch(x)), x)

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