3.103 \(\int \frac {\text {csch}^3(x)}{a+b \coth (x)} \, dx\)

Optimal. Leaf size=57 \[ -\frac {\sqrt {a^2-b^2} \tanh ^{-1}\left (\frac {\sinh (x) (a \coth (x)+b)}{\sqrt {a^2-b^2}}\right )}{b^2}+\frac {a \tanh ^{-1}(\cosh (x))}{b^2}-\frac {\text {csch}(x)}{b} \]

[Out]

a*arctanh(cosh(x))/b^2-csch(x)/b-arctanh((b+a*coth(x))*sinh(x)/(a^2-b^2)^(1/2))*(a^2-b^2)^(1/2)/b^2

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Rubi [A]  time = 0.11, antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {3510, 3486, 3770, 3509, 206} \[ -\frac {\sqrt {a^2-b^2} \tanh ^{-1}\left (\frac {\sinh (x) (a \coth (x)+b)}{\sqrt {a^2-b^2}}\right )}{b^2}+\frac {a \tanh ^{-1}(\cosh (x))}{b^2}-\frac {\text {csch}(x)}{b} \]

Antiderivative was successfully verified.

[In]

Int[Csch[x]^3/(a + b*Coth[x]),x]

[Out]

(a*ArcTanh[Cosh[x]])/b^2 - (Sqrt[a^2 - b^2]*ArcTanh[((b + a*Coth[x])*Sinh[x])/Sqrt[a^2 - b^2]])/b^2 - Csch[x]/
b

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 3486

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*(d*Sec[
e + f*x])^m)/(f*m), x] + Dist[a, Int[(d*Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f, m}, x] && (IntegerQ[2
*m] || NeQ[a^2 + b^2, 0])

Rule 3509

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

Rule 3510

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_)/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Dist[d^2/b^2, I
nt[(d*Sec[e + f*x])^(m - 2)*(a - b*Tan[e + f*x]), x], x] + Dist[(d^2*(a^2 + b^2))/b^2, Int[(d*Sec[e + f*x])^(m
- 2)/(a + b*Tan[e + f*x]), x], x] /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 + b^2, 0] && IGtQ[m, 1]

Rule 3770

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

Rubi steps

\begin {align*} \int \frac {\text {csch}^3(x)}{a+b \coth (x)} \, dx &=-\frac {\int (a-b \coth (x)) \text {csch}(x) \, dx}{b^2}+\frac {\left (a^2-b^2\right ) \int \frac {\text {csch}(x)}{a+b \coth (x)} \, dx}{b^2}\\ &=-\frac {\text {csch}(x)}{b}-\frac {a \int \text {csch}(x) \, dx}{b^2}-\frac {\left (a^2-b^2\right ) \operatorname {Subst}\left (\int \frac {1}{a^2-b^2-x^2} \, dx,x,i (-i b-i a \coth (x)) \sinh (x)\right )}{b^2}\\ &=\frac {a \tanh ^{-1}(\cosh (x))}{b^2}-\frac {\sqrt {a^2-b^2} \tanh ^{-1}\left (\frac {(b+a \coth (x)) \sinh (x)}{\sqrt {a^2-b^2}}\right )}{b^2}-\frac {\text {csch}(x)}{b}\\ \end {align*}

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Mathematica [A]  time = 0.11, size = 65, normalized size = 1.14 \[ -\frac {2 \sqrt {b-a} \sqrt {a+b} \tan ^{-1}\left (\frac {a+b \tanh \left (\frac {x}{2}\right )}{\sqrt {b-a} \sqrt {a+b}}\right )+a \log \left (\tanh \left (\frac {x}{2}\right )\right )+b \text {csch}(x)}{b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Csch[x]^3/(a + b*Coth[x]),x]

[Out]

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

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fricas [B]  time = 0.45, size = 384, normalized size = 6.74 \[ \left [\frac {\sqrt {a^{2} - b^{2}} {\left (\cosh \relax (x)^{2} + 2 \, \cosh \relax (x) \sinh \relax (x) + \sinh \relax (x)^{2} - 1\right )} \log \left (\frac {{\left (a + b\right )} \cosh \relax (x)^{2} + 2 \, {\left (a + b\right )} \cosh \relax (x) \sinh \relax (x) + {\left (a + b\right )} \sinh \relax (x)^{2} - 2 \, \sqrt {a^{2} - b^{2}} {\left (\cosh \relax (x) + \sinh \relax (x)\right )} + a - b}{{\left (a + b\right )} \cosh \relax (x)^{2} + 2 \, {\left (a + b\right )} \cosh \relax (x) \sinh \relax (x) + {\left (a + b\right )} \sinh \relax (x)^{2} - a + b}\right ) - 2 \, b \cosh \relax (x) + {\left (a \cosh \relax (x)^{2} + 2 \, a \cosh \relax (x) \sinh \relax (x) + a \sinh \relax (x)^{2} - a\right )} \log \left (\cosh \relax (x) + \sinh \relax (x) + 1\right ) - {\left (a \cosh \relax (x)^{2} + 2 \, a \cosh \relax (x) \sinh \relax (x) + a \sinh \relax (x)^{2} - a\right )} \log \left (\cosh \relax (x) + \sinh \relax (x) - 1\right ) - 2 \, b \sinh \relax (x)}{b^{2} \cosh \relax (x)^{2} + 2 \, b^{2} \cosh \relax (x) \sinh \relax (x) + b^{2} \sinh \relax (x)^{2} - b^{2}}, \frac {2 \, \sqrt {-a^{2} + b^{2}} {\left (\cosh \relax (x)^{2} + 2 \, \cosh \relax (x) \sinh \relax (x) + \sinh \relax (x)^{2} - 1\right )} \arctan \left (\frac {\sqrt {-a^{2} + b^{2}}}{{\left (a + b\right )} \cosh \relax (x) + {\left (a + b\right )} \sinh \relax (x)}\right ) - 2 \, b \cosh \relax (x) + {\left (a \cosh \relax (x)^{2} + 2 \, a \cosh \relax (x) \sinh \relax (x) + a \sinh \relax (x)^{2} - a\right )} \log \left (\cosh \relax (x) + \sinh \relax (x) + 1\right ) - {\left (a \cosh \relax (x)^{2} + 2 \, a \cosh \relax (x) \sinh \relax (x) + a \sinh \relax (x)^{2} - a\right )} \log \left (\cosh \relax (x) + \sinh \relax (x) - 1\right ) - 2 \, b \sinh \relax (x)}{b^{2} \cosh \relax (x)^{2} + 2 \, b^{2} \cosh \relax (x) \sinh \relax (x) + b^{2} \sinh \relax (x)^{2} - b^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(x)^3/(a+b*coth(x)),x, algorithm="fricas")

[Out]

[(sqrt(a^2 - b^2)*(cosh(x)^2 + 2*cosh(x)*sinh(x) + sinh(x)^2 - 1)*log(((a + b)*cosh(x)^2 + 2*(a + b)*cosh(x)*s
inh(x) + (a + b)*sinh(x)^2 - 2*sqrt(a^2 - b^2)*(cosh(x) + sinh(x)) + a - b)/((a + b)*cosh(x)^2 + 2*(a + b)*cos
h(x)*sinh(x) + (a + b)*sinh(x)^2 - a + b)) - 2*b*cosh(x) + (a*cosh(x)^2 + 2*a*cosh(x)*sinh(x) + a*sinh(x)^2 -
a)*log(cosh(x) + sinh(x) + 1) - (a*cosh(x)^2 + 2*a*cosh(x)*sinh(x) + a*sinh(x)^2 - a)*log(cosh(x) + sinh(x) -
1) - 2*b*sinh(x))/(b^2*cosh(x)^2 + 2*b^2*cosh(x)*sinh(x) + b^2*sinh(x)^2 - b^2), (2*sqrt(-a^2 + b^2)*(cosh(x)^
2 + 2*cosh(x)*sinh(x) + sinh(x)^2 - 1)*arctan(sqrt(-a^2 + b^2)/((a + b)*cosh(x) + (a + b)*sinh(x))) - 2*b*cosh
(x) + (a*cosh(x)^2 + 2*a*cosh(x)*sinh(x) + a*sinh(x)^2 - a)*log(cosh(x) + sinh(x) + 1) - (a*cosh(x)^2 + 2*a*co
sh(x)*sinh(x) + a*sinh(x)^2 - a)*log(cosh(x) + sinh(x) - 1) - 2*b*sinh(x))/(b^2*cosh(x)^2 + 2*b^2*cosh(x)*sinh
(x) + b^2*sinh(x)^2 - b^2)]

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giac [A]  time = 0.12, size = 85, normalized size = 1.49 \[ \frac {a \log \left (e^{x} + 1\right )}{b^{2}} - \frac {a \log \left ({\left | e^{x} - 1 \right |}\right )}{b^{2}} + \frac {2 \, {\left (a^{2} - b^{2}\right )} \arctan \left (\frac {a e^{x} + b e^{x}}{\sqrt {-a^{2} + b^{2}}}\right )}{\sqrt {-a^{2} + b^{2}} b^{2}} - \frac {2 \, e^{x}}{b {\left (e^{\left (2 \, x\right )} - 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(x)^3/(a+b*coth(x)),x, algorithm="giac")

[Out]

a*log(e^x + 1)/b^2 - a*log(abs(e^x - 1))/b^2 + 2*(a^2 - b^2)*arctan((a*e^x + b*e^x)/sqrt(-a^2 + b^2))/(sqrt(-a
^2 + b^2)*b^2) - 2*e^x/(b*(e^(2*x) - 1))

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maple [B]  time = 0.11, size = 115, normalized size = 2.02 \[ \frac {\tanh \left (\frac {x}{2}\right )}{2 b}+\frac {2 \arctan \left (\frac {2 \tanh \left (\frac {x}{2}\right ) b +2 a}{2 \sqrt {-a^{2}+b^{2}}}\right ) a^{2}}{b^{2} \sqrt {-a^{2}+b^{2}}}-\frac {2 \arctan \left (\frac {2 \tanh \left (\frac {x}{2}\right ) b +2 a}{2 \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}}-\frac {1}{2 b \tanh \left (\frac {x}{2}\right )}-\frac {a \ln \left (\tanh \left (\frac {x}{2}\right )\right )}{b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csch(x)^3/(a+b*coth(x)),x)

[Out]

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

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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(csch(x)^3/(a+b*coth(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 = 1.46, size = 230, normalized size = 4.04 \[ \frac {2\,{\mathrm {e}}^x}{b-b\,{\mathrm {e}}^{2\,x}}-\frac {a\,\ln \left (32\,a\,b^2-64\,a^2\,b+32\,a^3-32\,a^3\,{\mathrm {e}}^x-32\,a\,b^2\,{\mathrm {e}}^x+64\,a^2\,b\,{\mathrm {e}}^x\right )}{b^2}+\frac {a\,\ln \left (32\,a\,b^2-64\,a^2\,b+32\,a^3+32\,a^3\,{\mathrm {e}}^x+32\,a\,b^2\,{\mathrm {e}}^x-64\,a^2\,b\,{\mathrm {e}}^x\right )}{b^2}+\frac {\ln \left (32\,a\,\sqrt {a^2-b^2}-32\,b\,\sqrt {a^2-b^2}-32\,a^2\,{\mathrm {e}}^x+32\,b^2\,{\mathrm {e}}^x\right )\,\sqrt {a^2-b^2}}{b^2}-\frac {\ln \left (32\,a\,\sqrt {a^2-b^2}-32\,b\,\sqrt {a^2-b^2}+32\,a^2\,{\mathrm {e}}^x-32\,b^2\,{\mathrm {e}}^x\right )\,\sqrt {a^2-b^2}}{b^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sinh(x)^3*(a + b*coth(x))),x)

[Out]

(2*exp(x))/(b - b*exp(2*x)) - (a*log(32*a*b^2 - 64*a^2*b + 32*a^3 - 32*a^3*exp(x) - 32*a*b^2*exp(x) + 64*a^2*b
*exp(x)))/b^2 + (a*log(32*a*b^2 - 64*a^2*b + 32*a^3 + 32*a^3*exp(x) + 32*a*b^2*exp(x) - 64*a^2*b*exp(x)))/b^2
+ (log(32*a*(a^2 - b^2)^(1/2) - 32*b*(a^2 - b^2)^(1/2) - 32*a^2*exp(x) + 32*b^2*exp(x))*(a^2 - b^2)^(1/2))/b^2
- (log(32*a*(a^2 - b^2)^(1/2) - 32*b*(a^2 - b^2)^(1/2) + 32*a^2*exp(x) - 32*b^2*exp(x))*(a^2 - b^2)^(1/2))/b^
2

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {csch}^{3}{\relax (x )}}{a + b \coth {\relax (x )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(x)**3/(a+b*coth(x)),x)

[Out]

Integral(csch(x)**3/(a + b*coth(x)), x)

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