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

Optimal. Leaf size=59 \[ -\frac {2 a^2 \tanh ^{-1}\left (\frac {a-b \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2+b^2}}\right )}{b^2 \sqrt {a^2+b^2}}+\frac {a \tanh ^{-1}(\cosh (x))}{b^2}-\frac {\coth (x)}{b} \]

[Out]

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

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Rubi [A]  time = 0.16, antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.538, Rules used = {3790, 3789, 3770, 3831, 2660, 618, 206} \[ -\frac {2 a^2 \tanh ^{-1}\left (\frac {a-b \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2+b^2}}\right )}{b^2 \sqrt {a^2+b^2}}+\frac {a \tanh ^{-1}(\cosh (x))}{b^2}-\frac {\coth (x)}{b} \]

Antiderivative was successfully verified.

[In]

Int[Csch[x]^3/(a + b*Csch[x]),x]

[Out]

(a*ArcTanh[Cosh[x]])/b^2 - (2*a^2*ArcTanh[(a - b*Tanh[x/2])/Sqrt[a^2 + b^2]])/(b^2*Sqrt[a^2 + b^2]) - Coth[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 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 3770

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

Rule 3789

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 3790

Int[csc[(e_.) + (f_.)*(x_)]^3/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> -Simp[Cot[e + f*x]/(b*f), x
] - Dist[a/b, Int[Csc[e + f*x]^2/(a + b*Csc[e + f*x]), x], x] /; FreeQ[{a, b, e, f}, x]

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}^3(x)}{a+b \text {csch}(x)} \, dx &=-\frac {\coth (x)}{b}-\frac {a \int \frac {\text {csch}^2(x)}{a+b \text {csch}(x)} \, dx}{b}\\ &=-\frac {\coth (x)}{b}-\frac {a \int \text {csch}(x) \, dx}{b^2}+\frac {a^2 \int \frac {\text {csch}(x)}{a+b \text {csch}(x)} \, dx}{b^2}\\ &=\frac {a \tanh ^{-1}(\cosh (x))}{b^2}-\frac {\coth (x)}{b}+\frac {a^2 \int \frac {1}{1+\frac {a \sinh (x)}{b}} \, dx}{b^3}\\ &=\frac {a \tanh ^{-1}(\cosh (x))}{b^2}-\frac {\coth (x)}{b}+\frac {\left (2 a^2\right ) \operatorname {Subst}\left (\int \frac {1}{1+\frac {2 a x}{b}-x^2} \, dx,x,\tanh \left (\frac {x}{2}\right )\right )}{b^3}\\ &=\frac {a \tanh ^{-1}(\cosh (x))}{b^2}-\frac {\coth (x)}{b}-\frac {\left (4 a^2\right ) \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^3}\\ &=\frac {a \tanh ^{-1}(\cosh (x))}{b^2}-\frac {2 a^2 \tanh ^{-1}\left (\frac {b \left (\frac {a}{b}-\tanh \left (\frac {x}{2}\right )\right )}{\sqrt {a^2+b^2}}\right )}{b^2 \sqrt {a^2+b^2}}-\frac {\coth (x)}{b}\\ \end {align*}

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Mathematica [A]  time = 0.29, size = 71, normalized size = 1.20 \[ \frac {\frac {4 a^2 \tan ^{-1}\left (\frac {a-b \tanh \left (\frac {x}{2}\right )}{\sqrt {-a^2-b^2}}\right )}{\sqrt {-a^2-b^2}}-2 a \log \left (\tanh \left (\frac {x}{2}\right )\right )-2 b \coth (x)}{2 b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.43, size = 345, normalized size = 5.85 \[ \frac {2 \, a^{2} b + 2 \, b^{3} - {\left (a^{2} \cosh \relax (x)^{2} + 2 \, a^{2} \cosh \relax (x) \sinh \relax (x) + a^{2} \sinh \relax (x)^{2} - a^{2}\right )} \sqrt {a^{2} + b^{2}} \log \left (\frac {a^{2} \cosh \relax (x)^{2} + a^{2} \sinh \relax (x)^{2} + 2 \, a b \cosh \relax (x) + a^{2} + 2 \, b^{2} + 2 \, {\left (a^{2} \cosh \relax (x) + a b\right )} \sinh \relax (x) - 2 \, \sqrt {a^{2} + b^{2}} {\left (a \cosh \relax (x) + a \sinh \relax (x) + b\right )}}{a \cosh \relax (x)^{2} + a \sinh \relax (x)^{2} + 2 \, b \cosh \relax (x) + 2 \, {\left (a \cosh \relax (x) + b\right )} \sinh \relax (x) - a}\right ) + {\left (a^{3} + a b^{2} - {\left (a^{3} + a b^{2}\right )} \cosh \relax (x)^{2} - 2 \, {\left (a^{3} + a b^{2}\right )} \cosh \relax (x) \sinh \relax (x) - {\left (a^{3} + a b^{2}\right )} \sinh \relax (x)^{2}\right )} \log \left (\cosh \relax (x) + \sinh \relax (x) + 1\right ) - {\left (a^{3} + a b^{2} - {\left (a^{3} + a b^{2}\right )} \cosh \relax (x)^{2} - 2 \, {\left (a^{3} + a b^{2}\right )} \cosh \relax (x) \sinh \relax (x) - {\left (a^{3} + a b^{2}\right )} \sinh \relax (x)^{2}\right )} \log \left (\cosh \relax (x) + \sinh \relax (x) - 1\right )}{a^{2} b^{2} + b^{4} - {\left (a^{2} b^{2} + b^{4}\right )} \cosh \relax (x)^{2} - 2 \, {\left (a^{2} b^{2} + b^{4}\right )} \cosh \relax (x) \sinh \relax (x) - {\left (a^{2} b^{2} + b^{4}\right )} \sinh \relax (x)^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*a^2*b + 2*b^3 - (a^2*cosh(x)^2 + 2*a^2*cosh(x)*sinh(x) + a^2*sinh(x)^2 - a^2)*sqrt(a^2 + b^2)*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*cos
h(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)) + (a^3 + a*b
^2 - (a^3 + a*b^2)*cosh(x)^2 - 2*(a^3 + a*b^2)*cosh(x)*sinh(x) - (a^3 + a*b^2)*sinh(x)^2)*log(cosh(x) + sinh(x
) + 1) - (a^3 + a*b^2 - (a^3 + a*b^2)*cosh(x)^2 - 2*(a^3 + a*b^2)*cosh(x)*sinh(x) - (a^3 + a*b^2)*sinh(x)^2)*l
og(cosh(x) + sinh(x) - 1))/(a^2*b^2 + b^4 - (a^2*b^2 + b^4)*cosh(x)^2 - 2*(a^2*b^2 + b^4)*cosh(x)*sinh(x) - (a
^2*b^2 + b^4)*sinh(x)^2)

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giac [A]  time = 0.15, size = 98, normalized size = 1.66 \[ \frac {a^{2} \log \left (\frac {{\left | 2 \, a e^{x} + 2 \, b - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, a e^{x} + 2 \, b + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{\sqrt {a^{2} + b^{2}} b^{2}} + \frac {a \log \left (e^{x} + 1\right )}{b^{2}} - \frac {a \log \left ({\left | e^{x} - 1 \right |}\right )}{b^{2}} - \frac {2}{b {\left (e^{\left (2 \, x\right )} - 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.10, size = 73, normalized size = 1.24 \[ -\frac {\tanh \left (\frac {x}{2}\right )}{2 b}+\frac {2 a^{2} \arctanh \left (\frac {2 \tanh \left (\frac {x}{2}\right ) b -2 a}{2 \sqrt {a^{2}+b^{2}}}\right )}{b^{2} \sqrt {a^{2}+b^{2}}}-\frac {1}{2 b \tanh \left (\frac {x}{2}\right )}-\frac {a \ln \left (\tanh \left (\frac {x}{2}\right )\right )}{b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.42, size = 100, normalized size = 1.69 \[ \frac {a^{2} \log \left (\frac {a e^{\left (-x\right )} - b - \sqrt {a^{2} + b^{2}}}{a e^{\left (-x\right )} - b + \sqrt {a^{2} + b^{2}}}\right )}{\sqrt {a^{2} + b^{2}} b^{2}} + \frac {a \log \left (e^{\left (-x\right )} + 1\right )}{b^{2}} - \frac {a \log \left (e^{\left (-x\right )} - 1\right )}{b^{2}} + \frac {2}{b e^{\left (-2 \, x\right )} - b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.73, size = 292, normalized size = 4.95 \[ \frac {2}{b-b\,{\mathrm {e}}^{2\,x}}-\frac {a\,\ln \left (32\,{\mathrm {e}}^x-32\right )}{b^2}+\frac {a\,\ln \left (32\,{\mathrm {e}}^x+32\right )}{b^2}+\frac {a^2\,\ln \left (32\,a^4\,{\mathrm {e}}^x-64\,a\,b^3-64\,a^3\,b-32\,a^3\,\sqrt {a^2+b^2}+128\,b^4\,{\mathrm {e}}^x+128\,b^3\,{\mathrm {e}}^x\,\sqrt {a^2+b^2}+160\,a^2\,b^2\,{\mathrm {e}}^x-64\,a\,b^2\,\sqrt {a^2+b^2}+96\,a^2\,b\,{\mathrm {e}}^x\,\sqrt {a^2+b^2}\right )\,\sqrt {a^2+b^2}}{a^2\,b^2+b^4}-\frac {a^2\,\ln \left (32\,a^3\,\sqrt {a^2+b^2}-64\,a\,b^3-64\,a^3\,b+32\,a^4\,{\mathrm {e}}^x+128\,b^4\,{\mathrm {e}}^x-128\,b^3\,{\mathrm {e}}^x\,\sqrt {a^2+b^2}+160\,a^2\,b^2\,{\mathrm {e}}^x+64\,a\,b^2\,\sqrt {a^2+b^2}-96\,a^2\,b\,{\mathrm {e}}^x\,\sqrt {a^2+b^2}\right )\,\sqrt {a^2+b^2}}{a^2\,b^2+b^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sinh(x)^3*(a + b/sinh(x))),x)

[Out]

2/(b - b*exp(2*x)) - (a*log(32*exp(x) - 32))/b^2 + (a*log(32*exp(x) + 32))/b^2 + (a^2*log(32*a^4*exp(x) - 64*a
*b^3 - 64*a^3*b - 32*a^3*(a^2 + b^2)^(1/2) + 128*b^4*exp(x) + 128*b^3*exp(x)*(a^2 + b^2)^(1/2) + 160*a^2*b^2*e
xp(x) - 64*a*b^2*(a^2 + b^2)^(1/2) + 96*a^2*b*exp(x)*(a^2 + b^2)^(1/2))*(a^2 + b^2)^(1/2))/(b^4 + a^2*b^2) - (
a^2*log(32*a^3*(a^2 + b^2)^(1/2) - 64*a*b^3 - 64*a^3*b + 32*a^4*exp(x) + 128*b^4*exp(x) - 128*b^3*exp(x)*(a^2
+ b^2)^(1/2) + 160*a^2*b^2*exp(x) + 64*a*b^2*(a^2 + b^2)^(1/2) - 96*a^2*b*exp(x)*(a^2 + b^2)^(1/2))*(a^2 + b^2
)^(1/2))/(b^4 + a^2*b^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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