3.83 \(\int \frac {\text {csch}^4(x)}{a+b \text {csch}(x)} \, dx\)

Optimal. Leaf size=83 \[ -\frac {\left (2 a^2-b^2\right ) \tanh ^{-1}(\cosh (x))}{2 b^3}+\frac {2 a^3 \tanh ^{-1}\left (\frac {a-b \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2+b^2}}\right )}{b^3 \sqrt {a^2+b^2}}+\frac {a \coth (x)}{b^2}-\frac {\coth (x) \text {csch}(x)}{2 b} \]

[Out]

-1/2*(2*a^2-b^2)*arctanh(cosh(x))/b^3+a*coth(x)/b^2-1/2*coth(x)*csch(x)/b+2*a^3*arctanh((a-b*tanh(1/2*x))/(a^2
+b^2)^(1/2))/b^3/(a^2+b^2)^(1/2)

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Rubi [A]  time = 0.29, antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.615, Rules used = {3851, 4082, 3998, 3770, 3831, 2660, 618, 206} \[ \frac {2 a^3 \tanh ^{-1}\left (\frac {a-b \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2+b^2}}\right )}{b^3 \sqrt {a^2+b^2}}-\frac {\left (2 a^2-b^2\right ) \tanh ^{-1}(\cosh (x))}{2 b^3}+\frac {a \coth (x)}{b^2}-\frac {\coth (x) \text {csch}(x)}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[Csch[x]^4/(a + b*Csch[x]),x]

[Out]

-((2*a^2 - b^2)*ArcTanh[Cosh[x]])/(2*b^3) + (2*a^3*ArcTanh[(a - b*Tanh[x/2])/Sqrt[a^2 + b^2]])/(b^3*Sqrt[a^2 +
b^2]) + (a*Coth[x])/b^2 - (Coth[x]*Csch[x])/(2*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 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]

Rule 3851

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> -Simp[(d^3*Cot[e
+ f*x]*(d*Csc[e + f*x])^(n - 3))/(b*f*(n - 2)), x] + Dist[d^3/(b*(n - 2)), Int[((d*Csc[e + f*x])^(n - 3)*Simp
[a*(n - 3) + b*(n - 3)*Csc[e + f*x] - a*(n - 2)*Csc[e + f*x]^2, x])/(a + b*Csc[e + f*x]), x], x] /; FreeQ[{a,
b, d, e, f}, x] && NeQ[a^2 - b^2, 0] && GtQ[n, 3]

Rule 3998

Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x
_Symbol] :> Dist[B/b, Int[Csc[e + f*x], x], x] + Dist[(A*b - a*B)/b, Int[Csc[e + f*x]/(a + b*Csc[e + f*x]), x]
, x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[A*b - a*B, 0]

Rule 4082

Int[csc[(e_.) + (f_.)*(x_)]*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_
.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Simp[(C*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1))/(b*f*(m
+ 2)), x] + Dist[1/(b*(m + 2)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*Simp[b*A*(m + 2) + b*C*(m + 1) + (b*B*
(m + 2) - a*C)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

Rubi steps

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

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Mathematica [A]  time = 0.48, size = 124, normalized size = 1.49 \[ -\frac {-8 a^2 \log \left (\tanh \left (\frac {x}{2}\right )\right )+\frac {16 a^3 \tan ^{-1}\left (\frac {a-b \tanh \left (\frac {x}{2}\right )}{\sqrt {-a^2-b^2}}\right )}{\sqrt {-a^2-b^2}}-4 a b \tanh \left (\frac {x}{2}\right )-4 a b \coth \left (\frac {x}{2}\right )+b^2 \text {csch}^2\left (\frac {x}{2}\right )+b^2 \text {sech}^2\left (\frac {x}{2}\right )+4 b^2 \log \left (\tanh \left (\frac {x}{2}\right )\right )}{8 b^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Csch[x]^4/(a + b*Csch[x]),x]

[Out]

-1/8*((16*a^3*ArcTan[(a - b*Tanh[x/2])/Sqrt[-a^2 - b^2]])/Sqrt[-a^2 - b^2] - 4*a*b*Coth[x/2] + b^2*Csch[x/2]^2
- 8*a^2*Log[Tanh[x/2]] + 4*b^2*Log[Tanh[x/2]] + b^2*Sech[x/2]^2 - 4*a*b*Tanh[x/2])/b^3

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fricas [B]  time = 0.53, size = 947, normalized size = 11.41 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(x)^4/(a+b*csch(x)),x, algorithm="fricas")

[Out]

-1/2*(4*a^3*b + 4*a*b^3 + 2*(a^2*b^2 + b^4)*cosh(x)^3 + 2*(a^2*b^2 + b^4)*sinh(x)^3 - 4*(a^3*b + a*b^3)*cosh(x
)^2 - 2*(2*a^3*b + 2*a*b^3 - 3*(a^2*b^2 + b^4)*cosh(x))*sinh(x)^2 - 2*(a^3*cosh(x)^4 + 4*a^3*cosh(x)*sinh(x)^3
+ a^3*sinh(x)^4 - 2*a^3*cosh(x)^2 + a^3 + 2*(3*a^3*cosh(x)^2 - a^3)*sinh(x)^2 + 4*(a^3*cosh(x)^3 - a^3*cosh(x
))*sinh(x))*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*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)) + 2*(a^2*b^2 + b^4)*cosh(x) + ((2*a^4 + a^2*b^2 - b^4)*cosh(x)^4 + 4*(2*a^4 + a^2
*b^2 - b^4)*cosh(x)*sinh(x)^3 + (2*a^4 + a^2*b^2 - b^4)*sinh(x)^4 + 2*a^4 + a^2*b^2 - b^4 - 2*(2*a^4 + a^2*b^2
- b^4)*cosh(x)^2 - 2*(2*a^4 + a^2*b^2 - b^4 - 3*(2*a^4 + a^2*b^2 - b^4)*cosh(x)^2)*sinh(x)^2 + 4*((2*a^4 + a^
2*b^2 - b^4)*cosh(x)^3 - (2*a^4 + a^2*b^2 - b^4)*cosh(x))*sinh(x))*log(cosh(x) + sinh(x) + 1) - ((2*a^4 + a^2*
b^2 - b^4)*cosh(x)^4 + 4*(2*a^4 + a^2*b^2 - b^4)*cosh(x)*sinh(x)^3 + (2*a^4 + a^2*b^2 - b^4)*sinh(x)^4 + 2*a^4
+ a^2*b^2 - b^4 - 2*(2*a^4 + a^2*b^2 - b^4)*cosh(x)^2 - 2*(2*a^4 + a^2*b^2 - b^4 - 3*(2*a^4 + a^2*b^2 - b^4)*
cosh(x)^2)*sinh(x)^2 + 4*((2*a^4 + a^2*b^2 - b^4)*cosh(x)^3 - (2*a^4 + a^2*b^2 - b^4)*cosh(x))*sinh(x))*log(co
sh(x) + sinh(x) - 1) + 2*(a^2*b^2 + b^4 + 3*(a^2*b^2 + b^4)*cosh(x)^2 - 4*(a^3*b + a*b^3)*cosh(x))*sinh(x))/(a
^2*b^3 + b^5 + (a^2*b^3 + b^5)*cosh(x)^4 + 4*(a^2*b^3 + b^5)*cosh(x)*sinh(x)^3 + (a^2*b^3 + b^5)*sinh(x)^4 - 2
*(a^2*b^3 + b^5)*cosh(x)^2 - 2*(a^2*b^3 + b^5 - 3*(a^2*b^3 + b^5)*cosh(x)^2)*sinh(x)^2 + 4*((a^2*b^3 + b^5)*co
sh(x)^3 - (a^2*b^3 + b^5)*cosh(x))*sinh(x))

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giac [A]  time = 0.16, size = 141, normalized size = 1.70 \[ -\frac {a^{3} \log \left (\frac {{\left | 2 \, a e^{x} + 2 \, b - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, a e^{x} + 2 \, b + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{\sqrt {a^{2} + b^{2}} b^{3}} - \frac {{\left (2 \, a^{2} - b^{2}\right )} \log \left (e^{x} + 1\right )}{2 \, b^{3}} + \frac {{\left (2 \, a^{2} - b^{2}\right )} \log \left ({\left | e^{x} - 1 \right |}\right )}{2 \, b^{3}} - \frac {b e^{\left (3 \, x\right )} - 2 \, a e^{\left (2 \, x\right )} + b e^{x} + 2 \, a}{b^{2} {\left (e^{\left (2 \, x\right )} - 1\right )}^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(x)^4/(a+b*csch(x)),x, algorithm="giac")

[Out]

-a^3*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^3)
- 1/2*(2*a^2 - b^2)*log(e^x + 1)/b^3 + 1/2*(2*a^2 - b^2)*log(abs(e^x - 1))/b^3 - (b*e^(3*x) - 2*a*e^(2*x) + b*
e^x + 2*a)/(b^2*(e^(2*x) - 1)^2)

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maple [A]  time = 0.10, size = 108, normalized size = 1.30 \[ \frac {\tanh ^{2}\left (\frac {x}{2}\right )}{8 b}+\frac {a \tanh \left (\frac {x}{2}\right )}{2 b^{2}}-\frac {2 a^{3} \arctanh \left (\frac {2 \tanh \left (\frac {x}{2}\right ) b -2 a}{2 \sqrt {a^{2}+b^{2}}}\right )}{b^{3} \sqrt {a^{2}+b^{2}}}-\frac {1}{8 b \tanh \left (\frac {x}{2}\right )^{2}}+\frac {\ln \left (\tanh \left (\frac {x}{2}\right )\right ) a^{2}}{b^{3}}-\frac {\ln \left (\tanh \left (\frac {x}{2}\right )\right )}{2 b}+\frac {a}{2 b^{2} \tanh \left (\frac {x}{2}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csch(x)^4/(a+b*csch(x)),x)

[Out]

1/8/b*tanh(1/2*x)^2+1/2/b^2*a*tanh(1/2*x)-2/b^3*a^3/(a^2+b^2)^(1/2)*arctanh(1/2*(2*tanh(1/2*x)*b-2*a)/(a^2+b^2
)^(1/2))-1/8/b/tanh(1/2*x)^2+1/b^3*ln(tanh(1/2*x))*a^2-1/2/b*ln(tanh(1/2*x))+1/2*a/b^2/tanh(1/2*x)

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maxima [B]  time = 0.42, size = 158, normalized size = 1.90 \[ -\frac {a^{3} \log \left (\frac {a e^{\left (-x\right )} - b - \sqrt {a^{2} + b^{2}}}{a e^{\left (-x\right )} - b + \sqrt {a^{2} + b^{2}}}\right )}{\sqrt {a^{2} + b^{2}} b^{3}} + \frac {b e^{\left (-x\right )} + 2 \, a e^{\left (-2 \, x\right )} + b e^{\left (-3 \, x\right )} - 2 \, a}{2 \, b^{2} e^{\left (-2 \, x\right )} - b^{2} e^{\left (-4 \, x\right )} - b^{2}} - \frac {{\left (2 \, a^{2} - b^{2}\right )} \log \left (e^{\left (-x\right )} + 1\right )}{2 \, b^{3}} + \frac {{\left (2 \, a^{2} - b^{2}\right )} \log \left (e^{\left (-x\right )} - 1\right )}{2 \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(x)^4/(a+b*csch(x)),x, algorithm="maxima")

[Out]

-a^3*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^3) + (b*e^(-x)
+ 2*a*e^(-2*x) + b*e^(-3*x) - 2*a)/(2*b^2*e^(-2*x) - b^2*e^(-4*x) - b^2) - 1/2*(2*a^2 - b^2)*log(e^(-x) + 1)/b
^3 + 1/2*(2*a^2 - b^2)*log(e^(-x) - 1)/b^3

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mupad [B]  time = 2.09, size = 617, normalized size = 7.43 \[ \frac {{\mathrm {e}}^x}{b-b\,{\mathrm {e}}^{2\,x}}-\frac {2\,{\mathrm {e}}^x}{b-2\,b\,{\mathrm {e}}^{2\,x}+b\,{\mathrm {e}}^{4\,x}}-\frac {\ln \left (24\,a^4+4\,b^4-20\,a^2\,b^2-24\,a^4\,{\mathrm {e}}^x-4\,b^4\,{\mathrm {e}}^x+20\,a^2\,b^2\,{\mathrm {e}}^x\right )}{2\,b}+\frac {\ln \left (24\,a^4+4\,b^4-20\,a^2\,b^2+24\,a^4\,{\mathrm {e}}^x+4\,b^4\,{\mathrm {e}}^x-20\,a^2\,b^2\,{\mathrm {e}}^x\right )}{2\,b}+\frac {2\,a}{b^2\,{\mathrm {e}}^{2\,x}-b^2}+\frac {a^2\,\ln \left (24\,a^4+4\,b^4-20\,a^2\,b^2-24\,a^4\,{\mathrm {e}}^x-4\,b^4\,{\mathrm {e}}^x+20\,a^2\,b^2\,{\mathrm {e}}^x\right )}{b^3}-\frac {a^2\,\ln \left (24\,a^4+4\,b^4-20\,a^2\,b^2+24\,a^4\,{\mathrm {e}}^x+4\,b^4\,{\mathrm {e}}^x-20\,a^2\,b^2\,{\mathrm {e}}^x\right )}{b^3}-\frac {a^3\,\ln \left (16\,a\,b^5-24\,a^5\,\sqrt {a^2+b^2}-48\,a^5\,b-32\,a^3\,b^3+24\,a^6\,{\mathrm {e}}^x-32\,b^6\,{\mathrm {e}}^x-40\,a^3\,b^2\,\sqrt {a^2+b^2}-32\,b^5\,{\mathrm {e}}^x\,\sqrt {a^2+b^2}+56\,a^2\,b^4\,{\mathrm {e}}^x+112\,a^4\,b^2\,{\mathrm {e}}^x+16\,a\,b^4\,\sqrt {a^2+b^2}+72\,a^4\,b\,{\mathrm {e}}^x\,\sqrt {a^2+b^2}+72\,a^2\,b^3\,{\mathrm {e}}^x\,\sqrt {a^2+b^2}\right )\,\sqrt {a^2+b^2}}{a^2\,b^3+b^5}+\frac {a^3\,\ln \left (24\,a^5\,\sqrt {a^2+b^2}+16\,a\,b^5-48\,a^5\,b-32\,a^3\,b^3+24\,a^6\,{\mathrm {e}}^x-32\,b^6\,{\mathrm {e}}^x+40\,a^3\,b^2\,\sqrt {a^2+b^2}+32\,b^5\,{\mathrm {e}}^x\,\sqrt {a^2+b^2}+56\,a^2\,b^4\,{\mathrm {e}}^x+112\,a^4\,b^2\,{\mathrm {e}}^x-16\,a\,b^4\,\sqrt {a^2+b^2}-72\,a^4\,b\,{\mathrm {e}}^x\,\sqrt {a^2+b^2}-72\,a^2\,b^3\,{\mathrm {e}}^x\,\sqrt {a^2+b^2}\right )\,\sqrt {a^2+b^2}}{a^2\,b^3+b^5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sinh(x)^4*(a + b/sinh(x))),x)

[Out]

exp(x)/(b - b*exp(2*x)) - (2*exp(x))/(b - 2*b*exp(2*x) + b*exp(4*x)) - log(24*a^4 + 4*b^4 - 20*a^2*b^2 - 24*a^
4*exp(x) - 4*b^4*exp(x) + 20*a^2*b^2*exp(x))/(2*b) + log(24*a^4 + 4*b^4 - 20*a^2*b^2 + 24*a^4*exp(x) + 4*b^4*e
xp(x) - 20*a^2*b^2*exp(x))/(2*b) + (2*a)/(b^2*exp(2*x) - b^2) + (a^2*log(24*a^4 + 4*b^4 - 20*a^2*b^2 - 24*a^4*
exp(x) - 4*b^4*exp(x) + 20*a^2*b^2*exp(x)))/b^3 - (a^2*log(24*a^4 + 4*b^4 - 20*a^2*b^2 + 24*a^4*exp(x) + 4*b^4
*exp(x) - 20*a^2*b^2*exp(x)))/b^3 - (a^3*log(16*a*b^5 - 24*a^5*(a^2 + b^2)^(1/2) - 48*a^5*b - 32*a^3*b^3 + 24*
a^6*exp(x) - 32*b^6*exp(x) - 40*a^3*b^2*(a^2 + b^2)^(1/2) - 32*b^5*exp(x)*(a^2 + b^2)^(1/2) + 56*a^2*b^4*exp(x
) + 112*a^4*b^2*exp(x) + 16*a*b^4*(a^2 + b^2)^(1/2) + 72*a^4*b*exp(x)*(a^2 + b^2)^(1/2) + 72*a^2*b^3*exp(x)*(a
^2 + b^2)^(1/2))*(a^2 + b^2)^(1/2))/(b^5 + a^2*b^3) + (a^3*log(24*a^5*(a^2 + b^2)^(1/2) + 16*a*b^5 - 48*a^5*b
- 32*a^3*b^3 + 24*a^6*exp(x) - 32*b^6*exp(x) + 40*a^3*b^2*(a^2 + b^2)^(1/2) + 32*b^5*exp(x)*(a^2 + b^2)^(1/2)
+ 56*a^2*b^4*exp(x) + 112*a^4*b^2*exp(x) - 16*a*b^4*(a^2 + b^2)^(1/2) - 72*a^4*b*exp(x)*(a^2 + b^2)^(1/2) - 72
*a^2*b^3*exp(x)*(a^2 + b^2)^(1/2))*(a^2 + b^2)^(1/2))/(b^5 + a^2*b^3)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {csch}^{4}{\relax (x )}}{a + b \operatorname {csch}{\relax (x )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(x)**4/(a+b*csch(x)),x)

[Out]

Integral(csch(x)**4/(a + b*csch(x)), x)

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