3.788 \(\int \frac {\text {csch}^2(x)}{a+b \coth (x)+c \text {csch}(x)} \, dx\)

Optimal. Leaf size=118 \[ -\frac {2 a c \tanh ^{-1}\left (\frac {a+(b-c) \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2+c^2}}\right )}{\left (b^2-c^2\right ) \sqrt {a^2-b^2+c^2}}-\frac {b \log \left (2 a \tanh \left (\frac {x}{2}\right )+(b-c) \tanh ^2\left (\frac {x}{2}\right )+b+c\right )}{b^2-c^2}+\frac {\log \left (\tanh \left (\frac {x}{2}\right )\right )}{b+c} \]

[Out]

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

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Rubi [A]  time = 0.58, antiderivative size = 118, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.412, Rules used = {4397, 12, 1628, 634, 618, 206, 628} \[ -\frac {2 a c \tanh ^{-1}\left (\frac {a+(b-c) \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2+c^2}}\right )}{\left (b^2-c^2\right ) \sqrt {a^2-b^2+c^2}}-\frac {b \log \left (2 a \tanh \left (\frac {x}{2}\right )+(b-c) \tanh ^2\left (\frac {x}{2}\right )+b+c\right )}{b^2-c^2}+\frac {\log \left (\tanh \left (\frac {x}{2}\right )\right )}{b+c} \]

Antiderivative was successfully verified.

[In]

Int[Csch[x]^2/(a + b*Coth[x] + c*Csch[x]),x]

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 634

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 1628

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegra
nd[(d + e*x)^m*Pq*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rule 4397

Int[u_, x_Symbol] :> Int[TrigSimplify[u], x] /; TrigSimplifyQ[u]

Rubi steps

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

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Mathematica [A]  time = 0.24, size = 97, normalized size = 0.82 \[ \frac {-\frac {2 a c \tan ^{-1}\left (\frac {a+(b-c) \tanh \left (\frac {x}{2}\right )}{\sqrt {-a^2+b^2-c^2}}\right )}{\sqrt {-a^2+b^2-c^2}}+b \log (a \sinh (x)+b \cosh (x)+c)-b \log (\sinh (x))+c \log \left (\tanh \left (\frac {x}{2}\right )\right )}{c^2-b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Csch[x]^2/(a + b*Coth[x] + c*Csch[x]),x]

[Out]

((-2*a*c*ArcTan[(a + (b - c)*Tanh[x/2])/Sqrt[-a^2 + b^2 - c^2]])/Sqrt[-a^2 + b^2 - c^2] - b*Log[Sinh[x]] + b*L
og[c + b*Cosh[x] + a*Sinh[x]] + c*Log[Tanh[x/2]])/(-b^2 + c^2)

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fricas [B]  time = 1.55, size = 546, normalized size = 4.63 \[ \left [-\frac {\sqrt {a^{2} - b^{2} + c^{2}} a c \log \left (\frac {2 \, {\left (a + b\right )} c \cosh \relax (x) + {\left (a^{2} + 2 \, a b + b^{2}\right )} \cosh \relax (x)^{2} + {\left (a^{2} + 2 \, a b + b^{2}\right )} \sinh \relax (x)^{2} + a^{2} - b^{2} + 2 \, c^{2} + 2 \, {\left ({\left (a + b\right )} c + {\left (a^{2} + 2 \, a b + b^{2}\right )} \cosh \relax (x)\right )} \sinh \relax (x) + 2 \, \sqrt {a^{2} - b^{2} + c^{2}} {\left ({\left (a + b\right )} \cosh \relax (x) + {\left (a + b\right )} \sinh \relax (x) + c\right )}}{{\left (a + b\right )} \cosh \relax (x)^{2} + {\left (a + b\right )} \sinh \relax (x)^{2} + 2 \, c \cosh \relax (x) + 2 \, {\left ({\left (a + b\right )} \cosh \relax (x) + c\right )} \sinh \relax (x) - a + b}\right ) + {\left (a^{2} b - b^{3} + b c^{2}\right )} \log \left (\frac {2 \, {\left (b \cosh \relax (x) + a \sinh \relax (x) + c\right )}}{\cosh \relax (x) - \sinh \relax (x)}\right ) - {\left (a^{2} b - b^{3} + b c^{2} + c^{3} + {\left (a^{2} - b^{2}\right )} c\right )} \log \left (\cosh \relax (x) + \sinh \relax (x) + 1\right ) - {\left (a^{2} b - b^{3} + b c^{2} - c^{3} - {\left (a^{2} - b^{2}\right )} c\right )} \log \left (\cosh \relax (x) + \sinh \relax (x) - 1\right )}{a^{2} b^{2} - b^{4} - c^{4} - {\left (a^{2} - 2 \, b^{2}\right )} c^{2}}, \frac {2 \, \sqrt {-a^{2} + b^{2} - c^{2}} a c \arctan \left (\frac {\sqrt {-a^{2} + b^{2} - c^{2}} {\left ({\left (a + b\right )} \cosh \relax (x) + {\left (a + b\right )} \sinh \relax (x) + c\right )}}{a^{2} - b^{2} + c^{2}}\right ) - {\left (a^{2} b - b^{3} + b c^{2}\right )} \log \left (\frac {2 \, {\left (b \cosh \relax (x) + a \sinh \relax (x) + c\right )}}{\cosh \relax (x) - \sinh \relax (x)}\right ) + {\left (a^{2} b - b^{3} + b c^{2} + c^{3} + {\left (a^{2} - b^{2}\right )} c\right )} \log \left (\cosh \relax (x) + \sinh \relax (x) + 1\right ) + {\left (a^{2} b - b^{3} + b c^{2} - c^{3} - {\left (a^{2} - b^{2}\right )} c\right )} \log \left (\cosh \relax (x) + \sinh \relax (x) - 1\right )}{a^{2} b^{2} - b^{4} - c^{4} - {\left (a^{2} - 2 \, b^{2}\right )} c^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(x)^2/(a+b*coth(x)+c*csch(x)),x, algorithm="fricas")

[Out]

[-(sqrt(a^2 - b^2 + c^2)*a*c*log((2*(a + b)*c*cosh(x) + (a^2 + 2*a*b + b^2)*cosh(x)^2 + (a^2 + 2*a*b + b^2)*si
nh(x)^2 + a^2 - b^2 + 2*c^2 + 2*((a + b)*c + (a^2 + 2*a*b + b^2)*cosh(x))*sinh(x) + 2*sqrt(a^2 - b^2 + c^2)*((
a + b)*cosh(x) + (a + b)*sinh(x) + c))/((a + b)*cosh(x)^2 + (a + b)*sinh(x)^2 + 2*c*cosh(x) + 2*((a + b)*cosh(
x) + c)*sinh(x) - a + b)) + (a^2*b - b^3 + b*c^2)*log(2*(b*cosh(x) + a*sinh(x) + c)/(cosh(x) - sinh(x))) - (a^
2*b - b^3 + b*c^2 + c^3 + (a^2 - b^2)*c)*log(cosh(x) + sinh(x) + 1) - (a^2*b - b^3 + b*c^2 - c^3 - (a^2 - b^2)
*c)*log(cosh(x) + sinh(x) - 1))/(a^2*b^2 - b^4 - c^4 - (a^2 - 2*b^2)*c^2), (2*sqrt(-a^2 + b^2 - c^2)*a*c*arcta
n(sqrt(-a^2 + b^2 - c^2)*((a + b)*cosh(x) + (a + b)*sinh(x) + c)/(a^2 - b^2 + c^2)) - (a^2*b - b^3 + b*c^2)*lo
g(2*(b*cosh(x) + a*sinh(x) + c)/(cosh(x) - sinh(x))) + (a^2*b - b^3 + b*c^2 + c^3 + (a^2 - b^2)*c)*log(cosh(x)
+ sinh(x) + 1) + (a^2*b - b^3 + b*c^2 - c^3 - (a^2 - b^2)*c)*log(cosh(x) + sinh(x) - 1))/(a^2*b^2 - b^4 - c^4
- (a^2 - 2*b^2)*c^2)]

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giac [A]  time = 0.12, size = 122, normalized size = 1.03 \[ \frac {2 \, a c \arctan \left (\frac {a e^{x} + b e^{x} + c}{\sqrt {-a^{2} + b^{2} - c^{2}}}\right )}{\sqrt {-a^{2} + b^{2} - c^{2}} {\left (b^{2} - c^{2}\right )}} - \frac {b \log \left (a e^{\left (2 \, x\right )} + b e^{\left (2 \, x\right )} + 2 \, c e^{x} - a + b\right )}{b^{2} - c^{2}} + \frac {\log \left (e^{x} + 1\right )}{b - c} + \frac {\log \left ({\left | e^{x} - 1 \right |}\right )}{b + c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(x)^2/(a+b*coth(x)+c*csch(x)),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.20, size = 180, normalized size = 1.53 \[ -\frac {b \ln \left (\left (\tanh ^{2}\left (\frac {x}{2}\right )\right ) b -\left (\tanh ^{2}\left (\frac {x}{2}\right )\right ) c +2 a \tanh \left (\frac {x}{2}\right )+b +c \right )}{\left (b +c \right ) \left (b -c \right )}-\frac {2 \arctan \left (\frac {2 \left (b -c \right ) \tanh \left (\frac {x}{2}\right )+2 a}{2 \sqrt {-a^{2}+b^{2}-c^{2}}}\right ) a}{\left (b +c \right ) \sqrt {-a^{2}+b^{2}-c^{2}}}+\frac {2 \arctan \left (\frac {2 \left (b -c \right ) \tanh \left (\frac {x}{2}\right )+2 a}{2 \sqrt {-a^{2}+b^{2}-c^{2}}}\right ) b a}{\left (b +c \right ) \sqrt {-a^{2}+b^{2}-c^{2}}\, \left (b -c \right )}+\frac {\ln \left (\tanh \left (\frac {x}{2}\right )\right )}{b +c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csch(x)^2/(a+b*coth(x)+c*csch(x)),x)

[Out]

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

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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)^2/(a+b*coth(x)+c*csch(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(c^2-b^2+a^2>0)', see `assume?`
for more details)Is c^2-b^2+a^2 positive or negative?

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mupad [B]  time = 7.18, size = 1069, normalized size = 9.06 \[ \frac {\ln \left ({\mathrm {e}}^x-1\right )}{b+c}+\frac {\ln \left ({\mathrm {e}}^x+1\right )}{b-c}+\frac {\ln \left (-\frac {64\,\left (b-a+2\,c\,{\mathrm {e}}^x\right )}{{\left (a+b\right )}^4}-\frac {\left (\frac {32\,\left (-2\,a^3+3\,{\mathrm {e}}^x\,a^2\,c+2\,a\,b^2+6\,{\mathrm {e}}^x\,a\,b\,c-2\,a\,c^2+3\,{\mathrm {e}}^x\,b^2\,c+2\,b\,c^2+4\,{\mathrm {e}}^x\,c^3\right )}{{\left (a+b\right )}^5}+\frac {\left (\frac {32\,\left (a-b\right )\,\left (2\,b^3+6\,{\mathrm {e}}^x\,b^2\,c+2\,a\,b^2+b\,c^2+6\,a\,{\mathrm {e}}^x\,b\,c-3\,{\mathrm {e}}^x\,c^3+2\,a\,c^2\right )}{{\left (a+b\right )}^5}-\frac {32\,\left (a^2\,b+b\,c^2-b^3+a\,c\,\sqrt {a^2-b^2+c^2}\right )\,\left (-2\,a^3\,b^2-4\,{\mathrm {e}}^x\,a^3\,b\,c-2\,a^3\,c^2+{\mathrm {e}}^x\,a^2\,b^2\,c+4\,a^2\,b\,c^2+3\,{\mathrm {e}}^x\,a^2\,c^3+2\,a\,b^4+6\,{\mathrm {e}}^x\,a\,b^3\,c+a\,b^2\,c^2-6\,{\mathrm {e}}^x\,a\,b\,c^3-3\,a\,c^4+{\mathrm {e}}^x\,b^4\,c-3\,b^3\,c^2-5\,{\mathrm {e}}^x\,b^2\,c^3+3\,b\,c^4+4\,{\mathrm {e}}^x\,c^5\right )}{{\left (a+b\right )}^5\,\left (b^2-c^2\right )\,\left (a^2-b^2+c^2\right )}\right )\,\left (a^2\,b+b\,c^2-b^3+a\,c\,\sqrt {a^2-b^2+c^2}\right )}{\left (b^2-c^2\right )\,\left (a^2-b^2+c^2\right )}\right )\,\left (a^2\,b+b\,c^2-b^3+a\,c\,\sqrt {a^2-b^2+c^2}\right )}{\left (b^2-c^2\right )\,\left (a^2-b^2+c^2\right )}\right )\,\left (a^2\,b+b\,c^2-b^3+a\,c\,\sqrt {a^2-b^2+c^2}\right )}{-a^2\,b^2+a^2\,c^2+b^4-2\,b^2\,c^2+c^4}+\frac {\ln \left (-\frac {64\,\left (b-a+2\,c\,{\mathrm {e}}^x\right )}{{\left (a+b\right )}^4}-\frac {\left (\frac {32\,\left (-2\,a^3+3\,{\mathrm {e}}^x\,a^2\,c+2\,a\,b^2+6\,{\mathrm {e}}^x\,a\,b\,c-2\,a\,c^2+3\,{\mathrm {e}}^x\,b^2\,c+2\,b\,c^2+4\,{\mathrm {e}}^x\,c^3\right )}{{\left (a+b\right )}^5}+\frac {\left (\frac {32\,\left (a-b\right )\,\left (2\,b^3+6\,{\mathrm {e}}^x\,b^2\,c+2\,a\,b^2+b\,c^2+6\,a\,{\mathrm {e}}^x\,b\,c-3\,{\mathrm {e}}^x\,c^3+2\,a\,c^2\right )}{{\left (a+b\right )}^5}-\frac {32\,\left (a^2\,b+b\,c^2-b^3-a\,c\,\sqrt {a^2-b^2+c^2}\right )\,\left (-2\,a^3\,b^2-4\,{\mathrm {e}}^x\,a^3\,b\,c-2\,a^3\,c^2+{\mathrm {e}}^x\,a^2\,b^2\,c+4\,a^2\,b\,c^2+3\,{\mathrm {e}}^x\,a^2\,c^3+2\,a\,b^4+6\,{\mathrm {e}}^x\,a\,b^3\,c+a\,b^2\,c^2-6\,{\mathrm {e}}^x\,a\,b\,c^3-3\,a\,c^4+{\mathrm {e}}^x\,b^4\,c-3\,b^3\,c^2-5\,{\mathrm {e}}^x\,b^2\,c^3+3\,b\,c^4+4\,{\mathrm {e}}^x\,c^5\right )}{{\left (a+b\right )}^5\,\left (b^2-c^2\right )\,\left (a^2-b^2+c^2\right )}\right )\,\left (a^2\,b+b\,c^2-b^3-a\,c\,\sqrt {a^2-b^2+c^2}\right )}{\left (b^2-c^2\right )\,\left (a^2-b^2+c^2\right )}\right )\,\left (a^2\,b+b\,c^2-b^3-a\,c\,\sqrt {a^2-b^2+c^2}\right )}{\left (b^2-c^2\right )\,\left (a^2-b^2+c^2\right )}\right )\,\left (a^2\,b+b\,c^2-b^3-a\,c\,\sqrt {a^2-b^2+c^2}\right )}{-a^2\,b^2+a^2\,c^2+b^4-2\,b^2\,c^2+c^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sinh(x)^2*(a + c/sinh(x) + b*coth(x))),x)

[Out]

log(exp(x) - 1)/(b + c) + log(exp(x) + 1)/(b - c) + (log(- (64*(b - a + 2*c*exp(x)))/(a + b)^4 - (((32*(2*a*b^
2 - 2*a*c^2 + 2*b*c^2 - 2*a^3 + 4*c^3*exp(x) + 3*a^2*c*exp(x) + 3*b^2*c*exp(x) + 6*a*b*c*exp(x)))/(a + b)^5 +
(((32*(a - b)*(2*a*b^2 + 2*a*c^2 + b*c^2 + 2*b^3 - 3*c^3*exp(x) + 6*b^2*c*exp(x) + 6*a*b*c*exp(x)))/(a + b)^5
- (32*(a^2*b + b*c^2 - b^3 + a*c*(a^2 - b^2 + c^2)^(1/2))*(2*a*b^4 - 3*a*c^4 + 3*b*c^4 - 2*a^3*b^2 - 2*a^3*c^2
- 3*b^3*c^2 + 4*c^5*exp(x) + a*b^2*c^2 + 4*a^2*b*c^2 + b^4*c*exp(x) + 3*a^2*c^3*exp(x) - 5*b^2*c^3*exp(x) + a
^2*b^2*c*exp(x) - 6*a*b*c^3*exp(x) + 6*a*b^3*c*exp(x) - 4*a^3*b*c*exp(x)))/((a + b)^5*(b^2 - c^2)*(a^2 - b^2 +
c^2)))*(a^2*b + b*c^2 - b^3 + a*c*(a^2 - b^2 + c^2)^(1/2)))/((b^2 - c^2)*(a^2 - b^2 + c^2)))*(a^2*b + b*c^2 -
b^3 + a*c*(a^2 - b^2 + c^2)^(1/2)))/((b^2 - c^2)*(a^2 - b^2 + c^2)))*(a^2*b + b*c^2 - b^3 + a*c*(a^2 - b^2 +
c^2)^(1/2)))/(b^4 + c^4 - a^2*b^2 + a^2*c^2 - 2*b^2*c^2) + (log(- (64*(b - a + 2*c*exp(x)))/(a + b)^4 - (((32*
(2*a*b^2 - 2*a*c^2 + 2*b*c^2 - 2*a^3 + 4*c^3*exp(x) + 3*a^2*c*exp(x) + 3*b^2*c*exp(x) + 6*a*b*c*exp(x)))/(a +
b)^5 + (((32*(a - b)*(2*a*b^2 + 2*a*c^2 + b*c^2 + 2*b^3 - 3*c^3*exp(x) + 6*b^2*c*exp(x) + 6*a*b*c*exp(x)))/(a
+ b)^5 - (32*(a^2*b + b*c^2 - b^3 - a*c*(a^2 - b^2 + c^2)^(1/2))*(2*a*b^4 - 3*a*c^4 + 3*b*c^4 - 2*a^3*b^2 - 2*
a^3*c^2 - 3*b^3*c^2 + 4*c^5*exp(x) + a*b^2*c^2 + 4*a^2*b*c^2 + b^4*c*exp(x) + 3*a^2*c^3*exp(x) - 5*b^2*c^3*exp
(x) + a^2*b^2*c*exp(x) - 6*a*b*c^3*exp(x) + 6*a*b^3*c*exp(x) - 4*a^3*b*c*exp(x)))/((a + b)^5*(b^2 - c^2)*(a^2
- b^2 + c^2)))*(a^2*b + b*c^2 - b^3 - a*c*(a^2 - b^2 + c^2)^(1/2)))/((b^2 - c^2)*(a^2 - b^2 + c^2)))*(a^2*b +
b*c^2 - b^3 - a*c*(a^2 - b^2 + c^2)^(1/2)))/((b^2 - c^2)*(a^2 - b^2 + c^2)))*(a^2*b + b*c^2 - b^3 - a*c*(a^2 -
b^2 + c^2)^(1/2)))/(b^4 + c^4 - a^2*b^2 + a^2*c^2 - 2*b^2*c^2)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {csch}^{2}{\relax (x )}}{a + b \coth {\relax (x )} + c \operatorname {csch}{\relax (x )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(x)**2/(a+b*coth(x)+c*csch(x)),x)

[Out]

Integral(csch(x)**2/(a + b*coth(x) + c*csch(x)), x)

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