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

Optimal. Leaf size=82 \[ -\frac {b^2 \tanh ^{-1}(\cosh (x))}{a^3}+\frac {b \text {csch}(x)}{a^2}+\frac {b \sqrt {a^2-b^2} \tan ^{-1}\left (\frac {a \sinh (x)+b \cosh (x)}{\sqrt {a^2-b^2}}\right )}{a^3}+\frac {\tanh ^{-1}(\cosh (x))}{2 a}-\frac {\coth (x) \text {csch}(x)}{2 a} \]

[Out]

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

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Rubi [A]  time = 0.29, antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 11, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.846, Rules used = {3518, 3110, 3768, 3770, 2621, 321, 207, 2622, 3104, 3074, 206} \[ -\frac {b^2 \tanh ^{-1}(\cosh (x))}{a^3}+\frac {b \sqrt {a^2-b^2} \tan ^{-1}\left (\frac {a \sinh (x)+b \cosh (x)}{\sqrt {a^2-b^2}}\right )}{a^3}+\frac {b \text {csch}(x)}{a^2}+\frac {\tanh ^{-1}(\cosh (x))}{2 a}-\frac {\coth (x) \text {csch}(x)}{2 a} \]

Antiderivative was successfully verified.

[In]

Int[Csch[x]^3/(a + b*Tanh[x]),x]

[Out]

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

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 207

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 2621

Int[(csc[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*sec[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> -Dist[(f*a^n)^(-1), Subst
[Int[x^(m + n - 1)/(-1 + x^2/a^2)^((n + 1)/2), x], x, a*Csc[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && Integer
Q[(n + 1)/2] &&  !(IntegerQ[(m + 1)/2] && LtQ[0, m, n])

Rule 2622

Int[csc[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Dist[1/(f*a^n), Subst[Int
[x^(m + n - 1)/(-1 + x^2/a^2)^((n + 1)/2), x], x, a*Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n
+ 1)/2] &&  !(IntegerQ[(m + 1)/2] && LtQ[0, m, n])

Rule 3074

Int[(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> -Dist[d^(-1), Subst[Int
[1/(a^2 + b^2 - x^2), x], x, b*Cos[c + d*x] - a*Sin[c + d*x]], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 + b^2,
0]

Rule 3104

Int[cos[(c_.) + (d_.)*(x_)]^(m_)/(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)]), x_Symbol] :>
-Simp[Cos[c + d*x]^(m + 1)/(b*d*(m + 1)), x] + (-Dist[a/b^2, Int[Cos[c + d*x]^(m + 1), x], x] + Dist[(a^2 + b
^2)/b^2, Int[Cos[c + d*x]^(m + 2)/(a*Cos[c + d*x] + b*Sin[c + d*x]), x], x]) /; FreeQ[{a, b, c, d}, x] && NeQ[
a^2 + b^2, 0] && LtQ[m, -1]

Rule 3110

Int[(cos[(c_.) + (d_.)*(x_)]^(m_.)*sin[(c_.) + (d_.)*(x_)]^(n_.))/(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(
c_.) + (d_.)*(x_)]), x_Symbol] :> Int[ExpandTrig[(cos[c + d*x]^m*sin[c + d*x]^n)/(a*cos[c + d*x] + b*sin[c + d
*x]), x], x] /; FreeQ[{a, b, c, d, m, n}, x] && IntegersQ[m, n]

Rule 3518

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Int[(Sin[e + f*x]
^m*(a*Cos[e + f*x] + b*Sin[e + f*x])^n)/Cos[e + f*x]^n, x] /; FreeQ[{a, b, e, f}, x] && IntegerQ[(m - 1)/2] &&
ILtQ[n, 0] && ((LtQ[m, 5] && GtQ[n, -4]) || (EqQ[m, 5] && EqQ[n, -1]))

Rule 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]

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

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Mathematica [A]  time = 0.37, size = 123, normalized size = 1.50 \[ -\frac {a^2 \text {csch}^2\left (\frac {x}{2}\right )+a^2 \text {sech}^2\left (\frac {x}{2}\right )+4 a^2 \log \left (\tanh \left (\frac {x}{2}\right )\right )+4 a b \tanh \left (\frac {x}{2}\right )-4 a b \coth \left (\frac {x}{2}\right )-16 b \sqrt {a-b} \sqrt {a+b} \tan ^{-1}\left (\frac {a \tanh \left (\frac {x}{2}\right )+b}{\sqrt {a-b} \sqrt {a+b}}\right )-8 b^2 \log \left (\tanh \left (\frac {x}{2}\right )\right )}{8 a^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.59, size = 1165, normalized size = 14.21 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2*(2*(a^2 - 2*a*b)*cosh(x)^3 + 6*(a^2 - 2*a*b)*cosh(x)*sinh(x)^2 + 2*(a^2 - 2*a*b)*sinh(x)^3 - 2*(b*cosh(x
)^4 + 4*b*cosh(x)*sinh(x)^3 + b*sinh(x)^4 - 2*b*cosh(x)^2 + 2*(3*b*cosh(x)^2 - b)*sinh(x)^2 + 4*(b*cosh(x)^3 -
b*cosh(x))*sinh(x) + b)*sqrt(-a^2 + b^2)*log(((a + b)*cosh(x)^2 + 2*(a + b)*cosh(x)*sinh(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)*cosh(x)*sinh(x) + (a + b)*
sinh(x)^2 + a - b)) + 2*(a^2 + 2*a*b)*cosh(x) - ((a^2 - 2*b^2)*cosh(x)^4 + 4*(a^2 - 2*b^2)*cosh(x)*sinh(x)^3 +
(a^2 - 2*b^2)*sinh(x)^4 - 2*(a^2 - 2*b^2)*cosh(x)^2 + 2*(3*(a^2 - 2*b^2)*cosh(x)^2 - a^2 + 2*b^2)*sinh(x)^2 +
a^2 - 2*b^2 + 4*((a^2 - 2*b^2)*cosh(x)^3 - (a^2 - 2*b^2)*cosh(x))*sinh(x))*log(cosh(x) + sinh(x) + 1) + ((a^2
- 2*b^2)*cosh(x)^4 + 4*(a^2 - 2*b^2)*cosh(x)*sinh(x)^3 + (a^2 - 2*b^2)*sinh(x)^4 - 2*(a^2 - 2*b^2)*cosh(x)^2
+ 2*(3*(a^2 - 2*b^2)*cosh(x)^2 - a^2 + 2*b^2)*sinh(x)^2 + a^2 - 2*b^2 + 4*((a^2 - 2*b^2)*cosh(x)^3 - (a^2 - 2*
b^2)*cosh(x))*sinh(x))*log(cosh(x) + sinh(x) - 1) + 2*(3*(a^2 - 2*a*b)*cosh(x)^2 + a^2 + 2*a*b)*sinh(x))/(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)), -1/2*(2*(a^2 - 2*a*b)*cosh(x)^3 + 6*(a^2 - 2*a*b)*cosh(x)*sinh
(x)^2 + 2*(a^2 - 2*a*b)*sinh(x)^3 + 4*(b*cosh(x)^4 + 4*b*cosh(x)*sinh(x)^3 + b*sinh(x)^4 - 2*b*cosh(x)^2 + 2*(
3*b*cosh(x)^2 - b)*sinh(x)^2 + 4*(b*cosh(x)^3 - b*cosh(x))*sinh(x) + b)*sqrt(a^2 - b^2)*arctan(sqrt(a^2 - b^2)
/((a + b)*cosh(x) + (a + b)*sinh(x))) + 2*(a^2 + 2*a*b)*cosh(x) - ((a^2 - 2*b^2)*cosh(x)^4 + 4*(a^2 - 2*b^2)*c
osh(x)*sinh(x)^3 + (a^2 - 2*b^2)*sinh(x)^4 - 2*(a^2 - 2*b^2)*cosh(x)^2 + 2*(3*(a^2 - 2*b^2)*cosh(x)^2 - a^2 +
2*b^2)*sinh(x)^2 + a^2 - 2*b^2 + 4*((a^2 - 2*b^2)*cosh(x)^3 - (a^2 - 2*b^2)*cosh(x))*sinh(x))*log(cosh(x) + si
nh(x) + 1) + ((a^2 - 2*b^2)*cosh(x)^4 + 4*(a^2 - 2*b^2)*cosh(x)*sinh(x)^3 + (a^2 - 2*b^2)*sinh(x)^4 - 2*(a^2 -
2*b^2)*cosh(x)^2 + 2*(3*(a^2 - 2*b^2)*cosh(x)^2 - a^2 + 2*b^2)*sinh(x)^2 + a^2 - 2*b^2 + 4*((a^2 - 2*b^2)*cos
h(x)^3 - (a^2 - 2*b^2)*cosh(x))*sinh(x))*log(cosh(x) + sinh(x) - 1) + 2*(3*(a^2 - 2*a*b)*cosh(x)^2 + a^2 + 2*a
*b)*sinh(x))/(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))]

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giac [A]  time = 0.12, size = 125, normalized size = 1.52 \[ \frac {{\left (a^{2} - 2 \, b^{2}\right )} \log \left (e^{x} + 1\right )}{2 \, a^{3}} - \frac {{\left (a^{2} - 2 \, b^{2}\right )} \log \left ({\left | e^{x} - 1 \right |}\right )}{2 \, a^{3}} + \frac {2 \, {\left (a^{2} b - b^{3}\right )} \arctan \left (\frac {a e^{x} + b e^{x}}{\sqrt {a^{2} - b^{2}}}\right )}{\sqrt {a^{2} - b^{2}} a^{3}} - \frac {a e^{\left (3 \, x\right )} - 2 \, b e^{\left (3 \, x\right )} + a e^{x} + 2 \, b e^{x}}{a^{2} {\left (e^{\left (2 \, x\right )} - 1\right )}^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.13, size = 110, normalized size = 1.34 \[ \frac {\tanh ^{2}\left (\frac {x}{2}\right )}{8 a}-\frac {\tanh \left (\frac {x}{2}\right ) b}{2 a^{2}}+\frac {2 b \sqrt {a^{2}-b^{2}}\, \arctan \left (\frac {2 a \tanh \left (\frac {x}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{a^{3}}-\frac {1}{8 a \tanh \left (\frac {x}{2}\right )^{2}}-\frac {\ln \left (\tanh \left (\frac {x}{2}\right )\right )}{2 a}+\frac {\ln \left (\tanh \left (\frac {x}{2}\right )\right ) b^{2}}{a^{3}}+\frac {b}{2 a^{2} \tanh \left (\frac {x}{2}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8/a*tanh(1/2*x)^2-1/2/a^2*tanh(1/2*x)*b+2*b*(a^2-b^2)^(1/2)/a^3*arctan(1/2*(2*a*tanh(1/2*x)+2*b)/(a^2-b^2)^(
1/2))-1/8/a/tanh(1/2*x)^2-1/2/a*ln(tanh(1/2*x))+1/a^3*ln(tanh(1/2*x))*b^2+1/2/a^2*b/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*tanh(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*b^2-4*a^2>0)', see `assume?`
for more details)Is 4*b^2-4*a^2 positive or negative?

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mupad [B]  time = 1.51, size = 506, normalized size = 6.17 \[ \frac {\ln \left (8\,a^3\,b-16\,a\,b^3-4\,a^4+8\,b^4+4\,a^2\,b^2-4\,a^4\,{\mathrm {e}}^x+8\,b^4\,{\mathrm {e}}^x-16\,a\,b^3\,{\mathrm {e}}^x+8\,a^3\,b\,{\mathrm {e}}^x+4\,a^2\,b^2\,{\mathrm {e}}^x\right )}{2\,a}-\frac {2\,{\mathrm {e}}^x}{a-2\,a\,{\mathrm {e}}^{2\,x}+a\,{\mathrm {e}}^{4\,x}}-\frac {\ln \left (16\,a\,b^3-8\,a^3\,b+4\,a^4-8\,b^4-4\,a^2\,b^2-4\,a^4\,{\mathrm {e}}^x+8\,b^4\,{\mathrm {e}}^x-16\,a\,b^3\,{\mathrm {e}}^x+8\,a^3\,b\,{\mathrm {e}}^x+4\,a^2\,b^2\,{\mathrm {e}}^x\right )}{2\,a}-\frac {b^2\,\ln \left (8\,a^3\,b-16\,a\,b^3-4\,a^4+8\,b^4+4\,a^2\,b^2-4\,a^4\,{\mathrm {e}}^x+8\,b^4\,{\mathrm {e}}^x-16\,a\,b^3\,{\mathrm {e}}^x+8\,a^3\,b\,{\mathrm {e}}^x+4\,a^2\,b^2\,{\mathrm {e}}^x\right )}{a^3}+\frac {b^2\,\ln \left (16\,a\,b^3-8\,a^3\,b+4\,a^4-8\,b^4-4\,a^2\,b^2-4\,a^4\,{\mathrm {e}}^x+8\,b^4\,{\mathrm {e}}^x-16\,a\,b^3\,{\mathrm {e}}^x+8\,a^3\,b\,{\mathrm {e}}^x+4\,a^2\,b^2\,{\mathrm {e}}^x\right )}{a^3}-\frac {a\,{\mathrm {e}}^x}{a^2\,{\mathrm {e}}^{2\,x}-a^2}+\frac {2\,b\,{\mathrm {e}}^x}{a^2\,{\mathrm {e}}^{2\,x}-a^2}-\frac {b\,\ln \left (8\,b^2\,\sqrt {b^2-a^2}-8\,b^3\,{\mathrm {e}}^x+8\,a^2\,b\,{\mathrm {e}}^x-8\,a\,b\,\sqrt {b^2-a^2}\right )\,\sqrt {b^2-a^2}}{a^3}+\frac {b\,\ln \left (8\,b^2\,\sqrt {b^2-a^2}+8\,b^3\,{\mathrm {e}}^x-8\,a^2\,b\,{\mathrm {e}}^x-8\,a\,b\,\sqrt {b^2-a^2}\right )\,\sqrt {b^2-a^2}}{a^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sinh(x)^3*(a + b*tanh(x))),x)

[Out]

log(8*a^3*b - 16*a*b^3 - 4*a^4 + 8*b^4 + 4*a^2*b^2 - 4*a^4*exp(x) + 8*b^4*exp(x) - 16*a*b^3*exp(x) + 8*a^3*b*e
xp(x) + 4*a^2*b^2*exp(x))/(2*a) - (2*exp(x))/(a - 2*a*exp(2*x) + a*exp(4*x)) - log(16*a*b^3 - 8*a^3*b + 4*a^4
- 8*b^4 - 4*a^2*b^2 - 4*a^4*exp(x) + 8*b^4*exp(x) - 16*a*b^3*exp(x) + 8*a^3*b*exp(x) + 4*a^2*b^2*exp(x))/(2*a)
- (b^2*log(8*a^3*b - 16*a*b^3 - 4*a^4 + 8*b^4 + 4*a^2*b^2 - 4*a^4*exp(x) + 8*b^4*exp(x) - 16*a*b^3*exp(x) + 8
*a^3*b*exp(x) + 4*a^2*b^2*exp(x)))/a^3 + (b^2*log(16*a*b^3 - 8*a^3*b + 4*a^4 - 8*b^4 - 4*a^2*b^2 - 4*a^4*exp(x
) + 8*b^4*exp(x) - 16*a*b^3*exp(x) + 8*a^3*b*exp(x) + 4*a^2*b^2*exp(x)))/a^3 - (a*exp(x))/(a^2*exp(2*x) - a^2)
+ (2*b*exp(x))/(a^2*exp(2*x) - a^2) - (b*log(8*b^2*(b^2 - a^2)^(1/2) - 8*b^3*exp(x) + 8*a^2*b*exp(x) - 8*a*b*
(b^2 - a^2)^(1/2))*(b^2 - a^2)^(1/2))/a^3 + (b*log(8*b^2*(b^2 - a^2)^(1/2) + 8*b^3*exp(x) - 8*a^2*b*exp(x) - 8
*a*b*(b^2 - a^2)^(1/2))*(b^2 - a^2)^(1/2))/a^3

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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