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

Optimal. Leaf size=81 \[ \frac {b \coth (x)}{a^2}+\frac {\left (a^2-2 b^2\right ) \tanh ^{-1}(\cosh (x))}{2 a^3}+\frac {2 b^3 \tanh ^{-1}\left (\frac {b-a \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2+b^2}}\right )}{a^3 \sqrt {a^2+b^2}}-\frac {\coth (x) \text {csch}(x)}{2 a} \]

[Out]

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

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Rubi [A]  time = 0.32, antiderivative size = 81, 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 = {2802, 3055, 3001, 3770, 2660, 618, 206} \[ \frac {2 b^3 \tanh ^{-1}\left (\frac {b-a \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2+b^2}}\right )}{a^3 \sqrt {a^2+b^2}}+\frac {\left (a^2-2 b^2\right ) \tanh ^{-1}(\cosh (x))}{2 a^3}+\frac {b \coth (x)}{a^2}-\frac {\coth (x) \text {csch}(x)}{2 a} \]

Antiderivative was successfully verified.

[In]

Int[Csch[x]^3/(a + b*Sinh[x]),x]

[Out]

((a^2 - 2*b^2)*ArcTanh[Cosh[x]])/(2*a^3) + (2*b^3*ArcTanh[(b - a*Tanh[x/2])/Sqrt[a^2 + b^2]])/(a^3*Sqrt[a^2 +
b^2]) + (b*Coth[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 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 2802

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -S
imp[(b^2*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 -
b^2)), x] + Dist[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n
*Simp[a*(b*c - a*d)*(m + 1) + b^2*d*(m + n + 2) - (b^2*c + b*(b*c - a*d)*(m + 1))*Sin[e + f*x] - b^2*d*(m + n
+ 3)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] &
& NeQ[c^2 - d^2, 0] && LtQ[m, -1] && IntegersQ[2*m, 2*n] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !
(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] &&  !IntegerQ[m]) || EqQ[a, 0])))

Rule 3001

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)])), x_Symbol] :> Dist[(A*b - a*B)/(b*c - a*d), Int[1/(a + b*Sin[e + f*x]), x], x] + Dist[(B*c - A
*d)/(b*c - a*d), Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0]
&& NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]

Rule 3055

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*s
in[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((A*b^2 - a*b*B + a^2*C)*Cos[e +
f*x]*(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2)), x] + Dis
t[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[(m + 1)*(b
*c - a*d)*(a*A - b*B + a*C) + d*(A*b^2 - a*b*B + a^2*C)*(m + n + 2) - (c*(A*b^2 - a*b*B + a^2*C) + (m + 1)*(b*
c - a*d)*(A*b - a*B + b*C))*Sin[e + f*x] - d*(A*b^2 - a*b*B + a^2*C)*(m + n + 3)*Sin[e + f*x]^2, x], x], x] /;
FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && Lt
Q[m, -1] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] &&
!IntegerQ[m]) || EqQ[a, 0])))

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

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Mathematica [A]  time = 0.54, size = 118, normalized size = 1.46 \[ -\frac {4 \left (a^2-2 b^2\right ) \log \left (\tanh \left (\frac {x}{2}\right )\right )+\frac {16 b^3 \tan ^{-1}\left (\frac {b-a \tanh \left (\frac {x}{2}\right )}{\sqrt {-a^2-b^2}}\right )}{\sqrt {-a^2-b^2}}+a^2 \text {csch}^2\left (\frac {x}{2}\right )+a^2 \text {sech}^2\left (\frac {x}{2}\right )-4 a b \tanh \left (\frac {x}{2}\right )-4 a b \coth \left (\frac {x}{2}\right )}{8 a^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/8*((16*b^3*ArcTan[(b - a*Tanh[x/2])/Sqrt[-a^2 - b^2]])/Sqrt[-a^2 - b^2] - 4*a*b*Coth[x/2] + a^2*Csch[x/2]^2
+ 4*(a^2 - 2*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 = 1.06, size = 929, normalized size = 11.47 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.21, size = 137, normalized size = 1.69 \[ -\frac {b^{3} \log \left (\frac {{\left | 2 \, b e^{x} + 2 \, a - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, b e^{x} + 2 \, a + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{\sqrt {a^{2} + b^{2}} a^{3}} + \frac {{\left (a^{2} - 2 \, b^{2}\right )} \log \left (e^{x} + 1\right )}{2 \, a^{3}} - \frac {{\left (a^{2} - 2 \, b^{2}\right )} \log \left ({\left | e^{x} - 1 \right |}\right )}{2 \, a^{3}} - \frac {a e^{\left (3 \, x\right )} - 2 \, b e^{\left (2 \, x\right )} + a e^{x} + 2 \, b}{a^{2} {\left (e^{\left (2 \, x\right )} - 1\right )}^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.05, size = 108, normalized size = 1.33 \[ \frac {\tanh ^{2}\left (\frac {x}{2}\right )}{8 a}+\frac {\tanh \left (\frac {x}{2}\right ) b}{2 a^{2}}-\frac {2 b^{3} \arctanh \left (\frac {2 a \tanh \left (\frac {x}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{a^{3} \sqrt {a^{2}+b^{2}}}-\frac {1}{8 a \tanh \left (\frac {x}{2}\right )^{2}}-\frac {\ln \left (\tanh \left (\frac {x}{2}\right )\right )}{2 a}+\frac {\ln \left (\tanh \left (\frac {x}{2}\right )\right ) b^{2}}{a^{3}}+\frac {b}{2 a^{2} \tanh \left (\frac {x}{2}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.41, size = 154, normalized size = 1.90 \[ -\frac {b^{3} \log \left (\frac {b e^{\left (-x\right )} - a - \sqrt {a^{2} + b^{2}}}{b e^{\left (-x\right )} - a + \sqrt {a^{2} + b^{2}}}\right )}{\sqrt {a^{2} + b^{2}} a^{3}} + \frac {a e^{\left (-x\right )} + 2 \, b e^{\left (-2 \, x\right )} + a e^{\left (-3 \, x\right )} - 2 \, b}{2 \, a^{2} e^{\left (-2 \, x\right )} - a^{2} e^{\left (-4 \, x\right )} - a^{2}} + \frac {{\left (a^{2} - 2 \, b^{2}\right )} \log \left (e^{\left (-x\right )} + 1\right )}{2 \, a^{3}} - \frac {{\left (a^{2} - 2 \, b^{2}\right )} \log \left (e^{\left (-x\right )} - 1\right )}{2 \, a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.00, size = 617, normalized size = 7.62 \[ \frac {{\mathrm {e}}^x}{a-a\,{\mathrm {e}}^{2\,x}}-\frac {2\,{\mathrm {e}}^x}{a-2\,a\,{\mathrm {e}}^{2\,x}+a\,{\mathrm {e}}^{4\,x}}-\frac {\ln \left (4\,a^4+24\,b^4-20\,a^2\,b^2-4\,a^4\,{\mathrm {e}}^x-24\,b^4\,{\mathrm {e}}^x+20\,a^2\,b^2\,{\mathrm {e}}^x\right )}{2\,a}+\frac {\ln \left (4\,a^4+24\,b^4-20\,a^2\,b^2+4\,a^4\,{\mathrm {e}}^x+24\,b^4\,{\mathrm {e}}^x-20\,a^2\,b^2\,{\mathrm {e}}^x\right )}{2\,a}+\frac {2\,b}{a^2\,{\mathrm {e}}^{2\,x}-a^2}+\frac {b^2\,\ln \left (4\,a^4+24\,b^4-20\,a^2\,b^2-4\,a^4\,{\mathrm {e}}^x-24\,b^4\,{\mathrm {e}}^x+20\,a^2\,b^2\,{\mathrm {e}}^x\right )}{a^3}-\frac {b^2\,\ln \left (4\,a^4+24\,b^4-20\,a^2\,b^2+4\,a^4\,{\mathrm {e}}^x+24\,b^4\,{\mathrm {e}}^x-20\,a^2\,b^2\,{\mathrm {e}}^x\right )}{a^3}-\frac {b^3\,\ln \left (16\,a^5\,b-48\,a\,b^5-24\,b^5\,\sqrt {a^2+b^2}-32\,a^3\,b^3-32\,a^6\,{\mathrm {e}}^x+24\,b^6\,{\mathrm {e}}^x-40\,a^2\,b^3\,\sqrt {a^2+b^2}-32\,a^5\,{\mathrm {e}}^x\,\sqrt {a^2+b^2}+112\,a^2\,b^4\,{\mathrm {e}}^x+56\,a^4\,b^2\,{\mathrm {e}}^x+16\,a^4\,b\,\sqrt {a^2+b^2}+72\,a\,b^4\,{\mathrm {e}}^x\,\sqrt {a^2+b^2}+72\,a^3\,b^2\,{\mathrm {e}}^x\,\sqrt {a^2+b^2}\right )\,\sqrt {a^2+b^2}}{a^5+a^3\,b^2}+\frac {b^3\,\ln \left (24\,b^5\,\sqrt {a^2+b^2}-48\,a\,b^5+16\,a^5\,b-32\,a^3\,b^3-32\,a^6\,{\mathrm {e}}^x+24\,b^6\,{\mathrm {e}}^x+40\,a^2\,b^3\,\sqrt {a^2+b^2}+32\,a^5\,{\mathrm {e}}^x\,\sqrt {a^2+b^2}+112\,a^2\,b^4\,{\mathrm {e}}^x+56\,a^4\,b^2\,{\mathrm {e}}^x-16\,a^4\,b\,\sqrt {a^2+b^2}-72\,a\,b^4\,{\mathrm {e}}^x\,\sqrt {a^2+b^2}-72\,a^3\,b^2\,{\mathrm {e}}^x\,\sqrt {a^2+b^2}\right )\,\sqrt {a^2+b^2}}{a^5+a^3\,b^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sinh(x)^3*(a + b*sinh(x))),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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