3.235 \(\int \frac {\coth ^4(x)}{a+b \sinh (x)} \, dx\)

Optimal. Leaf size=108 \[ \frac {b \coth (x) \text {csch}(x)}{2 a^2}-\frac {2 \left (a^2+b^2\right )^{3/2} \tanh ^{-1}\left (\frac {b-a \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2+b^2}}\right )}{a^4}+\frac {b \left (3 a^2+2 b^2\right ) \tanh ^{-1}(\cosh (x))}{2 a^4}-\frac {\left (4 a^2+3 b^2\right ) \coth (x)}{3 a^3}-\frac {\coth (x) \text {csch}^2(x)}{3 a} \]

[Out]

1/2*b*(3*a^2+2*b^2)*arctanh(cosh(x))/a^4-2*(a^2+b^2)^(3/2)*arctanh((b-a*tanh(1/2*x))/(a^2+b^2)^(1/2))/a^4-1/3*
(4*a^2+3*b^2)*coth(x)/a^3+1/2*b*coth(x)*csch(x)/a^2-1/3*coth(x)*csch(x)^2/a

________________________________________________________________________________________

Rubi [A]  time = 0.41, antiderivative size = 108, 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 = {2725, 3055, 3001, 3770, 2660, 618, 206} \[ -\frac {2 \left (a^2+b^2\right )^{3/2} \tanh ^{-1}\left (\frac {b-a \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2+b^2}}\right )}{a^4}-\frac {\left (4 a^2+3 b^2\right ) \coth (x)}{3 a^3}+\frac {b \left (3 a^2+2 b^2\right ) \tanh ^{-1}(\cosh (x))}{2 a^4}+\frac {b \coth (x) \text {csch}(x)}{2 a^2}-\frac {\coth (x) \text {csch}^2(x)}{3 a} \]

Antiderivative was successfully verified.

[In]

Int[Coth[x]^4/(a + b*Sinh[x]),x]

[Out]

(b*(3*a^2 + 2*b^2)*ArcTanh[Cosh[x]])/(2*a^4) - (2*(a^2 + b^2)^(3/2)*ArcTanh[(b - a*Tanh[x/2])/Sqrt[a^2 + b^2]]
)/a^4 - ((4*a^2 + 3*b^2)*Coth[x])/(3*a^3) + (b*Coth[x]*Csch[x])/(2*a^2) - (Coth[x]*Csch[x]^2)/(3*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 2725

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)/tan[(e_.) + (f_.)*(x_)]^4, x_Symbol] :> -Simp[(Cos[e + f*x]*(a
+ b*Sin[e + f*x])^(m + 1))/(3*a*f*Sin[e + f*x]^3), x] + (-Dist[1/(6*a^2), Int[((a + b*Sin[e + f*x])^m*Simp[8*
a^2 - b^2*(m - 1)*(m - 2) + a*b*m*Sin[e + f*x] - (6*a^2 - b^2*m*(m - 2))*Sin[e + f*x]^2, x])/Sin[e + f*x]^2, x
], x] - Simp[(b*(m - 2)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(6*a^2*f*Sin[e + f*x]^2), x]) /; FreeQ[{a,
b, e, f, m}, x] && NeQ[a^2 - b^2, 0] &&  !LtQ[m, -1] && IntegerQ[2*m]

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 {\coth ^4(x)}{a+b \sinh (x)} \, dx &=\frac {b \coth (x) \text {csch}(x)}{2 a^2}-\frac {\coth (x) \text {csch}^2(x)}{3 a}+\frac {\int \frac {\text {csch}^2(x) \left (2 \left (4 a^2+3 b^2\right )-a b \sinh (x)+3 \left (2 a^2+b^2\right ) \sinh ^2(x)\right )}{a+b \sinh (x)} \, dx}{6 a^2}\\ &=-\frac {\left (4 a^2+3 b^2\right ) \coth (x)}{3 a^3}+\frac {b \coth (x) \text {csch}(x)}{2 a^2}-\frac {\coth (x) \text {csch}^2(x)}{3 a}+\frac {i \int \frac {\text {csch}(x) \left (3 i b \left (3 a^2+2 b^2\right )-3 i a \left (2 a^2+b^2\right ) \sinh (x)\right )}{a+b \sinh (x)} \, dx}{6 a^3}\\ &=-\frac {\left (4 a^2+3 b^2\right ) \coth (x)}{3 a^3}+\frac {b \coth (x) \text {csch}(x)}{2 a^2}-\frac {\coth (x) \text {csch}^2(x)}{3 a}+\frac {\left (a^2+b^2\right )^2 \int \frac {1}{a+b \sinh (x)} \, dx}{a^4}-\frac {\left (b \left (3 a^2+2 b^2\right )\right ) \int \text {csch}(x) \, dx}{2 a^4}\\ &=\frac {b \left (3 a^2+2 b^2\right ) \tanh ^{-1}(\cosh (x))}{2 a^4}-\frac {\left (4 a^2+3 b^2\right ) \coth (x)}{3 a^3}+\frac {b \coth (x) \text {csch}(x)}{2 a^2}-\frac {\coth (x) \text {csch}^2(x)}{3 a}+\frac {\left (2 \left (a^2+b^2\right )^2\right ) \operatorname {Subst}\left (\int \frac {1}{a+2 b x-a x^2} \, dx,x,\tanh \left (\frac {x}{2}\right )\right )}{a^4}\\ &=\frac {b \left (3 a^2+2 b^2\right ) \tanh ^{-1}(\cosh (x))}{2 a^4}-\frac {\left (4 a^2+3 b^2\right ) \coth (x)}{3 a^3}+\frac {b \coth (x) \text {csch}(x)}{2 a^2}-\frac {\coth (x) \text {csch}^2(x)}{3 a}-\frac {\left (4 \left (a^2+b^2\right )^2\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^4}\\ &=\frac {b \left (3 a^2+2 b^2\right ) \tanh ^{-1}(\cosh (x))}{2 a^4}-\frac {2 \left (a^2+b^2\right )^{3/2} \tanh ^{-1}\left (\frac {b-a \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2+b^2}}\right )}{a^4}-\frac {\left (4 a^2+3 b^2\right ) \coth (x)}{3 a^3}+\frac {b \coth (x) \text {csch}(x)}{2 a^2}-\frac {\coth (x) \text {csch}^2(x)}{3 a}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.43, size = 176, normalized size = 1.63 \[ \frac {-\frac {1}{2} a^3 \sinh (x) \text {csch}^4\left (\frac {x}{2}\right )+8 a^3 \sinh ^4\left (\frac {x}{2}\right ) \text {csch}^3(x)-4 a \left (4 a^2+3 b^2\right ) \tanh \left (\frac {x}{2}\right )-4 a \left (4 a^2+3 b^2\right ) \coth \left (\frac {x}{2}\right )-12 b \left (3 a^2+2 b^2\right ) \log \left (\tanh \left (\frac {x}{2}\right )\right )+48 \left (-a^2-b^2\right )^{3/2} \tan ^{-1}\left (\frac {b-a \tanh \left (\frac {x}{2}\right )}{\sqrt {-a^2-b^2}}\right )+3 a^2 b \text {csch}^2\left (\frac {x}{2}\right )+3 a^2 b \text {sech}^2\left (\frac {x}{2}\right )}{24 a^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(48*(-a^2 - b^2)^(3/2)*ArcTan[(b - a*Tanh[x/2])/Sqrt[-a^2 - b^2]] - 4*a*(4*a^2 + 3*b^2)*Coth[x/2] + 3*a^2*b*Cs
ch[x/2]^2 - 12*b*(3*a^2 + 2*b^2)*Log[Tanh[x/2]] + 3*a^2*b*Sech[x/2]^2 + 8*a^3*Csch[x]^3*Sinh[x/2]^4 - (a^3*Csc
h[x/2]^4*Sinh[x])/2 - 4*a*(4*a^2 + 3*b^2)*Tanh[x/2])/(24*a^4)

________________________________________________________________________________________

fricas [B]  time = 2.02, size = 1303, normalized size = 12.06 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(6*a^2*b*cosh(x)^5 + 6*a^2*b*sinh(x)^5 - 12*(2*a^3 + a*b^2)*cosh(x)^4 + 6*(5*a^2*b*cosh(x) - 4*a^3 - 2*a*b
^2)*sinh(x)^4 - 6*a^2*b*cosh(x) + 12*(5*a^2*b*cosh(x)^2 - 4*(2*a^3 + a*b^2)*cosh(x))*sinh(x)^3 - 16*a^3 - 12*a
*b^2 + 24*(a^3 + a*b^2)*cosh(x)^2 + 12*(5*a^2*b*cosh(x)^3 + 2*a^3 + 2*a*b^2 - 6*(2*a^3 + a*b^2)*cosh(x)^2)*sin
h(x)^2 + 6*((a^2 + b^2)*cosh(x)^6 + 6*(a^2 + b^2)*cosh(x)*sinh(x)^5 + (a^2 + b^2)*sinh(x)^6 - 3*(a^2 + b^2)*co
sh(x)^4 + 3*(5*(a^2 + b^2)*cosh(x)^2 - a^2 - b^2)*sinh(x)^4 + 4*(5*(a^2 + b^2)*cosh(x)^3 - 3*(a^2 + b^2)*cosh(
x))*sinh(x)^3 + 3*(a^2 + b^2)*cosh(x)^2 + 3*(5*(a^2 + b^2)*cosh(x)^4 - 6*(a^2 + b^2)*cosh(x)^2 + a^2 + b^2)*si
nh(x)^2 - a^2 - b^2 + 6*((a^2 + b^2)*cosh(x)^5 - 2*(a^2 + b^2)*cosh(x)^3 + (a^2 + b^2)*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)*si
nh(x) - b)) + 3*((3*a^2*b + 2*b^3)*cosh(x)^6 + 6*(3*a^2*b + 2*b^3)*cosh(x)*sinh(x)^5 + (3*a^2*b + 2*b^3)*sinh(
x)^6 - 3*(3*a^2*b + 2*b^3)*cosh(x)^4 - 3*(3*a^2*b + 2*b^3 - 5*(3*a^2*b + 2*b^3)*cosh(x)^2)*sinh(x)^4 + 4*(5*(3
*a^2*b + 2*b^3)*cosh(x)^3 - 3*(3*a^2*b + 2*b^3)*cosh(x))*sinh(x)^3 - 3*a^2*b - 2*b^3 + 3*(3*a^2*b + 2*b^3)*cos
h(x)^2 + 3*(5*(3*a^2*b + 2*b^3)*cosh(x)^4 + 3*a^2*b + 2*b^3 - 6*(3*a^2*b + 2*b^3)*cosh(x)^2)*sinh(x)^2 + 6*((3
*a^2*b + 2*b^3)*cosh(x)^5 - 2*(3*a^2*b + 2*b^3)*cosh(x)^3 + (3*a^2*b + 2*b^3)*cosh(x))*sinh(x))*log(cosh(x) +
sinh(x) + 1) - 3*((3*a^2*b + 2*b^3)*cosh(x)^6 + 6*(3*a^2*b + 2*b^3)*cosh(x)*sinh(x)^5 + (3*a^2*b + 2*b^3)*sinh
(x)^6 - 3*(3*a^2*b + 2*b^3)*cosh(x)^4 - 3*(3*a^2*b + 2*b^3 - 5*(3*a^2*b + 2*b^3)*cosh(x)^2)*sinh(x)^4 + 4*(5*(
3*a^2*b + 2*b^3)*cosh(x)^3 - 3*(3*a^2*b + 2*b^3)*cosh(x))*sinh(x)^3 - 3*a^2*b - 2*b^3 + 3*(3*a^2*b + 2*b^3)*co
sh(x)^2 + 3*(5*(3*a^2*b + 2*b^3)*cosh(x)^4 + 3*a^2*b + 2*b^3 - 6*(3*a^2*b + 2*b^3)*cosh(x)^2)*sinh(x)^2 + 6*((
3*a^2*b + 2*b^3)*cosh(x)^5 - 2*(3*a^2*b + 2*b^3)*cosh(x)^3 + (3*a^2*b + 2*b^3)*cosh(x))*sinh(x))*log(cosh(x) +
sinh(x) - 1) + 6*(5*a^2*b*cosh(x)^4 - 8*(2*a^3 + a*b^2)*cosh(x)^3 - a^2*b + 8*(a^3 + a*b^2)*cosh(x))*sinh(x))
/(a^4*cosh(x)^6 + 6*a^4*cosh(x)*sinh(x)^5 + a^4*sinh(x)^6 - 3*a^4*cosh(x)^4 + 3*a^4*cosh(x)^2 + 3*(5*a^4*cosh(
x)^2 - a^4)*sinh(x)^4 - a^4 + 4*(5*a^4*cosh(x)^3 - 3*a^4*cosh(x))*sinh(x)^3 + 3*(5*a^4*cosh(x)^4 - 6*a^4*cosh(
x)^2 + a^4)*sinh(x)^2 + 6*(a^4*cosh(x)^5 - 2*a^4*cosh(x)^3 + a^4*cosh(x))*sinh(x))

________________________________________________________________________________________

giac [B]  time = 0.46, size = 194, normalized size = 1.80 \[ \frac {{\left (3 \, a^{2} b + 2 \, b^{3}\right )} \log \left (e^{x} + 1\right )}{2 \, a^{4}} - \frac {{\left (3 \, a^{2} b + 2 \, b^{3}\right )} \log \left ({\left | e^{x} - 1 \right |}\right )}{2 \, a^{4}} + \frac {{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \log \left (\frac {{\left | 2 \, b e^{x} + 2 \, a - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, b e^{x} + 2 \, a + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{\sqrt {a^{2} + b^{2}} a^{4}} + \frac {3 \, a b e^{\left (5 \, x\right )} - 12 \, a^{2} e^{\left (4 \, x\right )} - 6 \, b^{2} e^{\left (4 \, x\right )} + 12 \, a^{2} e^{\left (2 \, x\right )} + 12 \, b^{2} e^{\left (2 \, x\right )} - 3 \, a b e^{x} - 8 \, a^{2} - 6 \, b^{2}}{3 \, a^{3} {\left (e^{\left (2 \, x\right )} - 1\right )}^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(3*a^2*b + 2*b^3)*log(e^x + 1)/a^4 - 1/2*(3*a^2*b + 2*b^3)*log(abs(e^x - 1))/a^4 + (a^4 + 2*a^2*b^2 + b^4)
*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^4) + 1/
3*(3*a*b*e^(5*x) - 12*a^2*e^(4*x) - 6*b^2*e^(4*x) + 12*a^2*e^(2*x) + 12*b^2*e^(2*x) - 3*a*b*e^x - 8*a^2 - 6*b^
2)/(a^3*(e^(2*x) - 1)^3)

________________________________________________________________________________________

maple [B]  time = 0.06, size = 232, normalized size = 2.15 \[ -\frac {\tanh ^{3}\left (\frac {x}{2}\right )}{24 a}-\frac {b \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )}{8 a^{2}}-\frac {5 \tanh \left (\frac {x}{2}\right )}{8 a}-\frac {b^{2} \tanh \left (\frac {x}{2}\right )}{2 a^{3}}+\frac {2 \arctanh \left (\frac {2 a \tanh \left (\frac {x}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{\sqrt {a^{2}+b^{2}}}+\frac {4 b^{2} \arctanh \left (\frac {2 a \tanh \left (\frac {x}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{a^{2} \sqrt {a^{2}+b^{2}}}+\frac {2 b^{4} \arctanh \left (\frac {2 a \tanh \left (\frac {x}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{a^{4} \sqrt {a^{2}+b^{2}}}-\frac {1}{24 a \tanh \left (\frac {x}{2}\right )^{3}}-\frac {5}{8 a \tanh \left (\frac {x}{2}\right )}-\frac {b^{2}}{2 a^{3} \tanh \left (\frac {x}{2}\right )}+\frac {b}{8 a^{2} \tanh \left (\frac {x}{2}\right )^{2}}-\frac {3 b \ln \left (\tanh \left (\frac {x}{2}\right )\right )}{2 a^{2}}-\frac {b^{3} \ln \left (\tanh \left (\frac {x}{2}\right )\right )}{a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24/a*tanh(1/2*x)^3-1/8/a^2*b*tanh(1/2*x)^2-5/8/a*tanh(1/2*x)-1/2/a^3*b^2*tanh(1/2*x)+2/(a^2+b^2)^(1/2)*arct
anh(1/2*(2*a*tanh(1/2*x)-2*b)/(a^2+b^2)^(1/2))+4*b^2/a^2/(a^2+b^2)^(1/2)*arctanh(1/2*(2*a*tanh(1/2*x)-2*b)/(a^
2+b^2)^(1/2))+2/a^4*b^4/(a^2+b^2)^(1/2)*arctanh(1/2*(2*a*tanh(1/2*x)-2*b)/(a^2+b^2)^(1/2))-1/24/a/tanh(1/2*x)^
3-5/8/a/tanh(1/2*x)-1/2/a^3/tanh(1/2*x)*b^2+1/8/a^2*b/tanh(1/2*x)^2-3/2/a^2*b*ln(tanh(1/2*x))-1/a^4*b^3*ln(tan
h(1/2*x))

________________________________________________________________________________________

maxima [B]  time = 0.46, size = 212, normalized size = 1.96 \[ -\frac {3 \, a b e^{\left (-x\right )} - 3 \, a b e^{\left (-5 \, x\right )} - 8 \, a^{2} - 6 \, b^{2} + 12 \, {\left (a^{2} + b^{2}\right )} e^{\left (-2 \, x\right )} - 6 \, {\left (2 \, a^{2} + b^{2}\right )} e^{\left (-4 \, x\right )}}{3 \, {\left (3 \, a^{3} e^{\left (-2 \, x\right )} - 3 \, a^{3} e^{\left (-4 \, x\right )} + a^{3} e^{\left (-6 \, x\right )} - a^{3}\right )}} + \frac {{\left (3 \, a^{2} b + 2 \, b^{3}\right )} \log \left (e^{\left (-x\right )} + 1\right )}{2 \, a^{4}} - \frac {{\left (3 \, a^{2} b + 2 \, b^{3}\right )} \log \left (e^{\left (-x\right )} - 1\right )}{2 \, a^{4}} + \frac {{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \log \left (\frac {b e^{\left (-x\right )} - a - \sqrt {a^{2} + b^{2}}}{b e^{\left (-x\right )} - a + \sqrt {a^{2} + b^{2}}}\right )}{\sqrt {a^{2} + b^{2}} a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(3*a*b*e^(-x) - 3*a*b*e^(-5*x) - 8*a^2 - 6*b^2 + 12*(a^2 + b^2)*e^(-2*x) - 6*(2*a^2 + b^2)*e^(-4*x))/(3*a
^3*e^(-2*x) - 3*a^3*e^(-4*x) + a^3*e^(-6*x) - a^3) + 1/2*(3*a^2*b + 2*b^3)*log(e^(-x) + 1)/a^4 - 1/2*(3*a^2*b
+ 2*b^3)*log(e^(-x) - 1)/a^4 + (a^4 + 2*a^2*b^2 + b^4)*log((b*e^(-x) - a - sqrt(a^2 + b^2))/(b*e^(-x) - a + sq
rt(a^2 + b^2)))/(sqrt(a^2 + b^2)*a^4)

________________________________________________________________________________________

mupad [B]  time = 1.41, size = 778, normalized size = 7.20 \[ \frac {\ln \left (-\frac {8\,\left (-30\,{\mathrm {e}}^x\,a^9+18\,a^8\,b-101\,{\mathrm {e}}^x\,a^7\,b^2+60\,a^6\,b^3-126\,{\mathrm {e}}^x\,a^5\,b^4+74\,a^4\,b^5-69\,{\mathrm {e}}^x\,a^3\,b^6+40\,a^2\,b^7-14\,{\mathrm {e}}^x\,a\,b^8+8\,b^9\right )}{a^9\,b^3}-\frac {\sqrt {{\left (a^2+b^2\right )}^3}\,\left (\frac {8\,\left (4\,a^8-36\,{\mathrm {e}}^x\,a^7\,b+34\,a^6\,b^2-75\,{\mathrm {e}}^x\,a^5\,b^3+57\,a^4\,b^4-52\,{\mathrm {e}}^x\,a^3\,b^5+36\,a^2\,b^6-12\,{\mathrm {e}}^x\,a\,b^7+8\,b^8\right )}{a^6\,b^4}-\frac {\left (\frac {16\,\left (-8\,{\mathrm {e}}^x\,a^5+4\,a^4\,b-15\,{\mathrm {e}}^x\,a^3\,b^2+8\,a^2\,b^3-7\,{\mathrm {e}}^x\,a\,b^4+4\,b^5\right )}{a\,b^5}+\frac {32\,\sqrt {{\left (a^2+b^2\right )}^3}\,\left (-4\,{\mathrm {e}}^x\,a^5+3\,a^4\,b-3\,{\mathrm {e}}^x\,a^3\,b^2+2\,a^2\,b^3\right )}{a^4\,b^5}\right )\,\sqrt {{\left (a^2+b^2\right )}^3}}{a^4}\right )}{a^4}\right )\,\sqrt {{\left (a^2+b^2\right )}^3}}{a^4}-\frac {\frac {2\,\left (2\,a^2+b^2\right )}{a^3}-\frac {b\,{\mathrm {e}}^x}{a^2}}{{\mathrm {e}}^{2\,x}-1}-\frac {\frac {4}{a}-\frac {2\,b\,{\mathrm {e}}^x}{a^2}}{{\mathrm {e}}^{4\,x}-2\,{\mathrm {e}}^{2\,x}+1}-\frac {\ln \left (\frac {\sqrt {{\left (a^2+b^2\right )}^3}\,\left (\frac {8\,\left (4\,a^8-36\,{\mathrm {e}}^x\,a^7\,b+34\,a^6\,b^2-75\,{\mathrm {e}}^x\,a^5\,b^3+57\,a^4\,b^4-52\,{\mathrm {e}}^x\,a^3\,b^5+36\,a^2\,b^6-12\,{\mathrm {e}}^x\,a\,b^7+8\,b^8\right )}{a^6\,b^4}+\frac {\left (\frac {16\,\left (-8\,{\mathrm {e}}^x\,a^5+4\,a^4\,b-15\,{\mathrm {e}}^x\,a^3\,b^2+8\,a^2\,b^3-7\,{\mathrm {e}}^x\,a\,b^4+4\,b^5\right )}{a\,b^5}-\frac {32\,\sqrt {{\left (a^2+b^2\right )}^3}\,\left (-4\,{\mathrm {e}}^x\,a^5+3\,a^4\,b-3\,{\mathrm {e}}^x\,a^3\,b^2+2\,a^2\,b^3\right )}{a^4\,b^5}\right )\,\sqrt {{\left (a^2+b^2\right )}^3}}{a^4}\right )}{a^4}-\frac {8\,\left (-30\,{\mathrm {e}}^x\,a^9+18\,a^8\,b-101\,{\mathrm {e}}^x\,a^7\,b^2+60\,a^6\,b^3-126\,{\mathrm {e}}^x\,a^5\,b^4+74\,a^4\,b^5-69\,{\mathrm {e}}^x\,a^3\,b^6+40\,a^2\,b^7-14\,{\mathrm {e}}^x\,a\,b^8+8\,b^9\right )}{a^9\,b^3}\right )\,\sqrt {{\left (a^2+b^2\right )}^3}}{a^4}-\frac {8}{3\,a\,\left (3\,{\mathrm {e}}^{2\,x}-3\,{\mathrm {e}}^{4\,x}+{\mathrm {e}}^{6\,x}-1\right )}-\frac {\ln \left ({\mathrm {e}}^x-1\right )\,\left (3\,a^2\,b+2\,b^3\right )}{2\,a^4}+\frac {\ln \left ({\mathrm {e}}^x+1\right )\,\left (3\,a^2\,b+2\,b^3\right )}{2\,a^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(coth(x)^4/(a + b*sinh(x)),x)

[Out]

(log(- (8*(18*a^8*b + 8*b^9 + 40*a^2*b^7 + 74*a^4*b^5 + 60*a^6*b^3 - 30*a^9*exp(x) - 14*a*b^8*exp(x) - 69*a^3*
b^6*exp(x) - 126*a^5*b^4*exp(x) - 101*a^7*b^2*exp(x)))/(a^9*b^3) - (((a^2 + b^2)^3)^(1/2)*((8*(4*a^8 + 8*b^8 +
36*a^2*b^6 + 57*a^4*b^4 + 34*a^6*b^2 - 12*a*b^7*exp(x) - 36*a^7*b*exp(x) - 52*a^3*b^5*exp(x) - 75*a^5*b^3*exp
(x)))/(a^6*b^4) - (((16*(4*a^4*b + 4*b^5 + 8*a^2*b^3 - 8*a^5*exp(x) - 7*a*b^4*exp(x) - 15*a^3*b^2*exp(x)))/(a*
b^5) + (32*((a^2 + b^2)^3)^(1/2)*(3*a^4*b + 2*a^2*b^3 - 4*a^5*exp(x) - 3*a^3*b^2*exp(x)))/(a^4*b^5))*((a^2 + b
^2)^3)^(1/2))/a^4))/a^4)*((a^2 + b^2)^3)^(1/2))/a^4 - ((2*(2*a^2 + b^2))/a^3 - (b*exp(x))/a^2)/(exp(2*x) - 1)
- (4/a - (2*b*exp(x))/a^2)/(exp(4*x) - 2*exp(2*x) + 1) - (log((((a^2 + b^2)^3)^(1/2)*((8*(4*a^8 + 8*b^8 + 36*a
^2*b^6 + 57*a^4*b^4 + 34*a^6*b^2 - 12*a*b^7*exp(x) - 36*a^7*b*exp(x) - 52*a^3*b^5*exp(x) - 75*a^5*b^3*exp(x)))
/(a^6*b^4) + (((16*(4*a^4*b + 4*b^5 + 8*a^2*b^3 - 8*a^5*exp(x) - 7*a*b^4*exp(x) - 15*a^3*b^2*exp(x)))/(a*b^5)
- (32*((a^2 + b^2)^3)^(1/2)*(3*a^4*b + 2*a^2*b^3 - 4*a^5*exp(x) - 3*a^3*b^2*exp(x)))/(a^4*b^5))*((a^2 + b^2)^3
)^(1/2))/a^4))/a^4 - (8*(18*a^8*b + 8*b^9 + 40*a^2*b^7 + 74*a^4*b^5 + 60*a^6*b^3 - 30*a^9*exp(x) - 14*a*b^8*ex
p(x) - 69*a^3*b^6*exp(x) - 126*a^5*b^4*exp(x) - 101*a^7*b^2*exp(x)))/(a^9*b^3))*((a^2 + b^2)^3)^(1/2))/a^4 - 8
/(3*a*(3*exp(2*x) - 3*exp(4*x) + exp(6*x) - 1)) - (log(exp(x) - 1)*(3*a^2*b + 2*b^3))/(2*a^4) + (log(exp(x) +
1)*(3*a^2*b + 2*b^3))/(2*a^4)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(x)**4/(a+b*sinh(x)),x)

[Out]

Integral(coth(x)**4/(a + b*sinh(x)), x)

________________________________________________________________________________________