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

Optimal. Leaf size=110 \[ \frac {\text {csch}^3(x) (b-a \cosh (x))}{3 \left (a^2-b^2\right )}+\frac {\text {csch}(x) \left (a \left (2 a^2-5 b^2\right ) \cosh (x)+3 b^3\right )}{3 \left (a^2-b^2\right )^2}+\frac {2 b^4 \tanh ^{-1}\left (\frac {\sqrt {a-b} \tanh \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{(a-b)^{5/2} (a+b)^{5/2}} \]

[Out]

2*b^4*arctanh((a-b)^(1/2)*tanh(1/2*x)/(a+b)^(1/2))/(a-b)^(5/2)/(a+b)^(5/2)+1/3*(3*b^3+a*(2*a^2-5*b^2)*cosh(x))
*csch(x)/(a^2-b^2)^2+1/3*(b-a*cosh(x))*csch(x)^3/(a^2-b^2)

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Rubi [A]  time = 0.25, antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {2696, 2866, 12, 2659, 208} \[ \frac {\text {csch}^3(x) (b-a \cosh (x))}{3 \left (a^2-b^2\right )}+\frac {\text {csch}(x) \left (a \left (2 a^2-5 b^2\right ) \cosh (x)+3 b^3\right )}{3 \left (a^2-b^2\right )^2}+\frac {2 b^4 \tanh ^{-1}\left (\frac {\sqrt {a-b} \tanh \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{(a-b)^{5/2} (a+b)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[Csch[x]^4/(a + b*Cosh[x]),x]

[Out]

(2*b^4*ArcTanh[(Sqrt[a - b]*Tanh[x/2])/Sqrt[a + b]])/((a - b)^(5/2)*(a + b)^(5/2)) + ((3*b^3 + a*(2*a^2 - 5*b^
2)*Cosh[x])*Csch[x])/(3*(a^2 - b^2)^2) + ((b - a*Cosh[x])*Csch[x]^3)/(3*(a^2 - b^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 208

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

Rule 2659

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x
]}, Dist[(2*e)/d, Subst[Int[1/(a + b + (a - b)*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 2696

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[((g*Co
s[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m + 1)*(b - a*Sin[e + f*x]))/(f*g*(a^2 - b^2)*(p + 1)), x] + Dist[1/
(g^2*(a^2 - b^2)*(p + 1)), Int[(g*Cos[e + f*x])^(p + 2)*(a + b*Sin[e + f*x])^m*(a^2*(p + 2) - b^2*(m + p + 2)
+ a*b*(m + p + 3)*Sin[e + f*x]), x], x] /; FreeQ[{a, b, e, f, g, m}, x] && NeQ[a^2 - b^2, 0] && LtQ[p, -1] &&
IntegersQ[2*m, 2*p]

Rule 2866

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> Simp[((g*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m + 1)*(b*c - a*d - (a*c -
b*d)*Sin[e + f*x]))/(f*g*(a^2 - b^2)*(p + 1)), x] + Dist[1/(g^2*(a^2 - b^2)*(p + 1)), Int[(g*Cos[e + f*x])^(p
+ 2)*(a + b*Sin[e + f*x])^m*Simp[c*(a^2*(p + 2) - b^2*(m + p + 2)) + a*b*d*m + b*(a*c - b*d)*(m + p + 3)*Sin[e
+ f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && NeQ[a^2 - b^2, 0] && LtQ[p, -1] && IntegerQ[2*m]

Rubi steps

\begin {align*} \int \frac {\text {csch}^4(x)}{a+b \cosh (x)} \, dx &=\frac {(b-a \cosh (x)) \text {csch}^3(x)}{3 \left (a^2-b^2\right )}+\frac {\int \frac {\left (-2 a^2+3 b^2-2 a b \cosh (x)\right ) \text {csch}^2(x)}{a+b \cosh (x)} \, dx}{3 \left (a^2-b^2\right )}\\ &=\frac {\left (3 b^3+a \left (2 a^2-5 b^2\right ) \cosh (x)\right ) \text {csch}(x)}{3 \left (a^2-b^2\right )^2}+\frac {(b-a \cosh (x)) \text {csch}^3(x)}{3 \left (a^2-b^2\right )}+\frac {\int \frac {3 b^4}{a+b \cosh (x)} \, dx}{3 \left (a^2-b^2\right )^2}\\ &=\frac {\left (3 b^3+a \left (2 a^2-5 b^2\right ) \cosh (x)\right ) \text {csch}(x)}{3 \left (a^2-b^2\right )^2}+\frac {(b-a \cosh (x)) \text {csch}^3(x)}{3 \left (a^2-b^2\right )}+\frac {b^4 \int \frac {1}{a+b \cosh (x)} \, dx}{\left (a^2-b^2\right )^2}\\ &=\frac {\left (3 b^3+a \left (2 a^2-5 b^2\right ) \cosh (x)\right ) \text {csch}(x)}{3 \left (a^2-b^2\right )^2}+\frac {(b-a \cosh (x)) \text {csch}^3(x)}{3 \left (a^2-b^2\right )}+\frac {\left (2 b^4\right ) \operatorname {Subst}\left (\int \frac {1}{a+b-(a-b) x^2} \, dx,x,\tanh \left (\frac {x}{2}\right )\right )}{\left (a^2-b^2\right )^2}\\ &=\frac {2 b^4 \tanh ^{-1}\left (\frac {\sqrt {a-b} \tanh \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{(a-b)^{5/2} (a+b)^{5/2}}+\frac {\left (3 b^3+a \left (2 a^2-5 b^2\right ) \cosh (x)\right ) \text {csch}(x)}{3 \left (a^2-b^2\right )^2}+\frac {(b-a \cosh (x)) \text {csch}^3(x)}{3 \left (a^2-b^2\right )}\\ \end {align*}

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Mathematica [A]  time = 0.57, size = 141, normalized size = 1.28 \[ \frac {1}{24} \left (-\frac {48 b^4 \tan ^{-1}\left (\frac {(a-b) \tanh \left (\frac {x}{2}\right )}{\sqrt {b^2-a^2}}\right )}{\left (b^2-a^2\right )^{5/2}}-\frac {14 b \tanh \left (\frac {x}{2}\right )}{(a-b)^2}+\frac {8 a \tanh \left (\frac {x}{2}\right )}{(a-b)^2}+\frac {2 (4 a+7 b) \coth \left (\frac {x}{2}\right )}{(a+b)^2}-\frac {\sinh (x) \text {csch}^4\left (\frac {x}{2}\right )}{2 (a+b)}+\frac {8 \sinh ^4\left (\frac {x}{2}\right ) \text {csch}^3(x)}{a-b}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-48*b^4*ArcTan[((a - b)*Tanh[x/2])/Sqrt[-a^2 + b^2]])/(-a^2 + b^2)^(5/2) + (2*(4*a + 7*b)*Coth[x/2])/(a + b)
^2 + (8*Csch[x]^3*Sinh[x/2]^4)/(a - b) - (Csch[x/2]^4*Sinh[x])/(2*(a + b)) + (8*a*Tanh[x/2])/(a - b)^2 - (14*b
*Tanh[x/2])/(a - b)^2)/24

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fricas [B]  time = 0.50, size = 2339, normalized size = 21.26 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/3*(6*(a^2*b^3 - b^5)*cosh(x)^5 + 6*(a^2*b^3 - b^5)*sinh(x)^5 + 4*a^5 - 14*a^3*b^2 + 10*a*b^4 - 6*(a^3*b^2 -
a*b^4)*cosh(x)^4 - 6*(a^3*b^2 - a*b^4 - 5*(a^2*b^3 - b^5)*cosh(x))*sinh(x)^4 + 4*(2*a^4*b - 7*a^2*b^3 + 5*b^5
)*cosh(x)^3 + 4*(2*a^4*b - 7*a^2*b^3 + 5*b^5 + 15*(a^2*b^3 - b^5)*cosh(x)^2 - 6*(a^3*b^2 - a*b^4)*cosh(x))*sin
h(x)^3 - 12*(a^5 - 3*a^3*b^2 + 2*a*b^4)*cosh(x)^2 - 12*(a^5 - 3*a^3*b^2 + 2*a*b^4 - 5*(a^2*b^3 - b^5)*cosh(x)^
3 + 3*(a^3*b^2 - a*b^4)*cosh(x)^2 - (2*a^4*b - 7*a^2*b^3 + 5*b^5)*cosh(x))*sinh(x)^2 + 3*(b^4*cosh(x)^6 + 6*b^
4*cosh(x)*sinh(x)^5 + b^4*sinh(x)^6 - 3*b^4*cosh(x)^4 + 3*b^4*cosh(x)^2 + 3*(5*b^4*cosh(x)^2 - b^4)*sinh(x)^4
- b^4 + 4*(5*b^4*cosh(x)^3 - 3*b^4*cosh(x))*sinh(x)^3 + 3*(5*b^4*cosh(x)^4 - 6*b^4*cosh(x)^2 + b^4)*sinh(x)^2
+ 6*(b^4*cosh(x)^5 - 2*b^4*cosh(x)^3 + b^4*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)) + 6*(a^2*b^3 - b^5)*cosh(x) + 6*
(a^2*b^3 - b^5 + 5*(a^2*b^3 - b^5)*cosh(x)^4 - 4*(a^3*b^2 - a*b^4)*cosh(x)^3 + 2*(2*a^4*b - 7*a^2*b^3 + 5*b^5)
*cosh(x)^2 - 4*(a^5 - 3*a^3*b^2 + 2*a*b^4)*cosh(x))*sinh(x))/((a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*cosh(x)^6 +
6*(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*cosh(x)*sinh(x)^5 + (a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*sinh(x)^6 - a^6
+ 3*a^4*b^2 - 3*a^2*b^4 + b^6 - 3*(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*cosh(x)^4 - 3*(a^6 - 3*a^4*b^2 + 3*a^2*b
^4 - b^6 - 5*(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*cosh(x)^2)*sinh(x)^4 + 4*(5*(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^
6)*cosh(x)^3 - 3*(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*cosh(x))*sinh(x)^3 + 3*(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6
)*cosh(x)^2 + 3*(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6 + 5*(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*cosh(x)^4 - 6*(a^6
- 3*a^4*b^2 + 3*a^2*b^4 - b^6)*cosh(x)^2)*sinh(x)^2 + 6*((a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*cosh(x)^5 - 2*(a^
6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*cosh(x)^3 + (a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*cosh(x))*sinh(x)), 2/3*(3*(a^
2*b^3 - b^5)*cosh(x)^5 + 3*(a^2*b^3 - b^5)*sinh(x)^5 + 2*a^5 - 7*a^3*b^2 + 5*a*b^4 - 3*(a^3*b^2 - a*b^4)*cosh(
x)^4 - 3*(a^3*b^2 - a*b^4 - 5*(a^2*b^3 - b^5)*cosh(x))*sinh(x)^4 + 2*(2*a^4*b - 7*a^2*b^3 + 5*b^5)*cosh(x)^3 +
2*(2*a^4*b - 7*a^2*b^3 + 5*b^5 + 15*(a^2*b^3 - b^5)*cosh(x)^2 - 6*(a^3*b^2 - a*b^4)*cosh(x))*sinh(x)^3 - 6*(a
^5 - 3*a^3*b^2 + 2*a*b^4)*cosh(x)^2 - 6*(a^5 - 3*a^3*b^2 + 2*a*b^4 - 5*(a^2*b^3 - b^5)*cosh(x)^3 + 3*(a^3*b^2
- a*b^4)*cosh(x)^2 - (2*a^4*b - 7*a^2*b^3 + 5*b^5)*cosh(x))*sinh(x)^2 - 3*(b^4*cosh(x)^6 + 6*b^4*cosh(x)*sinh(
x)^5 + b^4*sinh(x)^6 - 3*b^4*cosh(x)^4 + 3*b^4*cosh(x)^2 + 3*(5*b^4*cosh(x)^2 - b^4)*sinh(x)^4 - b^4 + 4*(5*b^
4*cosh(x)^3 - 3*b^4*cosh(x))*sinh(x)^3 + 3*(5*b^4*cosh(x)^4 - 6*b^4*cosh(x)^2 + b^4)*sinh(x)^2 + 6*(b^4*cosh(x
)^5 - 2*b^4*cosh(x)^3 + b^4*cosh(x))*sinh(x))*sqrt(-a^2 + b^2)*arctan(-sqrt(-a^2 + b^2)*(b*cosh(x) + b*sinh(x)
+ a)/(a^2 - b^2)) + 3*(a^2*b^3 - b^5)*cosh(x) + 3*(a^2*b^3 - b^5 + 5*(a^2*b^3 - b^5)*cosh(x)^4 - 4*(a^3*b^2 -
a*b^4)*cosh(x)^3 + 2*(2*a^4*b - 7*a^2*b^3 + 5*b^5)*cosh(x)^2 - 4*(a^5 - 3*a^3*b^2 + 2*a*b^4)*cosh(x))*sinh(x)
)/((a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*cosh(x)^6 + 6*(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*cosh(x)*sinh(x)^5 + (
a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*sinh(x)^6 - a^6 + 3*a^4*b^2 - 3*a^2*b^4 + b^6 - 3*(a^6 - 3*a^4*b^2 + 3*a^2*
b^4 - b^6)*cosh(x)^4 - 3*(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6 - 5*(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*cosh(x)^2)
*sinh(x)^4 + 4*(5*(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*cosh(x)^3 - 3*(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*cosh(x
))*sinh(x)^3 + 3*(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*cosh(x)^2 + 3*(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6 + 5*(a^6
- 3*a^4*b^2 + 3*a^2*b^4 - b^6)*cosh(x)^4 - 6*(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*cosh(x)^2)*sinh(x)^2 + 6*((a
^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*cosh(x)^5 - 2*(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*cosh(x)^3 + (a^6 - 3*a^4*b
^2 + 3*a^2*b^4 - b^6)*cosh(x))*sinh(x))]

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giac [A]  time = 0.13, size = 156, normalized size = 1.42 \[ \frac {2 \, b^{4} \arctan \left (\frac {b e^{x} + a}{\sqrt {-a^{2} + b^{2}}}\right )}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {-a^{2} + b^{2}}} + \frac {2 \, {\left (3 \, b^{3} e^{\left (5 \, x\right )} - 3 \, a b^{2} e^{\left (4 \, x\right )} + 4 \, a^{2} b e^{\left (3 \, x\right )} - 10 \, b^{3} e^{\left (3 \, x\right )} - 6 \, a^{3} e^{\left (2 \, x\right )} + 12 \, a b^{2} e^{\left (2 \, x\right )} + 3 \, b^{3} e^{x} + 2 \, a^{3} - 5 \, a b^{2}\right )}}{3 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} {\left (e^{\left (2 \, x\right )} - 1\right )}^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*b^4*arctan((b*e^x + a)/sqrt(-a^2 + b^2))/((a^4 - 2*a^2*b^2 + b^4)*sqrt(-a^2 + b^2)) + 2/3*(3*b^3*e^(5*x) - 3
*a*b^2*e^(4*x) + 4*a^2*b*e^(3*x) - 10*b^3*e^(3*x) - 6*a^3*e^(2*x) + 12*a*b^2*e^(2*x) + 3*b^3*e^x + 2*a^3 - 5*a
*b^2)/((a^4 - 2*a^2*b^2 + b^4)*(e^(2*x) - 1)^3)

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maple [A]  time = 0.09, size = 127, normalized size = 1.15 \[ -\frac {\frac {a \left (\tanh ^{3}\left (\frac {x}{2}\right )\right )}{3}-\frac {\left (\tanh ^{3}\left (\frac {x}{2}\right )\right ) b}{3}-3 a \tanh \left (\frac {x}{2}\right )+5 \tanh \left (\frac {x}{2}\right ) b}{8 \left (a -b \right )^{2}}+\frac {2 b^{4} \arctanh \left (\frac {\left (a -b \right ) \tanh \left (\frac {x}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a -b \right )^{2} \left (a +b \right )^{2} \sqrt {\left (a +b \right ) \left (a -b \right )}}-\frac {1}{24 \left (a +b \right ) \tanh \left (\frac {x}{2}\right )^{3}}-\frac {-3 a -5 b}{8 \left (a +b \right )^{2} \tanh \left (\frac {x}{2}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8/(a-b)^2*(1/3*a*tanh(1/2*x)^3-1/3*tanh(1/2*x)^3*b-3*a*tanh(1/2*x)+5*tanh(1/2*x)*b)+2/(a-b)^2/(a+b)^2*b^4/(
(a+b)*(a-b))^(1/2)*arctanh((a-b)*tanh(1/2*x)/((a+b)*(a-b))^(1/2))-1/24/(a+b)/tanh(1/2*x)^3-1/8/(a+b)^2*(-3*a-5
*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)^4/(a+b*cosh(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*a^2-4*b^2>0)', see `assume?`
for more details)Is 4*a^2-4*b^2 positive or negative?

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mupad [B]  time = 1.97, size = 642, normalized size = 5.84 \[ \frac {\frac {4\,\left (a\,b^2-a^3\right )}{{\left (a^2-b^2\right )}^2}+\frac {8\,{\mathrm {e}}^x\,\left (a^2\,b-b^3\right )}{3\,{\left (a^2-b^2\right )}^2}}{{\mathrm {e}}^{4\,x}-2\,{\mathrm {e}}^{2\,x}+1}-\frac {\frac {2\,a\,b^2}{{\left (a^2-b^2\right )}^2}-\frac {2\,b^3\,{\mathrm {e}}^x}{{\left (a^2-b^2\right )}^2}}{{\mathrm {e}}^{2\,x}-1}-\frac {\frac {8\,a}{3\,\left (a^2-b^2\right )}-\frac {8\,b\,{\mathrm {e}}^x}{3\,\left (a^2-b^2\right )}}{3\,{\mathrm {e}}^{2\,x}-3\,{\mathrm {e}}^{4\,x}+{\mathrm {e}}^{6\,x}-1}+\frac {2\,\mathrm {atan}\left (\left ({\mathrm {e}}^x\,\left (\frac {2\,b^2}{{\left (a^2-b^2\right )}^2\,\sqrt {b^8}\,\left (a^4-2\,a^2\,b^2+b^4\right )}+\frac {2\,a\,\left (a^5\,\sqrt {b^8}-2\,a^3\,b^2\,\sqrt {b^8}+a\,b^4\,\sqrt {b^8}\right )}{b^6\,\sqrt {-{\left (a^2-b^2\right )}^5}\,\left (a^4-2\,a^2\,b^2+b^4\right )\,\sqrt {-a^{10}+5\,a^8\,b^2-10\,a^6\,b^4+10\,a^4\,b^6-5\,a^2\,b^8+b^{10}}}\right )+\frac {2\,a\,\left (b^5\,\sqrt {b^8}-2\,a^2\,b^3\,\sqrt {b^8}+a^4\,b\,\sqrt {b^8}\right )}{b^6\,\sqrt {-{\left (a^2-b^2\right )}^5}\,\left (a^4-2\,a^2\,b^2+b^4\right )\,\sqrt {-a^{10}+5\,a^8\,b^2-10\,a^6\,b^4+10\,a^4\,b^6-5\,a^2\,b^8+b^{10}}}\right )\,\left (\frac {b^5\,\sqrt {-a^{10}+5\,a^8\,b^2-10\,a^6\,b^4+10\,a^4\,b^6-5\,a^2\,b^8+b^{10}}}{2}-a^2\,b^3\,\sqrt {-a^{10}+5\,a^8\,b^2-10\,a^6\,b^4+10\,a^4\,b^6-5\,a^2\,b^8+b^{10}}+\frac {a^4\,b\,\sqrt {-a^{10}+5\,a^8\,b^2-10\,a^6\,b^4+10\,a^4\,b^6-5\,a^2\,b^8+b^{10}}}{2}\right )\right )\,\sqrt {b^8}}{\sqrt {-a^{10}+5\,a^8\,b^2-10\,a^6\,b^4+10\,a^4\,b^6-5\,a^2\,b^8+b^{10}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sinh(x)^4*(a + b*cosh(x))),x)

[Out]

((4*(a*b^2 - a^3))/(a^2 - b^2)^2 + (8*exp(x)*(a^2*b - b^3))/(3*(a^2 - b^2)^2))/(exp(4*x) - 2*exp(2*x) + 1) - (
(2*a*b^2)/(a^2 - b^2)^2 - (2*b^3*exp(x))/(a^2 - b^2)^2)/(exp(2*x) - 1) - ((8*a)/(3*(a^2 - b^2)) - (8*b*exp(x))
/(3*(a^2 - b^2)))/(3*exp(2*x) - 3*exp(4*x) + exp(6*x) - 1) + (2*atan((exp(x)*((2*b^2)/((a^2 - b^2)^2*(b^8)^(1/
2)*(a^4 + b^4 - 2*a^2*b^2)) + (2*a*(a^5*(b^8)^(1/2) - 2*a^3*b^2*(b^8)^(1/2) + a*b^4*(b^8)^(1/2)))/(b^6*(-(a^2
- b^2)^5)^(1/2)*(a^4 + b^4 - 2*a^2*b^2)*(b^10 - a^10 - 5*a^2*b^8 + 10*a^4*b^6 - 10*a^6*b^4 + 5*a^8*b^2)^(1/2))
) + (2*a*(b^5*(b^8)^(1/2) - 2*a^2*b^3*(b^8)^(1/2) + a^4*b*(b^8)^(1/2)))/(b^6*(-(a^2 - b^2)^5)^(1/2)*(a^4 + b^4
- 2*a^2*b^2)*(b^10 - a^10 - 5*a^2*b^8 + 10*a^4*b^6 - 10*a^6*b^4 + 5*a^8*b^2)^(1/2)))*((b^5*(b^10 - a^10 - 5*a
^2*b^8 + 10*a^4*b^6 - 10*a^6*b^4 + 5*a^8*b^2)^(1/2))/2 - a^2*b^3*(b^10 - a^10 - 5*a^2*b^8 + 10*a^4*b^6 - 10*a^
6*b^4 + 5*a^8*b^2)^(1/2) + (a^4*b*(b^10 - a^10 - 5*a^2*b^8 + 10*a^4*b^6 - 10*a^6*b^4 + 5*a^8*b^2)^(1/2))/2))*(
b^8)^(1/2))/(b^10 - a^10 - 5*a^2*b^8 + 10*a^4*b^6 - 10*a^6*b^4 + 5*a^8*b^2)^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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