3.123 \(\int \frac {\coth ^4(x)}{a+b \text {sech}(x)} \, dx\)

Optimal. Leaf size=207 \[ -\frac {a b^2 x}{\left (a^2-b^2\right )^2}+\frac {a x}{a^2-b^2}-\frac {a \coth ^3(x)}{3 \left (a^2-b^2\right )}+\frac {a b^2 \coth (x)}{\left (a^2-b^2\right )^2}-\frac {a \coth (x)}{a^2-b^2}+\frac {b \text {csch}^3(x)}{3 \left (a^2-b^2\right )}+\frac {b \text {csch}(x)}{a^2-b^2}+\frac {b^4 x}{a \left (a^2-b^2\right )^2}-\frac {b^3 \text {csch}(x)}{\left (a^2-b^2\right )^2}-\frac {2 b^5 \tan ^{-1}\left (\frac {\sqrt {a-b} \tanh \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{a (a-b)^{5/2} (a+b)^{5/2}} \]

[Out]

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

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Rubi [A]  time = 0.33, antiderivative size = 207, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 8, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.615, Rules used = {3898, 2902, 2606, 3473, 8, 2735, 2659, 205} \[ \frac {b^4 x}{a \left (a^2-b^2\right )^2}-\frac {a b^2 x}{\left (a^2-b^2\right )^2}+\frac {a x}{a^2-b^2}-\frac {a \coth ^3(x)}{3 \left (a^2-b^2\right )}+\frac {a b^2 \coth (x)}{\left (a^2-b^2\right )^2}-\frac {a \coth (x)}{a^2-b^2}+\frac {b \text {csch}^3(x)}{3 \left (a^2-b^2\right )}-\frac {b^3 \text {csch}(x)}{\left (a^2-b^2\right )^2}+\frac {b \text {csch}(x)}{a^2-b^2}-\frac {2 b^5 \tan ^{-1}\left (\frac {\sqrt {a-b} \tanh \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{a (a-b)^{5/2} (a+b)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[Coth[x]^4/(a + b*Sech[x]),x]

[Out]

-((a*b^2*x)/(a^2 - b^2)^2) + (b^4*x)/(a*(a^2 - b^2)^2) + (a*x)/(a^2 - b^2) - (2*b^5*ArcTan[(Sqrt[a - b]*Tanh[x
/2])/Sqrt[a + b]])/(a*(a - b)^(5/2)*(a + b)^(5/2)) + (a*b^2*Coth[x])/(a^2 - b^2)^2 - (a*Coth[x])/(a^2 - b^2) -
(a*Coth[x]^3)/(3*(a^2 - b^2)) - (b^3*Csch[x])/(a^2 - b^2)^2 + (b*Csch[x])/(a^2 - b^2) + (b*Csch[x]^3)/(3*(a^2
- b^2))

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 205

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

Rule 2606

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

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 2735

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

Rule 2902

Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_))/((a_) + (b_.)*sin[(e_.) + (f_.
)*(x_)]), x_Symbol] :> Dist[(a*d^2)/(a^2 - b^2), Int[(g*Cos[e + f*x])^p*(d*Sin[e + f*x])^(n - 2), x], x] + (-D
ist[(b*d)/(a^2 - b^2), Int[(g*Cos[e + f*x])^p*(d*Sin[e + f*x])^(n - 1), x], x] - Dist[(a^2*d^2)/(g^2*(a^2 - b^
2)), Int[((g*Cos[e + f*x])^(p + 2)*(d*Sin[e + f*x])^(n - 2))/(a + b*Sin[e + f*x]), x], x]) /; FreeQ[{a, b, d,
e, f, g}, x] && NeQ[a^2 - b^2, 0] && IntegersQ[2*n, 2*p] && LtQ[p, -1] && GtQ[n, 1]

Rule 3473

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(b*Tan[c + d*x])^(n - 1))/(d*(n - 1)), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]

Rule 3898

Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Int[(Cos[c + d*x]^
m*(b + a*Sin[c + d*x])^n)/Sin[c + d*x]^(m + n), x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && IntegerQ[
n] && IntegerQ[m] && (IntegerQ[m/2] || LeQ[m, 1])

Rubi steps

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

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Mathematica [A]  time = 0.78, size = 166, normalized size = 0.80 \[ \frac {\text {sech}(x) (a \cosh (x)+b) \left (\frac {48 b^5 \tan ^{-1}\left (\frac {(b-a) \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{a \left (a^2-b^2\right )^{5/2}}+\frac {22 b \tanh \left (\frac {x}{2}\right )}{(a-b)^2}-\frac {16 a \tanh \left (\frac {x}{2}\right )}{(a-b)^2}-\frac {2 (8 a+11 b) \coth \left (\frac {x}{2}\right )}{(a+b)^2}-\frac {\sinh (x) \text {csch}^4\left (\frac {x}{2}\right )}{2 (a+b)}+\frac {8 \sinh ^4\left (\frac {x}{2}\right ) \text {csch}^3(x)}{a-b}+\frac {24 x}{a}\right )}{24 (a+b \text {sech}(x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((b + a*Cosh[x])*Sech[x]*((24*x)/a + (48*b^5*ArcTan[((-a + b)*Tanh[x/2])/Sqrt[a^2 - b^2]])/(a*(a^2 - b^2)^(5/2
)) - (2*(8*a + 11*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)) - (16*a*Tanh[x/2])/(a - b)^2 + (22*b*Tanh[x/2])/(a - b)^2))/(24*(a + b*Sech[x]))

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fricas [B]  time = 0.49, size = 3530, normalized size = 17.05 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/3*(3*(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*x*cosh(x)^6 + 3*(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*x*sinh(x)^6 -
8*a^6 + 22*a^4*b^2 - 14*a^2*b^4 + 6*(a^5*b - 3*a^3*b^3 + 2*a*b^5)*cosh(x)^5 + 6*(a^5*b - 3*a^3*b^3 + 2*a*b^5
+ 3*(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*x*cosh(x))*sinh(x)^5 - 3*(4*a^6 - 10*a^4*b^2 + 6*a^2*b^4 + 3*(a^6 - 3*
a^4*b^2 + 3*a^2*b^4 - b^6)*x)*cosh(x)^4 - 3*(4*a^6 - 10*a^4*b^2 + 6*a^2*b^4 - 15*(a^6 - 3*a^4*b^2 + 3*a^2*b^4
- b^6)*x*cosh(x)^2 + 3*(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*x - 10*(a^5*b - 3*a^3*b^3 + 2*a*b^5)*cosh(x))*sinh(
x)^4 - 4*(a^5*b - 5*a^3*b^3 + 4*a*b^5)*cosh(x)^3 - 4*(a^5*b - 5*a^3*b^3 + 4*a*b^5 - 15*(a^6 - 3*a^4*b^2 + 3*a^
2*b^4 - b^6)*x*cosh(x)^3 - 15*(a^5*b - 3*a^3*b^3 + 2*a*b^5)*cosh(x)^2 + 3*(4*a^6 - 10*a^4*b^2 + 6*a^2*b^4 + 3*
(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*x)*cosh(x))*sinh(x)^3 + 3*(4*a^6 - 12*a^4*b^2 + 8*a^2*b^4 + 3*(a^6 - 3*a^4
*b^2 + 3*a^2*b^4 - b^6)*x)*cosh(x)^2 + 3*(4*a^6 - 12*a^4*b^2 + 8*a^2*b^4 + 15*(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b
^6)*x*cosh(x)^4 + 20*(a^5*b - 3*a^3*b^3 + 2*a*b^5)*cosh(x)^3 - 6*(4*a^6 - 10*a^4*b^2 + 6*a^2*b^4 + 3*(a^6 - 3*
a^4*b^2 + 3*a^2*b^4 - b^6)*x)*cosh(x)^2 + 3*(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*x - 4*(a^5*b - 5*a^3*b^3 + 4*a
*b^5)*cosh(x))*sinh(x)^2 - 3*(b^5*cosh(x)^6 + 6*b^5*cosh(x)*sinh(x)^5 + b^5*sinh(x)^6 - 3*b^5*cosh(x)^4 + 3*b^
5*cosh(x)^2 - b^5 + 3*(5*b^5*cosh(x)^2 - b^5)*sinh(x)^4 + 4*(5*b^5*cosh(x)^3 - 3*b^5*cosh(x))*sinh(x)^3 + 3*(5
*b^5*cosh(x)^4 - 6*b^5*cosh(x)^2 + b^5)*sinh(x)^2 + 6*(b^5*cosh(x)^5 - 2*b^5*cosh(x)^3 + b^5*cosh(x))*sinh(x))
*sqrt(-a^2 + b^2)*log((a^2*cosh(x)^2 + a^2*sinh(x)^2 + 2*a*b*cosh(x) - a^2 + 2*b^2 + 2*(a^2*cosh(x) + a*b)*sin
h(x) + 2*sqrt(-a^2 + b^2)*(a*cosh(x) + a*sinh(x) + b))/(a*cosh(x)^2 + a*sinh(x)^2 + 2*b*cosh(x) + 2*(a*cosh(x)
+ b)*sinh(x) + a)) - 3*(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*x + 6*(a^5*b - 3*a^3*b^3 + 2*a*b^5)*cosh(x) + 6*(3
*(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*x*cosh(x)^5 + a^5*b - 3*a^3*b^3 + 2*a*b^5 + 5*(a^5*b - 3*a^3*b^3 + 2*a*b^
5)*cosh(x)^4 - 2*(4*a^6 - 10*a^4*b^2 + 6*a^2*b^4 + 3*(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*x)*cosh(x)^3 - 2*(a^5
*b - 5*a^3*b^3 + 4*a*b^5)*cosh(x)^2 + (4*a^6 - 12*a^4*b^2 + 8*a^2*b^4 + 3*(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*
x)*cosh(x))*sinh(x))/(a^7 - 3*a^5*b^2 + 3*a^3*b^4 - a*b^6 - (a^7 - 3*a^5*b^2 + 3*a^3*b^4 - a*b^6)*cosh(x)^6 -
6*(a^7 - 3*a^5*b^2 + 3*a^3*b^4 - a*b^6)*cosh(x)*sinh(x)^5 - (a^7 - 3*a^5*b^2 + 3*a^3*b^4 - a*b^6)*sinh(x)^6 +
3*(a^7 - 3*a^5*b^2 + 3*a^3*b^4 - a*b^6)*cosh(x)^4 + 3*(a^7 - 3*a^5*b^2 + 3*a^3*b^4 - a*b^6 - 5*(a^7 - 3*a^5*b^
2 + 3*a^3*b^4 - a*b^6)*cosh(x)^2)*sinh(x)^4 - 4*(5*(a^7 - 3*a^5*b^2 + 3*a^3*b^4 - a*b^6)*cosh(x)^3 - 3*(a^7 -
3*a^5*b^2 + 3*a^3*b^4 - a*b^6)*cosh(x))*sinh(x)^3 - 3*(a^7 - 3*a^5*b^2 + 3*a^3*b^4 - a*b^6)*cosh(x)^2 - 3*(a^7
- 3*a^5*b^2 + 3*a^3*b^4 - a*b^6 + 5*(a^7 - 3*a^5*b^2 + 3*a^3*b^4 - a*b^6)*cosh(x)^4 - 6*(a^7 - 3*a^5*b^2 + 3*
a^3*b^4 - a*b^6)*cosh(x)^2)*sinh(x)^2 - 6*((a^7 - 3*a^5*b^2 + 3*a^3*b^4 - a*b^6)*cosh(x)^5 - 2*(a^7 - 3*a^5*b^
2 + 3*a^3*b^4 - a*b^6)*cosh(x)^3 + (a^7 - 3*a^5*b^2 + 3*a^3*b^4 - a*b^6)*cosh(x))*sinh(x)), -1/3*(3*(a^6 - 3*a
^4*b^2 + 3*a^2*b^4 - b^6)*x*cosh(x)^6 + 3*(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*x*sinh(x)^6 - 8*a^6 + 22*a^4*b^2
- 14*a^2*b^4 + 6*(a^5*b - 3*a^3*b^3 + 2*a*b^5)*cosh(x)^5 + 6*(a^5*b - 3*a^3*b^3 + 2*a*b^5 + 3*(a^6 - 3*a^4*b^
2 + 3*a^2*b^4 - b^6)*x*cosh(x))*sinh(x)^5 - 3*(4*a^6 - 10*a^4*b^2 + 6*a^2*b^4 + 3*(a^6 - 3*a^4*b^2 + 3*a^2*b^4
- b^6)*x)*cosh(x)^4 - 3*(4*a^6 - 10*a^4*b^2 + 6*a^2*b^4 - 15*(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*x*cosh(x)^2
+ 3*(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*x - 10*(a^5*b - 3*a^3*b^3 + 2*a*b^5)*cosh(x))*sinh(x)^4 - 4*(a^5*b - 5
*a^3*b^3 + 4*a*b^5)*cosh(x)^3 - 4*(a^5*b - 5*a^3*b^3 + 4*a*b^5 - 15*(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*x*cosh
(x)^3 - 15*(a^5*b - 3*a^3*b^3 + 2*a*b^5)*cosh(x)^2 + 3*(4*a^6 - 10*a^4*b^2 + 6*a^2*b^4 + 3*(a^6 - 3*a^4*b^2 +
3*a^2*b^4 - b^6)*x)*cosh(x))*sinh(x)^3 + 3*(4*a^6 - 12*a^4*b^2 + 8*a^2*b^4 + 3*(a^6 - 3*a^4*b^2 + 3*a^2*b^4 -
b^6)*x)*cosh(x)^2 + 3*(4*a^6 - 12*a^4*b^2 + 8*a^2*b^4 + 15*(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*x*cosh(x)^4 + 2
0*(a^5*b - 3*a^3*b^3 + 2*a*b^5)*cosh(x)^3 - 6*(4*a^6 - 10*a^4*b^2 + 6*a^2*b^4 + 3*(a^6 - 3*a^4*b^2 + 3*a^2*b^4
- b^6)*x)*cosh(x)^2 + 3*(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*x - 4*(a^5*b - 5*a^3*b^3 + 4*a*b^5)*cosh(x))*sinh
(x)^2 + 6*(b^5*cosh(x)^6 + 6*b^5*cosh(x)*sinh(x)^5 + b^5*sinh(x)^6 - 3*b^5*cosh(x)^4 + 3*b^5*cosh(x)^2 - b^5 +
3*(5*b^5*cosh(x)^2 - b^5)*sinh(x)^4 + 4*(5*b^5*cosh(x)^3 - 3*b^5*cosh(x))*sinh(x)^3 + 3*(5*b^5*cosh(x)^4 - 6*
b^5*cosh(x)^2 + b^5)*sinh(x)^2 + 6*(b^5*cosh(x)^5 - 2*b^5*cosh(x)^3 + b^5*cosh(x))*sinh(x))*sqrt(a^2 - b^2)*ar
ctan(-(a*cosh(x) + a*sinh(x) + b)/sqrt(a^2 - b^2)) - 3*(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*x + 6*(a^5*b - 3*a^
3*b^3 + 2*a*b^5)*cosh(x) + 6*(3*(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*x*cosh(x)^5 + a^5*b - 3*a^3*b^3 + 2*a*b^5
+ 5*(a^5*b - 3*a^3*b^3 + 2*a*b^5)*cosh(x)^4 - 2*(4*a^6 - 10*a^4*b^2 + 6*a^2*b^4 + 3*(a^6 - 3*a^4*b^2 + 3*a^2*b
^4 - b^6)*x)*cosh(x)^3 - 2*(a^5*b - 5*a^3*b^3 + 4*a*b^5)*cosh(x)^2 + (4*a^6 - 12*a^4*b^2 + 8*a^2*b^4 + 3*(a^6
- 3*a^4*b^2 + 3*a^2*b^4 - b^6)*x)*cosh(x))*sinh(x))/(a^7 - 3*a^5*b^2 + 3*a^3*b^4 - a*b^6 - (a^7 - 3*a^5*b^2 +
3*a^3*b^4 - a*b^6)*cosh(x)^6 - 6*(a^7 - 3*a^5*b^2 + 3*a^3*b^4 - a*b^6)*cosh(x)*sinh(x)^5 - (a^7 - 3*a^5*b^2 +
3*a^3*b^4 - a*b^6)*sinh(x)^6 + 3*(a^7 - 3*a^5*b^2 + 3*a^3*b^4 - a*b^6)*cosh(x)^4 + 3*(a^7 - 3*a^5*b^2 + 3*a^3*
b^4 - a*b^6 - 5*(a^7 - 3*a^5*b^2 + 3*a^3*b^4 - a*b^6)*cosh(x)^2)*sinh(x)^4 - 4*(5*(a^7 - 3*a^5*b^2 + 3*a^3*b^4
- a*b^6)*cosh(x)^3 - 3*(a^7 - 3*a^5*b^2 + 3*a^3*b^4 - a*b^6)*cosh(x))*sinh(x)^3 - 3*(a^7 - 3*a^5*b^2 + 3*a^3*
b^4 - a*b^6)*cosh(x)^2 - 3*(a^7 - 3*a^5*b^2 + 3*a^3*b^4 - a*b^6 + 5*(a^7 - 3*a^5*b^2 + 3*a^3*b^4 - a*b^6)*cosh
(x)^4 - 6*(a^7 - 3*a^5*b^2 + 3*a^3*b^4 - a*b^6)*cosh(x)^2)*sinh(x)^2 - 6*((a^7 - 3*a^5*b^2 + 3*a^3*b^4 - a*b^6
)*cosh(x)^5 - 2*(a^7 - 3*a^5*b^2 + 3*a^3*b^4 - a*b^6)*cosh(x)^3 + (a^7 - 3*a^5*b^2 + 3*a^3*b^4 - a*b^6)*cosh(x
))*sinh(x))]

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giac [A]  time = 0.14, size = 190, normalized size = 0.92 \[ -\frac {2 \, b^{5} \arctan \left (\frac {a e^{x} + b}{\sqrt {a^{2} - b^{2}}}\right )}{{\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} \sqrt {a^{2} - b^{2}}} + \frac {x}{a} + \frac {2 \, {\left (3 \, a^{2} b e^{\left (5 \, x\right )} - 6 \, b^{3} e^{\left (5 \, x\right )} - 6 \, a^{3} e^{\left (4 \, x\right )} + 9 \, a b^{2} e^{\left (4 \, x\right )} - 2 \, a^{2} b e^{\left (3 \, x\right )} + 8 \, b^{3} e^{\left (3 \, x\right )} + 6 \, a^{3} e^{\left (2 \, x\right )} - 12 \, a b^{2} e^{\left (2 \, x\right )} + 3 \, a^{2} b e^{x} - 6 \, b^{3} e^{x} - 4 \, a^{3} + 7 \, a b^{2}\right )}}{3 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} {\left (e^{\left (2 \, x\right )} - 1\right )}^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*b^5*arctan((a*e^x + b)/sqrt(a^2 - b^2))/((a^5 - 2*a^3*b^2 + a*b^4)*sqrt(a^2 - b^2)) + x/a + 2/3*(3*a^2*b*e^
(5*x) - 6*b^3*e^(5*x) - 6*a^3*e^(4*x) + 9*a*b^2*e^(4*x) - 2*a^2*b*e^(3*x) + 8*b^3*e^(3*x) + 6*a^3*e^(2*x) - 12
*a*b^2*e^(2*x) + 3*a^2*b*e^x - 6*b^3*e^x - 4*a^3 + 7*a*b^2)/((a^4 - 2*a^2*b^2 + b^4)*(e^(2*x) - 1)^3)

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maple [A]  time = 0.18, size = 179, normalized size = 0.86 \[ -\frac {a \left (\tanh ^{3}\left (\frac {x}{2}\right )\right )}{24 \left (a -b \right )^{2}}+\frac {\left (\tanh ^{3}\left (\frac {x}{2}\right )\right ) b}{24 \left (a -b \right )^{2}}-\frac {5 a \tanh \left (\frac {x}{2}\right )}{8 \left (a -b \right )^{2}}+\frac {7 \tanh \left (\frac {x}{2}\right ) b}{8 \left (a -b \right )^{2}}-\frac {\ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{a}-\frac {2 b^{5} \arctan \left (\frac {\left (a -b \right ) \tanh \left (\frac {x}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a -b \right )^{2} \left (a +b \right )^{2} a \sqrt {\left (a +b \right ) \left (a -b \right )}}+\frac {\ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{a}-\frac {1}{24 \left (a +b \right ) \tanh \left (\frac {x}{2}\right )^{3}}-\frac {5 a}{8 \left (a +b \right )^{2} \tanh \left (\frac {x}{2}\right )}-\frac {7 b}{8 \left (a +b \right )^{2} \tanh \left (\frac {x}{2}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24/(a-b)^2*a*tanh(1/2*x)^3+1/24/(a-b)^2*tanh(1/2*x)^3*b-5/8/(a-b)^2*a*tanh(1/2*x)+7/8/(a-b)^2*tanh(1/2*x)*b
-1/a*ln(tanh(1/2*x)-1)-2/(a-b)^2/(a+b)^2/a*b^5/((a+b)*(a-b))^(1/2)*arctan((a-b)*tanh(1/2*x)/((a+b)*(a-b))^(1/2
))+1/a*ln(tanh(1/2*x)+1)-1/24/(a+b)/tanh(1/2*x)^3-5/8/(a+b)^2/tanh(1/2*x)*a-7/8/(a+b)^2/tanh(1/2*x)*b

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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(coth(x)^4/(a+b*sech(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 = 1.83, size = 713, normalized size = 3.44 \[ \frac {x}{a}-\frac {\frac {8\,a}{3\,\left (a^2-b^2\right )}-\frac {8\,b\,{\mathrm {e}}^x}{3\,\left (a^2-b^2\right )}}{3\,{\mathrm {e}}^{2\,x}-3\,{\mathrm {e}}^{4\,x}+{\mathrm {e}}^{6\,x}-1}-\frac {\frac {2\,\left (2\,a^4-3\,a^2\,b^2\right )}{a\,{\left (a^2-b^2\right )}^2}-\frac {2\,{\mathrm {e}}^x\,\left (a^2\,b-2\,b^3\right )}{{\left (a^2-b^2\right )}^2}}{{\mathrm {e}}^{2\,x}-1}-\frac {\frac {4\,\left (a^4-a^2\,b^2\right )}{a\,{\left (a^2-b^2\right )}^2}-\frac {8\,{\mathrm {e}}^x\,\left (a^2\,b-b^3\right )}{3\,{\left (a^2-b^2\right )}^2}}{{\mathrm {e}}^{4\,x}-2\,{\mathrm {e}}^{2\,x}+1}-\frac {2\,\mathrm {atan}\left (\left ({\mathrm {e}}^x\,\left (\frac {2\,b^5}{a^3\,{\left (a^2-b^2\right )}^2\,\sqrt {b^{10}}\,\left (a^5-2\,a^3\,b^2+a\,b^4\right )}+\frac {2\,\left (a\,b^5\,\sqrt {b^{10}}-2\,a^3\,b^3\,\sqrt {b^{10}}+a^5\,b\,\sqrt {b^{10}}\right )}{a^2\,b^4\,\sqrt {a^2\,{\left (a^2-b^2\right )}^5}\,\left (a^5-2\,a^3\,b^2+a\,b^4\right )\,\sqrt {a^{12}-5\,a^{10}\,b^2+10\,a^8\,b^4-10\,a^6\,b^6+5\,a^4\,b^8-a^2\,b^{10}}}\right )+\frac {2\,\left (a^6\,\sqrt {b^{10}}+a^2\,b^4\,\sqrt {b^{10}}-2\,a^4\,b^2\,\sqrt {b^{10}}\right )}{a^2\,b^4\,\sqrt {a^2\,{\left (a^2-b^2\right )}^5}\,\left (a^5-2\,a^3\,b^2+a\,b^4\right )\,\sqrt {a^{12}-5\,a^{10}\,b^2+10\,a^8\,b^4-10\,a^6\,b^6+5\,a^4\,b^8-a^2\,b^{10}}}\right )\,\left (\frac {a^6\,\sqrt {a^{12}-5\,a^{10}\,b^2+10\,a^8\,b^4-10\,a^6\,b^6+5\,a^4\,b^8-a^2\,b^{10}}}{2}+\frac {a^2\,b^4\,\sqrt {a^{12}-5\,a^{10}\,b^2+10\,a^8\,b^4-10\,a^6\,b^6+5\,a^4\,b^8-a^2\,b^{10}}}{2}-a^4\,b^2\,\sqrt {a^{12}-5\,a^{10}\,b^2+10\,a^8\,b^4-10\,a^6\,b^6+5\,a^4\,b^8-a^2\,b^{10}}\right )\right )\,\sqrt {b^{10}}}{\sqrt {a^{12}-5\,a^{10}\,b^2+10\,a^8\,b^4-10\,a^6\,b^6+5\,a^4\,b^8-a^2\,b^{10}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(coth(x)^4/(a + b/cosh(x)),x)

[Out]

x/a - ((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*(2
*a^4 - 3*a^2*b^2))/(a*(a^2 - b^2)^2) - (2*exp(x)*(a^2*b - 2*b^3))/(a^2 - b^2)^2)/(exp(2*x) - 1) - ((4*(a^4 - a
^2*b^2))/(a*(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*atan
((exp(x)*((2*b^5)/(a^3*(a^2 - b^2)^2*(b^10)^(1/2)*(a*b^4 + a^5 - 2*a^3*b^2)) + (2*(a*b^5*(b^10)^(1/2) - 2*a^3*
b^3*(b^10)^(1/2) + a^5*b*(b^10)^(1/2)))/(a^2*b^4*(a^2*(a^2 - b^2)^5)^(1/2)*(a*b^4 + a^5 - 2*a^3*b^2)*(a^12 - a
^2*b^10 + 5*a^4*b^8 - 10*a^6*b^6 + 10*a^8*b^4 - 5*a^10*b^2)^(1/2))) + (2*(a^6*(b^10)^(1/2) + a^2*b^4*(b^10)^(1
/2) - 2*a^4*b^2*(b^10)^(1/2)))/(a^2*b^4*(a^2*(a^2 - b^2)^5)^(1/2)*(a*b^4 + a^5 - 2*a^3*b^2)*(a^12 - a^2*b^10 +
5*a^4*b^8 - 10*a^6*b^6 + 10*a^8*b^4 - 5*a^10*b^2)^(1/2)))*((a^6*(a^12 - a^2*b^10 + 5*a^4*b^8 - 10*a^6*b^6 + 1
0*a^8*b^4 - 5*a^10*b^2)^(1/2))/2 + (a^2*b^4*(a^12 - a^2*b^10 + 5*a^4*b^8 - 10*a^6*b^6 + 10*a^8*b^4 - 5*a^10*b^
2)^(1/2))/2 - a^4*b^2*(a^12 - a^2*b^10 + 5*a^4*b^8 - 10*a^6*b^6 + 10*a^8*b^4 - 5*a^10*b^2)^(1/2)))*(b^10)^(1/2
))/(a^12 - a^2*b^10 + 5*a^4*b^8 - 10*a^6*b^6 + 10*a^8*b^4 - 5*a^10*b^2)^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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