3.186 \(\int \coth ^5(x) \sqrt {a+b \text {sech}^2(x)} \, dx\)

Optimal. Leaf size=125 \[ -\frac {\left (8 a^2+12 a b+3 b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a+b \text {sech}^2(x)}}{\sqrt {a+b}}\right )}{8 (a+b)^{3/2}}-\frac {1}{4} \coth ^4(x) \sqrt {a+b \text {sech}^2(x)}-\frac {(4 a+3 b) \coth ^2(x) \sqrt {a+b \text {sech}^2(x)}}{8 (a+b)}+\sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a+b \text {sech}^2(x)}}{\sqrt {a}}\right ) \]

[Out]

-1/8*(8*a^2+12*a*b+3*b^2)*arctanh((a+b*sech(x)^2)^(1/2)/(a+b)^(1/2))/(a+b)^(3/2)+arctanh((a+b*sech(x)^2)^(1/2)
/a^(1/2))*a^(1/2)-1/8*(4*a+3*b)*coth(x)^2*(a+b*sech(x)^2)^(1/2)/(a+b)-1/4*coth(x)^4*(a+b*sech(x)^2)^(1/2)

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Rubi [A]  time = 0.22, antiderivative size = 125, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.412, Rules used = {4139, 446, 99, 151, 156, 63, 208} \[ -\frac {\left (8 a^2+12 a b+3 b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a+b \text {sech}^2(x)}}{\sqrt {a+b}}\right )}{8 (a+b)^{3/2}}-\frac {1}{4} \coth ^4(x) \sqrt {a+b \text {sech}^2(x)}-\frac {(4 a+3 b) \coth ^2(x) \sqrt {a+b \text {sech}^2(x)}}{8 (a+b)}+\sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a+b \text {sech}^2(x)}}{\sqrt {a}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[Coth[x]^5*Sqrt[a + b*Sech[x]^2],x]

[Out]

Sqrt[a]*ArcTanh[Sqrt[a + b*Sech[x]^2]/Sqrt[a]] - ((8*a^2 + 12*a*b + 3*b^2)*ArcTanh[Sqrt[a + b*Sech[x]^2]/Sqrt[
a + b]])/(8*(a + b)^(3/2)) - ((4*a + 3*b)*Coth[x]^2*Sqrt[a + b*Sech[x]^2])/(8*(a + b)) - (Coth[x]^4*Sqrt[a + b
*Sech[x]^2])/4

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 99

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((a + b
*x)^(m + 1)*(c + d*x)^n*(e + f*x)^(p + 1))/((m + 1)*(b*e - a*f)), x] - Dist[1/((m + 1)*(b*e - a*f)), Int[(a +
b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p*Simp[d*e*n + c*f*(m + p + 2) + d*f*(m + n + p + 2)*x, x], x], x] /;
FreeQ[{a, b, c, d, e, f, p}, x] && LtQ[m, -1] && GtQ[n, 0] && (IntegersQ[2*m, 2*n, 2*p] || IntegersQ[m, n + p
] || IntegersQ[p, m + n])

Rule 151

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[((b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*
f)), x] + Dist[1/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[(a*d*f*
g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a*h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p
+ 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && LtQ[m, -1] && IntegerQ[m]

Rule 156

Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :>
Dist[(b*g - a*h)/(b*c - a*d), Int[(e + f*x)^p/(a + b*x), x], x] - Dist[(d*g - c*h)/(b*c - a*d), Int[(e + f*x)
^p/(c + d*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, 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 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 4139

Int[((a_) + (b_.)*((c_.)*sec[(e_.) + (f_.)*(x_)])^(n_))^(p_.)*tan[(e_.) + (f_.)*(x_)]^(m_.), x_Symbol] :> With
[{ff = FreeFactors[Sec[e + f*x], x]}, Dist[1/f, Subst[Int[((-1 + ff^2*x^2)^((m - 1)/2)*(a + b*(c*ff*x)^n)^p)/x
, x], x, Sec[e + f*x]/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[(m - 1)/2] && (GtQ[m, 0] || EqQ[
n, 2] || EqQ[n, 4] || IGtQ[p, 0] || IntegersQ[2*n, p])

Rubi steps

\begin {align*} \int \coth ^5(x) \sqrt {a+b \text {sech}^2(x)} \, dx &=\operatorname {Subst}\left (\int \frac {\sqrt {a+b x^2}}{x \left (-1+x^2\right )^3} \, dx,x,\text {sech}(x)\right )\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {\sqrt {a+b x}}{(-1+x)^3 x} \, dx,x,\text {sech}^2(x)\right )\\ &=-\frac {1}{4} \coth ^4(x) \sqrt {a+b \text {sech}^2(x)}+\frac {1}{4} \operatorname {Subst}\left (\int \frac {-2 a-\frac {3 b x}{2}}{(-1+x)^2 x \sqrt {a+b x}} \, dx,x,\text {sech}^2(x)\right )\\ &=-\frac {(4 a+3 b) \coth ^2(x) \sqrt {a+b \text {sech}^2(x)}}{8 (a+b)}-\frac {1}{4} \coth ^4(x) \sqrt {a+b \text {sech}^2(x)}-\frac {\operatorname {Subst}\left (\int \frac {-2 a (a+b)-\frac {1}{4} b (4 a+3 b) x}{(-1+x) x \sqrt {a+b x}} \, dx,x,\text {sech}^2(x)\right )}{4 (a+b)}\\ &=-\frac {(4 a+3 b) \coth ^2(x) \sqrt {a+b \text {sech}^2(x)}}{8 (a+b)}-\frac {1}{4} \coth ^4(x) \sqrt {a+b \text {sech}^2(x)}-\frac {1}{2} a \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,\text {sech}^2(x)\right )+\frac {\left (8 a^2+12 a b+3 b^2\right ) \operatorname {Subst}\left (\int \frac {1}{(-1+x) \sqrt {a+b x}} \, dx,x,\text {sech}^2(x)\right )}{16 (a+b)}\\ &=-\frac {(4 a+3 b) \coth ^2(x) \sqrt {a+b \text {sech}^2(x)}}{8 (a+b)}-\frac {1}{4} \coth ^4(x) \sqrt {a+b \text {sech}^2(x)}-\frac {a \operatorname {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b \text {sech}^2(x)}\right )}{b}+\frac {\left (8 a^2+12 a b+3 b^2\right ) \operatorname {Subst}\left (\int \frac {1}{-1-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b \text {sech}^2(x)}\right )}{8 b (a+b)}\\ &=\sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a+b \text {sech}^2(x)}}{\sqrt {a}}\right )-\frac {\left (8 a^2+12 a b+3 b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a+b \text {sech}^2(x)}}{\sqrt {a+b}}\right )}{8 (a+b)^{3/2}}-\frac {(4 a+3 b) \coth ^2(x) \sqrt {a+b \text {sech}^2(x)}}{8 (a+b)}-\frac {1}{4} \coth ^4(x) \sqrt {a+b \text {sech}^2(x)}\\ \end {align*}

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Mathematica [A]  time = 1.02, size = 191, normalized size = 1.53 \[ -\frac {\cosh (x) \sqrt {a+b \text {sech}^2(x)} \left (\sqrt {2} \left (8 a^2+12 a b+3 b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {a+b} \cosh (x)}{\sqrt {a \cosh (2 x)+a+2 b}}\right )+\sqrt {a+b} \left (\frac {1}{2} \coth (x) \text {csch}^3(x) \sqrt {a \cosh (2 x)+a+2 b} ((6 a+5 b) \cosh (2 x)-2 a-b)-8 \sqrt {2} \sqrt {a} (a+b) \log \left (\sqrt {a \cosh (2 x)+a+2 b}+\sqrt {2} \sqrt {a} \cosh (x)\right )\right )\right )}{8 (a+b)^{3/2} \sqrt {a \cosh (2 x)+a+2 b}} \]

Antiderivative was successfully verified.

[In]

Integrate[Coth[x]^5*Sqrt[a + b*Sech[x]^2],x]

[Out]

-1/8*(Cosh[x]*(Sqrt[2]*(8*a^2 + 12*a*b + 3*b^2)*ArcTanh[(Sqrt[2]*Sqrt[a + b]*Cosh[x])/Sqrt[a + 2*b + a*Cosh[2*
x]]] + Sqrt[a + b]*((Sqrt[a + 2*b + a*Cosh[2*x]]*(-2*a - b + (6*a + 5*b)*Cosh[2*x])*Coth[x]*Csch[x]^3)/2 - 8*S
qrt[2]*Sqrt[a]*(a + b)*Log[Sqrt[2]*Sqrt[a]*Cosh[x] + Sqrt[a + 2*b + a*Cosh[2*x]]]))*Sqrt[a + b*Sech[x]^2])/((a
+ b)^(3/2)*Sqrt[a + 2*b + a*Cosh[2*x]])

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(x)^5*(a+b*sech(x)^2)^(1/2),x, algorithm="fricas")

[Out]

Timed out

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(x)^5*(a+b*sech(x)^2)^(1/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Erro
r: Bad Argument Type

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maple [F]  time = 0.42, size = 0, normalized size = 0.00 \[ \int \left (\coth ^{5}\relax (x )\right ) \sqrt {a +b \mathrm {sech}\relax (x )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(coth(x)^5*(a+b*sech(x)^2)^(1/2),x)

[Out]

int(coth(x)^5*(a+b*sech(x)^2)^(1/2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {b \operatorname {sech}\relax (x)^{2} + a} \coth \relax (x)^{5}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(x)^5*(a+b*sech(x)^2)^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(b*sech(x)^2 + a)*coth(x)^5, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\mathrm {coth}\relax (x)}^5\,\sqrt {a+\frac {b}{{\mathrm {cosh}\relax (x)}^2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(coth(x)^5*(a + b/cosh(x)^2)^(1/2),x)

[Out]

int(coth(x)^5*(a + b/cosh(x)^2)^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {a + b \operatorname {sech}^{2}{\relax (x )}} \coth ^{5}{\relax (x )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(x)**5*(a+b*sech(x)**2)**(1/2),x)

[Out]

Integral(sqrt(a + b*sech(x)**2)*coth(x)**5, x)

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