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]

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

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Rubi [A]  time = 0.220621, antiderivative size = 125, normalized size of antiderivative = 1., 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 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])

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 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 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 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]

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*}

Mathematica [A]  time = 0.909891, 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]

-(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*Sqrt[
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])/(8*(a +
b)^(3/2)*Sqrt[a + 2*b + a*Cosh[2*x]])

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Maple [F]  time = 0.128, size = 0, normalized size = 0. \begin{align*} \int \left ({\rm coth} \left (x\right ) \right ) ^{5}\sqrt{a+b \left ({\rm sech} \left (x\right ) \right ) ^{2}}\, dx \end{align*}

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., size = 0, normalized size = 0. \begin{align*} \int \sqrt{b \operatorname{sech}\left (x\right )^{2} + a} \coth \left (x\right )^{5}\,{d x} \end{align*}

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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Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{b \operatorname{sech}\left (x\right )^{2} + a} \coth \left (x\right )^{5}\,{d x} \end{align*}

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]

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