3.218 \(\int \frac {\coth (x)}{(a+b \text {sech}^2(x))^{5/2}} \, dx\)

Optimal. Leaf size=109 \[ \frac {\tanh ^{-1}\left (\frac {\sqrt {a+b \text {sech}^2(x)}}{\sqrt {a}}\right )}{a^{5/2}}-\frac {b (2 a+b)}{a^2 (a+b)^2 \sqrt {a+b \text {sech}^2(x)}}-\frac {b}{3 a (a+b) \left (a+b \text {sech}^2(x)\right )^{3/2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {a+b \text {sech}^2(x)}}{\sqrt {a+b}}\right )}{(a+b)^{5/2}} \]

[Out]

arctanh((a+b*sech(x)^2)^(1/2)/a^(1/2))/a^(5/2)-arctanh((a+b*sech(x)^2)^(1/2)/(a+b)^(1/2))/(a+b)^(5/2)-1/3*b/a/
(a+b)/(a+b*sech(x)^2)^(3/2)-b*(2*a+b)/a^2/(a+b)^2/(a+b*sech(x)^2)^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.20, antiderivative size = 109, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.467, Rules used = {4139, 446, 85, 152, 156, 63, 208} \[ -\frac {b (2 a+b)}{a^2 (a+b)^2 \sqrt {a+b \text {sech}^2(x)}}+\frac {\tanh ^{-1}\left (\frac {\sqrt {a+b \text {sech}^2(x)}}{\sqrt {a}}\right )}{a^{5/2}}-\frac {b}{3 a (a+b) \left (a+b \text {sech}^2(x)\right )^{3/2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {a+b \text {sech}^2(x)}}{\sqrt {a+b}}\right )}{(a+b)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[Coth[x]/(a + b*Sech[x]^2)^(5/2),x]

[Out]

ArcTanh[Sqrt[a + b*Sech[x]^2]/Sqrt[a]]/a^(5/2) - ArcTanh[Sqrt[a + b*Sech[x]^2]/Sqrt[a + b]]/(a + b)^(5/2) - b/
(3*a*(a + b)*(a + b*Sech[x]^2)^(3/2)) - (b*(2*a + b))/(a^2*(a + b)^2*Sqrt[a + b*Sech[x]^2])

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 85

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

Rule 152

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] && IntegersQ[2*m, 2*n, 2*p]

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

________________________________________________________________________________________

Mathematica [B]  time = 1.11, size = 242, normalized size = 2.22 \[ \frac {\text {sech}^5(x) \left (-\frac {2 b \cosh (x) \left (7 a^2+a (7 a+4 b) \cosh (2 x)+16 a b+6 b^2\right ) (a \cosh (2 x)+a+2 b)}{3 a^2 (a+b)^2}-\frac {(a \cosh (2 x)+a+2 b)^{5/2} \left (\sqrt {a} \left (a^2-2 a b-b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {a+b} \cosh (x)}{\sqrt {a \cosh (2 x)+a+2 b}}\right )+(a+b)^2 \left (\sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {2 a+2 b} \cosh (x)}{\sqrt {a \cosh (2 x)+a+2 b}}\right )-2 \sqrt {a+b} \log \left (\sqrt {a \cosh (2 x)+a+2 b}+\sqrt {2} \sqrt {a} \cosh (x)\right )\right )\right )}{\sqrt {2} a^{5/2} (a+b)^{5/2}}\right )}{8 \left (a+b \text {sech}^2(x)\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Coth[x]/(a + b*Sech[x]^2)^(5/2),x]

[Out]

(((-2*b*Cosh[x]*(a + 2*b + a*Cosh[2*x])*(7*a^2 + 16*a*b + 6*b^2 + a*(7*a + 4*b)*Cosh[2*x]))/(3*a^2*(a + b)^2)
- ((a + 2*b + a*Cosh[2*x])^(5/2)*(Sqrt[a]*(a^2 - 2*a*b - b^2)*ArcTanh[(Sqrt[2]*Sqrt[a + b]*Cosh[x])/Sqrt[a + 2
*b + a*Cosh[2*x]]] + (a + b)^2*(Sqrt[a]*ArcTanh[(Sqrt[2*a + 2*b]*Cosh[x])/Sqrt[a + 2*b + a*Cosh[2*x]]] - 2*Sqr
t[a + b]*Log[Sqrt[2]*Sqrt[a]*Cosh[x] + Sqrt[a + 2*b + a*Cosh[2*x]]])))/(Sqrt[2]*a^(5/2)*(a + b)^(5/2)))*Sech[x
]^5)/(8*(a + b*Sech[x]^2)^(5/2))

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

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

________________________________________________________________________________________

maple [F]  time = 0.38, size = 0, normalized size = 0.00 \[ \int \frac {\coth \relax (x )}{\left (a +b \mathrm {sech}\relax (x )^{2}\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________