Integrand size = 15, antiderivative size = 109 \[ \int \frac {\coth (x)}{\left (a+b \text {sech}^2(x)\right )^{5/2}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {a+b \text {sech}^2(x)}}{\sqrt {a}}\right )}{a^{5/2}}-\frac {\text {arctanh}\left (\frac {\sqrt {a+b \text {sech}^2(x)}}{\sqrt {a+b}}\right )}{(a+b)^{5/2}}-\frac {b}{3 a (a+b) \left (a+b \text {sech}^2(x)\right )^{3/2}}-\frac {b (2 a+b)}{a^2 (a+b)^2 \sqrt {a+b \text {sech}^2(x)}} \]
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)
Leaf count is larger than twice the leaf count of optimal. \(242\) vs. \(2(109)=218\).
Time = 0.96 (sec) , antiderivative size = 242, normalized size of antiderivative = 2.22 \[ \int \frac {\coth (x)}{\left (a+b \text {sech}^2(x)\right )^{5/2}} \, dx=\frac {\left (-\frac {2 b \cosh (x) (a+2 b+a \cosh (2 x)) \left (7 a^2+16 a b+6 b^2+a (7 a+4 b) \cosh (2 x)\right )}{3 a^2 (a+b)^2}-\frac {(a+2 b+a \cosh (2 x))^{5/2} \left (\sqrt {a} \left (a^2-2 a b-b^2\right ) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {a+b} \cosh (x)}{\sqrt {a+2 b+a \cosh (2 x)}}\right )+(a+b)^2 \left (\sqrt {a} \text {arctanh}\left (\frac {\sqrt {2 a+2 b} \cosh (x)}{\sqrt {a+2 b+a \cosh (2 x)}}\right )-2 \sqrt {a+b} \log \left (\sqrt {2} \sqrt {a} \cosh (x)+\sqrt {a+2 b+a \cosh (2 x)}\right )\right )\right )}{\sqrt {2} a^{5/2} (a+b)^{5/2}}\right ) \text {sech}^5(x)}{8 \left (a+b \text {sech}^2(x)\right )^{5/2}} \]
(((-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])/Sq rt[a + 2*b + a*Cosh[2*x]]] - 2*Sqrt[a + b]*Log[Sqrt[2]*Sqrt[a]*Cosh[x] + S qrt[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))
Time = 0.37 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.31, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.800, Rules used = {3042, 26, 4627, 25, 354, 96, 25, 169, 27, 174, 73, 221}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\coth (x)}{\left (a+b \text {sech}^2(x)\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {i}{\tan (i x) \left (a+b \sec (i x)^2\right )^{5/2}}dx\) |
\(\Big \downarrow \) 26 |
\(\displaystyle i \int \frac {1}{\left (b \sec (i x)^2+a\right )^{5/2} \tan (i x)}dx\) |
\(\Big \downarrow \) 4627 |
\(\displaystyle \int -\frac {\cosh (x)}{\left (1-\text {sech}^2(x)\right ) \left (a+b \text {sech}^2(x)\right )^{5/2}}d\text {sech}(x)\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \frac {\cosh (x)}{\left (1-\text {sech}^2(x)\right ) \left (b \text {sech}^2(x)+a\right )^{5/2}}d\text {sech}(x)\) |
\(\Big \downarrow \) 354 |
\(\displaystyle -\frac {1}{2} \int \frac {\cosh (x)}{\left (1-\text {sech}^2(x)\right ) \left (b \text {sech}^2(x)+a\right )^{5/2}}d\text {sech}^2(x)\) |
\(\Big \downarrow \) 96 |
\(\displaystyle \frac {1}{2} \left (\frac {\int -\frac {\cosh (x) \left (-b \text {sech}^2(x)+a+b\right )}{\left (1-\text {sech}^2(x)\right ) \left (b \text {sech}^2(x)+a\right )^{3/2}}d\text {sech}^2(x)}{a (a+b)}-\frac {2 b}{3 a (a+b) \left (a+b \text {sech}^2(x)\right )^{3/2}}\right )\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{2} \left (-\frac {\int \frac {\cosh (x) \left (-b \text {sech}^2(x)+a+b\right )}{\left (1-\text {sech}^2(x)\right ) \left (b \text {sech}^2(x)+a\right )^{3/2}}d\text {sech}^2(x)}{a (a+b)}-\frac {2 b}{3 a (a+b) \left (a+b \text {sech}^2(x)\right )^{3/2}}\right )\) |
\(\Big \downarrow \) 169 |
\(\displaystyle \frac {1}{2} \left (-\frac {\frac {2 \int \frac {\cosh (x) \left ((a+b)^2-b (2 a+b) \text {sech}^2(x)\right )}{2 \left (1-\text {sech}^2(x)\right ) \sqrt {b \text {sech}^2(x)+a}}d\text {sech}^2(x)}{a (a+b)}+\frac {2 b (2 a+b)}{a (a+b) \sqrt {a+b \text {sech}^2(x)}}}{a (a+b)}-\frac {2 b}{3 a (a+b) \left (a+b \text {sech}^2(x)\right )^{3/2}}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{2} \left (-\frac {\frac {\int \frac {\cosh (x) \left ((a+b)^2-b (2 a+b) \text {sech}^2(x)\right )}{\left (1-\text {sech}^2(x)\right ) \sqrt {b \text {sech}^2(x)+a}}d\text {sech}^2(x)}{a (a+b)}+\frac {2 b (2 a+b)}{a (a+b) \sqrt {a+b \text {sech}^2(x)}}}{a (a+b)}-\frac {2 b}{3 a (a+b) \left (a+b \text {sech}^2(x)\right )^{3/2}}\right )\) |
\(\Big \downarrow \) 174 |
\(\displaystyle \frac {1}{2} \left (-\frac {\frac {a^2 \int \frac {1}{\left (1-\text {sech}^2(x)\right ) \sqrt {b \text {sech}^2(x)+a}}d\text {sech}^2(x)+(a+b)^2 \int \frac {\cosh (x)}{\sqrt {b \text {sech}^2(x)+a}}d\text {sech}^2(x)}{a (a+b)}+\frac {2 b (2 a+b)}{a (a+b) \sqrt {a+b \text {sech}^2(x)}}}{a (a+b)}-\frac {2 b}{3 a (a+b) \left (a+b \text {sech}^2(x)\right )^{3/2}}\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{2} \left (-\frac {\frac {\frac {2 a^2 \int \frac {1}{\frac {a+b}{b}-\frac {\text {sech}^4(x)}{b}}d\sqrt {b \text {sech}^2(x)+a}}{b}+\frac {2 (a+b)^2 \int \frac {1}{\frac {\text {sech}^4(x)}{b}-\frac {a}{b}}d\sqrt {b \text {sech}^2(x)+a}}{b}}{a (a+b)}+\frac {2 b (2 a+b)}{a (a+b) \sqrt {a+b \text {sech}^2(x)}}}{a (a+b)}-\frac {2 b}{3 a (a+b) \left (a+b \text {sech}^2(x)\right )^{3/2}}\right )\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {1}{2} \left (-\frac {\frac {\frac {2 a^2 \text {arctanh}\left (\frac {\sqrt {a+b \text {sech}^2(x)}}{\sqrt {a+b}}\right )}{\sqrt {a+b}}-\frac {2 (a+b)^2 \text {arctanh}\left (\frac {\sqrt {a+b \text {sech}^2(x)}}{\sqrt {a}}\right )}{\sqrt {a}}}{a (a+b)}+\frac {2 b (2 a+b)}{a (a+b) \sqrt {a+b \text {sech}^2(x)}}}{a (a+b)}-\frac {2 b}{3 a (a+b) \left (a+b \text {sech}^2(x)\right )^{3/2}}\right )\) |
((-2*b)/(3*a*(a + b)*(a + b*Sech[x]^2)^(3/2)) - (((-2*(a + b)^2*ArcTanh[Sq rt[a + b*Sech[x]^2]/Sqrt[a]])/Sqrt[a] + (2*a^2*ArcTanh[Sqrt[a + b*Sech[x]^ 2]/Sqrt[a + b]])/Sqrt[a + b])/(a*(a + b)) + (2*b*(2*a + b))/(a*(a + b)*Sqr t[a + b*Sech[x]^2]))/(a*(a + b)))/2
3.3.18.3.1 Defintions of rubi rules used
Int[(Complex[0, a_])*(Fx_), x_Symbol] :> Simp[(Complex[Identity[0], a]) I nt[Fx, x], x] /; FreeQ[a, x] && EqQ[a^2, 1]
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[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] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((e_.) + (f_.)*(x_))^(p_)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_] :> Simp[f*((e + f*x)^(p + 1)/((p + 1)*(b*e - a*f)*(d*e - c*f))), x] + S imp[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]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_)*((g_.) + (h_.)*(x_)), x_] :> 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] + S imp[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]
Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))* ((c_.) + (d_.)*(x_))), x_] :> Simp[(b*g - a*h)/(b*c - a*d) Int[(e + f*x)^ p/(a + b*x), x], x] - Simp[(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]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_S ymbol] :> Simp[1/2 Subst[Int[x^((m - 1)/2)*(a + b*x)^p*(c + d*x)^q, x], x , x^2], x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IntegerQ [(m - 1)/2]
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]}, Si mp[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] || Integers Q[2*n, p])
\[\int \frac {\coth \left (x \right )}{\left (a +\operatorname {sech}\left (x \right )^{2} b \right )^{\frac {5}{2}}}d x\]
Leaf count of result is larger than twice the leaf count of optimal. 4236 vs. \(2 (91) = 182\).
Time = 0.92 (sec) , antiderivative size = 18563, normalized size of antiderivative = 170.30 \[ \int \frac {\coth (x)}{\left (a+b \text {sech}^2(x)\right )^{5/2}} \, dx=\text {Too large to display} \]
Timed out. \[ \int \frac {\coth (x)}{\left (a+b \text {sech}^2(x)\right )^{5/2}} \, dx=\text {Timed out} \]
\[ \int \frac {\coth (x)}{\left (a+b \text {sech}^2(x)\right )^{5/2}} \, dx=\int { \frac {\coth \left (x\right )}{{\left (b \operatorname {sech}\left (x\right )^{2} + a\right )}^{\frac {5}{2}}} \,d x } \]
Exception generated. \[ \int \frac {\coth (x)}{\left (a+b \text {sech}^2(x)\right )^{5/2}} \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Error: Bad Argument Type
Timed out. \[ \int \frac {\coth (x)}{\left (a+b \text {sech}^2(x)\right )^{5/2}} \, dx=\int \frac {\mathrm {coth}\left (x\right )}{{\left (a+\frac {b}{{\mathrm {cosh}\left (x\right )}^2}\right )}^{5/2}} \,d x \]