Integrand size = 23, antiderivative size = 262 \[ \int \frac {\coth ^3(c+d x)}{\sqrt {a+b \text {sech}(c+d x)}} \, dx=\frac {2 \text {arctanh}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a}}\right )}{\sqrt {a} d}-\frac {\text {arctanh}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a-b}}\right )}{\sqrt {a-b} d}+\frac {b \text {arctanh}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a-b}}\right )}{4 (a-b)^{3/2} d}-\frac {b \text {arctanh}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a+b}}\right )}{4 (a+b)^{3/2} d}-\frac {\text {arctanh}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a+b}}\right )}{\sqrt {a+b} d}-\frac {\sqrt {a+b \text {sech}(c+d x)}}{4 (a+b) d (1-\text {sech}(c+d x))}-\frac {\sqrt {a+b \text {sech}(c+d x)}}{4 (a-b) d (1+\text {sech}(c+d x))} \]
1/4*b*arctanh((a+b*sech(d*x+c))^(1/2)/(a-b)^(1/2))/(a-b)^(3/2)/d-1/4*b*arc tanh((a+b*sech(d*x+c))^(1/2)/(a+b)^(1/2))/(a+b)^(3/2)/d+2*arctanh((a+b*sec h(d*x+c))^(1/2)/a^(1/2))/d/a^(1/2)-arctanh((a+b*sech(d*x+c))^(1/2)/(a-b)^( 1/2))/d/(a-b)^(1/2)-arctanh((a+b*sech(d*x+c))^(1/2)/(a+b)^(1/2))/d/(a+b)^( 1/2)-1/4*(a+b*sech(d*x+c))^(1/2)/(a+b)/d/(1-sech(d*x+c))-1/4*(a+b*sech(d*x +c))^(1/2)/(a-b)/d/(1+sech(d*x+c))
Time = 3.13 (sec) , antiderivative size = 280, normalized size of antiderivative = 1.07 \[ \int \frac {\coth ^3(c+d x)}{\sqrt {a+b \text {sech}(c+d x)}} \, dx=-\frac {-\frac {b^2 \arctan \left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {-a+b}}\right )}{(-a+b)^{3/2}}-\frac {8 b \text {arctanh}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a}}\right )}{\sqrt {a}}+\frac {4 b \text {arctanh}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a-b}}\right )}{\sqrt {a-b}}+\frac {b^2 \text {arctanh}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a+b}}\right )}{(a+b)^{3/2}}-\frac {4 a \text {arctanh}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a+b}}\right )}{\sqrt {a+b}}+4 \sqrt {a+b} \text {arctanh}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a+b}}\right )-\frac {b \sqrt {a+b \text {sech}(c+d x)}}{(a+b) (-1+\text {sech}(c+d x))}+\frac {b \sqrt {a+b \text {sech}(c+d x)}}{(a-b) (1+\text {sech}(c+d x))}}{4 b d} \]
-1/4*(-((b^2*ArcTan[Sqrt[a + b*Sech[c + d*x]]/Sqrt[-a + b]])/(-a + b)^(3/2 )) - (8*b*ArcTanh[Sqrt[a + b*Sech[c + d*x]]/Sqrt[a]])/Sqrt[a] + (4*b*ArcTa nh[Sqrt[a + b*Sech[c + d*x]]/Sqrt[a - b]])/Sqrt[a - b] + (b^2*ArcTanh[Sqrt [a + b*Sech[c + d*x]]/Sqrt[a + b]])/(a + b)^(3/2) - (4*a*ArcTanh[Sqrt[a + b*Sech[c + d*x]]/Sqrt[a + b]])/Sqrt[a + b] + 4*Sqrt[a + b]*ArcTanh[Sqrt[a + b*Sech[c + d*x]]/Sqrt[a + b]] - (b*Sqrt[a + b*Sech[c + d*x]])/((a + b)*( -1 + Sech[c + d*x])) + (b*Sqrt[a + b*Sech[c + d*x]])/((a - b)*(1 + Sech[c + d*x])))/(b*d)
Time = 0.48 (sec) , antiderivative size = 272, normalized size of antiderivative = 1.04, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3042, 26, 4373, 561, 25, 1567, 2009}
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.
\(\displaystyle \int \frac {\coth ^3(c+d x)}{\sqrt {a+b \text {sech}(c+d x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {i}{\cot \left (i c+i d x+\frac {\pi }{2}\right )^3 \sqrt {a+b \csc \left (i c+i d x+\frac {\pi }{2}\right )}}dx\) |
\(\Big \downarrow \) 26 |
\(\displaystyle i \int \frac {1}{\cot \left (\frac {1}{2} (2 i c+\pi )+i d x\right )^3 \sqrt {a+b \csc \left (\frac {1}{2} (2 i c+\pi )+i d x\right )}}dx\) |
\(\Big \downarrow \) 4373 |
\(\displaystyle -\frac {b^4 \int \frac {\cosh (c+d x)}{b \sqrt {a+b \text {sech}(c+d x)} \left (b^2-b^2 \text {sech}^2(c+d x)\right )^2}d(b \text {sech}(c+d x))}{d}\) |
\(\Big \downarrow \) 561 |
\(\displaystyle -\frac {2 b^4 \int -\frac {1}{\left (a-b^2 \text {sech}^2(c+d x)\right ) \left (b^4 \text {sech}^4(c+d x)-2 a b^2 \text {sech}^2(c+d x)+a^2-b^2\right )^2}d\sqrt {a+b \text {sech}(c+d x)}}{d}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {2 b^4 \int \frac {1}{\left (a-b^2 \text {sech}^2(c+d x)\right ) \left (b^4 \text {sech}^4(c+d x)-2 a b^2 \text {sech}^2(c+d x)+a^2-b^2\right )^2}d\sqrt {a+b \text {sech}(c+d x)}}{d}\) |
\(\Big \downarrow \) 1567 |
\(\displaystyle \frac {2 b^4 \int \left (-\frac {1}{2 b^4 \left (-b^2 \text {sech}^2(c+d x)+a+b\right )}+\frac {1}{2 b^4 \left (b^2 \text {sech}^2(c+d x)-a+b\right )}-\frac {1}{4 b^3 \left (-b^2 \text {sech}^2(c+d x)+a+b\right )^2}+\frac {1}{4 b^3 \left (b^2 \text {sech}^2(c+d x)-a+b\right )^2}+\frac {1}{b^4 \left (a-b^2 \text {sech}^2(c+d x)\right )}\right )d\sqrt {a+b \text {sech}(c+d x)}}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {2 b^4 \left (-\frac {\text {arctanh}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a}}\right )}{\sqrt {a} b^4}+\frac {\text {arctanh}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a-b}}\right )}{2 b^4 \sqrt {a-b}}+\frac {\text {arctanh}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a+b}}\right )}{2 b^4 \sqrt {a+b}}-\frac {\text {arctanh}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a-b}}\right )}{8 b^3 (a-b)^{3/2}}+\frac {\text {arctanh}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a+b}}\right )}{8 b^3 (a+b)^{3/2}}-\frac {\text {sech}(c+d x)}{8 b^2 (a-b) \left (a-b^2 \text {sech}^2(c+d x)-b\right )}+\frac {\text {sech}(c+d x)}{8 b^2 (a+b) \left (a-b^2 \text {sech}^2(c+d x)+b\right )}\right )}{d}\) |
(-2*b^4*(-(ArcTanh[Sqrt[a + b*Sech[c + d*x]]/Sqrt[a]]/(Sqrt[a]*b^4)) + Arc Tanh[Sqrt[a + b*Sech[c + d*x]]/Sqrt[a - b]]/(2*Sqrt[a - b]*b^4) - ArcTanh[ Sqrt[a + b*Sech[c + d*x]]/Sqrt[a - b]]/(8*(a - b)^(3/2)*b^3) + ArcTanh[Sqr t[a + b*Sech[c + d*x]]/Sqrt[a + b]]/(8*b^3*(a + b)^(3/2)) + ArcTanh[Sqrt[a + b*Sech[c + d*x]]/Sqrt[a + b]]/(2*b^4*Sqrt[a + b]) - Sech[c + d*x]/(8*(a - b)*b^2*(a - b - b^2*Sech[c + d*x]^2)) + Sech[c + d*x]/(8*b^2*(a + b)*(a + b - b^2*Sech[c + d*x]^2))))/d
3.2.37.3.1 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[(x_)^(m_.)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbo l] :> With[{k = Denominator[n]}, Simp[k/d Subst[Int[x^(k*(n + 1) - 1)*(-c /d + x^k/d)^m*Simp[(b*c^2 + a*d^2)/d^2 - 2*b*c*(x^k/d^2) + b*(x^(2*k)/d^2), x]^p, x], x, (c + d*x)^(1/k)], x]] /; FreeQ[{a, b, c, d, m, p}, x] && Frac tionQ[n] && IntegerQ[p] && IntegerQ[m]
Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x _Symbol] :> Int[ExpandIntegrand[(d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e, p, q}, x] && NeQ[b^2 - 4*a*c, 0] && ((IntegerQ[p] && IntegerQ[q]) || IGtQ[p, 0] || IGtQ[q, 0])
Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n _), x_Symbol] :> Simp[-(-1)^((m - 1)/2)/(d*b^(m - 1)) Subst[Int[(b^2 - x^ 2)^((m - 1)/2)*((a + x)^n/x), x], x, b*Csc[c + d*x]], x] /; FreeQ[{a, b, c, d, n}, x] && IntegerQ[(m - 1)/2] && NeQ[a^2 - b^2, 0]
\[\int \frac {\coth \left (d x +c \right )^{3}}{\sqrt {a +b \,\operatorname {sech}\left (d x +c \right )}}d x\]
Leaf count of result is larger than twice the leaf count of optimal. 1870 vs. \(2 (218) = 436\).
Time = 4.30 (sec) , antiderivative size = 20300, normalized size of antiderivative = 77.48 \[ \int \frac {\coth ^3(c+d x)}{\sqrt {a+b \text {sech}(c+d x)}} \, dx=\text {Too large to display} \]
\[ \int \frac {\coth ^3(c+d x)}{\sqrt {a+b \text {sech}(c+d x)}} \, dx=\int \frac {\coth ^{3}{\left (c + d x \right )}}{\sqrt {a + b \operatorname {sech}{\left (c + d x \right )}}}\, dx \]
\[ \int \frac {\coth ^3(c+d x)}{\sqrt {a+b \text {sech}(c+d x)}} \, dx=\int { \frac {\coth \left (d x + c\right )^{3}}{\sqrt {b \operatorname {sech}\left (d x + c\right ) + a}} \,d x } \]
\[ \int \frac {\coth ^3(c+d x)}{\sqrt {a+b \text {sech}(c+d x)}} \, dx=\int { \frac {\coth \left (d x + c\right )^{3}}{\sqrt {b \operatorname {sech}\left (d x + c\right ) + a}} \,d x } \]
Timed out. \[ \int \frac {\coth ^3(c+d x)}{\sqrt {a+b \text {sech}(c+d x)}} \, dx=\int \frac {{\mathrm {coth}\left (c+d\,x\right )}^3}{\sqrt {a+\frac {b}{\mathrm {cosh}\left (c+d\,x\right )}}} \,d x \]