\(\int \frac {\coth (c+d x)}{\sqrt {a+b \text {sech}(c+d x)}} \, dx\) [136]

Optimal result

Integrand size = 21, antiderivative size = 106 \[ \int \frac {\coth (c+d x)}{\sqrt {a+b \text {sech}(c+d x)}} \, dx=\frac {2 \text {arctanh}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a}}\right )}{\sqrt {a} d}-\frac {\text {arctanh}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a-b}}\right )}{\sqrt {a-b} d}-\frac {\text {arctanh}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a+b}}\right )}{\sqrt {a+b} d} \]

[Out]

2*arctanh((a+b*sech(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)

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {3970, 912, 1184, 212, 213} \[ \int \frac {\coth (c+d x)}{\sqrt {a+b \text {sech}(c+d x)}} \, dx=\frac {2 \text {arctanh}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a}}\right )}{\sqrt {a} d}-\frac {\text {arctanh}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a-b}}\right )}{d \sqrt {a-b}}-\frac {\text {arctanh}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a+b}}\right )}{d \sqrt {a+b}} \]

[In]

Int[Coth[c + d*x]/Sqrt[a + b*Sech[c + d*x]],x]

[Out]

(2*ArcTanh[Sqrt[a + b*Sech[c + d*x]]/Sqrt[a]])/(Sqrt[a]*d) - ArcTanh[Sqrt[a + b*Sech[c + d*x]]/Sqrt[a - b]]/(S
qrt[a - b]*d) - ArcTanh[Sqrt[a + b*Sech[c + d*x]]/Sqrt[a + b]]/(Sqrt[a + b]*d)

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 213

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[b, 2])^(-1))*ArcTanh[Rt[b, 2]*(x/Rt[-a, 2])]
, x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rule 912

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> With[{q = De
nominator[m]}, Dist[q/e, Subst[Int[x^(q*(m + 1) - 1)*((e*f - d*g)/e + g*(x^q/e))^n*((c*d^2 + a*e^2)/e^2 - 2*c*
d*(x^q/e^2) + c*(x^(2*q)/e^2))^p, x], x, (d + e*x)^(1/q)], x]] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*
g, 0] && NeQ[c*d^2 + a*e^2, 0] && IntegersQ[n, p] && FractionQ[m]

Rule 1184

Int[((d_) + (e_.)*(x_)^2)^(q_)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> Int[ExpandIntegrand[(d + e*x
^2)^q/(a + b*x^2 + c*x^4), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a
*e^2, 0] && IntegerQ[q]

Rule 3970

Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Dist[-(-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]

Rubi steps \begin{align*} \text {integral}& = -\frac {b^2 \text {Subst}\left (\int \frac {1}{x \sqrt {a+x} \left (b^2-x^2\right )} \, dx,x,b \text {sech}(c+d x)\right )}{d} \\ & = -\frac {\left (2 b^2\right ) \text {Subst}\left (\int \frac {1}{\left (-a+x^2\right ) \left (-a^2+b^2+2 a x^2-x^4\right )} \, dx,x,\sqrt {a+b \text {sech}(c+d x)}\right )}{d} \\ & = -\frac {\left (2 b^2\right ) \text {Subst}\left (\int \left (-\frac {1}{b^2 \left (a-x^2\right )}+\frac {1}{2 b^2 \left (a+b-x^2\right )}-\frac {1}{2 b^2 \left (-a+b+x^2\right )}\right ) \, dx,x,\sqrt {a+b \text {sech}(c+d x)}\right )}{d} \\ & = -\frac {\text {Subst}\left (\int \frac {1}{a+b-x^2} \, dx,x,\sqrt {a+b \text {sech}(c+d x)}\right )}{d}+\frac {\text {Subst}\left (\int \frac {1}{-a+b+x^2} \, dx,x,\sqrt {a+b \text {sech}(c+d x)}\right )}{d}+\frac {2 \text {Subst}\left (\int \frac {1}{a-x^2} \, dx,x,\sqrt {a+b \text {sech}(c+d x)}\right )}{d} \\ & = \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 {\text {arctanh}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a+b}}\right )}{\sqrt {a+b} d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.44 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.94 \[ \int \frac {\coth (c+d x)}{\sqrt {a+b \text {sech}(c+d x)}} \, dx=-\frac {-\frac {2 \text {arctanh}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a}}\right )}{\sqrt {a}}+\frac {\text {arctanh}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a-b}}\right )}{\sqrt {a-b}}+\frac {\text {arctanh}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a+b}}\right )}{\sqrt {a+b}}}{d} \]

[In]

Integrate[Coth[c + d*x]/Sqrt[a + b*Sech[c + d*x]],x]

[Out]

-(((-2*ArcTanh[Sqrt[a + b*Sech[c + d*x]]/Sqrt[a]])/Sqrt[a] + ArcTanh[Sqrt[a + b*Sech[c + d*x]]/Sqrt[a - b]]/Sq
rt[a - b] + ArcTanh[Sqrt[a + b*Sech[c + d*x]]/Sqrt[a + b]]/Sqrt[a + b])/d)

Maple [F]

\[\int \frac {\coth \left (d x +c \right )}{\sqrt {a +b \,\operatorname {sech}\left (d x +c \right )}}d x\]

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 446 vs. \(2 (88) = 176\).

Time = 0.86 (sec) , antiderivative size = 8908, normalized size of antiderivative = 84.04 \[ \int \frac {\coth (c+d x)}{\sqrt {a+b \text {sech}(c+d x)}} \, dx=\text {Too large to display} \]

[In]

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

[Out]

Too large to include

Sympy [F]

\[ \int \frac {\coth (c+d x)}{\sqrt {a+b \text {sech}(c+d x)}} \, dx=\int \frac {\coth {\left (c + d x \right )}}{\sqrt {a + b \operatorname {sech}{\left (c + d x \right )}}}\, dx \]

[In]

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

[Out]

Integral(coth(c + d*x)/sqrt(a + b*sech(c + d*x)), x)

Maxima [F]

\[ \int \frac {\coth (c+d x)}{\sqrt {a+b \text {sech}(c+d x)}} \, dx=\int { \frac {\coth \left (d x + c\right )}{\sqrt {b \operatorname {sech}\left (d x + c\right ) + a}} \,d x } \]

[In]

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

[Out]

integrate(coth(d*x + c)/sqrt(b*sech(d*x + c) + a), x)

Giac [F]

\[ \int \frac {\coth (c+d x)}{\sqrt {a+b \text {sech}(c+d x)}} \, dx=\int { \frac {\coth \left (d x + c\right )}{\sqrt {b \operatorname {sech}\left (d x + c\right ) + a}} \,d x } \]

[In]

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

[Out]

integrate(coth(d*x + c)/sqrt(b*sech(d*x + c) + a), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\coth (c+d x)}{\sqrt {a+b \text {sech}(c+d x)}} \, dx=\int \frac {\mathrm {coth}\left (c+d\,x\right )}{\sqrt {a+\frac {b}{\mathrm {cosh}\left (c+d\,x\right )}}} \,d x \]

[In]

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

[Out]

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