Integrand size = 15, antiderivative size = 56 \[ \int \frac {\coth (x)}{\sqrt {a+b \tanh ^2(x)}} \, dx=-\frac {\text {arctanh}\left (\frac {\sqrt {a+b \tanh ^2(x)}}{\sqrt {a}}\right )}{\sqrt {a}}+\frac {\text {arctanh}\left (\frac {\sqrt {a+b \tanh ^2(x)}}{\sqrt {a+b}}\right )}{\sqrt {a+b}} \] Output:
-arctanh((a+b*tanh(x)^2)^(1/2)/a^(1/2))/a^(1/2)+arctanh((a+b*tanh(x)^2)^(1 /2)/(a+b)^(1/2))/(a+b)^(1/2)
Time = 0.03 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.00 \[ \int \frac {\coth (x)}{\sqrt {a+b \tanh ^2(x)}} \, dx=-\frac {\text {arctanh}\left (\frac {\sqrt {a+b \tanh ^2(x)}}{\sqrt {a}}\right )}{\sqrt {a}}+\frac {\text {arctanh}\left (\frac {\sqrt {a+b \tanh ^2(x)}}{\sqrt {a+b}}\right )}{\sqrt {a+b}} \] Input:
Integrate[Coth[x]/Sqrt[a + b*Tanh[x]^2],x]
Output:
-(ArcTanh[Sqrt[a + b*Tanh[x]^2]/Sqrt[a]]/Sqrt[a]) + ArcTanh[Sqrt[a + b*Tan h[x]^2]/Sqrt[a + b]]/Sqrt[a + b]
Time = 0.29 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.09, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.533, Rules used = {3042, 26, 4153, 26, 354, 97, 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.
\(\displaystyle \int \frac {\coth (x)}{\sqrt {a+b \tanh ^2(x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {i}{\tan (i x) \sqrt {a-b \tan (i x)^2}}dx\) |
\(\Big \downarrow \) 26 |
\(\displaystyle i \int \frac {1}{\tan (i x) \sqrt {a-b \tan (i x)^2}}dx\) |
\(\Big \downarrow \) 4153 |
\(\displaystyle i \int -\frac {i \coth (x)}{\left (1-\tanh ^2(x)\right ) \sqrt {b \tanh ^2(x)+a}}d\tanh (x)\) |
\(\Big \downarrow \) 26 |
\(\displaystyle \int \frac {\coth (x)}{\left (1-\tanh ^2(x)\right ) \sqrt {a+b \tanh ^2(x)}}d\tanh (x)\) |
\(\Big \downarrow \) 354 |
\(\displaystyle \frac {1}{2} \int \frac {\coth (x)}{\left (1-\tanh ^2(x)\right ) \sqrt {b \tanh ^2(x)+a}}d\tanh ^2(x)\) |
\(\Big \downarrow \) 97 |
\(\displaystyle \frac {1}{2} \left (\int \frac {1}{\left (1-\tanh ^2(x)\right ) \sqrt {b \tanh ^2(x)+a}}d\tanh ^2(x)+\int \frac {\coth (x)}{\sqrt {b \tanh ^2(x)+a}}d\tanh ^2(x)\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{2} \left (\frac {2 \int \frac {1}{\frac {a+b}{b}-\frac {\tanh ^4(x)}{b}}d\sqrt {b \tanh ^2(x)+a}}{b}+\frac {2 \int \frac {1}{\frac {\tanh ^4(x)}{b}-\frac {a}{b}}d\sqrt {b \tanh ^2(x)+a}}{b}\right )\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {1}{2} \left (\frac {2 \text {arctanh}\left (\frac {\sqrt {a+b \tanh ^2(x)}}{\sqrt {a+b}}\right )}{\sqrt {a+b}}-\frac {2 \text {arctanh}\left (\frac {\sqrt {a+b \tanh ^2(x)}}{\sqrt {a}}\right )}{\sqrt {a}}\right )\) |
Input:
Int[Coth[x]/Sqrt[a + b*Tanh[x]^2],x]
Output:
((-2*ArcTanh[Sqrt[a + b*Tanh[x]^2]/Sqrt[a]])/Sqrt[a] + (2*ArcTanh[Sqrt[a + b*Tanh[x]^2]/Sqrt[a + b]])/Sqrt[a + b])/2
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_.) + (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[b/(b*c - a*d) Int[(e + f*x)^p/(a + b*x), x], x] - Simp[d/(b*c - a*d) Int[(e + f*x)^p/(c + d*x), x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && !IntegerQ[p]
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[((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[c*(ff/f) Subst[Int[(d*ff*(x/c))^m*((a + b*(ff*x)^n)^p/(c^2 + f f^2*x^2)), x], x, c*(Tan[e + f*x]/ff)], x]] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && (IGtQ[p, 0] || EqQ[n, 2] || EqQ[n, 4] || (IntegerQ[p] && Ratio nalQ[n]))
\[\int \frac {\coth \left (x \right )}{\sqrt {a +b \tanh \left (x \right )^{2}}}d x\]
Input:
int(coth(x)/(a+b*tanh(x)^2)^(1/2),x)
Output:
int(coth(x)/(a+b*tanh(x)^2)^(1/2),x)
Leaf count of result is larger than twice the leaf count of optimal. 488 vs. \(2 (44) = 88\).
Time = 0.22 (sec) , antiderivative size = 3371, normalized size of antiderivative = 60.20 \[ \int \frac {\coth (x)}{\sqrt {a+b \tanh ^2(x)}} \, dx=\text {Too large to display} \] Input:
integrate(coth(x)/(a+b*tanh(x)^2)^(1/2),x, algorithm="fricas")
Output:
Too large to include
\[ \int \frac {\coth (x)}{\sqrt {a+b \tanh ^2(x)}} \, dx=\int \frac {\coth {\left (x \right )}}{\sqrt {a + b \tanh ^{2}{\left (x \right )}}}\, dx \] Input:
integrate(coth(x)/(a+b*tanh(x)**2)**(1/2),x)
Output:
Integral(coth(x)/sqrt(a + b*tanh(x)**2), x)
\[ \int \frac {\coth (x)}{\sqrt {a+b \tanh ^2(x)}} \, dx=\int { \frac {\coth \left (x\right )}{\sqrt {b \tanh \left (x\right )^{2} + a}} \,d x } \] Input:
integrate(coth(x)/(a+b*tanh(x)^2)^(1/2),x, algorithm="maxima")
Output:
integrate(coth(x)/sqrt(b*tanh(x)^2 + a), x)
Leaf count of result is larger than twice the leaf count of optimal. 254 vs. \(2 (44) = 88\).
Time = 0.32 (sec) , antiderivative size = 254, normalized size of antiderivative = 4.54 \[ \int \frac {\coth (x)}{\sqrt {a+b \tanh ^2(x)}} \, dx=\frac {2 \, \arctan \left (-\frac {\sqrt {a + b} e^{\left (2 \, x\right )} - \sqrt {a e^{\left (4 \, x\right )} + b e^{\left (4 \, x\right )} + 2 \, a e^{\left (2 \, x\right )} - 2 \, b e^{\left (2 \, x\right )} + a + b} - \sqrt {a + b}}{2 \, \sqrt {-a}}\right )}{\sqrt {-a}} - \frac {\log \left ({\left | -{\left (\sqrt {a + b} e^{\left (2 \, x\right )} - \sqrt {a e^{\left (4 \, x\right )} + b e^{\left (4 \, x\right )} + 2 \, a e^{\left (2 \, x\right )} - 2 \, b e^{\left (2 \, x\right )} + a + b}\right )} {\left (a + b\right )} - \sqrt {a + b} {\left (a - b\right )} \right |}\right )}{2 \, \sqrt {a + b}} + \frac {\log \left ({\left | -\sqrt {a + b} e^{\left (2 \, x\right )} + \sqrt {a e^{\left (4 \, x\right )} + b e^{\left (4 \, x\right )} + 2 \, a e^{\left (2 \, x\right )} - 2 \, b e^{\left (2 \, x\right )} + a + b} + \sqrt {a + b} \right |}\right )}{2 \, \sqrt {a + b}} - \frac {\log \left ({\left | -\sqrt {a + b} e^{\left (2 \, x\right )} + \sqrt {a e^{\left (4 \, x\right )} + b e^{\left (4 \, x\right )} + 2 \, a e^{\left (2 \, x\right )} - 2 \, b e^{\left (2 \, x\right )} + a + b} - \sqrt {a + b} \right |}\right )}{2 \, \sqrt {a + b}} \] Input:
integrate(coth(x)/(a+b*tanh(x)^2)^(1/2),x, algorithm="giac")
Output:
2*arctan(-1/2*(sqrt(a + b)*e^(2*x) - sqrt(a*e^(4*x) + b*e^(4*x) + 2*a*e^(2 *x) - 2*b*e^(2*x) + a + b) - sqrt(a + b))/sqrt(-a))/sqrt(-a) - 1/2*log(abs (-(sqrt(a + b)*e^(2*x) - sqrt(a*e^(4*x) + b*e^(4*x) + 2*a*e^(2*x) - 2*b*e^ (2*x) + a + b))*(a + b) - sqrt(a + b)*(a - b)))/sqrt(a + b) + 1/2*log(abs( -sqrt(a + b)*e^(2*x) + sqrt(a*e^(4*x) + b*e^(4*x) + 2*a*e^(2*x) - 2*b*e^(2 *x) + a + b) + sqrt(a + b)))/sqrt(a + b) - 1/2*log(abs(-sqrt(a + b)*e^(2*x ) + sqrt(a*e^(4*x) + b*e^(4*x) + 2*a*e^(2*x) - 2*b*e^(2*x) + a + b) - sqrt (a + b)))/sqrt(a + b)
Timed out. \[ \int \frac {\coth (x)}{\sqrt {a+b \tanh ^2(x)}} \, dx=\int \frac {\mathrm {coth}\left (x\right )}{\sqrt {b\,{\mathrm {tanh}\left (x\right )}^2+a}} \,d x \] Input:
int(coth(x)/(a + b*tanh(x)^2)^(1/2),x)
Output:
int(coth(x)/(a + b*tanh(x)^2)^(1/2), x)
\[ \int \frac {\coth (x)}{\sqrt {a+b \tanh ^2(x)}} \, dx=\int \frac {\sqrt {\tanh \left (x \right )^{2} b +a}\, \coth \left (x \right )}{\tanh \left (x \right )^{2} b +a}d x \] Input:
int(coth(x)/(a+b*tanh(x)^2)^(1/2),x)
Output:
int((sqrt(tanh(x)**2*b + a)*coth(x))/(tanh(x)**2*b + a),x)