Integrand size = 15, antiderivative size = 53 \[ \int \frac {1}{\sqrt {x} \text {arctanh}(\tanh (a+b x))} \, dx=-\frac {2 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {b x-\text {arctanh}(\tanh (a+b x))}}\right )}{\sqrt {b} \sqrt {b x-\text {arctanh}(\tanh (a+b x))}} \] Output:
-2*arctanh(b^(1/2)*x^(1/2)/(b*x-arctanh(tanh(b*x+a)))^(1/2))/b^(1/2)/(b*x- arctanh(tanh(b*x+a)))^(1/2)
Time = 0.02 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.96 \[ \int \frac {1}{\sqrt {x} \text {arctanh}(\tanh (a+b x))} \, dx=\frac {2 \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {-b x+\text {arctanh}(\tanh (a+b x))}}\right )}{\sqrt {b} \sqrt {-b x+\text {arctanh}(\tanh (a+b x))}} \] Input:
Integrate[1/(Sqrt[x]*ArcTanh[Tanh[a + b*x]]),x]
Output:
(2*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[-(b*x) + ArcTanh[Tanh[a + b*x]]]])/(Sqrt[ b]*Sqrt[-(b*x) + ArcTanh[Tanh[a + b*x]]])
Time = 0.29 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {2593}
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.
\(\displaystyle \int \frac {1}{\sqrt {x} \text {arctanh}(\tanh (a+b x))} \, dx\) |
\(\Big \downarrow \) 2593 |
\(\displaystyle -\frac {2 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {b x-\text {arctanh}(\tanh (a+b x))}}\right )}{\sqrt {b} \sqrt {b x-\text {arctanh}(\tanh (a+b x))}}\) |
Input:
Int[1/(Sqrt[x]*ArcTanh[Tanh[a + b*x]]),x]
Output:
(-2*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[b*x - ArcTanh[Tanh[a + b*x]]]])/(Sqrt[b ]*Sqrt[b*x - ArcTanh[Tanh[a + b*x]]])
Int[1/((u_)*Sqrt[v_]), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simpli fy[D[v, x]]}, Simp[-2*(ArcTanh[Sqrt[v]/Rt[-(b*u - a*v)/a, 2]]/(a*Rt[-(b*u - a*v)/a, 2])), x] /; NeQ[b*u - a*v, 0] && NegQ[(b*u - a*v)/a]] /; Piecewise LinearQ[u, v, x]
Time = 0.26 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.77
method | result | size |
derivativedivides | \(\frac {2 \arctan \left (\frac {b \sqrt {x}}{\sqrt {\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) b}}\right )}{\sqrt {\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) b}}\) | \(41\) |
default | \(\frac {2 \arctan \left (\frac {b \sqrt {x}}{\sqrt {\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) b}}\right )}{\sqrt {\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) b}}\) | \(41\) |
Input:
int(1/x^(1/2)/arctanh(tanh(b*x+a)),x,method=_RETURNVERBOSE)
Output:
2/((arctanh(tanh(b*x+a))-b*x)*b)^(1/2)*arctan(b*x^(1/2)/((arctanh(tanh(b*x +a))-b*x)*b)^(1/2))
Time = 0.08 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.28 \[ \int \frac {1}{\sqrt {x} \text {arctanh}(\tanh (a+b x))} \, dx=\left [-\frac {\sqrt {-a b} \log \left (\frac {b x - a - 2 \, \sqrt {-a b} \sqrt {x}}{b x + a}\right )}{a b}, -\frac {2 \, \sqrt {a b} \arctan \left (\frac {\sqrt {a b}}{b \sqrt {x}}\right )}{a b}\right ] \] Input:
integrate(1/x^(1/2)/arctanh(tanh(b*x+a)),x, algorithm="fricas")
Output:
[-sqrt(-a*b)*log((b*x - a - 2*sqrt(-a*b)*sqrt(x))/(b*x + a))/(a*b), -2*sqr t(a*b)*arctan(sqrt(a*b)/(b*sqrt(x)))/(a*b)]
\[ \int \frac {1}{\sqrt {x} \text {arctanh}(\tanh (a+b x))} \, dx=\int \frac {1}{\sqrt {x} \operatorname {atanh}{\left (\tanh {\left (a + b x \right )} \right )}}\, dx \] Input:
integrate(1/x**(1/2)/atanh(tanh(b*x+a)),x)
Output:
Integral(1/(sqrt(x)*atanh(tanh(a + b*x))), x)
Time = 0.12 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.34 \[ \int \frac {1}{\sqrt {x} \text {arctanh}(\tanh (a+b x))} \, dx=\frac {2 \, \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b}} \] Input:
integrate(1/x^(1/2)/arctanh(tanh(b*x+a)),x, algorithm="maxima")
Output:
2*arctan(b*sqrt(x)/sqrt(a*b))/sqrt(a*b)
Time = 0.11 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.34 \[ \int \frac {1}{\sqrt {x} \text {arctanh}(\tanh (a+b x))} \, dx=\frac {2 \, \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b}} \] Input:
integrate(1/x^(1/2)/arctanh(tanh(b*x+a)),x, algorithm="giac")
Output:
2*arctan(b*sqrt(x)/sqrt(a*b))/sqrt(a*b)
Time = 5.91 (sec) , antiderivative size = 347, normalized size of antiderivative = 6.55 \[ \int \frac {1}{\sqrt {x} \text {arctanh}(\tanh (a+b x))} \, dx=\frac {\sqrt {2}\,\ln \left (\frac {b^2\,\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )\,\left (2\,\sqrt {2}\,a+4\,\sqrt {x}\,\sqrt {b\,\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}-\sqrt {2}\,\left (2\,a-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )-2\,\sqrt {2}\,b\,x\right )}{2\,\sqrt {b\,\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}\,\left (\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )\right )}\right )}{\sqrt {b\,\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-b\,\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b^2\,x}} \] Input:
int(1/(x^(1/2)*atanh(tanh(a + b*x))),x)
Output:
(2^(1/2)*log((b^2*(log(2/(exp(2*a)*exp(2*b*x) + 1)) - log((2*exp(2*a)*exp( 2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x)*(2*2^(1/2)*a + 4*x^(1/2)*(b*(l og(2/(exp(2*a)*exp(2*b*x) + 1)) - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*ex p(2*b*x) + 1)) + 2*b*x))^(1/2) - 2^(1/2)*(2*a - log((2*exp(2*a)*exp(2*b*x) )/(exp(2*a)*exp(2*b*x) + 1)) + log(2/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x) - 2*2^(1/2)*b*x))/(2*(b*(log(2/(exp(2*a)*exp(2*b*x) + 1)) - log((2*exp(2*a) *exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x))^(1/2)*(log((2*exp(2*a)*e xp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) - log(2/(exp(2*a)*exp(2*b*x) + 1)))) ))/(b*log(1/(exp(2*a)*exp(2*b*x) + 1)) - b*log((exp(2*a)*exp(2*b*x))/(exp( 2*a)*exp(2*b*x) + 1)) + 2*b^2*x)^(1/2)
\[ \int \frac {1}{\sqrt {x} \text {arctanh}(\tanh (a+b x))} \, dx=\int \frac {1}{\sqrt {x}\, \mathit {atanh} \left (\tanh \left (b x +a \right )\right )}d x \] Input:
int(1/x^(1/2)/atanh(tanh(b*x+a)),x)
Output:
int(1/(sqrt(x)*atanh(tanh(a + b*x))),x)