Integrand size = 17, antiderivative size = 61 \[ \int \frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{\sqrt {x}} \, dx=-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {\text {arctanh}(\tanh (a+b x))}}\right ) (b x-\text {arctanh}(\tanh (a+b x)))}{\sqrt {b}}+\sqrt {x} \sqrt {\text {arctanh}(\tanh (a+b x))} \] Output:
-arctanh(b^(1/2)*x^(1/2)/arctanh(tanh(b*x+a))^(1/2))*(b*x-arctanh(tanh(b*x +a)))/b^(1/2)+x^(1/2)*arctanh(tanh(b*x+a))^(1/2)
Time = 0.03 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.02 \[ \int \frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{\sqrt {x}} \, dx=\sqrt {x} \sqrt {\text {arctanh}(\tanh (a+b x))}+\frac {(-b x+\text {arctanh}(\tanh (a+b x))) \log \left (b \sqrt {x}+\sqrt {b} \sqrt {\text {arctanh}(\tanh (a+b x))}\right )}{\sqrt {b}} \] Input:
Integrate[Sqrt[ArcTanh[Tanh[a + b*x]]]/Sqrt[x],x]
Output:
Sqrt[x]*Sqrt[ArcTanh[Tanh[a + b*x]]] + ((-(b*x) + ArcTanh[Tanh[a + b*x]])* Log[b*Sqrt[x] + Sqrt[b]*Sqrt[ArcTanh[Tanh[a + b*x]]]])/Sqrt[b]
Time = 0.36 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {2600, 2596}
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.
\(\displaystyle \int \frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{\sqrt {x}} \, dx\) |
\(\Big \downarrow \) 2600 |
\(\displaystyle \sqrt {x} \sqrt {\text {arctanh}(\tanh (a+b x))}-\frac {1}{2} (b x-\text {arctanh}(\tanh (a+b x))) \int \frac {1}{\sqrt {x} \sqrt {\text {arctanh}(\tanh (a+b x))}}dx\) |
\(\Big \downarrow \) 2596 |
\(\displaystyle \sqrt {x} \sqrt {\text {arctanh}(\tanh (a+b x))}-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {\text {arctanh}(\tanh (a+b x))}}\right ) (b x-\text {arctanh}(\tanh (a+b x)))}{\sqrt {b}}\) |
Input:
Int[Sqrt[ArcTanh[Tanh[a + b*x]]]/Sqrt[x],x]
Output:
-((ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[ArcTanh[Tanh[a + b*x]]]]*(b*x - ArcTanh[ Tanh[a + b*x]]))/Sqrt[b]) + Sqrt[x]*Sqrt[ArcTanh[Tanh[a + b*x]]]
Int[1/(Sqrt[u_]*Sqrt[v_]), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Si mplify[D[v, x]]}, Simp[(2/Rt[a*b, 2])*ArcTanh[Rt[a*b, 2]*(Sqrt[u]/(a*Sqrt[v ]))], x] /; NeQ[b*u - a*v, 0] && PosQ[a*b]] /; PiecewiseLinearQ[u, v, x]
Int[(u_)^(m_)*(v_)^(n_.), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Sim plify[D[v, x]]}, Simp[u^(m + 1)*(v^n/(a*(m + n + 1))), x] - Simp[n*((b*u - a*v)/(a*(m + n + 1))) Int[u^m*v^(n - 1), x], x] /; NeQ[b*u - a*v, 0]] /; PiecewiseLinearQ[u, v, x] && NeQ[m + n + 2, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !IntegerQ[n] || LtQ[0, m, n])) && !ILtQ[m + n, -2]
Time = 0.41 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.80
method | result | size |
derivativedivides | \(\sqrt {x}\, \sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}+\frac {\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) \ln \left (\sqrt {b}\, \sqrt {x}+\sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}\right )}{\sqrt {b}}\) | \(49\) |
default | \(\sqrt {x}\, \sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}+\frac {\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) \ln \left (\sqrt {b}\, \sqrt {x}+\sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}\right )}{\sqrt {b}}\) | \(49\) |
Input:
int(arctanh(tanh(b*x+a))^(1/2)/x^(1/2),x,method=_RETURNVERBOSE)
Output:
x^(1/2)*arctanh(tanh(b*x+a))^(1/2)+1/b^(1/2)*(arctanh(tanh(b*x+a))-b*x)*ln (b^(1/2)*x^(1/2)+arctanh(tanh(b*x+a))^(1/2))
Time = 0.09 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.48 \[ \int \frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{\sqrt {x}} \, dx=\left [\frac {a \sqrt {b} \log \left (2 \, b x + 2 \, \sqrt {b x + a} \sqrt {b} \sqrt {x} + a\right ) + 2 \, \sqrt {b x + a} b \sqrt {x}}{2 \, b}, -\frac {a \sqrt {-b} \arctan \left (\frac {\sqrt {-b} \sqrt {x}}{\sqrt {b x + a}}\right ) - \sqrt {b x + a} b \sqrt {x}}{b}\right ] \] Input:
integrate(arctanh(tanh(b*x+a))^(1/2)/x^(1/2),x, algorithm="fricas")
Output:
[1/2*(a*sqrt(b)*log(2*b*x + 2*sqrt(b*x + a)*sqrt(b)*sqrt(x) + a) + 2*sqrt( b*x + a)*b*sqrt(x))/b, -(a*sqrt(-b)*arctan(sqrt(-b)*sqrt(x)/sqrt(b*x + a)) - sqrt(b*x + a)*b*sqrt(x))/b]
\[ \int \frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{\sqrt {x}} \, dx=\int \frac {\sqrt {\operatorname {atanh}{\left (\tanh {\left (a + b x \right )} \right )}}}{\sqrt {x}}\, dx \] Input:
integrate(atanh(tanh(b*x+a))**(1/2)/x**(1/2),x)
Output:
Integral(sqrt(atanh(tanh(a + b*x)))/sqrt(x), x)
\[ \int \frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{\sqrt {x}} \, dx=\int { \frac {\sqrt {\operatorname {artanh}\left (\tanh \left (b x + a\right )\right )}}{\sqrt {x}} \,d x } \] Input:
integrate(arctanh(tanh(b*x+a))^(1/2)/x^(1/2),x, algorithm="maxima")
Output:
integrate(sqrt(arctanh(tanh(b*x + a)))/sqrt(x), x)
Time = 0.12 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.59 \[ \int \frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{\sqrt {x}} \, dx=-\frac {a \log \left ({\left | -\sqrt {b} \sqrt {x} + \sqrt {b x + a} \right |}\right )}{\sqrt {b}} + \sqrt {b x + a} \sqrt {x} \] Input:
integrate(arctanh(tanh(b*x+a))^(1/2)/x^(1/2),x, algorithm="giac")
Output:
-a*log(abs(-sqrt(b)*sqrt(x) + sqrt(b*x + a)))/sqrt(b) + sqrt(b*x + a)*sqrt (x)
Timed out. \[ \int \frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{\sqrt {x}} \, dx=\int \frac {\sqrt {\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}}{\sqrt {x}} \,d x \] Input:
int(atanh(tanh(a + b*x))^(1/2)/x^(1/2),x)
Output:
int(atanh(tanh(a + b*x))^(1/2)/x^(1/2), x)
\[ \int \frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{\sqrt {x}} \, dx=\int \frac {\sqrt {x}\, \sqrt {\mathit {atanh} \left (\tanh \left (b x +a \right )\right )}}{x}d x \] Input:
int(atanh(tanh(b*x+a))^(1/2)/x^(1/2),x)
Output:
int((sqrt(x)*sqrt(atanh(tanh(a + b*x))))/x,x)