Integrand size = 17, antiderivative size = 49 \[ \int \frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{x^{3/2}} \, dx=2 \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {\text {arctanh}(\tanh (a+b x))}}\right )-\frac {2 \sqrt {\text {arctanh}(\tanh (a+b x))}}{\sqrt {x}} \] Output:
2*b^(1/2)*arctanh(b^(1/2)*x^(1/2)/arctanh(tanh(b*x+a))^(1/2))-2*arctanh(ta nh(b*x+a))^(1/2)/x^(1/2)
Time = 0.03 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.06 \[ \int \frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{x^{3/2}} \, dx=-\frac {2 \sqrt {\text {arctanh}(\tanh (a+b x))}}{\sqrt {x}}+2 \sqrt {b} \log \left (b \sqrt {x}+\sqrt {b} \sqrt {\text {arctanh}(\tanh (a+b x))}\right ) \] Input:
Integrate[Sqrt[ArcTanh[Tanh[a + b*x]]]/x^(3/2),x]
Output:
(-2*Sqrt[ArcTanh[Tanh[a + b*x]]])/Sqrt[x] + 2*Sqrt[b]*Log[b*Sqrt[x] + Sqrt [b]*Sqrt[ArcTanh[Tanh[a + b*x]]]]
Time = 0.33 (sec) , antiderivative size = 49, 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 = {2599, 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.
\(\displaystyle \int \frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{x^{3/2}} \, dx\) |
\(\Big \downarrow \) 2599 |
\(\displaystyle b \int \frac {1}{\sqrt {x} \sqrt {\text {arctanh}(\tanh (a+b x))}}dx-\frac {2 \sqrt {\text {arctanh}(\tanh (a+b x))}}{\sqrt {x}}\) |
\(\Big \downarrow \) 2596 |
\(\displaystyle 2 \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {\text {arctanh}(\tanh (a+b x))}}\right )-\frac {2 \sqrt {\text {arctanh}(\tanh (a+b x))}}{\sqrt {x}}\) |
Input:
Int[Sqrt[ArcTanh[Tanh[a + b*x]]]/x^(3/2),x]
Output:
2*Sqrt[b]*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[ArcTanh[Tanh[a + b*x]]]] - (2*Sqr t[ArcTanh[Tanh[a + b*x]]])/Sqrt[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 + 1))), x] - Simp[b*(n/(a*(m + 1 ))) Int[u^(m + 1)*v^(n - 1), x], x] /; NeQ[b*u - a*v, 0]] /; FreeQ[{m, n} , x] && PiecewiseLinearQ[u, v, x] && NeQ[m, -1] && ((LtQ[m, -1] && GtQ[n, 0 ] && !(ILtQ[m + n, -2] && (FractionQ[m] || GeQ[2*n + m + 1, 0]))) || (IGtQ [n, 0] && IGtQ[m, 0] && LeQ[n, m]) || (IGtQ[n, 0] && !IntegerQ[m]) || (ILt Q[m, 0] && !IntegerQ[n]))
Leaf count of result is larger than twice the leaf count of optimal. \(96\) vs. \(2(37)=74\).
Time = 0.40 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.98
method | result | size |
derivativedivides | \(-\frac {2 \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}{\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) \sqrt {x}}+\frac {4 b \left (\frac {\sqrt {x}\, \sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}}{2}+\frac {\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) \ln \left (\sqrt {b}\, \sqrt {x}+\sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}\right )}{2 \sqrt {b}}\right )}{\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x}\) | \(97\) |
default | \(-\frac {2 \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}{\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) \sqrt {x}}+\frac {4 b \left (\frac {\sqrt {x}\, \sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}}{2}+\frac {\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) \ln \left (\sqrt {b}\, \sqrt {x}+\sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}\right )}{2 \sqrt {b}}\right )}{\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x}\) | \(97\) |
Input:
int(arctanh(tanh(b*x+a))^(1/2)/x^(3/2),x,method=_RETURNVERBOSE)
Output:
-2/(arctanh(tanh(b*x+a))-b*x)/x^(1/2)*arctanh(tanh(b*x+a))^(3/2)+4*b/(arct anh(tanh(b*x+a))-b*x)*(1/2*x^(1/2)*arctanh(tanh(b*x+a))^(1/2)+1/2/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 = 0.07 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.76 \[ \int \frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{x^{3/2}} \, dx=\left [\frac {\sqrt {b} x \log \left (2 \, b x + 2 \, \sqrt {b x + a} \sqrt {b} \sqrt {x} + a\right ) - 2 \, \sqrt {b x + a} \sqrt {x}}{x}, -\frac {2 \, {\left (\sqrt {-b} x \arctan \left (\frac {\sqrt {-b} \sqrt {x}}{\sqrt {b x + a}}\right ) + \sqrt {b x + a} \sqrt {x}\right )}}{x}\right ] \] Input:
integrate(arctanh(tanh(b*x+a))^(1/2)/x^(3/2),x, algorithm="fricas")
Output:
[(sqrt(b)*x*log(2*b*x + 2*sqrt(b*x + a)*sqrt(b)*sqrt(x) + a) - 2*sqrt(b*x + a)*sqrt(x))/x, -2*(sqrt(-b)*x*arctan(sqrt(-b)*sqrt(x)/sqrt(b*x + a)) + s qrt(b*x + a)*sqrt(x))/x]
\[ \int \frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{x^{3/2}} \, dx=\int \frac {\sqrt {\operatorname {atanh}{\left (\tanh {\left (a + b x \right )} \right )}}}{x^{\frac {3}{2}}}\, dx \] Input:
integrate(atanh(tanh(b*x+a))**(1/2)/x**(3/2),x)
Output:
Integral(sqrt(atanh(tanh(a + b*x)))/x**(3/2), x)
Time = 0.15 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.82 \[ \int \frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{x^{3/2}} \, dx=2 \, \sqrt {b} \log \left (\frac {b \sqrt {x}}{\sqrt {a b}} + \sqrt {\frac {b x}{a} + 1}\right ) - \frac {2 \, \sqrt {b x + a}}{\sqrt {x}} \] Input:
integrate(arctanh(tanh(b*x+a))^(1/2)/x^(3/2),x, algorithm="maxima")
Output:
2*sqrt(b)*log(b*sqrt(x)/sqrt(a*b) + sqrt(b*x/a + 1)) - 2*sqrt(b*x + a)/sqr t(x)
Time = 0.12 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.16 \[ \int \frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{x^{3/2}} \, dx=-\sqrt {b} \log \left ({\left (\sqrt {b} \sqrt {x} - \sqrt {b x + a}\right )}^{2}\right ) + \frac {4 \, a \sqrt {b}}{{\left (\sqrt {b} \sqrt {x} - \sqrt {b x + a}\right )}^{2} - a} \] Input:
integrate(arctanh(tanh(b*x+a))^(1/2)/x^(3/2),x, algorithm="giac")
Output:
-sqrt(b)*log((sqrt(b)*sqrt(x) - sqrt(b*x + a))^2) + 4*a*sqrt(b)/((sqrt(b)* sqrt(x) - sqrt(b*x + a))^2 - a)
Timed out. \[ \int \frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{x^{3/2}} \, dx=\int \frac {\sqrt {\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}}{x^{3/2}} \,d x \] Input:
int(atanh(tanh(a + b*x))^(1/2)/x^(3/2),x)
Output:
int(atanh(tanh(a + b*x))^(1/2)/x^(3/2), x)
\[ \int \frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{x^{3/2}} \, dx=\frac {-2 \sqrt {\mathit {atanh} \left (\tanh \left (b x +a \right )\right )}+\sqrt {x}\, \left (\int \frac {\sqrt {x}\, \sqrt {\mathit {atanh} \left (\tanh \left (b x +a \right )\right )}}{\mathit {atanh} \left (\tanh \left (b x +a \right )\right ) x}d x \right ) b}{\sqrt {x}} \] Input:
int(atanh(tanh(b*x+a))^(1/2)/x^(3/2),x)
Output:
( - 2*sqrt(atanh(tanh(a + b*x))) + sqrt(x)*int((sqrt(x)*sqrt(atanh(tanh(a + b*x))))/(atanh(tanh(a + b*x))*x),x)*b)/sqrt(x)