Integrand size = 17, antiderivative size = 136 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{\sqrt {x}} \, dx=-\frac {5 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {\text {arctanh}(\tanh (a+b x))}}\right ) (b x-\text {arctanh}(\tanh (a+b x)))^3}{8 \sqrt {b}}+\frac {5}{8} \sqrt {x} (b x-\text {arctanh}(\tanh (a+b x)))^2 \sqrt {\text {arctanh}(\tanh (a+b x))}-\frac {5}{12} \sqrt {x} (b x-\text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))^{3/2}+\frac {1}{3} \sqrt {x} \text {arctanh}(\tanh (a+b x))^{5/2} \]
-5/8*arctanh(b^(1/2)*x^(1/2)/arctanh(tanh(b*x+a))^(1/2))*(b*x-arctanh(tanh (b*x+a)))^3/b^(1/2)-5/12*(b*x-arctanh(tanh(b*x+a)))*arctanh(tanh(b*x+a))^( 3/2)*x^(1/2)+1/3*x^(1/2)*arctanh(tanh(b*x+a))^(5/2)+5/8*(b*x-arctanh(tanh( b*x+a)))^2*x^(1/2)*arctanh(tanh(b*x+a))^(1/2)
Time = 0.05 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.74 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{\sqrt {x}} \, dx=\frac {1}{24} \sqrt {x} \sqrt {\text {arctanh}(\tanh (a+b x))} \left (15 b^2 x^2-40 b x \text {arctanh}(\tanh (a+b x))+33 \text {arctanh}(\tanh (a+b x))^2\right )+\frac {5 (-b x+\text {arctanh}(\tanh (a+b x)))^3 \log \left (b \sqrt {x}+\sqrt {b} \sqrt {\text {arctanh}(\tanh (a+b x))}\right )}{8 \sqrt {b}} \]
(Sqrt[x]*Sqrt[ArcTanh[Tanh[a + b*x]]]*(15*b^2*x^2 - 40*b*x*ArcTanh[Tanh[a + b*x]] + 33*ArcTanh[Tanh[a + b*x]]^2))/24 + (5*(-(b*x) + ArcTanh[Tanh[a + b*x]])^3*Log[b*Sqrt[x] + Sqrt[b]*Sqrt[ArcTanh[Tanh[a + b*x]]]])/(8*Sqrt[b ])
Time = 0.32 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.01, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {2600, 2600, 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.
| \(\displaystyle \int \frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{\sqrt {x}} \, dx\) |
| \(\Big \downarrow \) 2600 |
| \(\displaystyle \frac {1}{3} \sqrt {x} \text {arctanh}(\tanh (a+b x))^{5/2}-\frac {5}{6} (b x-\text {arctanh}(\tanh (a+b x))) \int \frac {\text {arctanh}(\tanh (a+b x))^{3/2}}{\sqrt {x}}dx\) |
| \(\Big \downarrow \) 2600 |
| \(\displaystyle \frac {1}{3} \sqrt {x} \text {arctanh}(\tanh (a+b x))^{5/2}-\frac {5}{6} (b x-\text {arctanh}(\tanh (a+b x))) \left (\frac {1}{2} \sqrt {x} \text {arctanh}(\tanh (a+b x))^{3/2}-\frac {3}{4} (b x-\text {arctanh}(\tanh (a+b x))) \int \frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{\sqrt {x}}dx\right )\) |
| \(\Big \downarrow \) 2600 |
| \(\displaystyle \frac {1}{3} \sqrt {x} \text {arctanh}(\tanh (a+b x))^{5/2}-\frac {5}{6} (b x-\text {arctanh}(\tanh (a+b x))) \left (\frac {1}{2} \sqrt {x} \text {arctanh}(\tanh (a+b x))^{3/2}-\frac {3}{4} (b x-\text {arctanh}(\tanh (a+b x))) \left (\sqrt {x} \sqrt {\text {arctanh}(\tanh (a+b x))}-\frac {1}{2} (b x-\text {arctanh}(\tanh (a+b x))) \int \frac {1}{\sqrt {x} \sqrt {\text {arctanh}(\tanh (a+b x))}}dx\right )\right )\) |
| \(\Big \downarrow \) 2596 |
| \(\displaystyle \frac {1}{3} \sqrt {x} \text {arctanh}(\tanh (a+b x))^{5/2}-\frac {5}{6} (b x-\text {arctanh}(\tanh (a+b x))) \left (\frac {1}{2} \sqrt {x} \text {arctanh}(\tanh (a+b x))^{3/2}-\frac {3}{4} \left (\sqrt {x} \sqrt {\text {arctanh}(\tanh (a+b x))}-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {\text {arctanh}(\tanh (a+b x))}}\right ) (b x-\text {arctanh}(\tanh (a+b x)))}{\sqrt {b}}\right ) (b x-\text {arctanh}(\tanh (a+b x)))\right )\) |
(Sqrt[x]*ArcTanh[Tanh[a + b*x]]^(5/2))/3 - (5*(b*x - ArcTanh[Tanh[a + b*x] ])*((-3*(-((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]]]) *(b*x - ArcTanh[Tanh[a + b*x]]))/4 + (Sqrt[x]*ArcTanh[Tanh[a + b*x]]^(3/2) )/2))/6
3.3.33.3.1 Defintions of rubi rules used
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 = 0.18 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.80
| method | result | size |
| derivativedivides | \(\frac {\sqrt {x}\, \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {5}{2}}}{3}+\frac {5 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) \left (\frac {\sqrt {x}\, \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}{4}+\frac {3 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) \left (\frac {\sqrt {x}\, \sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}}{2}+\frac {\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) \ln \left (\sqrt {b}\, \sqrt {x}+\sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}\right )}{2 \sqrt {b}}\right )}{4}\right )}{3}\) | \(109\) |
| default | \(\frac {\sqrt {x}\, \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {5}{2}}}{3}+\frac {5 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) \left (\frac {\sqrt {x}\, \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}{4}+\frac {3 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) \left (\frac {\sqrt {x}\, \sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}}{2}+\frac {\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) \ln \left (\sqrt {b}\, \sqrt {x}+\sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}\right )}{2 \sqrt {b}}\right )}{4}\right )}{3}\) | \(109\) |
1/3*x^(1/2)*arctanh(tanh(b*x+a))^(5/2)+5/3*(arctanh(tanh(b*x+a))-b*x)*(1/4 *x^(1/2)*arctanh(tanh(b*x+a))^(3/2)+3/4*(arctanh(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 = 0.27 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.04 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{\sqrt {x}} \, dx=\left [\frac {15 \, a^{3} \sqrt {b} \log \left (2 \, b x + 2 \, \sqrt {b x + a} \sqrt {b} \sqrt {x} + a\right ) + 2 \, {\left (8 \, b^{3} x^{2} + 26 \, a b^{2} x + 33 \, a^{2} b\right )} \sqrt {b x + a} \sqrt {x}}{48 \, b}, -\frac {15 \, a^{3} \sqrt {-b} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-b}}{b \sqrt {x}}\right ) - {\left (8 \, b^{3} x^{2} + 26 \, a b^{2} x + 33 \, a^{2} b\right )} \sqrt {b x + a} \sqrt {x}}{24 \, b}\right ] \]
[1/48*(15*a^3*sqrt(b)*log(2*b*x + 2*sqrt(b*x + a)*sqrt(b)*sqrt(x) + a) + 2 *(8*b^3*x^2 + 26*a*b^2*x + 33*a^2*b)*sqrt(b*x + a)*sqrt(x))/b, -1/24*(15*a ^3*sqrt(-b)*arctan(sqrt(b*x + a)*sqrt(-b)/(b*sqrt(x))) - (8*b^3*x^2 + 26*a *b^2*x + 33*a^2*b)*sqrt(b*x + a)*sqrt(x))/b]
\[ \int \frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{\sqrt {x}} \, dx=\int \frac {\operatorname {atanh}^{\frac {5}{2}}{\left (\tanh {\left (a + b x \right )} \right )}}{\sqrt {x}}\, dx \]
\[ \int \frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{\sqrt {x}} \, dx=\int { \frac {\operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{\frac {5}{2}}}{\sqrt {x}} \,d x } \]
Time = 74.70 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.85 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{\sqrt {x}} \, dx=-\frac {\sqrt {2} {\left (\frac {15 \, \sqrt {2} a^{3} \log \left ({\left | -\sqrt {b x + a} \sqrt {b} + \sqrt {{\left (b x + a\right )} b - a b} \right |}\right )}{\sqrt {b}} - \sqrt {{\left (b x + a\right )} b - a b} \sqrt {b x + a} {\left (2 \, {\left (b x + a\right )} {\left (\frac {4 \, \sqrt {2} {\left (b x + a\right )}}{b} + \frac {5 \, \sqrt {2} a}{b}\right )} + \frac {15 \, \sqrt {2} a^{2}}{b}\right )}\right )} b}{48 \, {\left | b \right |}} \]
-1/48*sqrt(2)*(15*sqrt(2)*a^3*log(abs(-sqrt(b*x + a)*sqrt(b) + sqrt((b*x + a)*b - a*b)))/sqrt(b) - sqrt((b*x + a)*b - a*b)*sqrt(b*x + a)*(2*(b*x + a )*(4*sqrt(2)*(b*x + a)/b + 5*sqrt(2)*a/b) + 15*sqrt(2)*a^2/b))*b/abs(b)
Timed out. \[ \int \frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{\sqrt {x}} \, dx=\int \frac {{\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}^{5/2}}{\sqrt {x}} \,d x \]