Integrand size = 17, antiderivative size = 142 \[ \int x^{3/2} \sqrt {\text {arctanh}(\tanh (a+b x))} \, dx=-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {\text {arctanh}(\tanh (a+b x))}}\right ) (b x-\text {arctanh}(\tanh (a+b x)))^3}{8 b^{5/2}}+\frac {1}{3} x^{5/2} \sqrt {\text {arctanh}(\tanh (a+b x))}-\frac {x^{3/2} (b x-\text {arctanh}(\tanh (a+b x))) \sqrt {\text {arctanh}(\tanh (a+b x))}}{12 b}-\frac {\sqrt {x} (b x-\text {arctanh}(\tanh (a+b x)))^2 \sqrt {\text {arctanh}(\tanh (a+b x))}}{8 b^2} \]
-1/8*arctanh(b^(1/2)*x^(1/2)/arctanh(tanh(b*x+a))^(1/2))*(b*x-arctanh(tanh (b*x+a)))^3/b^(5/2)+1/3*x^(5/2)*arctanh(tanh(b*x+a))^(1/2)-1/12*x^(3/2)*(b *x-arctanh(tanh(b*x+a)))*arctanh(tanh(b*x+a))^(1/2)/b-1/8*(b*x-arctanh(tan h(b*x+a)))^2*x^(1/2)*arctanh(tanh(b*x+a))^(1/2)/b^2
Time = 0.07 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.73 \[ \int x^{3/2} \sqrt {\text {arctanh}(\tanh (a+b x))} \, dx=\frac {\sqrt {x} \sqrt {\text {arctanh}(\tanh (a+b x))} \left (3 b^2 x^2+8 b x \text {arctanh}(\tanh (a+b x))-3 \text {arctanh}(\tanh (a+b x))^2\right )}{24 b^2}+\frac {(-b x+\text {arctanh}(\tanh (a+b x)))^3 \log \left (b \sqrt {x}+\sqrt {b} \sqrt {\text {arctanh}(\tanh (a+b x))}\right )}{8 b^{5/2}} \]
(Sqrt[x]*Sqrt[ArcTanh[Tanh[a + b*x]]]*(3*b^2*x^2 + 8*b*x*ArcTanh[Tanh[a + b*x]] - 3*ArcTanh[Tanh[a + b*x]]^2))/(24*b^2) + ((-(b*x) + ArcTanh[Tanh[a + b*x]])^3*Log[b*Sqrt[x] + Sqrt[b]*Sqrt[ArcTanh[Tanh[a + b*x]]]])/(8*b^(5/ 2))
Time = 0.34 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.02, 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.
\(\displaystyle \int x^{3/2} \sqrt {\text {arctanh}(\tanh (a+b x))} \, dx\) |
\(\Big \downarrow \) 2600 |
\(\displaystyle \frac {1}{3} x^{5/2} \sqrt {\text {arctanh}(\tanh (a+b x))}-\frac {1}{6} (b x-\text {arctanh}(\tanh (a+b x))) \int \frac {x^{3/2}}{\sqrt {\text {arctanh}(\tanh (a+b x))}}dx\) |
\(\Big \downarrow \) 2600 |
\(\displaystyle \frac {1}{3} x^{5/2} \sqrt {\text {arctanh}(\tanh (a+b x))}-\frac {1}{6} (b x-\text {arctanh}(\tanh (a+b x))) \left (\frac {3 (b x-\text {arctanh}(\tanh (a+b x))) \int \frac {\sqrt {x}}{\sqrt {\text {arctanh}(\tanh (a+b x))}}dx}{4 b}+\frac {x^{3/2} \sqrt {\text {arctanh}(\tanh (a+b x))}}{2 b}\right )\) |
\(\Big \downarrow \) 2600 |
\(\displaystyle \frac {1}{3} x^{5/2} \sqrt {\text {arctanh}(\tanh (a+b x))}-\frac {1}{6} (b x-\text {arctanh}(\tanh (a+b x))) \left (\frac {3 (b x-\text {arctanh}(\tanh (a+b x))) \left (\frac {(b x-\text {arctanh}(\tanh (a+b x))) \int \frac {1}{\sqrt {x} \sqrt {\text {arctanh}(\tanh (a+b x))}}dx}{2 b}+\frac {\sqrt {x} \sqrt {\text {arctanh}(\tanh (a+b x))}}{b}\right )}{4 b}+\frac {x^{3/2} \sqrt {\text {arctanh}(\tanh (a+b x))}}{2 b}\right )\) |
\(\Big \downarrow \) 2596 |
\(\displaystyle \frac {1}{3} x^{5/2} \sqrt {\text {arctanh}(\tanh (a+b x))}-\frac {1}{6} \left (\frac {3 \left (\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {\text {arctanh}(\tanh (a+b x))}}\right ) (b x-\text {arctanh}(\tanh (a+b x)))}{b^{3/2}}+\frac {\sqrt {x} \sqrt {\text {arctanh}(\tanh (a+b x))}}{b}\right ) (b x-\text {arctanh}(\tanh (a+b x)))}{4 b}+\frac {x^{3/2} \sqrt {\text {arctanh}(\tanh (a+b x))}}{2 b}\right ) (b x-\text {arctanh}(\tanh (a+b x)))\) |
-1/6*(((3*((ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[ArcTanh[Tanh[a + b*x]]]]*(b*x - ArcTanh[Tanh[a + b*x]]))/b^(3/2) + (Sqrt[x]*Sqrt[ArcTanh[Tanh[a + b*x]]]) /b)*(b*x - ArcTanh[Tanh[a + b*x]]))/(4*b) + (x^(3/2)*Sqrt[ArcTanh[Tanh[a + b*x]]])/(2*b))*(b*x - ArcTanh[Tanh[a + b*x]])) + (x^(5/2)*Sqrt[ArcTanh[Ta nh[a + b*x]]])/3
3.3.15.3.1 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 = 0.20 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.85
method | result | size |
derivativedivides | \(\frac {x^{\frac {3}{2}} \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}{3 b}-\frac {\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) \left (\frac {\sqrt {x}\, \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}{4 b}-\frac {\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) \left (\frac {\sqrt {x}\, \sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}}{2}+\frac {\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) \ln \left (\sqrt {b}\, \sqrt {x}+\sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}\right )}{2 \sqrt {b}}\right )}{4 b}\right )}{b}\) | \(121\) |
default | \(\frac {x^{\frac {3}{2}} \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}{3 b}-\frac {\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) \left (\frac {\sqrt {x}\, \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}{4 b}-\frac {\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) \left (\frac {\sqrt {x}\, \sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}}{2}+\frac {\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) \ln \left (\sqrt {b}\, \sqrt {x}+\sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}\right )}{2 \sqrt {b}}\right )}{4 b}\right )}{b}\) | \(121\) |
1/3*x^(3/2)*arctanh(tanh(b*x+a))^(3/2)/b-(arctanh(tanh(b*x+a))-b*x)/b*(1/4 *x^(1/2)*arctanh(tanh(b*x+a))^(3/2)/b-1/4*(arctanh(tanh(b*x+a))-b*x)/b*(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 = 0.26 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.99 \[ \int x^{3/2} \sqrt {\text {arctanh}(\tanh (a+b x))} \, dx=\left [\frac {3 \, a^{3} \sqrt {b} \log \left (2 \, b x + 2 \, \sqrt {b x + a} \sqrt {b} \sqrt {x} + a\right ) + 2 \, {\left (8 \, b^{3} x^{2} + 2 \, a b^{2} x - 3 \, a^{2} b\right )} \sqrt {b x + a} \sqrt {x}}{48 \, b^{3}}, -\frac {3 \, a^{3} \sqrt {-b} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-b}}{b \sqrt {x}}\right ) - {\left (8 \, b^{3} x^{2} + 2 \, a b^{2} x - 3 \, a^{2} b\right )} \sqrt {b x + a} \sqrt {x}}{24 \, b^{3}}\right ] \]
[1/48*(3*a^3*sqrt(b)*log(2*b*x + 2*sqrt(b*x + a)*sqrt(b)*sqrt(x) + a) + 2* (8*b^3*x^2 + 2*a*b^2*x - 3*a^2*b)*sqrt(b*x + a)*sqrt(x))/b^3, -1/24*(3*a^3 *sqrt(-b)*arctan(sqrt(b*x + a)*sqrt(-b)/(b*sqrt(x))) - (8*b^3*x^2 + 2*a*b^ 2*x - 3*a^2*b)*sqrt(b*x + a)*sqrt(x))/b^3]
\[ \int x^{3/2} \sqrt {\text {arctanh}(\tanh (a+b x))} \, dx=\int x^{\frac {3}{2}} \sqrt {\operatorname {atanh}{\left (\tanh {\left (a + b x \right )} \right )}}\, dx \]
\[ \int x^{3/2} \sqrt {\text {arctanh}(\tanh (a+b x))} \, dx=\int { x^{\frac {3}{2}} \sqrt {\operatorname {artanh}\left (\tanh \left (b x + a\right )\right )} \,d x } \]
Time = 0.28 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.42 \[ \int x^{3/2} \sqrt {\text {arctanh}(\tanh (a+b x))} \, dx=\frac {1}{24} \, \sqrt {b x + a} {\left (2 \, {\left (4 \, x + \frac {a}{b}\right )} x - \frac {3 \, a^{2}}{b^{2}}\right )} \sqrt {x} - \frac {a^{3} \log \left ({\left | -\sqrt {b} \sqrt {x} + \sqrt {b x + a} \right |}\right )}{8 \, b^{\frac {5}{2}}} \]
1/24*sqrt(b*x + a)*(2*(4*x + a/b)*x - 3*a^2/b^2)*sqrt(x) - 1/8*a^3*log(abs (-sqrt(b)*sqrt(x) + sqrt(b*x + a)))/b^(5/2)
Timed out. \[ \int x^{3/2} \sqrt {\text {arctanh}(\tanh (a+b x))} \, dx=\int x^{3/2}\,\sqrt {\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )} \,d x \]