Integrand size = 15, antiderivative size = 63 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^3}{x^{3/2}} \, dx=\frac {32}{5} b^3 x^{5/2}-16 b^2 x^{3/2} \text {arctanh}(\tanh (a+b x))+12 b \sqrt {x} \text {arctanh}(\tanh (a+b x))^2-\frac {2 \text {arctanh}(\tanh (a+b x))^3}{\sqrt {x}} \] Output:
32/5*b^3*x^(5/2)-16*b^2*x^(3/2)*arctanh(tanh(b*x+a))+12*b*x^(1/2)*arctanh( tanh(b*x+a))^2-2*arctanh(tanh(b*x+a))^3/x^(1/2)
Time = 0.02 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.90 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^3}{x^{3/2}} \, dx=\frac {2 \left (16 b^3 x^3-40 b^2 x^2 \text {arctanh}(\tanh (a+b x))+30 b x \text {arctanh}(\tanh (a+b x))^2-5 \text {arctanh}(\tanh (a+b x))^3\right )}{5 \sqrt {x}} \] Input:
Integrate[ArcTanh[Tanh[a + b*x]]^3/x^(3/2),x]
Output:
(2*(16*b^3*x^3 - 40*b^2*x^2*ArcTanh[Tanh[a + b*x]] + 30*b*x*ArcTanh[Tanh[a + b*x]]^2 - 5*ArcTanh[Tanh[a + b*x]]^3))/(5*Sqrt[x])
Time = 0.24 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.06, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {2599, 2599, 2599, 15}
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.
\(\displaystyle \int \frac {\text {arctanh}(\tanh (a+b x))^3}{x^{3/2}} \, dx\) |
\(\Big \downarrow \) 2599 |
\(\displaystyle 6 b \int \frac {\text {arctanh}(\tanh (a+b x))^2}{\sqrt {x}}dx-\frac {2 \text {arctanh}(\tanh (a+b x))^3}{\sqrt {x}}\) |
\(\Big \downarrow \) 2599 |
\(\displaystyle 6 b \left (2 \sqrt {x} \text {arctanh}(\tanh (a+b x))^2-4 b \int \sqrt {x} \text {arctanh}(\tanh (a+b x))dx\right )-\frac {2 \text {arctanh}(\tanh (a+b x))^3}{\sqrt {x}}\) |
\(\Big \downarrow \) 2599 |
\(\displaystyle 6 b \left (2 \sqrt {x} \text {arctanh}(\tanh (a+b x))^2-4 b \left (\frac {2}{3} x^{3/2} \text {arctanh}(\tanh (a+b x))-\frac {2}{3} b \int x^{3/2}dx\right )\right )-\frac {2 \text {arctanh}(\tanh (a+b x))^3}{\sqrt {x}}\) |
\(\Big \downarrow \) 15 |
\(\displaystyle 6 b \left (2 \sqrt {x} \text {arctanh}(\tanh (a+b x))^2-4 b \left (\frac {2}{3} x^{3/2} \text {arctanh}(\tanh (a+b x))-\frac {4}{15} b x^{5/2}\right )\right )-\frac {2 \text {arctanh}(\tanh (a+b x))^3}{\sqrt {x}}\) |
Input:
Int[ArcTanh[Tanh[a + b*x]]^3/x^(3/2),x]
Output:
(-2*ArcTanh[Tanh[a + b*x]]^3)/Sqrt[x] + 6*b*(2*Sqrt[x]*ArcTanh[Tanh[a + b* x]]^2 - 4*b*((-4*b*x^(5/2))/15 + (2*x^(3/2)*ArcTanh[Tanh[a + b*x]])/3))
Int[(a_.)*(x_)^(m_.), x_Symbol] :> Simp[a*(x^(m + 1)/(m + 1)), x] /; FreeQ[ {a, m}, x] && NeQ[m, -1]
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]))
Time = 1.06 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.02
method | result | size |
derivativedivides | \(-\frac {2 \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{3}}{\sqrt {x}}+12 b \left (\frac {b^{2} x^{\frac {5}{2}}}{5}+\frac {2 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) b \,x^{\frac {3}{2}}}{3}+\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )^{2} \sqrt {x}\right )\) | \(64\) |
default | \(-\frac {2 \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{3}}{\sqrt {x}}+12 b \left (\frac {b^{2} x^{\frac {5}{2}}}{5}+\frac {2 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) b \,x^{\frac {3}{2}}}{3}+\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )^{2} \sqrt {x}\right )\) | \(64\) |
risch | \(\text {Expression too large to display}\) | \(7814\) |
Input:
int(arctanh(tanh(b*x+a))^3/x^(3/2),x,method=_RETURNVERBOSE)
Output:
-2*arctanh(tanh(b*x+a))^3/x^(1/2)+12*b*(1/5*b^2*x^(5/2)+2/3*(arctanh(tanh( b*x+a))-b*x)*b*x^(3/2)+(arctanh(tanh(b*x+a))-b*x)^2*x^(1/2))
Time = 0.07 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.54 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^3}{x^{3/2}} \, dx=\frac {2 \, {\left (b^{3} x^{3} + 5 \, a b^{2} x^{2} + 15 \, a^{2} b x - 5 \, a^{3}\right )}}{5 \, \sqrt {x}} \] Input:
integrate(arctanh(tanh(b*x+a))^3/x^(3/2),x, algorithm="fricas")
Output:
2/5*(b^3*x^3 + 5*a*b^2*x^2 + 15*a^2*b*x - 5*a^3)/sqrt(x)
\[ \int \frac {\text {arctanh}(\tanh (a+b x))^3}{x^{3/2}} \, dx=\int \frac {\operatorname {atanh}^{3}{\left (\tanh {\left (a + b x \right )} \right )}}{x^{\frac {3}{2}}}\, dx \] Input:
integrate(atanh(tanh(b*x+a))**3/x**(3/2),x)
Output:
Integral(atanh(tanh(a + b*x))**3/x**(3/2), x)
Time = 0.06 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.87 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^3}{x^{3/2}} \, dx=12 \, b \sqrt {x} \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{2} - \frac {2 \, \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{3}}{\sqrt {x}} + \frac {16}{5} \, {\left (2 \, b^{2} x^{\frac {5}{2}} - 5 \, b x^{\frac {3}{2}} \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )\right )} b \] Input:
integrate(arctanh(tanh(b*x+a))^3/x^(3/2),x, algorithm="maxima")
Output:
12*b*sqrt(x)*arctanh(tanh(b*x + a))^2 - 2*arctanh(tanh(b*x + a))^3/sqrt(x) + 16/5*(2*b^2*x^(5/2) - 5*b*x^(3/2)*arctanh(tanh(b*x + a)))*b
Time = 0.12 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.56 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^3}{x^{3/2}} \, dx=\frac {2}{5} \, b^{3} x^{\frac {5}{2}} + 2 \, a b^{2} x^{\frac {3}{2}} + 6 \, a^{2} b \sqrt {x} - \frac {2 \, a^{3}}{\sqrt {x}} \] Input:
integrate(arctanh(tanh(b*x+a))^3/x^(3/2),x, algorithm="giac")
Output:
2/5*b^3*x^(5/2) + 2*a*b^2*x^(3/2) + 6*a^2*b*sqrt(x) - 2*a^3/sqrt(x)
Time = 3.15 (sec) , antiderivative size = 182, normalized size of antiderivative = 2.89 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^3}{x^{3/2}} \, dx=\frac {{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^3}{4\,\sqrt {x}}+\frac {2\,b^3\,x^{5/2}}{5}+\frac {3\,b\,\sqrt {x}\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^2}{2}-b^2\,x^{3/2}\,\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right ) \] Input:
int(atanh(tanh(a + b*x))^3/x^(3/2),x)
Output:
(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)^3/(4*x^(1/2)) + (2*b^3*x^(5/2))/5 + (3*b*x^(1/2) *(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 - b^2*x^(3/2)*(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)
\[ \int \frac {\text {arctanh}(\tanh (a+b x))^3}{x^{3/2}} \, dx=\int \frac {\mathit {atanh} \left (\tanh \left (b x +a \right )\right )^{3}}{\sqrt {x}\, x}d x \] Input:
int(atanh(tanh(b*x+a))^3/x^(3/2),x)
Output:
int(atanh(tanh(a + b*x))**3/(sqrt(x)*x),x)