Integrand size = 15, antiderivative size = 69 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^3}{x^{7/2}} \, dx=\frac {32 b^3 \sqrt {x}}{5}-\frac {16 b^2 \text {arctanh}(\tanh (a+b x))}{5 \sqrt {x}}-\frac {4 b \text {arctanh}(\tanh (a+b x))^2}{5 x^{3/2}}-\frac {2 \text {arctanh}(\tanh (a+b x))^3}{5 x^{5/2}} \] Output:
32/5*b^3*x^(1/2)-16/5*b^2*arctanh(tanh(b*x+a))/x^(1/2)-4/5*b*arctanh(tanh( b*x+a))^2/x^(3/2)-2/5*arctanh(tanh(b*x+a))^3/x^(5/2)
Time = 0.02 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.83 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^3}{x^{7/2}} \, dx=\frac {2 \left (16 b^3 x^3-8 b^2 x^2 \text {arctanh}(\tanh (a+b x))-2 b x \text {arctanh}(\tanh (a+b x))^2-\text {arctanh}(\tanh (a+b x))^3\right )}{5 x^{5/2}} \] Input:
Integrate[ArcTanh[Tanh[a + b*x]]^3/x^(7/2),x]
Output:
(2*(16*b^3*x^3 - 8*b^2*x^2*ArcTanh[Tanh[a + b*x]] - 2*b*x*ArcTanh[Tanh[a + b*x]]^2 - ArcTanh[Tanh[a + b*x]]^3))/(5*x^(5/2))
Time = 0.25 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.03, 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.
\(\displaystyle \int \frac {\text {arctanh}(\tanh (a+b x))^3}{x^{7/2}} \, dx\) |
\(\Big \downarrow \) 2599 |
\(\displaystyle \frac {6}{5} b \int \frac {\text {arctanh}(\tanh (a+b x))^2}{x^{5/2}}dx-\frac {2 \text {arctanh}(\tanh (a+b x))^3}{5 x^{5/2}}\) |
\(\Big \downarrow \) 2599 |
\(\displaystyle \frac {6}{5} b \left (\frac {4}{3} b \int \frac {\text {arctanh}(\tanh (a+b x))}{x^{3/2}}dx-\frac {2 \text {arctanh}(\tanh (a+b x))^2}{3 x^{3/2}}\right )-\frac {2 \text {arctanh}(\tanh (a+b x))^3}{5 x^{5/2}}\) |
\(\Big \downarrow \) 2599 |
\(\displaystyle \frac {6}{5} b \left (\frac {4}{3} b \left (2 b \int \frac {1}{\sqrt {x}}dx-\frac {2 \text {arctanh}(\tanh (a+b x))}{\sqrt {x}}\right )-\frac {2 \text {arctanh}(\tanh (a+b x))^2}{3 x^{3/2}}\right )-\frac {2 \text {arctanh}(\tanh (a+b x))^3}{5 x^{5/2}}\) |
\(\Big \downarrow \) 15 |
\(\displaystyle \frac {6}{5} b \left (\frac {4}{3} b \left (4 b \sqrt {x}-\frac {2 \text {arctanh}(\tanh (a+b x))}{\sqrt {x}}\right )-\frac {2 \text {arctanh}(\tanh (a+b x))^2}{3 x^{3/2}}\right )-\frac {2 \text {arctanh}(\tanh (a+b x))^3}{5 x^{5/2}}\) |
Input:
Int[ArcTanh[Tanh[a + b*x]]^3/x^(7/2),x]
Output:
(-2*ArcTanh[Tanh[a + b*x]]^3)/(5*x^(5/2)) + (6*b*((-2*ArcTanh[Tanh[a + b*x ]]^2)/(3*x^(3/2)) + (4*b*(4*b*Sqrt[x] - (2*ArcTanh[Tanh[a + b*x]])/Sqrt[x] ))/3))/5
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 = 0.97 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.81
method | result | size |
derivativedivides | \(-\frac {2 \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{3}}{5 x^{\frac {5}{2}}}+\frac {12 b \left (-\frac {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{2}}{3 x^{\frac {3}{2}}}+\frac {4 b \left (-\frac {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}{\sqrt {x}}+2 b \sqrt {x}\right )}{3}\right )}{5}\) | \(56\) |
default | \(-\frac {2 \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{3}}{5 x^{\frac {5}{2}}}+\frac {12 b \left (-\frac {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{2}}{3 x^{\frac {3}{2}}}+\frac {4 b \left (-\frac {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}{\sqrt {x}}+2 b \sqrt {x}\right )}{3}\right )}{5}\) | \(56\) |
risch | \(\text {Expression too large to display}\) | \(7814\) |
Input:
int(arctanh(tanh(b*x+a))^3/x^(7/2),x,method=_RETURNVERBOSE)
Output:
-2/5*arctanh(tanh(b*x+a))^3/x^(5/2)+12/5*b*(-1/3*arctanh(tanh(b*x+a))^2/x^ (3/2)+4/3*b*(-arctanh(tanh(b*x+a))/x^(1/2)+2*b*x^(1/2)))
Time = 0.08 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.51 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^3}{x^{7/2}} \, dx=\frac {2 \, {\left (5 \, b^{3} x^{3} - 15 \, a b^{2} x^{2} - 5 \, a^{2} b x - a^{3}\right )}}{5 \, x^{\frac {5}{2}}} \] Input:
integrate(arctanh(tanh(b*x+a))^3/x^(7/2),x, algorithm="fricas")
Output:
2/5*(5*b^3*x^3 - 15*a*b^2*x^2 - 5*a^2*b*x - a^3)/x^(5/2)
Time = 12.81 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.01 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^3}{x^{7/2}} \, dx=\frac {32 b^{3} \sqrt {x}}{5} - \frac {16 b^{2} \operatorname {atanh}{\left (\tanh {\left (a + b x \right )} \right )}}{5 \sqrt {x}} - \frac {4 b \operatorname {atanh}^{2}{\left (\tanh {\left (a + b x \right )} \right )}}{5 x^{\frac {3}{2}}} - \frac {2 \operatorname {atanh}^{3}{\left (\tanh {\left (a + b x \right )} \right )}}{5 x^{\frac {5}{2}}} \] Input:
integrate(atanh(tanh(b*x+a))**3/x**(7/2),x)
Output:
32*b**3*sqrt(x)/5 - 16*b**2*atanh(tanh(a + b*x))/(5*sqrt(x)) - 4*b*atanh(t anh(a + b*x))**2/(5*x**(3/2)) - 2*atanh(tanh(a + b*x))**3/(5*x**(5/2))
Time = 0.06 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.80 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^3}{x^{7/2}} \, dx=\frac {16}{5} \, {\left (2 \, b^{2} \sqrt {x} - \frac {b \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )}{\sqrt {x}}\right )} b - \frac {4 \, b \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{2}}{5 \, x^{\frac {3}{2}}} - \frac {2 \, \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{3}}{5 \, x^{\frac {5}{2}}} \] Input:
integrate(arctanh(tanh(b*x+a))^3/x^(7/2),x, algorithm="maxima")
Output:
16/5*(2*b^2*sqrt(x) - b*arctanh(tanh(b*x + a))/sqrt(x))*b - 4/5*b*arctanh( tanh(b*x + a))^2/x^(3/2) - 2/5*arctanh(tanh(b*x + a))^3/x^(5/2)
Time = 0.11 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.49 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^3}{x^{7/2}} \, dx=2 \, b^{3} \sqrt {x} - \frac {2 \, {\left (15 \, a b^{2} x^{2} + 5 \, a^{2} b x + a^{3}\right )}}{5 \, x^{\frac {5}{2}}} \] Input:
integrate(arctanh(tanh(b*x+a))^3/x^(7/2),x, algorithm="giac")
Output:
2*b^3*sqrt(x) - 2/5*(15*a*b^2*x^2 + 5*a^2*b*x + a^3)/x^(5/2)
Time = 3.16 (sec) , antiderivative size = 182, normalized size of antiderivative = 2.64 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^3}{x^{7/2}} \, dx=\frac {{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^3}{20\,x^{5/2}}+2\,b^3\,\sqrt {x}+\frac {3\,b^2\,\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}{\sqrt {x}}-\frac {b\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^2}{2\,x^{3/2}} \] Input:
int(atanh(tanh(a + b*x))^3/x^(7/2),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/(20*x^(5/2)) + 2*b^3*x^(1/2) + (3*b^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))/x^(1/2) - (b*(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*x^(3/2))
Time = 0.17 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.83 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^3}{x^{7/2}} \, dx=\frac {-\frac {2 \mathit {atanh} \left (\tanh \left (b x +a \right )\right )^{3}}{5}-\frac {4 \mathit {atanh} \left (\tanh \left (b x +a \right )\right )^{2} b x}{5}-\frac {16 \mathit {atanh} \left (\tanh \left (b x +a \right )\right ) b^{2} x^{2}}{5}+\frac {32 b^{3} x^{3}}{5}}{\sqrt {x}\, x^{2}} \] Input:
int(atanh(tanh(b*x+a))^3/x^(7/2),x)
Output:
(2*( - atanh(tanh(a + b*x))**3 - 2*atanh(tanh(a + b*x))**2*b*x - 8*atanh(t anh(a + b*x))*b**2*x**2 + 16*b**3*x**3))/(5*sqrt(x)*x**2)