Integrand size = 15, antiderivative size = 59 \[ \int \frac {x^2}{\text {arctanh}(\tanh (a+b x))^{5/2}} \, dx=-\frac {2 x^2}{3 b \text {arctanh}(\tanh (a+b x))^{3/2}}-\frac {8 x}{3 b^2 \sqrt {\text {arctanh}(\tanh (a+b x))}}+\frac {16 \sqrt {\text {arctanh}(\tanh (a+b x))}}{3 b^3} \] Output:
-2/3*x^2/b/arctanh(tanh(b*x+a))^(3/2)-8/3*x/b^2/arctanh(tanh(b*x+a))^(1/2) +16/3*arctanh(tanh(b*x+a))^(1/2)/b^3
Time = 0.03 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.81 \[ \int \frac {x^2}{\text {arctanh}(\tanh (a+b x))^{5/2}} \, dx=-\frac {2 \left (b^2 x^2+4 b x \text {arctanh}(\tanh (a+b x))-8 \text {arctanh}(\tanh (a+b x))^2\right )}{3 b^3 \text {arctanh}(\tanh (a+b x))^{3/2}} \] Input:
Integrate[x^2/ArcTanh[Tanh[a + b*x]]^(5/2),x]
Output:
(-2*(b^2*x^2 + 4*b*x*ArcTanh[Tanh[a + b*x]] - 8*ArcTanh[Tanh[a + b*x]]^2)) /(3*b^3*ArcTanh[Tanh[a + b*x]]^(3/2))
Time = 0.33 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.07, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {2599, 2599, 2588, 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.
\(\displaystyle \int \frac {x^2}{\text {arctanh}(\tanh (a+b x))^{5/2}} \, dx\) |
\(\Big \downarrow \) 2599 |
\(\displaystyle \frac {4 \int \frac {x}{\text {arctanh}(\tanh (a+b x))^{3/2}}dx}{3 b}-\frac {2 x^2}{3 b \text {arctanh}(\tanh (a+b x))^{3/2}}\) |
\(\Big \downarrow \) 2599 |
\(\displaystyle \frac {4 \left (\frac {2 \int \frac {1}{\sqrt {\text {arctanh}(\tanh (a+b x))}}dx}{b}-\frac {2 x}{b \sqrt {\text {arctanh}(\tanh (a+b x))}}\right )}{3 b}-\frac {2 x^2}{3 b \text {arctanh}(\tanh (a+b x))^{3/2}}\) |
\(\Big \downarrow \) 2588 |
\(\displaystyle \frac {4 \left (\frac {2 \int \frac {1}{\sqrt {\text {arctanh}(\tanh (a+b x))}}d\text {arctanh}(\tanh (a+b x))}{b^2}-\frac {2 x}{b \sqrt {\text {arctanh}(\tanh (a+b x))}}\right )}{3 b}-\frac {2 x^2}{3 b \text {arctanh}(\tanh (a+b x))^{3/2}}\) |
\(\Big \downarrow \) 15 |
\(\displaystyle \frac {4 \left (\frac {4 \sqrt {\text {arctanh}(\tanh (a+b x))}}{b^2}-\frac {2 x}{b \sqrt {\text {arctanh}(\tanh (a+b x))}}\right )}{3 b}-\frac {2 x^2}{3 b \text {arctanh}(\tanh (a+b x))^{3/2}}\) |
Input:
Int[x^2/ArcTanh[Tanh[a + b*x]]^(5/2),x]
Output:
(4*((-2*x)/(b*Sqrt[ArcTanh[Tanh[a + b*x]]]) + (4*Sqrt[ArcTanh[Tanh[a + b*x ]]])/b^2))/(3*b) - (2*x^2)/(3*b*ArcTanh[Tanh[a + b*x]]^(3/2))
Int[(a_.)*(x_)^(m_.), x_Symbol] :> Simp[a*(x^(m + 1)/(m + 1)), x] /; FreeQ[ {a, m}, x] && NeQ[m, -1]
Int[(u_)^(m_.), x_Symbol] :> With[{c = Simplify[D[u, x]]}, Simp[1/c Subst [Int[x^m, x], x, u], x]] /; FreeQ[m, x] && PiecewiseLinearQ[u, 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]))
Time = 0.20 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.54
method | result | size |
default | \(\frac {2 \sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}-\frac {2 \left (2 b x -2 \,\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )\right )}{\sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}}-\frac {2 \left (a^{2}+2 a \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )+\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2}\right )}{3 \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}}{b^{3}}\) | \(91\) |
Input:
int(x^2/arctanh(tanh(b*x+a))^(5/2),x,method=_RETURNVERBOSE)
Output:
2/b^3*(arctanh(tanh(b*x+a))^(1/2)-(2*b*x-2*arctanh(tanh(b*x+a)))/arctanh(t anh(b*x+a))^(1/2)-1/3*(a^2+2*a*(arctanh(tanh(b*x+a))-b*x-a)+(arctanh(tanh( b*x+a))-b*x-a)^2)/arctanh(tanh(b*x+a))^(3/2))
Time = 0.07 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.88 \[ \int \frac {x^2}{\text {arctanh}(\tanh (a+b x))^{5/2}} \, dx=\frac {2 \, {\left (3 \, b^{2} x^{2} + 12 \, a b x + 8 \, a^{2}\right )} \sqrt {b x + a}}{3 \, {\left (b^{5} x^{2} + 2 \, a b^{4} x + a^{2} b^{3}\right )}} \] Input:
integrate(x^2/arctanh(tanh(b*x+a))^(5/2),x, algorithm="fricas")
Output:
2/3*(3*b^2*x^2 + 12*a*b*x + 8*a^2)*sqrt(b*x + a)/(b^5*x^2 + 2*a*b^4*x + a^ 2*b^3)
Time = 2.38 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.20 \[ \int \frac {x^2}{\text {arctanh}(\tanh (a+b x))^{5/2}} \, dx=\begin {cases} - \frac {2 x^{2}}{3 b \operatorname {atanh}^{\frac {3}{2}}{\left (\tanh {\left (a + b x \right )} \right )}} - \frac {8 x}{3 b^{2} \sqrt {\operatorname {atanh}{\left (\tanh {\left (a + b x \right )} \right )}}} + \frac {16 \sqrt {\operatorname {atanh}{\left (\tanh {\left (a + b x \right )} \right )}}}{3 b^{3}} & \text {for}\: b \neq 0 \\\frac {x^{3}}{3 \operatorname {atanh}^{\frac {5}{2}}{\left (\tanh {\left (a \right )} \right )}} & \text {otherwise} \end {cases} \] Input:
integrate(x**2/atanh(tanh(b*x+a))**(5/2),x)
Output:
Piecewise((-2*x**2/(3*b*atanh(tanh(a + b*x))**(3/2)) - 8*x/(3*b**2*sqrt(at anh(tanh(a + b*x)))) + 16*sqrt(atanh(tanh(a + b*x)))/(3*b**3), Ne(b, 0)), (x**3/(3*atanh(tanh(a))**(5/2)), True))
Time = 0.19 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.71 \[ \int \frac {x^2}{\text {arctanh}(\tanh (a+b x))^{5/2}} \, dx=\frac {2 \, {\left (3 \, b^{3} x^{3} + 15 \, a b^{2} x^{2} + 20 \, a^{2} b x + 8 \, a^{3}\right )}}{3 \, {\left (b x + a\right )}^{\frac {5}{2}} b^{3}} \] Input:
integrate(x^2/arctanh(tanh(b*x+a))^(5/2),x, algorithm="maxima")
Output:
2/3*(3*b^3*x^3 + 15*a*b^2*x^2 + 20*a^2*b*x + 8*a^3)/((b*x + a)^(5/2)*b^3)
Time = 0.12 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.73 \[ \int \frac {x^2}{\text {arctanh}(\tanh (a+b x))^{5/2}} \, dx=\frac {2 \, {\left (\frac {3 \, \sqrt {b x + a}}{b} + \frac {6 \, {\left (b x + a\right )} a - a^{2}}{{\left (b x + a\right )}^{\frac {3}{2}} b}\right )}}{3 \, b^{2}} \] Input:
integrate(x^2/arctanh(tanh(b*x+a))^(5/2),x, algorithm="giac")
Output:
2/3*(3*sqrt(b*x + a)/b + (6*(b*x + a)*a - a^2)/((b*x + a)^(3/2)*b))/b^2
Time = 3.30 (sec) , antiderivative size = 259, normalized size of antiderivative = 4.39 \[ \int \frac {x^2}{\text {arctanh}(\tanh (a+b x))^{5/2}} \, dx=\frac {8\,\sqrt {\frac {\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}-\frac {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}}\,\left (-b^2\,x^2-2\,b\,x\,\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\,\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,{\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}^2-4\,\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )\,\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,{\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}^2\right )}{3\,b^3\,{\left (\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )\right )}^2} \] Input:
int(x^2/atanh(tanh(a + b*x))^(5/2),x)
Output:
(8*(log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/2 - log(2/(exp( 2*a)*exp(2*b*x) + 1))/2)^(1/2)*(2*log(2/(exp(2*a)*exp(2*b*x) + 1))^2 - 4*l og((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))*log(2/(exp(2*a)*exp( 2*b*x) + 1)) + 2*log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))^2 - b^2*x^2 - 2*b*x*log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x*log(2/(exp(2*a)*exp(2*b*x) + 1))))/(3*b^3*(log((2*exp(2*a)*exp(2*b* x))/(exp(2*a)*exp(2*b*x) + 1)) - log(2/(exp(2*a)*exp(2*b*x) + 1)))^2)
\[ \int \frac {x^2}{\text {arctanh}(\tanh (a+b x))^{5/2}} \, dx=\int \frac {\sqrt {\mathit {atanh} \left (\tanh \left (b x +a \right )\right )}\, x^{2}}{\mathit {atanh} \left (\tanh \left (b x +a \right )\right )^{3}}d x \] Input:
int(x^2/atanh(tanh(b*x+a))^(5/2),x)
Output:
int((sqrt(atanh(tanh(a + b*x)))*x**2)/atanh(tanh(a + b*x))**3,x)