Integrand size = 15, antiderivative size = 57 \[ \int \frac {x^2}{\sqrt {\text {arctanh}(\tanh (a+b x))}} \, dx=\frac {2 x^2 \sqrt {\text {arctanh}(\tanh (a+b x))}}{b}-\frac {8 x \text {arctanh}(\tanh (a+b x))^{3/2}}{3 b^2}+\frac {16 \text {arctanh}(\tanh (a+b x))^{5/2}}{15 b^3} \] Output:
2*x^2*arctanh(tanh(b*x+a))^(1/2)/b-8/3*x*arctanh(tanh(b*x+a))^(3/2)/b^2+16 /15*arctanh(tanh(b*x+a))^(5/2)/b^3
Time = 0.02 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.86 \[ \int \frac {x^2}{\sqrt {\text {arctanh}(\tanh (a+b x))}} \, dx=\frac {2 \sqrt {\text {arctanh}(\tanh (a+b x))} \left (15 b^2 x^2-20 b x \text {arctanh}(\tanh (a+b x))+8 \text {arctanh}(\tanh (a+b x))^2\right )}{15 b^3} \] Input:
Integrate[x^2/Sqrt[ArcTanh[Tanh[a + b*x]]],x]
Output:
(2*Sqrt[ArcTanh[Tanh[a + b*x]]]*(15*b^2*x^2 - 20*b*x*ArcTanh[Tanh[a + b*x] ] + 8*ArcTanh[Tanh[a + b*x]]^2))/(15*b^3)
Time = 0.24 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.11, 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.
\(\displaystyle \int \frac {x^2}{\sqrt {\text {arctanh}(\tanh (a+b x))}} \, dx\) |
\(\Big \downarrow \) 2599 |
\(\displaystyle \frac {2 x^2 \sqrt {\text {arctanh}(\tanh (a+b x))}}{b}-\frac {4 \int x \sqrt {\text {arctanh}(\tanh (a+b x))}dx}{b}\) |
\(\Big \downarrow \) 2599 |
\(\displaystyle \frac {2 x^2 \sqrt {\text {arctanh}(\tanh (a+b x))}}{b}-\frac {4 \left (\frac {2 x \text {arctanh}(\tanh (a+b x))^{3/2}}{3 b}-\frac {2 \int \text {arctanh}(\tanh (a+b x))^{3/2}dx}{3 b}\right )}{b}\) |
\(\Big \downarrow \) 2588 |
\(\displaystyle \frac {2 x^2 \sqrt {\text {arctanh}(\tanh (a+b x))}}{b}-\frac {4 \left (\frac {2 x \text {arctanh}(\tanh (a+b x))^{3/2}}{3 b}-\frac {2 \int \text {arctanh}(\tanh (a+b x))^{3/2}d\text {arctanh}(\tanh (a+b x))}{3 b^2}\right )}{b}\) |
\(\Big \downarrow \) 15 |
\(\displaystyle \frac {2 x^2 \sqrt {\text {arctanh}(\tanh (a+b x))}}{b}-\frac {4 \left (\frac {2 x \text {arctanh}(\tanh (a+b x))^{3/2}}{3 b}-\frac {4 \text {arctanh}(\tanh (a+b x))^{5/2}}{15 b^2}\right )}{b}\) |
Input:
Int[x^2/Sqrt[ArcTanh[Tanh[a + b*x]]],x]
Output:
(2*x^2*Sqrt[ArcTanh[Tanh[a + b*x]]])/b - (4*((2*x*ArcTanh[Tanh[a + b*x]]^( 3/2))/(3*b) - (4*ArcTanh[Tanh[a + b*x]]^(5/2))/(15*b^2)))/b
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 = 0.23 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.19
method | result | size |
default | \(\frac {\frac {2 \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {5}{2}}}{5}+\frac {2 \left (2 b x -2 \,\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )\right ) \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}{3}+2 \left (b x -\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )\right )^{2} \sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}}{b^{3}}\) | \(68\) |
Input:
int(x^2/arctanh(tanh(b*x+a))^(1/2),x,method=_RETURNVERBOSE)
Output:
2/b^3*(1/5*arctanh(tanh(b*x+a))^(5/2)+1/3*(2*b*x-2*arctanh(tanh(b*x+a)))*a rctanh(tanh(b*x+a))^(3/2)+(b*x-arctanh(tanh(b*x+a)))^2*arctanh(tanh(b*x+a) )^(1/2))
Time = 0.08 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.54 \[ \int \frac {x^2}{\sqrt {\text {arctanh}(\tanh (a+b x))}} \, dx=\frac {2 \, {\left (3 \, b^{2} x^{2} - 4 \, a b x + 8 \, a^{2}\right )} \sqrt {b x + a}}{15 \, b^{3}} \] Input:
integrate(x^2/arctanh(tanh(b*x+a))^(1/2),x, algorithm="fricas")
Output:
2/15*(3*b^2*x^2 - 4*a*b*x + 8*a^2)*sqrt(b*x + a)/b^3
\[ \int \frac {x^2}{\sqrt {\text {arctanh}(\tanh (a+b x))}} \, dx=\int \frac {x^{2}}{\sqrt {\operatorname {atanh}{\left (\tanh {\left (a + b x \right )} \right )}}}\, dx \] Input:
integrate(x**2/atanh(tanh(b*x+a))**(1/2),x)
Output:
Integral(x**2/sqrt(atanh(tanh(a + b*x))), x)
Time = 0.19 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.74 \[ \int \frac {x^2}{\sqrt {\text {arctanh}(\tanh (a+b x))}} \, dx=\frac {2 \, {\left (3 \, b^{3} x^{3} - a b^{2} x^{2} + 4 \, a^{2} b x + 8 \, a^{3}\right )}}{15 \, \sqrt {b x + a} b^{3}} \] Input:
integrate(x^2/arctanh(tanh(b*x+a))^(1/2),x, algorithm="maxima")
Output:
2/15*(3*b^3*x^3 - a*b^2*x^2 + 4*a^2*b*x + 8*a^3)/(sqrt(b*x + a)*b^3)
Time = 0.12 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.65 \[ \int \frac {x^2}{\sqrt {\text {arctanh}(\tanh (a+b x))}} \, dx=\frac {2 \, {\left (3 \, {\left (b x + a\right )}^{\frac {5}{2}} - 10 \, {\left (b x + a\right )}^{\frac {3}{2}} a + 15 \, \sqrt {b x + a} a^{2}\right )}}{15 \, b^{3}} \] Input:
integrate(x^2/arctanh(tanh(b*x+a))^(1/2),x, algorithm="giac")
Output:
2/15*(3*(b*x + a)^(5/2) - 10*(b*x + a)^(3/2)*a + 15*sqrt(b*x + a)*a^2)/b^3
Time = 3.05 (sec) , antiderivative size = 211, normalized size of antiderivative = 3.70 \[ \int \frac {x^2}{\sqrt {\text {arctanh}(\tanh (a+b x))}} \, dx=\frac {2\,\sqrt {\frac {\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}-\frac {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}}\,\left (15\,b^2\,x^2-10\,b\,x\,\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+10\,b\,x\,\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,{\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}^2-4\,\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )\,\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,{\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}^2\right )}{15\,b^3} \] Input:
int(x^2/atanh(tanh(a + b*x))^(1/2),x)
Output:
(2*(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 + 15*b^2*x^2 - 10*b*x*log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1 )) + 10*b*x*log(2/(exp(2*a)*exp(2*b*x) + 1))))/(15*b^3)
\[ \int \frac {x^2}{\sqrt {\text {arctanh}(\tanh (a+b x))}} \, dx=\int \frac {\sqrt {\mathit {atanh} \left (\tanh \left (b x +a \right )\right )}\, x^{2}}{\mathit {atanh} \left (\tanh \left (b x +a \right )\right )}d x \] Input:
int(x^2/atanh(tanh(b*x+a))^(1/2),x)
Output:
int((sqrt(atanh(tanh(a + b*x)))*x**2)/atanh(tanh(a + b*x)),x)