Integrand size = 17, antiderivative size = 68 \[ \int \frac {1}{x^{3/2} \text {arctanh}(\tanh (a+b x))^{3/2}} \, dx=-\frac {4 b \sqrt {x}}{(b x-\text {arctanh}(\tanh (a+b x)))^2 \sqrt {\text {arctanh}(\tanh (a+b x))}}+\frac {2}{\sqrt {x} (b x-\text {arctanh}(\tanh (a+b x))) \sqrt {\text {arctanh}(\tanh (a+b x))}} \] Output:
-4*b*x^(1/2)/(b*x-arctanh(tanh(b*x+a)))^2/arctanh(tanh(b*x+a))^(1/2)+2/x^( 1/2)/(b*x-arctanh(tanh(b*x+a)))/arctanh(tanh(b*x+a))^(1/2)
Time = 0.04 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.63 \[ \int \frac {1}{x^{3/2} \text {arctanh}(\tanh (a+b x))^{3/2}} \, dx=-\frac {2 (b x+\text {arctanh}(\tanh (a+b x)))}{\sqrt {x} \sqrt {\text {arctanh}(\tanh (a+b x))} (-b x+\text {arctanh}(\tanh (a+b x)))^2} \] Input:
Integrate[1/(x^(3/2)*ArcTanh[Tanh[a + b*x]]^(3/2)),x]
Output:
(-2*(b*x + ArcTanh[Tanh[a + b*x]]))/(Sqrt[x]*Sqrt[ArcTanh[Tanh[a + b*x]]]* (-(b*x) + ArcTanh[Tanh[a + b*x]])^2)
Time = 0.22 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {2602, 2598}
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.
\(\displaystyle \int \frac {1}{x^{3/2} \text {arctanh}(\tanh (a+b x))^{3/2}} \, dx\) |
\(\Big \downarrow \) 2602 |
\(\displaystyle \frac {2 b \int \frac {1}{\sqrt {x} \text {arctanh}(\tanh (a+b x))^{3/2}}dx}{b x-\text {arctanh}(\tanh (a+b x))}+\frac {2}{\sqrt {x} (b x-\text {arctanh}(\tanh (a+b x))) \sqrt {\text {arctanh}(\tanh (a+b x))}}\) |
\(\Big \downarrow \) 2598 |
\(\displaystyle \frac {2}{\sqrt {x} (b x-\text {arctanh}(\tanh (a+b x))) \sqrt {\text {arctanh}(\tanh (a+b x))}}-\frac {4 b \sqrt {x}}{(b x-\text {arctanh}(\tanh (a+b x)))^2 \sqrt {\text {arctanh}(\tanh (a+b x))}}\) |
Input:
Int[1/(x^(3/2)*ArcTanh[Tanh[a + b*x]]^(3/2)),x]
Output:
(-4*b*Sqrt[x])/((b*x - ArcTanh[Tanh[a + b*x]])^2*Sqrt[ArcTanh[Tanh[a + b*x ]]]) + 2/(Sqrt[x]*(b*x - ArcTanh[Tanh[a + b*x]])*Sqrt[ArcTanh[Tanh[a + b*x ]]])
Int[(u_)^(m_)*(v_)^(n_), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simp lify[D[v, x]]}, Simp[(-u^(m + 1))*(v^(n + 1)/((m + 1)*(b*u - a*v))), x] /; NeQ[b*u - a*v, 0]] /; FreeQ[{m, n}, x] && PiecewiseLinearQ[u, v, x] && EqQ[ m + n + 2, 0] && NeQ[m, -1]
Int[(u_)^(m_)*(v_)^(n_), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simp lify[D[v, x]]}, Simp[(-u^(m + 1))*(v^(n + 1)/((m + 1)*(b*u - a*v))), x] + S imp[b*((m + n + 2)/((m + 1)*(b*u - a*v))) Int[u^(m + 1)*v^n, x], x] /; Ne Q[b*u - a*v, 0]] /; PiecewiseLinearQ[u, v, x] && NeQ[m + n + 2, 0] && LtQ[m , -1]
Time = 0.40 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.87
method | result | size |
derivativedivides | \(-\frac {2}{\sqrt {x}\, \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) \sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}}-\frac {4 b \sqrt {x}}{\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )^{2} \sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}}\) | \(59\) |
default | \(-\frac {2}{\sqrt {x}\, \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) \sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}}-\frac {4 b \sqrt {x}}{\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )^{2} \sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}}\) | \(59\) |
Input:
int(1/x^(3/2)/arctanh(tanh(b*x+a))^(3/2),x,method=_RETURNVERBOSE)
Output:
-2/x^(1/2)/(arctanh(tanh(b*x+a))-b*x)/arctanh(tanh(b*x+a))^(1/2)-4*b/(arct anh(tanh(b*x+a))-b*x)^2*x^(1/2)/arctanh(tanh(b*x+a))^(1/2)
Time = 0.09 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.50 \[ \int \frac {1}{x^{3/2} \text {arctanh}(\tanh (a+b x))^{3/2}} \, dx=-\frac {2 \, {\left (2 \, b x + a\right )} \sqrt {b x + a} \sqrt {x}}{a^{2} b x^{2} + a^{3} x} \] Input:
integrate(1/x^(3/2)/arctanh(tanh(b*x+a))^(3/2),x, algorithm="fricas")
Output:
-2*(2*b*x + a)*sqrt(b*x + a)*sqrt(x)/(a^2*b*x^2 + a^3*x)
\[ \int \frac {1}{x^{3/2} \text {arctanh}(\tanh (a+b x))^{3/2}} \, dx=\int \frac {1}{x^{\frac {3}{2}} \operatorname {atanh}^{\frac {3}{2}}{\left (\tanh {\left (a + b x \right )} \right )}}\, dx \] Input:
integrate(1/x**(3/2)/atanh(tanh(b*x+a))**(3/2),x)
Output:
Integral(1/(x**(3/2)*atanh(tanh(a + b*x))**(3/2)), x)
Time = 0.13 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.47 \[ \int \frac {1}{x^{3/2} \text {arctanh}(\tanh (a+b x))^{3/2}} \, dx=-\frac {2 \, {\left (2 \, b^{2} x^{2} + 3 \, a b x + a^{2}\right )}}{{\left (b x + a\right )}^{\frac {3}{2}} a^{2} \sqrt {x}} \] Input:
integrate(1/x^(3/2)/arctanh(tanh(b*x+a))^(3/2),x, algorithm="maxima")
Output:
-2*(2*b^2*x^2 + 3*a*b*x + a^2)/((b*x + a)^(3/2)*a^2*sqrt(x))
Time = 0.12 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.74 \[ \int \frac {1}{x^{3/2} \text {arctanh}(\tanh (a+b x))^{3/2}} \, dx=-\frac {2 \, b \sqrt {x}}{\sqrt {b x + a} a^{2}} + \frac {4 \, \sqrt {b}}{{\left ({\left (\sqrt {b} \sqrt {x} - \sqrt {b x + a}\right )}^{2} - a\right )} a} \] Input:
integrate(1/x^(3/2)/arctanh(tanh(b*x+a))^(3/2),x, algorithm="giac")
Output:
-2*b*sqrt(x)/(sqrt(b*x + a)*a^2) + 4*sqrt(b)/(((sqrt(b)*sqrt(x) - sqrt(b*x + a))^2 - a)*a)
Time = 3.44 (sec) , antiderivative size = 281, normalized size of antiderivative = 4.13 \[ \int \frac {1}{x^{3/2} \text {arctanh}(\tanh (a+b x))^{3/2}} \, dx=-\frac {\sqrt {\frac {\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}-\frac {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}}\,\left (\frac {16\,x}{{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^2}-\frac {8\,\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-8\,\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+16\,b\,x}{2\,b\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^2}\right )}{x^{3/2}-\frac {\sqrt {x}\,\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}{2\,b}} \] Input:
int(1/(x^(3/2)*atanh(tanh(a + b*x))^(3/2)),x)
Output:
-((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)*((16*x)/(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 - (8*log (2/(exp(2*a)*exp(2*b*x) + 1)) - 8*log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*ex p(2*b*x) + 1)) + 16*b*x)/(2*b*(log(2/(exp(2*a)*exp(2*b*x) + 1)) - log((2*e xp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x)^2)))/(x^(3/2) - (x ^(1/2)*(log(2/(exp(2*a)*exp(2*b*x) + 1)) - log((2*exp(2*a)*exp(2*b*x))/(ex p(2*a)*exp(2*b*x) + 1)) + 2*b*x))/(2*b))
\[ \int \frac {1}{x^{3/2} \text {arctanh}(\tanh (a+b x))^{3/2}} \, dx=\int \frac {\sqrt {x}\, \sqrt {\mathit {atanh} \left (\tanh \left (b x +a \right )\right )}}{\mathit {atanh} \left (\tanh \left (b x +a \right )\right )^{2} x^{2}}d x \] Input:
int(1/x^(3/2)/atanh(tanh(b*x+a))^(3/2),x)
Output:
int((sqrt(x)*sqrt(atanh(tanh(a + b*x))))/(atanh(tanh(a + b*x))**2*x**2),x)