Integrand size = 15, antiderivative size = 125 \[ \int \frac {\sqrt {x}}{\text {arctanh}(\tanh (a+b x))^3} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {b x-\text {arctanh}(\tanh (a+b x))}}\right )}{4 b^{3/2} (b x-\text {arctanh}(\tanh (a+b x)))^{3/2}}-\frac {1}{4 b^2 \sqrt {x} (b x-\text {arctanh}(\tanh (a+b x)))}-\frac {\sqrt {x}}{2 b \text {arctanh}(\tanh (a+b x))^2}-\frac {1}{4 b^2 \sqrt {x} \text {arctanh}(\tanh (a+b x))} \] Output:
1/4*arctanh(b^(1/2)*x^(1/2)/(b*x-arctanh(tanh(b*x+a)))^(1/2))/b^(3/2)/(b*x -arctanh(tanh(b*x+a)))^(3/2)-1/4/b^2/x^(1/2)/(b*x-arctanh(tanh(b*x+a)))-1/ 2*x^(1/2)/b/arctanh(tanh(b*x+a))^2-1/4/b^2/x^(1/2)/arctanh(tanh(b*x+a))
Time = 0.10 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.86 \[ \int \frac {\sqrt {x}}{\text {arctanh}(\tanh (a+b x))^3} \, dx=\frac {1}{4} \left (-\frac {2 \sqrt {x}}{b \text {arctanh}(\tanh (a+b x))^2}+\frac {\arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {-b x+\text {arctanh}(\tanh (a+b x))}}\right )}{b^{3/2} (-b x+\text {arctanh}(\tanh (a+b x)))^{3/2}}+\frac {\sqrt {x}}{-b^2 x \text {arctanh}(\tanh (a+b x))+b \text {arctanh}(\tanh (a+b x))^2}\right ) \] Input:
Integrate[Sqrt[x]/ArcTanh[Tanh[a + b*x]]^3,x]
Output:
((-2*Sqrt[x])/(b*ArcTanh[Tanh[a + b*x]]^2) + ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt [-(b*x) + ArcTanh[Tanh[a + b*x]]]]/(b^(3/2)*(-(b*x) + ArcTanh[Tanh[a + b*x ]])^(3/2)) + Sqrt[x]/(-(b^2*x*ArcTanh[Tanh[a + b*x]]) + b*ArcTanh[Tanh[a + b*x]]^2))/4
Time = 0.47 (sec) , antiderivative size = 132, 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, 2594, 2593}
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.
\(\displaystyle \int \frac {\sqrt {x}}{\text {arctanh}(\tanh (a+b x))^3} \, dx\) |
\(\Big \downarrow \) 2599 |
\(\displaystyle \frac {\int \frac {1}{\sqrt {x} \text {arctanh}(\tanh (a+b x))^2}dx}{4 b}-\frac {\sqrt {x}}{2 b \text {arctanh}(\tanh (a+b x))^2}\) |
\(\Big \downarrow \) 2599 |
\(\displaystyle \frac {-\frac {\int \frac {1}{x^{3/2} \text {arctanh}(\tanh (a+b x))}dx}{2 b}-\frac {1}{b \sqrt {x} \text {arctanh}(\tanh (a+b x))}}{4 b}-\frac {\sqrt {x}}{2 b \text {arctanh}(\tanh (a+b x))^2}\) |
\(\Big \downarrow \) 2594 |
\(\displaystyle \frac {-\frac {\frac {b \int \frac {1}{\sqrt {x} \text {arctanh}(\tanh (a+b x))}dx}{b x-\text {arctanh}(\tanh (a+b x))}+\frac {2}{\sqrt {x} (b x-\text {arctanh}(\tanh (a+b x)))}}{2 b}-\frac {1}{b \sqrt {x} \text {arctanh}(\tanh (a+b x))}}{4 b}-\frac {\sqrt {x}}{2 b \text {arctanh}(\tanh (a+b x))^2}\) |
\(\Big \downarrow \) 2593 |
\(\displaystyle \frac {-\frac {\frac {2}{\sqrt {x} (b x-\text {arctanh}(\tanh (a+b x)))}-\frac {2 \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {b x-\text {arctanh}(\tanh (a+b x))}}\right )}{(b x-\text {arctanh}(\tanh (a+b x)))^{3/2}}}{2 b}-\frac {1}{b \sqrt {x} \text {arctanh}(\tanh (a+b x))}}{4 b}-\frac {\sqrt {x}}{2 b \text {arctanh}(\tanh (a+b x))^2}\) |
Input:
Int[Sqrt[x]/ArcTanh[Tanh[a + b*x]]^3,x]
Output:
(-1/2*((-2*Sqrt[b]*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[b*x - ArcTanh[Tanh[a + b *x]]]])/(b*x - ArcTanh[Tanh[a + b*x]])^(3/2) + 2/(Sqrt[x]*(b*x - ArcTanh[T anh[a + b*x]])))/b - 1/(b*Sqrt[x]*ArcTanh[Tanh[a + b*x]]))/(4*b) - Sqrt[x] /(2*b*ArcTanh[Tanh[a + b*x]]^2)
Int[1/((u_)*Sqrt[v_]), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simpli fy[D[v, x]]}, Simp[-2*(ArcTanh[Sqrt[v]/Rt[-(b*u - a*v)/a, 2]]/(a*Rt[-(b*u - a*v)/a, 2])), x] /; NeQ[b*u - a*v, 0] && NegQ[(b*u - a*v)/a]] /; Piecewise LinearQ[u, v, x]
Int[(v_)^(n_)/(u_), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[ D[v, x]]}, Simp[v^(n + 1)/((n + 1)*(b*u - a*v)), x] - Simp[a*((n + 1)/((n + 1)*(b*u - a*v))) Int[v^(n + 1)/u, x], x] /; NeQ[b*u - a*v, 0]] /; Piecew iseLinearQ[u, v, x] && LtQ[n, -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 = 1.23 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.78
method | result | size |
derivativedivides | \(\frac {\frac {2 x^{\frac {3}{2}}}{8 \,\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-8 b x}-\frac {\sqrt {x}}{4 b}}{\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{2}}+\frac {\arctan \left (\frac {b \sqrt {x}}{\sqrt {\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) b}}\right )}{4 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) b \sqrt {\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) b}}\) | \(98\) |
default | \(\frac {\frac {2 x^{\frac {3}{2}}}{8 \,\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-8 b x}-\frac {\sqrt {x}}{4 b}}{\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{2}}+\frac {\arctan \left (\frac {b \sqrt {x}}{\sqrt {\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) b}}\right )}{4 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) b \sqrt {\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) b}}\) | \(98\) |
risch | \(\text {Expression too large to display}\) | \(1595\) |
Input:
int(x^(1/2)/arctanh(tanh(b*x+a))^3,x,method=_RETURNVERBOSE)
Output:
2*(1/8/(arctanh(tanh(b*x+a))-b*x)*x^(3/2)-1/8*x^(1/2)/b)/arctanh(tanh(b*x+ a))^2+1/4/(arctanh(tanh(b*x+a))-b*x)/b/((arctanh(tanh(b*x+a))-b*x)*b)^(1/2 )*arctan(b*x^(1/2)/((arctanh(tanh(b*x+a))-b*x)*b)^(1/2))
Time = 0.09 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.49 \[ \int \frac {\sqrt {x}}{\text {arctanh}(\tanh (a+b x))^3} \, dx=\left [-\frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} \sqrt {-a b} \log \left (\frac {b x - a - 2 \, \sqrt {-a b} \sqrt {x}}{b x + a}\right ) - 2 \, {\left (a b^{2} x - a^{2} b\right )} \sqrt {x}}{8 \, {\left (a^{2} b^{4} x^{2} + 2 \, a^{3} b^{3} x + a^{4} b^{2}\right )}}, -\frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b}}{b \sqrt {x}}\right ) - {\left (a b^{2} x - a^{2} b\right )} \sqrt {x}}{4 \, {\left (a^{2} b^{4} x^{2} + 2 \, a^{3} b^{3} x + a^{4} b^{2}\right )}}\right ] \] Input:
integrate(x^(1/2)/arctanh(tanh(b*x+a))^3,x, algorithm="fricas")
Output:
[-1/8*((b^2*x^2 + 2*a*b*x + a^2)*sqrt(-a*b)*log((b*x - a - 2*sqrt(-a*b)*sq rt(x))/(b*x + a)) - 2*(a*b^2*x - a^2*b)*sqrt(x))/(a^2*b^4*x^2 + 2*a^3*b^3* x + a^4*b^2), -1/4*((b^2*x^2 + 2*a*b*x + a^2)*sqrt(a*b)*arctan(sqrt(a*b)/( b*sqrt(x))) - (a*b^2*x - a^2*b)*sqrt(x))/(a^2*b^4*x^2 + 2*a^3*b^3*x + a^4* b^2)]
\[ \int \frac {\sqrt {x}}{\text {arctanh}(\tanh (a+b x))^3} \, dx=\int \frac {\sqrt {x}}{\operatorname {atanh}^{3}{\left (\tanh {\left (a + b x \right )} \right )}}\, dx \] Input:
integrate(x**(1/2)/atanh(tanh(b*x+a))**3,x)
Output:
Integral(sqrt(x)/atanh(tanh(a + b*x))**3, x)
Time = 0.12 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.51 \[ \int \frac {\sqrt {x}}{\text {arctanh}(\tanh (a+b x))^3} \, dx=\frac {b x^{\frac {3}{2}} - a \sqrt {x}}{4 \, {\left (a b^{3} x^{2} + 2 \, a^{2} b^{2} x + a^{3} b\right )}} + \frac {\arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \, \sqrt {a b} a b} \] Input:
integrate(x^(1/2)/arctanh(tanh(b*x+a))^3,x, algorithm="maxima")
Output:
1/4*(b*x^(3/2) - a*sqrt(x))/(a*b^3*x^2 + 2*a^2*b^2*x + a^3*b) + 1/4*arctan (b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a*b)
Time = 0.11 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.42 \[ \int \frac {\sqrt {x}}{\text {arctanh}(\tanh (a+b x))^3} \, dx=\frac {\arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \, \sqrt {a b} a b} + \frac {b x^{\frac {3}{2}} - a \sqrt {x}}{4 \, {\left (b x + a\right )}^{2} a b} \] Input:
integrate(x^(1/2)/arctanh(tanh(b*x+a))^3,x, algorithm="giac")
Output:
1/4*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a*b) + 1/4*(b*x^(3/2) - a*sqrt( x))/((b*x + a)^2*a*b)
Time = 4.04 (sec) , antiderivative size = 580, normalized size of antiderivative = 4.64 \[ \int \frac {\sqrt {x}}{\text {arctanh}(\tanh (a+b x))^3} \, dx =\text {Too large to display} \] Input:
int(x^(1/2)/atanh(tanh(a + b*x))^3,x)
Output:
(2^(1/2)*log(-(4*b^(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)^(1/2)*(2^(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) + 4*b^(1/2)*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)^(1/2) + 2* 2^(1/2)*b*x)*(b^3*(2*a - 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*b*x)^2 + 4*a^2*b^3 - 4*a*b^3* (2*a - 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*b*x)))/(log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*e xp(2*b*x) + 1)) - log(2/(exp(2*a)*exp(2*b*x) + 1)))))/(4*b^(3/2)*(log(2/(e xp(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)^(3/2)) - (2*x^(1/2))/(b*(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) - x^(1/2)/(b *(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)))*(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))
\[ \int \frac {\sqrt {x}}{\text {arctanh}(\tanh (a+b x))^3} \, dx=\frac {\mathit {atanh} \left (\tanh \left (b x +a \right )\right )^{2} \left (\int \frac {1}{\sqrt {x}\, \mathit {atanh} \left (\tanh \left (b x +a \right )\right )^{2}}d x \right )-2 \sqrt {x}}{4 \mathit {atanh} \left (\tanh \left (b x +a \right )\right )^{2} b} \] Input:
int(x^(1/2)/atanh(tanh(b*x+a))^3,x)
Output:
(atanh(tanh(a + b*x))**2*int(1/(sqrt(x)*atanh(tanh(a + b*x))**2),x) - 2*sq rt(x))/(4*atanh(tanh(a + b*x))**2*b)