Integrand size = 15, antiderivative size = 69 \[ \int \sqrt {x} \text {arctanh}(\tanh (a+b x))^3 \, dx=-\frac {32}{315} b^3 x^{9/2}+\frac {16}{35} b^2 x^{7/2} \text {arctanh}(\tanh (a+b x))-\frac {4}{5} b x^{5/2} \text {arctanh}(\tanh (a+b x))^2+\frac {2}{3} x^{3/2} \text {arctanh}(\tanh (a+b x))^3 \] Output:
-32/315*b^3*x^(9/2)+16/35*b^2*x^(7/2)*arctanh(tanh(b*x+a))-4/5*b*x^(5/2)*a rctanh(tanh(b*x+a))^2+2/3*x^(3/2)*arctanh(tanh(b*x+a))^3
Time = 0.02 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.83 \[ \int \sqrt {x} \text {arctanh}(\tanh (a+b x))^3 \, dx=-\frac {2}{315} x^{3/2} \left (16 b^3 x^3-72 b^2 x^2 \text {arctanh}(\tanh (a+b x))+126 b x \text {arctanh}(\tanh (a+b x))^2-105 \text {arctanh}(\tanh (a+b x))^3\right ) \] Input:
Integrate[Sqrt[x]*ArcTanh[Tanh[a + b*x]]^3,x]
Output:
(-2*x^(3/2)*(16*b^3*x^3 - 72*b^2*x^2*ArcTanh[Tanh[a + b*x]] + 126*b*x*ArcT anh[Tanh[a + b*x]]^2 - 105*ArcTanh[Tanh[a + b*x]]^3))/315
Time = 0.24 (sec) , antiderivative size = 73, 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, 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.
\(\displaystyle \int \sqrt {x} \text {arctanh}(\tanh (a+b x))^3 \, dx\) |
\(\Big \downarrow \) 2599 |
\(\displaystyle \frac {2}{3} x^{3/2} \text {arctanh}(\tanh (a+b x))^3-2 b \int x^{3/2} \text {arctanh}(\tanh (a+b x))^2dx\) |
\(\Big \downarrow \) 2599 |
\(\displaystyle \frac {2}{3} x^{3/2} \text {arctanh}(\tanh (a+b x))^3-2 b \left (\frac {2}{5} x^{5/2} \text {arctanh}(\tanh (a+b x))^2-\frac {4}{5} b \int x^{5/2} \text {arctanh}(\tanh (a+b x))dx\right )\) |
\(\Big \downarrow \) 2599 |
\(\displaystyle \frac {2}{3} x^{3/2} \text {arctanh}(\tanh (a+b x))^3-2 b \left (\frac {2}{5} x^{5/2} \text {arctanh}(\tanh (a+b x))^2-\frac {4}{5} b \left (\frac {2}{7} x^{7/2} \text {arctanh}(\tanh (a+b x))-\frac {2}{7} b \int x^{7/2}dx\right )\right )\) |
\(\Big \downarrow \) 15 |
\(\displaystyle \frac {2}{3} x^{3/2} \text {arctanh}(\tanh (a+b x))^3-2 b \left (\frac {2}{5} x^{5/2} \text {arctanh}(\tanh (a+b x))^2-\frac {4}{5} b \left (\frac {2}{7} x^{7/2} \text {arctanh}(\tanh (a+b x))-\frac {4}{63} b x^{9/2}\right )\right )\) |
Input:
Int[Sqrt[x]*ArcTanh[Tanh[a + b*x]]^3,x]
Output:
(2*x^(3/2)*ArcTanh[Tanh[a + b*x]]^3)/3 - 2*b*((2*x^(5/2)*ArcTanh[Tanh[a + b*x]]^2)/5 - (4*b*((-4*b*x^(9/2))/63 + (2*x^(7/2)*ArcTanh[Tanh[a + b*x]])/ 7))/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 = 0.98 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.81
method | result | size |
derivativedivides | \(\frac {2 x^{\frac {3}{2}} \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{3}}{3}-4 b \left (\frac {x^{\frac {5}{2}} \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{2}}{5}-\frac {4 b \left (\frac {x^{\frac {7}{2}} \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}{7}-\frac {2 b \,x^{\frac {9}{2}}}{63}\right )}{5}\right )\) | \(56\) |
default | \(\frac {2 x^{\frac {3}{2}} \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{3}}{3}-4 b \left (\frac {x^{\frac {5}{2}} \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{2}}{5}-\frac {4 b \left (\frac {x^{\frac {7}{2}} \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}{7}-\frac {2 b \,x^{\frac {9}{2}}}{63}\right )}{5}\right )\) | \(56\) |
risch | \(\text {Expression too large to display}\) | \(7981\) |
Input:
int(x^(1/2)*arctanh(tanh(b*x+a))^3,x,method=_RETURNVERBOSE)
Output:
2/3*x^(3/2)*arctanh(tanh(b*x+a))^3-4*b*(1/5*x^(5/2)*arctanh(tanh(b*x+a))^2 -4/5*b*(1/7*x^(7/2)*arctanh(tanh(b*x+a))-2/63*b*x^(9/2)))
Time = 0.07 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.55 \[ \int \sqrt {x} \text {arctanh}(\tanh (a+b x))^3 \, dx=\frac {2}{315} \, {\left (35 \, b^{3} x^{4} + 135 \, a b^{2} x^{3} + 189 \, a^{2} b x^{2} + 105 \, a^{3} x\right )} \sqrt {x} \] Input:
integrate(x^(1/2)*arctanh(tanh(b*x+a))^3,x, algorithm="fricas")
Output:
2/315*(35*b^3*x^4 + 135*a*b^2*x^3 + 189*a^2*b*x^2 + 105*a^3*x)*sqrt(x)
\[ \int \sqrt {x} \text {arctanh}(\tanh (a+b x))^3 \, dx=\int \sqrt {x} \operatorname {atanh}^{3}{\left (\tanh {\left (a + b x \right )} \right )}\, dx \] Input:
integrate(x**(1/2)*atanh(tanh(b*x+a))**3,x)
Output:
Integral(sqrt(x)*atanh(tanh(a + b*x))**3, x)
Time = 0.06 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.80 \[ \int \sqrt {x} \text {arctanh}(\tanh (a+b x))^3 \, dx=-\frac {4}{5} \, b x^{\frac {5}{2}} \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{2} + \frac {2}{3} \, x^{\frac {3}{2}} \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{3} - \frac {16}{315} \, {\left (2 \, b^{2} x^{\frac {9}{2}} - 9 \, b x^{\frac {7}{2}} \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )\right )} b \] Input:
integrate(x^(1/2)*arctanh(tanh(b*x+a))^3,x, algorithm="maxima")
Output:
-4/5*b*x^(5/2)*arctanh(tanh(b*x + a))^2 + 2/3*x^(3/2)*arctanh(tanh(b*x + a ))^3 - 16/315*(2*b^2*x^(9/2) - 9*b*x^(7/2)*arctanh(tanh(b*x + a)))*b
Time = 0.11 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.51 \[ \int \sqrt {x} \text {arctanh}(\tanh (a+b x))^3 \, dx=\frac {2}{9} \, b^{3} x^{\frac {9}{2}} + \frac {6}{7} \, a b^{2} x^{\frac {7}{2}} + \frac {6}{5} \, a^{2} b x^{\frac {5}{2}} + \frac {2}{3} \, a^{3} x^{\frac {3}{2}} \] Input:
integrate(x^(1/2)*arctanh(tanh(b*x+a))^3,x, algorithm="giac")
Output:
2/9*b^3*x^(9/2) + 6/7*a*b^2*x^(7/2) + 6/5*a^2*b*x^(5/2) + 2/3*a^3*x^(3/2)
Time = 3.14 (sec) , antiderivative size = 182, normalized size of antiderivative = 2.64 \[ \int \sqrt {x} \text {arctanh}(\tanh (a+b x))^3 \, dx=\frac {2\,b^3\,x^{9/2}}{9}-\frac {x^{3/2}\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^3}{12}+\frac {3\,b\,x^{5/2}\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^2}{10}-\frac {3\,b^2\,x^{7/2}\,\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}{7} \] Input:
int(x^(1/2)*atanh(tanh(a + b*x))^3,x)
Output:
(2*b^3*x^(9/2))/9 - (x^(3/2)*(log(2/(exp(2*a)*exp(2*b*x) + 1)) - log((2*ex p(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x)^3)/12 + (3*b*x^(5/2 )*(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)^2)/10 - (3*b^2*x^(7/2)*(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))/7
\[ \int \sqrt {x} \text {arctanh}(\tanh (a+b x))^3 \, dx=\int \sqrt {x}\, \mathit {atanh} \left (\tanh \left (b x +a \right )\right )^{3}d x \] Input:
int(x^(1/2)*atanh(tanh(b*x+a))^3,x)
Output:
int(sqrt(x)*atanh(tanh(a + b*x))**3,x)