Integrand size = 15, antiderivative size = 69 \[ \int x^{5/2} \text {arctanh}(\tanh (a+b x))^3 \, dx=-\frac {32 b^3 x^{13/2}}{3003}+\frac {16}{231} b^2 x^{11/2} \text {arctanh}(\tanh (a+b x))-\frac {4}{21} b x^{9/2} \text {arctanh}(\tanh (a+b x))^2+\frac {2}{7} x^{7/2} \text {arctanh}(\tanh (a+b x))^3 \] Output:
-32/3003*b^3*x^(13/2)+16/231*b^2*x^(11/2)*arctanh(tanh(b*x+a))-4/21*b*x^(9 /2)*arctanh(tanh(b*x+a))^2+2/7*x^(7/2)*arctanh(tanh(b*x+a))^3
Time = 0.03 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.83 \[ \int x^{5/2} \text {arctanh}(\tanh (a+b x))^3 \, dx=\frac {2 x^{7/2} \left (-16 b^3 x^3+104 b^2 x^2 \text {arctanh}(\tanh (a+b x))-286 b x \text {arctanh}(\tanh (a+b x))^2+429 \text {arctanh}(\tanh (a+b x))^3\right )}{3003} \] Input:
Integrate[x^(5/2)*ArcTanh[Tanh[a + b*x]]^3,x]
Output:
(2*x^(7/2)*(-16*b^3*x^3 + 104*b^2*x^2*ArcTanh[Tanh[a + b*x]] - 286*b*x*Arc Tanh[Tanh[a + b*x]]^2 + 429*ArcTanh[Tanh[a + b*x]]^3))/3003
Time = 0.24 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.09, 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.
\(\displaystyle \int x^{5/2} \text {arctanh}(\tanh (a+b x))^3 \, dx\) |
\(\Big \downarrow \) 2599 |
\(\displaystyle \frac {2}{7} x^{7/2} \text {arctanh}(\tanh (a+b x))^3-\frac {6}{7} b \int x^{7/2} \text {arctanh}(\tanh (a+b x))^2dx\) |
\(\Big \downarrow \) 2599 |
\(\displaystyle \frac {2}{7} x^{7/2} \text {arctanh}(\tanh (a+b x))^3-\frac {6}{7} b \left (\frac {2}{9} x^{9/2} \text {arctanh}(\tanh (a+b x))^2-\frac {4}{9} b \int x^{9/2} \text {arctanh}(\tanh (a+b x))dx\right )\) |
\(\Big \downarrow \) 2599 |
\(\displaystyle \frac {2}{7} x^{7/2} \text {arctanh}(\tanh (a+b x))^3-\frac {6}{7} b \left (\frac {2}{9} x^{9/2} \text {arctanh}(\tanh (a+b x))^2-\frac {4}{9} b \left (\frac {2}{11} x^{11/2} \text {arctanh}(\tanh (a+b x))-\frac {2}{11} b \int x^{11/2}dx\right )\right )\) |
\(\Big \downarrow \) 15 |
\(\displaystyle \frac {2}{7} x^{7/2} \text {arctanh}(\tanh (a+b x))^3-\frac {6}{7} b \left (\frac {2}{9} x^{9/2} \text {arctanh}(\tanh (a+b x))^2-\frac {4}{9} b \left (\frac {2}{11} x^{11/2} \text {arctanh}(\tanh (a+b x))-\frac {4}{143} b x^{13/2}\right )\right )\) |
Input:
Int[x^(5/2)*ArcTanh[Tanh[a + b*x]]^3,x]
Output:
(2*x^(7/2)*ArcTanh[Tanh[a + b*x]]^3)/7 - (6*b*((2*x^(9/2)*ArcTanh[Tanh[a + b*x]]^2)/9 - (4*b*((-4*b*x^(13/2))/143 + (2*x^(11/2)*ArcTanh[Tanh[a + b*x ]])/11))/9))/7
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 = 1.05 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.81
method | result | size |
derivativedivides | \(\frac {2 x^{\frac {7}{2}} \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{3}}{7}-\frac {12 b \left (\frac {x^{\frac {9}{2}} \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{2}}{9}-\frac {4 b \left (\frac {x^{\frac {11}{2}} \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}{11}-\frac {2 x^{\frac {13}{2}} b}{143}\right )}{9}\right )}{7}\) | \(56\) |
default | \(\frac {2 x^{\frac {7}{2}} \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{3}}{7}-\frac {12 b \left (\frac {x^{\frac {9}{2}} \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{2}}{9}-\frac {4 b \left (\frac {x^{\frac {11}{2}} \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}{11}-\frac {2 x^{\frac {13}{2}} b}{143}\right )}{9}\right )}{7}\) | \(56\) |
risch | \(\text {Expression too large to display}\) | \(8179\) |
Input:
int(x^(5/2)*arctanh(tanh(b*x+a))^3,x,method=_RETURNVERBOSE)
Output:
2/7*x^(7/2)*arctanh(tanh(b*x+a))^3-12/7*b*(1/9*x^(9/2)*arctanh(tanh(b*x+a) )^2-4/9*b*(1/11*x^(11/2)*arctanh(tanh(b*x+a))-2/143*x^(13/2)*b))
Time = 0.07 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.58 \[ \int x^{5/2} \text {arctanh}(\tanh (a+b x))^3 \, dx=\frac {2}{3003} \, {\left (231 \, b^{3} x^{6} + 819 \, a b^{2} x^{5} + 1001 \, a^{2} b x^{4} + 429 \, a^{3} x^{3}\right )} \sqrt {x} \] Input:
integrate(x^(5/2)*arctanh(tanh(b*x+a))^3,x, algorithm="fricas")
Output:
2/3003*(231*b^3*x^6 + 819*a*b^2*x^5 + 1001*a^2*b*x^4 + 429*a^3*x^3)*sqrt(x )
Time = 41.58 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.01 \[ \int x^{5/2} \text {arctanh}(\tanh (a+b x))^3 \, dx=- \frac {32 b^{3} x^{\frac {13}{2}}}{3003} + \frac {16 b^{2} x^{\frac {11}{2}} \operatorname {atanh}{\left (\tanh {\left (a + b x \right )} \right )}}{231} - \frac {4 b x^{\frac {9}{2}} \operatorname {atanh}^{2}{\left (\tanh {\left (a + b x \right )} \right )}}{21} + \frac {2 x^{\frac {7}{2}} \operatorname {atanh}^{3}{\left (\tanh {\left (a + b x \right )} \right )}}{7} \] Input:
integrate(x**(5/2)*atanh(tanh(b*x+a))**3,x)
Output:
-32*b**3*x**(13/2)/3003 + 16*b**2*x**(11/2)*atanh(tanh(a + b*x))/231 - 4*b *x**(9/2)*atanh(tanh(a + b*x))**2/21 + 2*x**(7/2)*atanh(tanh(a + b*x))**3/ 7
Time = 0.06 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.80 \[ \int x^{5/2} \text {arctanh}(\tanh (a+b x))^3 \, dx=-\frac {4}{21} \, b x^{\frac {9}{2}} \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{2} + \frac {2}{7} \, x^{\frac {7}{2}} \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{3} - \frac {16}{3003} \, {\left (2 \, b^{2} x^{\frac {13}{2}} - 13 \, b x^{\frac {11}{2}} \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )\right )} b \] Input:
integrate(x^(5/2)*arctanh(tanh(b*x+a))^3,x, algorithm="maxima")
Output:
-4/21*b*x^(9/2)*arctanh(tanh(b*x + a))^2 + 2/7*x^(7/2)*arctanh(tanh(b*x + a))^3 - 16/3003*(2*b^2*x^(13/2) - 13*b*x^(11/2)*arctanh(tanh(b*x + a)))*b
Time = 0.11 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.51 \[ \int x^{5/2} \text {arctanh}(\tanh (a+b x))^3 \, dx=\frac {2}{13} \, b^{3} x^{\frac {13}{2}} + \frac {6}{11} \, a b^{2} x^{\frac {11}{2}} + \frac {2}{3} \, a^{2} b x^{\frac {9}{2}} + \frac {2}{7} \, a^{3} x^{\frac {7}{2}} \] Input:
integrate(x^(5/2)*arctanh(tanh(b*x+a))^3,x, algorithm="giac")
Output:
2/13*b^3*x^(13/2) + 6/11*a*b^2*x^(11/2) + 2/3*a^2*b*x^(9/2) + 2/7*a^3*x^(7 /2)
Time = 3.13 (sec) , antiderivative size = 182, normalized size of antiderivative = 2.64 \[ \int x^{5/2} \text {arctanh}(\tanh (a+b x))^3 \, dx=\frac {2\,b^3\,x^{13/2}}{13}-\frac {x^{7/2}\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^3}{28}+\frac {b\,x^{9/2}\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^2}{6}-\frac {3\,b^2\,x^{11/2}\,\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}{11} \] Input:
int(x^(5/2)*atanh(tanh(a + b*x))^3,x)
Output:
(2*b^3*x^(13/2))/13 - (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)^3)/28 + (b*x^(9/2 )*(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)/6 - (3*b^2*x^(11/2)*(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))/11
\[ \int x^{5/2} \text {arctanh}(\tanh (a+b x))^3 \, dx=\int \sqrt {x}\, \mathit {atanh} \left (\tanh \left (b x +a \right )\right )^{3} x^{2}d x \] Input:
int(x^(5/2)*atanh(tanh(b*x+a))^3,x)
Output:
int(sqrt(x)*atanh(tanh(a + b*x))**3*x**2,x)