Integrand size = 15, antiderivative size = 59 \[ \int x^2 \text {arctanh}(\tanh (a+b x))^{3/2} \, dx=\frac {2 x^2 \text {arctanh}(\tanh (a+b x))^{5/2}}{5 b}-\frac {8 x \text {arctanh}(\tanh (a+b x))^{7/2}}{35 b^2}+\frac {16 \text {arctanh}(\tanh (a+b x))^{9/2}}{315 b^3} \] Output:
2/5*x^2*arctanh(tanh(b*x+a))^(5/2)/b-8/35*x*arctanh(tanh(b*x+a))^(7/2)/b^2 +16/315*arctanh(tanh(b*x+a))^(9/2)/b^3
Time = 0.02 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.83 \[ \int x^2 \text {arctanh}(\tanh (a+b x))^{3/2} \, dx=\frac {2 \text {arctanh}(\tanh (a+b x))^{5/2} \left (63 b^2 x^2-36 b x \text {arctanh}(\tanh (a+b x))+8 \text {arctanh}(\tanh (a+b x))^2\right )}{315 b^3} \] Input:
Integrate[x^2*ArcTanh[Tanh[a + b*x]]^(3/2),x]
Output:
(2*ArcTanh[Tanh[a + b*x]]^(5/2)*(63*b^2*x^2 - 36*b*x*ArcTanh[Tanh[a + b*x] ] + 8*ArcTanh[Tanh[a + b*x]]^2))/(315*b^3)
Time = 0.33 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.14, 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.
\(\displaystyle \int x^2 \text {arctanh}(\tanh (a+b x))^{3/2} \, dx\) |
\(\Big \downarrow \) 2599 |
\(\displaystyle \frac {2 x^2 \text {arctanh}(\tanh (a+b x))^{5/2}}{5 b}-\frac {4 \int x \text {arctanh}(\tanh (a+b x))^{5/2}dx}{5 b}\) |
\(\Big \downarrow \) 2599 |
\(\displaystyle \frac {2 x^2 \text {arctanh}(\tanh (a+b x))^{5/2}}{5 b}-\frac {4 \left (\frac {2 x \text {arctanh}(\tanh (a+b x))^{7/2}}{7 b}-\frac {2 \int \text {arctanh}(\tanh (a+b x))^{7/2}dx}{7 b}\right )}{5 b}\) |
\(\Big \downarrow \) 2588 |
\(\displaystyle \frac {2 x^2 \text {arctanh}(\tanh (a+b x))^{5/2}}{5 b}-\frac {4 \left (\frac {2 x \text {arctanh}(\tanh (a+b x))^{7/2}}{7 b}-\frac {2 \int \text {arctanh}(\tanh (a+b x))^{7/2}d\text {arctanh}(\tanh (a+b x))}{7 b^2}\right )}{5 b}\) |
\(\Big \downarrow \) 15 |
\(\displaystyle \frac {2 x^2 \text {arctanh}(\tanh (a+b x))^{5/2}}{5 b}-\frac {4 \left (\frac {2 x \text {arctanh}(\tanh (a+b x))^{7/2}}{7 b}-\frac {4 \text {arctanh}(\tanh (a+b x))^{9/2}}{63 b^2}\right )}{5 b}\) |
Input:
Int[x^2*ArcTanh[Tanh[a + b*x]]^(3/2),x]
Output:
(2*x^2*ArcTanh[Tanh[a + b*x]]^(5/2))/(5*b) - (4*((2*x*ArcTanh[Tanh[a + b*x ]]^(7/2))/(7*b) - (4*ArcTanh[Tanh[a + b*x]]^(9/2))/(63*b^2)))/(5*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 = 0.24 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.17
method | result | size |
default | \(\frac {\frac {2 \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {9}{2}}}{9}+\frac {2 \left (2 b x -2 \,\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )\right ) \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {7}{2}}}{7}+\frac {2 \left (b x -\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )\right )^{2} \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {5}{2}}}{5}}{b^{3}}\) | \(69\) |
Input:
int(x^2*arctanh(tanh(b*x+a))^(3/2),x,method=_RETURNVERBOSE)
Output:
2/b^3*(1/9*arctanh(tanh(b*x+a))^(9/2)+1/7*(2*b*x-2*arctanh(tanh(b*x+a)))*a rctanh(tanh(b*x+a))^(7/2)+1/5*(b*x-arctanh(tanh(b*x+a)))^2*arctanh(tanh(b* x+a))^(5/2))
Time = 0.08 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.90 \[ \int x^2 \text {arctanh}(\tanh (a+b x))^{3/2} \, dx=\frac {2 \, {\left (35 \, b^{4} x^{4} + 50 \, a b^{3} x^{3} + 3 \, a^{2} b^{2} x^{2} - 4 \, a^{3} b x + 8 \, a^{4}\right )} \sqrt {b x + a}}{315 \, b^{3}} \] Input:
integrate(x^2*arctanh(tanh(b*x+a))^(3/2),x, algorithm="fricas")
Output:
2/315*(35*b^4*x^4 + 50*a*b^3*x^3 + 3*a^2*b^2*x^2 - 4*a^3*b*x + 8*a^4)*sqrt (b*x + a)/b^3
\[ \int x^2 \text {arctanh}(\tanh (a+b x))^{3/2} \, dx=\int x^{2} \operatorname {atanh}^{\frac {3}{2}}{\left (\tanh {\left (a + b x \right )} \right )}\, dx \] Input:
integrate(x**2*atanh(tanh(b*x+a))**(3/2),x)
Output:
Integral(x**2*atanh(tanh(a + b*x))**(3/2), x)
Time = 0.19 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.71 \[ \int x^2 \text {arctanh}(\tanh (a+b x))^{3/2} \, dx=\frac {2 \, {\left (35 \, b^{3} x^{3} + 15 \, a b^{2} x^{2} - 12 \, a^{2} b x + 8 \, a^{3}\right )} {\left (b x + a\right )}^{\frac {3}{2}}}{315 \, b^{3}} \] Input:
integrate(x^2*arctanh(tanh(b*x+a))^(3/2),x, algorithm="maxima")
Output:
2/315*(35*b^3*x^3 + 15*a*b^2*x^2 - 12*a^2*b*x + 8*a^3)*(b*x + a)^(3/2)/b^3
Leaf count of result is larger than twice the leaf count of optimal. 168 vs. \(2 (47) = 94\).
Time = 0.12 (sec) , antiderivative size = 168, normalized size of antiderivative = 2.85 \[ \int x^2 \text {arctanh}(\tanh (a+b x))^{3/2} \, dx=\frac {\sqrt {2} {\left (\frac {21 \, \sqrt {2} {\left (3 \, {\left (b x + a\right )}^{\frac {5}{2}} - 10 \, {\left (b x + a\right )}^{\frac {3}{2}} a + 15 \, \sqrt {b x + a} a^{2}\right )} a^{2}}{b^{2}} + \frac {18 \, \sqrt {2} {\left (5 \, {\left (b x + a\right )}^{\frac {7}{2}} - 21 \, {\left (b x + a\right )}^{\frac {5}{2}} a + 35 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{2} - 35 \, \sqrt {b x + a} a^{3}\right )} a}{b^{2}} + \frac {\sqrt {2} {\left (35 \, {\left (b x + a\right )}^{\frac {9}{2}} - 180 \, {\left (b x + a\right )}^{\frac {7}{2}} a + 378 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{2} - 420 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{3} + 315 \, \sqrt {b x + a} a^{4}\right )}}{b^{2}}\right )}}{315 \, b} \] Input:
integrate(x^2*arctanh(tanh(b*x+a))^(3/2),x, algorithm="giac")
Output:
1/315*sqrt(2)*(21*sqrt(2)*(3*(b*x + a)^(5/2) - 10*(b*x + a)^(3/2)*a + 15*s qrt(b*x + a)*a^2)*a^2/b^2 + 18*sqrt(2)*(5*(b*x + a)^(7/2) - 21*(b*x + a)^( 5/2)*a + 35*(b*x + a)^(3/2)*a^2 - 35*sqrt(b*x + a)*a^3)*a/b^2 + sqrt(2)*(3 5*(b*x + a)^(9/2) - 180*(b*x + a)^(7/2)*a + 378*(b*x + a)^(5/2)*a^2 - 420* (b*x + a)^(3/2)*a^3 + 315*sqrt(b*x + a)*a^4)/b^2)/b
Time = 3.18 (sec) , antiderivative size = 1153, normalized size of antiderivative = 19.54 \[ \int x^2 \text {arctanh}(\tanh (a+b x))^{3/2} \, dx=\text {Too large to display} \] Input:
int(x^2*atanh(tanh(a + b*x))^(3/2),x)
Output:
(2*b*x^4*(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))/9 + (x^3*(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*b*(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 + b*x))/9 - 2*b*(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*b) + (x^2*(log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/2 - lo g(2/(exp(2*a)*exp(2*b*x) + 1))/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)^2/2 + ( 6*((16*b*(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 + b*x))/9 - 2*b*(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))*( 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 + b*x))/(7*b)))/(5*b) + (8*(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) *((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)^2/2 + (6*((16*b*(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 + b*x))/9 - 2*b*(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))*(log(2/(exp(2*a)*exp(2*b*x) + 1))/2 -...
\[ \int x^2 \text {arctanh}(\tanh (a+b x))^{3/2} \, dx=\int \sqrt {\mathit {atanh} \left (\tanh \left (b x +a \right )\right )}\, \mathit {atanh} \left (\tanh \left (b x +a \right )\right ) x^{2}d x \] Input:
int(x^2*atanh(tanh(b*x+a))^(3/2),x)
Output:
int(sqrt(atanh(tanh(a + b*x)))*atanh(tanh(a + b*x))*x**2,x)