Integrand size = 15, antiderivative size = 59 \[ \int x^2 \text {arctanh}(\tanh (a+b x))^{5/2} \, dx=\frac {2 x^2 \text {arctanh}(\tanh (a+b x))^{7/2}}{7 b}-\frac {8 x \text {arctanh}(\tanh (a+b x))^{9/2}}{63 b^2}+\frac {16 \text {arctanh}(\tanh (a+b x))^{11/2}}{693 b^3} \] Output:
2/7*x^2*arctanh(tanh(b*x+a))^(7/2)/b-8/63*x*arctanh(tanh(b*x+a))^(9/2)/b^2 +16/693*arctanh(tanh(b*x+a))^(11/2)/b^3
Time = 0.02 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.83 \[ \int x^2 \text {arctanh}(\tanh (a+b x))^{5/2} \, dx=\frac {2 \text {arctanh}(\tanh (a+b x))^{7/2} \left (99 b^2 x^2-44 b x \text {arctanh}(\tanh (a+b x))+8 \text {arctanh}(\tanh (a+b x))^2\right )}{693 b^3} \] Input:
Integrate[x^2*ArcTanh[Tanh[a + b*x]]^(5/2),x]
Output:
(2*ArcTanh[Tanh[a + b*x]]^(7/2)*(99*b^2*x^2 - 44*b*x*ArcTanh[Tanh[a + b*x] ] + 8*ArcTanh[Tanh[a + b*x]]^2))/(693*b^3)
Time = 0.23 (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.
\(\displaystyle \int x^2 \text {arctanh}(\tanh (a+b x))^{5/2} \, dx\) |
\(\Big \downarrow \) 2599 |
\(\displaystyle \frac {2 x^2 \text {arctanh}(\tanh (a+b x))^{7/2}}{7 b}-\frac {4 \int x \text {arctanh}(\tanh (a+b x))^{7/2}dx}{7 b}\) |
\(\Big \downarrow \) 2599 |
\(\displaystyle \frac {2 x^2 \text {arctanh}(\tanh (a+b x))^{7/2}}{7 b}-\frac {4 \left (\frac {2 x \text {arctanh}(\tanh (a+b x))^{9/2}}{9 b}-\frac {2 \int \text {arctanh}(\tanh (a+b x))^{9/2}dx}{9 b}\right )}{7 b}\) |
\(\Big \downarrow \) 2588 |
\(\displaystyle \frac {2 x^2 \text {arctanh}(\tanh (a+b x))^{7/2}}{7 b}-\frac {4 \left (\frac {2 x \text {arctanh}(\tanh (a+b x))^{9/2}}{9 b}-\frac {2 \int \text {arctanh}(\tanh (a+b x))^{9/2}d\text {arctanh}(\tanh (a+b x))}{9 b^2}\right )}{7 b}\) |
\(\Big \downarrow \) 15 |
\(\displaystyle \frac {2 x^2 \text {arctanh}(\tanh (a+b x))^{7/2}}{7 b}-\frac {4 \left (\frac {2 x \text {arctanh}(\tanh (a+b x))^{9/2}}{9 b}-\frac {4 \text {arctanh}(\tanh (a+b x))^{11/2}}{99 b^2}\right )}{7 b}\) |
Input:
Int[x^2*ArcTanh[Tanh[a + b*x]]^(5/2),x]
Output:
(2*x^2*ArcTanh[Tanh[a + b*x]]^(7/2))/(7*b) - (4*((2*x*ArcTanh[Tanh[a + b*x ]]^(9/2))/(9*b) - (4*ArcTanh[Tanh[a + b*x]]^(11/2))/(99*b^2)))/(7*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 = 0.26 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.17
method | result | size |
default | \(\frac {\frac {2 \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {11}{2}}}{11}+\frac {2 \left (2 b x -2 \,\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )\right ) \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {9}{2}}}{9}+\frac {2 \left (b x -\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )\right )^{2} \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {7}{2}}}{7}}{b^{3}}\) | \(69\) |
Input:
int(x^2*arctanh(tanh(b*x+a))^(5/2),x,method=_RETURNVERBOSE)
Output:
2/b^3*(1/11*arctanh(tanh(b*x+a))^(11/2)+1/9*(2*b*x-2*arctanh(tanh(b*x+a))) *arctanh(tanh(b*x+a))^(9/2)+1/7*(b*x-arctanh(tanh(b*x+a)))^2*arctanh(tanh( b*x+a))^(7/2))
Time = 0.09 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.08 \[ \int x^2 \text {arctanh}(\tanh (a+b x))^{5/2} \, dx=\frac {2 \, {\left (63 \, b^{5} x^{5} + 161 \, a b^{4} x^{4} + 113 \, a^{2} b^{3} x^{3} + 3 \, a^{3} b^{2} x^{2} - 4 \, a^{4} b x + 8 \, a^{5}\right )} \sqrt {b x + a}}{693 \, b^{3}} \] Input:
integrate(x^2*arctanh(tanh(b*x+a))^(5/2),x, algorithm="fricas")
Output:
2/693*(63*b^5*x^5 + 161*a*b^4*x^4 + 113*a^2*b^3*x^3 + 3*a^3*b^2*x^2 - 4*a^ 4*b*x + 8*a^5)*sqrt(b*x + a)/b^3
Time = 40.76 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.20 \[ \int x^2 \text {arctanh}(\tanh (a+b x))^{5/2} \, dx=\begin {cases} \frac {2 x^{2} \operatorname {atanh}^{\frac {7}{2}}{\left (\tanh {\left (a + b x \right )} \right )}}{7 b} - \frac {8 x \operatorname {atanh}^{\frac {9}{2}}{\left (\tanh {\left (a + b x \right )} \right )}}{63 b^{2}} + \frac {16 \operatorname {atanh}^{\frac {11}{2}}{\left (\tanh {\left (a + b x \right )} \right )}}{693 b^{3}} & \text {for}\: b \neq 0 \\\frac {x^{3} \operatorname {atanh}^{\frac {5}{2}}{\left (\tanh {\left (a \right )} \right )}}{3} & \text {otherwise} \end {cases} \] Input:
integrate(x**2*atanh(tanh(b*x+a))**(5/2),x)
Output:
Piecewise((2*x**2*atanh(tanh(a + b*x))**(7/2)/(7*b) - 8*x*atanh(tanh(a + b *x))**(9/2)/(63*b**2) + 16*atanh(tanh(a + b*x))**(11/2)/(693*b**3), Ne(b, 0)), (x**3*atanh(tanh(a))**(5/2)/3, True))
Time = 0.20 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.71 \[ \int x^2 \text {arctanh}(\tanh (a+b x))^{5/2} \, dx=\frac {2 \, {\left (63 \, b^{3} x^{3} + 35 \, a b^{2} x^{2} - 20 \, a^{2} b x + 8 \, a^{3}\right )} {\left (b x + a\right )}^{\frac {5}{2}}}{693 \, b^{3}} \] Input:
integrate(x^2*arctanh(tanh(b*x+a))^(5/2),x, algorithm="maxima")
Output:
2/693*(63*b^3*x^3 + 35*a*b^2*x^2 - 20*a^2*b*x + 8*a^3)*(b*x + a)^(5/2)/b^3
Leaf count of result is larger than twice the leaf count of optimal. 248 vs. \(2 (47) = 94\).
Time = 0.12 (sec) , antiderivative size = 248, normalized size of antiderivative = 4.20 \[ \int x^2 \text {arctanh}(\tanh (a+b x))^{5/2} \, dx=\frac {\sqrt {2} {\left (\frac {231 \, \sqrt {2} {\left (3 \, {\left (b x + a\right )}^{\frac {5}{2}} - 10 \, {\left (b x + a\right )}^{\frac {3}{2}} a + 15 \, \sqrt {b x + a} a^{2}\right )} a^{3}}{b^{2}} + \frac {297 \, \sqrt {2} {\left (5 \, {\left (b x + a\right )}^{\frac {7}{2}} - 21 \, {\left (b x + a\right )}^{\frac {5}{2}} a + 35 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{2} - 35 \, \sqrt {b x + a} a^{3}\right )} a^{2}}{b^{2}} + \frac {33 \, \sqrt {2} {\left (35 \, {\left (b x + a\right )}^{\frac {9}{2}} - 180 \, {\left (b x + a\right )}^{\frac {7}{2}} a + 378 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{2} - 420 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{3} + 315 \, \sqrt {b x + a} a^{4}\right )} a}{b^{2}} + \frac {5 \, \sqrt {2} {\left (63 \, {\left (b x + a\right )}^{\frac {11}{2}} - 385 \, {\left (b x + a\right )}^{\frac {9}{2}} a + 990 \, {\left (b x + a\right )}^{\frac {7}{2}} a^{2} - 1386 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{3} + 1155 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{4} - 693 \, \sqrt {b x + a} a^{5}\right )}}{b^{2}}\right )}}{3465 \, b} \] Input:
integrate(x^2*arctanh(tanh(b*x+a))^(5/2),x, algorithm="giac")
Output:
1/3465*sqrt(2)*(231*sqrt(2)*(3*(b*x + a)^(5/2) - 10*(b*x + a)^(3/2)*a + 15 *sqrt(b*x + a)*a^2)*a^3/b^2 + 297*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^2/b^2 + 33*sq rt(2)*(35*(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)*a/b^2 + 5*sqrt(2)*(63 *(b*x + a)^(11/2) - 385*(b*x + a)^(9/2)*a + 990*(b*x + a)^(7/2)*a^2 - 1386 *(b*x + a)^(5/2)*a^3 + 1155*(b*x + a)^(3/2)*a^4 - 693*sqrt(b*x + a)*a^5)/b ^2)/b
Time = 3.25 (sec) , antiderivative size = 1789, normalized size of antiderivative = 30.32 \[ \int x^2 \text {arctanh}(\tanh (a+b x))^{5/2} \, dx=\text {Too large to display} \] Input:
int(x^2*atanh(tanh(a + b*x))^(5/2),x)
Output:
(2*b^2*x^5*(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))/11 + (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)*((3*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)^2)/2 - (8*(3*b^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) - (20*b^2*(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))/11)*(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*b)))/(7*b) - (x^4*(3*b^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) - (20*b^2*(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))/11)*(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*b) - (x^2*(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)^3/4 - (6*((3*b*(log(2/(ex p(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 - (8*(3*b^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) - (20*b^2*(l...
\[ \int x^2 \text {arctanh}(\tanh (a+b x))^{5/2} \, dx=\int \sqrt {\mathit {atanh} \left (\tanh \left (b x +a \right )\right )}\, \mathit {atanh} \left (\tanh \left (b x +a \right )\right )^{2} x^{2}d x \] Input:
int(x^2*atanh(tanh(b*x+a))^(5/2),x)
Output:
int(sqrt(atanh(tanh(a + b*x)))*atanh(tanh(a + b*x))**2*x**2,x)