Integrand size = 15, antiderivative size = 80 \[ \int x^3 \sqrt {\text {arctanh}(\tanh (a+b x))} \, dx=\frac {2 x^3 \text {arctanh}(\tanh (a+b x))^{3/2}}{3 b}-\frac {4 x^2 \text {arctanh}(\tanh (a+b x))^{5/2}}{5 b^2}+\frac {16 x \text {arctanh}(\tanh (a+b x))^{7/2}}{35 b^3}-\frac {32 \text {arctanh}(\tanh (a+b x))^{9/2}}{315 b^4} \] Output:
2/3*x^3*arctanh(tanh(b*x+a))^(3/2)/b-4/5*x^2*arctanh(tanh(b*x+a))^(5/2)/b^ 2+16/35*x*arctanh(tanh(b*x+a))^(7/2)/b^3-32/315*arctanh(tanh(b*x+a))^(9/2) /b^4
Time = 0.02 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.82 \[ \int x^3 \sqrt {\text {arctanh}(\tanh (a+b x))} \, dx=\frac {2 \text {arctanh}(\tanh (a+b x))^{3/2} \left (105 b^3 x^3-126 b^2 x^2 \text {arctanh}(\tanh (a+b x))+72 b x \text {arctanh}(\tanh (a+b x))^2-16 \text {arctanh}(\tanh (a+b x))^3\right )}{315 b^4} \] Input:
Integrate[x^3*Sqrt[ArcTanh[Tanh[a + b*x]]],x]
Output:
(2*ArcTanh[Tanh[a + b*x]]^(3/2)*(105*b^3*x^3 - 126*b^2*x^2*ArcTanh[Tanh[a + b*x]] + 72*b*x*ArcTanh[Tanh[a + b*x]]^2 - 16*ArcTanh[Tanh[a + b*x]]^3))/ (315*b^4)
Time = 0.39 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.18, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {2599, 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.
\(\displaystyle \int x^3 \sqrt {\text {arctanh}(\tanh (a+b x))} \, dx\) |
\(\Big \downarrow \) 2599 |
\(\displaystyle \frac {2 x^3 \text {arctanh}(\tanh (a+b x))^{3/2}}{3 b}-\frac {2 \int x^2 \text {arctanh}(\tanh (a+b x))^{3/2}dx}{b}\) |
\(\Big \downarrow \) 2599 |
\(\displaystyle \frac {2 x^3 \text {arctanh}(\tanh (a+b x))^{3/2}}{3 b}-\frac {2 \left (\frac {2 x^2 \text {arctanh}(\tanh (a+b x))^{5/2}}{5 b}-\frac {4 \int x \text {arctanh}(\tanh (a+b x))^{5/2}dx}{5 b}\right )}{b}\) |
\(\Big \downarrow \) 2599 |
\(\displaystyle \frac {2 x^3 \text {arctanh}(\tanh (a+b x))^{3/2}}{3 b}-\frac {2 \left (\frac {2 x^2 \text {arctanh}(\tanh (a+b x))^{5/2}}{5 b}-\frac {4 \left (\frac {2 x \text {arctanh}(\tanh (a+b x))^{7/2}}{7 b}-\frac {2 \int \text {arctanh}(\tanh (a+b x))^{7/2}dx}{7 b}\right )}{5 b}\right )}{b}\) |
\(\Big \downarrow \) 2588 |
\(\displaystyle \frac {2 x^3 \text {arctanh}(\tanh (a+b x))^{3/2}}{3 b}-\frac {2 \left (\frac {2 x^2 \text {arctanh}(\tanh (a+b x))^{5/2}}{5 b}-\frac {4 \left (\frac {2 x \text {arctanh}(\tanh (a+b x))^{7/2}}{7 b}-\frac {2 \int \text {arctanh}(\tanh (a+b x))^{7/2}d\text {arctanh}(\tanh (a+b x))}{7 b^2}\right )}{5 b}\right )}{b}\) |
\(\Big \downarrow \) 15 |
\(\displaystyle \frac {2 x^3 \text {arctanh}(\tanh (a+b x))^{3/2}}{3 b}-\frac {2 \left (\frac {2 x^2 \text {arctanh}(\tanh (a+b x))^{5/2}}{5 b}-\frac {4 \left (\frac {2 x \text {arctanh}(\tanh (a+b x))^{7/2}}{7 b}-\frac {4 \text {arctanh}(\tanh (a+b x))^{9/2}}{63 b^2}\right )}{5 b}\right )}{b}\) |
Input:
Int[x^3*Sqrt[ArcTanh[Tanh[a + b*x]]],x]
Output:
(2*x^3*ArcTanh[Tanh[a + b*x]]^(3/2))/(3*b) - (2*((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*ArcT anh[Tanh[a + b*x]]^(9/2))/(63*b^2)))/(5*b)))/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 = 0.25 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.55
method | result | size |
default | \(\frac {\frac {2 \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {9}{2}}}{9}+\frac {2 \left (-3 \,\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )+3 b x \right ) \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {7}{2}}}{7}+\frac {2 \left (\left (b x -\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )\right ) \left (2 b x -2 \,\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )\right )+\left (b x -\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )\right )^{2}\right ) \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {5}{2}}}{5}+\frac {2 \left (b x -\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )\right )^{3} \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}{3}}{b^{4}}\) | \(124\) |
Input:
int(x^3*arctanh(tanh(b*x+a))^(1/2),x,method=_RETURNVERBOSE)
Output:
2/b^4*(1/9*arctanh(tanh(b*x+a))^(9/2)+1/7*(-3*arctanh(tanh(b*x+a))+3*b*x)* arctanh(tanh(b*x+a))^(7/2)+1/5*((b*x-arctanh(tanh(b*x+a)))*(2*b*x-2*arctan h(tanh(b*x+a)))+(b*x-arctanh(tanh(b*x+a)))^2)*arctanh(tanh(b*x+a))^(5/2)+1 /3*(b*x-arctanh(tanh(b*x+a)))^3*arctanh(tanh(b*x+a))^(3/2))
Time = 0.08 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.66 \[ \int x^3 \sqrt {\text {arctanh}(\tanh (a+b x))} \, dx=\frac {2 \, {\left (35 \, b^{4} x^{4} + 5 \, a b^{3} x^{3} - 6 \, a^{2} b^{2} x^{2} + 8 \, a^{3} b x - 16 \, a^{4}\right )} \sqrt {b x + a}}{315 \, b^{4}} \] Input:
integrate(x^3*arctanh(tanh(b*x+a))^(1/2),x, algorithm="fricas")
Output:
2/315*(35*b^4*x^4 + 5*a*b^3*x^3 - 6*a^2*b^2*x^2 + 8*a^3*b*x - 16*a^4)*sqrt (b*x + a)/b^4
\[ \int x^3 \sqrt {\text {arctanh}(\tanh (a+b x))} \, dx=\int x^{3} \sqrt {\operatorname {atanh}{\left (\tanh {\left (a + b x \right )} \right )}}\, dx \] Input:
integrate(x**3*atanh(tanh(b*x+a))**(1/2),x)
Output:
Integral(x**3*sqrt(atanh(tanh(a + b*x))), x)
Time = 0.19 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.66 \[ \int x^3 \sqrt {\text {arctanh}(\tanh (a+b x))} \, dx=\frac {2 \, {\left (35 \, b^{4} x^{4} + 5 \, a b^{3} x^{3} - 6 \, a^{2} b^{2} x^{2} + 8 \, a^{3} b x - 16 \, a^{4}\right )} \sqrt {b x + a}}{315 \, b^{4}} \] Input:
integrate(x^3*arctanh(tanh(b*x+a))^(1/2),x, algorithm="maxima")
Output:
2/315*(35*b^4*x^4 + 5*a*b^3*x^3 - 6*a^2*b^2*x^2 + 8*a^3*b*x - 16*a^4)*sqrt (b*x + a)/b^4
Time = 0.12 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.56 \[ \int x^3 \sqrt {\text {arctanh}(\tanh (a+b x))} \, dx=\frac {\sqrt {2} {\left (\frac {9 \, \sqrt {2} {\left (5 \, {\left (b x + a\right )}^{\frac {7}{2}} - 21 \, {\left (b x + a\right )}^{\frac {5}{2}} a + 35 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{2} - 35 \, \sqrt {b x + a} a^{3}\right )} a}{b^{3}} + \frac {\sqrt {2} {\left (35 \, {\left (b x + a\right )}^{\frac {9}{2}} - 180 \, {\left (b x + a\right )}^{\frac {7}{2}} a + 378 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{2} - 420 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{3} + 315 \, \sqrt {b x + a} a^{4}\right )}}{b^{3}}\right )}}{315 \, b} \] Input:
integrate(x^3*arctanh(tanh(b*x+a))^(1/2),x, algorithm="giac")
Output:
1/315*sqrt(2)*(9*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^3 + sqrt(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)/b^3)/b
Time = 3.08 (sec) , antiderivative size = 648, normalized size of antiderivative = 8.10 \[ \int x^3 \sqrt {\text {arctanh}(\tanh (a+b x))} \, dx =\text {Too large to display} \] Input:
int(x^3*atanh(tanh(a + b*x))^(1/2),x)
Output:
(2*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)*( log(2/(exp(2*a)*exp(2*b*x) + 1))/9 - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a) *exp(2*b*x) + 1))/9 + (2*b*x)/9))/(7*b) - (16*(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))/2 - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a )*exp(2*b*x) + 1))/2 + b*x)^3*(log(2/(exp(2*a)*exp(2*b*x) + 1))/9 - log((2 *exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/9 + (2*b*x)/9))/(35*b^4) - (6*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))/2 - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/2 + b*x)*(log(2/(e xp(2*a)*exp(2*b*x) + 1))/9 - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b *x) + 1))/9 + (2*b*x)/9))/(35*b^2) - (8*x*(log((2*exp(2*a)*exp(2*b*x))/(ex p(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))/2 - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*ex p(2*b*x) + 1))/2 + b*x)^2*(log(2/(exp(2*a)*exp(2*b*x) + 1))/9 - log((2*exp (2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/9 + (2*b*x)/9))/(35*b^3)
\[ \int x^3 \sqrt {\text {arctanh}(\tanh (a+b x))} \, dx=\int \sqrt {\mathit {atanh} \left (\tanh \left (b x +a \right )\right )}\, x^{3}d x \] Input:
int(x^3*atanh(tanh(b*x+a))^(1/2),x)
Output:
int(sqrt(atanh(tanh(a + b*x)))*x**3,x)