Integrand size = 13, antiderivative size = 71 \[ \int \frac {x^3}{\text {arctanh}(\tanh (a+b x))^3} \, dx=\frac {3 x}{b^3}-\frac {x^3}{2 b \text {arctanh}(\tanh (a+b x))^2}-\frac {3 x^2}{2 b^2 \text {arctanh}(\tanh (a+b x))}+\frac {3 (b x-\text {arctanh}(\tanh (a+b x))) \log (\text {arctanh}(\tanh (a+b x)))}{b^4} \] Output:
3*x/b^3-1/2*x^3/b/arctanh(tanh(b*x+a))^2-3/2*x^2/b^2/arctanh(tanh(b*x+a))+ 3*(b*x-arctanh(tanh(b*x+a)))*ln(arctanh(tanh(b*x+a)))/b^4
Time = 0.05 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.21 \[ \int \frac {x^3}{\text {arctanh}(\tanh (a+b x))^3} \, dx=-\frac {b^3 x^3+3 b^2 x^2 \text {arctanh}(\tanh (a+b x))+\text {arctanh}(\tanh (a+b x))^3 (5+6 \log (\text {arctanh}(\tanh (a+b x))))-b x \text {arctanh}(\tanh (a+b x))^2 (11+6 \log (\text {arctanh}(\tanh (a+b x))))}{2 b^4 \text {arctanh}(\tanh (a+b x))^2} \] Input:
Integrate[x^3/ArcTanh[Tanh[a + b*x]]^3,x]
Output:
-1/2*(b^3*x^3 + 3*b^2*x^2*ArcTanh[Tanh[a + b*x]] + ArcTanh[Tanh[a + b*x]]^ 3*(5 + 6*Log[ArcTanh[Tanh[a + b*x]]]) - b*x*ArcTanh[Tanh[a + b*x]]^2*(11 + 6*Log[ArcTanh[Tanh[a + b*x]]]))/(b^4*ArcTanh[Tanh[a + b*x]]^2)
Time = 0.30 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.14, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {2599, 2599, 2589, 2588, 14}
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.
\(\displaystyle \int \frac {x^3}{\text {arctanh}(\tanh (a+b x))^3} \, dx\) |
\(\Big \downarrow \) 2599 |
\(\displaystyle \frac {3 \int \frac {x^2}{\text {arctanh}(\tanh (a+b x))^2}dx}{2 b}-\frac {x^3}{2 b \text {arctanh}(\tanh (a+b x))^2}\) |
\(\Big \downarrow \) 2599 |
\(\displaystyle \frac {3 \left (\frac {2 \int \frac {x}{\text {arctanh}(\tanh (a+b x))}dx}{b}-\frac {x^2}{b \text {arctanh}(\tanh (a+b x))}\right )}{2 b}-\frac {x^3}{2 b \text {arctanh}(\tanh (a+b x))^2}\) |
\(\Big \downarrow \) 2589 |
\(\displaystyle \frac {3 \left (\frac {2 \left (\frac {(b x-\text {arctanh}(\tanh (a+b x))) \int \frac {1}{\text {arctanh}(\tanh (a+b x))}dx}{b}+\frac {x}{b}\right )}{b}-\frac {x^2}{b \text {arctanh}(\tanh (a+b x))}\right )}{2 b}-\frac {x^3}{2 b \text {arctanh}(\tanh (a+b x))^2}\) |
\(\Big \downarrow \) 2588 |
\(\displaystyle \frac {3 \left (\frac {2 \left (\frac {(b x-\text {arctanh}(\tanh (a+b x))) \int \frac {1}{\text {arctanh}(\tanh (a+b x))}d\text {arctanh}(\tanh (a+b x))}{b^2}+\frac {x}{b}\right )}{b}-\frac {x^2}{b \text {arctanh}(\tanh (a+b x))}\right )}{2 b}-\frac {x^3}{2 b \text {arctanh}(\tanh (a+b x))^2}\) |
\(\Big \downarrow \) 14 |
\(\displaystyle \frac {3 \left (\frac {2 \left (\frac {(b x-\text {arctanh}(\tanh (a+b x))) \log (\text {arctanh}(\tanh (a+b x)))}{b^2}+\frac {x}{b}\right )}{b}-\frac {x^2}{b \text {arctanh}(\tanh (a+b x))}\right )}{2 b}-\frac {x^3}{2 b \text {arctanh}(\tanh (a+b x))^2}\) |
Input:
Int[x^3/ArcTanh[Tanh[a + b*x]]^3,x]
Output:
-1/2*x^3/(b*ArcTanh[Tanh[a + b*x]]^2) + (3*(-(x^2/(b*ArcTanh[Tanh[a + b*x] ])) + (2*(x/b + ((b*x - ArcTanh[Tanh[a + b*x]])*Log[ArcTanh[Tanh[a + b*x]] ])/b^2))/b))/(2*b)
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[(v_)/(u_), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, Simp[b*(x/a), x] - Simp[(b*u - a*v)/a Int[1/u, x], x] /; NeQ[b*u - a*v, 0]] /; PiecewiseLinearQ[u, v, 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]))
Leaf count of result is larger than twice the leaf count of optimal. \(168\) vs. \(2(67)=134\).
Time = 0.23 (sec) , antiderivative size = 169, normalized size of antiderivative = 2.38
method | result | size |
default | \(\frac {x}{b^{3}}-\frac {3 a^{2}+6 a \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )+3 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2}}{b^{4} \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}-\frac {-a^{3}-3 a^{2} \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )-3 a \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2}-\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )^{3}}{2 b^{4} \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{2}}+\frac {\left (-3 \,\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )+3 b x \right ) \ln \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )\right )}{b^{4}}\) | \(169\) |
risch | \(\text {Expression too large to display}\) | \(3408\) |
Input:
int(x^3/arctanh(tanh(b*x+a))^3,x,method=_RETURNVERBOSE)
Output:
x/b^3-(3*a^2+6*a*(arctanh(tanh(b*x+a))-b*x-a)+3*(arctanh(tanh(b*x+a))-b*x- a)^2)/b^4/arctanh(tanh(b*x+a))-1/2*(-a^3-3*a^2*(arctanh(tanh(b*x+a))-b*x-a )-3*a*(arctanh(tanh(b*x+a))-b*x-a)^2-(arctanh(tanh(b*x+a))-b*x-a)^3)/b^4/a rctanh(tanh(b*x+a))^2+(-3*arctanh(tanh(b*x+a))+3*b*x)/b^4*ln(arctanh(tanh( b*x+a)))
Time = 0.09 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.17 \[ \int \frac {x^3}{\text {arctanh}(\tanh (a+b x))^3} \, dx=\frac {2 \, b^{3} x^{3} + 4 \, a b^{2} x^{2} - 4 \, a^{2} b x - 5 \, a^{3} - 6 \, {\left (a b^{2} x^{2} + 2 \, a^{2} b x + a^{3}\right )} \log \left (b x + a\right )}{2 \, {\left (b^{6} x^{2} + 2 \, a b^{5} x + a^{2} b^{4}\right )}} \] Input:
integrate(x^3/arctanh(tanh(b*x+a))^3,x, algorithm="fricas")
Output:
1/2*(2*b^3*x^3 + 4*a*b^2*x^2 - 4*a^2*b*x - 5*a^3 - 6*(a*b^2*x^2 + 2*a^2*b* x + a^3)*log(b*x + a))/(b^6*x^2 + 2*a*b^5*x + a^2*b^4)
\[ \int \frac {x^3}{\text {arctanh}(\tanh (a+b x))^3} \, dx=\int \frac {x^{3}}{\operatorname {atanh}^{3}{\left (\tanh {\left (a + b x \right )} \right )}}\, dx \] Input:
integrate(x**3/atanh(tanh(b*x+a))**3,x)
Output:
Integral(x**3/atanh(tanh(a + b*x))**3, x)
Time = 0.59 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.97 \[ \int \frac {x^3}{\text {arctanh}(\tanh (a+b x))^3} \, dx=\frac {2 \, b^{3} x^{3} + 4 \, a b^{2} x^{2} - 4 \, a^{2} b x - 5 \, a^{3}}{2 \, {\left (b^{6} x^{2} + 2 \, a b^{5} x + a^{2} b^{4}\right )}} - \frac {3 \, a \log \left (b x + a\right )}{b^{4}} \] Input:
integrate(x^3/arctanh(tanh(b*x+a))^3,x, algorithm="maxima")
Output:
1/2*(2*b^3*x^3 + 4*a*b^2*x^2 - 4*a^2*b*x - 5*a^3)/(b^6*x^2 + 2*a*b^5*x + a ^2*b^4) - 3*a*log(b*x + a)/b^4
Time = 0.12 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.62 \[ \int \frac {x^3}{\text {arctanh}(\tanh (a+b x))^3} \, dx=\frac {x}{b^{3}} - \frac {3 \, a \log \left ({\left | b x + a \right |}\right )}{b^{4}} - \frac {6 \, a^{2} b x + 5 \, a^{3}}{2 \, {\left (b x + a\right )}^{2} b^{4}} \] Input:
integrate(x^3/arctanh(tanh(b*x+a))^3,x, algorithm="giac")
Output:
x/b^3 - 3*a*log(abs(b*x + a))/b^4 - 1/2*(6*a^2*b*x + 5*a^3)/((b*x + a)^2*b ^4)
Time = 3.31 (sec) , antiderivative size = 620, normalized size of antiderivative = 8.73 \[ \int \frac {x^3}{\text {arctanh}(\tanh (a+b x))^3} \, dx =\text {Too large to display} \] Input:
int(x^3/atanh(tanh(a + b*x))^3,x)
Output:
x/b^3 - (x*(3*(2*a - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1) ) + log(2/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x)^2 - 12*a*(2*a - log((2*exp(2 *a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + log(2/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x) + 12*a^2) - (5*((2*a - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)* exp(2*b*x) + 1)) + log(2/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x)^3 - 8*a^3 - 6 *a*(2*a - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + log(2/( exp(2*a)*exp(2*b*x) + 1)) + 2*b*x)^2 + 12*a^2*(2*a - log((2*exp(2*a)*exp(2 *b*x))/(exp(2*a)*exp(2*b*x) + 1)) + log(2/(exp(2*a)*exp(2*b*x) + 1)) + 2*b *x)))/(4*b))/(b^3*(2*a - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + log(2/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x)^2 + x*(8*a*b^4 - 4*b^4*( 2*a - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + log(2/(exp( 2*a)*exp(2*b*x) + 1)) + 2*b*x)) + 4*a^2*b^3 + 4*b^5*x^2 - 4*a*b^3*(2*a - l og((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + log(2/(exp(2*a)*ex p(2*b*x) + 1)) + 2*b*x)) + (log(log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp( 2*b*x) + 1)) - log(2/(exp(2*a)*exp(2*b*x) + 1)))*(3*log(2/(exp(2*a)*exp(2* b*x) + 1)) - 3*log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 6* b*x))/(2*b^4)
\[ \int \frac {x^3}{\text {arctanh}(\tanh (a+b x))^3} \, dx=\int \frac {x^{3}}{\mathit {atanh} \left (\tanh \left (b x +a \right )\right )^{3}}d x \] Input:
int(x^3/atanh(tanh(b*x+a))^3,x)
Output:
int(x**3/atanh(tanh(a + b*x))**3,x)