Integrand size = 13, antiderivative size = 98 \[ \int \frac {x^4}{\text {arctanh}(\tanh (a+b x))^2} \, dx=\frac {4 x^3}{3 b^2}+\frac {2 x^2 (b x-\text {arctanh}(\tanh (a+b x)))}{b^3}+\frac {4 x (b x-\text {arctanh}(\tanh (a+b x)))^2}{b^4}-\frac {x^4}{b \text {arctanh}(\tanh (a+b x))}+\frac {4 (b x-\text {arctanh}(\tanh (a+b x)))^3 \log (\text {arctanh}(\tanh (a+b x)))}{b^5} \] Output:
4/3*x^3/b^2+2*x^2*(b*x-arctanh(tanh(b*x+a)))/b^3+4*x*(b*x-arctanh(tanh(b*x +a)))^2/b^4-x^4/b/arctanh(tanh(b*x+a))+4*(b*x-arctanh(tanh(b*x+a)))^3*ln(a rctanh(tanh(b*x+a)))/b^5
Time = 0.06 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.08 \[ \int \frac {x^4}{\text {arctanh}(\tanh (a+b x))^2} \, dx=\frac {x^3}{3 b^2}-\frac {x^2 (-b x+\text {arctanh}(\tanh (a+b x)))}{b^3}+\frac {3 x (-b x+\text {arctanh}(\tanh (a+b x)))^2}{b^4}-\frac {(-b x+\text {arctanh}(\tanh (a+b x)))^4}{b^5 \text {arctanh}(\tanh (a+b x))}-\frac {4 (-b x+\text {arctanh}(\tanh (a+b x)))^3 \log (\text {arctanh}(\tanh (a+b x)))}{b^5} \] Input:
Integrate[x^4/ArcTanh[Tanh[a + b*x]]^2,x]
Output:
x^3/(3*b^2) - (x^2*(-(b*x) + ArcTanh[Tanh[a + b*x]]))/b^3 + (3*x*(-(b*x) + ArcTanh[Tanh[a + b*x]])^2)/b^4 - (-(b*x) + ArcTanh[Tanh[a + b*x]])^4/(b^5 *ArcTanh[Tanh[a + b*x]]) - (4*(-(b*x) + ArcTanh[Tanh[a + b*x]])^3*Log[ArcT anh[Tanh[a + b*x]]])/b^5
Time = 0.61 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.12, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {2599, 2590, 2590, 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.
\(\displaystyle \int \frac {x^4}{\text {arctanh}(\tanh (a+b x))^2} \, dx\) |
\(\Big \downarrow \) 2599 |
\(\displaystyle \frac {4 \int \frac {x^3}{\text {arctanh}(\tanh (a+b x))}dx}{b}-\frac {x^4}{b \text {arctanh}(\tanh (a+b x))}\) |
\(\Big \downarrow \) 2590 |
\(\displaystyle \frac {4 \left (\frac {(b x-\text {arctanh}(\tanh (a+b x))) \int \frac {x^2}{\text {arctanh}(\tanh (a+b x))}dx}{b}+\frac {x^3}{3 b}\right )}{b}-\frac {x^4}{b \text {arctanh}(\tanh (a+b x))}\) |
\(\Big \downarrow \) 2590 |
\(\displaystyle \frac {4 \left (\frac {(b x-\text {arctanh}(\tanh (a+b x))) \left (\frac {(b x-\text {arctanh}(\tanh (a+b x))) \int \frac {x}{\text {arctanh}(\tanh (a+b x))}dx}{b}+\frac {x^2}{2 b}\right )}{b}+\frac {x^3}{3 b}\right )}{b}-\frac {x^4}{b \text {arctanh}(\tanh (a+b x))}\) |
\(\Big \downarrow \) 2589 |
\(\displaystyle \frac {4 \left (\frac {(b x-\text {arctanh}(\tanh (a+b x))) \left (\frac {(b x-\text {arctanh}(\tanh (a+b x))) \left (\frac {(b x-\text {arctanh}(\tanh (a+b x))) \int \frac {1}{\text {arctanh}(\tanh (a+b x))}dx}{b}+\frac {x}{b}\right )}{b}+\frac {x^2}{2 b}\right )}{b}+\frac {x^3}{3 b}\right )}{b}-\frac {x^4}{b \text {arctanh}(\tanh (a+b x))}\) |
\(\Big \downarrow \) 2588 |
\(\displaystyle \frac {4 \left (\frac {(b x-\text {arctanh}(\tanh (a+b x))) \left (\frac {(b x-\text {arctanh}(\tanh (a+b x))) \left (\frac {(b x-\text {arctanh}(\tanh (a+b x))) \int \frac {1}{\text {arctanh}(\tanh (a+b x))}d\text {arctanh}(\tanh (a+b x))}{b^2}+\frac {x}{b}\right )}{b}+\frac {x^2}{2 b}\right )}{b}+\frac {x^3}{3 b}\right )}{b}-\frac {x^4}{b \text {arctanh}(\tanh (a+b x))}\) |
\(\Big \downarrow \) 14 |
\(\displaystyle \frac {4 \left (\frac {(b x-\text {arctanh}(\tanh (a+b x))) \left (\frac {(b x-\text {arctanh}(\tanh (a+b x))) \left (\frac {(b x-\text {arctanh}(\tanh (a+b x))) \log (\text {arctanh}(\tanh (a+b x)))}{b^2}+\frac {x}{b}\right )}{b}+\frac {x^2}{2 b}\right )}{b}+\frac {x^3}{3 b}\right )}{b}-\frac {x^4}{b \text {arctanh}(\tanh (a+b x))}\) |
Input:
Int[x^4/ArcTanh[Tanh[a + b*x]]^2,x]
Output:
-(x^4/(b*ArcTanh[Tanh[a + b*x]])) + (4*(x^3/(3*b) + ((b*x - ArcTanh[Tanh[a + b*x]])*(x^2/(2*b) + ((b*x - ArcTanh[Tanh[a + b*x]])*(x/b + ((b*x - ArcT anh[Tanh[a + b*x]])*Log[ArcTanh[Tanh[a + b*x]]])/b^2))/b))/b))/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[(v_)^(n_)/(u_), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[ D[v, x]]}, Simp[v^n/(a*n), x] - Simp[(b*u - a*v)/a Int[v^(n - 1)/u, x], x ] /; NeQ[b*u - a*v, 0]] /; PiecewiseLinearQ[u, v, x] && GtQ[n, 0] && NeQ[n, 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]))
Leaf count of result is larger than twice the leaf count of optimal. \(261\) vs. \(2(96)=192\).
Time = 1.22 (sec) , antiderivative size = 262, normalized size of antiderivative = 2.67
method | result | size |
default | \(\frac {\frac {b^{2} x^{3}}{3}-x^{2} a b -b \,x^{2} \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )+3 x \,a^{2}+6 x a \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )+3 x \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2}}{b^{4}}-\frac {a^{4}+4 a^{3} \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )+6 a^{2} \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2}+4 a \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )^{3}+\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )^{4}}{b^{5} \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}+\frac {\left (-4 a^{3}-12 a^{2} \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )-12 a \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2}-4 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )^{3}\right ) \ln \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )\right )}{b^{5}}\) | \(262\) |
risch | \(\text {Expression too large to display}\) | \(51088\) |
Input:
int(x^4/arctanh(tanh(b*x+a))^2,x,method=_RETURNVERBOSE)
Output:
1/b^4*(1/3*b^2*x^3-x^2*a*b-b*x^2*(arctanh(tanh(b*x+a))-b*x-a)+3*x*a^2+6*x* a*(arctanh(tanh(b*x+a))-b*x-a)+3*x*(arctanh(tanh(b*x+a))-b*x-a)^2)-(a^4+4* a^3*(arctanh(tanh(b*x+a))-b*x-a)+6*a^2*(arctanh(tanh(b*x+a))-b*x-a)^2+4*a* (arctanh(tanh(b*x+a))-b*x-a)^3+(arctanh(tanh(b*x+a))-b*x-a)^4)/b^5/arctanh (tanh(b*x+a))+(-4*a^3-12*a^2*(arctanh(tanh(b*x+a))-b*x-a)-12*a*(arctanh(ta nh(b*x+a))-b*x-a)^2-4*(arctanh(tanh(b*x+a))-b*x-a)^3)/b^5*ln(arctanh(tanh( b*x+a)))
Time = 0.08 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.74 \[ \int \frac {x^4}{\text {arctanh}(\tanh (a+b x))^2} \, dx=\frac {b^{4} x^{4} - 2 \, a b^{3} x^{3} + 6 \, a^{2} b^{2} x^{2} + 9 \, a^{3} b x - 3 \, a^{4} - 12 \, {\left (a^{3} b x + a^{4}\right )} \log \left (b x + a\right )}{3 \, {\left (b^{6} x + a b^{5}\right )}} \] Input:
integrate(x^4/arctanh(tanh(b*x+a))^2,x, algorithm="fricas")
Output:
1/3*(b^4*x^4 - 2*a*b^3*x^3 + 6*a^2*b^2*x^2 + 9*a^3*b*x - 3*a^4 - 12*(a^3*b *x + a^4)*log(b*x + a))/(b^6*x + a*b^5)
\[ \int \frac {x^4}{\text {arctanh}(\tanh (a+b x))^2} \, dx=\int \frac {x^{4}}{\operatorname {atanh}^{2}{\left (\tanh {\left (a + b x \right )} \right )}}\, dx \] Input:
integrate(x**4/atanh(tanh(b*x+a))**2,x)
Output:
Integral(x**4/atanh(tanh(a + b*x))**2, x)
Time = 0.35 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.71 \[ \int \frac {x^4}{\text {arctanh}(\tanh (a+b x))^2} \, dx=\frac {b^{4} x^{4} - 2 \, a b^{3} x^{3} + 6 \, a^{2} b^{2} x^{2} + 9 \, a^{3} b x - 3 \, a^{4}}{3 \, {\left (b^{6} x + a b^{5}\right )}} - \frac {4 \, a^{3} \log \left (b x + a\right )}{b^{5}} \] Input:
integrate(x^4/arctanh(tanh(b*x+a))^2,x, algorithm="maxima")
Output:
1/3*(b^4*x^4 - 2*a*b^3*x^3 + 6*a^2*b^2*x^2 + 9*a^3*b*x - 3*a^4)/(b^6*x + a *b^5) - 4*a^3*log(b*x + a)/b^5
Time = 0.11 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.63 \[ \int \frac {x^4}{\text {arctanh}(\tanh (a+b x))^2} \, dx=-\frac {4 \, a^{3} \log \left ({\left | b x + a \right |}\right )}{b^{5}} - \frac {a^{4}}{{\left (b x + a\right )} b^{5}} + \frac {b^{4} x^{3} - 3 \, a b^{3} x^{2} + 9 \, a^{2} b^{2} x}{3 \, b^{6}} \] Input:
integrate(x^4/arctanh(tanh(b*x+a))^2,x, algorithm="giac")
Output:
-4*a^3*log(abs(b*x + a))/b^5 - a^4/((b*x + a)*b^5) + 1/3*(b^4*x^3 - 3*a*b^ 3*x^2 + 9*a^2*b^2*x)/b^6
Time = 3.26 (sec) , antiderivative size = 669, normalized size of antiderivative = 6.83 \[ \int \frac {x^4}{\text {arctanh}(\tanh (a+b x))^2} \, dx =\text {Too large to display} \] Input:
int(x^4/atanh(tanh(a + b*x))^2,x)
Output:
x^3/(3*b^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 + 24*a^2*(2*a - log((2*ex p(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 + 16*a^4 - 8*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)^3 - 32*a^ 3*(2*a - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + log(2/(e xp(2*a)*exp(2*b*x) + 1)) + 2*b*x))/(2*b*(8*a*b^4 + 8*b^5*x - 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)*e xp(2*b*x) + 1)) + 2*b*x))) + (x^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*b^3) + (3* x*(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)/(4*b^4) + (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)))*((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)))/(2*b^5)
\[ \int \frac {x^4}{\text {arctanh}(\tanh (a+b x))^2} \, dx=\int \frac {x^{4}}{\mathit {atanh} \left (\tanh \left (b x +a \right )\right )^{2}}d x \] Input:
int(x^4/atanh(tanh(b*x+a))^2,x)
Output:
int(x**4/atanh(tanh(a + b*x))**2,x)