Integrand size = 13, antiderivative size = 92 \[ \int \frac {x^4}{\text {arctanh}(\tanh (a+b x))^3} \, dx=\frac {3 x^2}{b^3}+\frac {6 x (b x-\text {arctanh}(\tanh (a+b x)))}{b^4}-\frac {x^4}{2 b \text {arctanh}(\tanh (a+b x))^2}-\frac {2 x^3}{b^2 \text {arctanh}(\tanh (a+b x))}+\frac {6 (b x-\text {arctanh}(\tanh (a+b x)))^2 \log (\text {arctanh}(\tanh (a+b x)))}{b^5} \]
3*x^2/b^3+6*x*(b*x-arctanh(tanh(b*x+a)))/b^4-1/2*x^4/b/arctanh(tanh(b*x+a) )^2-2*x^3/b^2/arctanh(tanh(b*x+a))+6*(b*x-arctanh(tanh(b*x+a)))^2*ln(arcta nh(tanh(b*x+a)))/b^5
Time = 0.03 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.24 \[ \int \frac {x^4}{\text {arctanh}(\tanh (a+b x))^3} \, dx=\frac {x^2}{2 b^3}-\frac {3 x (-b x+\text {arctanh}(\tanh (a+b x)))}{b^4}+\frac {4 (-b x+\text {arctanh}(\tanh (a+b x)))^3}{b^5 \text {arctanh}(\tanh (a+b x))}-\frac {(-b x+\text {arctanh}(\tanh (a+b x)))^4}{2 b^5 \text {arctanh}(\tanh (a+b x))^2}+\frac {6 (-b x+\text {arctanh}(\tanh (a+b x)))^2 \log (\text {arctanh}(\tanh (a+b x)))}{b^5} \]
x^2/(2*b^3) - (3*x*(-(b*x) + ArcTanh[Tanh[a + b*x]]))/b^4 + (4*(-(b*x) + A rcTanh[Tanh[a + b*x]])^3)/(b^5*ArcTanh[Tanh[a + b*x]]) - (-(b*x) + ArcTanh [Tanh[a + b*x]])^4/(2*b^5*ArcTanh[Tanh[a + b*x]]^2) + (6*(-(b*x) + ArcTanh [Tanh[a + b*x]])^2*Log[ArcTanh[Tanh[a + b*x]]])/b^5
Time = 0.36 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.16, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {2599, 2599, 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.
\(\displaystyle \int \frac {x^4}{\text {arctanh}(\tanh (a+b x))^3} \, dx\) |
\(\Big \downarrow \) 2599 |
\(\displaystyle \frac {2 \int \frac {x^3}{\text {arctanh}(\tanh (a+b x))^2}dx}{b}-\frac {x^4}{2 b \text {arctanh}(\tanh (a+b x))^2}\) |
\(\Big \downarrow \) 2599 |
\(\displaystyle \frac {2 \left (\frac {3 \int \frac {x^2}{\text {arctanh}(\tanh (a+b x))}dx}{b}-\frac {x^3}{b \text {arctanh}(\tanh (a+b x))}\right )}{b}-\frac {x^4}{2 b \text {arctanh}(\tanh (a+b x))^2}\) |
\(\Big \downarrow \) 2590 |
\(\displaystyle \frac {2 \left (\frac {3 \left (\frac {(b x-\text {arctanh}(\tanh (a+b x))) \int \frac {x}{\text {arctanh}(\tanh (a+b x))}dx}{b}+\frac {x^2}{2 b}\right )}{b}-\frac {x^3}{b \text {arctanh}(\tanh (a+b x))}\right )}{b}-\frac {x^4}{2 b \text {arctanh}(\tanh (a+b x))^2}\) |
\(\Big \downarrow \) 2589 |
\(\displaystyle \frac {2 \left (\frac {3 \left (\frac {(b x-\text {arctanh}(\tanh (a+b x))) \left (\frac {(b x-\text {arctanh}(\tanh (a+b x))) \int \frac {1}{\text {arctanh}(\tanh (a+b x))}dx}{b}+\frac {x}{b}\right )}{b}+\frac {x^2}{2 b}\right )}{b}-\frac {x^3}{b \text {arctanh}(\tanh (a+b x))}\right )}{b}-\frac {x^4}{2 b \text {arctanh}(\tanh (a+b x))^2}\) |
\(\Big \downarrow \) 2588 |
\(\displaystyle \frac {2 \left (\frac {3 \left (\frac {(b x-\text {arctanh}(\tanh (a+b x))) \left (\frac {(b x-\text {arctanh}(\tanh (a+b x))) \int \frac {1}{\text {arctanh}(\tanh (a+b x))}d\text {arctanh}(\tanh (a+b x))}{b^2}+\frac {x}{b}\right )}{b}+\frac {x^2}{2 b}\right )}{b}-\frac {x^3}{b \text {arctanh}(\tanh (a+b x))}\right )}{b}-\frac {x^4}{2 b \text {arctanh}(\tanh (a+b x))^2}\) |
\(\Big \downarrow \) 14 |
\(\displaystyle \frac {2 \left (\frac {3 \left (\frac {(b x-\text {arctanh}(\tanh (a+b x))) \left (\frac {(b x-\text {arctanh}(\tanh (a+b x))) \log (\text {arctanh}(\tanh (a+b x)))}{b^2}+\frac {x}{b}\right )}{b}+\frac {x^2}{2 b}\right )}{b}-\frac {x^3}{b \text {arctanh}(\tanh (a+b x))}\right )}{b}-\frac {x^4}{2 b \text {arctanh}(\tanh (a+b x))^2}\) |
-1/2*x^4/(b*ArcTanh[Tanh[a + b*x]]^2) + (2*(-(x^3/(b*ArcTanh[Tanh[a + b*x] ])) + (3*(x^2/(2*b) + ((b*x - ArcTanh[Tanh[a + b*x]])*(x/b + ((b*x - ArcTa nh[Tanh[a + b*x]])*Log[ArcTanh[Tanh[a + b*x]]])/b^2))/b))/b))/b
3.2.3.3.1 Defintions of rubi rules used
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. \(265\) vs. \(2(90)=180\).
Time = 0.29 (sec) , antiderivative size = 266, normalized size of antiderivative = 2.89
method | result | size |
default | \(\frac {\frac {b \,x^{2}}{2}-3 a x -3 x \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )}{b^{4}}-\frac {-4 a^{3}-12 a^{2} \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )-12 a \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2}-4 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )^{3}}{b^{5} \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}+\frac {\left (6 a^{2}+12 a \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )+6 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2}\right ) \ln \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )\right )}{b^{5}}-\frac {a^{4}+4 a^{3} \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )+6 a^{2} \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2}+4 a \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )^{3}+\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )^{4}}{2 b^{5} \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{2}}\) | \(266\) |
risch | \(\text {Expression too large to display}\) | \(15090\) |
1/b^4*(1/2*b*x^2-3*a*x-3*x*(arctanh(tanh(b*x+a))-b*x-a))-(-4*a^3-12*a^2*(a rctanh(tanh(b*x+a))-b*x-a)-12*a*(arctanh(tanh(b*x+a))-b*x-a)^2-4*(arctanh( tanh(b*x+a))-b*x-a)^3)/b^5/arctanh(tanh(b*x+a))+(6*a^2+12*a*(arctanh(tanh( b*x+a))-b*x-a)+6*(arctanh(tanh(b*x+a))-b*x-a)^2)/b^5*ln(arctanh(tanh(b*x+a )))-1/2*(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))^2
Time = 0.25 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.03 \[ \int \frac {x^4}{\text {arctanh}(\tanh (a+b x))^3} \, dx=\frac {b^{4} x^{4} - 4 \, a b^{3} x^{3} - 11 \, a^{2} b^{2} x^{2} + 2 \, a^{3} b x + 7 \, a^{4} + 12 \, {\left (a^{2} b^{2} x^{2} + 2 \, a^{3} b x + a^{4}\right )} \log \left (b x + a\right )}{2 \, {\left (b^{7} x^{2} + 2 \, a b^{6} x + a^{2} b^{5}\right )}} \]
1/2*(b^4*x^4 - 4*a*b^3*x^3 - 11*a^2*b^2*x^2 + 2*a^3*b*x + 7*a^4 + 12*(a^2* b^2*x^2 + 2*a^3*b*x + a^4)*log(b*x + a))/(b^7*x^2 + 2*a*b^6*x + a^2*b^5)
\[ \int \frac {x^4}{\text {arctanh}(\tanh (a+b x))^3} \, dx=\int \frac {x^{4}}{\operatorname {atanh}^{3}{\left (\tanh {\left (a + b x \right )} \right )}}\, dx \]
Time = 0.75 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.88 \[ \int \frac {x^4}{\text {arctanh}(\tanh (a+b x))^3} \, dx=\frac {b^{4} x^{4} - 4 \, a b^{3} x^{3} - 11 \, a^{2} b^{2} x^{2} + 2 \, a^{3} b x + 7 \, a^{4}}{2 \, {\left (b^{7} x^{2} + 2 \, a b^{6} x + a^{2} b^{5}\right )}} + \frac {6 \, a^{2} \log \left (b x + a\right )}{b^{5}} \]
1/2*(b^4*x^4 - 4*a*b^3*x^3 - 11*a^2*b^2*x^2 + 2*a^3*b*x + 7*a^4)/(b^7*x^2 + 2*a*b^6*x + a^2*b^5) + 6*a^2*log(b*x + a)/b^5
Time = 0.26 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.66 \[ \int \frac {x^4}{\text {arctanh}(\tanh (a+b x))^3} \, dx=\frac {6 \, a^{2} \log \left ({\left | b x + a \right |}\right )}{b^{5}} + \frac {b^{3} x^{2} - 6 \, a b^{2} x}{2 \, b^{6}} + \frac {8 \, a^{3} b x + 7 \, a^{4}}{2 \, {\left (b x + a\right )}^{2} b^{5}} \]
6*a^2*log(abs(b*x + a))/b^5 + 1/2*(b^3*x^2 - 6*a*b^2*x)/b^6 + 1/2*(8*a^3*b *x + 7*a^4)/((b*x + a)^2*b^5)
Time = 4.15 (sec) , antiderivative size = 867, normalized size of antiderivative = 9.42 \[ \int \frac {x^4}{\text {arctanh}(\tanh (a+b x))^3} \, dx=\text {Too large to display} \]
((7*((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*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 + 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 - 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)))/(4*b) - x*(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)^3 - 32*a^3 - 24*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 + 48*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*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)^2 + x*(16*a*b^5 - 8*b^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)) + 8*a^2*b^4 + 8*b^6*x^2 - 8*a*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)) + x^2/(2*b^3) + (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* (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...