Integrand size = 13, antiderivative size = 81 \[ \int \frac {x^3}{\text {arctanh}(\tanh (a+b x))} \, dx=\frac {x^3}{3 b}+\frac {x^2 (b x-\text {arctanh}(\tanh (a+b x)))}{2 b^2}+\frac {x (b x-\text {arctanh}(\tanh (a+b x)))^2}{b^3}+\frac {(b x-\text {arctanh}(\tanh (a+b x)))^3 \log (\text {arctanh}(\tanh (a+b x)))}{b^4} \]
1/3*x^3/b+1/2*x^2*(b*x-arctanh(tanh(b*x+a)))/b^2+x*(b*x-arctanh(tanh(b*x+a )))^2/b^3+(b*x-arctanh(tanh(b*x+a)))^3*ln(arctanh(tanh(b*x+a)))/b^4
Time = 0.04 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.98 \[ \int \frac {x^3}{\text {arctanh}(\tanh (a+b x))} \, dx=\frac {x^3}{3 b}-\frac {x^2 (-b x+\text {arctanh}(\tanh (a+b x)))}{2 b^2}+\frac {x (-b x+\text {arctanh}(\tanh (a+b x)))^2}{b^3}-\frac {(-b x+\text {arctanh}(\tanh (a+b x)))^3 \log (\text {arctanh}(\tanh (a+b x)))}{b^4} \]
x^3/(3*b) - (x^2*(-(b*x) + ArcTanh[Tanh[a + b*x]]))/(2*b^2) + (x*(-(b*x) + ArcTanh[Tanh[a + b*x]])^2)/b^3 - ((-(b*x) + ArcTanh[Tanh[a + b*x]])^3*Log [ArcTanh[Tanh[a + b*x]]])/b^4
Time = 0.30 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.07, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {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.
\(\displaystyle \int \frac {x^3}{\text {arctanh}(\tanh (a+b x))} \, dx\) |
\(\Big \downarrow \) 2590 |
\(\displaystyle \frac {(b x-\text {arctanh}(\tanh (a+b x))) \int \frac {x^2}{\text {arctanh}(\tanh (a+b x))}dx}{b}+\frac {x^3}{3 b}\) |
\(\Big \downarrow \) 2590 |
\(\displaystyle \frac {(b x-\text {arctanh}(\tanh (a+b x))) \left (\frac {(b x-\text {arctanh}(\tanh (a+b x))) \int \frac {x}{\text {arctanh}(\tanh (a+b x))}dx}{b}+\frac {x^2}{2 b}\right )}{b}+\frac {x^3}{3 b}\) |
\(\Big \downarrow \) 2589 |
\(\displaystyle \frac {(b x-\text {arctanh}(\tanh (a+b x))) \left (\frac {(b x-\text {arctanh}(\tanh (a+b x))) \left (\frac {(b x-\text {arctanh}(\tanh (a+b x))) \int \frac {1}{\text {arctanh}(\tanh (a+b x))}dx}{b}+\frac {x}{b}\right )}{b}+\frac {x^2}{2 b}\right )}{b}+\frac {x^3}{3 b}\) |
\(\Big \downarrow \) 2588 |
\(\displaystyle \frac {(b x-\text {arctanh}(\tanh (a+b x))) \left (\frac {(b x-\text {arctanh}(\tanh (a+b x))) \left (\frac {(b x-\text {arctanh}(\tanh (a+b x))) \int \frac {1}{\text {arctanh}(\tanh (a+b x))}d\text {arctanh}(\tanh (a+b x))}{b^2}+\frac {x}{b}\right )}{b}+\frac {x^2}{2 b}\right )}{b}+\frac {x^3}{3 b}\) |
\(\Big \downarrow \) 14 |
\(\displaystyle \frac {(b x-\text {arctanh}(\tanh (a+b x))) \left (\frac {(b x-\text {arctanh}(\tanh (a+b x))) \left (\frac {(b x-\text {arctanh}(\tanh (a+b x))) \log (\text {arctanh}(\tanh (a+b x)))}{b^2}+\frac {x}{b}\right )}{b}+\frac {x^2}{2 b}\right )}{b}+\frac {x^3}{3 b}\) |
x^3/(3*b) + ((b*x - ArcTanh[Tanh[a + b*x]])*(x^2/(2*b) + ((b*x - ArcTanh[T anh[a + b*x]])*(x/b + ((b*x - ArcTanh[Tanh[a + b*x]])*Log[ArcTanh[Tanh[a + b*x]]])/b^2))/b))/b
3.1.86.3.1 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]
Leaf count of result is larger than twice the leaf count of optimal. \(162\) vs. \(2(77)=154\).
Time = 0.95 (sec) , antiderivative size = 163, normalized size of antiderivative = 2.01
method | result | size |
default | \(\frac {\frac {b^{2} x^{3}}{3}-\frac {a b \,x^{2}}{2}-\frac {x^{2} b \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )}{2}+a^{2} x +2 x a \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )+x \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2}}{b^{3}}+\frac {\left (-a^{3}-3 a^{2} \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )-3 a \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2}-\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )^{3}\right ) \ln \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )\right )}{b^{4}}\) | \(163\) |
risch | \(\text {Expression too large to display}\) | \(50814\) |
1/b^3*(1/3*b^2*x^3-1/2*a*b*x^2-1/2*x^2*b*(arctanh(tanh(b*x+a))-b*x-a)+a^2* x+2*x*a*(arctanh(tanh(b*x+a))-b*x-a)+x*(arctanh(tanh(b*x+a))-b*x-a)^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*ln(arctanh(tanh(b*x+a)))
Time = 0.25 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.51 \[ \int \frac {x^3}{\text {arctanh}(\tanh (a+b x))} \, dx=\frac {2 \, b^{3} x^{3} - 3 \, a b^{2} x^{2} + 6 \, a^{2} b x - 6 \, a^{3} \log \left (b x + a\right )}{6 \, b^{4}} \]
\[ \int \frac {x^3}{\text {arctanh}(\tanh (a+b x))} \, dx=\int \frac {x^{3}}{\operatorname {atanh}{\left (\tanh {\left (a + b x \right )} \right )}}\, dx \]
Time = 0.35 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.52 \[ \int \frac {x^3}{\text {arctanh}(\tanh (a+b x))} \, dx=-\frac {a^{3} \log \left (b x + a\right )}{b^{4}} + \frac {2 \, b^{2} x^{3} - 3 \, a b x^{2} + 6 \, a^{2} x}{6 \, b^{3}} \]
Time = 0.27 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.53 \[ \int \frac {x^3}{\text {arctanh}(\tanh (a+b x))} \, dx=-\frac {a^{3} \log \left ({\left | b x + a \right |}\right )}{b^{4}} + \frac {2 \, b^{2} x^{3} - 3 \, a b x^{2} + 6 \, a^{2} x}{6 \, b^{3}} \]
Time = 0.14 (sec) , antiderivative size = 354, normalized size of antiderivative = 4.37 \[ \int \frac {x^3}{\text {arctanh}(\tanh (a+b x))} \, dx=\frac {x^3}{3\,b}+\frac {x^2\,\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}{4\,b^2}+\frac {x\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^2}{4\,b^3}+\frac {\ln \left (\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )\right )\,\left ({\left (2\,a-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^3-8\,a^3-6\,a\,{\left (2\,a-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^2+12\,a^2\,\left (2\,a-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )\right )}{8\,b^4} \]
x^3/(3*b) + (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))/(4*b^2) + (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^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)))*((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)))/(8*b^4)