Integrand size = 13, antiderivative size = 92 \[ \int \frac {x^4}{\coth ^{-1}(\tanh (a+b x))^3} \, dx=\frac {3 x^2}{b^3}+\frac {6 x \left (b x-\coth ^{-1}(\tanh (a+b x))\right )}{b^4}-\frac {x^4}{2 b \coth ^{-1}(\tanh (a+b x))^2}-\frac {2 x^3}{b^2 \coth ^{-1}(\tanh (a+b x))}+\frac {6 \left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2 \log \left (\coth ^{-1}(\tanh (a+b x))\right )}{b^5} \]
3*x^2/b^3+6*x*(b*x-arccoth(tanh(b*x+a)))/b^4-1/2*x^4/b/arccoth(tanh(b*x+a) )^2-2*x^3/b^2/arccoth(tanh(b*x+a))+6*(b*x-arccoth(tanh(b*x+a)))^2*ln(arcco th(tanh(b*x+a)))/b^5
Time = 0.04 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.24 \[ \int \frac {x^4}{\coth ^{-1}(\tanh (a+b x))^3} \, dx=\frac {x^2}{2 b^3}-\frac {3 x \left (-b x+\coth ^{-1}(\tanh (a+b x))\right )}{b^4}+\frac {4 \left (-b x+\coth ^{-1}(\tanh (a+b x))\right )^3}{b^5 \coth ^{-1}(\tanh (a+b x))}-\frac {\left (-b x+\coth ^{-1}(\tanh (a+b x))\right )^4}{2 b^5 \coth ^{-1}(\tanh (a+b x))^2}+\frac {6 \left (-b x+\coth ^{-1}(\tanh (a+b x))\right )^2 \log \left (\coth ^{-1}(\tanh (a+b x))\right )}{b^5} \]
x^2/(2*b^3) - (3*x*(-(b*x) + ArcCoth[Tanh[a + b*x]]))/b^4 + (4*(-(b*x) + A rcCoth[Tanh[a + b*x]])^3)/(b^5*ArcCoth[Tanh[a + b*x]]) - (-(b*x) + ArcCoth [Tanh[a + b*x]])^4/(2*b^5*ArcCoth[Tanh[a + b*x]]^2) + (6*(-(b*x) + ArcCoth [Tanh[a + b*x]])^2*Log[ArcCoth[Tanh[a + b*x]]])/b^5
Time = 0.35 (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.
\(\displaystyle \int \frac {x^4}{\coth ^{-1}(\tanh (a+b x))^3} \, dx\) |
\(\Big \downarrow \) 2599 |
\(\displaystyle \frac {2 \int \frac {x^3}{\coth ^{-1}(\tanh (a+b x))^2}dx}{b}-\frac {x^4}{2 b \coth ^{-1}(\tanh (a+b x))^2}\) |
\(\Big \downarrow \) 2599 |
\(\displaystyle \frac {2 \left (\frac {3 \int \frac {x^2}{\coth ^{-1}(\tanh (a+b x))}dx}{b}-\frac {x^3}{b \coth ^{-1}(\tanh (a+b x))}\right )}{b}-\frac {x^4}{2 b \coth ^{-1}(\tanh (a+b x))^2}\) |
\(\Big \downarrow \) 2590 |
\(\displaystyle \frac {2 \left (\frac {3 \left (\frac {\left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \int \frac {x}{\coth ^{-1}(\tanh (a+b x))}dx}{b}+\frac {x^2}{2 b}\right )}{b}-\frac {x^3}{b \coth ^{-1}(\tanh (a+b x))}\right )}{b}-\frac {x^4}{2 b \coth ^{-1}(\tanh (a+b x))^2}\) |
\(\Big \downarrow \) 2589 |
\(\displaystyle \frac {2 \left (\frac {3 \left (\frac {\left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \left (\frac {\left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \int \frac {1}{\coth ^{-1}(\tanh (a+b x))}dx}{b}+\frac {x}{b}\right )}{b}+\frac {x^2}{2 b}\right )}{b}-\frac {x^3}{b \coth ^{-1}(\tanh (a+b x))}\right )}{b}-\frac {x^4}{2 b \coth ^{-1}(\tanh (a+b x))^2}\) |
\(\Big \downarrow \) 2588 |
\(\displaystyle \frac {2 \left (\frac {3 \left (\frac {\left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \left (\frac {\left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \int \frac {1}{\coth ^{-1}(\tanh (a+b x))}d\coth ^{-1}(\tanh (a+b x))}{b^2}+\frac {x}{b}\right )}{b}+\frac {x^2}{2 b}\right )}{b}-\frac {x^3}{b \coth ^{-1}(\tanh (a+b x))}\right )}{b}-\frac {x^4}{2 b \coth ^{-1}(\tanh (a+b x))^2}\) |
\(\Big \downarrow \) 14 |
\(\displaystyle \frac {2 \left (\frac {3 \left (\frac {\left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \left (\frac {\left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \log \left (\coth ^{-1}(\tanh (a+b x))\right )}{b^2}+\frac {x}{b}\right )}{b}+\frac {x^2}{2 b}\right )}{b}-\frac {x^3}{b \coth ^{-1}(\tanh (a+b x))}\right )}{b}-\frac {x^4}{2 b \coth ^{-1}(\tanh (a+b x))^2}\) |
-1/2*x^4/(b*ArcCoth[Tanh[a + b*x]]^2) + (2*(-(x^3/(b*ArcCoth[Tanh[a + b*x] ])) + (3*(x^2/(2*b) + ((b*x - ArcCoth[Tanh[a + b*x]])*(x/b + ((b*x - ArcCo th[Tanh[a + b*x]])*Log[ArcCoth[Tanh[a + b*x]]])/b^2))/b))/b))/b
3.2.76.3.1 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]))
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.43 (sec) , antiderivative size = 29456, normalized size of antiderivative = 320.17
Result contains complex when optimal does not.
Time = 0.24 (sec) , antiderivative size = 265, normalized size of antiderivative = 2.88 \[ \int \frac {x^4}{\coth ^{-1}(\tanh (a+b x))^3} \, dx=\frac {16 \, b^{4} x^{4} - 64 \, a b^{3} x^{3} - 176 \, a^{2} b^{2} x^{2} + 32 \, a^{3} b x + 7 \, \pi ^{4} - 4 i \, \pi ^{3} {\left (b x + 14 \, a\right )} + 112 \, a^{4} + 4 \, \pi ^{2} {\left (11 \, b^{2} x^{2} - 6 \, a b x - 42 \, a^{2}\right )} - 16 i \, \pi {\left (2 \, b^{3} x^{3} + 11 \, a b^{2} x^{2} - 3 \, a^{2} b x - 14 \, a^{3}\right )} + 12 \, {\left (16 \, a^{2} b^{2} x^{2} + 32 \, a^{3} b x + \pi ^{4} - 4 i \, \pi ^{3} {\left (b x + 2 \, a\right )} + 16 \, a^{4} - 4 \, \pi ^{2} {\left (b^{2} x^{2} + 6 \, a b x + 6 \, a^{2}\right )} + 16 i \, \pi {\left (a b^{2} x^{2} + 3 \, a^{2} b x + 2 \, a^{3}\right )}\right )} \log \left (i \, \pi + 2 \, b x + 2 \, a\right )}{8 \, {\left (4 \, b^{7} x^{2} + 8 \, a b^{6} x - \pi ^{2} b^{5} + 4 \, a^{2} b^{5} + 4 i \, \pi {\left (b^{6} x + a b^{5}\right )}\right )}} \]
1/8*(16*b^4*x^4 - 64*a*b^3*x^3 - 176*a^2*b^2*x^2 + 32*a^3*b*x + 7*pi^4 - 4 *I*pi^3*(b*x + 14*a) + 112*a^4 + 4*pi^2*(11*b^2*x^2 - 6*a*b*x - 42*a^2) - 16*I*pi*(2*b^3*x^3 + 11*a*b^2*x^2 - 3*a^2*b*x - 14*a^3) + 12*(16*a^2*b^2*x ^2 + 32*a^3*b*x + pi^4 - 4*I*pi^3*(b*x + 2*a) + 16*a^4 - 4*pi^2*(b^2*x^2 + 6*a*b*x + 6*a^2) + 16*I*pi*(a*b^2*x^2 + 3*a^2*b*x + 2*a^3))*log(I*pi + 2* b*x + 2*a))/(4*b^7*x^2 + 8*a*b^6*x - pi^2*b^5 + 4*a^2*b^5 + 4*I*pi*(b^6*x + a*b^5))
\[ \int \frac {x^4}{\coth ^{-1}(\tanh (a+b x))^3} \, dx=\int \frac {x^{4}}{\operatorname {acoth}^{3}{\left (\tanh {\left (a + b x \right )} \right )}}\, dx \]
Result contains complex when optimal does not.
Time = 0.78 (sec) , antiderivative size = 198, normalized size of antiderivative = 2.15 \[ \int \frac {x^4}{\coth ^{-1}(\tanh (a+b x))^3} \, dx=\frac {16 \, b^{4} x^{4} + 7 \, \pi ^{4} + 56 i \, \pi ^{3} a - 168 \, \pi ^{2} a^{2} - 224 i \, \pi a^{3} + 112 \, a^{4} - 32 \, {\left (-i \, \pi b^{3} + 2 \, a b^{3}\right )} x^{3} + 44 \, {\left (\pi ^{2} b^{2} + 4 i \, \pi a b^{2} - 4 \, a^{2} b^{2}\right )} x^{2} - 4 \, {\left (-i \, \pi ^{3} b + 6 \, \pi ^{2} a b + 12 i \, \pi a^{2} b - 8 \, a^{3} b\right )} x}{8 \, {\left (4 \, b^{7} x^{2} - \pi ^{2} b^{5} - 4 i \, \pi a b^{5} + 4 \, a^{2} b^{5} - 4 \, {\left (i \, \pi b^{6} - 2 \, a b^{6}\right )} x\right )}} - \frac {3 \, {\left (\pi ^{2} + 4 i \, \pi a - 4 \, a^{2}\right )} \log \left (-i \, \pi + 2 \, b x + 2 \, a\right )}{2 \, b^{5}} \]
1/8*(16*b^4*x^4 + 7*pi^4 + 56*I*pi^3*a - 168*pi^2*a^2 - 224*I*pi*a^3 + 112 *a^4 - 32*(-I*pi*b^3 + 2*a*b^3)*x^3 + 44*(pi^2*b^2 + 4*I*pi*a*b^2 - 4*a^2* b^2)*x^2 - 4*(-I*pi^3*b + 6*pi^2*a*b + 12*I*pi*a^2*b - 8*a^3*b)*x)/(4*b^7* x^2 - pi^2*b^5 - 4*I*pi*a*b^5 + 4*a^2*b^5 - 4*(I*pi*b^6 - 2*a*b^6)*x) - 3/ 2*(pi^2 + 4*I*pi*a - 4*a^2)*log(-I*pi + 2*b*x + 2*a)/b^5
Result contains complex when optimal does not.
Time = 0.30 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.77 \[ \int \frac {x^4}{\coth ^{-1}(\tanh (a+b x))^3} \, dx=-\frac {16 \, \pi ^{3} b x - 96 i \, \pi ^{2} a b x - 192 \, \pi a^{2} b x + 128 i \, a^{3} b x + 7 i \, \pi ^{4} + 56 \, \pi ^{3} a - 168 i \, \pi ^{2} a^{2} - 224 \, \pi a^{3} + 112 i \, a^{4}}{-32 i \, b^{7} x^{2} + 32 \, \pi b^{6} x - 64 i \, a b^{6} x + 8 i \, \pi ^{2} b^{5} + 32 \, \pi a b^{5} - 32 i \, a^{2} b^{5}} + \frac {x^{2}}{2 \, b^{3}} - \frac {3 \, {\left (i \, \pi + 2 \, a\right )} x}{2 \, b^{4}} - \frac {3 \, {\left (\pi ^{2} - 4 i \, \pi a - 4 \, a^{2}\right )} \log \left (i \, \pi + 2 \, b x + 2 \, a\right )}{2 \, b^{5}} \]
-(16*pi^3*b*x - 96*I*pi^2*a*b*x - 192*pi*a^2*b*x + 128*I*a^3*b*x + 7*I*pi^ 4 + 56*pi^3*a - 168*I*pi^2*a^2 - 224*pi*a^3 + 112*I*a^4)/(-32*I*b^7*x^2 + 32*pi*b^6*x - 64*I*a*b^6*x + 8*I*pi^2*b^5 + 32*pi*a*b^5 - 32*I*a^2*b^5) + 1/2*x^2/b^3 - 3/2*(I*pi + 2*a)*x/b^4 - 3/2*(pi^2 - 4*I*pi*a - 4*a^2)*log(I *pi + 2*b*x + 2*a)/b^5
Time = 4.34 (sec) , antiderivative size = 867, normalized size of antiderivative = 9.42 \[ \int \frac {x^4}{\coth ^{-1}(\tanh (a+b x))^3} \, dx=\text {Too large to display} \]
((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)*ex p(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/(exp(2*a )*exp(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*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^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)) + l og(-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(...