Integrand size = 13, antiderivative size = 110 \[ \int x^m \coth ^{-1}(\tanh (a+b x))^3 \, dx=-\frac {6 b^3 x^{4+m}}{(1+m) \left (24+26 m+9 m^2+m^3\right )}+\frac {6 b^2 x^{3+m} \coth ^{-1}(\tanh (a+b x))}{6+11 m+6 m^2+m^3}-\frac {3 b x^{2+m} \coth ^{-1}(\tanh (a+b x))^2}{2+3 m+m^2}+\frac {x^{1+m} \coth ^{-1}(\tanh (a+b x))^3}{1+m} \]
-6*b^3*x^(4+m)/(1+m)/(m^3+9*m^2+26*m+24)+6*b^2*x^(3+m)*arccoth(tanh(b*x+a) )/(m^3+6*m^2+11*m+6)-3*b*x^(2+m)*arccoth(tanh(b*x+a))^2/(m^2+3*m+2)+x^(1+m )*arccoth(tanh(b*x+a))^3/(1+m)
Time = 0.11 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.88 \[ \int x^m \coth ^{-1}(\tanh (a+b x))^3 \, dx=\frac {x^{1+m} \left (-6 b^3 x^3+6 b^2 (4+m) x^2 \coth ^{-1}(\tanh (a+b x))-3 b \left (12+7 m+m^2\right ) x \coth ^{-1}(\tanh (a+b x))^2+\left (24+26 m+9 m^2+m^3\right ) \coth ^{-1}(\tanh (a+b x))^3\right )}{(1+m) (2+m) (3+m) (4+m)} \]
(x^(1 + m)*(-6*b^3*x^3 + 6*b^2*(4 + m)*x^2*ArcCoth[Tanh[a + b*x]] - 3*b*(1 2 + 7*m + m^2)*x*ArcCoth[Tanh[a + b*x]]^2 + (24 + 26*m + 9*m^2 + m^3)*ArcC oth[Tanh[a + b*x]]^3))/((1 + m)*(2 + m)*(3 + m)*(4 + m))
Time = 0.29 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.86, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {2599, 2599, 2599, 15}
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.
| \(\displaystyle \int x^m \coth ^{-1}(\tanh (a+b x))^3 \, dx\) |
| \(\Big \downarrow \) 2599 |
| \(\displaystyle \frac {x^{m+1} \coth ^{-1}(\tanh (a+b x))^3}{m+1}-\frac {3 b \int x^{m+1} \coth ^{-1}(\tanh (a+b x))^2dx}{m+1}\) |
| \(\Big \downarrow \) 2599 |
| \(\displaystyle \frac {x^{m+1} \coth ^{-1}(\tanh (a+b x))^3}{m+1}-\frac {3 b \left (\frac {x^{m+2} \coth ^{-1}(\tanh (a+b x))^2}{m+2}-\frac {2 b \int x^{m+2} \coth ^{-1}(\tanh (a+b x))dx}{m+2}\right )}{m+1}\) |
| \(\Big \downarrow \) 2599 |
| \(\displaystyle \frac {x^{m+1} \coth ^{-1}(\tanh (a+b x))^3}{m+1}-\frac {3 b \left (\frac {x^{m+2} \coth ^{-1}(\tanh (a+b x))^2}{m+2}-\frac {2 b \left (\frac {x^{m+3} \coth ^{-1}(\tanh (a+b x))}{m+3}-\frac {b \int x^{m+3}dx}{m+3}\right )}{m+2}\right )}{m+1}\) |
| \(\Big \downarrow \) 15 |
| \(\displaystyle \frac {x^{m+1} \coth ^{-1}(\tanh (a+b x))^3}{m+1}-\frac {3 b \left (\frac {x^{m+2} \coth ^{-1}(\tanh (a+b x))^2}{m+2}-\frac {2 b \left (\frac {x^{m+3} \coth ^{-1}(\tanh (a+b x))}{m+3}-\frac {b x^{m+4}}{(m+3) (m+4)}\right )}{m+2}\right )}{m+1}\) |
(x^(1 + m)*ArcCoth[Tanh[a + b*x]]^3)/(1 + m) - (3*b*((x^(2 + m)*ArcCoth[Ta nh[a + b*x]]^2)/(2 + m) - (2*b*(-((b*x^(4 + m))/((3 + m)*(4 + m))) + (x^(3 + m)*ArcCoth[Tanh[a + b*x]])/(3 + m)))/(2 + m)))/(1 + m)
3.2.46.3.1 Defintions of rubi rules used
Int[(a_.)*(x_)^(m_.), x_Symbol] :> Simp[a*(x^(m + 1)/(m + 1)), x] /; FreeQ[ {a, m}, x] && NeQ[m, -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]))
Time = 6.03 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.79
| method | result | size |
| parallelrisch | \(-\frac {36 b \operatorname {arccoth}\left (\tanh \left (b x +a \right )\right )^{2} x^{m} x^{2}-24 b^{2} \operatorname {arccoth}\left (\tanh \left (b x +a \right )\right ) x^{m} x^{3}-x \,x^{m} \operatorname {arccoth}\left (\tanh \left (b x +a \right )\right )^{3} m^{3}-9 x \,x^{m} \operatorname {arccoth}\left (\tanh \left (b x +a \right )\right )^{3} m^{2}-26 x \,x^{m} \operatorname {arccoth}\left (\tanh \left (b x +a \right )\right )^{3} m +6 b^{3} x^{m} x^{4}-24 \operatorname {arccoth}\left (\tanh \left (b x +a \right )\right )^{3} x \,x^{m}-6 x^{3} x^{m} \operatorname {arccoth}\left (\tanh \left (b x +a \right )\right ) b^{2} m +3 x^{2} x^{m} \operatorname {arccoth}\left (\tanh \left (b x +a \right )\right )^{2} b \,m^{2}+21 x^{2} x^{m} \operatorname {arccoth}\left (\tanh \left (b x +a \right )\right )^{2} b m}{\left (1+m \right ) \left (m^{3}+9 m^{2}+26 m +24\right )}\) | \(197\) |
| risch | \(\text {Expression too large to display}\) | \(63382\) |
-(36*b*arccoth(tanh(b*x+a))^2*x^m*x^2-24*b^2*arccoth(tanh(b*x+a))*x^m*x^3- x*x^m*arccoth(tanh(b*x+a))^3*m^3-9*x*x^m*arccoth(tanh(b*x+a))^3*m^2-26*x*x ^m*arccoth(tanh(b*x+a))^3*m+6*b^3*x^m*x^4-24*arccoth(tanh(b*x+a))^3*x*x^m- 6*x^3*x^m*arccoth(tanh(b*x+a))*b^2*m+3*x^2*x^m*arccoth(tanh(b*x+a))^2*b*m^ 2+21*x^2*x^m*arccoth(tanh(b*x+a))^2*b*m)/(1+m)/(m^3+9*m^2+26*m+24)
Result contains complex when optimal does not.
Time = 0.26 (sec) , antiderivative size = 627, normalized size of antiderivative = 5.70 \[ \int x^m \coth ^{-1}(\tanh (a+b x))^3 \, dx=\frac {{\left (-i \, \pi ^{3} {\left (m^{3} + 9 \, m^{2} + 26 \, m + 24\right )} x + 8 \, {\left (b^{3} m^{3} + 6 \, b^{3} m^{2} + 11 \, b^{3} m + 6 \, b^{3}\right )} x^{4} + 24 \, {\left (a b^{2} m^{3} + 7 \, a b^{2} m^{2} + 14 \, a b^{2} m + 8 \, a b^{2}\right )} x^{3} - 6 \, \pi ^{2} {\left ({\left (b m^{3} + 8 \, b m^{2} + 19 \, b m + 12 \, b\right )} x^{2} + {\left (a m^{3} + 9 \, a m^{2} + 26 \, a m + 24 \, a\right )} x\right )} + 24 \, {\left (a^{2} b m^{3} + 8 \, a^{2} b m^{2} + 19 \, a^{2} b m + 12 \, a^{2} b\right )} x^{2} + 12 i \, \pi {\left ({\left (b^{2} m^{3} + 7 \, b^{2} m^{2} + 14 \, b^{2} m + 8 \, b^{2}\right )} x^{3} + 2 \, {\left (a b m^{3} + 8 \, a b m^{2} + 19 \, a b m + 12 \, a b\right )} x^{2} + {\left (a^{2} m^{3} + 9 \, a^{2} m^{2} + 26 \, a^{2} m + 24 \, a^{2}\right )} x\right )} + 8 \, {\left (a^{3} m^{3} + 9 \, a^{3} m^{2} + 26 \, a^{3} m + 24 \, a^{3}\right )} x\right )} \cosh \left (m \log \left (x\right )\right ) + {\left (-i \, \pi ^{3} {\left (m^{3} + 9 \, m^{2} + 26 \, m + 24\right )} x + 8 \, {\left (b^{3} m^{3} + 6 \, b^{3} m^{2} + 11 \, b^{3} m + 6 \, b^{3}\right )} x^{4} + 24 \, {\left (a b^{2} m^{3} + 7 \, a b^{2} m^{2} + 14 \, a b^{2} m + 8 \, a b^{2}\right )} x^{3} - 6 \, \pi ^{2} {\left ({\left (b m^{3} + 8 \, b m^{2} + 19 \, b m + 12 \, b\right )} x^{2} + {\left (a m^{3} + 9 \, a m^{2} + 26 \, a m + 24 \, a\right )} x\right )} + 24 \, {\left (a^{2} b m^{3} + 8 \, a^{2} b m^{2} + 19 \, a^{2} b m + 12 \, a^{2} b\right )} x^{2} + 12 i \, \pi {\left ({\left (b^{2} m^{3} + 7 \, b^{2} m^{2} + 14 \, b^{2} m + 8 \, b^{2}\right )} x^{3} + 2 \, {\left (a b m^{3} + 8 \, a b m^{2} + 19 \, a b m + 12 \, a b\right )} x^{2} + {\left (a^{2} m^{3} + 9 \, a^{2} m^{2} + 26 \, a^{2} m + 24 \, a^{2}\right )} x\right )} + 8 \, {\left (a^{3} m^{3} + 9 \, a^{3} m^{2} + 26 \, a^{3} m + 24 \, a^{3}\right )} x\right )} \sinh \left (m \log \left (x\right )\right )}{8 \, {\left (m^{4} + 10 \, m^{3} + 35 \, m^{2} + 50 \, m + 24\right )}} \]
1/8*((-I*pi^3*(m^3 + 9*m^2 + 26*m + 24)*x + 8*(b^3*m^3 + 6*b^3*m^2 + 11*b^ 3*m + 6*b^3)*x^4 + 24*(a*b^2*m^3 + 7*a*b^2*m^2 + 14*a*b^2*m + 8*a*b^2)*x^3 - 6*pi^2*((b*m^3 + 8*b*m^2 + 19*b*m + 12*b)*x^2 + (a*m^3 + 9*a*m^2 + 26*a *m + 24*a)*x) + 24*(a^2*b*m^3 + 8*a^2*b*m^2 + 19*a^2*b*m + 12*a^2*b)*x^2 + 12*I*pi*((b^2*m^3 + 7*b^2*m^2 + 14*b^2*m + 8*b^2)*x^3 + 2*(a*b*m^3 + 8*a* b*m^2 + 19*a*b*m + 12*a*b)*x^2 + (a^2*m^3 + 9*a^2*m^2 + 26*a^2*m + 24*a^2) *x) + 8*(a^3*m^3 + 9*a^3*m^2 + 26*a^3*m + 24*a^3)*x)*cosh(m*log(x)) + (-I* pi^3*(m^3 + 9*m^2 + 26*m + 24)*x + 8*(b^3*m^3 + 6*b^3*m^2 + 11*b^3*m + 6*b ^3)*x^4 + 24*(a*b^2*m^3 + 7*a*b^2*m^2 + 14*a*b^2*m + 8*a*b^2)*x^3 - 6*pi^2 *((b*m^3 + 8*b*m^2 + 19*b*m + 12*b)*x^2 + (a*m^3 + 9*a*m^2 + 26*a*m + 24*a )*x) + 24*(a^2*b*m^3 + 8*a^2*b*m^2 + 19*a^2*b*m + 12*a^2*b)*x^2 + 12*I*pi* ((b^2*m^3 + 7*b^2*m^2 + 14*b^2*m + 8*b^2)*x^3 + 2*(a*b*m^3 + 8*a*b*m^2 + 1 9*a*b*m + 12*a*b)*x^2 + (a^2*m^3 + 9*a^2*m^2 + 26*a^2*m + 24*a^2)*x) + 8*( a^3*m^3 + 9*a^3*m^2 + 26*a^3*m + 24*a^3)*x)*sinh(m*log(x)))/(m^4 + 10*m^3 + 35*m^2 + 50*m + 24)
\[ \int x^m \coth ^{-1}(\tanh (a+b x))^3 \, dx=\begin {cases} b^{3} \log {\left (x \right )} - \frac {b^{2} \operatorname {acoth}{\left (\tanh {\left (a + b x \right )} \right )}}{x} - \frac {b \operatorname {acoth}^{2}{\left (\tanh {\left (a + b x \right )} \right )}}{2 x^{2}} - \frac {\operatorname {acoth}^{3}{\left (\tanh {\left (a + b x \right )} \right )}}{3 x^{3}} & \text {for}\: m = -4 \\\int \frac {\operatorname {acoth}^{3}{\left (\tanh {\left (a + b x \right )} \right )}}{x^{3}}\, dx & \text {for}\: m = -3 \\\int \frac {\operatorname {acoth}^{3}{\left (\tanh {\left (a + b x \right )} \right )}}{x^{2}}\, dx & \text {for}\: m = -2 \\\int \frac {\operatorname {acoth}^{3}{\left (\tanh {\left (a + b x \right )} \right )}}{x}\, dx & \text {for}\: m = -1 \\- \frac {6 b^{3} x^{4} x^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} + \frac {6 b^{2} m x^{3} x^{m} \operatorname {acoth}{\left (\tanh {\left (a + b x \right )} \right )}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} + \frac {24 b^{2} x^{3} x^{m} \operatorname {acoth}{\left (\tanh {\left (a + b x \right )} \right )}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} - \frac {3 b m^{2} x^{2} x^{m} \operatorname {acoth}^{2}{\left (\tanh {\left (a + b x \right )} \right )}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} - \frac {21 b m x^{2} x^{m} \operatorname {acoth}^{2}{\left (\tanh {\left (a + b x \right )} \right )}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} - \frac {36 b x^{2} x^{m} \operatorname {acoth}^{2}{\left (\tanh {\left (a + b x \right )} \right )}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} + \frac {m^{3} x x^{m} \operatorname {acoth}^{3}{\left (\tanh {\left (a + b x \right )} \right )}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} + \frac {9 m^{2} x x^{m} \operatorname {acoth}^{3}{\left (\tanh {\left (a + b x \right )} \right )}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} + \frac {26 m x x^{m} \operatorname {acoth}^{3}{\left (\tanh {\left (a + b x \right )} \right )}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} + \frac {24 x x^{m} \operatorname {acoth}^{3}{\left (\tanh {\left (a + b x \right )} \right )}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} & \text {otherwise} \end {cases} \]
Piecewise((b**3*log(x) - b**2*acoth(tanh(a + b*x))/x - b*acoth(tanh(a + b* x))**2/(2*x**2) - acoth(tanh(a + b*x))**3/(3*x**3), Eq(m, -4)), (Integral( acoth(tanh(a + b*x))**3/x**3, x), Eq(m, -3)), (Integral(acoth(tanh(a + b*x ))**3/x**2, x), Eq(m, -2)), (Integral(acoth(tanh(a + b*x))**3/x, x), Eq(m, -1)), (-6*b**3*x**4*x**m/(m**4 + 10*m**3 + 35*m**2 + 50*m + 24) + 6*b**2* m*x**3*x**m*acoth(tanh(a + b*x))/(m**4 + 10*m**3 + 35*m**2 + 50*m + 24) + 24*b**2*x**3*x**m*acoth(tanh(a + b*x))/(m**4 + 10*m**3 + 35*m**2 + 50*m + 24) - 3*b*m**2*x**2*x**m*acoth(tanh(a + b*x))**2/(m**4 + 10*m**3 + 35*m**2 + 50*m + 24) - 21*b*m*x**2*x**m*acoth(tanh(a + b*x))**2/(m**4 + 10*m**3 + 35*m**2 + 50*m + 24) - 36*b*x**2*x**m*acoth(tanh(a + b*x))**2/(m**4 + 10* m**3 + 35*m**2 + 50*m + 24) + m**3*x*x**m*acoth(tanh(a + b*x))**3/(m**4 + 10*m**3 + 35*m**2 + 50*m + 24) + 9*m**2*x*x**m*acoth(tanh(a + b*x))**3/(m* *4 + 10*m**3 + 35*m**2 + 50*m + 24) + 26*m*x*x**m*acoth(tanh(a + b*x))**3/ (m**4 + 10*m**3 + 35*m**2 + 50*m + 24) + 24*x*x**m*acoth(tanh(a + b*x))**3 /(m**4 + 10*m**3 + 35*m**2 + 50*m + 24), True))
Time = 0.31 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.99 \[ \int x^m \coth ^{-1}(\tanh (a+b x))^3 \, dx=-\frac {3 \, b x^{2} x^{m} \operatorname {arcoth}\left (\tanh \left (b x + a\right )\right )^{2}}{{\left (m + 2\right )} {\left (m + 1\right )}} + \frac {x^{m + 1} \operatorname {arcoth}\left (\tanh \left (b x + a\right )\right )^{3}}{m + 1} - \frac {6 \, {\left (\frac {b^{2} x^{4} x^{m}}{{\left (m + 4\right )} {\left (m + 3\right )} {\left (m + 2\right )}} - \frac {b x^{3} x^{m} \operatorname {arcoth}\left (\tanh \left (b x + a\right )\right )}{{\left (m + 3\right )} {\left (m + 2\right )}}\right )} b}{m + 1} \]
-3*b*x^2*x^m*arccoth(tanh(b*x + a))^2/((m + 2)*(m + 1)) + x^(m + 1)*arccot h(tanh(b*x + a))^3/(m + 1) - 6*(b^2*x^4*x^m/((m + 4)*(m + 3)*(m + 2)) - b* x^3*x^m*arccoth(tanh(b*x + a))/((m + 3)*(m + 2)))*b/(m + 1)
\[ \int x^m \coth ^{-1}(\tanh (a+b x))^3 \, dx=\int { x^{m} \operatorname {arcoth}\left (\tanh \left (b x + a\right )\right )^{3} \,d x } \]
Time = 4.04 (sec) , antiderivative size = 332, normalized size of antiderivative = 3.02 \[ \int x^m \coth ^{-1}(\tanh (a+b x))^3 \, dx=\frac {8\,b^3\,x^m\,x^4\,\left (m^3+6\,m^2+11\,m+6\right )}{8\,m^4+80\,m^3+280\,m^2+400\,m+192}-\frac {x\,x^m\,{\left (\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+2\,b\,x\right )}^3\,\left (m^3+9\,m^2+26\,m+24\right )}{8\,m^4+80\,m^3+280\,m^2+400\,m+192}-\frac {12\,b^2\,x^m\,x^3\,\left (\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+2\,b\,x\right )\,\left (m^3+7\,m^2+14\,m+8\right )}{8\,m^4+80\,m^3+280\,m^2+400\,m+192}+\frac {6\,b\,x^m\,x^2\,{\left (\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+2\,b\,x\right )}^2\,\left (m^3+8\,m^2+19\,m+12\right )}{8\,m^4+80\,m^3+280\,m^2+400\,m+192} \]
(8*b^3*x^m*x^4*(11*m + 6*m^2 + m^3 + 6))/(400*m + 280*m^2 + 80*m^3 + 8*m^4 + 192) - (x*x^m*(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)^3*(26*m + 9*m^2 + m^3 + 24))/( 400*m + 280*m^2 + 80*m^3 + 8*m^4 + 192) - (12*b^2*x^m*x^3*(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)*(14*m + 7*m^2 + m^3 + 8))/(400*m + 280*m^2 + 80*m^3 + 8*m^4 + 1 92) + (6*b*x^m*x^2*(log(-2/(exp(2*a)*exp(2*b*x) - 1)) - log((2*exp(2*a)*ex p(2*b*x))/(exp(2*a)*exp(2*b*x) - 1)) + 2*b*x)^2*(19*m + 8*m^2 + m^3 + 12)) /(400*m + 280*m^2 + 80*m^3 + 8*m^4 + 192)