\(\int x^2 \coth ^{-1}(\tanh (a+b x)) \, dx\) [129]

Optimal result

Integrand size = 11, antiderivative size = 23 \[ \int x^2 \coth ^{-1}(\tanh (a+b x)) \, dx=-\frac {b x^4}{12}+\frac {1}{3} x^3 \coth ^{-1}(\tanh (a+b x)) \]

[Out]

-1/12*b*x^4+1/3*x^3*arccoth(tanh(b*x+a))

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {2199, 30} \[ \int x^2 \coth ^{-1}(\tanh (a+b x)) \, dx=\frac {1}{3} x^3 \coth ^{-1}(\tanh (a+b x))-\frac {b x^4}{12} \]

[In]

Int[x^2*ArcCoth[Tanh[a + b*x]],x]

[Out]

-1/12*(b*x^4) + (x^3*ArcCoth[Tanh[a + b*x]])/3

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2199

Int[(u_)^(m_)*(v_)^(n_.), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, Simp[u^(m + 1)*(v^
n/(a*(m + 1))), x] - Dist[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] && (Fra
ctionQ[m] || GeQ[2*n + m + 1, 0]))) || (IGtQ[n, 0] && IGtQ[m, 0] && LeQ[n, m]) || (IGtQ[n, 0] &&  !IntegerQ[m]
) || (ILtQ[m, 0] &&  !IntegerQ[n]))

Rubi steps \begin{align*} \text {integral}& = \frac {1}{3} x^3 \coth ^{-1}(\tanh (a+b x))-\frac {1}{3} b \int x^3 \, dx \\ & = -\frac {b x^4}{12}+\frac {1}{3} x^3 \coth ^{-1}(\tanh (a+b x)) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87 \[ \int x^2 \coth ^{-1}(\tanh (a+b x)) \, dx=-\frac {1}{12} x^3 \left (b x-4 \coth ^{-1}(\tanh (a+b x))\right ) \]

[In]

Integrate[x^2*ArcCoth[Tanh[a + b*x]],x]

[Out]

-1/12*(x^3*(b*x - 4*ArcCoth[Tanh[a + b*x]]))

Maple [A] (verified)

Time = 0.49 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87

method	result	size
default	\(-\frac {b \,x^{4}}{12}+\frac {x^{3} \operatorname {arccoth}\left (\tanh \left (b x +a \right )\right )}{3}\)	\(20\)
parallelrisch	\(-\frac {b \,x^{4}}{12}+\frac {x^{3} \operatorname {arccoth}\left (\tanh \left (b x +a \right )\right )}{3}\)	\(20\)
parts	\(-\frac {b \,x^{4}}{12}+\frac {x^{3} \operatorname {arccoth}\left (\tanh \left (b x +a \right )\right )}{3}\)	\(20\)
risch	\(\frac {x^{3} \ln \left ({\mathrm e}^{b x +a}\right )}{3}-\frac {b \,x^{4}}{12}-\frac {i \pi \,x^{3}}{6}-\frac {i \pi \,x^{3} \operatorname {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right ) \operatorname {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )}{12}-\frac {i \pi \,x^{3} \operatorname {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right )^{3}}{6}-\frac {i \pi \,x^{3} \operatorname {csgn}\left (i {\mathrm e}^{b x +a}\right )^{2} \operatorname {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )}{12}+\frac {i \pi \,x^{3} \operatorname {csgn}\left (i {\mathrm e}^{b x +a}\right ) \operatorname {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )^{2}}{6}+\frac {i \pi \,x^{3} \operatorname {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{2}}{12}+\frac {i \pi \,x^{3} \operatorname {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right )^{2}}{6}-\frac {i \pi \,x^{3} \operatorname {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )^{3}}{12}+\frac {i \pi \,x^{3} \operatorname {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{2}}{12}-\frac {i \pi \,x^{3} \operatorname {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{3}}{12}\)	\(363\)

[In]

int(x^2*arccoth(tanh(b*x+a)),x,method=_RETURNVERBOSE)

[Out]

-1/12*b*x^4+1/3*x^3*arccoth(tanh(b*x+a))

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.25 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int x^2 \coth ^{-1}(\tanh (a+b x)) \, dx=\frac {1}{4} \, b x^{4} + \frac {1}{6} i \, \pi x^{3} + \frac {1}{3} \, a x^{3} \]

[In]

integrate(x^2*arccoth(tanh(b*x+a)),x, algorithm="fricas")

[Out]

1/4*b*x^4 + 1/6*I*pi*x^3 + 1/3*a*x^3

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int x^2 \coth ^{-1}(\tanh (a+b x)) \, dx=- \frac {b x^{4}}{12} + \frac {x^{3} \operatorname {acoth}{\left (\tanh {\left (a + b x \right )} \right )}}{3} \]

[In]

integrate(x**2*acoth(tanh(b*x+a)),x)

[Out]

-b*x**4/12 + x**3*acoth(tanh(a + b*x))/3

Maxima [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int x^2 \coth ^{-1}(\tanh (a+b x)) \, dx=-\frac {1}{12} \, b x^{4} + \frac {1}{3} \, x^{3} \operatorname {arcoth}\left (\tanh \left (b x + a\right )\right ) \]

[In]

integrate(x^2*arccoth(tanh(b*x+a)),x, algorithm="maxima")

[Out]

-1/12*b*x^4 + 1/3*x^3*arccoth(tanh(b*x + a))

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 71 vs. \(2 (19) = 38\).

Time = 0.29 (sec) , antiderivative size = 71, normalized size of antiderivative = 3.09 \[ \int x^2 \coth ^{-1}(\tanh (a+b x)) \, dx=-\frac {1}{12} \, b x^{4} + \frac {1}{6} \, x^{3} \log \left (-\frac {\frac {e^{\left (2 \, b x + 2 \, a\right )} + 1}{e^{\left (2 \, b x + 2 \, a\right )} - 1} + 1}{\frac {e^{\left (2 \, b x + 2 \, a\right )} + 1}{e^{\left (2 \, b x + 2 \, a\right )} - 1} - 1}\right ) \]

[In]

integrate(x^2*arccoth(tanh(b*x+a)),x, algorithm="giac")

[Out]

-1/12*b*x^4 + 1/6*x^3*log(-((e^(2*b*x + 2*a) + 1)/(e^(2*b*x + 2*a) - 1) + 1)/((e^(2*b*x + 2*a) + 1)/(e^(2*b*x
+ 2*a) - 1) - 1))

Mupad [B] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int x^2 \coth ^{-1}(\tanh (a+b x)) \, dx=\frac {x^3\,\mathrm {acoth}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}{3}-\frac {b\,x^4}{12} \]

[In]

int(x^2*acoth(tanh(a + b*x)),x)

[Out]

(x^3*acoth(tanh(a + b*x)))/3 - (b*x^4)/12