3.293 \(\int x^3 \coth ^{-1}(a+b x^4) \, dx\)

Optimal. Leaf size=44 \[ \frac{\log \left (1-\left (a+b x^4\right )^2\right )}{8 b}+\frac{\left (a+b x^4\right ) \coth ^{-1}\left (a+b x^4\right )}{4 b} \]

[Out]

((a + b*x^4)*ArcCoth[a + b*x^4])/(4*b) + Log[1 - (a + b*x^4)^2]/(8*b)

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Rubi [A]  time = 0.0523872, antiderivative size = 44, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {6715, 6104, 5911, 260} \[ \frac{\log \left (1-\left (a+b x^4\right )^2\right )}{8 b}+\frac{\left (a+b x^4\right ) \coth ^{-1}\left (a+b x^4\right )}{4 b} \]

Antiderivative was successfully verified.

[In]

Int[x^3*ArcCoth[a + b*x^4],x]

[Out]

((a + b*x^4)*ArcCoth[a + b*x^4])/(4*b) + Log[1 - (a + b*x^4)^2]/(8*b)

Rule 6715

Int[(u_)*(x_)^(m_.), x_Symbol] :> Dist[1/(m + 1), Subst[Int[SubstFor[x^(m + 1), u, x], x], x, x^(m + 1)], x] /
; FreeQ[m, x] && NeQ[m, -1] && FunctionOfQ[x^(m + 1), u, x]

Rule 6104

Int[((a_.) + ArcCoth[(c_) + (d_.)*(x_)]*(b_.))^(p_.), x_Symbol] :> Dist[1/d, Subst[Int[(a + b*ArcCoth[x])^p, x
], x, c + d*x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[p, 0]

Rule 5911

Int[((a_.) + ArcCoth[(c_.)*(x_)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*ArcCoth[c*x])^p, x] - Dist[b*c*p, In
t[(x*(a + b*ArcCoth[c*x])^(p - 1))/(1 - c^2*x^2), x], x] /; FreeQ[{a, b, c}, x] && IGtQ[p, 0]

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rubi steps

\begin{align*} \int x^3 \coth ^{-1}\left (a+b x^4\right ) \, dx &=\frac{1}{4} \operatorname{Subst}\left (\int \coth ^{-1}(a+b x) \, dx,x,x^4\right )\\ &=\frac{\operatorname{Subst}\left (\int \coth ^{-1}(x) \, dx,x,a+b x^4\right )}{4 b}\\ &=\frac{\left (a+b x^4\right ) \coth ^{-1}\left (a+b x^4\right )}{4 b}-\frac{\operatorname{Subst}\left (\int \frac{x}{1-x^2} \, dx,x,a+b x^4\right )}{4 b}\\ &=\frac{\left (a+b x^4\right ) \coth ^{-1}\left (a+b x^4\right )}{4 b}+\frac{\log \left (1-\left (a+b x^4\right )^2\right )}{8 b}\\ \end{align*}

Mathematica [A]  time = 0.016213, size = 39, normalized size = 0.89 \[ \frac{\log \left (1-\left (a+b x^4\right )^2\right )+2 \left (a+b x^4\right ) \coth ^{-1}\left (a+b x^4\right )}{8 b} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*ArcCoth[a + b*x^4],x]

[Out]

(2*(a + b*x^4)*ArcCoth[a + b*x^4] + Log[1 - (a + b*x^4)^2])/(8*b)

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Maple [A]  time = 0.078, size = 46, normalized size = 1.1 \begin{align*}{\frac{{\rm arccoth} \left (b{x}^{4}+a\right ){x}^{4}}{4}}+{\frac{{\rm arccoth} \left (b{x}^{4}+a\right )a}{4\,b}}+{\frac{\ln \left ( \left ( b{x}^{4}+a \right ) ^{2}-1 \right ) }{8\,b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*arccoth(b*x^4+a),x)

[Out]

1/4*arccoth(b*x^4+a)*x^4+1/4/b*arccoth(b*x^4+a)*a+1/8/b*ln((b*x^4+a)^2-1)

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Maxima [A]  time = 1.04667, size = 50, normalized size = 1.14 \begin{align*} \frac{2 \,{\left (b x^{4} + a\right )} \operatorname{arcoth}\left (b x^{4} + a\right ) + \log \left (-{\left (b x^{4} + a\right )}^{2} + 1\right )}{8 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arccoth(b*x^4+a),x, algorithm="maxima")

[Out]

1/8*(2*(b*x^4 + a)*arccoth(b*x^4 + a) + log(-(b*x^4 + a)^2 + 1))/b

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Fricas [A]  time = 1.82153, size = 149, normalized size = 3.39 \begin{align*} \frac{b x^{4} \log \left (\frac{b x^{4} + a + 1}{b x^{4} + a - 1}\right ) +{\left (a + 1\right )} \log \left (b x^{4} + a + 1\right ) -{\left (a - 1\right )} \log \left (b x^{4} + a - 1\right )}{8 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arccoth(b*x^4+a),x, algorithm="fricas")

[Out]

1/8*(b*x^4*log((b*x^4 + a + 1)/(b*x^4 + a - 1)) + (a + 1)*log(b*x^4 + a + 1) - (a - 1)*log(b*x^4 + a - 1))/b

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Sympy [A]  time = 9.03658, size = 60, normalized size = 1.36 \begin{align*} \begin{cases} \frac{a \operatorname{acoth}{\left (a + b x^{4} \right )}}{4 b} + \frac{x^{4} \operatorname{acoth}{\left (a + b x^{4} \right )}}{4} + \frac{\log{\left (a + b x^{4} + 1 \right )}}{4 b} - \frac{\operatorname{acoth}{\left (a + b x^{4} \right )}}{4 b} & \text{for}\: b \neq 0 \\\frac{x^{4} \operatorname{acoth}{\left (a \right )}}{4} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*acoth(b*x**4+a),x)

[Out]

Piecewise((a*acoth(a + b*x**4)/(4*b) + x**4*acoth(a + b*x**4)/4 + log(a + b*x**4 + 1)/(4*b) - acoth(a + b*x**4
)/(4*b), Ne(b, 0)), (x**4*acoth(a)/4, True))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{3} \operatorname{arcoth}\left (b x^{4} + a\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arccoth(b*x^4+a),x, algorithm="giac")

[Out]

integrate(x^3*arccoth(b*x^4 + a), x)