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]

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

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Rubi [A]  time = 0.05, antiderivative size = 44, normalized size of antiderivative = 1.00, 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 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]

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 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 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]

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*}

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Mathematica [A]  time = 0.02, 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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fricas [A]  time = 0.61, size = 58, normalized size = 1.32 \[ \frac {b x^{4} \log \left (\frac {b x^{4} + a + 1}{b x^{4} + a - 1}\right ) + {\left (a + 1\right )} \log \left (b x^{4} + a + 1\right ) - {\left (a - 1\right )} \log \left (b x^{4} + a - 1\right )}{8 \, b} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{3} \operatorname {arcoth}\left (b x^{4} + a\right )\,{d x} \]

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)

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maple [A]  time = 0.07, size = 46, normalized size = 1.05 \[ \frac {\mathrm {arccoth}\left (b \,x^{4}+a \right ) x^{4}}{4}+\frac {\mathrm {arccoth}\left (b \,x^{4}+a \right ) a}{4 b}+\frac {\ln \left (\left (b \,x^{4}+a \right )^{2}-1\right )}{8 b} \]

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 = 0.32, size = 37, normalized size = 0.84 \[ \frac {2 \, {\left (b x^{4} + a\right )} \operatorname {arcoth}\left (b x^{4} + a\right ) + \log \left (-{\left (b x^{4} + a\right )}^{2} + 1\right )}{8 \, b} \]

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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mupad [B]  time = 1.56, size = 107, normalized size = 2.43 \[ \frac {x^4\,\ln \left (\frac {b\,x^4+a+1}{b\,x^4+a}\right )}{8}-\frac {x^4\,\ln \left (\frac {b\,x^4+a-1}{b\,x^4+a}\right )}{8}+\frac {\ln \left (b\,x^4+a-1\right )}{8\,b}+\frac {\ln \left (b\,x^4+a+1\right )}{8\,b}-\frac {a\,\ln \left (b\,x^4+a-1\right )}{8\,b}+\frac {a\,\ln \left (b\,x^4+a+1\right )}{8\,b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*acoth(a + b*x^4),x)

[Out]

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

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sympy [A]  time = 4.22, size = 60, normalized size = 1.36 \[ \begin {cases} \frac {a \operatorname {acoth}{\left (a + b x^{4} \right )}}{4 b} + \frac {x^{4} \operatorname {acoth}{\left (a + b x^{4} \right )}}{4} + \frac {\log {\left (a + b x^{4} + 1 \right )}}{4 b} - \frac {\operatorname {acoth}{\left (a + b x^{4} \right )}}{4 b} & \text {for}\: b \neq 0 \\\frac {x^{4} \operatorname {acoth}{\relax (a )}}{4} & \text {otherwise} \end {cases} \]

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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