\(\int x^3 \coth ^{-1}(a+b x^4) \, dx\) [293]

Optimal result

Integrand size = 12, antiderivative size = 44 \[ \int x^3 \coth ^{-1}\left (a+b x^4\right ) \, dx=\frac {\left (a+b x^4\right ) \coth ^{-1}\left (a+b x^4\right )}{4 b}+\frac {\log \left (1-\left (a+b x^4\right )^2\right )}{8 b} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {6847, 6239, 6022, 266} \[ \int x^3 \coth ^{-1}\left (a+b x^4\right ) \, dx=\frac {\log \left (1-\left (a+b x^4\right )^2\right )}{8 b}+\frac {\left (a+b x^4\right ) \coth ^{-1}\left (a+b x^4\right )}{4 b} \]

[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 266

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 6022

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

Rule 6239

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 6847

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*} \text {integral}& = \frac {1}{4} \text {Subst}\left (\int \coth ^{-1}(a+b x) \, dx,x,x^4\right ) \\ & = \frac {\text {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 {\text {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] (verified)

Time = 0.01 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.89 \[ \int x^3 \coth ^{-1}\left (a+b x^4\right ) \, dx=\frac {2 \left (a+b x^4\right ) \coth ^{-1}\left (a+b x^4\right )+\log \left (1-\left (a+b x^4\right )^2\right )}{8 b} \]

[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)

Maple [A] (verified)

Time = 0.30 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.84

method	result	size
derivativedivides	\(\frac {\left (b \,x^{4}+a \right ) \operatorname {arccoth}\left (b \,x^{4}+a \right )+\frac {\ln \left (\left (b \,x^{4}+a \right )^{2}-1\right )}{2}}{4 b}\)	\(37\)
default	\(\frac {\left (b \,x^{4}+a \right ) \operatorname {arccoth}\left (b \,x^{4}+a \right )+\frac {\ln \left (\left (b \,x^{4}+a \right )^{2}-1\right )}{2}}{4 b}\)	\(37\)
parts	\(\frac {x^{4} \operatorname {arccoth}\left (b \,x^{4}+a \right )}{4}+b \left (\frac {\left (1-a \right ) \ln \left (b \,x^{4}+a -1\right )}{8 b^{2}}+\frac {\left (1+a \right ) \ln \left (b \,x^{4}+a +1\right )}{8 b^{2}}\right )\)	\(54\)
parallelrisch	\(-\frac {-\operatorname {arccoth}\left (b \,x^{4}+a \right ) x^{4} b^{2}-\operatorname {arccoth}\left (b \,x^{4}+a \right ) a b -\ln \left (b \,x^{4}+a -1\right ) b -b \,\operatorname {arccoth}\left (b \,x^{4}+a \right )}{4 b^{2}}\)	\(58\)
risch	\(\frac {x^{4} \ln \left (b \,x^{4}+a +1\right )}{8}-\frac {x^{4} \ln \left (b \,x^{4}+a -1\right )}{8}+\frac {\ln \left (b \,x^{4}+a +1\right ) a}{8 b}-\frac {\ln \left (-b \,x^{4}-a +1\right ) a}{8 b}+\frac {\ln \left (b \,x^{4}+a +1\right )}{8 b}+\frac {\ln \left (-b \,x^{4}-a +1\right )}{8 b}\)	\(94\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.32 \[ \int x^3 \coth ^{-1}\left (a+b x^4\right ) \, dx=\frac {b x^{4} \log \left (\frac {b x^{4} + a + 1}{b x^{4} + a - 1}\right ) + {\left (a + 1\right )} \log \left (b x^{4} + a + 1\right ) - {\left (a - 1\right )} \log \left (b x^{4} + a - 1\right )}{8 \, b} \]

[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

Sympy [A] (verification not implemented)

Time = 0.71 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.36 \[ \int x^3 \coth ^{-1}\left (a+b x^4\right ) \, dx=\begin {cases} \frac {a \operatorname {acoth}{\left (a + b x^{4} \right )}}{4 b} + \frac {x^{4} \operatorname {acoth}{\left (a + b x^{4} \right )}}{4} + \frac {\log {\left (a + b x^{4} + 1 \right )}}{4 b} - \frac {\operatorname {acoth}{\left (a + b x^{4} \right )}}{4 b} & \text {for}\: b \neq 0 \\\frac {x^{4} \operatorname {acoth}{\left (a \right )}}{4} & \text {otherwise} \end {cases} \]

[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))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.84 \[ \int x^3 \coth ^{-1}\left (a+b x^4\right ) \, dx=\frac {2 \, {\left (b x^{4} + a\right )} \operatorname {arcoth}\left (b x^{4} + a\right ) + \log \left (-{\left (b x^{4} + a\right )}^{2} + 1\right )}{8 \, b} \]

[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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 225 vs. \(2 (40) = 80\).

Time = 0.28 (sec) , antiderivative size = 225, normalized size of antiderivative = 5.11 \[ \int x^3 \coth ^{-1}\left (a+b x^4\right ) \, dx=\frac {1}{8} \, {\left ({\left (a + 1\right )} b - {\left (a - 1\right )} b\right )} {\left (\frac {\log \left (\frac {{\left | b x^{4} + a + 1 \right |}}{{\left | b x^{4} + a - 1 \right |}}\right )}{b^{2}} - \frac {\log \left ({\left | \frac {b x^{4} + a + 1}{b x^{4} + a - 1} - 1 \right |}\right )}{b^{2}} + \frac {\log \left (-\frac {\frac {1}{a - \frac {{\left (\frac {{\left (b x^{4} + a + 1\right )} {\left (a - 1\right )}}{b x^{4} + a - 1} - a - 1\right )} b}{\frac {{\left (b x^{4} + a + 1\right )} b}{b x^{4} + a - 1} - b}} + 1}{\frac {1}{a - \frac {{\left (\frac {{\left (b x^{4} + a + 1\right )} {\left (a - 1\right )}}{b x^{4} + a - 1} - a - 1\right )} b}{\frac {{\left (b x^{4} + a + 1\right )} b}{b x^{4} + a - 1} - b}} - 1}\right )}{b^{2} {\left (\frac {b x^{4} + a + 1}{b x^{4} + a - 1} - 1\right )}}\right )} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 4.42 (sec) , antiderivative size = 107, normalized size of antiderivative = 2.43 \[ \int x^3 \coth ^{-1}\left (a+b x^4\right ) \, dx=\frac {x^4\,\ln \left (\frac {b\,x^4+a+1}{b\,x^4+a}\right )}{8}-\frac {x^4\,\ln \left (\frac {b\,x^4+a-1}{b\,x^4+a}\right )}{8}+\frac {\ln \left (b\,x^4+a-1\right )}{8\,b}+\frac {\ln \left (b\,x^4+a+1\right )}{8\,b}-\frac {a\,\ln \left (b\,x^4+a-1\right )}{8\,b}+\frac {a\,\ln \left (b\,x^4+a+1\right )}{8\,b} \]

[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)