\(\int x^4 \coth ^{-1}(a x) \, dx\) [2]

Optimal result

Integrand size = 8, antiderivative size = 50 \[ \int x^4 \coth ^{-1}(a x) \, dx=\frac {x^2}{10 a^3}+\frac {x^4}{20 a}+\frac {1}{5} x^5 \coth ^{-1}(a x)+\frac {\log \left (1-a^2 x^2\right )}{10 a^5} \]

[Out]

1/10*x^2/a^3+1/20*x^4/a+1/5*x^5*arccoth(a*x)+1/10*ln(-a^2*x^2+1)/a^5

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {6038, 272, 45} \[ \int x^4 \coth ^{-1}(a x) \, dx=\frac {x^2}{10 a^3}+\frac {\log \left (1-a^2 x^2\right )}{10 a^5}+\frac {1}{5} x^5 \coth ^{-1}(a x)+\frac {x^4}{20 a} \]

[In]

Int[x^4*ArcCoth[a*x],x]

[Out]

x^2/(10*a^3) + x^4/(20*a) + (x^5*ArcCoth[a*x])/5 + Log[1 - a^2*x^2]/(10*a^5)

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 6038

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

Rubi steps \begin{align*} \text {integral}& = \frac {1}{5} x^5 \coth ^{-1}(a x)-\frac {1}{5} a \int \frac {x^5}{1-a^2 x^2} \, dx \\ & = \frac {1}{5} x^5 \coth ^{-1}(a x)-\frac {1}{10} a \text {Subst}\left (\int \frac {x^2}{1-a^2 x} \, dx,x,x^2\right ) \\ & = \frac {1}{5} x^5 \coth ^{-1}(a x)-\frac {1}{10} a \text {Subst}\left (\int \left (-\frac {1}{a^4}-\frac {x}{a^2}-\frac {1}{a^4 \left (-1+a^2 x\right )}\right ) \, dx,x,x^2\right ) \\ & = \frac {x^2}{10 a^3}+\frac {x^4}{20 a}+\frac {1}{5} x^5 \coth ^{-1}(a x)+\frac {\log \left (1-a^2 x^2\right )}{10 a^5} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.00 \[ \int x^4 \coth ^{-1}(a x) \, dx=\frac {x^2}{10 a^3}+\frac {x^4}{20 a}+\frac {1}{5} x^5 \coth ^{-1}(a x)+\frac {\log \left (1-a^2 x^2\right )}{10 a^5} \]

[In]

Integrate[x^4*ArcCoth[a*x],x]

[Out]

x^2/(10*a^3) + x^4/(20*a) + (x^5*ArcCoth[a*x])/5 + Log[1 - a^2*x^2]/(10*a^5)

Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.94

method	result	size
parts	\(\frac {x^{5} \operatorname {arccoth}\left (a x \right )}{5}+\frac {a \left (\frac {\frac {1}{2} x^{4} a^{2}+x^{2}}{2 a^{4}}+\frac {\ln \left (a^{2} x^{2}-1\right )}{2 a^{6}}\right )}{5}\)	\(47\)
derivativedivides	\(\frac {\frac {a^{5} x^{5} \operatorname {arccoth}\left (a x \right )}{5}+\frac {a^{4} x^{4}}{20}+\frac {a^{2} x^{2}}{10}+\frac {\ln \left (a x -1\right )}{10}+\frac {\ln \left (a x +1\right )}{10}}{a^{5}}\)	\(50\)
default	\(\frac {\frac {a^{5} x^{5} \operatorname {arccoth}\left (a x \right )}{5}+\frac {a^{4} x^{4}}{20}+\frac {a^{2} x^{2}}{10}+\frac {\ln \left (a x -1\right )}{10}+\frac {\ln \left (a x +1\right )}{10}}{a^{5}}\)	\(50\)
parallelrisch	\(-\frac {-4 a^{5} x^{5} \operatorname {arccoth}\left (a x \right )-a^{4} x^{4}-2-2 a^{2} x^{2}-4 \ln \left (a x -1\right )-4 \,\operatorname {arccoth}\left (a x \right )}{20 a^{5}}\)	\(50\)
risch	\(\frac {x^{5} \ln \left (a x +1\right )}{10}-\frac {\ln \left (a x -1\right ) x^{5}}{10}+\frac {x^{4}}{20 a}+\frac {x^{2}}{10 a^{3}}+\frac {\ln \left (a^{2} x^{2}-1\right )}{10 a^{5}}+\frac {1}{20 a^{5}}\)	\(60\)

[In]

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

[Out]

1/5*x^5*arccoth(a*x)+1/5*a*(1/2/a^4*(1/2*x^4*a^2+x^2)+1/2/a^6*ln(a^2*x^2-1))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.10 \[ \int x^4 \coth ^{-1}(a x) \, dx=\frac {2 \, a^{5} x^{5} \log \left (\frac {a x + 1}{a x - 1}\right ) + a^{4} x^{4} + 2 \, a^{2} x^{2} + 2 \, \log \left (a^{2} x^{2} - 1\right )}{20 \, a^{5}} \]

[In]

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

[Out]

1/20*(2*a^5*x^5*log((a*x + 1)/(a*x - 1)) + a^4*x^4 + 2*a^2*x^2 + 2*log(a^2*x^2 - 1))/a^5

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.30 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.08 \[ \int x^4 \coth ^{-1}(a x) \, dx=\begin {cases} \frac {x^{5} \operatorname {acoth}{\left (a x \right )}}{5} + \frac {x^{4}}{20 a} + \frac {x^{2}}{10 a^{3}} + \frac {\log {\left (a x + 1 \right )}}{5 a^{5}} - \frac {\operatorname {acoth}{\left (a x \right )}}{5 a^{5}} & \text {for}\: a \neq 0 \\\frac {i \pi x^{5}}{10} & \text {otherwise} \end {cases} \]

[In]

integrate(x**4*acoth(a*x),x)

[Out]

Piecewise((x**5*acoth(a*x)/5 + x**4/(20*a) + x**2/(10*a**3) + log(a*x + 1)/(5*a**5) - acoth(a*x)/(5*a**5), Ne(
a, 0)), (I*pi*x**5/10, True))

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.92 \[ \int x^4 \coth ^{-1}(a x) \, dx=\frac {1}{5} \, x^{5} \operatorname {arcoth}\left (a x\right ) + \frac {1}{20} \, a {\left (\frac {a^{2} x^{4} + 2 \, x^{2}}{a^{4}} + \frac {2 \, \log \left (a^{2} x^{2} - 1\right )}{a^{6}}\right )} \]

[In]

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

[Out]

1/5*x^5*arccoth(a*x) + 1/20*a*((a^2*x^4 + 2*x^2)/a^4 + 2*log(a^2*x^2 - 1)/a^6)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 255 vs. \(2 (42) = 84\).

Time = 0.28 (sec) , antiderivative size = 255, normalized size of antiderivative = 5.10 \[ \int x^4 \coth ^{-1}(a x) \, dx=\frac {1}{5} \, a {\left (\frac {\log \left (\frac {{\left | a x + 1 \right |}}{{\left | a x - 1 \right |}}\right )}{a^{6}} - \frac {\log \left ({\left | \frac {a x + 1}{a x - 1} - 1 \right |}\right )}{a^{6}} + \frac {4 \, {\left (\frac {{\left (a x + 1\right )}^{3}}{{\left (a x - 1\right )}^{3}} - \frac {{\left (a x + 1\right )}^{2}}{{\left (a x - 1\right )}^{2}} + \frac {a x + 1}{a x - 1}\right )}}{a^{6} {\left (\frac {a x + 1}{a x - 1} - 1\right )}^{4}} + \frac {{\left (\frac {5 \, {\left (a x + 1\right )}^{4}}{{\left (a x - 1\right )}^{4}} + \frac {10 \, {\left (a x + 1\right )}^{2}}{{\left (a x - 1\right )}^{2}} + 1\right )} \log \left (-\frac {\frac {\frac {{\left (a x + 1\right )} a}{a x - 1} - a}{a {\left (\frac {a x + 1}{a x - 1} + 1\right )}} + 1}{\frac {\frac {{\left (a x + 1\right )} a}{a x - 1} - a}{a {\left (\frac {a x + 1}{a x - 1} + 1\right )}} - 1}\right )}{a^{6} {\left (\frac {a x + 1}{a x - 1} - 1\right )}^{5}}\right )} \]

[In]

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

[Out]

1/5*a*(log(abs(a*x + 1)/abs(a*x - 1))/a^6 - log(abs((a*x + 1)/(a*x - 1) - 1))/a^6 + 4*((a*x + 1)^3/(a*x - 1)^3
- (a*x + 1)^2/(a*x - 1)^2 + (a*x + 1)/(a*x - 1))/(a^6*((a*x + 1)/(a*x - 1) - 1)^4) + (5*(a*x + 1)^4/(a*x - 1)
^4 + 10*(a*x + 1)^2/(a*x - 1)^2 + 1)*log(-(((a*x + 1)*a/(a*x - 1) - a)/(a*((a*x + 1)/(a*x - 1) + 1)) + 1)/(((a
*x + 1)*a/(a*x - 1) - a)/(a*((a*x + 1)/(a*x - 1) + 1)) - 1))/(a^6*((a*x + 1)/(a*x - 1) - 1)^5))

Mupad [B] (verification not implemented)

Time = 4.32 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.86 \[ \int x^4 \coth ^{-1}(a x) \, dx=\frac {\frac {\ln \left (a^2\,x^2-1\right )}{10}+\frac {a^2\,x^2}{10}+\frac {a^4\,x^4}{20}}{a^5}+\frac {x^5\,\mathrm {acoth}\left (a\,x\right )}{5} \]

[In]

int(x^4*acoth(a*x),x)

[Out]

(log(a^2*x^2 - 1)/10 + (a^2*x^2)/10 + (a^4*x^4)/20)/a^5 + (x^5*acoth(a*x))/5