\(\int x^{-1+n} \coth ^{-1}(a+b x^n) \, dx\) [294]

Optimal result

Integrand size = 14, antiderivative size = 47 \[ \int x^{-1+n} \coth ^{-1}\left (a+b x^n\right ) \, dx=\frac {\left (a+b x^n\right ) \coth ^{-1}\left (a+b x^n\right )}{b n}+\frac {\log \left (1-\left (a+b x^n\right )^2\right )}{2 b n} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {6847, 6239, 6022, 266} \[ \int x^{-1+n} \coth ^{-1}\left (a+b x^n\right ) \, dx=\frac {\log \left (1-\left (a+b x^n\right )^2\right )}{2 b n}+\frac {\left (a+b x^n\right ) \coth ^{-1}\left (a+b x^n\right )}{b n} \]

[In]

Int[x^(-1 + n)*ArcCoth[a + b*x^n],x]

[Out]

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

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 {\text {Subst}\left (\int \coth ^{-1}(a+b x) \, dx,x,x^n\right )}{n} \\ & = \frac {\text {Subst}\left (\int \coth ^{-1}(x) \, dx,x,a+b x^n\right )}{b n} \\ & = \frac {\left (a+b x^n\right ) \coth ^{-1}\left (a+b x^n\right )}{b n}-\frac {\text {Subst}\left (\int \frac {x}{1-x^2} \, dx,x,a+b x^n\right )}{b n} \\ & = \frac {\left (a+b x^n\right ) \coth ^{-1}\left (a+b x^n\right )}{b n}+\frac {\log \left (1-\left (a+b x^n\right )^2\right )}{2 b n} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.89 \[ \int x^{-1+n} \coth ^{-1}\left (a+b x^n\right ) \, dx=\frac {2 \left (a+b x^n\right ) \coth ^{-1}\left (a+b x^n\right )+\log \left (1-\left (a+b x^n\right )^2\right )}{2 b n} \]

[In]

Integrate[x^(-1 + n)*ArcCoth[a + b*x^n],x]

[Out]

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(117\) vs. \(2(45)=90\).

Time = 4.40 (sec) , antiderivative size = 118, normalized size of antiderivative = 2.51

method	result	size
risch	\(\frac {x^{n} \ln \left (a +b \,x^{n}+1\right )}{2 n}-\frac {x^{n} \ln \left (-1+a +b \,x^{n}\right )}{2 n}+\frac {\ln \left (x^{n}+\frac {1+a}{b}\right ) a}{2 b n}-\frac {\ln \left (x^{n}+\frac {-1+a}{b}\right ) a}{2 b n}+\frac {\ln \left (x^{n}+\frac {1+a}{b}\right )}{2 b n}+\frac {\ln \left (x^{n}+\frac {-1+a}{b}\right )}{2 b n}\)	\(118\)

[In]

int(x^(-1+n)*arccoth(a+b*x^n),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 108 vs. \(2 (45) = 90\).

Time = 0.26 (sec) , antiderivative size = 108, normalized size of antiderivative = 2.30 \[ \int x^{-1+n} \coth ^{-1}\left (a+b x^n\right ) \, dx=\frac {{\left (a + 1\right )} \log \left (b \cosh \left (n \log \left (x\right )\right ) + b \sinh \left (n \log \left (x\right )\right ) + a + 1\right ) - {\left (a - 1\right )} \log \left (b \cosh \left (n \log \left (x\right )\right ) + b \sinh \left (n \log \left (x\right )\right ) + a - 1\right ) + {\left (b \cosh \left (n \log \left (x\right )\right ) + b \sinh \left (n \log \left (x\right )\right )\right )} \log \left (\frac {b \cosh \left (n \log \left (x\right )\right ) + b \sinh \left (n \log \left (x\right )\right ) + a + 1}{b \cosh \left (n \log \left (x\right )\right ) + b \sinh \left (n \log \left (x\right )\right ) + a - 1}\right )}{2 \, b n} \]

[In]

integrate(x^(-1+n)*arccoth(a+b*x^n),x, algorithm="fricas")

[Out]

1/2*((a + 1)*log(b*cosh(n*log(x)) + b*sinh(n*log(x)) + a + 1) - (a - 1)*log(b*cosh(n*log(x)) + b*sinh(n*log(x)
) + a - 1) + (b*cosh(n*log(x)) + b*sinh(n*log(x)))*log((b*cosh(n*log(x)) + b*sinh(n*log(x)) + a + 1)/(b*cosh(n
*log(x)) + b*sinh(n*log(x)) + a - 1)))/(b*n)

Sympy [F(-2)]

Exception generated. \[ \int x^{-1+n} \coth ^{-1}\left (a+b x^n\right ) \, dx=\text {Exception raised: HeuristicGCDFailed} \]

[In]

integrate(x**(-1+n)*acoth(a+b*x**n),x)

[Out]

Exception raised: HeuristicGCDFailed >> no luck

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.85 \[ \int x^{-1+n} \coth ^{-1}\left (a+b x^n\right ) \, dx=\frac {2 \, {\left (b x^{n} + a\right )} \operatorname {arcoth}\left (b x^{n} + a\right ) + \log \left (-{\left (b x^{n} + a\right )}^{2} + 1\right )}{2 \, b n} \]

[In]

integrate(x^(-1+n)*arccoth(a+b*x^n),x, algorithm="maxima")

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 119 vs. \(2 (45) = 90\).

Time = 0.29 (sec) , antiderivative size = 119, normalized size of antiderivative = 2.53 \[ \int x^{-1+n} \coth ^{-1}\left (a+b x^n\right ) \, dx=\frac {{\left ({\left (a + 1\right )} b - {\left (a - 1\right )} b\right )} {\left (\frac {\log \left (\frac {{\left | b x^{n} + a + 1 \right |}}{{\left | b x^{n} + a - 1 \right |}}\right )}{b^{2}} - \frac {\log \left ({\left | \frac {b x^{n} + a + 1}{b x^{n} + a - 1} - 1 \right |}\right )}{b^{2}} + \frac {\log \left (\frac {b x^{n} + a + 1}{b x^{n} + a - 1}\right )}{b^{2} {\left (\frac {b x^{n} + a + 1}{b x^{n} + a - 1} - 1\right )}}\right )}}{2 \, n} \]

[In]

integrate(x^(-1+n)*arccoth(a+b*x^n),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 5.53 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.23 \[ \int x^{-1+n} \coth ^{-1}\left (a+b x^n\right ) \, dx=\frac {\frac {\ln \left (a^2+b^2\,x^{2\,n}+2\,a\,b\,x^n-1\right )}{2}+a\,\mathrm {acoth}\left (a+b\,x^n\right )}{b\,n}+\frac {x^n\,\mathrm {acoth}\left (a+b\,x^n\right )}{n} \]

[In]

int(x^(n - 1)*acoth(a + b*x^n),x)

[Out]

(log(a^2 + b^2*x^(2*n) + 2*a*b*x^n - 1)/2 + a*acoth(a + b*x^n))/(b*n) + (x^n*acoth(a + b*x^n))/n