\(\int x^{-1+n} \cot ^{-1}(a+b x^n) \, dx\) [150]

Optimal result

Integrand size = 14, antiderivative size = 45 \[ \int x^{-1+n} \cot ^{-1}\left (a+b x^n\right ) \, dx=\frac {\left (a+b x^n\right ) \cot ^{-1}\left (a+b x^n\right )}{b n}+\frac {\log \left (1+\left (a+b x^n\right )^2\right )}{2 b n} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 45, 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, 5148, 4931, 266} \[ \int x^{-1+n} \cot ^{-1}\left (a+b x^n\right ) \, dx=\frac {\log \left (\left (a+b x^n\right )^2+1\right )}{2 b n}+\frac {\left (a+b x^n\right ) \cot ^{-1}\left (a+b x^n\right )}{b n} \]

[In]

Int[x^(-1 + n)*ArcCot[a + b*x^n],x]

[Out]

((a + b*x^n)*ArcCot[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 4931

Int[((a_.) + ArcCot[(c_.)*(x_)^(n_.)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*ArcCot[c*x^n])^p, x] + Dist[b*c
*n*p, Int[x^n*((a + b*ArcCot[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 5148

Int[((a_.) + ArcCot[(c_) + (d_.)*(x_)]*(b_.))^(p_.), x_Symbol] :> Dist[1/d, Subst[Int[(a + b*ArcCot[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 \cot ^{-1}(a+b x) \, dx,x,x^n\right )}{n} \\ & = \frac {\text {Subst}\left (\int \cot ^{-1}(x) \, dx,x,a+b x^n\right )}{b n} \\ & = \frac {\left (a+b x^n\right ) \cot ^{-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 ) \cot ^{-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 = 40, normalized size of antiderivative = 0.89 \[ \int x^{-1+n} \cot ^{-1}\left (a+b x^n\right ) \, dx=\frac {2 \left (a+b x^n\right ) \cot ^{-1}\left (a+b x^n\right )+\log \left (1+\left (a+b x^n\right )^2\right )}{2 b n} \]

[In]

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

[Out]

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

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 6.30 (sec) , antiderivative size = 149, normalized size of antiderivative = 3.31

method	result	size
risch	\(\frac {i x^{n} \ln \left (1+i \left (a +b \,x^{n}\right )\right )}{2 n}-\frac {i x^{n} \ln \left (1-i \left (a +b \,x^{n}\right )\right )}{2 n}+\frac {\pi \,x^{n}}{2 n}+\frac {i \ln \left (x^{n}-\frac {i-a}{b}\right ) a}{2 b n}-\frac {i \ln \left (\frac {i+a}{b}+x^{n}\right ) a}{2 b n}+\frac {\ln \left (x^{n}-\frac {i-a}{b}\right )}{2 b n}+\frac {\ln \left (\frac {i+a}{b}+x^{n}\right )}{2 b n}\)	\(149\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.24 \[ \int x^{-1+n} \cot ^{-1}\left (a+b x^n\right ) \, dx=\frac {2 \, b x^{n} \operatorname {arccot}\left (b x^{n} + a\right ) - 2 \, a \arctan \left (b x^{n} + a\right ) + \log \left (b^{2} x^{2 \, n} + 2 \, a b x^{n} + a^{2} + 1\right )}{2 \, b n} \]

[In]

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

[Out]

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

Sympy [F(-1)]

Timed out. \[ \int x^{-1+n} \cot ^{-1}\left (a+b x^n\right ) \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.84 \[ \int x^{-1+n} \cot ^{-1}\left (a+b x^n\right ) \, dx=\frac {2 \, {\left (b x^{n} + a\right )} \operatorname {arccot}\left (b x^{n} + a\right ) + \log \left ({\left (b x^{n} + a\right )}^{2} + 1\right )}{2 \, b n} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.33 \[ \int x^{-1+n} \cot ^{-1}\left (a+b x^n\right ) \, dx=\frac {b {\left (\frac {2 \, {\left (b x^{n} + a\right )} \arctan \left (\frac {1}{b x^{n} + a}\right )}{b^{2}} + \frac {\log \left (\frac {1}{{\left (b x^{n} + a\right )}^{2}} + 1\right )}{b^{2}} - \frac {\log \left (\frac {1}{{\left (b x^{n} + a\right )}^{2}}\right )}{b^{2}}\right )}}{2 \, n} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 1.73 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.29 \[ \int x^{-1+n} \cot ^{-1}\left (a+b x^n\right ) \, dx=\frac {\frac {\ln \left (a^2+b^2\,x^{2\,n}+2\,a\,b\,x^n+1\right )}{2}+a\,\mathrm {acot}\left (a+b\,x^n\right )}{b\,n}+\frac {x^n\,\mathrm {acot}\left (a+b\,x^n\right )}{n} \]

[In]

int(x^(n - 1)*acot(a + b*x^n),x)

[Out]

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