3.2 \(\int x^{-1+n} \tan ^{-1}(a+b x^n) \, dx\)

Optimal. Leaf size=45 \[ \frac {\left (a+b x^n\right ) \tan ^{-1}\left (a+b x^n\right )}{b n}-\frac {\log \left (\left (a+b x^n\right )^2+1\right )}{2 b n} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 45, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {6715, 5039, 4846, 260} \[ \frac {\left (a+b x^n\right ) \tan ^{-1}\left (a+b x^n\right )}{b n}-\frac {\log \left (\left (a+b x^n\right )^2+1\right )}{2 b n} \]

Antiderivative was successfully verified.

[In]

Int[x^(-1 + n)*ArcTan[a + b*x^n],x]

[Out]

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

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 4846

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

Rule 5039

Int[((a_.) + ArcTan[(c_) + (d_.)*(x_)]*(b_.))^(p_.), x_Symbol] :> Dist[1/d, Subst[Int[(a + b*ArcTan[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^{-1+n} \tan ^{-1}\left (a+b x^n\right ) \, dx &=\frac {\operatorname {Subst}\left (\int \tan ^{-1}(a+b x) \, dx,x,x^n\right )}{n}\\ &=\frac {\operatorname {Subst}\left (\int \tan ^{-1}(x) \, dx,x,a+b x^n\right )}{b n}\\ &=\frac {\left (a+b x^n\right ) \tan ^{-1}\left (a+b x^n\right )}{b n}-\frac {\operatorname {Subst}\left (\int \frac {x}{1+x^2} \, dx,x,a+b x^n\right )}{b n}\\ &=\frac {\left (a+b x^n\right ) \tan ^{-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*}

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Mathematica [A]  time = 0.05, size = 40, normalized size = 0.89 \[ -\frac {\log \left (\left (a+b x^n\right )^2+1\right )-2 \left (a+b x^n\right ) \tan ^{-1}\left (a+b x^n\right )}{2 b n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.48, size = 58, normalized size = 1.29 \[ \frac {2 \, b x^{n} \arctan \left (b x^{n} + a\right ) + 2 \, a \arctan \left (b x^{n} + a\right ) - \log \left (b^{2} x^{2 \, n} + 2 \, a b x^{n} + a^{2} + 1\right )}{2 \, b n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*b*x^n*arctan(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)

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giac [A]  time = 0.12, size = 40, normalized size = 0.89 \[ \frac {2 \, {\left (b x^{n} + a\right )} \arctan \left (b x^{n} + a\right ) - \log \left ({\left (b x^{n} + a\right )}^{2} + 1\right )}{2 \, b n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.29, size = 140, normalized size = 3.11 \[ -\frac {i x^{n} \ln \left (1+i \left (a +b \,x^{n}\right )\right )}{2 n}+\frac {i x^{n} \ln \left (1-i \left (a +b \,x^{n}\right )\right )}{2 n}-\frac {\ln \left (\frac {i+a}{b}+x^{n}\right )}{2 b n}-\frac {\ln \left (x^{n}-\frac {i-a}{b}\right )}{2 b n}+\frac {i \ln \left (\frac {i+a}{b}+x^{n}\right ) a}{2 b n}-\frac {i \ln \left (x^{n}-\frac {i-a}{b}\right ) a}{2 b n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(n-1)*arctan(a+b*x^n),x)

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

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maxima [A]  time = 0.31, size = 40, normalized size = 0.89 \[ \frac {2 \, {\left (b x^{n} + a\right )} \arctan \left (b x^{n} + a\right ) - \log \left ({\left (b x^{n} + a\right )}^{2} + 1\right )}{2 \, b n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.63, size = 58, normalized size = 1.29 \[ \frac {x^n\,\mathrm {atan}\left (a+b\,x^n\right )}{n}-\frac {\ln \left (a^2+b^2\,x^{2\,n}+2\,a\,b\,x^n+1\right )-2\,a\,\mathrm {atan}\left (a+b\,x^n\right )}{2\,b\,n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(n - 1)*atan(a + b*x^n),x)

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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