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) - Log[1 + (a + b*x^n)^2]/(2*b*n)

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Rubi [A]  time = 0.0421264, antiderivative size = 45, normalized size of antiderivative = 1., 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 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]

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 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 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]

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*}

Mathematica [A]  time = 0.0361654, 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]

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

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Maple [C]  time = 0.115, size = 140, normalized size = 3.1 \begin{align*}{\frac{-{\frac{i}{2}}{x}^{n}\ln \left ( 1+i \left ( a+b{x}^{n} \right ) \right ) }{n}}+{\frac{{\frac{i}{2}}{x}^{n}\ln \left ( 1-i \left ( a+b{x}^{n} \right ) \right ) }{n}}-{\frac{1}{2\,bn}\ln \left ({\frac{i+a}{b}}+{x}^{n} \right ) }-{\frac{1}{2\,bn}\ln \left ({x}^{n}-{\frac{i-a}{b}} \right ) }+{\frac{{\frac{i}{2}}a}{bn}\ln \left ({\frac{i+a}{b}}+{x}^{n} \right ) }-{\frac{{\frac{i}{2}}a}{bn}\ln \left ({x}^{n}-{\frac{i-a}{b}} \right ) } \end{align*}

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 = 0.974863, size = 54, normalized size = 1.2 \begin{align*} \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} \end{align*}

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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Fricas [A]  time = 2.3383, size = 140, normalized size = 3.11 \begin{align*} \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} \end{align*}

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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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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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Giac [A]  time = 1.09973, size = 54, normalized size = 1.2 \begin{align*} \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} \end{align*}

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)