3.1 \(\int x^3 \tan ^{-1}(a+b x^4) \, dx\)

Optimal. Leaf size=42 \[ \frac{\left (a+b x^4\right ) \tan ^{-1}\left (a+b x^4\right )}{4 b}-\frac{\log \left (\left (a+b x^4\right )^2+1\right )}{8 b} \]

[Out]

((a + b*x^4)*ArcTan[a + b*x^4])/(4*b) - Log[1 + (a + b*x^4)^2]/(8*b)

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Rubi [A]  time = 0.0408917, antiderivative size = 42, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {6715, 5039, 4846, 260} \[ \frac{\left (a+b x^4\right ) \tan ^{-1}\left (a+b x^4\right )}{4 b}-\frac{\log \left (\left (a+b x^4\right )^2+1\right )}{8 b} \]

Antiderivative was successfully verified.

[In]

Int[x^3*ArcTan[a + b*x^4],x]

[Out]

((a + b*x^4)*ArcTan[a + b*x^4])/(4*b) - Log[1 + (a + b*x^4)^2]/(8*b)

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^3 \tan ^{-1}\left (a+b x^4\right ) \, dx &=\frac{1}{4} \operatorname{Subst}\left (\int \tan ^{-1}(a+b x) \, dx,x,x^4\right )\\ &=\frac{\operatorname{Subst}\left (\int \tan ^{-1}(x) \, dx,x,a+b x^4\right )}{4 b}\\ &=\frac{\left (a+b x^4\right ) \tan ^{-1}\left (a+b x^4\right )}{4 b}-\frac{\operatorname{Subst}\left (\int \frac{x}{1+x^2} \, dx,x,a+b x^4\right )}{4 b}\\ &=\frac{\left (a+b x^4\right ) \tan ^{-1}\left (a+b x^4\right )}{4 b}-\frac{\log \left (1+\left (a+b x^4\right )^2\right )}{8 b}\\ \end{align*}

Mathematica [A]  time = 0.0147019, size = 37, normalized size = 0.88 \[ -\frac{\log \left (\left (a+b x^4\right )^2+1\right )-2 \left (a+b x^4\right ) \tan ^{-1}\left (a+b x^4\right )}{8 b} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*ArcTan[a + b*x^4],x]

[Out]

-(-2*(a + b*x^4)*ArcTan[a + b*x^4] + Log[1 + (a + b*x^4)^2])/(8*b)

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Maple [A]  time = 0.035, size = 46, normalized size = 1.1 \begin{align*}{\frac{\arctan \left ( b{x}^{4}+a \right ){x}^{4}}{4}}+{\frac{\arctan \left ( b{x}^{4}+a \right ) a}{4\,b}}-{\frac{\ln \left ( 1+ \left ( b{x}^{4}+a \right ) ^{2} \right ) }{8\,b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*arctan(b*x^4+a),x)

[Out]

1/4*arctan(b*x^4+a)*x^4+1/4/b*arctan(b*x^4+a)*a-1/8*ln(1+(b*x^4+a)^2)/b

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Maxima [A]  time = 0.991128, size = 50, normalized size = 1.19 \begin{align*} \frac{2 \,{\left (b x^{4} + a\right )} \arctan \left (b x^{4} + a\right ) - \log \left ({\left (b x^{4} + a\right )}^{2} + 1\right )}{8 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arctan(b*x^4+a),x, algorithm="maxima")

[Out]

1/8*(2*(b*x^4 + a)*arctan(b*x^4 + a) - log((b*x^4 + a)^2 + 1))/b

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Fricas [A]  time = 2.36511, size = 105, normalized size = 2.5 \begin{align*} \frac{2 \,{\left (b x^{4} + a\right )} \arctan \left (b x^{4} + a\right ) - \log \left (b^{2} x^{8} + 2 \, a b x^{4} + a^{2} + 1\right )}{8 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arctan(b*x^4+a),x, algorithm="fricas")

[Out]

1/8*(2*(b*x^4 + a)*arctan(b*x^4 + a) - log(b^2*x^8 + 2*a*b*x^4 + a^2 + 1))/b

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Sympy [A]  time = 6.02931, size = 60, normalized size = 1.43 \begin{align*} \begin{cases} \frac{a \operatorname{atan}{\left (a + b x^{4} \right )}}{4 b} + \frac{x^{4} \operatorname{atan}{\left (a + b x^{4} \right )}}{4} - \frac{\log{\left (a^{2} + 2 a b x^{4} + b^{2} x^{8} + 1 \right )}}{8 b} & \text{for}\: b \neq 0 \\\frac{x^{4} \operatorname{atan}{\left (a \right )}}{4} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*atan(b*x**4+a),x)

[Out]

Piecewise((a*atan(a + b*x**4)/(4*b) + x**4*atan(a + b*x**4)/4 - log(a**2 + 2*a*b*x**4 + b**2*x**8 + 1)/(8*b),
Ne(b, 0)), (x**4*atan(a)/4, True))

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Giac [A]  time = 1.13338, size = 50, normalized size = 1.19 \begin{align*} \frac{2 \,{\left (b x^{4} + a\right )} \arctan \left (b x^{4} + a\right ) - \log \left ({\left (b x^{4} + a\right )}^{2} + 1\right )}{8 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arctan(b*x^4+a),x, algorithm="giac")

[Out]

1/8*(2*(b*x^4 + a)*arctan(b*x^4 + a) - log((b*x^4 + a)^2 + 1))/b