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

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

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 42, normalized size of antiderivative = 1.00, 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, 5040, 4847, 260} \[ \frac {\log \left (\left (a+b x^4\right )^2+1\right )}{8 b}+\frac {\left (a+b x^4\right ) \cot ^{-1}\left (a+b x^4\right )}{4 b} \]

Antiderivative was successfully verified.

[In]

Int[x^3*ArcCot[a + b*x^4],x]

[Out]

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

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 4847

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

Rule 5040

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 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^3 \cot ^{-1}\left (a+b x^4\right ) \, dx &=\frac {1}{4} \operatorname {Subst}\left (\int \cot ^{-1}(a+b x) \, dx,x,x^4\right )\\ &=\frac {\operatorname {Subst}\left (\int \cot ^{-1}(x) \, dx,x,a+b x^4\right )}{4 b}\\ &=\frac {\left (a+b x^4\right ) \cot ^{-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 ) \cot ^{-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*}

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Mathematica [A]  time = 0.02, size = 37, normalized size = 0.88 \[ \frac {\log \left (\left (a+b x^4\right )^2+1\right )+2 \left (a+b x^4\right ) \cot ^{-1}\left (a+b x^4\right )}{8 b} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*ArcCot[a + b*x^4],x]

[Out]

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

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fricas [A]  time = 0.68, size = 51, normalized size = 1.21 \[ \frac {2 \, b x^{4} \operatorname {arccot}\left (b x^{4} + a\right ) - 2 \, a \arctan \left (b x^{4} + a\right ) + \log \left (b^{2} x^{8} + 2 \, a b x^{4} + a^{2} + 1\right )}{8 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.23, size = 127, normalized size = 3.02 \[ -\frac {\arctan \left (\frac {1}{b x^{4} + a}\right ) \tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x^{4} + a}\right )\right )^{2} + \log \left (\frac {16 \, \tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x^{4} + a}\right )\right )^{2}}{\tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x^{4} + a}\right )\right )^{4} + 2 \, \tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x^{4} + a}\right )\right )^{2} + 1}\right ) \tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x^{4} + a}\right )\right ) - \arctan \left (\frac {1}{b x^{4} + a}\right )}{8 \, b \tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x^{4} + a}\right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(arctan(1/(b*x^4 + a))*tan(1/2*arctan(1/(b*x^4 + a)))^2 + log(16*tan(1/2*arctan(1/(b*x^4 + a)))^2/(tan(1/
2*arctan(1/(b*x^4 + a)))^4 + 2*tan(1/2*arctan(1/(b*x^4 + a)))^2 + 1))*tan(1/2*arctan(1/(b*x^4 + a))) - arctan(
1/(b*x^4 + a)))/(b*tan(1/2*arctan(1/(b*x^4 + a))))

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maple [A]  time = 0.04, size = 46, normalized size = 1.10 \[ \frac {\mathrm {arccot}\left (b \,x^{4}+a \right ) x^{4}}{4}+\frac {\mathrm {arccot}\left (b \,x^{4}+a \right ) a}{4 b}+\frac {\ln \left (1+\left (b \,x^{4}+a \right )^{2}\right )}{8 b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.32, size = 35, normalized size = 0.83 \[ \frac {2 \, {\left (b x^{4} + a\right )} \operatorname {arccot}\left (b x^{4} + a\right ) + \log \left ({\left (b x^{4} + a\right )}^{2} + 1\right )}{8 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.76, size = 230, normalized size = 5.48 \[ \frac {\ln \left (a^2+2\,a\,b\,x^4+b^2\,x^8+1\right )}{8\,b}+\frac {x^4\,\mathrm {acot}\left (b\,x^4+a\right )}{4}-\frac {a\,\mathrm {atan}\left (\frac {a}{a^6+3\,a^4+3\,a^2+1}+\frac {3\,a^3}{a^6+3\,a^4+3\,a^2+1}+\frac {3\,a^5}{a^6+3\,a^4+3\,a^2+1}+\frac {a^7}{a^6+3\,a^4+3\,a^2+1}+\frac {b\,x^4}{a^6+3\,a^4+3\,a^2+1}+\frac {3\,a^2\,b\,x^4}{a^6+3\,a^4+3\,a^2+1}+\frac {3\,a^4\,b\,x^4}{a^6+3\,a^4+3\,a^2+1}+\frac {a^6\,b\,x^4}{a^6+3\,a^4+3\,a^2+1}\right )}{4\,b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*acot(a + b*x^4),x)

[Out]

log(a^2 + b^2*x^8 + 2*a*b*x^4 + 1)/(8*b) + (x^4*acot(a + b*x^4))/4 - (a*atan(a/(3*a^2 + 3*a^4 + a^6 + 1) + (3*
a^3)/(3*a^2 + 3*a^4 + a^6 + 1) + (3*a^5)/(3*a^2 + 3*a^4 + a^6 + 1) + a^7/(3*a^2 + 3*a^4 + a^6 + 1) + (b*x^4)/(
3*a^2 + 3*a^4 + a^6 + 1) + (3*a^2*b*x^4)/(3*a^2 + 3*a^4 + a^6 + 1) + (3*a^4*b*x^4)/(3*a^2 + 3*a^4 + a^6 + 1) +
(a^6*b*x^4)/(3*a^2 + 3*a^4 + a^6 + 1)))/(4*b)

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sympy [A]  time = 3.34, size = 60, normalized size = 1.43 \[ \begin {cases} \frac {a \operatorname {acot}{\left (a + b x^{4} \right )}}{4 b} + \frac {x^{4} \operatorname {acot}{\left (a + b x^{4} \right )}}{4} + \frac {\log {\left (a^{2} + 2 a b x^{4} + b^{2} x^{8} + 1 \right )}}{8 b} & \text {for}\: b \neq 0 \\\frac {x^{4} \operatorname {acot}{\relax (a )}}{4} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a*acot(a + b*x**4)/(4*b) + x**4*acot(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*acot(a)/4, True))

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