3.75 \(\int x^3 \cot ^{-1}(a x^2) \, dx\)

Optimal. Leaf size=37 \[ -\frac {\tan ^{-1}\left (a x^2\right )}{4 a^2}+\frac {x^2}{4 a}+\frac {1}{4} x^4 \cot ^{-1}\left (a x^2\right ) \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 37, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {5034, 275, 321, 203} \[ -\frac {\tan ^{-1}\left (a x^2\right )}{4 a^2}+\frac {x^2}{4 a}+\frac {1}{4} x^4 \cot ^{-1}\left (a x^2\right ) \]

Antiderivative was successfully verified.

[In]

Int[x^3*ArcCot[a*x^2],x]

[Out]

x^2/(4*a) + (x^4*ArcCot[a*x^2])/4 - ArcTan[a*x^2]/(4*a^2)

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 275

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 5034

Int[((a_.) + ArcCot[(c_.)*(x_)^(n_)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcCot
[c*x^n]))/(d*(m + 1)), x] + Dist[(b*c*n)/(d*(m + 1)), Int[(x^(n - 1)*(d*x)^(m + 1))/(1 + c^2*x^(2*n)), x], x]
/; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1]

Rubi steps

\begin {align*} \int x^3 \cot ^{-1}\left (a x^2\right ) \, dx &=\frac {1}{4} x^4 \cot ^{-1}\left (a x^2\right )+\frac {1}{2} a \int \frac {x^5}{1+a^2 x^4} \, dx\\ &=\frac {1}{4} x^4 \cot ^{-1}\left (a x^2\right )+\frac {1}{4} a \operatorname {Subst}\left (\int \frac {x^2}{1+a^2 x^2} \, dx,x,x^2\right )\\ &=\frac {x^2}{4 a}+\frac {1}{4} x^4 \cot ^{-1}\left (a x^2\right )-\frac {\operatorname {Subst}\left (\int \frac {1}{1+a^2 x^2} \, dx,x,x^2\right )}{4 a}\\ &=\frac {x^2}{4 a}+\frac {1}{4} x^4 \cot ^{-1}\left (a x^2\right )-\frac {\tan ^{-1}\left (a x^2\right )}{4 a^2}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 37, normalized size = 1.00 \[ -\frac {\tan ^{-1}\left (a x^2\right )}{4 a^2}+\frac {x^2}{4 a}+\frac {1}{4} x^4 \cot ^{-1}\left (a x^2\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*ArcCot[a*x^2],x]

[Out]

x^2/(4*a) + (x^4*ArcCot[a*x^2])/4 - ArcTan[a*x^2]/(4*a^2)

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fricas [A]  time = 0.74, size = 27, normalized size = 0.73 \[ \frac {a x^{2} + {\left (a^{2} x^{4} + 1\right )} \operatorname {arccot}\left (a x^{2}\right )}{4 \, a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arccot(a*x^2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.11, size = 38, normalized size = 1.03 \[ \frac {1}{4} \, {\left (\frac {x^{4} \arctan \left (\frac {1}{a x^{2}}\right )}{a} + \frac {x^{2}}{a^{2}} + \frac {\arctan \left (\frac {1}{a x^{2}}\right )}{a^{3}}\right )} a \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arccot(a*x^2),x, algorithm="giac")

[Out]

1/4*(x^4*arctan(1/(a*x^2))/a + x^2/a^2 + arctan(1/(a*x^2))/a^3)*a

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maple [A]  time = 0.04, size = 32, normalized size = 0.86 \[ \frac {x^{2}}{4 a}+\frac {x^{4} \mathrm {arccot}\left (a \,x^{2}\right )}{4}-\frac {\arctan \left (a \,x^{2}\right )}{4 a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*arccot(a*x^2),x)

[Out]

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

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maxima [A]  time = 0.42, size = 34, normalized size = 0.92 \[ \frac {1}{4} \, x^{4} \operatorname {arccot}\left (a x^{2}\right ) + \frac {1}{4} \, a {\left (\frac {x^{2}}{a^{2}} - \frac {\arctan \left (a x^{2}\right )}{a^{3}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arccot(a*x^2),x, algorithm="maxima")

[Out]

1/4*x^4*arccot(a*x^2) + 1/4*a*(x^2/a^2 - arctan(a*x^2)/a^3)

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mupad [B]  time = 0.61, size = 31, normalized size = 0.84 \[ \frac {x^4\,\mathrm {acot}\left (a\,x^2\right )}{4}-\frac {\mathrm {atan}\left (a\,x^2\right )}{4\,a^2}+\frac {x^2}{4\,a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*acot(a*x^2),x)

[Out]

(x^4*acot(a*x^2))/4 - atan(a*x^2)/(4*a^2) + x^2/(4*a)

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sympy [A]  time = 1.03, size = 36, normalized size = 0.97 \[ \begin {cases} \frac {x^{4} \operatorname {acot}{\left (a x^{2} \right )}}{4} + \frac {x^{2}}{4 a} + \frac {\operatorname {acot}{\left (a x^{2} \right )}}{4 a^{2}} & \text {for}\: a \neq 0 \\\frac {\pi x^{4}}{8} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*acot(a*x**2),x)

[Out]

Piecewise((x**4*acot(a*x**2)/4 + x**2/(4*a) + acot(a*x**2)/(4*a**2), Ne(a, 0)), (pi*x**4/8, True))

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