3.1.16 \(\int x \cot ^{-1}(a x)^2 \, dx\) [16]

Optimal. Leaf size=53 \[ \frac {x \cot ^{-1}(a x)}{a}+\frac {\cot ^{-1}(a x)^2}{2 a^2}+\frac {1}{2} x^2 \cot ^{-1}(a x)^2+\frac {\log \left (1+a^2 x^2\right )}{2 a^2} \]

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.625, Rules used = {4947, 5037, 4931, 266, 5005} \begin {gather*} \frac {\log \left (a^2 x^2+1\right )}{2 a^2}+\frac {\cot ^{-1}(a x)^2}{2 a^2}+\frac {1}{2} x^2 \cot ^{-1}(a x)^2+\frac {x \cot ^{-1}(a x)}{a} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 266

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 4931

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

Rule 4947

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

Rule 5005

Int[((a_.) + ArcCot[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[-(a + b*ArcCot[c*x])^(p
+ 1)/(b*c*d*(p + 1)), x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[e, c^2*d] && NeQ[p, -1]

Rule 5037

Int[(((a_.) + ArcCot[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Dist[f^2/
e, Int[(f*x)^(m - 2)*(a + b*ArcCot[c*x])^p, x], x] - Dist[d*(f^2/e), Int[(f*x)^(m - 2)*((a + b*ArcCot[c*x])^p/
(d + e*x^2)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && GtQ[p, 0] && GtQ[m, 1]

Rubi steps

\begin {align*} \int x \cot ^{-1}(a x)^2 \, dx &=\frac {1}{2} x^2 \cot ^{-1}(a x)^2+a \int \frac {x^2 \cot ^{-1}(a x)}{1+a^2 x^2} \, dx\\ &=\frac {1}{2} x^2 \cot ^{-1}(a x)^2+\frac {\int \cot ^{-1}(a x) \, dx}{a}-\frac {\int \frac {\cot ^{-1}(a x)}{1+a^2 x^2} \, dx}{a}\\ &=\frac {x \cot ^{-1}(a x)}{a}+\frac {\cot ^{-1}(a x)^2}{2 a^2}+\frac {1}{2} x^2 \cot ^{-1}(a x)^2+\int \frac {x}{1+a^2 x^2} \, dx\\ &=\frac {x \cot ^{-1}(a x)}{a}+\frac {\cot ^{-1}(a x)^2}{2 a^2}+\frac {1}{2} x^2 \cot ^{-1}(a x)^2+\frac {\log \left (1+a^2 x^2\right )}{2 a^2}\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 42, normalized size = 0.79 \begin {gather*} \frac {2 a x \cot ^{-1}(a x)+\left (1+a^2 x^2\right ) \cot ^{-1}(a x)^2+\log \left (1+a^2 x^2\right )}{2 a^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.09, size = 57, normalized size = 1.08

method	result	size
derivativedivides	\(\frac {\frac {a^{2} x^{2} \mathrm {arccot}\left (a x \right )^{2}}{2}-\mathrm {arccot}\left (a x \right ) \arctan \left (a x \right )+a x \,\mathrm {arccot}\left (a x \right )+\frac {\ln \left (a^{2} x^{2}+1\right )}{2}-\frac {\arctan \left (a x \right )^{2}}{2}}{a^{2}}\)	\(57\)
default	\(\frac {\frac {a^{2} x^{2} \mathrm {arccot}\left (a x \right )^{2}}{2}-\mathrm {arccot}\left (a x \right ) \arctan \left (a x \right )+a x \,\mathrm {arccot}\left (a x \right )+\frac {\ln \left (a^{2} x^{2}+1\right )}{2}-\frac {\arctan \left (a x \right )^{2}}{2}}{a^{2}}\)	\(57\)
risch	\(-\frac {\left (a^{2} x^{2}+1\right ) \ln \left (i a x +1\right )^{2}}{8 a^{2}}+\frac {\left (i \pi \,a^{2} x^{2}+x^{2} \ln \left (-i a x +1\right ) a^{2}+2 i a x +\ln \left (-i a x +1\right )\right ) \ln \left (i a x +1\right )}{4 a^{2}}-\frac {i \pi \,x^{2} \ln \left (-i a x +1\right )}{4}+\frac {\pi ^{2} x^{2}}{8}-\frac {x^{2} \ln \left (-i a x +1\right )^{2}}{8}-\frac {i x \ln \left (-i a x +1\right )}{2 a}+\frac {\pi x}{2 a}-\frac {\pi \arctan \left (a x \right )}{2 a^{2}}-\frac {\ln \left (-i a x +1\right )^{2}}{8 a^{2}}+\frac {\ln \left (a^{2} x^{2}+1\right )}{2 a^{2}}\)	\(178\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*arccot(a*x)^2,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.49, size = 57, normalized size = 1.08 \begin {gather*} \frac {1}{2} \, x^{2} \operatorname {arccot}\left (a x\right )^{2} + a {\left (\frac {x}{a^{2}} - \frac {\arctan \left (a x\right )}{a^{3}}\right )} \operatorname {arccot}\left (a x\right ) - \frac {\arctan \left (a x\right )^{2} - \log \left (a^{2} x^{2} + 1\right )}{2 \, a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 2.23, size = 40, normalized size = 0.75 \begin {gather*} \frac {2 \, a x \operatorname {arccot}\left (a x\right ) + {\left (a^{2} x^{2} + 1\right )} \operatorname {arccot}\left (a x\right )^{2} + \log \left (a^{2} x^{2} + 1\right )}{2 \, a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 0.15, size = 54, normalized size = 1.02 \begin {gather*} \begin {cases} \frac {x^{2} \operatorname {acot}^{2}{\left (a x \right )}}{2} + \frac {x \operatorname {acot}{\left (a x \right )}}{a} + \frac {\log {\left (a^{2} x^{2} + 1 \right )}}{2 a^{2}} + \frac {\operatorname {acot}^{2}{\left (a x \right )}}{2 a^{2}} & \text {for}\: a \neq 0 \\\frac {\pi ^{2} x^{2}}{8} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((x**2*acot(a*x)**2/2 + x*acot(a*x)/a + log(a**2*x**2 + 1)/(2*a**2) + acot(a*x)**2/(2*a**2), Ne(a, 0)
), (pi**2*x**2/8, True))

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x*arccot(a*x)^2, x)

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Mupad [B]
time = 0.14, size = 44, normalized size = 0.83 \begin {gather*} \frac {x^2\,{\mathrm {acot}\left (a\,x\right )}^2}{2}+\frac {\frac {{\mathrm {acot}\left (a\,x\right )}^2}{2}+a\,x\,\mathrm {acot}\left (a\,x\right )+\frac {\ln \left (a^2\,x^2+1\right )}{2}}{a^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(x^2*acot(a*x)^2)/2 + (log(a^2*x^2 + 1)/2 + acot(a*x)^2/2 + a*x*acot(a*x))/a^2

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