3.100 \(\int x^2 \cot ^{-1}(a+b x) \, dx\)

Optimal. Leaf size=80 \[ -\frac{\left (1-3 a^2\right ) \log \left ((a+b x)^2+1\right )}{6 b^3}+\frac{a \left (3-a^2\right ) \tan ^{-1}(a+b x)}{3 b^3}-\frac{a x}{b^2}+\frac{(a+b x)^2}{6 b^3}+\frac{1}{3} x^3 \cot ^{-1}(a+b x) \]

[Out]

-((a*x)/b^2) + (a + b*x)^2/(6*b^3) + (x^3*ArcCot[a + b*x])/3 + (a*(3 - a^2)*ArcTan[a + b*x])/(3*b^3) - ((1 - 3
*a^2)*Log[1 + (a + b*x)^2])/(6*b^3)

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Rubi [A]  time = 0.0782526, antiderivative size = 80, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.6, Rules used = {5048, 4863, 702, 635, 203, 260} \[ -\frac{\left (1-3 a^2\right ) \log \left ((a+b x)^2+1\right )}{6 b^3}+\frac{a \left (3-a^2\right ) \tan ^{-1}(a+b x)}{3 b^3}-\frac{a x}{b^2}+\frac{(a+b x)^2}{6 b^3}+\frac{1}{3} x^3 \cot ^{-1}(a+b x) \]

Antiderivative was successfully verified.

[In]

Int[x^2*ArcCot[a + b*x],x]

[Out]

-((a*x)/b^2) + (a + b*x)^2/(6*b^3) + (x^3*ArcCot[a + b*x])/3 + (a*(3 - a^2)*ArcTan[a + b*x])/(3*b^3) - ((1 - 3
*a^2)*Log[1 + (a + b*x)^2])/(6*b^3)

Rule 5048

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

Rule 4863

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

Rule 702

Int[((d_) + (e_.)*(x_))^(m_)/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[PolynomialDivide[(d + e*x)^m, a + c*x^2,
x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && IGtQ[m, 1] && (NeQ[d, 0] || GtQ[m, 2])

Rule 635

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Dist[d, Int[1/(a + c*x^2), x], x] + Dist[e, Int[x/
(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] &&  !NiceSqrtQ[-(a*c)]

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 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^2 \cot ^{-1}(a+b x) \, dx &=\frac{\operatorname{Subst}\left (\int \left (-\frac{a}{b}+\frac{x}{b}\right )^2 \cot ^{-1}(x) \, dx,x,a+b x\right )}{b}\\ &=\frac{1}{3} x^3 \cot ^{-1}(a+b x)+\frac{1}{3} \operatorname{Subst}\left (\int \frac{\left (-\frac{a}{b}+\frac{x}{b}\right )^3}{1+x^2} \, dx,x,a+b x\right )\\ &=\frac{1}{3} x^3 \cot ^{-1}(a+b x)+\frac{1}{3} \operatorname{Subst}\left (\int \left (-\frac{3 a}{b^3}+\frac{x}{b^3}+\frac{a \left (3-a^2\right )-\left (1-3 a^2\right ) x}{b^3 \left (1+x^2\right )}\right ) \, dx,x,a+b x\right )\\ &=-\frac{a x}{b^2}+\frac{(a+b x)^2}{6 b^3}+\frac{1}{3} x^3 \cot ^{-1}(a+b x)+\frac{\operatorname{Subst}\left (\int \frac{a \left (3-a^2\right )-\left (1-3 a^2\right ) x}{1+x^2} \, dx,x,a+b x\right )}{3 b^3}\\ &=-\frac{a x}{b^2}+\frac{(a+b x)^2}{6 b^3}+\frac{1}{3} x^3 \cot ^{-1}(a+b x)-\frac{\left (1-3 a^2\right ) \operatorname{Subst}\left (\int \frac{x}{1+x^2} \, dx,x,a+b x\right )}{3 b^3}+\frac{\left (a \left (3-a^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,a+b x\right )}{3 b^3}\\ &=-\frac{a x}{b^2}+\frac{(a+b x)^2}{6 b^3}+\frac{1}{3} x^3 \cot ^{-1}(a+b x)+\frac{a \left (3-a^2\right ) \tan ^{-1}(a+b x)}{3 b^3}-\frac{\left (1-3 a^2\right ) \log \left (1+(a+b x)^2\right )}{6 b^3}\\ \end{align*}

Mathematica [C]  time = 0.0418469, size = 114, normalized size = 1.42 \[ \frac{\frac{1}{3} b \left (\frac{a+b x}{b}-\frac{a}{b}\right )^3 \cot ^{-1}(a+b x)+\frac{1}{3} b \left (\frac{(a+b x)^2}{2 b^3}-\frac{3 a x}{b^2}-\frac{(1-i a)^3 \log (a+b x+i)}{2 b^3}-\frac{(1+i a)^3 \log (-a-b x+i)}{2 b^3}\right )}{b} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*ArcCot[a + b*x],x]

[Out]

((b*(-(a/b) + (a + b*x)/b)^3*ArcCot[a + b*x])/3 + (b*((-3*a*x)/b^2 + (a + b*x)^2/(2*b^3) - ((1 + I*a)^3*Log[I
- a - b*x])/(2*b^3) - ((1 - I*a)^3*Log[I + a + b*x])/(2*b^3)))/3)/b

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Maple [A]  time = 0.041, size = 94, normalized size = 1.2 \begin{align*}{\frac{{x}^{3}{\rm arccot} \left (bx+a\right )}{3}}+{\frac{{x}^{2}}{6\,b}}-{\frac{2\,ax}{3\,{b}^{2}}}-{\frac{5\,{a}^{2}}{6\,{b}^{3}}}+{\frac{\ln \left ( 1+ \left ( bx+a \right ) ^{2} \right ){a}^{2}}{2\,{b}^{3}}}-{\frac{\ln \left ( 1+ \left ( bx+a \right ) ^{2} \right ) }{6\,{b}^{3}}}-{\frac{\arctan \left ( bx+a \right ){a}^{3}}{3\,{b}^{3}}}+{\frac{\arctan \left ( bx+a \right ) a}{{b}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*arccot(b*x+a),x)

[Out]

1/3*x^3*arccot(b*x+a)+1/6/b*x^2-2/3*a*x/b^2-5/6/b^3*a^2+1/2/b^3*ln(1+(b*x+a)^2)*a^2-1/6/b^3*ln(1+(b*x+a)^2)-1/
3/b^3*arctan(b*x+a)*a^3+1/b^3*arctan(b*x+a)*a

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Maxima [A]  time = 1.47509, size = 115, normalized size = 1.44 \begin{align*} \frac{1}{3} \, x^{3} \operatorname{arccot}\left (b x + a\right ) + \frac{1}{6} \, b{\left (\frac{b x^{2} - 4 \, a x}{b^{3}} - \frac{2 \,{\left (a^{3} - 3 \, a\right )} \arctan \left (\frac{b^{2} x + a b}{b}\right )}{b^{4}} + \frac{{\left (3 \, a^{2} - 1\right )} \log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}{b^{4}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*x^3*arccot(b*x + a) + 1/6*b*((b*x^2 - 4*a*x)/b^3 - 2*(a^3 - 3*a)*arctan((b^2*x + a*b)/b)/b^4 + (3*a^2 - 1)
*log(b^2*x^2 + 2*a*b*x + a^2 + 1)/b^4)

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Fricas [A]  time = 2.20814, size = 184, normalized size = 2.3 \begin{align*} \frac{2 \, b^{3} x^{3} \operatorname{arccot}\left (b x + a\right ) + b^{2} x^{2} - 4 \, a b x - 2 \,{\left (a^{3} - 3 \, a\right )} \arctan \left (b x + a\right ) +{\left (3 \, a^{2} - 1\right )} \log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}{6 \, b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(2*b^3*x^3*arccot(b*x + a) + b^2*x^2 - 4*a*b*x - 2*(a^3 - 3*a)*arctan(b*x + a) + (3*a^2 - 1)*log(b^2*x^2 +
2*a*b*x + a^2 + 1))/b^3

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Sympy [A]  time = 1.30885, size = 117, normalized size = 1.46 \begin{align*} \begin{cases} \frac{a^{3} \operatorname{acot}{\left (a + b x \right )}}{3 b^{3}} + \frac{a^{2} \log{\left (a^{2} + 2 a b x + b^{2} x^{2} + 1 \right )}}{2 b^{3}} - \frac{2 a x}{3 b^{2}} - \frac{a \operatorname{acot}{\left (a + b x \right )}}{b^{3}} + \frac{x^{3} \operatorname{acot}{\left (a + b x \right )}}{3} + \frac{x^{2}}{6 b} - \frac{\log{\left (a^{2} + 2 a b x + b^{2} x^{2} + 1 \right )}}{6 b^{3}} & \text{for}\: b \neq 0 \\\frac{x^{3} \operatorname{acot}{\left (a \right )}}{3} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a**3*acot(a + b*x)/(3*b**3) + a**2*log(a**2 + 2*a*b*x + b**2*x**2 + 1)/(2*b**3) - 2*a*x/(3*b**2) -
a*acot(a + b*x)/b**3 + x**3*acot(a + b*x)/3 + x**2/(6*b) - log(a**2 + 2*a*b*x + b**2*x**2 + 1)/(6*b**3), Ne(b,
0)), (x**3*acot(a)/3, True))

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Giac [A]  time = 1.12045, size = 113, normalized size = 1.41 \begin{align*} \frac{1}{3} \, x^{3} \arctan \left (\frac{1}{b x + a}\right ) - \frac{1}{6} \, b{\left (\frac{2 \,{\left (a^{3} - 3 \, a\right )} \arctan \left (b x + a\right )}{b^{4}} - \frac{{\left (3 \, a^{2} - 1\right )} \log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}{b^{4}} - \frac{b^{2} x^{2} - 4 \, a b x}{b^{4}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*x^3*arctan(1/(b*x + a)) - 1/6*b*(2*(a^3 - 3*a)*arctan(b*x + a)/b^4 - (3*a^2 - 1)*log(b^2*x^2 + 2*a*b*x + a
^2 + 1)/b^4 - (b^2*x^2 - 4*a*b*x)/b^4)