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

Optimal. Leaf size=60 \[ -\frac{\left (1-a^2\right ) \tan ^{-1}(a+b x)}{2 b^2}-\frac{a \log \left ((a+b x)^2+1\right )}{2 b^2}+\frac{1}{2} x^2 \cot ^{-1}(a+b x)+\frac{x}{2 b} \]

[Out]

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

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Rubi [A]  time = 0.0553805, antiderivative size = 60, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.75, Rules used = {5048, 4863, 702, 635, 203, 260} \[ -\frac{\left (1-a^2\right ) \tan ^{-1}(a+b x)}{2 b^2}-\frac{a \log \left ((a+b x)^2+1\right )}{2 b^2}+\frac{1}{2} x^2 \cot ^{-1}(a+b x)+\frac{x}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[x*ArcCot[a + b*x],x]

[Out]

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

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

Mathematica [C]  time = 0.0328707, size = 90, normalized size = 1.5 \[ \frac{i a^2 \log (a+b x+i)+2 b^2 x^2 \cot ^{-1}(a+b x)-2 a \log (a+b x+i)-i (a-i)^2 \log (-a-b x+i)-i \log (a+b x+i)+2 b x}{4 b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x*ArcCot[a + b*x],x]

[Out]

(2*b*x + 2*b^2*x^2*ArcCot[a + b*x] - I*(-I + a)^2*Log[I - a - b*x] - I*Log[I + a + b*x] - 2*a*Log[I + a + b*x]
+ I*a^2*Log[I + a + b*x])/(4*b^2)

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Maple [A]  time = 0.044, size = 66, normalized size = 1.1 \begin{align*}{\frac{{x}^{2}{\rm arccot} \left (bx+a\right )}{2}}-{\frac{{\rm arccot} \left (bx+a\right ){a}^{2}}{2\,{b}^{2}}}+{\frac{x}{2\,b}}+{\frac{a}{2\,{b}^{2}}}-{\frac{a\ln \left ( 1+ \left ( bx+a \right ) ^{2} \right ) }{2\,{b}^{2}}}-{\frac{\arctan \left ( bx+a \right ) }{2\,{b}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.47358, size = 92, normalized size = 1.53 \begin{align*} \frac{1}{2} \, x^{2} \operatorname{arccot}\left (b x + a\right ) + \frac{1}{2} \, b{\left (\frac{x}{b^{2}} + \frac{{\left (a^{2} - 1\right )} \arctan \left (\frac{b^{2} x + a b}{b}\right )}{b^{3}} - \frac{a \log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}{b^{3}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.2728, size = 143, normalized size = 2.38 \begin{align*} \frac{b^{2} x^{2} \operatorname{arccot}\left (b x + a\right ) + b x +{\left (a^{2} - 1\right )} \arctan \left (b x + a\right ) - a \log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}{2 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.833307, size = 78, normalized size = 1.3 \begin{align*} \begin{cases} - \frac{a^{2} \operatorname{acot}{\left (a + b x \right )}}{2 b^{2}} - \frac{a \log{\left (a^{2} + 2 a b x + b^{2} x^{2} + 1 \right )}}{2 b^{2}} + \frac{x^{2} \operatorname{acot}{\left (a + b x \right )}}{2} + \frac{x}{2 b} + \frac{\operatorname{acot}{\left (a + b x \right )}}{2 b^{2}} & \text{for}\: b \neq 0 \\\frac{x^{2} \operatorname{acot}{\left (a \right )}}{2} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.09278, size = 84, normalized size = 1.4 \begin{align*} \frac{1}{2} \, x^{2} \arctan \left (\frac{1}{b x + a}\right ) + \frac{1}{2} \, b{\left (\frac{x}{b^{2}} + \frac{{\left (a^{2} - 1\right )} \arctan \left (b x + a\right )}{b^{3}} - \frac{a \log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}{b^{3}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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