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

Optimal. Leaf size=39 \[ -\frac{\tan ^{-1}(a+b x)}{2 b}+\frac{(a+b x)^2 \cot ^{-1}(a+b x)}{2 b}+\frac{x}{2} \]

[Out]

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

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Rubi [A]  time = 0.0209446, antiderivative size = 39, normalized size of antiderivative = 1., 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 = {5044, 4853, 321, 203} \[ -\frac{\tan ^{-1}(a+b x)}{2 b}+\frac{(a+b x)^2 \cot ^{-1}(a+b x)}{2 b}+\frac{x}{2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x)*ArcCot[a + b*x],x]

[Out]

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

Rule 5044

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

Rule 4853

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

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 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])

Rubi steps

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

Mathematica [C]  time = 0.0548926, size = 141, normalized size = 3.62 \[ \frac{a \left (\log \left (a^2+2 a b x+b^2 x^2+1\right )-2 a \tan ^{-1}(a+b x)\right )}{2 b}+\frac{1}{2} b \left (-\frac{i (-a+i)^2 \log (-a-b x+i)}{2 b^2}+\frac{i (a+i)^2 \log (a+b x+i)}{2 b^2}+\frac{x}{b}\right )+\frac{1}{2} b \left (\frac{a+b x}{b}-\frac{a}{b}\right )^2 \cot ^{-1}(a+b x)+a x \cot ^{-1}(a+b x) \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)*ArcCot[a + b*x],x]

[Out]

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

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Maple [A]  time = 0.043, size = 57, normalized size = 1.5 \begin{align*}{\frac{b{\rm arccot} \left (bx+a\right ){x}^{2}}{2}}+{\rm arccot} \left (bx+a\right )xa+{\frac{{\rm arccot} \left (bx+a\right ){a}^{2}}{2\,b}}+{\frac{x}{2}}+{\frac{a}{2\,b}}-{\frac{\arctan \left ( bx+a \right ) }{2\,b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.45168, size = 70, normalized size = 1.79 \begin{align*} \frac{1}{2} \, b{\left (\frac{x}{b} - \frac{{\left (a^{2} + 1\right )} \arctan \left (\frac{b^{2} x + a b}{b}\right )}{b^{2}}\right )} + \frac{1}{2} \,{\left (b x^{2} + 2 \, a x\right )} \operatorname{arccot}\left (b x + a\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.10774, size = 82, normalized size = 2.1 \begin{align*} \frac{b x +{\left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )} \operatorname{arccot}\left (b x + a\right )}{2 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 1.07128, size = 56, normalized size = 1.44 \begin{align*} \begin{cases} \frac{a^{2} \operatorname{acot}{\left (a + b x \right )}}{2 b} + a x \operatorname{acot}{\left (a + b x \right )} + \frac{b x^{2} \operatorname{acot}{\left (a + b x \right )}}{2} + \frac{x}{2} + \frac{\operatorname{acot}{\left (a + b x \right )}}{2 b} & \text{for}\: b \neq 0 \\a x \operatorname{acot}{\left (a \right )} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.11124, size = 54, normalized size = 1.38 \begin{align*} \frac{1}{2} \,{\left (b x^{2} + 2 \, a x\right )} \arctan \left (\frac{1}{b x + a}\right ) + \frac{1}{2} \, x - \frac{{\left (a^{2} + 1\right )} \arctan \left (b x + a\right )}{2 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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