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

Optimal. Leaf size=52 \[ \frac{(a+b x)^2}{6 b}-\frac{\log \left ((a+b x)^2+1\right )}{6 b}+\frac{(a+b x)^3 \cot ^{-1}(a+b x)}{3 b} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0386925, antiderivative size = 52, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {5044, 4853, 266, 43} \[ \frac{(a+b x)^2}{6 b}-\frac{\log \left ((a+b x)^2+1\right )}{6 b}+\frac{(a+b x)^3 \cot ^{-1}(a+b x)}{3 b} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^2*ArcCot[a + b*x],x]

[Out]

(a + b*x)^2/(6*b) + ((a + b*x)^3*ArcCot[a + b*x])/(3*b) - Log[1 + (a + b*x)^2]/(6*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 266

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

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

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

Mathematica [A]  time = 0.0136469, size = 42, normalized size = 0.81 \[ \frac{(a+b x)^2-\log \left ((a+b x)^2+1\right )+2 (a+b x)^3 \cot ^{-1}(a+b x)}{6 b} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)^2*ArcCot[a + b*x],x]

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.042, size = 86, normalized size = 1.7 \begin{align*}{\frac{{b}^{2}{\rm arccot} \left (bx+a\right ){x}^{3}}{3}}+b{\rm arccot} \left (bx+a\right ){x}^{2}a+{\rm arccot} \left (bx+a\right )x{a}^{2}+{\frac{{\rm arccot} \left (bx+a\right ){a}^{3}}{3\,b}}+{\frac{b{x}^{2}}{6}}+{\frac{ax}{3}}+{\frac{{a}^{2}}{6\,b}}-{\frac{\ln \left ( 1+ \left ( bx+a \right ) ^{2} \right ) }{6\,b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [B]  time = 1.47172, size = 126, normalized size = 2.42 \begin{align*} -\frac{1}{6} \,{\left (\frac{2 \, a^{3} \arctan \left (\frac{b^{2} x + a b}{b}\right )}{b^{2}} - \frac{b x^{2} + 2 \, a x}{b} + \frac{\log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}{b^{2}}\right )} b + \frac{1}{3} \,{\left (b^{2} x^{3} + 3 \, a b x^{2} + 3 \, a^{2} 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)^2*arccot(b*x+a),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 2.1865, size = 192, normalized size = 3.69 \begin{align*} \frac{b^{2} x^{2} - 2 \, a^{3} \arctan \left (b x + a\right ) + 2 \, a b x + 2 \,{\left (b^{3} x^{3} + 3 \, a b^{2} x^{2} + 3 \, a^{2} b x\right )} \operatorname{arccot}\left (b x + a\right ) - \log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}{6 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Sympy [A]  time = 1.67135, size = 100, normalized size = 1.92 \begin{align*} \begin{cases} \frac{a^{3} \operatorname{acot}{\left (a + b x \right )}}{3 b} + a^{2} x \operatorname{acot}{\left (a + b x \right )} + a b x^{2} \operatorname{acot}{\left (a + b x \right )} + \frac{a x}{3} + \frac{b^{2} x^{3} \operatorname{acot}{\left (a + b x \right )}}{3} + \frac{b x^{2}}{6} - \frac{\log{\left (\frac{a^{2}}{b^{2}} + \frac{2 a x}{b} + x^{2} + \frac{1}{b^{2}} \right )}}{6 b} & \text{for}\: b \neq 0 \\a^{2} 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)**2*acot(b*x+a),x)

[Out]

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

________________________________________________________________________________________

Giac [A]  time = 1.10006, size = 86, normalized size = 1.65 \begin{align*} \frac{{\left (b x + a\right )}^{3} \arctan \left (\frac{1}{b x + a}\right )}{3 \, b} - \frac{\log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}{6 \, b} + \frac{b^{5} x^{2} + 2 \, a b^{4} x}{6 \, b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(b*x + a)^3*arctan(1/(b*x + a))/b - 1/6*log(b^2*x^2 + 2*a*b*x + a^2 + 1)/b + 1/6*(b^5*x^2 + 2*a*b^4*x)/b^4