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]

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

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Rubi [A]  time = 0.02, antiderivative size = 39, normalized size of antiderivative = 1.00, 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 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 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 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 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]

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*}

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Mathematica [C]  time = 0.06, 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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fricas [A]  time = 0.52, size = 33, normalized size = 0.85 \[ \frac {b x + {\left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )} \operatorname {arccot}\left (b x + a\right )}{2 \, b} \]

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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giac [B]  time = 0.18, size = 100, normalized size = 2.56 \[ \frac {\arctan \left (\frac {1}{b x + a}\right ) \tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right )^{4} + 2 \, \arctan \left (\frac {1}{b x + a}\right ) \tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right )^{2} - 2 \, \tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right )^{3} + \arctan \left (\frac {1}{b x + a}\right ) + 2 \, \tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right )}{8 \, b \tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(arctan(1/(b*x + a))*tan(1/2*arctan(1/(b*x + a)))^4 + 2*arctan(1/(b*x + a))*tan(1/2*arctan(1/(b*x + a)))^2
- 2*tan(1/2*arctan(1/(b*x + a)))^3 + arctan(1/(b*x + a)) + 2*tan(1/2*arctan(1/(b*x + a))))/(b*tan(1/2*arctan(
1/(b*x + a)))^2)

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maple [A]  time = 0.04, size = 57, normalized size = 1.46 \[ \frac {b \,\mathrm {arccot}\left (b x +a \right ) x^{2}}{2}+\mathrm {arccot}\left (b x +a \right ) x a +\frac {\mathrm {arccot}\left (b x +a \right ) a^{2}}{2 b}+\frac {x}{2}+\frac {a}{2 b}-\frac {\arctan \left (b x +a \right )}{2 b} \]

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 = 0.43, size = 52, normalized size = 1.33 \[ \frac {1}{2} \, b {\left (\frac {x}{b} - \frac {{\left (a^{2} + 1\right )} \arctan \left (\frac {b^{2} x + a b}{b}\right )}{b^{2}}\right )} + \frac {1}{2} \, {\left (b x^{2} + 2 \, a x\right )} \operatorname {arccot}\left (b x + a\right ) \]

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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mupad [B]  time = 1.51, size = 49, normalized size = 1.26 \[ \frac {x}{2}+\frac {\frac {\mathrm {acot}\left (a+b\,x\right )}{2}+\frac {a^2\,\mathrm {acot}\left (a+b\,x\right )}{2}}{b}+a\,x\,\mathrm {acot}\left (a+b\,x\right )+\frac {b\,x^2\,\mathrm {acot}\left (a+b\,x\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(acot(a + b*x)*(a + b*x),x)

[Out]

x/2 + (acot(a + b*x)/2 + (a^2*acot(a + b*x))/2)/b + a*x*acot(a + b*x) + (b*x^2*acot(a + b*x))/2

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sympy [A]  time = 1.77, size = 56, normalized size = 1.44 \[ \begin {cases} \frac {a^{2} \operatorname {acot}{\left (a + b x \right )}}{2 b} + a x \operatorname {acot}{\left (a + b x \right )} + \frac {b x^{2} \operatorname {acot}{\left (a + b x \right )}}{2} + \frac {x}{2} + \frac {\operatorname {acot}{\left (a + b x \right )}}{2 b} & \text {for}\: b \neq 0 \\a x \operatorname {acot}{\relax (a )} & \text {otherwise} \end {cases} \]

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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