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]

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

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Rubi [A]  time = 0.06, antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.750, 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 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]

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 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 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 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]

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*}

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Mathematica [C]  time = 0.04, size = 90, normalized size = 1.50 \[ \frac {i a^2 \log (a+b x+i)+2 b^2 x^2 \cot ^{-1}(a+b x)-2 a \log (a+b x+i)-i (a-i)^2 \log (-a-b x+i)-i \log (a+b x+i)+2 b x}{4 b^2} \]

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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fricas [A]  time = 1.05, size = 55, normalized size = 0.92 \[ \frac {b^{2} x^{2} \operatorname {arccot}\left (b x + a\right ) + b x + {\left (a^{2} - 1\right )} \arctan \left (b x + a\right ) - a \log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}{2 \, b^{2}} \]

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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giac [B]  time = 0.27, size = 210, normalized size = 3.50 \[ \frac {4 \, a \arctan \left (\frac {1}{b x + a}\right ) \tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right )^{3} + \arctan \left (\frac {1}{b x + a}\right ) \tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right )^{4} + 4 \, a \log \left (\frac {16 \, \tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right )^{2}}{\tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right )^{4} + 2 \, \tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right )^{2} + 1}\right ) \tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right )^{2} - 4 \, a \arctan \left (\frac {1}{b x + a}\right ) \tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right ) + 2 \, \arctan \left (\frac {1}{b x + a}\right ) \tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right )^{2} - 2 \, \tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right )^{3} + \arctan \left (\frac {1}{b x + a}\right ) + 2 \, \tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right )}{8 \, b^{2} \tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(4*a*arctan(1/(b*x + a))*tan(1/2*arctan(1/(b*x + a)))^3 + arctan(1/(b*x + a))*tan(1/2*arctan(1/(b*x + a)))
^4 + 4*a*log(16*tan(1/2*arctan(1/(b*x + a)))^2/(tan(1/2*arctan(1/(b*x + a)))^4 + 2*tan(1/2*arctan(1/(b*x + a))
)^2 + 1))*tan(1/2*arctan(1/(b*x + a)))^2 - 4*a*arctan(1/(b*x + a))*tan(1/2*arctan(1/(b*x + a))) + 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^2*tan(1/2*arctan(1/(b*x + a)))^2)

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maple [A]  time = 0.04, size = 66, normalized size = 1.10 \[ \frac {x^{2} \mathrm {arccot}\left (b x +a \right )}{2}-\frac {\mathrm {arccot}\left (b x +a \right ) a^{2}}{2 b^{2}}+\frac {x}{2 b}+\frac {a}{2 b^{2}}-\frac {a \ln \left (1+\left (b x +a \right )^{2}\right )}{2 b^{2}}-\frac {\arctan \left (b x +a \right )}{2 b^{2}} \]

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 = 0.43, size = 68, normalized size = 1.13 \[ \frac {1}{2} \, x^{2} \operatorname {arccot}\left (b x + a\right ) + \frac {1}{2} \, b {\left (\frac {x}{b^{2}} + \frac {{\left (a^{2} - 1\right )} \arctan \left (\frac {b^{2} x + a b}{b}\right )}{b^{3}} - \frac {a \log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}{b^{3}}\right )} \]

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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mupad [B]  time = 0.97, size = 61, normalized size = 1.02 \[ \frac {x^2\,\mathrm {acot}\left (a+b\,x\right )}{2}+\frac {\frac {\mathrm {acot}\left (a+b\,x\right )}{2}+\frac {b\,x}{2}-\frac {a^2\,\mathrm {acot}\left (a+b\,x\right )}{2}-\frac {a\,\ln \left (a^2+2\,a\,b\,x+b^2\,x^2+1\right )}{2}}{b^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*acot(a + b*x),x)

[Out]

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

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sympy [A]  time = 0.61, size = 78, normalized size = 1.30 \[ \begin {cases} - \frac {a^{2} \operatorname {acot}{\left (a + b x \right )}}{2 b^{2}} - \frac {a \log {\left (a^{2} + 2 a b x + b^{2} x^{2} + 1 \right )}}{2 b^{2}} + \frac {x^{2} \operatorname {acot}{\left (a + b x \right )}}{2} + \frac {x}{2 b} + \frac {\operatorname {acot}{\left (a + b x \right )}}{2 b^{2}} & \text {for}\: b \neq 0 \\\frac {x^{2} \operatorname {acot}{\relax (a )}}{2} & \text {otherwise} \end {cases} \]

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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