3.105 \(\int \frac{\cot ^{-1}(a+b x)}{x^3} \, dx\)

Optimal. Leaf size=95 \[ \frac{a b^2 \log (x)}{\left (a^2+1\right )^2}-\frac{a b^2 \log \left ((a+b x)^2+1\right )}{2 \left (a^2+1\right )^2}+\frac{\left (1-a^2\right ) b^2 \tan ^{-1}(a+b x)}{2 \left (a^2+1\right )^2}+\frac{b}{2 \left (a^2+1\right ) x}-\frac{\cot ^{-1}(a+b x)}{2 x^2} \]

[Out]

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

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Rubi [A]  time = 0.0821167, antiderivative size = 95, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.7, Rules used = {5046, 371, 710, 801, 635, 203, 260} \[ \frac{a b^2 \log (x)}{\left (a^2+1\right )^2}-\frac{a b^2 \log \left ((a+b x)^2+1\right )}{2 \left (a^2+1\right )^2}+\frac{\left (1-a^2\right ) b^2 \tan ^{-1}(a+b x)}{2 \left (a^2+1\right )^2}+\frac{b}{2 \left (a^2+1\right ) x}-\frac{\cot ^{-1}(a+b x)}{2 x^2} \]

Antiderivative was successfully verified.

[In]

Int[ArcCot[a + b*x]/x^3,x]

[Out]

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

Rule 5046

Int[((a_.) + ArcCot[(c_) + (d_.)*(x_)]*(b_.))^(p_.)*((e_.) + (f_.)*(x_))^(m_), x_Symbol] :> Simp[((e + f*x)^(m
+ 1)*(a + b*ArcCot[c + d*x])^p)/(f*(m + 1)), x] + Dist[(b*d*p)/(f*(m + 1)), Int[((e + f*x)^(m + 1)*(a + b*Arc
Cot[c + d*x])^(p - 1))/(1 + (c + d*x)^2), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[p, 0] && ILtQ[m, -1]

Rule 371

Int[((a_) + (b_.)*(v_)^(n_))^(p_.)*(x_)^(m_.), x_Symbol] :> With[{c = Coefficient[v, x, 0], d = Coefficient[v,
x, 1]}, Dist[1/d^(m + 1), Subst[Int[SimplifyIntegrand[(x - c)^m*(a + b*x^n)^p, x], x], x, v], x] /; NeQ[c, 0]
] /; FreeQ[{a, b, n, p}, x] && LinearQ[v, x] && IntegerQ[m]

Rule 710

Int[((d_) + (e_.)*(x_))^(m_)/((a_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(e*(d + e*x)^(m + 1))/((m + 1)*(c*d^2 +
a*e^2)), x] + Dist[c/(c*d^2 + a*e^2), Int[((d + e*x)^(m + 1)*(d - e*x))/(a + c*x^2), x], x] /; FreeQ[{a, c, d,
e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[m, -1]

Rule 801

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
(d + e*x)^m*(f + g*x))/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

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

Mathematica [C]  time = 0.0928639, size = 92, normalized size = 0.97 \[ \frac{-2 \cot ^{-1}(a+b x)+\frac{b x \left (i (a+i)^2 b x \log (-a-b x+i)+4 a b x \log (x)+(a-i) ((-1-i a) b x \log (a+b x+i)+2 (a+i))\right )}{\left (a^2+1\right )^2}}{4 x^2} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcCot[a + b*x]/x^3,x]

[Out]

(-2*ArcCot[a + b*x] + (b*x*(4*a*b*x*Log[x] + I*(I + a)^2*b*x*Log[I - a - b*x] + (-I + a)*(2*(I + a) + (-1 - I*
a)*b*x*Log[I + a + b*x])))/(1 + a^2)^2)/(4*x^2)

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Maple [A]  time = 0.048, size = 104, normalized size = 1.1 \begin{align*} -{\frac{{\rm arccot} \left (bx+a\right )}{2\,{x}^{2}}}-{\frac{{b}^{2}\arctan \left ( bx+a \right ){a}^{2}}{2\, \left ({a}^{2}+1 \right ) ^{2}}}-{\frac{a{b}^{2}\ln \left ( 1+ \left ( bx+a \right ) ^{2} \right ) }{2\, \left ({a}^{2}+1 \right ) ^{2}}}+{\frac{{b}^{2}\arctan \left ( bx+a \right ) }{2\, \left ({a}^{2}+1 \right ) ^{2}}}+{\frac{b}{ \left ( 2\,{a}^{2}+2 \right ) x}}+{\frac{a{b}^{2}\ln \left ( bx \right ) }{ \left ({a}^{2}+1 \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arccot(b*x+a)/x^3,x)

[Out]

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

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Maxima [A]  time = 1.47359, size = 151, normalized size = 1.59 \begin{align*} -\frac{1}{2} \,{\left (\frac{{\left (a^{2} - 1\right )} b \arctan \left (\frac{b^{2} x + a b}{b}\right )}{a^{4} + 2 \, a^{2} + 1} + \frac{a b \log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}{a^{4} + 2 \, a^{2} + 1} - \frac{2 \, a b \log \left (x\right )}{a^{4} + 2 \, a^{2} + 1} - \frac{1}{{\left (a^{2} + 1\right )} x}\right )} b - \frac{\operatorname{arccot}\left (b x + a\right )}{2 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccot(b*x+a)/x^3,x, algorithm="maxima")

[Out]

-1/2*((a^2 - 1)*b*arctan((b^2*x + a*b)/b)/(a^4 + 2*a^2 + 1) + a*b*log(b^2*x^2 + 2*a*b*x + a^2 + 1)/(a^4 + 2*a^
2 + 1) - 2*a*b*log(x)/(a^4 + 2*a^2 + 1) - 1/((a^2 + 1)*x))*b - 1/2*arccot(b*x + a)/x^2

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Fricas [A]  time = 2.30524, size = 248, normalized size = 2.61 \begin{align*} -\frac{{\left (a^{2} - 1\right )} b^{2} x^{2} \arctan \left (b x + a\right ) + a b^{2} x^{2} \log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right ) - 2 \, a b^{2} x^{2} \log \left (x\right ) -{\left (a^{2} + 1\right )} b x +{\left (a^{4} + 2 \, a^{2} + 1\right )} \operatorname{arccot}\left (b x + a\right )}{2 \,{\left (a^{4} + 2 \, a^{2} + 1\right )} x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccot(b*x+a)/x^3,x, algorithm="fricas")

[Out]

-1/2*((a^2 - 1)*b^2*x^2*arctan(b*x + a) + a*b^2*x^2*log(b^2*x^2 + 2*a*b*x + a^2 + 1) - 2*a*b^2*x^2*log(x) - (a
^2 + 1)*b*x + (a^4 + 2*a^2 + 1)*arccot(b*x + a))/((a^4 + 2*a^2 + 1)*x^2)

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Sympy [B]  time = 15.2081, size = 675, normalized size = 7.11 \begin{align*} \begin{cases} - \frac{2 b^{3} x^{3} \operatorname{acot}{\left (b x - i \right )}}{16 b x^{3} - 32 i x^{2}} - \frac{i b^{3} x^{3}}{16 b x^{3} - 32 i x^{2}} + \frac{4 i b^{2} x^{2} \operatorname{acot}{\left (b x - i \right )}}{16 b x^{3} - 32 i x^{2}} - \frac{8 b x \operatorname{acot}{\left (b x - i \right )}}{16 b x^{3} - 32 i x^{2}} - \frac{2 i b x}{16 b x^{3} - 32 i x^{2}} + \frac{16 i \operatorname{acot}{\left (b x - i \right )}}{16 b x^{3} - 32 i x^{2}} + \frac{4}{16 b x^{3} - 32 i x^{2}} & \text{for}\: a = - i \\- \frac{2 b^{3} x^{3} \operatorname{acot}{\left (b x + i \right )}}{16 b x^{3} + 32 i x^{2}} + \frac{i b^{3} x^{3}}{16 b x^{3} + 32 i x^{2}} - \frac{4 i b^{2} x^{2} \operatorname{acot}{\left (b x + i \right )}}{16 b x^{3} + 32 i x^{2}} - \frac{8 b x \operatorname{acot}{\left (b x + i \right )}}{16 b x^{3} + 32 i x^{2}} + \frac{2 i b x}{16 b x^{3} + 32 i x^{2}} - \frac{16 i \operatorname{acot}{\left (b x + i \right )}}{16 b x^{3} + 32 i x^{2}} + \frac{4}{16 b x^{3} + 32 i x^{2}} & \text{for}\: a = i \\- \frac{a^{4} \operatorname{acot}{\left (a + b x \right )}}{2 a^{4} x^{2} + 4 a^{2} x^{2} + 2 x^{2}} + \frac{a^{2} b^{2} x^{2} \operatorname{acot}{\left (a + b x \right )}}{2 a^{4} x^{2} + 4 a^{2} x^{2} + 2 x^{2}} + \frac{a^{2} b x}{2 a^{4} x^{2} + 4 a^{2} x^{2} + 2 x^{2}} - \frac{2 a^{2} \operatorname{acot}{\left (a + b x \right )}}{2 a^{4} x^{2} + 4 a^{2} x^{2} + 2 x^{2}} + \frac{2 a b^{2} x^{2} \log{\left (x \right )}}{2 a^{4} x^{2} + 4 a^{2} x^{2} + 2 x^{2}} - \frac{a b^{2} x^{2} \log{\left (a^{2} + 2 a b x + b^{2} x^{2} + 1 \right )}}{2 a^{4} x^{2} + 4 a^{2} x^{2} + 2 x^{2}} - \frac{2 a b^{2} x^{2}}{2 a^{4} x^{2} + 4 a^{2} x^{2} + 2 x^{2}} - \frac{b^{2} x^{2} \operatorname{acot}{\left (a + b x \right )}}{2 a^{4} x^{2} + 4 a^{2} x^{2} + 2 x^{2}} + \frac{b x}{2 a^{4} x^{2} + 4 a^{2} x^{2} + 2 x^{2}} - \frac{\operatorname{acot}{\left (a + b x \right )}}{2 a^{4} x^{2} + 4 a^{2} x^{2} + 2 x^{2}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(acot(b*x+a)/x**3,x)

[Out]

Piecewise((-2*b**3*x**3*acot(b*x - I)/(16*b*x**3 - 32*I*x**2) - I*b**3*x**3/(16*b*x**3 - 32*I*x**2) + 4*I*b**2
*x**2*acot(b*x - I)/(16*b*x**3 - 32*I*x**2) - 8*b*x*acot(b*x - I)/(16*b*x**3 - 32*I*x**2) - 2*I*b*x/(16*b*x**3
- 32*I*x**2) + 16*I*acot(b*x - I)/(16*b*x**3 - 32*I*x**2) + 4/(16*b*x**3 - 32*I*x**2), Eq(a, -I)), (-2*b**3*x
**3*acot(b*x + I)/(16*b*x**3 + 32*I*x**2) + I*b**3*x**3/(16*b*x**3 + 32*I*x**2) - 4*I*b**2*x**2*acot(b*x + I)/
(16*b*x**3 + 32*I*x**2) - 8*b*x*acot(b*x + I)/(16*b*x**3 + 32*I*x**2) + 2*I*b*x/(16*b*x**3 + 32*I*x**2) - 16*I
*acot(b*x + I)/(16*b*x**3 + 32*I*x**2) + 4/(16*b*x**3 + 32*I*x**2), Eq(a, I)), (-a**4*acot(a + b*x)/(2*a**4*x*
*2 + 4*a**2*x**2 + 2*x**2) + a**2*b**2*x**2*acot(a + b*x)/(2*a**4*x**2 + 4*a**2*x**2 + 2*x**2) + a**2*b*x/(2*a
**4*x**2 + 4*a**2*x**2 + 2*x**2) - 2*a**2*acot(a + b*x)/(2*a**4*x**2 + 4*a**2*x**2 + 2*x**2) + 2*a*b**2*x**2*l
og(x)/(2*a**4*x**2 + 4*a**2*x**2 + 2*x**2) - a*b**2*x**2*log(a**2 + 2*a*b*x + b**2*x**2 + 1)/(2*a**4*x**2 + 4*
a**2*x**2 + 2*x**2) - 2*a*b**2*x**2/(2*a**4*x**2 + 4*a**2*x**2 + 2*x**2) - b**2*x**2*acot(a + b*x)/(2*a**4*x**
2 + 4*a**2*x**2 + 2*x**2) + b*x/(2*a**4*x**2 + 4*a**2*x**2 + 2*x**2) - acot(a + b*x)/(2*a**4*x**2 + 4*a**2*x**
2 + 2*x**2), True))

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Giac [A]  time = 1.11415, size = 158, normalized size = 1.66 \begin{align*} -\frac{1}{2} \,{\left (\frac{a b \log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}{a^{4} + 2 \, a^{2} + 1} - \frac{2 \, a b \log \left ({\left | x \right |}\right )}{a^{4} + 2 \, a^{2} + 1} + \frac{{\left (a^{2} b^{2} - b^{2}\right )} \arctan \left (b x + a\right )}{{\left (a^{4} + 2 \, a^{2} + 1\right )} b} - \frac{1}{{\left (a^{2} + 1\right )} x}\right )} b - \frac{\arctan \left (\frac{1}{b x + a}\right )}{2 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccot(b*x+a)/x^3,x, algorithm="giac")

[Out]

-1/2*(a*b*log(b^2*x^2 + 2*a*b*x + a^2 + 1)/(a^4 + 2*a^2 + 1) - 2*a*b*log(abs(x))/(a^4 + 2*a^2 + 1) + (a^2*b^2
- b^2)*arctan(b*x + a)/((a^4 + 2*a^2 + 1)*b) - 1/((a^2 + 1)*x))*b - 1/2*arctan(1/(b*x + a))/x^2