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]

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

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Rubi [A]  time = 0.08, antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.700, 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 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 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 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 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 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]

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*}

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Mathematica [C]  time = 0.11, 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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fricas [A]  time = 0.74, size = 99, normalized size = 1.04 \[ -\frac {{\left (a^{2} - 1\right )} b^{2} x^{2} \arctan \left (b x + a\right ) + a b^{2} x^{2} \log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right ) - 2 \, a b^{2} x^{2} \log \relax (x) - {\left (a^{2} + 1\right )} b x + {\left (a^{4} + 2 \, a^{2} + 1\right )} \operatorname {arccot}\left (b x + a\right )}{2 \, {\left (a^{4} + 2 \, a^{2} + 1\right )} x^{2}} \]

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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giac [B]  time = 0.71, size = 1309, normalized size = 13.78 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(4*a^3*b*arctan(1/(b*x + a))*tan(1/2*arctan(1/(b*x + a)))^3 + a^2*b*arctan(1/(b*x + a))*tan(1/2*arctan(1/(
b*x + a)))^4 + 4*a^3*b*log(4*(4*a^2*tan(1/2*arctan(1/(b*x + a)))^2 + 4*a*tan(1/2*arctan(1/(b*x + a)))^3 + tan(
1/2*arctan(1/(b*x + a)))^4 - 4*a*tan(1/2*arctan(1/(b*x + a))) - 2*tan(1/2*arctan(1/(b*x + a)))^2 + 1)/(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^2*b*log(
4*(4*a^2*tan(1/2*arctan(1/(b*x + a)))^2 + 4*a*tan(1/2*arctan(1/(b*x + a)))^3 + tan(1/2*arctan(1/(b*x + a)))^4
- 4*a*tan(1/2*arctan(1/(b*x + a))) - 2*tan(1/2*arctan(1/(b*x + a)))^2 + 1)/(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)))^3 + a*b*log(4*(4*a^2*tan(1/2*arctan(1/(b*x
+ a)))^2 + 4*a*tan(1/2*arctan(1/(b*x + a)))^3 + tan(1/2*arctan(1/(b*x + a)))^4 - 4*a*tan(1/2*arctan(1/(b*x + a
))) - 2*tan(1/2*arctan(1/(b*x + a)))^2 + 1)/(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)))^4 - 4*a^3*b*arctan(1/(b*x + a))*tan(1/2*arctan(1/(b*x + a))) - 14*a^2*b*ar
ctan(1/(b*x + a))*tan(1/2*arctan(1/(b*x + a)))^2 + 2*a^2*b*tan(1/2*arctan(1/(b*x + a)))^3 - 4*a*b*arctan(1/(b*
x + a))*tan(1/2*arctan(1/(b*x + a)))^3 + a*b*tan(1/2*arctan(1/(b*x + a)))^4 - b*arctan(1/(b*x + a))*tan(1/2*ar
ctan(1/(b*x + a)))^4 - 4*a^2*b*log(4*(4*a^2*tan(1/2*arctan(1/(b*x + a)))^2 + 4*a*tan(1/2*arctan(1/(b*x + a)))^
3 + tan(1/2*arctan(1/(b*x + a)))^4 - 4*a*tan(1/2*arctan(1/(b*x + a))) - 2*tan(1/2*arctan(1/(b*x + a)))^2 + 1)/
(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*a*b*
log(4*(4*a^2*tan(1/2*arctan(1/(b*x + a)))^2 + 4*a*tan(1/2*arctan(1/(b*x + a)))^3 + tan(1/2*arctan(1/(b*x + a))
)^4 - 4*a*tan(1/2*arctan(1/(b*x + a))) - 2*tan(1/2*arctan(1/(b*x + a)))^2 + 1)/(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 + a^2*b*arctan(1/(b*x + a)) - 2*a^2*b
*tan(1/2*arctan(1/(b*x + a))) + 4*a*b*arctan(1/(b*x + a))*tan(1/2*arctan(1/(b*x + a))) - 6*a*b*tan(1/2*arctan(
1/(b*x + a)))^2 - 2*b*arctan(1/(b*x + a))*tan(1/2*arctan(1/(b*x + a)))^2 - 2*b*tan(1/2*arctan(1/(b*x + a)))^3
+ a*b*log(4*(4*a^2*tan(1/2*arctan(1/(b*x + a)))^2 + 4*a*tan(1/2*arctan(1/(b*x + a)))^3 + tan(1/2*arctan(1/(b*x
+ a)))^4 - 4*a*tan(1/2*arctan(1/(b*x + a))) - 2*tan(1/2*arctan(1/(b*x + a)))^2 + 1)/(tan(1/2*arctan(1/(b*x +
a)))^4 + 2*tan(1/2*arctan(1/(b*x + a)))^2 + 1)) + a*b - b*arctan(1/(b*x + a)) + 2*b*tan(1/2*arctan(1/(b*x + a)
)))*b/(4*a^6*tan(1/2*arctan(1/(b*x + a)))^2 + 4*a^5*tan(1/2*arctan(1/(b*x + a)))^3 + a^4*tan(1/2*arctan(1/(b*x
+ a)))^4 - 4*a^5*tan(1/2*arctan(1/(b*x + a))) + 6*a^4*tan(1/2*arctan(1/(b*x + a)))^2 + 8*a^3*tan(1/2*arctan(1
/(b*x + a)))^3 + 2*a^2*tan(1/2*arctan(1/(b*x + a)))^4 + a^4 - 8*a^3*tan(1/2*arctan(1/(b*x + a))) + 4*a*tan(1/2
*arctan(1/(b*x + a)))^3 + tan(1/2*arctan(1/(b*x + a)))^4 + 2*a^2 - 4*a*tan(1/2*arctan(1/(b*x + a))) - 2*tan(1/
2*arctan(1/(b*x + a)))^2 + 1)

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maple [A]  time = 0.05, size = 104, normalized size = 1.09 \[ -\frac {\mathrm {arccot}\left (b x +a \right )}{2 x^{2}}+\frac {b}{2 \left (a^{2}+1\right ) x}+\frac {b^{2} a \ln \left (b x \right )}{\left (a^{2}+1\right )^{2}}-\frac {b^{2} \arctan \left (b x +a \right ) a^{2}}{2 \left (a^{2}+1\right )^{2}}-\frac {a \,b^{2} \ln \left (1+\left (b x +a \right )^{2}\right )}{2 \left (a^{2}+1\right )^{2}}+\frac {b^{2} \arctan \left (b x +a \right )}{2 \left (a^{2}+1\right )^{2}} \]

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

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maxima [A]  time = 0.42, size = 112, normalized size = 1.18 \[ -\frac {1}{2} \, {\left (\frac {{\left (a^{2} - 1\right )} b \arctan \left (\frac {b^{2} x + a b}{b}\right )}{a^{4} + 2 \, a^{2} + 1} + \frac {a b \log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}{a^{4} + 2 \, a^{2} + 1} - \frac {2 \, a b \log \relax (x)}{a^{4} + 2 \, a^{2} + 1} - \frac {1}{{\left (a^{2} + 1\right )} x}\right )} b - \frac {\operatorname {arccot}\left (b x + a\right )}{2 \, x^{2}} \]

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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mupad [B]  time = 1.39, size = 230, normalized size = 2.42 \[ \frac {\mathrm {atan}\left (\frac {2\,x\,b^2+2\,a\,b}{2\,\sqrt {b^2\,\left (a^2+1\right )-a^2\,b^2}}\right )\,\left (b^3-a^2\,b^3\right )}{\sqrt {b^2}\,\left (2\,a^4+4\,a^2+2\right )}-\frac {a\,b^2\,\ln \left (a^2+2\,a\,b\,x+b^2\,x^2+1\right )}{2\,{\left (a^2+1\right )}^2}-\frac {\mathrm {acot}\left (a+b\,x\right )\,\left (\frac {a^2}{2}+\frac {1}{2}\right )-\frac {b\,x}{2}+\frac {b^2\,x^2\,\mathrm {acot}\left (a+b\,x\right )}{2}-\frac {x^3\,\left (b^3-3\,a^2\,b^3\right )}{2\,\left (a^4+2\,a^2+1\right )}+\frac {a\,b^4\,x^4}{{\left (a^2+1\right )}^2}+a\,b\,x\,\mathrm {acot}\left (a+b\,x\right )}{a^2\,x^2+2\,a\,b\,x^3+b^2\,x^4+x^2}+\frac {a\,b^2\,\ln \relax (x)}{{\left (a^2+1\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(acot(a + b*x)/x^3,x)

[Out]

(atan((2*a*b + 2*b^2*x)/(2*(b^2*(a^2 + 1) - a^2*b^2)^(1/2)))*(b^3 - a^2*b^3))/((b^2)^(1/2)*(4*a^2 + 2*a^4 + 2)
) - (a*b^2*log(a^2 + b^2*x^2 + 2*a*b*x + 1))/(2*(a^2 + 1)^2) - (acot(a + b*x)*(a^2/2 + 1/2) - (b*x)/2 + (b^2*x
^2*acot(a + b*x))/2 - (x^3*(b^3 - 3*a^2*b^3))/(2*(2*a^2 + a^4 + 1)) + (a*b^4*x^4)/(a^2 + 1)^2 + a*b*x*acot(a +
b*x))/(x^2 + a^2*x^2 + b^2*x^4 + 2*a*b*x^3) + (a*b^2*log(x))/(a^2 + 1)^2

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sympy [B]  time = 2.83, size = 381, normalized size = 4.01 \[ \begin {cases} - \frac {b^{2} \operatorname {acot}{\left (b x - i \right )}}{8} + \frac {b}{8 x} - \frac {\operatorname {acot}{\left (b x - i \right )}}{2 x^{2}} + \frac {i}{8 x^{2}} & \text {for}\: a = - i \\- \frac {b^{2} \operatorname {acot}{\left (b x + i \right )}}{8} + \frac {b}{8 x} - \frac {\operatorname {acot}{\left (b x + i \right )}}{2 x^{2}} - \frac {i}{8 x^{2}} & \text {for}\: a = i \\- \frac {a^{4} \operatorname {acot}{\left (a + b x \right )}}{2 a^{4} x^{2} + 4 a^{2} x^{2} + 2 x^{2}} + \frac {a^{2} b^{2} x^{2} \operatorname {acot}{\left (a + b x \right )}}{2 a^{4} x^{2} + 4 a^{2} x^{2} + 2 x^{2}} + \frac {a^{2} b x}{2 a^{4} x^{2} + 4 a^{2} x^{2} + 2 x^{2}} - \frac {2 a^{2} \operatorname {acot}{\left (a + b x \right )}}{2 a^{4} x^{2} + 4 a^{2} x^{2} + 2 x^{2}} + \frac {2 a b^{2} x^{2} \log {\relax (x )}}{2 a^{4} x^{2} + 4 a^{2} x^{2} + 2 x^{2}} - \frac {a b^{2} x^{2} \log {\left (a^{2} + 2 a b x + b^{2} x^{2} + 1 \right )}}{2 a^{4} x^{2} + 4 a^{2} x^{2} + 2 x^{2}} - \frac {b^{2} x^{2} \operatorname {acot}{\left (a + b x \right )}}{2 a^{4} x^{2} + 4 a^{2} x^{2} + 2 x^{2}} + \frac {b x}{2 a^{4} x^{2} + 4 a^{2} x^{2} + 2 x^{2}} - \frac {\operatorname {acot}{\left (a + b x \right )}}{2 a^{4} x^{2} + 4 a^{2} x^{2} + 2 x^{2}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-b**2*acot(b*x - I)/8 + b/(8*x) - acot(b*x - I)/(2*x**2) + I/(8*x**2), Eq(a, -I)), (-b**2*acot(b*x
+ I)/8 + b/(8*x) - acot(b*x + I)/(2*x**2) - I/(8*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*log(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) - 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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