3.66 \(\int \frac {\cot ^{-1}(x)}{\sqrt {a+a x^2}} \, dx\)

Optimal. Leaf size=155 \[ -\frac {i \sqrt {x^2+1} \text {Li}_2\left (-\frac {i \sqrt {i x+1}}{\sqrt {1-i x}}\right )}{\sqrt {a x^2+a}}+\frac {i \sqrt {x^2+1} \text {Li}_2\left (\frac {i \sqrt {i x+1}}{\sqrt {1-i x}}\right )}{\sqrt {a x^2+a}}-\frac {2 i \sqrt {x^2+1} \tan ^{-1}\left (\frac {\sqrt {1+i x}}{\sqrt {1-i x}}\right ) \cot ^{-1}(x)}{\sqrt {a x^2+a}} \]

[Out]

-2*I*arccot(x)*arctan((1+I*x)^(1/2)/(1-I*x)^(1/2))*(x^2+1)^(1/2)/(a*x^2+a)^(1/2)-I*polylog(2,-I*(1+I*x)^(1/2)/
(1-I*x)^(1/2))*(x^2+1)^(1/2)/(a*x^2+a)^(1/2)+I*polylog(2,I*(1+I*x)^(1/2)/(1-I*x)^(1/2))*(x^2+1)^(1/2)/(a*x^2+a
)^(1/2)

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Rubi [A]  time = 0.04, antiderivative size = 155, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {4891, 4887} \[ -\frac {i \sqrt {x^2+1} \text {PolyLog}\left (2,-\frac {i \sqrt {1+i x}}{\sqrt {1-i x}}\right )}{\sqrt {a x^2+a}}+\frac {i \sqrt {x^2+1} \text {PolyLog}\left (2,\frac {i \sqrt {1+i x}}{\sqrt {1-i x}}\right )}{\sqrt {a x^2+a}}-\frac {2 i \sqrt {x^2+1} \tan ^{-1}\left (\frac {\sqrt {1+i x}}{\sqrt {1-i x}}\right ) \cot ^{-1}(x)}{\sqrt {a x^2+a}} \]

Antiderivative was successfully verified.

[In]

Int[ArcCot[x]/Sqrt[a + a*x^2],x]

[Out]

((-2*I)*Sqrt[1 + x^2]*ArcCot[x]*ArcTan[Sqrt[1 + I*x]/Sqrt[1 - I*x]])/Sqrt[a + a*x^2] - (I*Sqrt[1 + x^2]*PolyLo
g[2, ((-I)*Sqrt[1 + I*x])/Sqrt[1 - I*x]])/Sqrt[a + a*x^2] + (I*Sqrt[1 + x^2]*PolyLog[2, (I*Sqrt[1 + I*x])/Sqrt
[1 - I*x]])/Sqrt[a + a*x^2]

Rule 4887

Int[((a_.) + ArcCot[(c_.)*(x_)]*(b_.))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[(-2*I*(a + b*ArcCot[c*x])*
ArcTan[Sqrt[1 + I*c*x]/Sqrt[1 - I*c*x]])/(c*Sqrt[d]), x] + (-Simp[(I*b*PolyLog[2, -((I*Sqrt[1 + I*c*x])/Sqrt[1
- I*c*x])])/(c*Sqrt[d]), x] + Simp[(I*b*PolyLog[2, (I*Sqrt[1 + I*c*x])/Sqrt[1 - I*c*x]])/(c*Sqrt[d]), x]) /;
FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && GtQ[d, 0]

Rule 4891

Int[((a_.) + ArcCot[(c_.)*(x_)]*(b_.))^(p_.)/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Dist[Sqrt[1 + c^2*x^2]/Sq
rt[d + e*x^2], Int[(a + b*ArcCot[c*x])^p/Sqrt[1 + c^2*x^2], x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*
d] && IGtQ[p, 0] &&  !GtQ[d, 0]

Rubi steps

\begin {align*} \int \frac {\cot ^{-1}(x)}{\sqrt {a+a x^2}} \, dx &=\frac {\sqrt {1+x^2} \int \frac {\cot ^{-1}(x)}{\sqrt {1+x^2}} \, dx}{\sqrt {a+a x^2}}\\ &=-\frac {2 i \sqrt {1+x^2} \cot ^{-1}(x) \tan ^{-1}\left (\frac {\sqrt {1+i x}}{\sqrt {1-i x}}\right )}{\sqrt {a+a x^2}}-\frac {i \sqrt {1+x^2} \text {Li}_2\left (-\frac {i \sqrt {1+i x}}{\sqrt {1-i x}}\right )}{\sqrt {a+a x^2}}+\frac {i \sqrt {1+x^2} \text {Li}_2\left (\frac {i \sqrt {1+i x}}{\sqrt {1-i x}}\right )}{\sqrt {a+a x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.12, size = 89, normalized size = 0.57 \[ -\frac {\sqrt {a \left (x^2+1\right )} \left (i \text {Li}_2\left (-e^{i \cot ^{-1}(x)}\right )-i \text {Li}_2\left (e^{i \cot ^{-1}(x)}\right )+\cot ^{-1}(x) \left (\log \left (1-e^{i \cot ^{-1}(x)}\right )-\log \left (1+e^{i \cot ^{-1}(x)}\right )\right )\right )}{a \sqrt {\frac {1}{x^2}+1} x} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[ArcCot[x]/Sqrt[a + a*x^2],x]

[Out]

-((Sqrt[a*(1 + x^2)]*(ArcCot[x]*(Log[1 - E^(I*ArcCot[x])] - Log[1 + E^(I*ArcCot[x])]) + I*PolyLog[2, -E^(I*Arc
Cot[x])] - I*PolyLog[2, E^(I*ArcCot[x])]))/(a*Sqrt[1 + x^(-2)]*x))

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fricas [F]  time = 0.58, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\operatorname {arccot}\relax (x)}{\sqrt {a x^{2} + a}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccot(x)/(a*x^2+a)^(1/2),x, algorithm="fricas")

[Out]

integral(arccot(x)/sqrt(a*x^2 + a), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {arccot}\relax (x)}{\sqrt {a x^{2} + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccot(x)/(a*x^2+a)^(1/2),x, algorithm="giac")

[Out]

integrate(arccot(x)/sqrt(a*x^2 + a), x)

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maple [A]  time = 0.92, size = 99, normalized size = 0.64 \[ \frac {i \left (i \mathrm {arccot}\relax (x ) \ln \left (1-\frac {x +i}{\sqrt {x^{2}+1}}\right )-i \mathrm {arccot}\relax (x ) \ln \left (1+\frac {x +i}{\sqrt {x^{2}+1}}\right )+\polylog \left (2, \frac {x +i}{\sqrt {x^{2}+1}}\right )-\polylog \left (2, -\frac {x +i}{\sqrt {x^{2}+1}}\right )\right ) \sqrt {a \left (x +i\right ) \left (x -i\right )}}{\sqrt {x^{2}+1}\, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arccot(x)/(a*x^2+a)^(1/2),x)

[Out]

I*(I*ln(1-(x+I)/(x^2+1)^(1/2))*arccot(x)-I*ln(1+(x+I)/(x^2+1)^(1/2))*arccot(x)+polylog(2,(x+I)/(x^2+1)^(1/2))-
polylog(2,-(x+I)/(x^2+1)^(1/2)))*(a*(x+I)*(x-I))^(1/2)/(x^2+1)^(1/2)/a

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {arccot}\relax (x)}{\sqrt {a x^{2} + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccot(x)/(a*x^2+a)^(1/2),x, algorithm="maxima")

[Out]

integrate(arccot(x)/sqrt(a*x^2 + a), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\mathrm {acot}\relax (x)}{\sqrt {a\,x^2+a}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(acot(x)/(a + a*x^2)^(1/2),x)

[Out]

int(acot(x)/(a + a*x^2)^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {acot}{\relax (x )}}{\sqrt {a \left (x^{2} + 1\right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(acot(x)/(a*x**2+a)**(1/2),x)

[Out]

Integral(acot(x)/sqrt(a*(x**2 + 1)), x)

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