3.1.66 \(\int \frac {\cot ^{-1}(x)}{\sqrt {a+a x^2}} \, dx\) [66]

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

[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 = 0.03, 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 = {5011, 5007} \begin {gather*} -\frac {2 i \sqrt {x^2+1} \text {ArcTan}\left (\frac {\sqrt {1+i x}}{\sqrt {1-i x}}\right ) \cot ^{-1}(x)}{\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 {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}} \end {gather*}

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 5007

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 5011

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 = 0.08, size = 89, normalized size = 0.57 \begin {gather*} -\frac {\sqrt {a \left (1+x^2\right )} \left (\cot ^{-1}(x) \left (\log \left (1-e^{i \cot ^{-1}(x)}\right )-\log \left (1+e^{i \cot ^{-1}(x)}\right )\right )+i \text {PolyLog}\left (2,-e^{i \cot ^{-1}(x)}\right )-i \text {PolyLog}\left (2,e^{i \cot ^{-1}(x)}\right )\right )}{a \sqrt {1+\frac {1}{x^2}} x} \end {gather*}

Antiderivative was successfully verified.

[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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Maple [A]
time = 0.20, size = 99, normalized size = 0.64

method	result	size
default	\(-\frac {i \left (i \mathrm {arccot}\left (x \right ) \ln \left (\frac {i+x}{\sqrt {x^{2}+1}}+1\right )-i \mathrm {arccot}\left (x \right ) \ln \left (1-\frac {i+x}{\sqrt {x^{2}+1}}\right )+\polylog \left (2, -\frac {i+x}{\sqrt {x^{2}+1}}\right )-\polylog \left (2, \frac {i+x}{\sqrt {x^{2}+1}}\right )\right ) \sqrt {a \left (i+x \right ) \left (x -i\right )}}{\sqrt {x^{2}+1}\, a}\)	\(99\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-I*(I*arccot(x)*ln((I+x)/(x^2+1)^(1/2)+1)-I*arccot(x)*ln(1-(I+x)/(x^2+1)^(1/2))+polylog(2,-(I+x)/(x^2+1)^(1/2)
)-polylog(2,(I+x)/(x^2+1)^(1/2)))*(a*(I+x)*(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 \begin {gather*} \text {Failed to integrate} \end {gather*}

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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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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

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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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\mathrm {acot}\left (x\right )}{\sqrt {a\,x^2+a}} \,d x \end {gather*}

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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