3.1.65 \(\int \sqrt {a+a x^2} \cot ^{-1}(x) \, dx\) [65]

Optimal. Leaf size=195 \[ \frac {1}{2} \sqrt {a+a x^2}+\frac {1}{2} x \sqrt {a+a x^2} \cot ^{-1}(x)-\frac {i a \sqrt {1+x^2} \cot ^{-1}(x) \text {ArcTan}\left (\frac {\sqrt {1+i x}}{\sqrt {1-i x}}\right )}{\sqrt {a+a x^2}}-\frac {i a \sqrt {1+x^2} \text {PolyLog}\left (2,-\frac {i \sqrt {1+i x}}{\sqrt {1-i x}}\right )}{2 \sqrt {a+a x^2}}+\frac {i a \sqrt {1+x^2} \text {PolyLog}\left (2,\frac {i \sqrt {1+i x}}{\sqrt {1-i x}}\right )}{2 \sqrt {a+a x^2}} \]

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 195, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {4999, 5011, 5007} \begin {gather*} -\frac {i a \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 a \sqrt {x^2+1} \text {Li}_2\left (-\frac {i \sqrt {i x+1}}{\sqrt {1-i x}}\right )}{2 \sqrt {a x^2+a}}+\frac {i a \sqrt {x^2+1} \text {Li}_2\left (\frac {i \sqrt {i x+1}}{\sqrt {1-i x}}\right )}{2 \sqrt {a x^2+a}}+\frac {1}{2} \sqrt {a x^2+a}+\frac {1}{2} x \sqrt {a x^2+a} \cot ^{-1}(x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + a*x^2]*ArcCot[x],x]

[Out]

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

Rule 4999

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

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

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Mathematica [A]
time = 0.89, size = 136, normalized size = 0.70 \begin {gather*} -\frac {\left (a \left (1+x^2\right )\right )^{3/2} \left (-2 \cot \left (\frac {1}{2} \cot ^{-1}(x)\right )-\cot ^{-1}(x) \csc ^2\left (\frac {1}{2} \cot ^{-1}(x)\right )+4 \cot ^{-1}(x) \log \left (1-e^{i \cot ^{-1}(x)}\right )-4 \cot ^{-1}(x) \log \left (1+e^{i \cot ^{-1}(x)}\right )+4 i \text {PolyLog}\left (2,-e^{i \cot ^{-1}(x)}\right )-4 i \text {PolyLog}\left (2,e^{i \cot ^{-1}(x)}\right )+\cot ^{-1}(x) \sec ^2\left (\frac {1}{2} \cot ^{-1}(x)\right )-2 \tan \left (\frac {1}{2} \cot ^{-1}(x)\right )\right )}{8 a \left (1+\frac {1}{x^2}\right )^{3/2} x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/8*((a*(1 + x^2))^(3/2)*(-2*Cot[ArcCot[x]/2] - ArcCot[x]*Csc[ArcCot[x]/2]^2 + 4*ArcCot[x]*Log[1 - E^(I*ArcCo
t[x])] - 4*ArcCot[x]*Log[1 + E^(I*ArcCot[x])] + (4*I)*PolyLog[2, -E^(I*ArcCot[x])] - (4*I)*PolyLog[2, E^(I*Arc
Cot[x])] + ArcCot[x]*Sec[ArcCot[x]/2]^2 - 2*Tan[ArcCot[x]/2]))/(a*(1 + x^(-2))^(3/2)*x^3)

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Maple [A]
time = 0.27, size = 117, normalized size = 0.60

method	result	size
default	\(\frac {\sqrt {a \left (i+x \right ) \left (x -i\right )}\, \left (x \,\mathrm {arccot}\left (x \right )+1\right )}{2}-\frac {i \sqrt {a \left (i+x \right ) \left (x -i\right )}\, \left (i \mathrm {arccot}\left (x \right ) \ln \left (\frac {i+x}{\sqrt {x^{2}+1}}+1\right )-i \mathrm {arccot}\left (x \right ) \ln \left (1-\frac {i+x}{\sqrt {x^{2}+1}}\right )+\polylog \left (2, -\frac {i+x}{\sqrt {x^{2}+1}}\right )-\polylog \left (2, \frac {i+x}{\sqrt {x^{2}+1}}\right )\right )}{2 \sqrt {x^{2}+1}}\)	\(117\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \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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