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

Optimal. Leaf size=132 \[ -\frac {i \text {Li}_2\left (-\frac {i \sqrt {i (a+b x)+1}}{\sqrt {1-i (a+b x)}}\right )}{b}+\frac {i \text {Li}_2\left (\frac {i \sqrt {i (a+b x)+1}}{\sqrt {1-i (a+b x)}}\right )}{b}-\frac {2 i \tan ^{-1}\left (\frac {\sqrt {1+i (a+b x)}}{\sqrt {1-i (a+b x)}}\right ) \cot ^{-1}(a+b x)}{b} \]

[Out]

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

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Rubi [A]  time = 0.10, antiderivative size = 132, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {5056, 4887} \[ -\frac {i \text {PolyLog}\left (2,-\frac {i \sqrt {1+i (a+b x)}}{\sqrt {1-i (a+b x)}}\right )}{b}+\frac {i \text {PolyLog}\left (2,\frac {i \sqrt {1+i (a+b x)}}{\sqrt {1-i (a+b x)}}\right )}{b}-\frac {2 i \tan ^{-1}\left (\frac {\sqrt {1+i (a+b x)}}{\sqrt {1-i (a+b x)}}\right ) \cot ^{-1}(a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

Int[ArcCot[a + b*x]/Sqrt[1 + a^2 + 2*a*b*x + b^2*x^2],x]

[Out]

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

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 5056

Int[((a_.) + ArcCot[(c_) + (d_.)*(x_)]*(b_.))^(p_.)*((A_.) + (B_.)*(x_) + (C_.)*(x_)^2)^(q_.), x_Symbol] :> Di
st[1/d, Subst[Int[(C/d^2 + (C*x^2)/d^2)^q*(a + b*ArcCot[x])^p, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, A, B,
C, p, q}, x] && EqQ[B*(1 + c^2) - 2*A*c*d, 0] && EqQ[2*c*C - B*d, 0]

Rubi steps

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

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Mathematica [A]  time = 0.16, size = 127, normalized size = 0.96 \[ -\frac {\sqrt {a^2+2 a b x+b^2 x^2+1} \left (i \text {Li}_2\left (-e^{i \cot ^{-1}(a+b x)}\right )-i \text {Li}_2\left (e^{i \cot ^{-1}(a+b x)}\right )+\cot ^{-1}(a+b x) \left (\log \left (1-e^{i \cot ^{-1}(a+b x)}\right )-\log \left (1+e^{i \cot ^{-1}(a+b x)}\right )\right )\right )}{b (a+b x) \sqrt {\frac {1}{(a+b x)^2}+1}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[ArcCot[a + b*x]/Sqrt[1 + a^2 + 2*a*b*x + b^2*x^2],x]

[Out]

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

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fricas [F]  time = 0.63, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\operatorname {arccot}\left (b x + a\right )}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(arccot(b*x + a)/sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {arccot}\left (b x + a\right )}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arccot(b*x + a)/sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1), x)

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maple [A]  time = 0.69, size = 125, normalized size = 0.95 \[ -\frac {\mathrm {arccot}\left (b x +a \right ) \ln \left (1-\frac {b x +a +i}{\sqrt {1+\left (b x +a \right )^{2}}}\right )}{b}+\frac {\mathrm {arccot}\left (b x +a \right ) \ln \left (1+\frac {b x +a +i}{\sqrt {1+\left (b x +a \right )^{2}}}\right )}{b}-\frac {i \dilog \left (1+\frac {b x +a +i}{\sqrt {1+\left (b x +a \right )^{2}}}\right )}{b}+\frac {i \dilog \left (1-\frac {b x +a +i}{\sqrt {1+\left (b x +a \right )^{2}}}\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {arccot}\left (b x + a\right )}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arccot(b*x + a)/sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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