3.2.15 \(\int \frac {\cot ^{-1}(a+b x)}{\sqrt {(1+a^2) c+2 a b c x+b^2 c x^2}} \, dx\) [115]

Optimal. Leaf size=216 \[ -\frac {2 i \sqrt {1+(a+b x)^2} \cot ^{-1}(a+b x) \text {ArcTan}\left (\frac {\sqrt {1+i (a+b x)}}{\sqrt {1-i (a+b x)}}\right )}{b \sqrt {c+c (a+b x)^2}}-\frac {i \sqrt {1+(a+b x)^2} \text {PolyLog}\left (2,-\frac {i \sqrt {1+i (a+b x)}}{\sqrt {1-i (a+b x)}}\right )}{b \sqrt {c+c (a+b x)^2}}+\frac {i \sqrt {1+(a+b x)^2} \text {PolyLog}\left (2,\frac {i \sqrt {1+i (a+b x)}}{\sqrt {1-i (a+b x)}}\right )}{b \sqrt {c+c (a+b x)^2}} \]

[Out]

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

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Rubi [A]
time = 0.11, antiderivative size = 216, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {5164, 5011, 5007} \begin {gather*} -\frac {2 i \sqrt {(a+b x)^2+1} \text {ArcTan}\left (\frac {\sqrt {1+i (a+b x)}}{\sqrt {1-i (a+b x)}}\right ) \cot ^{-1}(a+b x)}{b \sqrt {c (a+b x)^2+c}}-\frac {i \sqrt {(a+b x)^2+1} \text {Li}_2\left (-\frac {i \sqrt {i (a+b x)+1}}{\sqrt {1-i (a+b x)}}\right )}{b \sqrt {c (a+b x)^2+c}}+\frac {i \sqrt {(a+b x)^2+1} \text {Li}_2\left (\frac {i \sqrt {i (a+b x)+1}}{\sqrt {1-i (a+b x)}}\right )}{b \sqrt {c (a+b x)^2+c}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 5164

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

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Mathematica [A]
time = 0.07, size = 138, normalized size = 0.64 \begin {gather*} -\frac {\left (1+(a+b x)^2\right ) \left (\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 )+i \text {PolyLog}\left (2,-e^{i \cot ^{-1}(a+b x)}\right )-i \text {PolyLog}\left (2,e^{i \cot ^{-1}(a+b x)}\right )\right )}{b (a+b x) \sqrt {c \left (1+a^2+2 a b x+b^2 x^2\right )} \sqrt {1+\frac {1}{(a+b x)^2}}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.37, size = 156, normalized size = 0.72

method	result	size
default	\(\frac {i \left (i \mathrm {arccot}\left (b x +a \right ) \ln \left (1-\frac {b x +a +i}{\sqrt {1+\left (b x +a \right )^{2}}}\right )-i \mathrm {arccot}\left (b x +a \right ) \ln \left (\frac {b x +a +i}{\sqrt {1+\left (b x +a \right )^{2}}}+1\right )-\polylog \left (2, -\frac {b x +a +i}{\sqrt {1+\left (b x +a \right )^{2}}}\right )+\polylog \left (2, \frac {b x +a +i}{\sqrt {1+\left (b x +a \right )^{2}}}\right )\right ) \sqrt {c \left (b x +a -i\right ) \left (b x +a +i\right )}}{\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\, b c}\)	\(156\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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