3.144 \(\int \frac{(a+b \tan ^{-1}(\frac{c}{x}))^2}{x} \, dx\)

Optimal. Leaf size=148 \[ i b \text{PolyLog}\left (2,1-\frac{2}{1+\frac{i c}{x}}\right ) \left (a+b \cot ^{-1}\left (\frac{x}{c}\right )\right )-i b \text{PolyLog}\left (2,-1+\frac{2}{1+\frac{i c}{x}}\right ) \left (a+b \cot ^{-1}\left (\frac{x}{c}\right )\right )+\frac{1}{2} b^2 \text{PolyLog}\left (3,1-\frac{2}{1+\frac{i c}{x}}\right )-\frac{1}{2} b^2 \text{PolyLog}\left (3,-1+\frac{2}{1+\frac{i c}{x}}\right )-2 \tanh ^{-1}\left (1-\frac{2}{1+\frac{i c}{x}}\right ) \left (a+b \cot ^{-1}\left (\frac{x}{c}\right )\right )^2 \]

[Out]

-2*(a + b*ArcCot[x/c])^2*ArcTanh[1 - 2/(1 + (I*c)/x)] + I*b*(a + b*ArcCot[x/c])*PolyLog[2, 1 - 2/(1 + (I*c)/x)
] - I*b*(a + b*ArcCot[x/c])*PolyLog[2, -1 + 2/(1 + (I*c)/x)] + (b^2*PolyLog[3, 1 - 2/(1 + (I*c)/x)])/2 - (b^2*
PolyLog[3, -1 + 2/(1 + (I*c)/x)])/2

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Rubi [A]  time = 0.293581, antiderivative size = 148, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.375, Rules used = {5031, 4850, 4988, 4884, 4994, 6610} \[ i b \text{PolyLog}\left (2,1-\frac{2}{1+\frac{i c}{x}}\right ) \left (a+b \cot ^{-1}\left (\frac{x}{c}\right )\right )-i b \text{PolyLog}\left (2,-1+\frac{2}{1+\frac{i c}{x}}\right ) \left (a+b \cot ^{-1}\left (\frac{x}{c}\right )\right )+\frac{1}{2} b^2 \text{PolyLog}\left (3,1-\frac{2}{1+\frac{i c}{x}}\right )-\frac{1}{2} b^2 \text{PolyLog}\left (3,-1+\frac{2}{1+\frac{i c}{x}}\right )-2 \tanh ^{-1}\left (1-\frac{2}{1+\frac{i c}{x}}\right ) \left (a+b \cot ^{-1}\left (\frac{x}{c}\right )\right )^2 \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTan[c/x])^2/x,x]

[Out]

-2*(a + b*ArcCot[x/c])^2*ArcTanh[1 - 2/(1 + (I*c)/x)] + I*b*(a + b*ArcCot[x/c])*PolyLog[2, 1 - 2/(1 + (I*c)/x)
] - I*b*(a + b*ArcCot[x/c])*PolyLog[2, -1 + 2/(1 + (I*c)/x)] + (b^2*PolyLog[3, 1 - 2/(1 + (I*c)/x)])/2 - (b^2*
PolyLog[3, -1 + 2/(1 + (I*c)/x)])/2

Rule 5031

Int[((a_.) + ArcTan[(c_.)*(x_)^(n_)]*(b_.))^(p_.)/(x_), x_Symbol] :> Dist[1/n, Subst[Int[(a + b*ArcTan[c*x])^p
/x, x], x, x^n], x] /; FreeQ[{a, b, c, n}, x] && IGtQ[p, 0]

Rule 4850

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_)/(x_), x_Symbol] :> Simp[2*(a + b*ArcTan[c*x])^p*ArcTanh[1 - 2/(1 +
I*c*x)], x] - Dist[2*b*c*p, Int[((a + b*ArcTan[c*x])^(p - 1)*ArcTanh[1 - 2/(1 + I*c*x)])/(1 + c^2*x^2), x], x
] /; FreeQ[{a, b, c}, x] && IGtQ[p, 1]

Rule 4988

Int[(ArcTanh[u_]*((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Dist[1/2, Int[(
Log[1 + u]*(a + b*ArcTan[c*x])^p)/(d + e*x^2), x], x] - Dist[1/2, Int[(Log[1 - u]*(a + b*ArcTan[c*x])^p)/(d +
e*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && EqQ[e, c^2*d] && EqQ[u^2 - (1 - (2*I)/(I - c*x))^
2, 0]

Rule 4884

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

Rule 4994

Int[(Log[u_]*((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> -Simp[(I*(a + b*Arc
Tan[c*x])^p*PolyLog[2, 1 - u])/(2*c*d), x] + Dist[(b*p*I)/2, Int[((a + b*ArcTan[c*x])^(p - 1)*PolyLog[2, 1 - u
])/(d + e*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && EqQ[e, c^2*d] && EqQ[(1 - u)^2 - (1 - (2*
I)/(I - c*x))^2, 0]

Rule 6610

Int[(u_)*PolyLog[n_, v_], x_Symbol] :> With[{w = DerivativeDivides[v, u*v, x]}, Simp[w*PolyLog[n + 1, v], x] /
;  !FalseQ[w]] /; FreeQ[n, x]

Rubi steps

\begin{align*} \int \frac{\left (a+b \tan ^{-1}\left (\frac{c}{x}\right )\right )^2}{x} \, dx &=-\operatorname{Subst}\left (\int \frac{\left (a+b \tan ^{-1}(c x)\right )^2}{x} \, dx,x,\frac{1}{x}\right )\\ &=-2 \left (a+b \cot ^{-1}\left (\frac{x}{c}\right )\right )^2 \tanh ^{-1}\left (1-\frac{2}{1+\frac{i c}{x}}\right )+(4 b c) \operatorname{Subst}\left (\int \frac{\left (a+b \tan ^{-1}(c x)\right ) \tanh ^{-1}\left (1-\frac{2}{1+i c x}\right )}{1+c^2 x^2} \, dx,x,\frac{1}{x}\right )\\ &=-2 \left (a+b \cot ^{-1}\left (\frac{x}{c}\right )\right )^2 \tanh ^{-1}\left (1-\frac{2}{1+\frac{i c}{x}}\right )-(2 b c) \operatorname{Subst}\left (\int \frac{\left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2}{1+i c x}\right )}{1+c^2 x^2} \, dx,x,\frac{1}{x}\right )+(2 b c) \operatorname{Subst}\left (\int \frac{\left (a+b \tan ^{-1}(c x)\right ) \log \left (2-\frac{2}{1+i c x}\right )}{1+c^2 x^2} \, dx,x,\frac{1}{x}\right )\\ &=-2 \left (a+b \cot ^{-1}\left (\frac{x}{c}\right )\right )^2 \tanh ^{-1}\left (1-\frac{2}{1+\frac{i c}{x}}\right )+i b \left (a+b \cot ^{-1}\left (\frac{x}{c}\right )\right ) \text{Li}_2\left (1-\frac{2}{1+\frac{i c}{x}}\right )-i b \left (a+b \cot ^{-1}\left (\frac{x}{c}\right )\right ) \text{Li}_2\left (-1+\frac{2}{1+\frac{i c}{x}}\right )-\left (i b^2 c\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2\left (1-\frac{2}{1+i c x}\right )}{1+c^2 x^2} \, dx,x,\frac{1}{x}\right )+\left (i b^2 c\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2\left (-1+\frac{2}{1+i c x}\right )}{1+c^2 x^2} \, dx,x,\frac{1}{x}\right )\\ &=-2 \left (a+b \cot ^{-1}\left (\frac{x}{c}\right )\right )^2 \tanh ^{-1}\left (1-\frac{2}{1+\frac{i c}{x}}\right )+i b \left (a+b \cot ^{-1}\left (\frac{x}{c}\right )\right ) \text{Li}_2\left (1-\frac{2}{1+\frac{i c}{x}}\right )-i b \left (a+b \cot ^{-1}\left (\frac{x}{c}\right )\right ) \text{Li}_2\left (-1+\frac{2}{1+\frac{i c}{x}}\right )+\frac{1}{2} b^2 \text{Li}_3\left (1-\frac{2}{1+\frac{i c}{x}}\right )-\frac{1}{2} b^2 \text{Li}_3\left (-1+\frac{2}{1+\frac{i c}{x}}\right )\\ \end{align*}

Mathematica [A]  time = 0.0853304, size = 148, normalized size = 1. \[ \frac{1}{2} b \left (2 i \text{PolyLog}\left (2,\frac{c+i x}{c-i x}\right ) \left (a+b \tan ^{-1}\left (\frac{c}{x}\right )\right )-2 i \text{PolyLog}\left (2,\frac{x-i c}{x+i c}\right ) \left (a+b \tan ^{-1}\left (\frac{c}{x}\right )\right )+b \left (\text{PolyLog}\left (3,\frac{c+i x}{c-i x}\right )-\text{PolyLog}\left (3,\frac{x-i c}{x+i c}\right )\right )\right )-2 \tanh ^{-1}\left (\frac{c+i x}{c-i x}\right ) \left (a+b \tan ^{-1}\left (\frac{c}{x}\right )\right )^2 \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*ArcTan[c/x])^2/x,x]

[Out]

-2*(a + b*ArcTan[c/x])^2*ArcTanh[(c + I*x)/(c - I*x)] + (b*((2*I)*(a + b*ArcTan[c/x])*PolyLog[2, (c + I*x)/(c
- I*x)] - (2*I)*(a + b*ArcTan[c/x])*PolyLog[2, ((-I)*c + x)/(I*c + x)] + b*(PolyLog[3, (c + I*x)/(c - I*x)] -
PolyLog[3, ((-I)*c + x)/(I*c + x)])))/2

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Maple [C]  time = 0.394, size = 1249, normalized size = 8.4 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-b^2*ln(c/x)*arctan(c/x)^2+b^2*arctan(c/x)^2*ln((1+I*c/x)^2/(1+c^2/x^2)-1)-b^2*arctan(c/x)^2*ln(1-(1+I*c/x)/(1
+c^2/x^2)^(1/2))-b^2*arctan(c/x)^2*ln(1+(1+I*c/x)/(1+c^2/x^2)^(1/2))+1/2*I*b^2*Pi*csgn(I/((1+I*c/x)^2/(1+c^2/x
^2)+1))*csgn(I*((1+I*c/x)^2/(1+c^2/x^2)-1)/((1+I*c/x)^2/(1+c^2/x^2)+1))^2*arctan(c/x)^2+1/2*I*b^2*Pi*csgn(I*((
1+I*c/x)^2/(1+c^2/x^2)-1)/((1+I*c/x)^2/(1+c^2/x^2)+1))*csgn(((1+I*c/x)^2/(1+c^2/x^2)-1)/((1+I*c/x)^2/(1+c^2/x^
2)+1))^2*arctan(c/x)^2+1/2*b^2*polylog(3,-(1+I*c/x)^2/(1+c^2/x^2))-1/2*I*b^2*Pi*csgn(I*((1+I*c/x)^2/(1+c^2/x^2
)-1))*csgn(I/((1+I*c/x)^2/(1+c^2/x^2)+1))*csgn(I*((1+I*c/x)^2/(1+c^2/x^2)-1)/((1+I*c/x)^2/(1+c^2/x^2)+1))*arct
an(c/x)^2-1/2*I*b^2*Pi*arctan(c/x)^2-I*a*b*dilog(1+I*c/x)+I*a*b*dilog(1-I*c/x)-2*a*b*ln(c/x)*arctan(c/x)-I*b^2
*arctan(c/x)*polylog(2,-(1+I*c/x)^2/(1+c^2/x^2))+2*I*b^2*arctan(c/x)*polylog(2,(1+I*c/x)/(1+c^2/x^2)^(1/2))+2*
I*b^2*arctan(c/x)*polylog(2,-(1+I*c/x)/(1+c^2/x^2)^(1/2))-I*a*b*ln(c/x)*ln(1+I*c/x)+I*a*b*ln(c/x)*ln(1-I*c/x)-
1/2*I*b^2*Pi*csgn(I*((1+I*c/x)^2/(1+c^2/x^2)-1)/((1+I*c/x)^2/(1+c^2/x^2)+1))^3*arctan(c/x)^2+1/2*I*b^2*Pi*csgn
(((1+I*c/x)^2/(1+c^2/x^2)-1)/((1+I*c/x)^2/(1+c^2/x^2)+1))^2*arctan(c/x)^2-1/2*I*b^2*Pi*csgn(((1+I*c/x)^2/(1+c^
2/x^2)-1)/((1+I*c/x)^2/(1+c^2/x^2)+1))^3*arctan(c/x)^2-a^2*ln(c/x)-2*b^2*polylog(3,(1+I*c/x)/(1+c^2/x^2)^(1/2)
)-2*b^2*polylog(3,-(1+I*c/x)/(1+c^2/x^2)^(1/2))+1/2*I*b^2*Pi*csgn(I*((1+I*c/x)^2/(1+c^2/x^2)-1))*csgn(I*((1+I*
c/x)^2/(1+c^2/x^2)-1)/((1+I*c/x)^2/(1+c^2/x^2)+1))^2*arctan(c/x)^2-1/2*I*b^2*Pi*csgn(I*((1+I*c/x)^2/(1+c^2/x^2
)-1)/((1+I*c/x)^2/(1+c^2/x^2)+1))*csgn(((1+I*c/x)^2/(1+c^2/x^2)-1)/((1+I*c/x)^2/(1+c^2/x^2)+1))*arctan(c/x)^2

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} a^{2} \log \left (x\right ) + \frac{1}{16} \, \int \frac{12 \, b^{2} \arctan \left (c, x\right )^{2} + b^{2} \log \left (c^{2} + x^{2}\right )^{2} + 32 \, a b \arctan \left (c, x\right )}{x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^2*log(x) + 1/16*integrate((12*b^2*arctan2(c, x)^2 + b^2*log(c^2 + x^2)^2 + 32*a*b*arctan2(c, x))/x, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{b^{2} \arctan \left (\frac{c}{x}\right )^{2} + 2 \, a b \arctan \left (\frac{c}{x}\right ) + a^{2}}{x}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b^2*arctan(c/x)^2 + 2*a*b*arctan(c/x) + a^2)/x, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b \operatorname{atan}{\left (\frac{c}{x} \right )}\right )^{2}}{x}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*atan(c/x))**2/x, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \arctan \left (\frac{c}{x}\right ) + a\right )}^{2}}{x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*arctan(c/x) + a)^2/x, x)