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

Optimal. Leaf size=148 \[ i b \text {Li}_2\left (1-\frac {2}{\frac {i c}{x}+1}\right ) \left (a+b \cot ^{-1}\left (\frac {x}{c}\right )\right )-i b \text {Li}_2\left (\frac {2}{\frac {i c}{x}+1}-1\right ) \left (a+b \cot ^{-1}\left (\frac {x}{c}\right )\right )-2 \tanh ^{-1}\left (1-\frac {2}{1+\frac {i c}{x}}\right ) \left (a+b \cot ^{-1}\left (\frac {x}{c}\right )\right )^2+\frac {1}{2} b^2 \text {Li}_3\left (1-\frac {2}{\frac {i c}{x}+1}\right )-\frac {1}{2} b^2 \text {Li}_3\left (\frac {2}{\frac {i c}{x}+1}-1\right ) \]

[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))+1/2*b^2*polylog(3,1-2/(1+I*c/x))-1/2*b^2*polylog(3,-1+2/(1+I*c/x))

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Rubi [A]  time = 0.29, antiderivative size = 148, normalized size of antiderivative = 1.00, 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 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 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 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 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 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 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*}

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Mathematica [A]  time = 0.09, size = 148, normalized size = 1.00 \[ \frac {1}{2} b \left (2 i \text {Li}_2\left (\frac {c+i x}{c-i x}\right ) \left (a+b \tan ^{-1}\left (\frac {c}{x}\right )\right )-2 i \text {Li}_2\left (\frac {x-i c}{i c+x}\right ) \left (a+b \tan ^{-1}\left (\frac {c}{x}\right )\right )+b \left (\text {Li}_3\left (\frac {c+i x}{c-i x}\right )-\text {Li}_3\left (\frac {x-i c}{i c+x}\right )\right )\right )-2 \tanh ^{-1}\left (\frac {c+i x}{c-i x}\right ) \left (a+b \tan ^{-1}\left (\frac {c}{x}\right )\right )^2 \]

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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fricas [F]  time = 0.56, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b^{2} \arctan \left (\frac {c}{x}\right )^{2} + 2 \, a b \arctan \left (\frac {c}{x}\right ) + a^{2}}{x}, x\right ) \]

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

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)

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maple [C]  time = 0.34, size = 1249, normalized size = 8.44 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-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-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))*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))*arctan(c/x)^2+1/2*b^2*polylog(3,-(1
+I*c/x)^2/(1+c^2/x^2))-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))-a^2*ln(c/x)-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+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-b^2*ln(c/x)*arctan(c/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)/((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+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-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))-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)+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))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ a^{2} \log \relax (x) + \frac {1}{16} \, \int \frac {12 \, b^{2} \arctan \left (c, x\right )^{2} + b^{2} \log \left (c^{2} + x^{2}\right )^{2} + 32 \, a b \arctan \left (c, x\right )}{x}\,{d x} \]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atan(c/x))^2/x,x)

[Out]

int((a + b*atan(c/x))^2/x, x)

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

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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