3.151 \(\int \frac {(a+b \tan ^{-1}(\frac {c}{x}))^3}{x} \, dx\)

Optimal. Leaf size=230 \[ \frac {3}{2} b^2 \text {Li}_3\left (1-\frac {2}{\frac {i c}{x}+1}\right ) \left (a+b \cot ^{-1}\left (\frac {x}{c}\right )\right )-\frac {3}{2} b^2 \text {Li}_3\left (\frac {2}{\frac {i c}{x}+1}-1\right ) \left (a+b \cot ^{-1}\left (\frac {x}{c}\right )\right )+\frac {3}{2} i b \text {Li}_2\left (1-\frac {2}{\frac {i c}{x}+1}\right ) \left (a+b \cot ^{-1}\left (\frac {x}{c}\right )\right )^2-\frac {3}{2} i b \text {Li}_2\left (\frac {2}{\frac {i c}{x}+1}-1\right ) \left (a+b \cot ^{-1}\left (\frac {x}{c}\right )\right )^2-2 \tanh ^{-1}\left (1-\frac {2}{1+\frac {i c}{x}}\right ) \left (a+b \cot ^{-1}\left (\frac {x}{c}\right )\right )^3-\frac {3}{4} i b^3 \text {Li}_4\left (1-\frac {2}{\frac {i c}{x}+1}\right )+\frac {3}{4} i b^3 \text {Li}_4\left (\frac {2}{\frac {i c}{x}+1}-1\right ) \]

[Out]

2*(a+b*arccot(x/c))^3*arctanh(-1+2/(1+I*c/x))+3/2*I*b*(a+b*arccot(x/c))^2*polylog(2,1-2/(1+I*c/x))-3/2*I*b*(a+
b*arccot(x/c))^2*polylog(2,-1+2/(1+I*c/x))+3/2*b^2*(a+b*arccot(x/c))*polylog(3,1-2/(1+I*c/x))-3/2*b^2*(a+b*arc
cot(x/c))*polylog(3,-1+2/(1+I*c/x))-3/4*I*b^3*polylog(4,1-2/(1+I*c/x))+3/4*I*b^3*polylog(4,-1+2/(1+I*c/x))

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Rubi [A]  time = 0.49, antiderivative size = 230, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.438, Rules used = {5031, 4850, 4988, 4884, 4994, 4998, 6610} \[ \frac {3}{2} b^2 \text {PolyLog}\left (3,1-\frac {2}{1+\frac {i c}{x}}\right ) \left (a+b \cot ^{-1}\left (\frac {x}{c}\right )\right )-\frac {3}{2} b^2 \text {PolyLog}\left (3,-1+\frac {2}{1+\frac {i c}{x}}\right ) \left (a+b \cot ^{-1}\left (\frac {x}{c}\right )\right )+\frac {3}{2} i b \text {PolyLog}\left (2,1-\frac {2}{1+\frac {i c}{x}}\right ) \left (a+b \cot ^{-1}\left (\frac {x}{c}\right )\right )^2-\frac {3}{2} i b \text {PolyLog}\left (2,-1+\frac {2}{1+\frac {i c}{x}}\right ) \left (a+b \cot ^{-1}\left (\frac {x}{c}\right )\right )^2-\frac {3}{4} i b^3 \text {PolyLog}\left (4,1-\frac {2}{1+\frac {i c}{x}}\right )+\frac {3}{4} i b^3 \text {PolyLog}\left (4,-1+\frac {2}{1+\frac {i c}{x}}\right )-2 \tanh ^{-1}\left (1-\frac {2}{1+\frac {i c}{x}}\right ) \left (a+b \cot ^{-1}\left (\frac {x}{c}\right )\right )^3 \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTan[c/x])^3/x,x]

[Out]

-2*(a + b*ArcCot[x/c])^3*ArcTanh[1 - 2/(1 + (I*c)/x)] + ((3*I)/2)*b*(a + b*ArcCot[x/c])^2*PolyLog[2, 1 - 2/(1
+ (I*c)/x)] - ((3*I)/2)*b*(a + b*ArcCot[x/c])^2*PolyLog[2, -1 + 2/(1 + (I*c)/x)] + (3*b^2*(a + b*ArcCot[x/c])*
PolyLog[3, 1 - 2/(1 + (I*c)/x)])/2 - (3*b^2*(a + b*ArcCot[x/c])*PolyLog[3, -1 + 2/(1 + (I*c)/x)])/2 - ((3*I)/4
)*b^3*PolyLog[4, 1 - 2/(1 + (I*c)/x)] + ((3*I)/4)*b^3*PolyLog[4, -1 + 2/(1 + (I*c)/x)]

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 4998

Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*PolyLog[k_, u_])/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(I*(a
+ b*ArcTan[c*x])^p*PolyLog[k + 1, u])/(2*c*d), x] - Dist[(b*p*I)/2, Int[((a + b*ArcTan[c*x])^(p - 1)*PolyLog[k
+ 1, u])/(d + e*x^2), x], x] /; FreeQ[{a, b, c, d, e, k}, x] && IGtQ[p, 0] && EqQ[e, c^2*d] && EqQ[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 )^3}{x} \, dx &=-\operatorname {Subst}\left (\int \frac {\left (a+b \tan ^{-1}(c x)\right )^3}{x} \, dx,x,\frac {1}{x}\right )\\ &=-2 \left (a+b \cot ^{-1}\left (\frac {x}{c}\right )\right )^3 \tanh ^{-1}\left (1-\frac {2}{1+\frac {i c}{x}}\right )+(6 b c) \operatorname {Subst}\left (\int \frac {\left (a+b \tan ^{-1}(c x)\right )^2 \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 )^3 \tanh ^{-1}\left (1-\frac {2}{1+\frac {i c}{x}}\right )-(3 b c) \operatorname {Subst}\left (\int \frac {\left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx,x,\frac {1}{x}\right )+(3 b c) \operatorname {Subst}\left (\int \frac {\left (a+b \tan ^{-1}(c x)\right )^2 \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 )^3 \tanh ^{-1}\left (1-\frac {2}{1+\frac {i c}{x}}\right )+\frac {3}{2} i b \left (a+b \cot ^{-1}\left (\frac {x}{c}\right )\right )^2 \text {Li}_2\left (1-\frac {2}{1+\frac {i c}{x}}\right )-\frac {3}{2} i b \left (a+b \cot ^{-1}\left (\frac {x}{c}\right )\right )^2 \text {Li}_2\left (-1+\frac {2}{1+\frac {i c}{x}}\right )-\left (3 i b^2 c\right ) \operatorname {Subst}\left (\int \frac {\left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx,x,\frac {1}{x}\right )+\left (3 i b^2 c\right ) \operatorname {Subst}\left (\int \frac {\left (a+b \tan ^{-1}(c x)\right ) \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 )^3 \tanh ^{-1}\left (1-\frac {2}{1+\frac {i c}{x}}\right )+\frac {3}{2} i b \left (a+b \cot ^{-1}\left (\frac {x}{c}\right )\right )^2 \text {Li}_2\left (1-\frac {2}{1+\frac {i c}{x}}\right )-\frac {3}{2} i b \left (a+b \cot ^{-1}\left (\frac {x}{c}\right )\right )^2 \text {Li}_2\left (-1+\frac {2}{1+\frac {i c}{x}}\right )+\frac {3}{2} b^2 \left (a+b \cot ^{-1}\left (\frac {x}{c}\right )\right ) \text {Li}_3\left (1-\frac {2}{1+\frac {i c}{x}}\right )-\frac {3}{2} b^2 \left (a+b \cot ^{-1}\left (\frac {x}{c}\right )\right ) \text {Li}_3\left (-1+\frac {2}{1+\frac {i c}{x}}\right )-\frac {1}{2} \left (3 b^3 c\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_3\left (1-\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx,x,\frac {1}{x}\right )+\frac {1}{2} \left (3 b^3 c\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_3\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 )^3 \tanh ^{-1}\left (1-\frac {2}{1+\frac {i c}{x}}\right )+\frac {3}{2} i b \left (a+b \cot ^{-1}\left (\frac {x}{c}\right )\right )^2 \text {Li}_2\left (1-\frac {2}{1+\frac {i c}{x}}\right )-\frac {3}{2} i b \left (a+b \cot ^{-1}\left (\frac {x}{c}\right )\right )^2 \text {Li}_2\left (-1+\frac {2}{1+\frac {i c}{x}}\right )+\frac {3}{2} b^2 \left (a+b \cot ^{-1}\left (\frac {x}{c}\right )\right ) \text {Li}_3\left (1-\frac {2}{1+\frac {i c}{x}}\right )-\frac {3}{2} b^2 \left (a+b \cot ^{-1}\left (\frac {x}{c}\right )\right ) \text {Li}_3\left (-1+\frac {2}{1+\frac {i c}{x}}\right )-\frac {3}{4} i b^3 \text {Li}_4\left (1-\frac {2}{1+\frac {i c}{x}}\right )+\frac {3}{4} i b^3 \text {Li}_4\left (-1+\frac {2}{1+\frac {i c}{x}}\right )\\ \end {align*}

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Mathematica [A]  time = 0.22, size = 219, normalized size = 0.95 \[ \frac {3}{4} i b \left (2 \text {Li}_2\left (\frac {c+i x}{c-i x}\right ) \left (a+b \tan ^{-1}\left (\frac {c}{x}\right )\right )^2-2 \text {Li}_2\left (\frac {x-i c}{i c+x}\right ) \left (a+b \tan ^{-1}\left (\frac {c}{x}\right )\right )^2+b \left (-2 i \text {Li}_3\left (\frac {c+i x}{c-i x}\right ) \left (a+b \tan ^{-1}\left (\frac {c}{x}\right )\right )+2 i \text {Li}_3\left (\frac {x-i c}{i c+x}\right ) \left (a+b \tan ^{-1}\left (\frac {c}{x}\right )\right )+b \left (\text {Li}_4\left (\frac {x-i c}{i c+x}\right )-\text {Li}_4\left (\frac {c+i x}{c-i x}\right )\right )\right )\right )-2 \tanh ^{-1}\left (\frac {c+i x}{c-i x}\right ) \left (a+b \tan ^{-1}\left (\frac {c}{x}\right )\right )^3 \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*ArcTan[c/x])^3/x,x]

[Out]

-2*(a + b*ArcTan[c/x])^3*ArcTanh[(c + I*x)/(c - I*x)] + ((3*I)/4)*b*(2*(a + b*ArcTan[c/x])^2*PolyLog[2, (c + I
*x)/(c - I*x)] - 2*(a + b*ArcTan[c/x])^2*PolyLog[2, ((-I)*c + x)/(I*c + x)] + b*((-2*I)*(a + b*ArcTan[c/x])*Po
lyLog[3, (c + I*x)/(c - I*x)] + (2*I)*(a + b*ArcTan[c/x])*PolyLog[3, ((-I)*c + x)/(I*c + x)] + b*(-PolyLog[4,
(c + I*x)/(c - I*x)] + PolyLog[4, ((-I)*c + x)/(I*c + x)])))

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fricas [F]  time = 0.41, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b^{3} \arctan \left (\frac {c}{x}\right )^{3} + 3 \, a b^{2} \arctan \left (\frac {c}{x}\right )^{2} + 3 \, a^{2} b \arctan \left (\frac {c}{x}\right ) + a^{3}}{x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctan(c/x))^3/x,x, algorithm="fricas")

[Out]

integral((b^3*arctan(c/x)^3 + 3*a*b^2*arctan(c/x)^2 + 3*a^2*b*arctan(c/x) + a^3)/x, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \arctan \left (\frac {c}{x}\right ) + a\right )}^{3}}{x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctan(c/x))^3/x,x, algorithm="giac")

[Out]

integrate((b*arctan(c/x) + a)^3/x, x)

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maple [C]  time = 0.24, size = 2542, normalized size = 11.05 \[ \text {Expression too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctan(c/x))^3/x,x)

[Out]

-3/2*I*a*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)^2-3/2*I*a^2*b*dilog(1+I*c/x)+3/2*I*a^2*b*dilog(1-I*c/x)-
1/2*I*b^3*Pi*arctan(c/x)^3+3*I*b^3*arctan(c/x)^2*polylog(2,-(1+I*c/x)/(1+c^2/x^2)^(1/2))-3/2*I*b^3*arctan(c/x)
^2*polylog(2,-(1+I*c/x)^2/(1+c^2/x^2))+3*I*b^3*arctan(c/x)^2*polylog(2,(1+I*c/x)/(1+c^2/x^2)^(1/2))-3*a*b^2*ln
(c/x)*arctan(c/x)^2+3*a*b^2*arctan(c/x)^2*ln((1+I*c/x)^2/(1+c^2/x^2)-1)-3*a*b^2*arctan(c/x)^2*ln(1+(1+I*c/x)/(
1+c^2/x^2)^(1/2))-3*a*b^2*arctan(c/x)^2*ln(1-(1+I*c/x)/(1+c^2/x^2)^(1/2))-3*a^2*b*ln(c/x)*arctan(c/x)-a^3*ln(c
/x)-1/2*I*b^3*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)^3+3/2*I*a*b^2*Pi*csgn(I*((1+I*c/x)^2/(1+c^2/x^2)-1))*cs
gn(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+3/2*I*a*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+6*I*a*b^2*a
rctan(c/x)*polylog(2,-(1+I*c/x)/(1+c^2/x^2)^(1/2))-3/2*I*a*b^2*Pi*arctan(c/x)^2-3*I*a*b^2*arctan(c/x)*polylog(
2,-(1+I*c/x)^2/(1+c^2/x^2))-3/2*I*a^2*b*ln(c/x)*ln(1+I*c/x)+3/2*I*a^2*b*ln(c/x)*ln(1-I*c/x)+1/2*I*b^3*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)^3-1/2*I*b^3*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)^3-1/2*I*b^3*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)^3+6*I*a*b^2*arctan(c/x)*polylog(2,(1+I*c/x)/(1+c^2/x^2)^(1/2))+3/2*b^3*arct
an(c/x)*polylog(3,-(1+I*c/x)^2/(1+c^2/x^2))-3/2*I*a*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^3*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)^3+1/2*I*b^3*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)^3+1/2*I*b^3*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)^3+1/2*I*b^3*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)^3+3/2*I*a*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)^2-3/2*I*a*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*arc
tan(c/x)^2-6*I*b^3*polylog(4,(1+I*c/x)/(1+c^2/x^2)^(1/2))-6*I*b^3*polylog(4,-(1+I*c/x)/(1+c^2/x^2)^(1/2))+3/4*
I*b^3*polylog(4,-(1+I*c/x)^2/(1+c^2/x^2))+3/2*a*b^2*polylog(3,-(1+I*c/x)^2/(1+c^2/x^2))-6*a*b^2*polylog(3,-(1+
I*c/x)/(1+c^2/x^2)^(1/2))-6*a*b^2*polylog(3,(1+I*c/x)/(1+c^2/x^2)^(1/2))-b^3*ln(c/x)*arctan(c/x)^3+b^3*arctan(
c/x)^3*ln((1+I*c/x)^2/(1+c^2/x^2)-1)-b^3*arctan(c/x)^3*ln(1+(1+I*c/x)/(1+c^2/x^2)^(1/2))-6*b^3*arctan(c/x)*pol
ylog(3,-(1+I*c/x)/(1+c^2/x^2)^(1/2))-b^3*arctan(c/x)^3*ln(1-(1+I*c/x)/(1+c^2/x^2)^(1/2))-6*b^3*arctan(c/x)*pol
ylog(3,(1+I*c/x)/(1+c^2/x^2)^(1/2))+3/2*I*a*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)^2-3/2*I*a*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 = 0.00, size = 0, normalized size = 0.00 \[ a^{3} \log \relax (x) + \frac {1}{32} \, \int \frac {28 \, b^{3} \arctan \left (c, x\right )^{3} + 3 \, b^{3} \arctan \left (c, x\right ) \log \left (c^{2} + x^{2}\right )^{2} + 96 \, a b^{2} \arctan \left (c, x\right )^{2} + 96 \, a^{2} b \arctan \left (c, x\right )}{x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctan(c/x))^3/x,x, algorithm="maxima")

[Out]

a^3*log(x) + 1/32*integrate((28*b^3*arctan2(c, x)^3 + 3*b^3*arctan2(c, x)*log(c^2 + x^2)^2 + 96*a*b^2*arctan2(
c, x)^2 + 96*a^2*b*arctan2(c, x))/x, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (a+b\,\mathrm {atan}\left (\frac {c}{x}\right )\right )}^3}{x} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atan(c/x))^3/x,x)

[Out]

int((a + b*atan(c/x))^3/x, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a + b \operatorname {atan}{\left (\frac {c}{x} \right )}\right )^{3}}{x}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*atan(c/x))**3/x,x)

[Out]

Integral((a + b*atan(c/x))**3/x, x)

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