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

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

[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)]

________________________________________________________________________________________

Rubi [A]  time = 0.494497, antiderivative size = 230, normalized size of antiderivative = 1., 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 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 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 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*}

Mathematica [A]  time = 0.207111, size = 219, normalized size = 0.95 \[ \frac{3}{4} i b \left (2 \text{PolyLog}\left (2,\frac{c+i x}{c-i x}\right ) \left (a+b \tan ^{-1}\left (\frac{c}{x}\right )\right )^2-2 \text{PolyLog}\left (2,\frac{x-i c}{x+i c}\right ) \left (a+b \tan ^{-1}\left (\frac{c}{x}\right )\right )^2+b \left (-2 i \text{PolyLog}\left (3,\frac{c+i x}{c-i x}\right ) \left (a+b \tan ^{-1}\left (\frac{c}{x}\right )\right )+2 i \text{PolyLog}\left (3,\frac{x-i c}{x+i c}\right ) \left (a+b \tan ^{-1}\left (\frac{c}{x}\right )\right )+b \left (\text{PolyLog}\left (4,\frac{x-i c}{x+i c}\right )-\text{PolyLog}\left (4,\frac{c+i x}{c-i x}\right )\right )\right )\right )-2 \tanh ^{-1}\left (\frac{c+i x}{c-i x}\right ) \left (a+b \tan ^{-1}\left (\frac{c}{x}\right )\right )^3 \]

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)])))

________________________________________________________________________________________

Maple [C]  time = 0.276, size = 2542, normalized size = 11.1 \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))^3/x,x)

[Out]

-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))+3/2*a*b^2*poly
log(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*l
n(1-(1+I*c/x)/(1+c^2/x^2)^(1/2))-6*b^3*arctan(c/x)*polylog(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)*polylog(3,-(1+I*c/x)/(1+c^2/x^2)^(1/2))+3/2*b^3*arctan(c/x)*
polylog(3,-(1+I*c/x)^2/(1+c^2/x^2))+3/4*I*b^3*polylog(4,-(1+I*c/x)^2/(1+c^2/x^2))-a^3*ln(c/x)-3/2*I*a*b^2*Pi*c
sgn(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*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)^3-3/2*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+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))*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+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)^2-1/2*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+1/2*I*b^3*P
i*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+6*I*a*b^2*arctan(c/x)*polylog(
2,(1+I*c/x)/(1+c^2/x^2)^(1/2))-3*I*a*b^2*arctan(c/x)*polylog(2,-(1+I*c/x)^2/(1+c^2/x^2))+6*I*a*b^2*arctan(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)^2-3/2*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))^3*arctan(c/
x)^3-1/2*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*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*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(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+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+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))^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
)^3-3*a^2*b*ln(c/x)*arctan(c/x)-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))-3/2*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))

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} a^{3} \log \left (x\right ) + \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} \end{align*}

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)

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\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 ) \end{align*}

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)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b \operatorname{atan}{\left (\frac{c}{x} \right )}\right )^{3}}{x}\, dx \end{align*}

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)

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \arctan \left (\frac{c}{x}\right ) + a\right )}^{3}}{x}\,{d x} \end{align*}

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)