3.1253 \(\int \frac{(d+e x^2) (a+b \tan ^{-1}(c x))^2}{x^3} \, dx\)

Optimal. Leaf size=220 \[ -i b e \text{PolyLog}\left (2,1-\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )+i b e \text{PolyLog}\left (2,-1+\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )-\frac{1}{2} b^2 e \text{PolyLog}\left (3,1-\frac{2}{1+i c x}\right )+\frac{1}{2} b^2 e \text{PolyLog}\left (3,-1+\frac{2}{1+i c x}\right )-\frac{1}{2} c^2 d \left (a+b \tan ^{-1}(c x)\right )^2-\frac{d \left (a+b \tan ^{-1}(c x)\right )^2}{2 x^2}-\frac{b c d \left (a+b \tan ^{-1}(c x)\right )}{x}+2 e \tanh ^{-1}\left (1-\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2-\frac{1}{2} b^2 c^2 d \log \left (c^2 x^2+1\right )+b^2 c^2 d \log (x) \]

[Out]

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

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Rubi [A]  time = 0.460768, antiderivative size = 220, normalized size of antiderivative = 1., number of steps used = 16, number of rules used = 12, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.571, Rules used = {4980, 4852, 4918, 266, 36, 29, 31, 4884, 4850, 4988, 4994, 6610} \[ -i b e \text{PolyLog}\left (2,1-\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )+i b e \text{PolyLog}\left (2,-1+\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )-\frac{1}{2} b^2 e \text{PolyLog}\left (3,1-\frac{2}{1+i c x}\right )+\frac{1}{2} b^2 e \text{PolyLog}\left (3,-1+\frac{2}{1+i c x}\right )-\frac{1}{2} c^2 d \left (a+b \tan ^{-1}(c x)\right )^2-\frac{d \left (a+b \tan ^{-1}(c x)\right )^2}{2 x^2}-\frac{b c d \left (a+b \tan ^{-1}(c x)\right )}{x}+2 e \tanh ^{-1}\left (1-\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2-\frac{1}{2} b^2 c^2 d \log \left (c^2 x^2+1\right )+b^2 c^2 d \log (x) \]

Antiderivative was successfully verified.

[In]

Int[((d + e*x^2)*(a + b*ArcTan[c*x])^2)/x^3,x]

[Out]

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

Rule 4980

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> With
[{u = ExpandIntegrand[(a + b*ArcTan[c*x])^p, (f*x)^m*(d + e*x^2)^q, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b,
c, d, e, f, m}, x] && IntegerQ[q] && IGtQ[p, 0] && ((EqQ[p, 1] && GtQ[q, 0]) || IntegerQ[m])

Rule 4852

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcTa
n[c*x])^p)/(d*(m + 1)), x] - Dist[(b*c*p)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcTan[c*x])^(p - 1))/(1 + c^
2*x^2), x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[p, 0] && (EqQ[p, 1] || IntegerQ[m]) && NeQ[m, -1]

Rule 4918

Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Dist[1/d,
Int[(f*x)^m*(a + b*ArcTan[c*x])^p, x], x] - Dist[e/(d*f^2), Int[((f*x)^(m + 2)*(a + b*ArcTan[c*x])^p)/(d + e*
x^2), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && GtQ[p, 0] && LtQ[m, -1]

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

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 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 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 (d+e x^2\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{x^3} \, dx &=\int \left (\frac{d \left (a+b \tan ^{-1}(c x)\right )^2}{x^3}+\frac{e \left (a+b \tan ^{-1}(c x)\right )^2}{x}\right ) \, dx\\ &=d \int \frac{\left (a+b \tan ^{-1}(c x)\right )^2}{x^3} \, dx+e \int \frac{\left (a+b \tan ^{-1}(c x)\right )^2}{x} \, dx\\ &=-\frac{d \left (a+b \tan ^{-1}(c x)\right )^2}{2 x^2}+2 e \left (a+b \tan ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac{2}{1+i c x}\right )+(b c d) \int \frac{a+b \tan ^{-1}(c x)}{x^2 \left (1+c^2 x^2\right )} \, dx-(4 b c e) \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\\ &=-\frac{d \left (a+b \tan ^{-1}(c x)\right )^2}{2 x^2}+2 e \left (a+b \tan ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac{2}{1+i c x}\right )+(b c d) \int \frac{a+b \tan ^{-1}(c x)}{x^2} \, dx-\left (b c^3 d\right ) \int \frac{a+b \tan ^{-1}(c x)}{1+c^2 x^2} \, dx+(2 b c e) \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-(2 b c e) \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\\ &=-\frac{b c d \left (a+b \tan ^{-1}(c x)\right )}{x}-\frac{1}{2} c^2 d \left (a+b \tan ^{-1}(c x)\right )^2-\frac{d \left (a+b \tan ^{-1}(c x)\right )^2}{2 x^2}+2 e \left (a+b \tan ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac{2}{1+i c x}\right )-i b e \left (a+b \tan ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2}{1+i c x}\right )+i b e \left (a+b \tan ^{-1}(c x)\right ) \text{Li}_2\left (-1+\frac{2}{1+i c x}\right )+\left (b^2 c^2 d\right ) \int \frac{1}{x \left (1+c^2 x^2\right )} \, dx+\left (i b^2 c e\right ) \int \frac{\text{Li}_2\left (1-\frac{2}{1+i c x}\right )}{1+c^2 x^2} \, dx-\left (i b^2 c e\right ) \int \frac{\text{Li}_2\left (-1+\frac{2}{1+i c x}\right )}{1+c^2 x^2} \, dx\\ &=-\frac{b c d \left (a+b \tan ^{-1}(c x)\right )}{x}-\frac{1}{2} c^2 d \left (a+b \tan ^{-1}(c x)\right )^2-\frac{d \left (a+b \tan ^{-1}(c x)\right )^2}{2 x^2}+2 e \left (a+b \tan ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac{2}{1+i c x}\right )-i b e \left (a+b \tan ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2}{1+i c x}\right )+i b e \left (a+b \tan ^{-1}(c x)\right ) \text{Li}_2\left (-1+\frac{2}{1+i c x}\right )-\frac{1}{2} b^2 e \text{Li}_3\left (1-\frac{2}{1+i c x}\right )+\frac{1}{2} b^2 e \text{Li}_3\left (-1+\frac{2}{1+i c x}\right )+\frac{1}{2} \left (b^2 c^2 d\right ) \operatorname{Subst}\left (\int \frac{1}{x \left (1+c^2 x\right )} \, dx,x,x^2\right )\\ &=-\frac{b c d \left (a+b \tan ^{-1}(c x)\right )}{x}-\frac{1}{2} c^2 d \left (a+b \tan ^{-1}(c x)\right )^2-\frac{d \left (a+b \tan ^{-1}(c x)\right )^2}{2 x^2}+2 e \left (a+b \tan ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac{2}{1+i c x}\right )-i b e \left (a+b \tan ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2}{1+i c x}\right )+i b e \left (a+b \tan ^{-1}(c x)\right ) \text{Li}_2\left (-1+\frac{2}{1+i c x}\right )-\frac{1}{2} b^2 e \text{Li}_3\left (1-\frac{2}{1+i c x}\right )+\frac{1}{2} b^2 e \text{Li}_3\left (-1+\frac{2}{1+i c x}\right )+\frac{1}{2} \left (b^2 c^2 d\right ) \operatorname{Subst}\left (\int \frac{1}{x} \, dx,x,x^2\right )-\frac{1}{2} \left (b^2 c^4 d\right ) \operatorname{Subst}\left (\int \frac{1}{1+c^2 x} \, dx,x,x^2\right )\\ &=-\frac{b c d \left (a+b \tan ^{-1}(c x)\right )}{x}-\frac{1}{2} c^2 d \left (a+b \tan ^{-1}(c x)\right )^2-\frac{d \left (a+b \tan ^{-1}(c x)\right )^2}{2 x^2}+2 e \left (a+b \tan ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac{2}{1+i c x}\right )+b^2 c^2 d \log (x)-\frac{1}{2} b^2 c^2 d \log \left (1+c^2 x^2\right )-i b e \left (a+b \tan ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2}{1+i c x}\right )+i b e \left (a+b \tan ^{-1}(c x)\right ) \text{Li}_2\left (-1+\frac{2}{1+i c x}\right )-\frac{1}{2} b^2 e \text{Li}_3\left (1-\frac{2}{1+i c x}\right )+\frac{1}{2} b^2 e \text{Li}_3\left (-1+\frac{2}{1+i c x}\right )\\ \end{align*}

Mathematica [A]  time = 0.34584, size = 273, normalized size = 1.24 \[ i a b e (\text{PolyLog}(2,-i c x)-\text{PolyLog}(2,i c x))+\frac{1}{24} b^2 e \left (24 i \tan ^{-1}(c x) \text{PolyLog}\left (2,e^{-2 i \tan ^{-1}(c x)}\right )+24 i \tan ^{-1}(c x) \text{PolyLog}\left (2,-e^{2 i \tan ^{-1}(c x)}\right )+12 \text{PolyLog}\left (3,e^{-2 i \tan ^{-1}(c x)}\right )-12 \text{PolyLog}\left (3,-e^{2 i \tan ^{-1}(c x)}\right )+16 i \tan ^{-1}(c x)^3+24 \tan ^{-1}(c x)^2 \log \left (1-e^{-2 i \tan ^{-1}(c x)}\right )-24 \tan ^{-1}(c x)^2 \log \left (1+e^{2 i \tan ^{-1}(c x)}\right )-i \pi ^3\right )-\frac{a^2 d}{2 x^2}+a^2 e \log (x)-\frac{a b d \left (\tan ^{-1}(c x)+c x \left (c x \tan ^{-1}(c x)+1\right )\right )}{x^2}-\frac{b^2 d \left (-2 c^2 x^2 \log \left (\frac{c x}{\sqrt{c^2 x^2+1}}\right )+\left (c^2 x^2+1\right ) \tan ^{-1}(c x)^2+2 c x \tan ^{-1}(c x)\right )}{2 x^2} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[((d + e*x^2)*(a + b*ArcTan[c*x])^2)/x^3,x]

[Out]

-(a^2*d)/(2*x^2) - (a*b*d*(ArcTan[c*x] + c*x*(1 + c*x*ArcTan[c*x])))/x^2 + a^2*e*Log[x] - (b^2*d*(2*c*x*ArcTan
[c*x] + (1 + c^2*x^2)*ArcTan[c*x]^2 - 2*c^2*x^2*Log[(c*x)/Sqrt[1 + c^2*x^2]]))/(2*x^2) + I*a*b*e*(PolyLog[2, (
-I)*c*x] - PolyLog[2, I*c*x]) + (b^2*e*((-I)*Pi^3 + (16*I)*ArcTan[c*x]^3 + 24*ArcTan[c*x]^2*Log[1 - E^((-2*I)*
ArcTan[c*x])] - 24*ArcTan[c*x]^2*Log[1 + E^((2*I)*ArcTan[c*x])] + (24*I)*ArcTan[c*x]*PolyLog[2, E^((-2*I)*ArcT
an[c*x])] + (24*I)*ArcTan[c*x]*PolyLog[2, -E^((2*I)*ArcTan[c*x])] + 12*PolyLog[3, E^((-2*I)*ArcTan[c*x])] - 12
*PolyLog[3, -E^((2*I)*ArcTan[c*x])]))/24

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Maple [C]  time = 4.318, size = 1313, normalized size = 6. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x^2+d)*(a+b*arctan(c*x))^2/x^3,x)

[Out]

1/2*I*b^2*e*Pi*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)-1))*csgn(I/((1+I*c*x)^2/(c^2*x^2+1)+1))*csgn(I*((1+I*c*x)^2/(c^
2*x^2+1)-1)/((1+I*c*x)^2/(c^2*x^2+1)+1))*arctan(c*x)^2+I*a*b*e*ln(c*x)*ln(1+I*c*x)-I*a*b*e*ln(c*x)*ln(1-I*c*x)
+1/2*I*b^2*e*Pi*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)-1)/((1+I*c*x)^2/(c^2*x^2+1)+1))^3*arctan(c*x)^2-1/2*I*b^2*e*Pi
*csgn(((1+I*c*x)^2/(c^2*x^2+1)-1)/((1+I*c*x)^2/(c^2*x^2+1)+1))^2*arctan(c*x)^2+1/2*I*b^2*e*Pi*csgn(((1+I*c*x)^
2/(c^2*x^2+1)-1)/((1+I*c*x)^2/(c^2*x^2+1)+1))^3*arctan(c*x)^2-1/2*a^2*d/x^2-1/2*c^2*b^2*arctan(c*x)^2*d+c^2*b^
2*d*ln((1+I*c*x)/(c^2*x^2+1)^(1/2)-1)+c^2*b^2*d*ln(1+(1+I*c*x)/(c^2*x^2+1)^(1/2))-b^2*e*arctan(c*x)^2*ln((1+I*
c*x)^2/(c^2*x^2+1)-1)+b^2*e*arctan(c*x)^2*ln(1-(1+I*c*x)/(c^2*x^2+1)^(1/2))+b^2*e*arctan(c*x)^2*ln(1+(1+I*c*x)
/(c^2*x^2+1)^(1/2))+b^2*arctan(c*x)^2*e*ln(c*x)-1/2*b^2*arctan(c*x)^2*d/x^2-1/2*I*b^2*e*Pi*csgn(I*((1+I*c*x)^2
/(c^2*x^2+1)-1)/((1+I*c*x)^2/(c^2*x^2+1)+1))*csgn(((1+I*c*x)^2/(c^2*x^2+1)-1)/((1+I*c*x)^2/(c^2*x^2+1)+1))^2*a
rctan(c*x)^2-1/2*I*b^2*e*Pi*csgn(I/((1+I*c*x)^2/(c^2*x^2+1)+1))*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)-1)/((1+I*c*x)^
2/(c^2*x^2+1)+1))^2*arctan(c*x)^2+1/2*I*b^2*e*Pi*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)-1)/((1+I*c*x)^2/(c^2*x^2+1)+1
))*csgn(((1+I*c*x)^2/(c^2*x^2+1)-1)/((1+I*c*x)^2/(c^2*x^2+1)+1))*arctan(c*x)^2-1/2*I*b^2*e*Pi*csgn(I*((1+I*c*x
)^2/(c^2*x^2+1)-1))*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)-1)/((1+I*c*x)^2/(c^2*x^2+1)+1))^2*arctan(c*x)^2+a^2*e*ln(c
*x)-1/2*b^2*e*polylog(3,-(1+I*c*x)^2/(c^2*x^2+1))+2*b^2*e*polylog(3,(1+I*c*x)/(c^2*x^2+1)^(1/2))+2*b^2*e*polyl
og(3,-(1+I*c*x)/(c^2*x^2+1)^(1/2))-c*a*b*d/x-a*b*arctan(c*x)*d/x^2+I*a*b*e*dilog(1+I*c*x)+I*b^2*e*arctan(c*x)*
polylog(2,-(1+I*c*x)^2/(c^2*x^2+1))+2*a*b*arctan(c*x)*e*ln(c*x)-c^2*a*b*arctan(c*x)*d-c*b^2*arctan(c*x)*d/x-2*
I*b^2*e*arctan(c*x)*polylog(2,(1+I*c*x)/(c^2*x^2+1)^(1/2))-2*I*b^2*e*arctan(c*x)*polylog(2,-(1+I*c*x)/(c^2*x^2
+1)^(1/2))+1/2*I*b^2*e*Pi*arctan(c*x)^2-I*a*b*e*dilog(1-I*c*x)-I*c^2*b^2*arctan(c*x)*d

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)*(a+b*arctan(c*x))^2/x^3,x, algorithm="maxima")

[Out]

-((c*arctan(c*x) + 1/x)*c + arctan(c*x)/x^2)*a*b*d + a^2*e*log(x) - 1/2*a^2*d/x^2 - 1/96*(12*b^2*d*arctan(c*x)
^2 - 3*b^2*d*log(c^2*x^2 + 1)^2 - (1152*b^2*c^2*e*integrate(1/16*x^4*arctan(c*x)^2/(c^2*x^5 + x^3), x) + 3072*
a*b*c^2*e*integrate(1/16*x^4*arctan(c*x)/(c^2*x^5 + x^3), x) + 1152*b^2*c^2*d*integrate(1/16*x^2*arctan(c*x)^2
/(c^2*x^5 + x^3), x) + 96*b^2*c^2*d*integrate(1/16*x^2*log(c^2*x^2 + 1)^2/(c^2*x^5 + x^3), x) - 192*b^2*c^2*d*
integrate(1/16*x^2*log(c^2*x^2 + 1)/(c^2*x^5 + x^3), x) + b^2*e*log(c^2*x^2 + 1)^3 + 384*b^2*c*d*integrate(1/1
6*x*arctan(c*x)/(c^2*x^5 + x^3), x) + 1152*b^2*e*integrate(1/16*x^2*arctan(c*x)^2/(c^2*x^5 + x^3), x) + 96*b^2
*e*integrate(1/16*x^2*log(c^2*x^2 + 1)^2/(c^2*x^5 + x^3), x) + 3072*a*b*e*integrate(1/16*x^2*arctan(c*x)/(c^2*
x^5 + x^3), x) + 1152*b^2*d*integrate(1/16*arctan(c*x)^2/(c^2*x^5 + x^3), x) + 96*b^2*d*integrate(1/16*log(c^2
*x^2 + 1)^2/(c^2*x^5 + x^3), x))*x^2)/x^2

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{a^{2} e x^{2} + a^{2} d +{\left (b^{2} e x^{2} + b^{2} d\right )} \arctan \left (c x\right )^{2} + 2 \,{\left (a b e x^{2} + a b d\right )} \arctan \left (c x\right )}{x^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)*(a+b*arctan(c*x))^2/x^3,x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b \operatorname{atan}{\left (c x \right )}\right )^{2} \left (d + e x^{2}\right )}{x^{3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x**2+d)*(a+b*atan(c*x))**2/x**3,x)

[Out]

Integral((a + b*atan(c*x))**2*(d + e*x**2)/x**3, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (e x^{2} + d\right )}{\left (b \arctan \left (c x\right ) + a\right )}^{2}}{x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)*(a+b*arctan(c*x))^2/x^3,x, algorithm="giac")

[Out]

integrate((e*x^2 + d)*(b*arctan(c*x) + a)^2/x^3, x)