3.13.0 ∫ (d+ex2)(a+b arctan(cx))2 _____________________________________

Integrand size = 21, antiderivative size = 220 \[ \int \frac {\left (d+e x^2\right ) (a+b \arctan (c x))^2}{x^3} \, dx=-\frac {b c d (a+b \arctan (c x))}{x}-\frac {1}{2} c^2 d (a+b \arctan (c x))^2-\frac {d (a+b \arctan (c x))^2}{2 x^2}+2 e (a+b \arctan (c x))^2 \text {arctanh}\left (1-\frac {2}{1+i c x}\right )+b^2 c^2 d \log (x)-\frac {1}{2} b^2 c^2 d \log \left (1+c^2 x^2\right )-i b e (a+b \arctan (c x)) \operatorname {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )+i b e (a+b \arctan (c x)) \operatorname {PolyLog}\left (2,-1+\frac {2}{1+i c x}\right )-\frac {1}{2} b^2 e \operatorname {PolyLog}\left (3,1-\frac {2}{1+i c x}\right )+\frac {1}{2} b^2 e \operatorname {PolyLog}\left (3,-1+\frac {2}{1+i c x}\right ) \]

-b*c*d*(a+b*arctan(c*x))/x-1/2*c^2*d*(a+b*arctan(c*x))^2-1/2*d*(a+b*arctan(c*x))^2/x^2-2*e*(a+b*arctan(c*x))^2

*arctanh(-1+2/(1+I*c*x))+b^2*c^2*d*ln(x)-1/2*b^2*c^2*d*ln(c^2*x^2+1)-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))-1/2*b^2*e*polylog(3,1-2/(1+I*c*x))+1/2*b^2*e*polylog

Rubi [A] (veriﬁed)

Time = 0.32 (sec) , antiderivative size = 220, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {5100, 4946, 5038, 272, 36, 29, 31, 5004, 4942, 5108, 5114, 6745} \[ \int \frac {\left (d+e x^2\right ) (a+b \arctan (c x))^2}{x^3} \, dx=2 e \text {arctanh}\left (1-\frac {2}{1+i c x}\right ) (a+b \arctan (c x))^2-\frac {1}{2} c^2 d (a+b \arctan (c x))^2-\frac {d (a+b \arctan (c x))^2}{2 x^2}-\frac {b c d (a+b \arctan (c x))}{x}-i b e \operatorname {PolyLog}\left (2,1-\frac {2}{i c x+1}\right ) (a+b \arctan (c x))+i b e \operatorname {PolyLog}\left (2,\frac {2}{i c x+1}-1\right ) (a+b \arctan (c x))-\frac {1}{2} b^2 c^2 d \log \left (c^2 x^2+1\right )+b^2 c^2 d \log (x)-\frac {1}{2} b^2 e \operatorname {PolyLog}\left (3,1-\frac {2}{i c x+1}\right )+\frac {1}{2} b^2 e \operatorname {PolyLog}\left (3,\frac {2}{i c x+1}-1\right ) \]

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

-((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

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

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

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]

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]]

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

Int[((a_.) + ArcTan[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)*((a + b*ArcTan[c*x^

n])^p/(m + 1)), x] - Dist[b*c*n*(p/(m + 1)), Int[x^(m + n)*((a + b*ArcTan[c*x^n])^(p - 1)/(1 + c^2*x^(2*n))),

x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0] && (EqQ[p, 1] || (EqQ[n, 1] && IntegerQ[m])) && NeQ[m, -1]

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]

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]

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])

Int[(ArcTanh[u_]*((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Dist[1/2, Int[L

og[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)))^

Int[(Log[u_]*((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(-I)*(a + b*Ar

cTan[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

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

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {d (a+b \arctan (c x))^2}{x^3}+\frac {e (a+b \arctan (c x))^2}{x}\right ) \, dx \\ & = d \int \frac {(a+b \arctan (c x))^2}{x^3} \, dx+e \int \frac {(a+b \arctan (c x))^2}{x} \, dx \\ & = -\frac {d (a+b \arctan (c x))^2}{2 x^2}+2 e (a+b \arctan (c x))^2 \text {arctanh}\left (1-\frac {2}{1+i c x}\right )+(b c d) \int \frac {a+b \arctan (c x)}{x^2 \left (1+c^2 x^2\right )} \, dx-(4 b c e) \int \frac {(a+b \arctan (c x)) \text {arctanh}\left (1-\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx \\ & = -\frac {d (a+b \arctan (c x))^2}{2 x^2}+2 e (a+b \arctan (c x))^2 \text {arctanh}\left (1-\frac {2}{1+i c x}\right )+(b c d) \int \frac {a+b \arctan (c x)}{x^2} \, dx-\left (b c^3 d\right ) \int \frac {a+b \arctan (c x)}{1+c^2 x^2} \, dx+(2 b c e) \int \frac {(a+b \arctan (c x)) \log \left (\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx-(2 b c e) \int \frac {(a+b \arctan (c x)) \log \left (2-\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx \\ & = -\frac {b c d (a+b \arctan (c x))}{x}-\frac {1}{2} c^2 d (a+b \arctan (c x))^2-\frac {d (a+b \arctan (c x))^2}{2 x^2}+2 e (a+b \arctan (c x))^2 \text {arctanh}\left (1-\frac {2}{1+i c x}\right )-i b e (a+b \arctan (c x)) \operatorname {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )+i b e (a+b \arctan (c x)) \operatorname {PolyLog}\left (2,-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 {\operatorname {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx-\left (i b^2 c e\right ) \int \frac {\operatorname {PolyLog}\left (2,-1+\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx \\ & = -\frac {b c d (a+b \arctan (c x))}{x}-\frac {1}{2} c^2 d (a+b \arctan (c x))^2-\frac {d (a+b \arctan (c x))^2}{2 x^2}+2 e (a+b \arctan (c x))^2 \text {arctanh}\left (1-\frac {2}{1+i c x}\right )-i b e (a+b \arctan (c x)) \operatorname {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )+i b e (a+b \arctan (c x)) \operatorname {PolyLog}\left (2,-1+\frac {2}{1+i c x}\right )-\frac {1}{2} b^2 e \operatorname {PolyLog}\left (3,1-\frac {2}{1+i c x}\right )+\frac {1}{2} b^2 e \operatorname {PolyLog}\left (3,-1+\frac {2}{1+i c x}\right )+\frac {1}{2} \left (b^2 c^2 d\right ) \text {Subst}\left (\int \frac {1}{x \left (1+c^2 x\right )} \, dx,x,x^2\right ) \\ & = -\frac {b c d (a+b \arctan (c x))}{x}-\frac {1}{2} c^2 d (a+b \arctan (c x))^2-\frac {d (a+b \arctan (c x))^2}{2 x^2}+2 e (a+b \arctan (c x))^2 \text {arctanh}\left (1-\frac {2}{1+i c x}\right )-i b e (a+b \arctan (c x)) \operatorname {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )+i b e (a+b \arctan (c x)) \operatorname {PolyLog}\left (2,-1+\frac {2}{1+i c x}\right )-\frac {1}{2} b^2 e \operatorname {PolyLog}\left (3,1-\frac {2}{1+i c x}\right )+\frac {1}{2} b^2 e \operatorname {PolyLog}\left (3,-1+\frac {2}{1+i c x}\right )+\frac {1}{2} \left (b^2 c^2 d\right ) \text {Subst}\left (\int \frac {1}{x} \, dx,x,x^2\right )-\frac {1}{2} \left (b^2 c^4 d\right ) \text {Subst}\left (\int \frac {1}{1+c^2 x} \, dx,x,x^2\right ) \\ & = -\frac {b c d (a+b \arctan (c x))}{x}-\frac {1}{2} c^2 d (a+b \arctan (c x))^2-\frac {d (a+b \arctan (c x))^2}{2 x^2}+2 e (a+b \arctan (c x))^2 \text {arctanh}\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 (a+b \arctan (c x)) \operatorname {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )+i b e (a+b \arctan (c x)) \operatorname {PolyLog}\left (2,-1+\frac {2}{1+i c x}\right )-\frac {1}{2} b^2 e \operatorname {PolyLog}\left (3,1-\frac {2}{1+i c x}\right )+\frac {1}{2} b^2 e \operatorname {PolyLog}\left (3,-1+\frac {2}{1+i c x}\right ) \\ \end{align*}

Mathematica [A] (veriﬁed)

Time = 0.25 (sec) , antiderivative size = 273, normalized size of antiderivative = 1.24 \[ \int \frac {\left (d+e x^2\right ) (a+b \arctan (c x))^2}{x^3} \, dx=-\frac {a^2 d}{2 x^2}-\frac {a b d (\arctan (c x)+c x (1+c x \arctan (c x)))}{x^2}+a^2 e \log (x)-\frac {b^2 d \left (2 c x \arctan (c x)+\left (1+c^2 x^2\right ) \arctan (c x)^2-2 c^2 x^2 \log \left (\frac {c x}{\sqrt {1+c^2 x^2}}\right )\right )}{2 x^2}+i a b e (\operatorname {PolyLog}(2,-i c x)-\operatorname {PolyLog}(2,i c x))+\frac {1}{24} b^2 e \left (-i \pi ^3+16 i \arctan (c x)^3+24 \arctan (c x)^2 \log \left (1-e^{-2 i \arctan (c x)}\right )-24 \arctan (c x)^2 \log \left (1+e^{2 i \arctan (c x)}\right )+24 i \arctan (c x) \operatorname {PolyLog}\left (2,e^{-2 i \arctan (c x)}\right )+24 i \arctan (c x) \operatorname {PolyLog}\left (2,-e^{2 i \arctan (c x)}\right )+12 \operatorname {PolyLog}\left (3,e^{-2 i \arctan (c x)}\right )-12 \operatorname {PolyLog}\left (3,-e^{2 i \arctan (c x)}\right )\right ) \]

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

-1/2*(a^2*d)/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

Maple [C] (warning: unable to verify)

Time = 10.21 (sec) , antiderivative size = 1289, normalized size of antiderivative = 5.86


method	result	size

derivativedivides	\(\text {Expression too large to display}\)	\(1289\)
default	\(\text {Expression too large to display}\)	\(1289\)
parts	\(\text {Expression too large to display}\)	\(1318\)

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

c^2*(a^2/c^2*e*ln(c*x)-1/2*a^2*d/c^2/x^2+b^2/c^2*(-1/2*I*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+arctan(c*x)^2*e*ln(c*x)+d*c^2*ln(1+(1+I*c*

x)/(c^2*x^2+1)^(1/2))+d*c^2*ln((1+I*c*x)/(c^2*x^2+1)^(1/2)-1)+1/2*I*e*Pi*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)-1))*c

sgn(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)

^2-1/2*c*d*arctan(c*x)*(I*c*x-(c^2*x^2+1)^(1/2)+1)/x-2*I*e*arctan(c*x)*polylog(2,-(1+I*c*x)/(c^2*x^2+1)^(1/2))

+1/2*I*e*Pi*arctan(c*x)^2-2*I*e*arctan(c*x)*polylog(2,(1+I*c*x)/(c^2*x^2+1)^(1/2))+I*e*arctan(c*x)*polylog(2,-

(1+I*c*x)^2/(c^2*x^2+1))-1/2*I*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*c*d*arctan(c*x)*(I*c*x+(c^2*x^2+1)^(1/2)+1)/x-1/2*I*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*arctan(c*x)^2+1/2*I*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+e*arctan(c*x)^2*ln(1-(1+I*c*x)/(c^2*x^2+1)^(

1/2))-e*ln((1+I*c*x)^2/(c^2*x^2+1)-1)*arctan(c*x)^2+e*arctan(c*x)^2*ln(1+(1+I*c*x)/(c^2*x^2+1)^(1/2))-1/2*e*po

lylog(3,-(1+I*c*x)^2/(c^2*x^2+1))+2*e*polylog(3,-(1+I*c*x)/(c^2*x^2+1)^(1/2))+2*e*polylog(3,(1+I*c*x)/(c^2*x^2

+1)^(1/2))-1/2*I*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*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*arctan(c*x)^2*d/x^2+1/2*I

*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*arctan(c*x)^2*c^2*d)+2

*a*b/c^2*(arctan(c*x)*e*ln(c*x)-1/2*arctan(c*x)*d/x^2+1/2*I*e*ln(c*x)*ln(1+I*c*x)-1/2*I*e*ln(c*x)*ln(1-I*c*x)+

1/2*I*e*dilog(1+I*c*x)-1/2*I*e*dilog(1-I*c*x)+1/2*d*c^2*(-1/c/x-arctan(c*x))))

Fricas [F]

\[ \int \frac {\left (d+e x^2\right ) (a+b \arctan (c x))^2}{x^3} \, dx=\int { \frac {{\left (e x^{2} + d\right )} {\left (b \arctan \left (c x\right ) + a\right )}^{2}}{x^{3}} \,d x } \]

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

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)

Sympy [F]

\[ \int \frac {\left (d+e x^2\right ) (a+b \arctan (c x))^2}{x^3} \, dx=\int \frac {\left (a + b \operatorname {atan}{\left (c x \right )}\right )^{2} \left (d + e x^{2}\right )}{x^{3}}\, dx \]

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

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

Maxima [F]

\[ \int \frac {\left (d+e x^2\right ) (a+b \arctan (c x))^2}{x^3} \, dx=\int { \frac {{\left (e x^{2} + d\right )} {\left (b \arctan \left (c x\right ) + a\right )}^{2}}{x^{3}} \,d x } \]

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

-((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

Giac [F(-1)]

Timed out. \[ \int \frac {\left (d+e x^2\right ) (a+b \arctan (c x))^2}{x^3} \, dx=\text {Timed out} \]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (d+e x^2\right ) (a+b \arctan (c x))^2}{x^3} \, dx=\int \frac {{\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )}^2\,\left (e\,x^2+d\right )}{x^3} \,d x \]

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

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

\(\int \frac {(d+e x^2) (a+b \arctan (c x))^2}{x^3} \, dx\) [1253]

Optimal result