\(\int \frac {x^2 (a+b \arctan (c x))}{(d+e x^2)^2} \, dx\) [1161]

Optimal result

Integrand size = 21, antiderivative size = 1335 \[ \int \frac {x^2 (a+b \arctan (c x))}{\left (d+e x^2\right )^2} \, dx =\text {Too large to display} \] Output:

-1/2*x*(a+b*arctan(c*x))/e/(e*x^2+d)+a*arctan(e^(1/2)*x/d^(1/2))/d^(1/2)/e 
^(3/2)-1/2*(a+b*arctan(c*x))*arctan(e^(1/2)*x/d^(1/2))/d^(1/2)/e^(3/2)+1/8 
*I*b*c*ln(-e^(1/2)*(1+(-c^2)^(1/2)*x)/(I*(-c^2)^(1/2)*d^(1/2)-e^(1/2)))*ln 
(1-I*e^(1/2)*x/d^(1/2))/(-c^2)^(1/2)/d^(1/2)/e^(3/2)-1/4*I*b*ln(1+I*c*x)*l 
n(c*((-d)^(1/2)-e^(1/2)*x)/(c*(-d)^(1/2)-I*e^(1/2)))/(-d)^(1/2)/e^(3/2)-1/ 
8*I*b*c*polylog(2,(-c^2)^(1/2)*(d^(1/2)+I*e^(1/2)*x)/((-c^2)^(1/2)*d^(1/2) 
-I*e^(1/2)))/(-c^2)^(1/2)/d^(1/2)/e^(3/2)-1/4*I*b*polylog(2,e^(1/2)*(1+I*c 
*x)/(I*c*(-d)^(1/2)+e^(1/2)))/(-d)^(1/2)/e^(3/2)+1/4*I*b*polylog(2,e^(1/2) 
*(I+c*x)/(c*(-d)^(1/2)+I*e^(1/2)))/(-d)^(1/2)/e^(3/2)+1/4*I*b*ln(1-I*c*x)* 
ln(c*((-d)^(1/2)-e^(1/2)*x)/(c*(-d)^(1/2)+I*e^(1/2)))/(-d)^(1/2)/e^(3/2)-1 
/4*I*b*polylog(2,e^(1/2)*(1-I*c*x)/(I*c*(-d)^(1/2)+e^(1/2)))/(-d)^(1/2)/e^ 
(3/2)-1/8*I*b*c*polylog(2,(-c^2)^(1/2)*(d^(1/2)-I*e^(1/2)*x)/((-c^2)^(1/2) 
*d^(1/2)-I*e^(1/2)))/(-c^2)^(1/2)/d^(1/2)/e^(3/2)+1/4*b*c*ln(c^2*x^2+1)/(c 
^2*d-e)/e-1/4*b*c*ln(e*x^2+d)/(c^2*d-e)/e+1/4*I*b*ln(1+I*c*x)*ln(c*((-d)^( 
1/2)+e^(1/2)*x)/(c*(-d)^(1/2)+I*e^(1/2)))/(-d)^(1/2)/e^(3/2)-1/8*I*b*c*ln( 
e^(1/2)*(1+(-c^2)^(1/2)*x)/(I*(-c^2)^(1/2)*d^(1/2)+e^(1/2)))*ln(1+I*e^(1/2 
)*x/d^(1/2))/(-c^2)^(1/2)/d^(1/2)/e^(3/2)+1/8*I*b*c*polylog(2,(-c^2)^(1/2) 
*(d^(1/2)+I*e^(1/2)*x)/((-c^2)^(1/2)*d^(1/2)+I*e^(1/2)))/(-c^2)^(1/2)/d^(1 
/2)/e^(3/2)+1/8*I*b*c*ln(-e^(1/2)*(1-(-c^2)^(1/2)*x)/(I*(-c^2)^(1/2)*d^(1/ 
2)-e^(1/2)))*ln(1+I*e^(1/2)*x/d^(1/2))/(-c^2)^(1/2)/d^(1/2)/e^(3/2)-1/8...
 

Mathematica [A] (warning: unable to verify)

Time = 8.94 (sec) , antiderivative size = 877, normalized size of antiderivative = 0.66 \[ \int \frac {x^2 (a+b \arctan (c x))}{\left (d+e x^2\right )^2} \, dx =\text {Too large to display} \] Input:

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

Output:

-1/2*(a*x)/(e*(d + e*x^2)) + (a*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(2*Sqrt[d]*e^ 
(3/2)) + (b*c*((-2*Log[(c^2*d + e + (c^2*d - e)*Cos[2*ArcTan[c*x]])/(c^2*d 
 + e)])/(c^2*d - e) + (-4*ArcTan[c*x]*ArcTanh[Sqrt[-(c^2*d*e)]/(c*e*x)] + 
2*ArcCos[(c^2*d + e)/(-(c^2*d) + e)]*ArcTanh[(c*e*x)/Sqrt[-(c^2*d*e)]] + ( 
ArcCos[(c^2*d + e)/(-(c^2*d) + e)] - (2*I)*ArcTanh[(c*e*x)/Sqrt[-(c^2*d*e) 
]])*Log[(-2*c^2*d*(I*e + Sqrt[-(c^2*d*e)])*(-I + c*x))/((c^2*d - e)*(c^2*d 
 - c*Sqrt[-(c^2*d*e)]*x))] + (ArcCos[(c^2*d + e)/(-(c^2*d) + e)] + (2*I)*A 
rcTanh[(c*e*x)/Sqrt[-(c^2*d*e)]])*Log[((2*I)*c^2*d*(e + I*Sqrt[-(c^2*d*e)] 
)*(I + c*x))/((c^2*d - e)*(c^2*d - c*Sqrt[-(c^2*d*e)]*x))] - (ArcCos[(c^2* 
d + e)/(-(c^2*d) + e)] - (2*I)*ArcTanh[Sqrt[-(c^2*d*e)]/(c*e*x)] + (2*I)*A 
rcTanh[(c*e*x)/Sqrt[-(c^2*d*e)]])*Log[(Sqrt[2]*Sqrt[-(c^2*d*e)])/(Sqrt[-(c 
^2*d) + e]*E^(I*ArcTan[c*x])*Sqrt[-(c^2*d) - e + (-(c^2*d) + e)*Cos[2*ArcT 
an[c*x]]])] - (ArcCos[(c^2*d + e)/(-(c^2*d) + e)] + (2*I)*ArcTanh[Sqrt[-(c 
^2*d*e)]/(c*e*x)] - (2*I)*ArcTanh[(c*e*x)/Sqrt[-(c^2*d*e)]])*Log[(Sqrt[2]* 
Sqrt[-(c^2*d*e)]*E^(I*ArcTan[c*x]))/(Sqrt[-(c^2*d) + e]*Sqrt[-(c^2*d) - e 
+ (-(c^2*d) + e)*Cos[2*ArcTan[c*x]]])] + I*(PolyLog[2, ((c^2*d + e - (2*I) 
*Sqrt[-(c^2*d*e)])*(c^2*d + c*Sqrt[-(c^2*d*e)]*x))/((c^2*d - e)*(c^2*d - c 
*Sqrt[-(c^2*d*e)]*x))] - PolyLog[2, ((c^2*d + e + (2*I)*Sqrt[-(c^2*d*e)])* 
(c^2*d + c*Sqrt[-(c^2*d*e)]*x))/((c^2*d - e)*(c^2*d - c*Sqrt[-(c^2*d*e)]*x 
))]))/Sqrt[-(c^2*d*e)] - (4*ArcTan[c*x]*Sin[2*ArcTan[c*x]])/(c^2*d + e ...
 

Rubi [A] (verified)

Time = 2.27 (sec) , antiderivative size = 1335, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {5515, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 5515

\(\displaystyle \int \left (\frac {a+b \arctan (c x)}{e \left (d+e x^2\right )}-\frac {d (a+b \arctan (c x))}{e \left (d+e x^2\right )^2}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {\arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) (a+b \arctan (c x))}{2 \sqrt {d} e^{3/2}}-\frac {x (a+b \arctan (c x))}{2 e \left (e x^2+d\right )}+\frac {a \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {d} e^{3/2}}-\frac {i b \log (i c x+1) \log \left (\frac {c \left (\sqrt {-d}-\sqrt {e} x\right )}{c \sqrt {-d}-i \sqrt {e}}\right )}{4 \sqrt {-d} e^{3/2}}+\frac {i b \log (1-i c x) \log \left (\frac {c \left (\sqrt {-d}-\sqrt {e} x\right )}{\sqrt {-d} c+i \sqrt {e}}\right )}{4 \sqrt {-d} e^{3/2}}-\frac {i b \log (1-i c x) \log \left (\frac {c \left (\sqrt {e} x+\sqrt {-d}\right )}{c \sqrt {-d}-i \sqrt {e}}\right )}{4 \sqrt {-d} e^{3/2}}+\frac {i b \log (i c x+1) \log \left (\frac {c \left (\sqrt {e} x+\sqrt {-d}\right )}{\sqrt {-d} c+i \sqrt {e}}\right )}{4 \sqrt {-d} e^{3/2}}-\frac {i b c \log \left (\frac {\sqrt {e} \left (1-\sqrt {-c^2} x\right )}{i \sqrt {-c^2} \sqrt {d}+\sqrt {e}}\right ) \log \left (1-\frac {i \sqrt {e} x}{\sqrt {d}}\right )}{8 \sqrt {-c^2} \sqrt {d} e^{3/2}}+\frac {i b c \log \left (-\frac {\sqrt {e} \left (\sqrt {-c^2} x+1\right )}{i \sqrt {-c^2} \sqrt {d}-\sqrt {e}}\right ) \log \left (1-\frac {i \sqrt {e} x}{\sqrt {d}}\right )}{8 \sqrt {-c^2} \sqrt {d} e^{3/2}}+\frac {i b c \log \left (-\frac {\sqrt {e} \left (1-\sqrt {-c^2} x\right )}{i \sqrt {-c^2} \sqrt {d}-\sqrt {e}}\right ) \log \left (\frac {i \sqrt {e} x}{\sqrt {d}}+1\right )}{8 \sqrt {-c^2} \sqrt {d} e^{3/2}}-\frac {i b c \log \left (\frac {\sqrt {e} \left (\sqrt {-c^2} x+1\right )}{i \sqrt {-c^2} \sqrt {d}+\sqrt {e}}\right ) \log \left (\frac {i \sqrt {e} x}{\sqrt {d}}+1\right )}{8 \sqrt {-c^2} \sqrt {d} e^{3/2}}+\frac {b c \log \left (c^2 x^2+1\right )}{4 \left (c^2 d-e\right ) e}-\frac {b c \log \left (e x^2+d\right )}{4 \left (c^2 d-e\right ) e}+\frac {i b \operatorname {PolyLog}\left (2,\frac {\sqrt {e} (i-c x)}{\sqrt {-d} c+i \sqrt {e}}\right )}{4 \sqrt {-d} e^{3/2}}-\frac {i b \operatorname {PolyLog}\left (2,\frac {\sqrt {e} (1-i c x)}{i \sqrt {-d} c+\sqrt {e}}\right )}{4 \sqrt {-d} e^{3/2}}-\frac {i b \operatorname {PolyLog}\left (2,\frac {\sqrt {e} (i c x+1)}{i \sqrt {-d} c+\sqrt {e}}\right )}{4 \sqrt {-d} e^{3/2}}+\frac {i b \operatorname {PolyLog}\left (2,\frac {\sqrt {e} (c x+i)}{\sqrt {-d} c+i \sqrt {e}}\right )}{4 \sqrt {-d} e^{3/2}}-\frac {i b c \operatorname {PolyLog}\left (2,\frac {\sqrt {-c^2} \left (\sqrt {d}-i \sqrt {e} x\right )}{\sqrt {-c^2} \sqrt {d}-i \sqrt {e}}\right )}{8 \sqrt {-c^2} \sqrt {d} e^{3/2}}+\frac {i b c \operatorname {PolyLog}\left (2,\frac {\sqrt {-c^2} \left (\sqrt {d}-i \sqrt {e} x\right )}{\sqrt {-c^2} \sqrt {d}+i \sqrt {e}}\right )}{8 \sqrt {-c^2} \sqrt {d} e^{3/2}}-\frac {i b c \operatorname {PolyLog}\left (2,\frac {\sqrt {-c^2} \left (i \sqrt {e} x+\sqrt {d}\right )}{\sqrt {-c^2} \sqrt {d}-i \sqrt {e}}\right )}{8 \sqrt {-c^2} \sqrt {d} e^{3/2}}+\frac {i b c \operatorname {PolyLog}\left (2,\frac {\sqrt {-c^2} \left (i \sqrt {e} x+\sqrt {d}\right )}{\sqrt {-c^2} \sqrt {d}+i \sqrt {e}}\right )}{8 \sqrt {-c^2} \sqrt {d} e^{3/2}}\)

Input:

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

Output:

-1/2*(x*(a + b*ArcTan[c*x]))/(e*(d + e*x^2)) + (a*ArcTan[(Sqrt[e]*x)/Sqrt[ 
d]])/(Sqrt[d]*e^(3/2)) - ((a + b*ArcTan[c*x])*ArcTan[(Sqrt[e]*x)/Sqrt[d]]) 
/(2*Sqrt[d]*e^(3/2)) - ((I/4)*b*Log[1 + I*c*x]*Log[(c*(Sqrt[-d] - Sqrt[e]* 
x))/(c*Sqrt[-d] - I*Sqrt[e])])/(Sqrt[-d]*e^(3/2)) + ((I/4)*b*Log[1 - I*c*x 
]*Log[(c*(Sqrt[-d] - Sqrt[e]*x))/(c*Sqrt[-d] + I*Sqrt[e])])/(Sqrt[-d]*e^(3 
/2)) - ((I/4)*b*Log[1 - I*c*x]*Log[(c*(Sqrt[-d] + Sqrt[e]*x))/(c*Sqrt[-d] 
- I*Sqrt[e])])/(Sqrt[-d]*e^(3/2)) + ((I/4)*b*Log[1 + I*c*x]*Log[(c*(Sqrt[- 
d] + Sqrt[e]*x))/(c*Sqrt[-d] + I*Sqrt[e])])/(Sqrt[-d]*e^(3/2)) - ((I/8)*b* 
c*Log[(Sqrt[e]*(1 - Sqrt[-c^2]*x))/(I*Sqrt[-c^2]*Sqrt[d] + Sqrt[e])]*Log[1 
 - (I*Sqrt[e]*x)/Sqrt[d]])/(Sqrt[-c^2]*Sqrt[d]*e^(3/2)) + ((I/8)*b*c*Log[- 
((Sqrt[e]*(1 + Sqrt[-c^2]*x))/(I*Sqrt[-c^2]*Sqrt[d] - Sqrt[e]))]*Log[1 - ( 
I*Sqrt[e]*x)/Sqrt[d]])/(Sqrt[-c^2]*Sqrt[d]*e^(3/2)) + ((I/8)*b*c*Log[-((Sq 
rt[e]*(1 - Sqrt[-c^2]*x))/(I*Sqrt[-c^2]*Sqrt[d] - Sqrt[e]))]*Log[1 + (I*Sq 
rt[e]*x)/Sqrt[d]])/(Sqrt[-c^2]*Sqrt[d]*e^(3/2)) - ((I/8)*b*c*Log[(Sqrt[e]* 
(1 + Sqrt[-c^2]*x))/(I*Sqrt[-c^2]*Sqrt[d] + Sqrt[e])]*Log[1 + (I*Sqrt[e]*x 
)/Sqrt[d]])/(Sqrt[-c^2]*Sqrt[d]*e^(3/2)) + (b*c*Log[1 + c^2*x^2])/(4*(c^2* 
d - e)*e) - (b*c*Log[d + e*x^2])/(4*(c^2*d - e)*e) + ((I/4)*b*PolyLog[2, ( 
Sqrt[e]*(I - c*x))/(c*Sqrt[-d] + I*Sqrt[e])])/(Sqrt[-d]*e^(3/2)) - ((I/4)* 
b*PolyLog[2, (Sqrt[e]*(1 - I*c*x))/(I*c*Sqrt[-d] + Sqrt[e])])/(Sqrt[-d]*e^ 
(3/2)) - ((I/4)*b*PolyLog[2, (Sqrt[e]*(1 + I*c*x))/(I*c*Sqrt[-d] + Sqrt...
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 5515

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])
 

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 2304 vs. \(2 (991 ) = 1982\).

Time = 3.13 (sec) , antiderivative size = 2305, normalized size of antiderivative = 1.73

method	result	size
parts	\(\text {Expression too large to display}\)	\(2305\)
derivativedivides	\(\text {Expression too large to display}\)	\(2344\)
default	\(\text {Expression too large to display}\)	\(2344\)
risch	\(\text {Expression too large to display}\)	\(2391\)

Input:

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

Output:

3/4*I*b*c^3*d*ln(1-(c^2*d-e)*(1+I*c*x)^2/(c^2*x^2+1)/(-c^2*d+2*(c^2*d*e)^( 
1/2)-e))*arctan(c*x)/(c^2*d-e)/e/(c^4*d^2-2*c^2*d*e+e^2)*(c^2*d*e)^(1/2)+1 
/4*I*b/c*ln(1-(c^2*d-e)*(1+I*c*x)^2/(c^2*x^2+1)/(-c^2*d+2*(c^2*d*e)^(1/2)- 
e))*arctan(c*x)/d/(c^2*d-e)/(c^4*d^2-2*c^2*d*e+e^2)*e*(c^2*d*e)^(1/2)-1/4* 
I*b*c^5*d^2*ln(1-(c^2*d-e)*(1+I*c*x)^2/(c^2*x^2+1)/(-c^2*d+2*(c^2*d*e)^(1/ 
2)-e))*arctan(c*x)/(c^2*d-e)/e^2/(c^4*d^2-2*c^2*d*e+e^2)*(c^2*d*e)^(1/2)-3 
/8*b*c*polylog(2,(c^2*d-e)*(1+I*c*x)^2/(c^2*x^2+1)/(-c^2*d+2*(c^2*d*e)^(1/ 
2)-e))/(c^4*d^2-2*c^2*d*e+e^2)/(c^2*d-e)*(c^2*d*e)^(1/2)-1/4*b*(d*e)^(1/2) 
/d/e*arctanh(1/4*(2*(c^2*d-e)*(1+I*c*x)^2/(c^2*x^2+1)+2*c^2*d+2*e)/c/(d*e) 
^(1/2))/(c^2*d-e)+1/2*b*c^2*arctan(c*x)/(c^2*d-e)/(c^2*e*x^2+c^2*d)*x-1/4* 
b*c^2*(d*e)^(1/2)/e^2*arctanh(1/4*(2*(c^2*d-e)*(1+I*c*x)^2/(c^2*x^2+1)+2*c 
^2*d+2*e)/c/(d*e)^(1/2))/(c^2*d-e)+b*c^3/(c^2*d-e)^2/e*d*ln((1+I*c*x)/(c^2 
*x^2+1)^(1/2))-3/4*b*c*arctan(c*x)^2/(c^4*d^2-2*c^2*d*e+e^2)/(c^2*d-e)*(c^ 
2*d*e)^(1/2)+1/4*b*c*(c^2*d*e)^(1/2)/(c^2*d-e)/e^2*arctan(c*x)^2+1/8*b*c*( 
c^2*d*e)^(1/2)/(c^2*d-e)/e^2*polylog(2,(c^2*d-e)*(1+I*c*x)^2/(c^2*x^2+1)/( 
-c^2*d-2*(c^2*d*e)^(1/2)-e))-1/4*b*c^3/(c^2*d-e)^2/e*d*ln(c^2*d*(1+I*c*x)^ 
4/(c^2*x^2+1)^2+2*c^2*d*(1+I*c*x)^2/(c^2*x^2+1)-e*(1+I*c*x)^4/(c^2*x^2+1)^ 
2+c^2*d+2*e*(1+I*c*x)^2/(c^2*x^2+1)-e)+1/4*b*c/(c^2*d-e)^2*ln(c^2*d*(1+I*c 
*x)^4/(c^2*x^2+1)^2+2*c^2*d*(1+I*c*x)^2/(c^2*x^2+1)-e*(1+I*c*x)^4/(c^2*x^2 
+1)^2+c^2*d+2*e*(1+I*c*x)^2/(c^2*x^2+1)-e)-1/4*b*(d*e)^(1/2)/d*arctanh(...
 

Fricas [F]

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

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

Output:

integral((b*x^2*arctan(c*x) + a*x^2)/(e^2*x^4 + 2*d*e*x^2 + d^2), x)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {x^2 (a+b \arctan (c x))}{\left (d+e x^2\right )^2} \, dx=\text {Exception raised: ValueError} \] Input:

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

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(e>0)', see `assume?` for more de 
tails)Is e
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {x^2 (a+b \arctan (c x))}{\left (d+e x^2\right )^2} \, dx =\text {Too large to display} \] Input:

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

Output:

(sqrt(e)*sqrt(d)*atan((e*x)/(sqrt(e)*sqrt(d)))*a*c**4*d**3 + sqrt(e)*sqrt( 
d)*atan((e*x)/(sqrt(e)*sqrt(d)))*a*c**4*d**2*e*x**2 - sqrt(e)*sqrt(d)*atan 
((e*x)/(sqrt(e)*sqrt(d)))*a*d*e**2 - sqrt(e)*sqrt(d)*atan((e*x)/(sqrt(e)*s 
qrt(d)))*a*e**3*x**2 + atan(c*x)**2*b*c**3*d**3*e + atan(c*x)**2*b*c**3*d* 
*2*e**2*x**2 - atan(c*x)**2*b*c*d**2*e**2 - atan(c*x)**2*b*c*d*e**3*x**2 - 
 2*atan(c*x)*b*c**4*d**3*e*x + 2*atan(c*x)*b*c**2*d**2*e**2*x + 2*int((ata 
n(c*x)*x**2)/(c**4*d**3*x**2 + 2*c**4*d**2*e*x**4 + c**4*d*e**2*x**6 + c** 
2*d**3 + 3*c**2*d**2*e*x**2 + 3*c**2*d*e**2*x**4 + c**2*e**3*x**6 + d**2*e 
 + 2*d*e**2*x**2 + e**3*x**4),x)*b*c**8*d**6*e + 2*int((atan(c*x)*x**2)/(c 
**4*d**3*x**2 + 2*c**4*d**2*e*x**4 + c**4*d*e**2*x**6 + c**2*d**3 + 3*c**2 
*d**2*e*x**2 + 3*c**2*d*e**2*x**4 + c**2*e**3*x**6 + d**2*e + 2*d*e**2*x** 
2 + e**3*x**4),x)*b*c**8*d**5*e**2*x**2 - 4*int((atan(c*x)*x**2)/(c**4*d** 
3*x**2 + 2*c**4*d**2*e*x**4 + c**4*d*e**2*x**6 + c**2*d**3 + 3*c**2*d**2*e 
*x**2 + 3*c**2*d*e**2*x**4 + c**2*e**3*x**6 + d**2*e + 2*d*e**2*x**2 + e** 
3*x**4),x)*b*c**6*d**5*e**2 - 4*int((atan(c*x)*x**2)/(c**4*d**3*x**2 + 2*c 
**4*d**2*e*x**4 + c**4*d*e**2*x**6 + c**2*d**3 + 3*c**2*d**2*e*x**2 + 3*c* 
*2*d*e**2*x**4 + c**2*e**3*x**6 + d**2*e + 2*d*e**2*x**2 + e**3*x**4),x)*b 
*c**6*d**4*e**3*x**2 + 4*int((atan(c*x)*x**2)/(c**4*d**3*x**2 + 2*c**4*d** 
2*e*x**4 + c**4*d*e**2*x**6 + c**2*d**3 + 3*c**2*d**2*e*x**2 + 3*c**2*d*e* 
*2*x**4 + c**2*e**3*x**6 + d**2*e + 2*d*e**2*x**2 + e**3*x**4),x)*b*c**...