\(\int \frac {a+b \arctan (c x)}{x^2 (d+e x^2)^2} \, dx\) [1163]

Optimal result

Integrand size = 21, antiderivative size = 1382 \[ \int \frac {a+b \arctan (c x)}{x^2 \left (d+e x^2\right )^2} \, dx =\text {Too large to display} \] Output:

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

Mathematica [A] (verified)

Time = 12.68 (sec) , antiderivative size = 992, normalized size of antiderivative = 0.72 \[ \int \frac {a+b \arctan (c x)}{x^2 \left (d+e x^2\right )^2} \, dx =\text {Too large to display} \] Input:

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

Output:

-(a/(d^2*x)) - (a*e*x)/(2*d^2*(d + e*x^2)) - (3*a*Sqrt[e]*ArcTan[(Sqrt[e]* 
x)/Sqrt[d]])/(2*d^(5/2)) + b*c^5*(-(ArcTan[c*x]/(c^5*d^2*x)) + Log[(c*x)/S 
qrt[1 + c^2*x^2]]/(c^4*d^2) - (e*Log[1 - ((-(c^2*d) + e)*Cos[2*ArcTan[c*x] 
])/(c^2*d + e)])/(4*c^4*d^2*(c^2*d - e)) - (3*e*(4*ArcTan[c*x]*ArcTanh[(c* 
d)/(Sqrt[-(c^2*d*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)*ArcT 
anh[(c*e*x)/Sqrt[-(c^2*d*e)]])*Log[1 - ((c^2*d + e - (2*I)*Sqrt[-(c^2*d*e) 
])*(2*c^2*d - 2*c*Sqrt[-(c^2*d*e)]*x))/((c^2*d - e)*(2*c^2*d + 2*c*Sqrt[-( 
c^2*d*e)]*x))] + (-ArcCos[(-(c^2*d) - e)/(c^2*d - e)] - (2*I)*ArcTanh[(c*e 
*x)/Sqrt[-(c^2*d*e)]])*Log[1 - ((c^2*d + e + (2*I)*Sqrt[-(c^2*d*e)])*(2*c^ 
2*d - 2*c*Sqrt[-(c^2*d*e)]*x))/((c^2*d - e)*(2*c^2*d + 2*c*Sqrt[-(c^2*d*e) 
]*x))] + (ArcCos[(-(c^2*d) - e)/(c^2*d - e)] - (2*I)*(ArcTanh[(c*d)/(Sqrt[ 
-(c^2*d*e)]*x)] + ArcTanh[(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*ArcTan[c*x]]])] + (ArcCos[(-(c^2*d) - e)/(c^2*d - e)] + (2*I)*(ArcT 
anh[(c*d)/(Sqrt[-(c^2*d*e)]*x)] + 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)])*(2*c^2*d - 2*c*Sqrt[-(c^2*d*e)]*x))/((c^2*d - e)*(2*c^2 
*d + 2*c*Sqrt[-(c^2*d*e)]*x))] - PolyLog[2, ((c^2*d + e + (2*I)*Sqrt[-(...
 

Rubi [A] (verified)

Time = 2.06 (sec) , antiderivative size = 1382, 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.

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

\(\Big \downarrow \) 5515

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

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {\sqrt {e} \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) (a+b \arctan (c x))}{2 d^{5/2}}-\frac {a+b \arctan (c x)}{d^2 x}-\frac {e x (a+b \arctan (c x))}{2 d^2 \left (e x^2+d\right )}-\frac {a \sqrt {e} \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{d^{5/2}}+\frac {b c \log (x)}{d^2}+\frac {i b \sqrt {e} \log (i c x+1) \log \left (\frac {c \left (\sqrt {-d}-\sqrt {e} x\right )}{c \sqrt {-d}-i \sqrt {e}}\right )}{4 (-d)^{5/2}}-\frac {i b \sqrt {e} \log (1-i c x) \log \left (\frac {c \left (\sqrt {-d}-\sqrt {e} x\right )}{\sqrt {-d} c+i \sqrt {e}}\right )}{4 (-d)^{5/2}}+\frac {i b \sqrt {e} \log (1-i c x) \log \left (\frac {c \left (\sqrt {e} x+\sqrt {-d}\right )}{c \sqrt {-d}-i \sqrt {e}}\right )}{4 (-d)^{5/2}}-\frac {i b \sqrt {e} \log (i c x+1) \log \left (\frac {c \left (\sqrt {e} x+\sqrt {-d}\right )}{\sqrt {-d} c+i \sqrt {e}}\right )}{4 (-d)^{5/2}}-\frac {i b c \sqrt {e} \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} d^{5/2}}+\frac {i b c \sqrt {e} \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} d^{5/2}}+\frac {i b c \sqrt {e} \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} d^{5/2}}-\frac {i b c \sqrt {e} \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} d^{5/2}}+\frac {b c e \log \left (c^2 x^2+1\right )}{4 d^2 \left (c^2 d-e\right )}-\frac {b c \log \left (c^2 x^2+1\right )}{2 d^2}-\frac {b c e \log \left (e x^2+d\right )}{4 d^2 \left (c^2 d-e\right )}-\frac {i b \sqrt {e} \operatorname {PolyLog}\left (2,\frac {\sqrt {e} (i-c x)}{\sqrt {-d} c+i \sqrt {e}}\right )}{4 (-d)^{5/2}}+\frac {i b \sqrt {e} \operatorname {PolyLog}\left (2,\frac {\sqrt {e} (1-i c x)}{i \sqrt {-d} c+\sqrt {e}}\right )}{4 (-d)^{5/2}}+\frac {i b \sqrt {e} \operatorname {PolyLog}\left (2,\frac {\sqrt {e} (i c x+1)}{i \sqrt {-d} c+\sqrt {e}}\right )}{4 (-d)^{5/2}}-\frac {i b \sqrt {e} \operatorname {PolyLog}\left (2,\frac {\sqrt {e} (c x+i)}{\sqrt {-d} c+i \sqrt {e}}\right )}{4 (-d)^{5/2}}-\frac {i b c \sqrt {e} \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} d^{5/2}}+\frac {i b c \sqrt {e} \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} d^{5/2}}-\frac {i b c \sqrt {e} \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} d^{5/2}}+\frac {i b c \sqrt {e} \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} d^{5/2}}\)

Input:

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

Output:

-((a + b*ArcTan[c*x])/(d^2*x)) - (e*x*(a + b*ArcTan[c*x]))/(2*d^2*(d + e*x 
^2)) - (a*Sqrt[e]*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/d^(5/2) - (Sqrt[e]*(a + b*A 
rcTan[c*x])*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(2*d^(5/2)) + (b*c*Log[x])/d^2 + 
((I/4)*b*Sqrt[e]*Log[1 + I*c*x]*Log[(c*(Sqrt[-d] - Sqrt[e]*x))/(c*Sqrt[-d] 
 - I*Sqrt[e])])/(-d)^(5/2) - ((I/4)*b*Sqrt[e]*Log[1 - I*c*x]*Log[(c*(Sqrt[ 
-d] - Sqrt[e]*x))/(c*Sqrt[-d] + I*Sqrt[e])])/(-d)^(5/2) + ((I/4)*b*Sqrt[e] 
*Log[1 - I*c*x]*Log[(c*(Sqrt[-d] + Sqrt[e]*x))/(c*Sqrt[-d] - I*Sqrt[e])])/ 
(-d)^(5/2) - ((I/4)*b*Sqrt[e]*Log[1 + I*c*x]*Log[(c*(Sqrt[-d] + Sqrt[e]*x) 
)/(c*Sqrt[-d] + I*Sqrt[e])])/(-d)^(5/2) - ((I/8)*b*c*Sqrt[e]*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]*d^(5/2)) + ((I/8)*b*c*Sqrt[e]*Log[-((Sqrt[e]*(1 + 
Sqrt[-c^2]*x))/(I*Sqrt[-c^2]*Sqrt[d] - Sqrt[e]))]*Log[1 - (I*Sqrt[e]*x)/Sq 
rt[d]])/(Sqrt[-c^2]*d^(5/2)) + ((I/8)*b*c*Sqrt[e]*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]*d^(5/2)) - ((I/8)*b*c*Sqrt[e]*Log[(Sqrt[e]*(1 + Sqrt[-c^2] 
*x))/(I*Sqrt[-c^2]*Sqrt[d] + Sqrt[e])]*Log[1 + (I*Sqrt[e]*x)/Sqrt[d]])/(Sq 
rt[-c^2]*d^(5/2)) - (b*c*Log[1 + c^2*x^2])/(2*d^2) + (b*c*e*Log[1 + c^2*x^ 
2])/(4*d^2*(c^2*d - e)) - (b*c*e*Log[d + e*x^2])/(4*d^2*(c^2*d - e)) - ((I 
/4)*b*Sqrt[e]*PolyLog[2, (Sqrt[e]*(I - c*x))/(c*Sqrt[-d] + I*Sqrt[e])])/(- 
d)^(5/2) + ((I/4)*b*Sqrt[e]*PolyLog[2, (Sqrt[e]*(1 - I*c*x))/(I*c*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. 2567 vs. \(2 (1036 ) = 2072\).

Time = 1.86 (sec) , antiderivative size = 2568, normalized size of antiderivative = 1.86

method	result	size
risch	\(\text {Expression too large to display}\)	\(2568\)
parts	\(\text {Expression too large to display}\)	\(3558\)
derivativedivides	\(\text {Expression too large to display}\)	\(3591\)
default	\(\text {Expression too large to display}\)	\(3591\)

Input:

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

Output:

-3/8*b/d^2*e/(d*e)^(1/2)*dilog((c*(d*e)^(1/2)-(1+I*c*x)*e+e)/(c*(d*e)^(1/2 
)+e))+3/8*b/d^2*e/(d*e)^(1/2)*dilog((c*(d*e)^(1/2)+(1+I*c*x)*e-e)/(c*(d*e) 
^(1/2)-e))+1/2*I*b/d^2*ln(1+I*c*x)/x-1/4*c^3*b/d*e*ln(1-I*c*x)/(c^2*d-e)/( 
-c^2*e*x^2-c^2*d)-1/4*c^2*b/d*e/(c^2*d-e)/(d*e)^(1/2)*arctanh(1/2*(2*(1-I* 
c*x)*e-2*e)/c/(d*e)^(1/2))+1/8*c^2*b/d*e^2*ln(1-I*c*x)/(c^2*d-e)/(-c^2*e*x 
^2-c^2*d)/(d*e)^(1/2)*ln((c*(d*e)^(1/2)+(1-I*c*x)*e-e)/(c*(d*e)^(1/2)-e))+ 
1/4*I*c^4*b*ln(1-I*c*x)/d/(c^2*d-e)/(-c^2*e*x^2-c^2*d)*e*x-1/4*I*c^2*b/d^2 
*e^2*ln(1-I*c*x)/(c^2*d-e)/(-c^2*e*x^2-c^2*d)*x-1/8*c^2*b/d*e^2*ln(1-I*c*x 
)/(c^2*d-e)/(-c^2*e*x^2-c^2*d)/(d*e)^(1/2)*ln((c*(d*e)^(1/2)-(1-I*c*x)*e+e 
)/(c*(d*e)^(1/2)+e))-1/8*b*c^2/d*e^2*ln(1+I*c*x)/(c^2*d-e)/(-c^2*e*x^2-c^2 
*d)/(d*e)^(1/2)*ln((c*(d*e)^(1/2)-(1+I*c*x)*e+e)/(c*(d*e)^(1/2)+e))-1/4*c^ 
3*b/d^2*e^2*ln(1-I*c*x)/(c^2*d-e)/(-c^2*e*x^2-c^2*d)*x^2+1/8*c^4*b*ln(1-I* 
c*x)/(c^2*d-e)/(-c^2*e*x^2-c^2*d)/(d*e)^(1/2)*ln((c*(d*e)^(1/2)-(1-I*c*x)* 
e+e)/(c*(d*e)^(1/2)+e))*e-1/8*c^4*b*ln(1-I*c*x)/(c^2*d-e)/(-c^2*e*x^2-c^2* 
d)/(d*e)^(1/2)*ln((c*(d*e)^(1/2)+(1-I*c*x)*e-e)/(c*(d*e)^(1/2)-e))*e-1/4*b 
*c^3/d^2*e^2*ln(1+I*c*x)/(c^2*d-e)/(-c^2*e*x^2-c^2*d)*x^2+1/8*b*c^4*e*ln(1 
+I*c*x)/(c^2*d-e)/(-c^2*e*x^2-c^2*d)/(d*e)^(1/2)*ln((c*(d*e)^(1/2)-(1+I*c* 
x)*e+e)/(c*(d*e)^(1/2)+e))-1/8*b*c^4*e*ln(1+I*c*x)/(c^2*d-e)/(-c^2*e*x^2-c 
^2*d)/(d*e)^(1/2)*ln((c*(d*e)^(1/2)+(1+I*c*x)*e-e)/(c*(d*e)^(1/2)-e))-3/2* 
I*a/d^2*e/(d*e)^(1/2)*arctanh(1/2*(2*(1-I*c*x)*e-2*e)/c/(d*e)^(1/2))-1/...
 

Fricas [F]

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

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

Output:

integral((b*arctan(c*x) + a)/(e^2*x^6 + 2*d*e*x^4 + d^2*x^2), x)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F(-2)]

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

integrate((a+b*arctan(c*x))/x^2/(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]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

( - 9*sqrt(e)*sqrt(d)*atan((e*x)/(sqrt(e)*sqrt(d)))*a*c**4*d**3*x - 9*sqrt 
(e)*sqrt(d)*atan((e*x)/(sqrt(e)*sqrt(d)))*a*c**4*d**2*e*x**3 + 12*sqrt(e)* 
sqrt(d)*atan((e*x)/(sqrt(e)*sqrt(d)))*a*c**2*d**2*e*x + 12*sqrt(e)*sqrt(d) 
*atan((e*x)/(sqrt(e)*sqrt(d)))*a*c**2*d*e**2*x**3 - 3*sqrt(e)*sqrt(d)*atan 
((e*x)/(sqrt(e)*sqrt(d)))*a*d*e**2*x - 3*sqrt(e)*sqrt(d)*atan((e*x)/(sqrt( 
e)*sqrt(d)))*a*e**3*x**3 - 3*atan(c*x)**2*b*c**5*d**4*x - 3*atan(c*x)**2*b 
*c**5*d**3*e*x**3 + 3*atan(c*x)**2*b*c**3*d**3*e*x + 3*atan(c*x)**2*b*c**3 
*d**2*e**2*x**3 - 6*atan(c*x)*b*c**4*d**4 - 6*atan(c*x)*b*c**4*d**3*e*x**2 
 + 8*atan(c*x)*b*c**2*d**3*e + 6*atan(c*x)*b*c**2*d**2*e**2*x**2 - 2*atan( 
c*x)*b*d**2*e**2 + 18*int(atan(c*x)/(3*c**4*d**3*x**2 + 6*c**4*d**2*e*x**4 
 + 3*c**4*d*e**2*x**6 + 3*c**2*d**3 + 5*c**2*d**2*e*x**2 + 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**7*x 
+ 18*int(atan(c*x)/(3*c**4*d**3*x**2 + 6*c**4*d**2*e*x**4 + 3*c**4*d*e**2* 
x**6 + 3*c**2*d**3 + 5*c**2*d**2*e*x**2 + 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*x**3 - 60*int(ata 
n(c*x)/(3*c**4*d**3*x**2 + 6*c**4*d**2*e*x**4 + 3*c**4*d*e**2*x**6 + 3*c** 
2*d**3 + 5*c**2*d**2*e*x**2 + 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**6*e*x - 60*int(atan(c*x)/(3*c**4* 
d**3*x**2 + 6*c**4*d**2*e*x**4 + 3*c**4*d*e**2*x**6 + 3*c**2*d**3 + 5*c**2 
*d**2*e*x**2 + c**2*d*e**2*x**4 - c**2*e**3*x**6 - d**2*e - 2*d*e**2*x*...