\(\int \frac {x (a+b \arctan (c x))^2}{(d+e x^2)^2} \, dx\) [1268]

Optimal result

Integrand size = 21, antiderivative size = 457 \[ \int \frac {x (a+b \arctan (c x))^2}{\left (d+e x^2\right )^2} \, dx=\frac {c^2 (a+b \arctan (c x))^2}{2 \left (c^2 d-e\right ) e}-\frac {(a+b \arctan (c x))^2}{4 d e \left (1-\frac {\sqrt {e} x}{\sqrt {-d}}\right )}-\frac {(a+b \arctan (c x))^2}{4 d e \left (1+\frac {\sqrt {e} x}{\sqrt {-d}}\right )}-\frac {b c (a+b \arctan (c x)) \log \left (\frac {2 c \left (\sqrt {-d}-\sqrt {e} x\right )}{\left (c \sqrt {-d}-i \sqrt {e}\right ) (1-i c x)}\right )}{2 \sqrt {-d} \left (c^2 d-e\right ) \sqrt {e}}+\frac {b c (a+b \arctan (c x)) \log \left (\frac {2 c \left (\sqrt {-d}+\sqrt {e} x\right )}{\left (c \sqrt {-d}+i \sqrt {e}\right ) (1-i c x)}\right )}{2 \sqrt {-d} \left (c^2 d-e\right ) \sqrt {e}}+\frac {i b^2 c \operatorname {PolyLog}\left (2,1-\frac {2 c \left (\sqrt {-d}-\sqrt {e} x\right )}{\left (c \sqrt {-d}-i \sqrt {e}\right ) (1-i c x)}\right )}{4 \sqrt {-d} \left (c^2 d-e\right ) \sqrt {e}}-\frac {i b^2 c \operatorname {PolyLog}\left (2,1-\frac {2 c \left (\sqrt {-d}+\sqrt {e} x\right )}{\left (c \sqrt {-d}+i \sqrt {e}\right ) (1-i c x)}\right )}{4 \sqrt {-d} \left (c^2 d-e\right ) \sqrt {e}} \] Output:

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

Mathematica [A] (warning: unable to verify)

Time = 6.42 (sec) , antiderivative size = 836, normalized size of antiderivative = 1.83 \[ \int \frac {x (a+b \arctan (c x))^2}{\left (d+e x^2\right )^2} \, dx =\text {Too large to display} \] Input:

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

Output:

((-2*a^2)/(e*(d + e*x^2)) + (4*a*b*(-(((1 + c^2*x^2)*ArcTan[c*x])/(d + e*x 
^2)) + (c*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(Sqrt[d]*Sqrt[e])))/(-(c^2*d) + e) 
+ (b^2*c^2*((4*ArcTan[c*x]^2)/(c^2*d + e + (c^2*d - e)*Cos[2*ArcTan[c*x]]) 
 + (-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)*ArcTanh[(c*e*x)/Sqrt[-(c^2*d*e)]])*Log[(2*c*d*(( 
-I)*e + Sqrt[-(c^2*d*e)])*(-I + c*x))/((c^2*d - e)*(c*d + 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[(2*c*d*(I*e + Sqrt[-(c^2*d*e)])*(I + c*x))/((c^2*d - e)*( 
c*d + 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)*(ArcTanh[(c*d)/(Sqrt[-(c^2*d*e)]*x)] + ArcTanh[(c*e*x)/Sqr 
t[-(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*d - Sqrt[-(c^2*d*e)]*x))/((c^2* 
d - e)*(c*d + Sqrt[-(c^2*d*e)]*x))] - PolyLog[2, ((c^2*d + e + (2*I)*Sqrt[ 
-(c^2*d*e)])*(c*d - Sqrt[-(c^2*d*e)]*x))/((c^2*d - e)*(c*d + Sqrt[-(c^2*d* 
e)]*x))]))/Sqrt[-(c^2*d*e)]))/(c^2*d - e))/4
 

Rubi [B] (verified)

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(996\) vs. \(2(457)=914\).

Time = 1.56 (sec) , antiderivative size = 996, normalized size of antiderivative = 2.18, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {5513, 5389, 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.

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

\(\Big \downarrow \) 5513

\(\displaystyle \frac {\int \frac {(a+b \arctan (c x))^2}{\left (1-\frac {\sqrt {e} x}{\sqrt {-d}}\right )^2}dx}{4 (-d)^{3/2} \sqrt {e}}-\frac {\int \frac {(a+b \arctan (c x))^2}{\left (\frac {\sqrt {e} x}{\sqrt {-d}}+1\right )^2}dx}{4 (-d)^{3/2} \sqrt {e}}\)

\(\Big \downarrow \) 5389

\(\displaystyle \frac {\frac {\sqrt {-d} (a+b \arctan (c x))^2}{\sqrt {e} \left (1-\frac {\sqrt {e} x}{\sqrt {-d}}\right )}-\frac {2 b c \sqrt {-d} \int \left (-\frac {\sqrt {-d} \left (\sqrt {e} x+\sqrt {-d}\right ) (a+b \arctan (c x)) c^2}{\left (c^2 d-e\right ) \left (c^2 x^2+1\right )}-\frac {\sqrt {-d} e (a+b \arctan (c x))}{\left (c^2 d-e\right ) \left (\sqrt {-d}-\sqrt {e} x\right )}\right )dx}{\sqrt {e}}}{4 (-d)^{3/2} \sqrt {e}}-\frac {\frac {2 b c \sqrt {-d} \int \left (\frac {c^2 \left (d+\sqrt {-d} \sqrt {e} x\right ) (a+b \arctan (c x))}{\left (c^2 d-e\right ) \left (c^2 x^2+1\right )}-\frac {\sqrt {-d} e (a+b \arctan (c x))}{\left (c^2 d-e\right ) \left (\sqrt {e} x+\sqrt {-d}\right )}\right )dx}{\sqrt {e}}-\frac {\sqrt {-d} (a+b \arctan (c x))^2}{\sqrt {e} \left (\frac {\sqrt {e} x}{\sqrt {-d}}+1\right )}}{4 (-d)^{3/2} \sqrt {e}}\)

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 2009

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

rule 5389

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_)*((d_) + (e_.)*(x_))^(q_.), x_Sy 
mbol] :> Simp[(d + e*x)^(q + 1)*((a + b*ArcTan[c*x])^p/(e*(q + 1))), x] - S 
imp[b*c*(p/(e*(q + 1)))   Int[ExpandIntegrand[(a + b*ArcTan[c*x])^(p - 1), 
(d + e*x)^(q + 1)/(1 + c^2*x^2), x], x], x] /; FreeQ[{a, b, c, d, e}, x] && 
 IGtQ[p, 1] && IntegerQ[q] && NeQ[q, -1]
 

rule 5513

Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*(x_))/((d_) + (e_.)*(x_)^2)^2 
, x_Symbol] :> Simp[1/(4*d^2*Rt[-e/d, 2])   Int[(a + b*ArcTan[c*x])^p/(1 - 
Rt[-e/d, 2]*x)^2, x], x] - Simp[1/(4*d^2*Rt[-e/d, 2])   Int[(a + b*ArcTan[c 
*x])^p/(1 + Rt[-e/d, 2]*x)^2, x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p 
, 0]
 

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1174 vs. \(2 (377 ) = 754\).

Time = 3.71 (sec) , antiderivative size = 1175, normalized size of antiderivative = 2.57

method	result	size
parts	\(\text {Expression too large to display}\)	\(1175\)
derivativedivides	\(\text {Expression too large to display}\)	\(1209\)
default	\(\text {Expression too large to display}\)	\(1209\)

Input:

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

Output:

-1/2*a^2/e/(e*x^2+d)-1/2*b^2*c^2*arctan(c*x)^2/e/(c^2*e*x^2+c^2*d)+1/2*b^2 
*c^2/(c^2*d-e)/(c^4*d^2-2*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))*(c^2*d*e)^(1/2)-1/2*b^2*c^4/e/(c^2* 
d-e)/(c^4*d^2-2*c^2*d*e+e^2)*arctan(c*x)^2*(c^2*d*e)^(1/2)*d-1/4*b^2*c^4/e 
/(c^2*d-e)/(c^4*d^2-2*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))*(c^2*d*e)^(1/2)*d-1/2*I*b^2*c^4/e*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)/(c^4*d^2-2*c^2*d*e+e^2)*(c^2*d*e)^(1/2)*d-1/2*I*b^2*e*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)*(c^2*d*e)^(1/2)-1/2*b^2*e/d/(c^2*d-e)/( 
c^4*d^2-2*c^2*d*e+e^2)*arctan(c*x)^2*(c^2*d*e)^(1/2)-1/4*b^2*e/d/(c^2*d-e) 
/(c^4*d^2-2*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))*(c^2*d*e)^(1/2)+1/2*I*b^2/e*(c^2*d*e)^(1/2)/d/(c^ 
2*d-e)*arctan(c*x)*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))+1/2*b^2/e*(c^2*d*e)^(1/2)/d/(c^2*d-e)*arctan(c*x)^2+1/2*b^2* 
c^2/e/(c^2*d-e)*arctan(c*x)^2+b^2*c^2/(c^2*d-e)/(c^4*d^2-2*c^2*d*e+e^2)*ar 
ctan(c*x)^2*(c^2*d*e)^(1/2)+1/4*b^2/e*(c^2*d*e)^(1/2)/d/(c^2*d-e)*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))+I*b^2*c^ 
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))*arc 
tan(c*x)/(c^2*d-e)/(c^4*d^2-2*c^2*d*e+e^2)*(c^2*d*e)^(1/2)-a*b*c^2*arct...
 

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F(-2)]

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

integrate(x*(a+b*arctan(c*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]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

( - 2*sqrt(e)*sqrt(d)*atan((e*x)/(sqrt(e)*sqrt(d)))*a*b*c**3*d**2 - 2*sqrt 
(e)*sqrt(d)*atan((e*x)/(sqrt(e)*sqrt(d)))*a*b*c**3*d*e*x**2 - 2*sqrt(e)*sq 
rt(d)*atan((e*x)/(sqrt(e)*sqrt(d)))*a*b*c*d*e - 2*sqrt(e)*sqrt(d)*atan((e* 
x)/(sqrt(e)*sqrt(d)))*a*b*c*e**2*x**2 + atan(c*x)**2*b**2*c**4*d**2*e*x**2 
 - atan(c*x)**2*b**2*c**2*d**2*e - atan(c*x)**2*b**2*c**2*d*e**2*x**2 + at 
an(c*x)**2*b**2*d*e**2 + 2*atan(c*x)*a*b*c**4*d**2*e*x**2 + 2*atan(c*x)*a* 
b*c**2*d**2*e + 2*atan(c*x)*a*b*c**2*d*e**2*x**2 + 2*atan(c*x)*a*b*d*e**2 
+ 2*atan(c*x)*b**2*c**3*d**2*e*x - 2*atan(c*x)*b**2*c*d*e**2*x - 4*int((at 
an(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**2*c**7*d**5*e - 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**2*c**7*d**4*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**2*c**5*d**4*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**2*c**5*d**3*e**3*x**2 + 4*int((atan(c*x)*x**2)/(c**4*d**3*x...