3.13.67 \(\int \frac {(a+b \arctan (c x))^2}{x^3 (d+e x^2)} \, dx\) [1267]

3.13.67.1 Optimal result

Integrand size = 23, antiderivative size = 745 \[ \int \frac {(a+b \arctan (c x))^2}{x^3 \left (d+e x^2\right )} \, dx=-\frac {b c (a+b \arctan (c x))}{d x}-\frac {c^2 (a+b \arctan (c x))^2}{2 d}-\frac {(a+b \arctan (c x))^2}{2 d x^2}-\frac {2 e (a+b \arctan (c x))^2 \text {arctanh}\left (1-\frac {2}{1+i c x}\right )}{d^2}+\frac {b^2 c^2 \log (x)}{d}-\frac {e (a+b \arctan (c x))^2 \log \left (\frac {2}{1-i c x}\right )}{d^2}+\frac {e (a+b \arctan (c x))^2 \log \left (\frac {2 c \left (\sqrt {-d}-\sqrt {e} x\right )}{\left (c \sqrt {-d}-i \sqrt {e}\right ) (1-i c x)}\right )}{2 d^2}+\frac {e (a+b \arctan (c x))^2 \log \left (\frac {2 c \left (\sqrt {-d}+\sqrt {e} x\right )}{\left (c \sqrt {-d}+i \sqrt {e}\right ) (1-i c x)}\right )}{2 d^2}-\frac {b^2 c^2 \log \left (1+c^2 x^2\right )}{2 d}+\frac {i b e (a+b \arctan (c x)) \operatorname {PolyLog}\left (2,1-\frac {2}{1-i c x}\right )}{d^2}+\frac {i b e (a+b \arctan (c x)) \operatorname {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )}{d^2}-\frac {i b e (a+b \arctan (c x)) \operatorname {PolyLog}\left (2,-1+\frac {2}{1+i c x}\right )}{d^2}-\frac {i b e (a+b \arctan (c x)) \operatorname {PolyLog}\left (2,1-\frac {2 c \left (\sqrt {-d}-\sqrt {e} x\right )}{\left (c \sqrt {-d}-i \sqrt {e}\right ) (1-i c x)}\right )}{2 d^2}-\frac {i b e (a+b \arctan (c x)) \operatorname {PolyLog}\left (2,1-\frac {2 c \left (\sqrt {-d}+\sqrt {e} x\right )}{\left (c \sqrt {-d}+i \sqrt {e}\right ) (1-i c x)}\right )}{2 d^2}-\frac {b^2 e \operatorname {PolyLog}\left (3,1-\frac {2}{1-i c x}\right )}{2 d^2}+\frac {b^2 e \operatorname {PolyLog}\left (3,1-\frac {2}{1+i c x}\right )}{2 d^2}-\frac {b^2 e \operatorname {PolyLog}\left (3,-1+\frac {2}{1+i c x}\right )}{2 d^2}+\frac {b^2 e \operatorname {PolyLog}\left (3,1-\frac {2 c \left (\sqrt {-d}-\sqrt {e} x\right )}{\left (c \sqrt {-d}-i \sqrt {e}\right ) (1-i c x)}\right )}{4 d^2}+\frac {b^2 e \operatorname {PolyLog}\left (3,1-\frac {2 c \left (\sqrt {-d}+\sqrt {e} x\right )}{\left (c \sqrt {-d}+i \sqrt {e}\right ) (1-i c x)}\right )}{4 d^2} \]

output

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

3.13.67.2 Mathematica [A] (warning: unable to verify)

Time = 14.31 (sec) , antiderivative size = 1412, normalized size of antiderivative = 1.90 \[ \int \frac {(a+b \arctan (c x))^2}{x^3 \left (d+e x^2\right )} \, dx =\text {Too large to display} \]

input

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

output

-1/24*((12*a^2*d)/x^2 + (24*a*b*c*d)/x + (24*a*b*d*(1 + c^2*x^2)*ArcTan[c* 
x])/x^2 + 24*a^2*e*Log[x] - 12*a^2*e*Log[d + e*x^2] - (24*I)*a*b*e*(ArcTan 
[c*x]*(ArcTan[c*x] + (2*I)*Log[1 - E^((2*I)*ArcTan[c*x])]) + PolyLog[2, E^ 
((2*I)*ArcTan[c*x])]) - (48*a*b*(c^2*d - e)*e*((-I)*ArcTan[c*x]^2 + (2*I)* 
ArcSin[Sqrt[(c^2*d)/(c^2*d - e)]]*ArcTan[(c*e*x)/Sqrt[c^2*d*e]] + (-ArcSin 
[Sqrt[(c^2*d)/(c^2*d - e)]] + ArcTan[c*x])*Log[1 + ((c^2*d + e + 2*Sqrt[c^ 
2*d*e])*E^((2*I)*ArcTan[c*x]))/(c^2*d - e)] + (ArcSin[Sqrt[(c^2*d)/(c^2*d 
- e)]] + ArcTan[c*x])*Log[(-2*Sqrt[c^2*d*e]*E^((2*I)*ArcTan[c*x]) + e*(-1 
+ E^((2*I)*ArcTan[c*x])) + c^2*d*(1 + E^((2*I)*ArcTan[c*x])))/(c^2*d - e)] 
 - (I/2)*(PolyLog[2, ((-(c^2*d) - e + 2*Sqrt[c^2*d*e])*E^((2*I)*ArcTan[c*x 
]))/(c^2*d - e)] + PolyLog[2, -(((c^2*d + e + 2*Sqrt[c^2*d*e])*E^((2*I)*Ar 
cTan[c*x]))/(c^2*d - e))])))/(2*c^2*d - 2*e) + b^2*((-I)*e*Pi^3 + (24*c*d* 
ArcTan[c*x])/x + (12*d*(1 + c^2*x^2)*ArcTan[c*x]^2)/x^2 + (8*I)*e*ArcTan[c 
*x]^3 + 24*e*ArcTan[c*x]^2*Log[1 - E^((-2*I)*ArcTan[c*x])] - 24*c^2*d*Log[ 
c*x] + 12*c^2*d*Log[1 + c^2*x^2] + (24*I)*e*ArcTan[c*x]*PolyLog[2, E^((-2* 
I)*ArcTan[c*x])] + 12*e*PolyLog[3, E^((-2*I)*ArcTan[c*x])]) + 2*b^2*e*((4* 
I)*ArcTan[c*x]^3 + 12*ArcSin[Sqrt[(c^2*d)/(c^2*d - e)]]*ArcTan[c*x]*Log[1 
+ ((c^2*d + e + 2*Sqrt[c^2*d*e])*E^((2*I)*ArcTan[c*x]))/(c^2*d - e)] - 12* 
ArcTan[c*x]^2*Log[1 + ((c^2*d + e + 2*Sqrt[c^2*d*e])*E^((2*I)*ArcTan[c*x]) 
)/(c^2*d - e)] - 12*ArcSin[Sqrt[(c^2*d)/(c^2*d - e)]]*ArcTan[c*x]*Log[(...
 

3.13.67.3 Rubi [A] (verified)

Time = 1.73 (sec) , antiderivative size = 724, normalized size of antiderivative = 0.97, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.478, Rules used = {5453, 5361, 5453, 5361, 243, 47, 14, 16, 5419, 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.

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

\(\Big \downarrow \) 5453

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

\(\Big \downarrow \) 5361

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

\(\Big \downarrow \) 5453

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

\(\Big \downarrow \) 5361

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

\(\Big \downarrow \) 243

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

\(\Big \downarrow \) 47

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

\(\Big \downarrow \) 14

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

\(\Big \downarrow \) 16

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

\(\Big \downarrow \) 5419

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

\(\Big \downarrow \) 5515

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

\(\Big \downarrow \) 2009

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

input

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

output

(-1/2*(a + b*ArcTan[c*x])^2/x^2 + b*c*(-((a + b*ArcTan[c*x])/x) - (c*(a + 
b*ArcTan[c*x])^2)/(2*b) + (b*c*(Log[x^2] - Log[1 + c^2*x^2]))/2))/d - (e*( 
(2*(a + b*ArcTan[c*x])^2*ArcTanh[1 - 2/(1 + I*c*x)])/d + ((a + b*ArcTan[c* 
x])^2*Log[2/(1 - I*c*x)])/d - ((a + b*ArcTan[c*x])^2*Log[(2*c*(Sqrt[-d] - 
Sqrt[e]*x))/((c*Sqrt[-d] - I*Sqrt[e])*(1 - I*c*x))])/(2*d) - ((a + b*ArcTa 
n[c*x])^2*Log[(2*c*(Sqrt[-d] + Sqrt[e]*x))/((c*Sqrt[-d] + I*Sqrt[e])*(1 - 
I*c*x))])/(2*d) - (I*b*(a + b*ArcTan[c*x])*PolyLog[2, 1 - 2/(1 - I*c*x)])/ 
d - (I*b*(a + b*ArcTan[c*x])*PolyLog[2, 1 - 2/(1 + I*c*x)])/d + (I*b*(a + 
b*ArcTan[c*x])*PolyLog[2, -1 + 2/(1 + I*c*x)])/d + ((I/2)*b*(a + b*ArcTan[ 
c*x])*PolyLog[2, 1 - (2*c*(Sqrt[-d] - Sqrt[e]*x))/((c*Sqrt[-d] - I*Sqrt[e] 
)*(1 - I*c*x))])/d + ((I/2)*b*(a + b*ArcTan[c*x])*PolyLog[2, 1 - (2*c*(Sqr 
t[-d] + Sqrt[e]*x))/((c*Sqrt[-d] + I*Sqrt[e])*(1 - I*c*x))])/d + (b^2*Poly 
Log[3, 1 - 2/(1 - I*c*x)])/(2*d) - (b^2*PolyLog[3, 1 - 2/(1 + I*c*x)])/(2* 
d) + (b^2*PolyLog[3, -1 + 2/(1 + I*c*x)])/(2*d) - (b^2*PolyLog[3, 1 - (2*c 
*(Sqrt[-d] - Sqrt[e]*x))/((c*Sqrt[-d] - I*Sqrt[e])*(1 - I*c*x))])/(4*d) - 
(b^2*PolyLog[3, 1 - (2*c*(Sqrt[-d] + Sqrt[e]*x))/((c*Sqrt[-d] + I*Sqrt[e]) 
*(1 - I*c*x))])/(4*d)))/d
 

3.13.67.3.1 Defintions of rubi rules used

rule 14

Int[(a_.)/(x_), x_Symbol] :> Simp[a*Log[x], x] /; FreeQ[a, x]
 

rule 16

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

rule 47

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

rule 243

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2   Subst[In 
t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I 
ntegerQ[(m - 1)/2]
 

rule 2009

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

rule 5361

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] - Simp[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]
 

rule 5419

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)^2), x_Symbo 
l] :> 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 5453

Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_))/((d_) + (e 
_.)*(x_)^2), x_Symbol] :> Simp[1/d   Int[(f*x)^m*(a + b*ArcTan[c*x])^p, x], 
 x] - Simp[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 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])
 

3.13.67.4 Maple [F]

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

input

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

output

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

3.13.67.5 Fricas [F]

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

input

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

output

integral((b^2*arctan(c*x)^2 + 2*a*b*arctan(c*x) + a^2)/(e*x^5 + d*x^3), x)
 

3.13.67.6 Sympy [F]

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

input

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

output

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

3.13.67.7 Maxima [F]

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

input

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

output

1/2*a^2*(e*log(e*x^2 + d)/d^2 - 2*e*log(x)/d^2 - 1/(d*x^2)) + integrate((b 
^2*arctan(c*x)^2 + 2*a*b*arctan(c*x))/(e*x^5 + d*x^3), x)
 

3.13.67.8 Giac [F]

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

input

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

output

sage0*x
 

3.13.67.9 Mupad [F(-1)]

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

input

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

output

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