3.12.57 \(\int \frac {a+b \arctan (c x)}{x^2 (d+e x^2)} \, dx\) [1157]

3.12.57.1 Optimal result

Integrand size = 21, antiderivative size = 561 \[ \int \frac {a+b \arctan (c x)}{x^2 \left (d+e x^2\right )} \, dx=-\frac {a+b \arctan (c x)}{d x}-\frac {a \sqrt {e} \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{d^{3/2}}+\frac {b c \log (x)}{d}-\frac {i b \sqrt {e} \log (1+i c x) \log \left (\frac {c \left (\sqrt {-d}-\sqrt {e} x\right )}{c \sqrt {-d}-i \sqrt {e}}\right )}{4 (-d)^{3/2}}+\frac {i b \sqrt {e} \log (1-i c x) \log \left (\frac {c \left (\sqrt {-d}-\sqrt {e} x\right )}{c \sqrt {-d}+i \sqrt {e}}\right )}{4 (-d)^{3/2}}-\frac {i b \sqrt {e} \log (1-i c x) \log \left (\frac {c \left (\sqrt {-d}+\sqrt {e} x\right )}{c \sqrt {-d}-i \sqrt {e}}\right )}{4 (-d)^{3/2}}+\frac {i b \sqrt {e} \log (1+i c x) \log \left (\frac {c \left (\sqrt {-d}+\sqrt {e} x\right )}{c \sqrt {-d}+i \sqrt {e}}\right )}{4 (-d)^{3/2}}-\frac {b c \log \left (1+c^2 x^2\right )}{2 d}+\frac {i b \sqrt {e} \operatorname {PolyLog}\left (2,\frac {\sqrt {e} (i-c x)}{c \sqrt {-d}+i \sqrt {e}}\right )}{4 (-d)^{3/2}}-\frac {i b \sqrt {e} \operatorname {PolyLog}\left (2,\frac {\sqrt {e} (1-i c x)}{i c \sqrt {-d}+\sqrt {e}}\right )}{4 (-d)^{3/2}}-\frac {i b \sqrt {e} \operatorname {PolyLog}\left (2,\frac {\sqrt {e} (1+i c x)}{i c \sqrt {-d}+\sqrt {e}}\right )}{4 (-d)^{3/2}}+\frac {i b \sqrt {e} \operatorname {PolyLog}\left (2,\frac {\sqrt {e} (i+c x)}{c \sqrt {-d}+i \sqrt {e}}\right )}{4 (-d)^{3/2}} \]

output

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

3.12.57.2 Mathematica [A] (verified)

Time = 0.54 (sec) , antiderivative size = 467, normalized size of antiderivative = 0.83 \[ \int \frac {a+b \arctan (c x)}{x^2 \left (d+e x^2\right )} \, dx=\frac {-\frac {a+b \arctan (c x)}{x}+b c \left (\log (x)-\frac {1}{2} \log \left (1+c^2 x^2\right )\right )-\frac {\sqrt {e} \left (4 a \sqrt {-d} \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )+i b \sqrt {d} \left (\log (1+i c x) \log \left (\frac {c \left (\sqrt {-d}+\sqrt {e} x\right )}{c \sqrt {-d}+i \sqrt {e}}\right )+\operatorname {PolyLog}\left (2,\frac {\sqrt {e} (i-c x)}{c \sqrt {-d}+i \sqrt {e}}\right )\right )-i b \sqrt {d} \left (\log (1-i c x) \log \left (\frac {c \left (\sqrt {-d}+\sqrt {e} x\right )}{c \sqrt {-d}-i \sqrt {e}}\right )+\operatorname {PolyLog}\left (2,\frac {\sqrt {e} (1-i c x)}{i c \sqrt {-d}+\sqrt {e}}\right )\right )-i b \sqrt {d} \left (\log (1+i c x) \log \left (\frac {c \left (\sqrt {-d}-\sqrt {e} x\right )}{c \sqrt {-d}-i \sqrt {e}}\right )+\operatorname {PolyLog}\left (2,\frac {\sqrt {e} (1+i c x)}{i c \sqrt {-d}+\sqrt {e}}\right )\right )+i b \sqrt {d} \left (\log (1-i c x) \log \left (\frac {c \left (\sqrt {-d}-\sqrt {e} x\right )}{c \sqrt {-d}+i \sqrt {e}}\right )+\operatorname {PolyLog}\left (2,\frac {\sqrt {e} (i+c x)}{c \sqrt {-d}+i \sqrt {e}}\right )\right )\right )}{4 \sqrt {-d^2}}}{d} \]

input

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

output

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

3.12.57.3 Rubi [A] (verified)

Time = 0.93 (sec) , antiderivative size = 558, normalized size of antiderivative = 0.99, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.524, Rules used = {5453, 5361, 243, 47, 14, 16, 5445, 218, 5443, 2856, 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.

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

\(\Big \downarrow \) 5453

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

\(\Big \downarrow \) 5361

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

\(\Big \downarrow \) 243

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

\(\Big \downarrow \) 47

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

\(\Big \downarrow \) 14

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

\(\Big \downarrow \) 16

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

\(\Big \downarrow \) 5445

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

\(\Big \downarrow \) 218

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

\(\Big \downarrow \) 5443

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

\(\Big \downarrow \) 2856

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

\(\Big \downarrow \) 2009

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

input

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

output

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

3.12.57.3.1 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 218

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R 
t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
 

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 2856

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_) + (g_. 
)*(x_)^(r_))^(q_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*Log[c*(d + e*x) 
^n])^p, (f + g*x^r)^q, x], x] /; FreeQ[{a, b, c, d, e, f, g, n, r}, x] && I 
GtQ[p, 0] && IntegerQ[q] && (GtQ[q, 0] || (IntegerQ[r] && NeQ[r, 1]))
 

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 5443

Int[ArcTan[(c_.)*(x_)]/((d_.) + (e_.)*(x_)^2), x_Symbol] :> Simp[I/2   Int[ 
Log[1 - I*c*x]/(d + e*x^2), x], x] - Simp[I/2   Int[Log[1 + I*c*x]/(d + e*x 
^2), x], x] /; FreeQ[{c, d, e}, x]
 

rule 5445

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

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]
 

3.12.57.4 Maple [A] (verified)

Time = 0.89 (sec) , antiderivative size = 537, normalized size of antiderivative = 0.96

method	result	size
risch	\(-\frac {b e \ln \left (-i c x +1\right ) \ln \left (\frac {c \sqrt {e d}-\left (-i c x +1\right ) e +e}{c \sqrt {e d}+e}\right )}{4 d \sqrt {e d}}+\frac {b e \ln \left (-i c x +1\right ) \ln \left (\frac {c \sqrt {e d}+\left (-i c x +1\right ) e -e}{c \sqrt {e d}-e}\right )}{4 d \sqrt {e d}}-\frac {b e \operatorname {dilog}\left (\frac {c \sqrt {e d}-\left (-i c x +1\right ) e +e}{c \sqrt {e d}+e}\right )}{4 d \sqrt {e d}}+\frac {b e \operatorname {dilog}\left (\frac {c \sqrt {e d}+\left (-i c x +1\right ) e -e}{c \sqrt {e d}-e}\right )}{4 d \sqrt {e d}}+\frac {c b \ln \left (-i c x \right )}{2 d}-\frac {c b \ln \left (-i c x +1\right )}{2 d}+\frac {i b \ln \left (i c x +1\right )}{2 d x}-\frac {i b \ln \left (-i c x +1\right )}{2 d x}-\frac {a}{d x}-\frac {b e \ln \left (i c x +1\right ) \ln \left (\frac {c \sqrt {e d}-\left (i c x +1\right ) e +e}{c \sqrt {e d}+e}\right )}{4 d \sqrt {e d}}+\frac {b e \ln \left (i c x +1\right ) \ln \left (\frac {c \sqrt {e d}+\left (i c x +1\right ) e -e}{c \sqrt {e d}-e}\right )}{4 d \sqrt {e d}}-\frac {b e \operatorname {dilog}\left (\frac {c \sqrt {e d}-\left (i c x +1\right ) e +e}{c \sqrt {e d}+e}\right )}{4 d \sqrt {e d}}+\frac {b e \operatorname {dilog}\left (\frac {c \sqrt {e d}+\left (i c x +1\right ) e -e}{c \sqrt {e d}-e}\right )}{4 d \sqrt {e d}}+\frac {b c \ln \left (i c x \right )}{2 d}-\frac {b c \ln \left (i c x +1\right )}{2 d}-\frac {i a e \,\operatorname {arctanh}\left (\frac {2 \left (-i c x +1\right ) e -2 e}{2 c \sqrt {e d}}\right )}{d \sqrt {e d}}\)	\(537\)
parts	\(\text {Expression too large to display}\)	\(2411\)
derivativedivides	\(\text {Expression too large to display}\)	\(2432\)
default	\(\text {Expression too large to display}\)	\(2432\)

input

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

output

-1/4*b*e/d*ln(1-I*c*x)/(e*d)^(1/2)*ln((c*(e*d)^(1/2)-(1-I*c*x)*e+e)/(c*(e* 
d)^(1/2)+e))+1/4*b*e/d*ln(1-I*c*x)/(e*d)^(1/2)*ln((c*(e*d)^(1/2)+(1-I*c*x) 
*e-e)/(c*(e*d)^(1/2)-e))-1/4*b*e/d/(e*d)^(1/2)*dilog((c*(e*d)^(1/2)-(1-I*c 
*x)*e+e)/(c*(e*d)^(1/2)+e))+1/4*b*e/d/(e*d)^(1/2)*dilog((c*(e*d)^(1/2)+(1- 
I*c*x)*e-e)/(c*(e*d)^(1/2)-e))+1/2*c*b/d*ln(-I*c*x)-1/2*c*b/d*ln(1-I*c*x)+ 
1/2*I*b/d*ln(1+I*c*x)/x-1/2*I*b/d*ln(1-I*c*x)/x-a/d/x-1/4*b*e/d*ln(1+I*c*x 
)/(e*d)^(1/2)*ln((c*(e*d)^(1/2)-(1+I*c*x)*e+e)/(c*(e*d)^(1/2)+e))+1/4*b*e/ 
d*ln(1+I*c*x)/(e*d)^(1/2)*ln((c*(e*d)^(1/2)+(1+I*c*x)*e-e)/(c*(e*d)^(1/2)- 
e))-1/4*b*e/d/(e*d)^(1/2)*dilog((c*(e*d)^(1/2)-(1+I*c*x)*e+e)/(c*(e*d)^(1/ 
2)+e))+1/4*b*e/d/(e*d)^(1/2)*dilog((c*(e*d)^(1/2)+(1+I*c*x)*e-e)/(c*(e*d)^ 
(1/2)-e))+1/2*b*c/d*ln(I*c*x)-1/2*b*c/d*ln(1+I*c*x)-I*a*e/d/(e*d)^(1/2)*ar 
ctanh(1/2*(2*(1-I*c*x)*e-2*e)/c/(e*d)^(1/2))
 

3.12.57.5 Fricas [F]

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

input

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

output

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

3.12.57.6 Sympy [F(-1)]

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

input

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

output

Timed out
 

3.12.57.7 Maxima [F(-2)]

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

input

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

3.12.57.8 Giac [F]

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

input

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

output

sage0*x
 

3.12.57.9 Mupad [F(-1)]

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

input

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

output

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