\(\int \frac {a+b \arctan (c x)}{(d+e x)^2} \, dx\) [6]

Optimal result

Integrand size = 16, antiderivative size = 98 \[ \int \frac {a+b \arctan (c x)}{(d+e x)^2} \, dx=\frac {b c^2 d \arctan (c x)}{e \left (c^2 d^2+e^2\right )}-\frac {a+b \arctan (c x)}{e (d+e x)}+\frac {b c \log (d+e x)}{c^2 d^2+e^2}-\frac {b c \log \left (1+c^2 x^2\right )}{2 \left (c^2 d^2+e^2\right )} \] Output:

b*c^2*d*arctan(c*x)/e/(c^2*d^2+e^2)-(a+b*arctan(c*x))/e/(e*x+d)+b*c*ln(e*x 
+d)/(c^2*d^2+e^2)-b*c*ln(c^2*x^2+1)/(2*c^2*d^2+2*e^2)
 

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.13 \[ \int \frac {a+b \arctan (c x)}{(d+e x)^2} \, dx=\frac {-\frac {a+b \arctan (c x)}{d+e x}+\frac {b c \left (\left (\sqrt {-c^2} d-e\right ) \log \left (1-\sqrt {-c^2} x\right )-\left (\sqrt {-c^2} d+e\right ) \log \left (1+\sqrt {-c^2} x\right )+2 e \log (d+e x)\right )}{2 \left (c^2 d^2+e^2\right )}}{e} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.96, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {5387, 479, 452, 216, 240}

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.

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

\(\Big \downarrow \) 5387

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

\(\Big \downarrow \) 479

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

\(\Big \downarrow \) 452

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

\(\Big \downarrow \) 216

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

\(\Big \downarrow \) 240

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

Input:

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

Output:

-((a + b*ArcTan[c*x])/(e*(d + e*x))) + (b*c*((e*Log[d + e*x])/(c^2*d^2 + e 
^2) + (c^2*((d*ArcTan[c*x])/c - (e*Log[1 + c^2*x^2])/(2*c^2)))/(c^2*d^2 + 
e^2)))/e
 

Defintions of rubi rules used

rule 216

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[b, 0])
 

rule 240

Int[(x_)/((a_) + (b_.)*(x_)^2), x_Symbol] :> Simp[Log[RemoveContent[a + b*x 
^2, x]]/(2*b), x] /; FreeQ[{a, b}, x]
 

rule 452

Int[((c_) + (d_.)*(x_))/((a_) + (b_.)*(x_)^2), x_Symbol] :> Simp[c   Int[1/ 
(a + b*x^2), x], x] + Simp[d   Int[x/(a + b*x^2), x], x] /; FreeQ[{a, b, c, 
 d}, x] && NeQ[b*c^2 + a*d^2, 0]
 

rule 479

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

rule 5387

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

Maple [A] (verified)

Time = 0.24 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.12

method	result	size
parts	\(-\frac {a}{\left (e x +d \right ) e}+\frac {b \left (-\frac {c^{2} \arctan \left (c x \right )}{\left (c e x +c d \right ) e}+\frac {c^{2} \left (\frac {e \ln \left (c e x +c d \right )}{c^{2} d^{2}+e^{2}}+\frac {-\frac {e \ln \left (c^{2} x^{2}+1\right )}{2}+d c \arctan \left (c x \right )}{c^{2} d^{2}+e^{2}}\right )}{e}\right )}{c}\)	\(110\)
derivativedivides	\(\frac {-\frac {a \,c^{2}}{\left (c e x +c d \right ) e}+b \,c^{2} \left (-\frac {\arctan \left (c x \right )}{\left (c e x +c d \right ) e}+\frac {\frac {e \ln \left (c e x +c d \right )}{c^{2} d^{2}+e^{2}}+\frac {-\frac {e \ln \left (c^{2} x^{2}+1\right )}{2}+d c \arctan \left (c x \right )}{c^{2} d^{2}+e^{2}}}{e}\right )}{c}\)	\(114\)
default	\(\frac {-\frac {a \,c^{2}}{\left (c e x +c d \right ) e}+b \,c^{2} \left (-\frac {\arctan \left (c x \right )}{\left (c e x +c d \right ) e}+\frac {\frac {e \ln \left (c e x +c d \right )}{c^{2} d^{2}+e^{2}}+\frac {-\frac {e \ln \left (c^{2} x^{2}+1\right )}{2}+d c \arctan \left (c x \right )}{c^{2} d^{2}+e^{2}}}{e}\right )}{c}\)	\(114\)
parallelrisch	\(-\frac {-2 b \,c^{4} d \arctan \left (c x \right ) x e +\ln \left (c^{2} x^{2}+1\right ) x b \,c^{3} e^{2}-2 \ln \left (e x +d \right ) x b \,c^{3} e^{2}+\ln \left (c^{2} x^{2}+1\right ) b \,c^{3} d e -2 \ln \left (e x +d \right ) b \,c^{3} d e +2 a \,c^{4} d^{2}+2 e^{2} b \arctan \left (c x \right ) c^{2}+2 a \,c^{2} e^{2}}{2 \left (e x +d \right ) c^{2} e \left (c^{2} d^{2}+e^{2}\right )}\)	\(140\)
risch	\(\frac {i b \ln \left (i c x +1\right )}{2 e \left (e x +d \right )}+\frac {-i b \,c^{2} d^{2} \ln \left (-i c x +1\right )-i b \,e^{2} \ln \left (-i c x +1\right )+2 \ln \left (-e x -d \right ) b c \,e^{2} x +2 \ln \left (-e x -d \right ) b c d e -2 a \,c^{2} d^{2}-2 e^{2} a -\ln \left (\left (-i c^{2} d +3 c e \right ) x +3 i e +c d \right ) b c \,e^{2} x -\ln \left (\left (-i c^{2} d +3 c e \right ) x +3 i e +c d \right ) b c d e +i \ln \left (\left (-i c^{2} d +3 c e \right ) x +3 i e +c d \right ) b \,c^{2} d e x +i \ln \left (\left (-i c^{2} d +3 c e \right ) x +3 i e +c d \right ) b \,c^{2} d^{2}-\ln \left (\left (-i c^{2} d -3 c e \right ) x +3 i e -c d \right ) b c \,e^{2} x -\ln \left (\left (-i c^{2} d -3 c e \right ) x +3 i e -c d \right ) b c d e -i \ln \left (\left (-i c^{2} d -3 c e \right ) x +3 i e -c d \right ) b \,c^{2} d e x -i \ln \left (\left (-i c^{2} d -3 c e \right ) x +3 i e -c d \right ) b \,c^{2} d^{2}}{2 \left (e x +d \right ) \left (i c d +e \right ) \left (-i c d +e \right ) e}\)	\(391\)

Input:

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

Output:

-a/(e*x+d)/e+b/c*(-c^2/(c*e*x+c*d)/e*arctan(c*x)+c^2/e*(e/(c^2*d^2+e^2)*ln 
(c*e*x+c*d)+1/(c^2*d^2+e^2)*(-1/2*e*ln(c^2*x^2+1)+d*c*arctan(c*x))))
 

Fricas [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.18 \[ \int \frac {a+b \arctan (c x)}{(d+e x)^2} \, dx=-\frac {2 \, a c^{2} d^{2} + 2 \, a e^{2} - 2 \, {\left (b c^{2} d e x - b e^{2}\right )} \arctan \left (c x\right ) + {\left (b c e^{2} x + b c d e\right )} \log \left (c^{2} x^{2} + 1\right ) - 2 \, {\left (b c e^{2} x + b c d e\right )} \log \left (e x + d\right )}{2 \, {\left (c^{2} d^{3} e + d e^{3} + {\left (c^{2} d^{2} e^{2} + e^{4}\right )} x\right )}} \] Input:

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

Output:

-1/2*(2*a*c^2*d^2 + 2*a*e^2 - 2*(b*c^2*d*e*x - b*e^2)*arctan(c*x) + (b*c*e 
^2*x + b*c*d*e)*log(c^2*x^2 + 1) - 2*(b*c*e^2*x + b*c*d*e)*log(e*x + d))/( 
c^2*d^3*e + d*e^3 + (c^2*d^2*e^2 + e^4)*x)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.94 (sec) , antiderivative size = 658, normalized size of antiderivative = 6.71 \[ \int \frac {a+b \arctan (c x)}{(d+e x)^2} \, dx=\begin {cases} \frac {a x}{d^{2}} & \text {for}\: c = 0 \wedge e = 0 \\\frac {a x + b x \operatorname {atan}{\left (c x \right )} - \frac {b \log {\left (x^{2} + \frac {1}{c^{2}} \right )}}{2 c}}{d^{2}} & \text {for}\: e = 0 \\- \frac {a}{d e + e^{2} x} & \text {for}\: c = 0 \\- \frac {2 a d}{2 d^{2} e + 2 d e^{2} x} + \frac {i b d \operatorname {atanh}{\left (\frac {e x}{d} \right )}}{2 d^{2} e + 2 d e^{2} x} + \frac {i b d}{2 d^{2} e + 2 d e^{2} x} - \frac {i b e x \operatorname {atanh}{\left (\frac {e x}{d} \right )}}{2 d^{2} e + 2 d e^{2} x} & \text {for}\: c = - \frac {i e}{d} \\- \frac {2 a d}{2 d^{2} e + 2 d e^{2} x} - \frac {i b d \operatorname {atanh}{\left (\frac {e x}{d} \right )}}{2 d^{2} e + 2 d e^{2} x} - \frac {i b d}{2 d^{2} e + 2 d e^{2} x} + \frac {i b e x \operatorname {atanh}{\left (\frac {e x}{d} \right )}}{2 d^{2} e + 2 d e^{2} x} & \text {for}\: c = \frac {i e}{d} \\- \frac {2 a c^{2} d^{2}}{2 c^{2} d^{3} e + 2 c^{2} d^{2} e^{2} x + 2 d e^{3} + 2 e^{4} x} - \frac {2 a e^{2}}{2 c^{2} d^{3} e + 2 c^{2} d^{2} e^{2} x + 2 d e^{3} + 2 e^{4} x} + \frac {2 b c^{2} d e x \operatorname {atan}{\left (c x \right )}}{2 c^{2} d^{3} e + 2 c^{2} d^{2} e^{2} x + 2 d e^{3} + 2 e^{4} x} - \frac {b c d e \log {\left (x^{2} + \frac {1}{c^{2}} \right )}}{2 c^{2} d^{3} e + 2 c^{2} d^{2} e^{2} x + 2 d e^{3} + 2 e^{4} x} + \frac {2 b c d e \log {\left (\frac {d}{e} + x \right )}}{2 c^{2} d^{3} e + 2 c^{2} d^{2} e^{2} x + 2 d e^{3} + 2 e^{4} x} - \frac {b c e^{2} x \log {\left (x^{2} + \frac {1}{c^{2}} \right )}}{2 c^{2} d^{3} e + 2 c^{2} d^{2} e^{2} x + 2 d e^{3} + 2 e^{4} x} + \frac {2 b c e^{2} x \log {\left (\frac {d}{e} + x \right )}}{2 c^{2} d^{3} e + 2 c^{2} d^{2} e^{2} x + 2 d e^{3} + 2 e^{4} x} - \frac {2 b e^{2} \operatorname {atan}{\left (c x \right )}}{2 c^{2} d^{3} e + 2 c^{2} d^{2} e^{2} x + 2 d e^{3} + 2 e^{4} x} & \text {otherwise} \end {cases} \] Input:

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

Output:

Piecewise((a*x/d**2, Eq(c, 0) & Eq(e, 0)), ((a*x + b*x*atan(c*x) - b*log(x 
**2 + c**(-2))/(2*c))/d**2, Eq(e, 0)), (-a/(d*e + e**2*x), Eq(c, 0)), (-2* 
a*d/(2*d**2*e + 2*d*e**2*x) + I*b*d*atanh(e*x/d)/(2*d**2*e + 2*d*e**2*x) + 
 I*b*d/(2*d**2*e + 2*d*e**2*x) - I*b*e*x*atanh(e*x/d)/(2*d**2*e + 2*d*e**2 
*x), Eq(c, -I*e/d)), (-2*a*d/(2*d**2*e + 2*d*e**2*x) - I*b*d*atanh(e*x/d)/ 
(2*d**2*e + 2*d*e**2*x) - I*b*d/(2*d**2*e + 2*d*e**2*x) + I*b*e*x*atanh(e* 
x/d)/(2*d**2*e + 2*d*e**2*x), Eq(c, I*e/d)), (-2*a*c**2*d**2/(2*c**2*d**3* 
e + 2*c**2*d**2*e**2*x + 2*d*e**3 + 2*e**4*x) - 2*a*e**2/(2*c**2*d**3*e + 
2*c**2*d**2*e**2*x + 2*d*e**3 + 2*e**4*x) + 2*b*c**2*d*e*x*atan(c*x)/(2*c* 
*2*d**3*e + 2*c**2*d**2*e**2*x + 2*d*e**3 + 2*e**4*x) - b*c*d*e*log(x**2 + 
 c**(-2))/(2*c**2*d**3*e + 2*c**2*d**2*e**2*x + 2*d*e**3 + 2*e**4*x) + 2*b 
*c*d*e*log(d/e + x)/(2*c**2*d**3*e + 2*c**2*d**2*e**2*x + 2*d*e**3 + 2*e** 
4*x) - b*c*e**2*x*log(x**2 + c**(-2))/(2*c**2*d**3*e + 2*c**2*d**2*e**2*x 
+ 2*d*e**3 + 2*e**4*x) + 2*b*c*e**2*x*log(d/e + x)/(2*c**2*d**3*e + 2*c**2 
*d**2*e**2*x + 2*d*e**3 + 2*e**4*x) - 2*b*e**2*atan(c*x)/(2*c**2*d**3*e + 
2*c**2*d**2*e**2*x + 2*d*e**3 + 2*e**4*x), True))
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.09 \[ \int \frac {a+b \arctan (c x)}{(d+e x)^2} \, dx=\frac {1}{2} \, {\left ({\left (\frac {2 \, c d \arctan \left (c x\right )}{c^{2} d^{2} e + e^{3}} - \frac {\log \left (c^{2} x^{2} + 1\right )}{c^{2} d^{2} + e^{2}} + \frac {2 \, \log \left (e x + d\right )}{c^{2} d^{2} + e^{2}}\right )} c - \frac {2 \, \arctan \left (c x\right )}{e^{2} x + d e}\right )} b - \frac {a}{e^{2} x + d e} \] Input:

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

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1152 vs. \(2 (96) = 192\).

Time = 0.32 (sec) , antiderivative size = 1152, normalized size of antiderivative = 11.76 \[ \int \frac {a+b \arctan (c x)}{(d+e x)^2} \, dx=\text {Too large to display} \] Input:

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

Output:

1/2*(c^3*d*log(4*(c^2*d^2*tan(1/2*arctan(-(e*x + d)*(c - c*d/(e*x + d))/e) 
)^4 - 2*c^2*d^2*tan(1/2*arctan(-(e*x + d)*(c - c*d/(e*x + d))/e))^2 + 4*c* 
d*e*tan(1/2*arctan(-(e*x + d)*(c - c*d/(e*x + d))/e))^3 + c^2*d^2 - 4*c*d* 
e*tan(1/2*arctan(-(e*x + d)*(c - c*d/(e*x + d))/e)) + 4*e^2*tan(1/2*arctan 
(-(e*x + d)*(c - c*d/(e*x + d))/e))^2)/(tan(1/2*arctan(-(e*x + d)*(c - c*d 
/(e*x + d))/e))^4 + 2*tan(1/2*arctan(-(e*x + d)*(c - c*d/(e*x + d))/e))^2 
+ 1))*tan(1/2*arctan(-(e*x + d)*(c - c*d/(e*x + d))/e))^2 - 4*c^3*d*arctan 
(-(e*x + d)*(c - c*d/(e*x + d))/e)*tan(1/2*arctan(-(e*x + d)*(c - c*d/(e*x 
 + d))/e)) + 2*c^2*e*arctan(-(e*x + d)*(c - c*d/(e*x + d))/e)*tan(1/2*arct 
an(-(e*x + d)*(c - c*d/(e*x + d))/e))^2 - c^3*d*log(4*(c^2*d^2*tan(1/2*arc 
tan(-(e*x + d)*(c - c*d/(e*x + d))/e))^4 - 2*c^2*d^2*tan(1/2*arctan(-(e*x 
+ d)*(c - c*d/(e*x + d))/e))^2 + 4*c*d*e*tan(1/2*arctan(-(e*x + d)*(c - c* 
d/(e*x + d))/e))^3 + c^2*d^2 - 4*c*d*e*tan(1/2*arctan(-(e*x + d)*(c - c*d/ 
(e*x + d))/e)) + 4*e^2*tan(1/2*arctan(-(e*x + d)*(c - c*d/(e*x + d))/e))^2 
)/(tan(1/2*arctan(-(e*x + d)*(c - c*d/(e*x + d))/e))^4 + 2*tan(1/2*arctan( 
-(e*x + d)*(c - c*d/(e*x + d))/e))^2 + 1)) + 2*c^2*e*log(4*(c^2*d^2*tan(1/ 
2*arctan(-(e*x + d)*(c - c*d/(e*x + d))/e))^4 - 2*c^2*d^2*tan(1/2*arctan(- 
(e*x + d)*(c - c*d/(e*x + d))/e))^2 + 4*c*d*e*tan(1/2*arctan(-(e*x + d)*(c 
 - c*d/(e*x + d))/e))^3 + c^2*d^2 - 4*c*d*e*tan(1/2*arctan(-(e*x + d)*(c - 
 c*d/(e*x + d))/e)) + 4*e^2*tan(1/2*arctan(-(e*x + d)*(c - c*d/(e*x + d...
 

Mupad [B] (verification not implemented)

Time = 4.10 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.14 \[ \int \frac {a+b \arctan (c x)}{(d+e x)^2} \, dx=\frac {d^2\,\left (b\,c\,\ln \left (d+e\,x\right )-\frac {b\,c\,\ln \left (c^2\,x^2+1\right )}{2}+a\,c^2\,x+b\,c^2\,x\,\mathrm {atan}\left (c\,x\right )\right )-d\,e\,\left (b\,\mathrm {atan}\left (c\,x\right )-b\,c\,x\,\ln \left (d+e\,x\right )+\frac {b\,c\,x\,\ln \left (c^2\,x^2+1\right )}{2}\right )+a\,e^2\,x}{d\,\left (c^2\,d^2+e^2\right )\,\left (d+e\,x\right )} \] Input:

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

Output:

(d^2*(b*c*log(d + e*x) - (b*c*log(c^2*x^2 + 1))/2 + a*c^2*x + b*c^2*x*atan 
(c*x)) - d*e*(b*atan(c*x) - b*c*x*log(d + e*x) + (b*c*x*log(c^2*x^2 + 1))/ 
2) + a*e^2*x)/(d*(e^2 + c^2*d^2)*(d + e*x))
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.38 \[ \int \frac {a+b \arctan (c x)}{(d+e x)^2} \, dx=\frac {2 \mathit {atan} \left (c x \right ) b \,c^{2} d^{2} x -2 \mathit {atan} \left (c x \right ) b d e -\mathrm {log}\left (c^{2} x^{2}+1\right ) b c \,d^{2}-\mathrm {log}\left (c^{2} x^{2}+1\right ) b c d e x +2 \,\mathrm {log}\left (e x +d \right ) b c \,d^{2}+2 \,\mathrm {log}\left (e x +d \right ) b c d e x +2 a \,c^{2} d^{2} x +2 a \,e^{2} x}{2 d \left (c^{2} d^{2} e x +c^{2} d^{3}+e^{3} x +d \,e^{2}\right )} \] Input:

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

Output:

(2*atan(c*x)*b*c**2*d**2*x - 2*atan(c*x)*b*d*e - log(c**2*x**2 + 1)*b*c*d* 
*2 - log(c**2*x**2 + 1)*b*c*d*e*x + 2*log(d + e*x)*b*c*d**2 + 2*log(d + e* 
x)*b*c*d*e*x + 2*a*c**2*d**2*x + 2*a*e**2*x)/(2*d*(c**2*d**3 + c**2*d**2*e 
*x + d*e**2 + e**3*x))