3.4.20 \(\int \frac {(d+e x) (2+x+3 x^2-5 x^3+4 x^4)}{(3+2 x+5 x^2)^3} \, dx\) [320]

3.4.20.1 Optimal result

Integrand size = 36, antiderivative size = 103 \[ \int \frac {(d+e x) \left (2+x+3 x^2-5 x^3+4 x^4\right )}{\left (3+2 x+5 x^2\right )^3} \, dx=-\frac {(1367+423 x) (d+e x)}{7000 \left (3+2 x+5 x^2\right )^2}+\frac {34347 d-6511 e+(11015 d+36353 e) x}{196000 \left (3+2 x+5 x^2\right )}+\frac {(42375 d-34207 e) \arctan \left (\frac {1+5 x}{\sqrt {14}}\right )}{196000 \sqrt {14}}+\frac {2}{125} e \log \left (3+2 x+5 x^2\right ) \]

output

-1/7000*(1367+423*x)*(e*x+d)/(5*x^2+2*x+3)^2+1/196000*(34347*d-6511*e+(110 
15*d+36353*e)*x)/(5*x^2+2*x+3)+2/125*e*ln(5*x^2+2*x+3)+1/2744000*(42375*d- 
34207*e)*arctan(1/14*(1+5*x)*14^(1/2))*14^(1/2)
 

3.4.20.2 Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.04 \[ \int \frac {(d+e x) \left (2+x+3 x^2-5 x^3+4 x^4\right )}{\left (3+2 x+5 x^2\right )^3} \, dx=\frac {-6835 d+1269 e-2115 d x-5989 e x}{35000 \left (3+2 x+5 x^2\right )^2}+\frac {171735 d-44399 e+55075 d x+181765 e x}{980000 \left (3+2 x+5 x^2\right )}+\frac {(42375 d-34207 e) \arctan \left (\frac {1+5 x}{\sqrt {14}}\right )}{196000 \sqrt {14}}+\frac {2}{125} e \log \left (3+2 x+5 x^2\right ) \]

input

Integrate[((d + e*x)*(2 + x + 3*x^2 - 5*x^3 + 4*x^4))/(3 + 2*x + 5*x^2)^3, 
x]
 

output

(-6835*d + 1269*e - 2115*d*x - 5989*e*x)/(35000*(3 + 2*x + 5*x^2)^2) + (17 
1735*d - 44399*e + 55075*d*x + 181765*e*x)/(980000*(3 + 2*x + 5*x^2)) + (( 
42375*d - 34207*e)*ArcTan[(1 + 5*x)/Sqrt[14]])/(196000*Sqrt[14]) + (2*e*Lo 
g[3 + 2*x + 5*x^2])/125
 

3.4.20.3 Rubi [A] (verified)

Time = 0.39 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.13, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2175, 27, 2191, 27, 1142, 27, 1083, 217, 1103}

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.

\(\displaystyle \int \frac {\left (4 x^4-5 x^3+3 x^2+x+2\right ) (d+e x)}{\left (5 x^2+2 x+3\right )^3} \, dx\)

\(\Big \downarrow \) 2175

\(\displaystyle \frac {1}{112} \int \frac {2 \left (5600 e x^3+280 (20 d-33 e) x^2-30 (308 d-123 e) x+3267 d+1367 e\right )}{125 \left (5 x^2+2 x+3\right )^2}dx-\frac {(423 x+1367) (d+e x)}{7000 \left (5 x^2+2 x+3\right )^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {5600 e x^3+280 (20 d-33 e) x^2-30 (308 d-123 e) x+3267 d+1367 e}{\left (5 x^2+2 x+3\right )^2}dx}{7000}-\frac {(423 x+1367) (d+e x)}{7000 \left (5 x^2+2 x+3\right )^2}\)

\(\Big \downarrow \) 2191

\(\displaystyle \frac {\frac {1}{56} \int \frac {10 (8475 d-5587 e+6272 e x)}{5 x^2+2 x+3}dx+\frac {x (11015 d+36353 e)+34347 d-6511 e}{28 \left (5 x^2+2 x+3\right )}}{7000}-\frac {(423 x+1367) (d+e x)}{7000 \left (5 x^2+2 x+3\right )^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {5}{28} \int \frac {8475 d-5587 e+6272 e x}{5 x^2+2 x+3}dx+\frac {x (11015 d+36353 e)+34347 d-6511 e}{28 \left (5 x^2+2 x+3\right )}}{7000}-\frac {(423 x+1367) (d+e x)}{7000 \left (5 x^2+2 x+3\right )^2}\)

\(\Big \downarrow \) 1142

\(\displaystyle \frac {\frac {5}{28} \left (\frac {1}{5} (42375 d-34207 e) \int \frac {1}{5 x^2+2 x+3}dx+\frac {3136}{5} e \int \frac {2 (5 x+1)}{5 x^2+2 x+3}dx\right )+\frac {x (11015 d+36353 e)+34347 d-6511 e}{28 \left (5 x^2+2 x+3\right )}}{7000}-\frac {(423 x+1367) (d+e x)}{7000 \left (5 x^2+2 x+3\right )^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {5}{28} \left (\frac {1}{5} (42375 d-34207 e) \int \frac {1}{5 x^2+2 x+3}dx+\frac {6272}{5} e \int \frac {5 x+1}{5 x^2+2 x+3}dx\right )+\frac {x (11015 d+36353 e)+34347 d-6511 e}{28 \left (5 x^2+2 x+3\right )}}{7000}-\frac {(423 x+1367) (d+e x)}{7000 \left (5 x^2+2 x+3\right )^2}\)

\(\Big \downarrow \) 1083

\(\displaystyle \frac {\frac {5}{28} \left (\frac {6272}{5} e \int \frac {5 x+1}{5 x^2+2 x+3}dx-\frac {2}{5} (42375 d-34207 e) \int \frac {1}{-(10 x+2)^2-56}d(10 x+2)\right )+\frac {x (11015 d+36353 e)+34347 d-6511 e}{28 \left (5 x^2+2 x+3\right )}}{7000}-\frac {(423 x+1367) (d+e x)}{7000 \left (5 x^2+2 x+3\right )^2}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {\frac {5}{28} \left (\frac {6272}{5} e \int \frac {5 x+1}{5 x^2+2 x+3}dx+\frac {\arctan \left (\frac {10 x+2}{2 \sqrt {14}}\right ) (42375 d-34207 e)}{5 \sqrt {14}}\right )+\frac {x (11015 d+36353 e)+34347 d-6511 e}{28 \left (5 x^2+2 x+3\right )}}{7000}-\frac {(423 x+1367) (d+e x)}{7000 \left (5 x^2+2 x+3\right )^2}\)

\(\Big \downarrow \) 1103

\(\displaystyle \frac {\frac {5}{28} \left (\frac {\arctan \left (\frac {10 x+2}{2 \sqrt {14}}\right ) (42375 d-34207 e)}{5 \sqrt {14}}+\frac {3136}{5} e \log \left (5 x^2+2 x+3\right )\right )+\frac {x (11015 d+36353 e)+34347 d-6511 e}{28 \left (5 x^2+2 x+3\right )}}{7000}-\frac {(423 x+1367) (d+e x)}{7000 \left (5 x^2+2 x+3\right )^2}\)

input

Int[((d + e*x)*(2 + x + 3*x^2 - 5*x^3 + 4*x^4))/(3 + 2*x + 5*x^2)^3,x]
 

output

-1/7000*((1367 + 423*x)*(d + e*x))/(3 + 2*x + 5*x^2)^2 + ((34347*d - 6511* 
e + (11015*d + 36353*e)*x)/(28*(3 + 2*x + 5*x^2)) + (5*(((42375*d - 34207* 
e)*ArcTan[(2 + 10*x)/(2*Sqrt[14])])/(5*Sqrt[14]) + (3136*e*Log[3 + 2*x + 5 
*x^2])/5))/28)/7000
 

3.4.20.3.1 Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 217

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

rule 1083

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[-2   Subst[I 
nt[1/Simp[b^2 - 4*a*c - x^2, x], x], x, b + 2*c*x], x] /; FreeQ[{a, b, c}, 
x]
 

rule 1103

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, 
e}, x] && EqQ[2*c*d - b*e, 0]
 

rule 1142

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

rule 2175

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p 
_), x_Symbol] :> With[{Qx = PolynomialQuotient[Pq, a + b*x + c*x^2, x], R = 
 Coeff[PolynomialRemainder[Pq, a + b*x + c*x^2, x], x, 0], S = Coeff[Polyno 
mialRemainder[Pq, a + b*x + c*x^2, x], x, 1]}, Simp[(d + e*x)^m*(a + b*x + 
c*x^2)^(p + 1)*((R*b - 2*a*S + (2*c*R - b*S)*x)/((p + 1)*(b^2 - 4*a*c))), x 
] + Simp[1/((p + 1)*(b^2 - 4*a*c))   Int[(d + e*x)^(m - 1)*(a + b*x + c*x^2 
)^(p + 1)*ExpandToSum[(p + 1)*(b^2 - 4*a*c)*(d + e*x)*Qx + S*(2*a*e*m + b*d 
*(2*p + 3)) - R*(b*e*m + 2*c*d*(2*p + 3)) - e*(2*c*R - b*S)*(m + 2*p + 3)*x 
, x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && PolyQ[Pq, x] && NeQ[b^2 - 4*a 
*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && LtQ[p, -1] && GtQ[m, 0] && (Inte 
gerQ[p] ||  !IntegerQ[m] ||  !RationalQ[a, b, c, d, e]) &&  !(IGtQ[m, 0] && 
 RationalQ[a, b, c, d, e] && (IntegerQ[p] || ILtQ[p + 1/2, 0]))
 

rule 2191

Int[(Pq_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = 
PolynomialQuotient[Pq, a + b*x + c*x^2, x], f = Coeff[PolynomialRemainder[P 
q, a + b*x + c*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b*x + 
c*x^2, x], x, 1]}, Simp[(b*f - 2*a*g + (2*c*f - b*g)*x)*((a + b*x + c*x^2)^ 
(p + 1)/((p + 1)*(b^2 - 4*a*c))), x] + Simp[1/((p + 1)*(b^2 - 4*a*c))   Int 
[(a + b*x + c*x^2)^(p + 1)*ExpandToSum[(p + 1)*(b^2 - 4*a*c)*Q - (2*p + 3)* 
(2*c*f - b*g), x], x], x]] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && NeQ[b^ 
2 - 4*a*c, 0] && LtQ[p, -1]
 

3.4.20.4 Maple [A] (verified)

Time = 0.76 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.88

method	result	size
default	\(\frac {25 \left (\frac {36353 e}{980000}+\frac {2203 d}{196000}\right ) x^{3}+25 \left (\frac {28307 e}{4900000}+\frac {38753 d}{980000}\right ) x^{2}+25 \left (\frac {57761 e}{4900000}+\frac {17979 d}{980000}\right ) x +\frac {12953 d}{39200}-\frac {19533 e}{196000}}{\left (5 x^{2}+2 x +3\right )^{2}}+\frac {2 e \ln \left (5 x^{2}+2 x +3\right )}{125}+\frac {\left (8475 d -\frac {34207 e}{5}\right ) \sqrt {14}\, \arctan \left (\frac {\left (10 x +2\right ) \sqrt {14}}{28}\right )}{548800}\)	\(91\)
risch	\(\frac {25 \left (\frac {36353 e}{980000}+\frac {2203 d}{196000}\right ) x^{3}+25 \left (\frac {28307 e}{4900000}+\frac {38753 d}{980000}\right ) x^{2}+25 \left (\frac {57761 e}{4900000}+\frac {17979 d}{980000}\right ) x +\frac {12953 d}{39200}-\frac {19533 e}{196000}}{\left (5 x^{2}+2 x +3\right )^{2}}+\frac {2 e \ln \left (350 x^{2}+140 x +210\right )}{125}+\frac {339 \sqrt {14}\, d \arctan \left (\frac {5 \sqrt {14}\, x}{14}+\frac {\sqrt {14}}{14}\right )}{21952}-\frac {34207 \sqrt {14}\, e \arctan \left (\frac {5 \sqrt {14}\, x}{14}+\frac {\sqrt {14}}{14}\right )}{2744000}\)	\(106\)

input

int((e*x+d)*(4*x^4-5*x^3+3*x^2+x+2)/(5*x^2+2*x+3)^3,x,method=_RETURNVERBOS 
E)
 

output

25*((36353/980000*e+2203/196000*d)*x^3+(28307/4900000*e+38753/980000*d)*x^ 
2+(57761/4900000*e+17979/980000*d)*x+12953/980000*d-19533/4900000*e)/(5*x^ 
2+2*x+3)^2+2/125*e*ln(5*x^2+2*x+3)+1/548800*(8475*d-34207/5*e)*14^(1/2)*ar 
ctan(1/28*(10*x+2)*14^(1/2))
 

3.4.20.5 Fricas [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.67 \[ \int \frac {(d+e x) \left (2+x+3 x^2-5 x^3+4 x^4\right )}{\left (3+2 x+5 x^2\right )^3} \, dx=\frac {70 \, {\left (11015 \, d + 36353 \, e\right )} x^{3} + 14 \, {\left (193765 \, d + 28307 \, e\right )} x^{2} + \sqrt {14} {\left (25 \, {\left (42375 \, d - 34207 \, e\right )} x^{4} + 20 \, {\left (42375 \, d - 34207 \, e\right )} x^{3} + 34 \, {\left (42375 \, d - 34207 \, e\right )} x^{2} + 12 \, {\left (42375 \, d - 34207 \, e\right )} x + 381375 \, d - 307863 \, e\right )} \arctan \left (\frac {1}{14} \, \sqrt {14} {\left (5 \, x + 1\right )}\right ) + 14 \, {\left (89895 \, d + 57761 \, e\right )} x + 43904 \, {\left (25 \, e x^{4} + 20 \, e x^{3} + 34 \, e x^{2} + 12 \, e x + 9 \, e\right )} \log \left (5 \, x^{2} + 2 \, x + 3\right ) + 906710 \, d - 273462 \, e}{2744000 \, {\left (25 \, x^{4} + 20 \, x^{3} + 34 \, x^{2} + 12 \, x + 9\right )}} \]

input

integrate((e*x+d)*(4*x^4-5*x^3+3*x^2+x+2)/(5*x^2+2*x+3)^3,x, algorithm="fr 
icas")
 

output

1/2744000*(70*(11015*d + 36353*e)*x^3 + 14*(193765*d + 28307*e)*x^2 + sqrt 
(14)*(25*(42375*d - 34207*e)*x^4 + 20*(42375*d - 34207*e)*x^3 + 34*(42375* 
d - 34207*e)*x^2 + 12*(42375*d - 34207*e)*x + 381375*d - 307863*e)*arctan( 
1/14*sqrt(14)*(5*x + 1)) + 14*(89895*d + 57761*e)*x + 43904*(25*e*x^4 + 20 
*e*x^3 + 34*e*x^2 + 12*e*x + 9*e)*log(5*x^2 + 2*x + 3) + 906710*d - 273462 
*e)/(25*x^4 + 20*x^3 + 34*x^2 + 12*x + 9)
 

3.4.20.6 Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.88 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.58 \[ \int \frac {(d+e x) \left (2+x+3 x^2-5 x^3+4 x^4\right )}{\left (3+2 x+5 x^2\right )^3} \, dx=\left (\frac {2 e}{125} - \frac {\sqrt {14} i \left (42375 d - 34207 e\right )}{5488000}\right ) \log {\left (x + \frac {8475 d - \frac {34207 e}{5} - \frac {\sqrt {14} i \left (42375 d - 34207 e\right )}{5}}{42375 d - 34207 e} \right )} + \left (\frac {2 e}{125} + \frac {\sqrt {14} i \left (42375 d - 34207 e\right )}{5488000}\right ) \log {\left (x + \frac {8475 d - \frac {34207 e}{5} + \frac {\sqrt {14} i \left (42375 d - 34207 e\right )}{5}}{42375 d - 34207 e} \right )} + \frac {64765 d - 19533 e + x^{3} \cdot \left (55075 d + 181765 e\right ) + x^{2} \cdot \left (193765 d + 28307 e\right ) + x \left (89895 d + 57761 e\right )}{4900000 x^{4} + 3920000 x^{3} + 6664000 x^{2} + 2352000 x + 1764000} \]

input

integrate((e*x+d)*(4*x**4-5*x**3+3*x**2+x+2)/(5*x**2+2*x+3)**3,x)
 

output

(2*e/125 - sqrt(14)*I*(42375*d - 34207*e)/5488000)*log(x + (8475*d - 34207 
*e/5 - sqrt(14)*I*(42375*d - 34207*e)/5)/(42375*d - 34207*e)) + (2*e/125 + 
 sqrt(14)*I*(42375*d - 34207*e)/5488000)*log(x + (8475*d - 34207*e/5 + sqr 
t(14)*I*(42375*d - 34207*e)/5)/(42375*d - 34207*e)) + (64765*d - 19533*e + 
 x**3*(55075*d + 181765*e) + x**2*(193765*d + 28307*e) + x*(89895*d + 5776 
1*e))/(4900000*x**4 + 3920000*x**3 + 6664000*x**2 + 2352000*x + 1764000)
 

3.4.20.7 Maxima [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.98 \[ \int \frac {(d+e x) \left (2+x+3 x^2-5 x^3+4 x^4\right )}{\left (3+2 x+5 x^2\right )^3} \, dx=\frac {1}{2744000} \, \sqrt {14} {\left (42375 \, d - 34207 \, e\right )} \arctan \left (\frac {1}{14} \, \sqrt {14} {\left (5 \, x + 1\right )}\right ) + \frac {2}{125} \, e \log \left (5 \, x^{2} + 2 \, x + 3\right ) + \frac {5 \, {\left (11015 \, d + 36353 \, e\right )} x^{3} + {\left (193765 \, d + 28307 \, e\right )} x^{2} + {\left (89895 \, d + 57761 \, e\right )} x + 64765 \, d - 19533 \, e}{196000 \, {\left (25 \, x^{4} + 20 \, x^{3} + 34 \, x^{2} + 12 \, x + 9\right )}} \]

input

integrate((e*x+d)*(4*x^4-5*x^3+3*x^2+x+2)/(5*x^2+2*x+3)^3,x, algorithm="ma 
xima")
 

output

1/2744000*sqrt(14)*(42375*d - 34207*e)*arctan(1/14*sqrt(14)*(5*x + 1)) + 2 
/125*e*log(5*x^2 + 2*x + 3) + 1/196000*(5*(11015*d + 36353*e)*x^3 + (19376 
5*d + 28307*e)*x^2 + (89895*d + 57761*e)*x + 64765*d - 19533*e)/(25*x^4 + 
20*x^3 + 34*x^2 + 12*x + 9)
 

3.4.20.8 Giac [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.88 \[ \int \frac {(d+e x) \left (2+x+3 x^2-5 x^3+4 x^4\right )}{\left (3+2 x+5 x^2\right )^3} \, dx=\frac {1}{2744000} \, \sqrt {14} {\left (42375 \, d - 34207 \, e\right )} \arctan \left (\frac {1}{14} \, \sqrt {14} {\left (5 \, x + 1\right )}\right ) + \frac {2}{125} \, e \log \left (5 \, x^{2} + 2 \, x + 3\right ) + \frac {5 \, {\left (11015 \, d + 36353 \, e\right )} x^{3} + {\left (193765 \, d + 28307 \, e\right )} x^{2} + {\left (89895 \, d + 57761 \, e\right )} x + 64765 \, d - 19533 \, e}{196000 \, {\left (5 \, x^{2} + 2 \, x + 3\right )}^{2}} \]

input

integrate((e*x+d)*(4*x^4-5*x^3+3*x^2+x+2)/(5*x^2+2*x+3)^3,x, algorithm="gi 
ac")
 

output

1/2744000*sqrt(14)*(42375*d - 34207*e)*arctan(1/14*sqrt(14)*(5*x + 1)) + 2 
/125*e*log(5*x^2 + 2*x + 3) + 1/196000*(5*(11015*d + 36353*e)*x^3 + (19376 
5*d + 28307*e)*x^2 + (89895*d + 57761*e)*x + 64765*d - 19533*e)/(5*x^2 + 2 
*x + 3)^2
 

3.4.20.9 Mupad [B] (verification not implemented)

Time = 13.55 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.21 \[ \int \frac {(d+e x) \left (2+x+3 x^2-5 x^3+4 x^4\right )}{\left (3+2 x+5 x^2\right )^3} \, dx=\frac {\left (\frac {2203\,d}{7840}+\frac {36353\,e}{39200}\right )\,x^3+\left (\frac {38753\,d}{39200}+\frac {28307\,e}{196000}\right )\,x^2+\left (\frac {17979\,d}{39200}+\frac {57761\,e}{196000}\right )\,x+\frac {12953\,d}{39200}-\frac {19533\,e}{196000}}{25\,x^4+20\,x^3+34\,x^2+12\,x+9}+\frac {2\,e\,\ln \left (5\,x^2+2\,x+3\right )}{125}+\frac {\sqrt {14}\,\mathrm {atan}\left (\frac {\frac {\sqrt {14}\,\left (42375\,d-34207\,e\right )}{2744000}+\frac {\sqrt {14}\,x\,\left (42375\,d-34207\,e\right )}{548800}}{\frac {339\,d}{1568}-\frac {34207\,e}{196000}}\right )\,\left (42375\,d-34207\,e\right )}{2744000} \]

input

int(((d + e*x)*(x + 3*x^2 - 5*x^3 + 4*x^4 + 2))/(2*x + 5*x^2 + 3)^3,x)
 

output

((12953*d)/39200 - (19533*e)/196000 + x^3*((2203*d)/7840 + (36353*e)/39200 
) + x^2*((38753*d)/39200 + (28307*e)/196000) + x*((17979*d)/39200 + (57761 
*e)/196000))/(12*x + 34*x^2 + 20*x^3 + 25*x^4 + 9) + (2*e*log(2*x + 5*x^2 
+ 3))/125 + (14^(1/2)*atan(((14^(1/2)*(42375*d - 34207*e))/2744000 + (14^( 
1/2)*x*(42375*d - 34207*e))/548800)/((339*d)/1568 - (34207*e)/196000))*(42 
375*d - 34207*e))/2744000