\(\int \frac {(b+2 c x) (d+e x)^2}{(a+b x+c x^2)^3} \, dx\) [563]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [A] (verified)
Fricas [B] (verification not implemented)
Sympy [B] (verification not implemented)
Maxima [F(-2)]
Giac [A] (verification not implemented)
Mupad [B] (verification not implemented)
Reduce [B] (verification not implemented)

Optimal result

Integrand size = 26, antiderivative size = 112 \[ \int \frac {(b+2 c x) (d+e x)^2}{\left (a+b x+c x^2\right )^3} \, dx=-\frac {(d+e x)^2}{2 \left (a+b x+c x^2\right )^2}-\frac {e (b d-2 a e+(2 c d-b e) x)}{\left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}+\frac {2 e (2 c d-b e) \text {arctanh}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{3/2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.28 \[ \int \frac {(b+2 c x) (d+e x)^2}{\left (a+b x+c x^2\right )^3} \, dx=\frac {1}{2} \left (\frac {e \left (b^2 e+4 c (-2 a e+c d x)+2 b c (d-e x)\right )}{c \left (-b^2+4 a c\right ) (a+x (b+c x))}+\frac {e^2 (a+b x)-c d (d+2 e x)}{c (a+x (b+c x))^2}-\frac {4 e (-2 c d+b e) \arctan \left (\frac {b+2 c x}{\sqrt {-b^2+4 a c}}\right )}{\left (-b^2+4 a c\right )^{3/2}}\right ) \] Input:

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

Output:

((e*(b^2*e + 4*c*(-2*a*e + c*d*x) + 2*b*c*(d - e*x)))/(c*(-b^2 + 4*a*c)*(a 
 + x*(b + c*x))) + (e^2*(a + b*x) - c*d*(d + 2*e*x))/(c*(a + x*(b + c*x))^ 
2) - (4*e*(-2*c*d + b*e)*ArcTan[(b + 2*c*x)/Sqrt[-b^2 + 4*a*c]])/(-b^2 + 4 
*a*c)^(3/2))/2
 

Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.01, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {1222, 1159, 1083, 219}

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.

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

\(\Big \downarrow \) 1222

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

\(\Big \downarrow \) 1159

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

\(\Big \downarrow \) 1083

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

\(\Big \downarrow \) 219

\(\displaystyle e \left (\frac {2 (2 c d-b e) \text {arctanh}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{3/2}}-\frac {-2 a e+x (2 c d-b e)+b d}{\left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}\right )-\frac {(d+e x)^2}{2 \left (a+b x+c x^2\right )^2}\)

Input:

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

Output:

-1/2*(d + e*x)^2/(a + b*x + c*x^2)^2 + e*(-((b*d - 2*a*e + (2*c*d - b*e)*x 
)/((b^2 - 4*a*c)*(a + b*x + c*x^2))) + (2*(2*c*d - b*e)*ArcTanh[(b + 2*c*x 
)/Sqrt[b^2 - 4*a*c]])/(b^2 - 4*a*c)^(3/2))
 

Defintions of rubi rules used

rule 219
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[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 1159
Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol 
] :> Simp[((b*d - 2*a*e + (2*c*d - b*e)*x)/((p + 1)*(b^2 - 4*a*c)))*(a + b* 
x + c*x^2)^(p + 1), x] - Simp[(2*p + 3)*((2*c*d - b*e)/((p + 1)*(b^2 - 4*a* 
c)))   Int[(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] & 
& LtQ[p, -1] && NeQ[p, -3/2]
 

rule 1222
Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + ( 
c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[g*(d + e*x)^m*((a + b*x + c*x^2)^(p + 
1)/(2*c*(p + 1))), x] - Simp[e*g*(m/(2*c*(p + 1)))   Int[(d + e*x)^(m - 1)* 
(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && EqQ 
[2*c*f - b*g, 0] && LtQ[p, -1] && GtQ[m, 0]
 
Maple [A] (verified)

Time = 1.33 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.76

method result size
default \(\frac {-\frac {\left (b e -2 c d \right ) c e \,x^{3}}{4 a c -b^{2}}-\frac {e \left (8 a c e +e \,b^{2}-6 d b c \right ) x^{2}}{2 \left (4 a c -b^{2}\right )}-\frac {e \left (3 a b e +2 a c d -2 b^{2} d \right ) x}{4 a c -b^{2}}-\frac {4 e^{2} a^{2}-2 a b d e +4 a \,d^{2} c -b^{2} d^{2}}{2 \left (4 a c -b^{2}\right )}}{\left (c \,x^{2}+b x +a \right )^{2}}-\frac {2 \left (b e -2 c d \right ) e \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\left (4 a c -b^{2}\right )^{\frac {3}{2}}}\) \(197\)
risch \(\frac {-\frac {\left (b e -2 c d \right ) c e \,x^{3}}{4 a c -b^{2}}-\frac {e \left (8 a c e +e \,b^{2}-6 d b c \right ) x^{2}}{2 \left (4 a c -b^{2}\right )}-\frac {e \left (3 a b e +2 a c d -2 b^{2} d \right ) x}{4 a c -b^{2}}-\frac {4 e^{2} a^{2}-2 a b d e +4 a \,d^{2} c -b^{2} d^{2}}{2 \left (4 a c -b^{2}\right )}}{\left (c \,x^{2}+b x +a \right )^{2}}+\frac {e^{2} \ln \left (\left (-8 a \,c^{2}+2 b^{2} c \right ) x -\left (-4 a c +b^{2}\right )^{\frac {3}{2}}-4 a b c +b^{3}\right ) b}{\left (-4 a c +b^{2}\right )^{\frac {3}{2}}}-\frac {2 e \ln \left (\left (-8 a \,c^{2}+2 b^{2} c \right ) x -\left (-4 a c +b^{2}\right )^{\frac {3}{2}}-4 a b c +b^{3}\right ) c d}{\left (-4 a c +b^{2}\right )^{\frac {3}{2}}}-\frac {e^{2} \ln \left (\left (8 a \,c^{2}-2 b^{2} c \right ) x -\left (-4 a c +b^{2}\right )^{\frac {3}{2}}+4 a b c -b^{3}\right ) b}{\left (-4 a c +b^{2}\right )^{\frac {3}{2}}}+\frac {2 e \ln \left (\left (8 a \,c^{2}-2 b^{2} c \right ) x -\left (-4 a c +b^{2}\right )^{\frac {3}{2}}+4 a b c -b^{3}\right ) c d}{\left (-4 a c +b^{2}\right )^{\frac {3}{2}}}\) \(367\)

Input:

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

Output:

(-(b*e-2*c*d)*c*e/(4*a*c-b^2)*x^3-1/2*e*(8*a*c*e+b^2*e-6*b*c*d)/(4*a*c-b^2 
)*x^2-e*(3*a*b*e+2*a*c*d-2*b^2*d)/(4*a*c-b^2)*x-1/2*(4*a^2*e^2-2*a*b*d*e+4 
*a*c*d^2-b^2*d^2)/(4*a*c-b^2))/(c*x^2+b*x+a)^2-2*(b*e-2*c*d)*e/(4*a*c-b^2) 
^(3/2)*arctan((2*c*x+b)/(4*a*c-b^2)^(1/2))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 492 vs. \(2 (106) = 212\).

Time = 0.09 (sec) , antiderivative size = 1004, normalized size of antiderivative = 8.96 \[ \int \frac {(b+2 c x) (d+e x)^2}{\left (a+b x+c x^2\right )^3} \, dx =\text {Too large to display} \] Input:

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

Output:

[-1/2*(2*(2*(b^2*c^2 - 4*a*c^3)*d*e - (b^3*c - 4*a*b*c^2)*e^2)*x^3 + (b^4 
- 8*a*b^2*c + 16*a^2*c^2)*d^2 + 2*(a*b^3 - 4*a^2*b*c)*d*e - 4*(a^2*b^2 - 4 
*a^3*c)*e^2 + (6*(b^3*c - 4*a*b*c^2)*d*e - (b^4 + 4*a*b^2*c - 32*a^2*c^2)* 
e^2)*x^2 - 2*(2*a^2*c*d*e - a^2*b*e^2 + (2*c^3*d*e - b*c^2*e^2)*x^4 + 2*(2 
*b*c^2*d*e - b^2*c*e^2)*x^3 + (2*(b^2*c + 2*a*c^2)*d*e - (b^3 + 2*a*b*c)*e 
^2)*x^2 + 2*(2*a*b*c*d*e - a*b^2*e^2)*x)*sqrt(b^2 - 4*a*c)*log((2*c^2*x^2 
+ 2*b*c*x + b^2 - 2*a*c + sqrt(b^2 - 4*a*c)*(2*c*x + b))/(c*x^2 + b*x + a) 
) + 2*(2*(b^4 - 5*a*b^2*c + 4*a^2*c^2)*d*e - 3*(a*b^3 - 4*a^2*b*c)*e^2)*x) 
/(a^2*b^4 - 8*a^3*b^2*c + 16*a^4*c^2 + (b^4*c^2 - 8*a*b^2*c^3 + 16*a^2*c^4 
)*x^4 + 2*(b^5*c - 8*a*b^3*c^2 + 16*a^2*b*c^3)*x^3 + (b^6 - 6*a*b^4*c + 32 
*a^3*c^3)*x^2 + 2*(a*b^5 - 8*a^2*b^3*c + 16*a^3*b*c^2)*x), -1/2*(2*(2*(b^2 
*c^2 - 4*a*c^3)*d*e - (b^3*c - 4*a*b*c^2)*e^2)*x^3 + (b^4 - 8*a*b^2*c + 16 
*a^2*c^2)*d^2 + 2*(a*b^3 - 4*a^2*b*c)*d*e - 4*(a^2*b^2 - 4*a^3*c)*e^2 + (6 
*(b^3*c - 4*a*b*c^2)*d*e - (b^4 + 4*a*b^2*c - 32*a^2*c^2)*e^2)*x^2 - 4*(2* 
a^2*c*d*e - a^2*b*e^2 + (2*c^3*d*e - b*c^2*e^2)*x^4 + 2*(2*b*c^2*d*e - b^2 
*c*e^2)*x^3 + (2*(b^2*c + 2*a*c^2)*d*e - (b^3 + 2*a*b*c)*e^2)*x^2 + 2*(2*a 
*b*c*d*e - a*b^2*e^2)*x)*sqrt(-b^2 + 4*a*c)*arctan(-sqrt(-b^2 + 4*a*c)*(2* 
c*x + b)/(b^2 - 4*a*c)) + 2*(2*(b^4 - 5*a*b^2*c + 4*a^2*c^2)*d*e - 3*(a*b^ 
3 - 4*a^2*b*c)*e^2)*x)/(a^2*b^4 - 8*a^3*b^2*c + 16*a^4*c^2 + (b^4*c^2 - 8* 
a*b^2*c^3 + 16*a^2*c^4)*x^4 + 2*(b^5*c - 8*a*b^3*c^2 + 16*a^2*b*c^3)*x^...
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 530 vs. \(2 (102) = 204\).

Time = 5.26 (sec) , antiderivative size = 530, normalized size of antiderivative = 4.73 \[ \int \frac {(b+2 c x) (d+e x)^2}{\left (a+b x+c x^2\right )^3} \, dx=e \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{3}}} \left (b e - 2 c d\right ) \log {\left (x + \frac {- 16 a^{2} c^{2} e \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{3}}} \left (b e - 2 c d\right ) + 8 a b^{2} c e \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{3}}} \left (b e - 2 c d\right ) - b^{4} e \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{3}}} \left (b e - 2 c d\right ) + b^{2} e^{2} - 2 b c d e}{2 b c e^{2} - 4 c^{2} d e} \right )} - e \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{3}}} \left (b e - 2 c d\right ) \log {\left (x + \frac {16 a^{2} c^{2} e \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{3}}} \left (b e - 2 c d\right ) - 8 a b^{2} c e \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{3}}} \left (b e - 2 c d\right ) + b^{4} e \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{3}}} \left (b e - 2 c d\right ) + b^{2} e^{2} - 2 b c d e}{2 b c e^{2} - 4 c^{2} d e} \right )} + \frac {- 4 a^{2} e^{2} + 2 a b d e - 4 a c d^{2} + b^{2} d^{2} + x^{3} \left (- 2 b c e^{2} + 4 c^{2} d e\right ) + x^{2} \left (- 8 a c e^{2} - b^{2} e^{2} + 6 b c d e\right ) + x \left (- 6 a b e^{2} - 4 a c d e + 4 b^{2} d e\right )}{8 a^{3} c - 2 a^{2} b^{2} + x^{4} \cdot \left (8 a c^{3} - 2 b^{2} c^{2}\right ) + x^{3} \cdot \left (16 a b c^{2} - 4 b^{3} c\right ) + x^{2} \cdot \left (16 a^{2} c^{2} + 4 a b^{2} c - 2 b^{4}\right ) + x \left (16 a^{2} b c - 4 a b^{3}\right )} \] Input:

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

Output:

e*sqrt(-1/(4*a*c - b**2)**3)*(b*e - 2*c*d)*log(x + (-16*a**2*c**2*e*sqrt(- 
1/(4*a*c - b**2)**3)*(b*e - 2*c*d) + 8*a*b**2*c*e*sqrt(-1/(4*a*c - b**2)** 
3)*(b*e - 2*c*d) - b**4*e*sqrt(-1/(4*a*c - b**2)**3)*(b*e - 2*c*d) + b**2* 
e**2 - 2*b*c*d*e)/(2*b*c*e**2 - 4*c**2*d*e)) - e*sqrt(-1/(4*a*c - b**2)**3 
)*(b*e - 2*c*d)*log(x + (16*a**2*c**2*e*sqrt(-1/(4*a*c - b**2)**3)*(b*e - 
2*c*d) - 8*a*b**2*c*e*sqrt(-1/(4*a*c - b**2)**3)*(b*e - 2*c*d) + b**4*e*sq 
rt(-1/(4*a*c - b**2)**3)*(b*e - 2*c*d) + b**2*e**2 - 2*b*c*d*e)/(2*b*c*e** 
2 - 4*c**2*d*e)) + (-4*a**2*e**2 + 2*a*b*d*e - 4*a*c*d**2 + b**2*d**2 + x* 
*3*(-2*b*c*e**2 + 4*c**2*d*e) + x**2*(-8*a*c*e**2 - b**2*e**2 + 6*b*c*d*e) 
 + x*(-6*a*b*e**2 - 4*a*c*d*e + 4*b**2*d*e))/(8*a**3*c - 2*a**2*b**2 + x** 
4*(8*a*c**3 - 2*b**2*c**2) + x**3*(16*a*b*c**2 - 4*b**3*c) + x**2*(16*a**2 
*c**2 + 4*a*b**2*c - 2*b**4) + x*(16*a**2*b*c - 4*a*b**3))
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {(b+2 c x) (d+e x)^2}{\left (a+b x+c x^2\right )^3} \, dx=\text {Exception raised: ValueError} \] Input:

integrate((2*c*x+b)*(e*x+d)^2/(c*x^2+b*x+a)^3,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(4*a*c-b^2>0)', see `assume?` for 
 more deta
 

Giac [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.63 \[ \int \frac {(b+2 c x) (d+e x)^2}{\left (a+b x+c x^2\right )^3} \, dx=-\frac {2 \, {\left (2 \, c d e - b e^{2}\right )} \arctan \left (\frac {2 \, c x + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{{\left (b^{2} - 4 \, a c\right )} \sqrt {-b^{2} + 4 \, a c}} - \frac {4 \, c^{2} d e x^{3} - 2 \, b c e^{2} x^{3} + 6 \, b c d e x^{2} - b^{2} e^{2} x^{2} - 8 \, a c e^{2} x^{2} + 4 \, b^{2} d e x - 4 \, a c d e x - 6 \, a b e^{2} x + b^{2} d^{2} - 4 \, a c d^{2} + 2 \, a b d e - 4 \, a^{2} e^{2}}{2 \, {\left (c x^{2} + b x + a\right )}^{2} {\left (b^{2} - 4 \, a c\right )}} \] Input:

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

Output:

-2*(2*c*d*e - b*e^2)*arctan((2*c*x + b)/sqrt(-b^2 + 4*a*c))/((b^2 - 4*a*c) 
*sqrt(-b^2 + 4*a*c)) - 1/2*(4*c^2*d*e*x^3 - 2*b*c*e^2*x^3 + 6*b*c*d*e*x^2 
- b^2*e^2*x^2 - 8*a*c*e^2*x^2 + 4*b^2*d*e*x - 4*a*c*d*e*x - 6*a*b*e^2*x + 
b^2*d^2 - 4*a*c*d^2 + 2*a*b*d*e - 4*a^2*e^2)/((c*x^2 + b*x + a)^2*(b^2 - 4 
*a*c))
 

Mupad [B] (verification not implemented)

Time = 12.00 (sec) , antiderivative size = 284, normalized size of antiderivative = 2.54 \[ \int \frac {(b+2 c x) (d+e x)^2}{\left (a+b x+c x^2\right )^3} \, dx=\frac {2\,e\,\mathrm {atan}\left (\frac {\left (4\,a\,c-b^2\right )\,\left (\frac {e\,\left (b^3-4\,a\,b\,c\right )\,\left (b\,e-2\,c\,d\right )}{{\left (4\,a\,c-b^2\right )}^{5/2}}-\frac {2\,c\,e\,x\,\left (b\,e-2\,c\,d\right )}{{\left (4\,a\,c-b^2\right )}^{3/2}}\right )}{b\,e^2-2\,c\,d\,e}\right )\,\left (b\,e-2\,c\,d\right )}{{\left (4\,a\,c-b^2\right )}^{3/2}}-\frac {\frac {4\,a^2\,e^2-2\,a\,b\,d\,e+4\,c\,a\,d^2-b^2\,d^2}{2\,\left (4\,a\,c-b^2\right )}-\frac {e\,x^3\,\left (2\,c^2\,d-b\,c\,e\right )}{4\,a\,c-b^2}+\frac {e\,x\,\left (-2\,d\,b^2+3\,a\,e\,b+2\,a\,c\,d\right )}{4\,a\,c-b^2}+\frac {e\,x^2\,\left (e\,b^2-6\,c\,d\,b+8\,a\,c\,e\right )}{2\,\left (4\,a\,c-b^2\right )}}{x^2\,\left (b^2+2\,a\,c\right )+a^2+c^2\,x^4+2\,a\,b\,x+2\,b\,c\,x^3} \] Input:

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

Output:

(2*e*atan(((4*a*c - b^2)*((e*(b^3 - 4*a*b*c)*(b*e - 2*c*d))/(4*a*c - b^2)^ 
(5/2) - (2*c*e*x*(b*e - 2*c*d))/(4*a*c - b^2)^(3/2)))/(b*e^2 - 2*c*d*e))*( 
b*e - 2*c*d))/(4*a*c - b^2)^(3/2) - ((4*a^2*e^2 - b^2*d^2 + 4*a*c*d^2 - 2* 
a*b*d*e)/(2*(4*a*c - b^2)) - (e*x^3*(2*c^2*d - b*c*e))/(4*a*c - b^2) + (e* 
x*(3*a*b*e - 2*b^2*d + 2*a*c*d))/(4*a*c - b^2) + (e*x^2*(b^2*e + 8*a*c*e - 
 6*b*c*d))/(2*(4*a*c - b^2)))/(x^2*(2*a*c + b^2) + a^2 + c^2*x^4 + 2*a*b*x 
 + 2*b*c*x^3)
 

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 925, normalized size of antiderivative = 8.26 \[ \int \frac {(b+2 c x) (d+e x)^2}{\left (a+b x+c x^2\right )^3} \, dx =\text {Too large to display} \] Input:

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

Output:

( - 4*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2))*a**2*b**2*e* 
*2 + 8*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2))*a**2*b*c*d* 
e - 8*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2))*a*b**3*e**2* 
x + 16*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2))*a*b**2*c*d* 
e*x - 8*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2))*a*b**2*c*e 
**2*x**2 + 16*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2))*a*b* 
c**2*d*e*x**2 - 4*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2))* 
b**4*e**2*x**2 + 8*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2)) 
*b**3*c*d*e*x**2 - 8*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2 
))*b**3*c*e**2*x**3 + 16*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a*c - 
b**2))*b**2*c**2*d*e*x**3 - 4*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a 
*c - b**2))*b**2*c**2*e**2*x**4 + 8*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sq 
rt(4*a*c - b**2))*b*c**3*d*e*x**4 - 12*a**3*b*c*e**2 - 8*a**3*c**2*d*e + 3 
*a**2*b**3*e**2 + 10*a**2*b**2*c*d*e - 16*a**2*b**2*c*e**2*x - 16*a**2*b*c 
**2*d**2 - 32*a**2*b*c**2*d*e*x - 24*a**2*b*c**2*e**2*x**2 - 16*a**2*c**3* 
d*e*x**2 - 2*a*b**4*d*e + 4*a*b**4*e**2*x + 8*a*b**3*c*d**2 + 24*a*b**3*c* 
d*e*x + 6*a*b**3*c*e**2*x**2 + 20*a*b**2*c**2*d*e*x**2 + 4*a*b*c**3*e**2*x 
**4 - 8*a*c**4*d*e*x**4 - b**5*d**2 - 4*b**5*d*e*x - 4*b**4*c*d*e*x**2 - b 
**3*c**2*e**2*x**4 + 2*b**2*c**3*d*e*x**4)/(2*b*(16*a**4*c**2 - 8*a**3*b** 
2*c + 32*a**3*b*c**2*x + 32*a**3*c**3*x**2 + a**2*b**4 - 16*a**2*b**3*c...