\(\int \frac {1}{(d+e x)^3 (a+b x+c x^2)^2} \, dx\) [484]

Optimal result

Integrand size = 20, antiderivative size = 493 \[ \int \frac {1}{(d+e x)^3 \left (a+b x+c x^2\right )^2} \, dx=-\frac {e^3}{2 \left (c d^2-b d e+a e^2\right )^2 (d+e x)^2}-\frac {2 e^3 (2 c d-b e)}{\left (c d^2-b d e+a e^2\right )^3 (d+e x)}-\frac {3 b^3 c d e^2-b^4 e^3+b c^2 d \left (c d^2-9 a e^2\right )-b^2 c e \left (3 c d^2-4 a e^2\right )+2 a c^2 e \left (3 c d^2-a e^2\right )+c (2 c d-b e) \left (c^2 d^2+b^2 e^2-c e (b d+3 a e)\right ) x}{\left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right )^3 \left (a+b x+c x^2\right )}+\frac {(2 c d-b e) \left (2 c^4 d^4-3 b^4 e^4-4 c^3 d^2 e (b d-5 a e)+4 b^2 c e^3 (b d+5 a e)-2 c^2 e^2 \left (b^2 d^2+10 a b d e+15 a^2 e^2\right )\right ) \text {arctanh}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{3/2} \left (c d^2-b d e+a e^2\right )^4}+\frac {e^3 \left (10 c^2 d^2+3 b^2 e^2-2 c e (5 b d+a e)\right ) \log (d+e x)}{\left (c d^2-b d e+a e^2\right )^4}-\frac {e^3 \left (10 c^2 d^2+3 b^2 e^2-2 c e (5 b d+a e)\right ) \log \left (a+b x+c x^2\right )}{2 \left (c d^2-b d e+a e^2\right )^4} \] Output:

-1/2*e^3/(a*e^2-b*d*e+c*d^2)^2/(e*x+d)^2-2*e^3*(-b*e+2*c*d)/(a*e^2-b*d*e+c 
*d^2)^3/(e*x+d)-(3*b^3*c*d*e^2-b^4*e^3+b*c^2*d*(-9*a*e^2+c*d^2)-b^2*c*e*(- 
4*a*e^2+3*c*d^2)+2*a*c^2*e*(-a*e^2+3*c*d^2)+c*(-b*e+2*c*d)*(c^2*d^2+b^2*e^ 
2-c*e*(3*a*e+b*d))*x)/(-4*a*c+b^2)/(a*e^2-b*d*e+c*d^2)^3/(c*x^2+b*x+a)+(-b 
*e+2*c*d)*(2*c^4*d^4-3*b^4*e^4-4*c^3*d^2*e*(-5*a*e+b*d)+4*b^2*c*e^3*(5*a*e 
+b*d)-2*c^2*e^2*(15*a^2*e^2+10*a*b*d*e+b^2*d^2))*arctanh((2*c*x+b)/(-4*a*c 
+b^2)^(1/2))/(-4*a*c+b^2)^(3/2)/(a*e^2-b*d*e+c*d^2)^4+e^3*(10*c^2*d^2+3*b^ 
2*e^2-2*c*e*(a*e+5*b*d))*ln(e*x+d)/(a*e^2-b*d*e+c*d^2)^4-1/2*e^3*(10*c^2*d 
^2+3*b^2*e^2-2*c*e*(a*e+5*b*d))*ln(c*x^2+b*x+a)/(a*e^2-b*d*e+c*d^2)^4
 

Mathematica [A] (verified)

Time = 0.72 (sec) , antiderivative size = 489, normalized size of antiderivative = 0.99 \[ \int \frac {1}{(d+e x)^3 \left (a+b x+c x^2\right )^2} \, dx=-\frac {e^3}{2 \left (c d^2+e (-b d+a e)\right )^2 (d+e x)^2}+\frac {2 e^3 (-2 c d+b e)}{\left (c d^2+e (-b d+a e)\right )^3 (d+e x)}+\frac {-b^4 e^3+b^3 c e^2 (3 d-e x)+b^2 c e \left (4 a e^2-3 c d (d-e x)\right )+2 c^2 \left (-a^2 e^3+c^2 d^3 x+3 a c d e (d-e x)\right )+b c^2 \left (c d^2 (d-3 e x)+3 a e^2 (-3 d+e x)\right )}{\left (b^2-4 a c\right ) \left (-c d^2+e (b d-a e)\right )^3 (a+x (b+c x))}+\frac {(-2 c d+b e) \left (-2 c^4 d^4+3 b^4 e^4+4 c^3 d^2 e (b d-5 a e)-4 b^2 c e^3 (b d+5 a e)+2 c^2 e^2 \left (b^2 d^2+10 a b d e+15 a^2 e^2\right )\right ) \arctan \left (\frac {b+2 c x}{\sqrt {-b^2+4 a c}}\right )}{\left (-b^2+4 a c\right )^{3/2} \left (c d^2+e (-b d+a e)\right )^4}+\frac {e^3 \left (10 c^2 d^2+3 b^2 e^2-2 c e (5 b d+a e)\right ) \log (d+e x)}{\left (c d^2+e (-b d+a e)\right )^4}-\frac {e^3 \left (10 c^2 d^2+3 b^2 e^2-2 c e (5 b d+a e)\right ) \log (a+x (b+c x))}{2 \left (c d^2+e (-b d+a e)\right )^4} \] Input:

Integrate[1/((d + e*x)^3*(a + b*x + c*x^2)^2),x]
 

Output:

-1/2*e^3/((c*d^2 + e*(-(b*d) + a*e))^2*(d + e*x)^2) + (2*e^3*(-2*c*d + b*e 
))/((c*d^2 + e*(-(b*d) + a*e))^3*(d + e*x)) + (-(b^4*e^3) + b^3*c*e^2*(3*d 
 - e*x) + b^2*c*e*(4*a*e^2 - 3*c*d*(d - e*x)) + 2*c^2*(-(a^2*e^3) + c^2*d^ 
3*x + 3*a*c*d*e*(d - e*x)) + b*c^2*(c*d^2*(d - 3*e*x) + 3*a*e^2*(-3*d + e* 
x)))/((b^2 - 4*a*c)*(-(c*d^2) + e*(b*d - a*e))^3*(a + x*(b + c*x))) + ((-2 
*c*d + b*e)*(-2*c^4*d^4 + 3*b^4*e^4 + 4*c^3*d^2*e*(b*d - 5*a*e) - 4*b^2*c* 
e^3*(b*d + 5*a*e) + 2*c^2*e^2*(b^2*d^2 + 10*a*b*d*e + 15*a^2*e^2))*ArcTan[ 
(b + 2*c*x)/Sqrt[-b^2 + 4*a*c]])/((-b^2 + 4*a*c)^(3/2)*(c*d^2 + e*(-(b*d) 
+ a*e))^4) + (e^3*(10*c^2*d^2 + 3*b^2*e^2 - 2*c*e*(5*b*d + a*e))*Log[d + e 
*x])/(c*d^2 + e*(-(b*d) + a*e))^4 - (e^3*(10*c^2*d^2 + 3*b^2*e^2 - 2*c*e*( 
5*b*d + a*e))*Log[a + x*(b + c*x)])/(2*(c*d^2 + e*(-(b*d) + a*e))^4)
 

Rubi [A] (verified)

Time = 1.16 (sec) , antiderivative size = 513, normalized size of antiderivative = 1.04, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {1165, 1200, 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.

\(\displaystyle \int \frac {1}{(d+e x)^3 \left (a+b x+c x^2\right )^2} \, dx\)

\(\Big \downarrow \) 1165

\(\displaystyle -\frac {\int \frac {2 c^2 d^2-3 b^2 e^2+c e (b d+8 a e)+3 c e (2 c d-b e) x}{(d+e x)^3 \left (c x^2+b x+a\right )}dx}{\left (b^2-4 a c\right ) \left (a e^2-b d e+c d^2\right )}-\frac {2 a c e+b^2 (-e)+c x (2 c d-b e)+b c d}{\left (b^2-4 a c\right ) (d+e x)^2 \left (a+b x+c x^2\right ) \left (a e^2-b d e+c d^2\right )}\)

\(\Big \downarrow \) 1200

\(\displaystyle -\frac {\int \left (\frac {\left (b^2-4 a c\right ) \left (-10 c^2 d^2-3 b^2 e^2+2 c e (5 b d+a e)\right ) e^4}{\left (c d^2-b e d+a e^2\right )^3 (d+e x)}+\frac {(2 c d-b e) \left (-c^2 d^2-3 b^2 e^2+c e (b d+11 a e)\right ) e^2}{\left (c d^2-b e d+a e^2\right )^2 (d+e x)^2}+\frac {\left (-4 c^2 d^2-3 b^2 e^2+4 c e (b d+2 a e)\right ) e^2}{\left (c d^2-b e d+a e^2\right ) (d+e x)^3}+\frac {2 c^5 d^5-5 c^4 e (b d-4 a e) d^3-10 a c^3 e^3 (5 b d+3 a e) d+3 b^5 e^5-b^3 c e^4 (10 b d+17 a e)+b c^2 e^3 \left (10 b^2 d^2+50 a b e d+19 a^2 e^2\right )+c \left (b^2-4 a c\right ) e^3 \left (10 c^2 d^2+3 b^2 e^2-2 c e (5 b d+a e)\right ) x}{\left (c d^2-b e d+a e^2\right )^3 \left (c x^2+b x+a\right )}\right )dx}{\left (b^2-4 a c\right ) \left (a e^2-b d e+c d^2\right )}-\frac {2 a c e+b^2 (-e)+c x (2 c d-b e)+b c d}{\left (b^2-4 a c\right ) (d+e x)^2 \left (a+b x+c x^2\right ) \left (a e^2-b d e+c d^2\right )}\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {-\frac {(2 c d-b e) \text {arctanh}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right ) \left (-2 c^2 e^2 \left (15 a^2 e^2+10 a b d e+b^2 d^2\right )+4 b^2 c e^3 (5 a e+b d)-4 c^3 d^2 e (b d-5 a e)-3 b^4 e^4+2 c^4 d^4\right )}{\sqrt {b^2-4 a c} \left (a e^2-b d e+c d^2\right )^3}+\frac {e (2 c d-b e) \left (-c e (11 a e+b d)+3 b^2 e^2+c^2 d^2\right )}{(d+e x) \left (a e^2-b d e+c d^2\right )^2}+\frac {e \left (-4 c e (2 a e+b d)+3 b^2 e^2+4 c^2 d^2\right )}{2 (d+e x)^2 \left (a e^2-b d e+c d^2\right )}+\frac {e^3 \left (b^2-4 a c\right ) \left (-2 c e (a e+5 b d)+3 b^2 e^2+10 c^2 d^2\right ) \log \left (a+b x+c x^2\right )}{2 \left (a e^2-b d e+c d^2\right )^3}-\frac {e^3 \left (b^2-4 a c\right ) \log (d+e x) \left (-2 c e (a e+5 b d)+3 b^2 e^2+10 c^2 d^2\right )}{\left (a e^2-b d e+c d^2\right )^3}}{\left (b^2-4 a c\right ) \left (a e^2-b d e+c d^2\right )}-\frac {2 a c e+b^2 (-e)+c x (2 c d-b e)+b c d}{\left (b^2-4 a c\right ) (d+e x)^2 \left (a+b x+c x^2\right ) \left (a e^2-b d e+c d^2\right )}\)

Input:

Int[1/((d + e*x)^3*(a + b*x + c*x^2)^2),x]
 

Output:

-((b*c*d - b^2*e + 2*a*c*e + c*(2*c*d - b*e)*x)/((b^2 - 4*a*c)*(c*d^2 - b* 
d*e + a*e^2)*(d + e*x)^2*(a + b*x + c*x^2))) - ((e*(4*c^2*d^2 + 3*b^2*e^2 
- 4*c*e*(b*d + 2*a*e)))/(2*(c*d^2 - b*d*e + a*e^2)*(d + e*x)^2) + (e*(2*c* 
d - b*e)*(c^2*d^2 + 3*b^2*e^2 - c*e*(b*d + 11*a*e)))/((c*d^2 - b*d*e + a*e 
^2)^2*(d + e*x)) - ((2*c*d - b*e)*(2*c^4*d^4 - 3*b^4*e^4 - 4*c^3*d^2*e*(b* 
d - 5*a*e) + 4*b^2*c*e^3*(b*d + 5*a*e) - 2*c^2*e^2*(b^2*d^2 + 10*a*b*d*e + 
 15*a^2*e^2))*ArcTanh[(b + 2*c*x)/Sqrt[b^2 - 4*a*c]])/(Sqrt[b^2 - 4*a*c]*( 
c*d^2 - b*d*e + a*e^2)^3) - ((b^2 - 4*a*c)*e^3*(10*c^2*d^2 + 3*b^2*e^2 - 2 
*c*e*(5*b*d + a*e))*Log[d + e*x])/(c*d^2 - b*d*e + a*e^2)^3 + ((b^2 - 4*a* 
c)*e^3*(10*c^2*d^2 + 3*b^2*e^2 - 2*c*e*(5*b*d + a*e))*Log[a + b*x + c*x^2] 
)/(2*(c*d^2 - b*d*e + a*e^2)^3))/((b^2 - 4*a*c)*(c*d^2 - b*d*e + a*e^2))
 

Defintions of rubi rules used

rule 1165

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S 
ymbol] :> Simp[(d + e*x)^(m + 1)*(b*c*d - b^2*e + 2*a*c*e + c*(2*c*d - b*e) 
*x)*((a + b*x + c*x^2)^(p + 1)/((p + 1)*(b^2 - 4*a*c)*(c*d^2 - b*d*e + a*e^ 
2))), x] + Simp[1/((p + 1)*(b^2 - 4*a*c)*(c*d^2 - b*d*e + a*e^2))   Int[(d 
+ e*x)^m*Simp[b*c*d*e*(2*p - m + 2) + b^2*e^2*(m + p + 2) - 2*c^2*d^2*(2*p 
+ 3) - 2*a*c*e^2*(m + 2*p + 3) - c*e*(2*c*d - b*e)*(m + 2*p + 4)*x, x]*(a + 
 b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && LtQ[p, -1] 
 && IntQuadraticQ[a, b, c, d, e, m, p, x]
 

rule 1200

Int[(((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.))/((a_.) + (b_.)* 
(x_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*((f + g* 
x)^n/(a + b*x + c*x^2)), x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && In 
tegersQ[n]
 

rule 2009

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

Maple [A] (verified)

Time = 1.30 (sec) , antiderivative size = 821, normalized size of antiderivative = 1.67

method	result	size
default	\(-\frac {e^{3}}{2 \left (a \,e^{2}-b d e +c \,d^{2}\right )^{2} \left (e x +d \right )^{2}}+\frac {2 e^{3} \left (b e -2 c d \right )}{\left (a \,e^{2}-b d e +c \,d^{2}\right )^{3} \left (e x +d \right )}-\frac {e^{3} \left (2 a c \,e^{2}-3 b^{2} e^{2}+10 b c d e -10 c^{2} d^{2}\right ) \ln \left (e x +d \right )}{\left (a \,e^{2}-b d e +c \,d^{2}\right )^{4}}+\frac {\frac {\frac {c \left (3 a^{2} b \,e^{5} c -6 a^{2} c^{2} d \,e^{4}-a \,b^{3} e^{5}+6 a b \,c^{2} d^{2} e^{3}-4 a \,c^{3} d^{3} e^{2}+b^{4} d \,e^{4}-4 b^{3} c \,d^{2} e^{3}+6 b^{2} c^{2} d^{3} e^{2}-5 b \,c^{3} d^{4} e +2 c^{4} d^{5}\right ) x}{4 a c -b^{2}}-\frac {2 a^{3} c^{2} e^{5}-4 a^{2} b^{2} c \,e^{5}+7 a^{2} b \,c^{2} d \,e^{4}-4 a^{2} c^{3} d^{2} e^{3}+a \,b^{4} e^{5}+a \,b^{3} c d \,e^{4}-10 a \,b^{2} c^{2} d^{2} e^{3}+14 a b \,c^{3} d^{3} e^{2}-6 a \,c^{4} d^{4} e -b^{5} d \,e^{4}+4 b^{4} c \,d^{2} e^{3}-6 b^{3} c^{2} d^{3} e^{2}+4 b^{2} c^{3} d^{4} e -b \,c^{4} d^{5}}{4 a c -b^{2}}}{c \,x^{2}+b x +a}+\frac {\frac {\left (8 a^{2} c^{3} e^{5}-14 a \,b^{2} c^{2} e^{5}+40 a b \,c^{3} d \,e^{4}-40 a \,c^{4} d^{2} e^{3}+3 b^{4} c \,e^{5}-10 b^{3} c^{2} d \,e^{4}+10 b^{2} c^{3} d^{2} e^{3}\right ) \ln \left (c \,x^{2}+b x +a \right )}{2 c}+\frac {2 \left (19 a^{2} b \,c^{2} e^{5}-30 a^{2} c^{3} d \,e^{4}-17 a \,b^{3} c \,e^{5}+50 a \,b^{2} c^{2} d \,e^{4}-50 a b \,c^{3} d^{2} e^{3}+20 a \,c^{4} d^{3} e^{2}+3 b^{5} e^{5}-10 b^{4} c d \,e^{4}+10 b^{3} d^{2} e^{3} c^{2}-5 b \,c^{4} d^{4} e +2 d^{5} c^{5}-\frac {\left (8 a^{2} c^{3} e^{5}-14 a \,b^{2} c^{2} e^{5}+40 a b \,c^{3} d \,e^{4}-40 a \,c^{4} d^{2} e^{3}+3 b^{4} c \,e^{5}-10 b^{3} c^{2} d \,e^{4}+10 b^{2} c^{3} d^{2} e^{3}\right ) b}{2 c}\right ) \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}}}{4 a c -b^{2}}}{\left (a \,e^{2}-b d e +c \,d^{2}\right )^{4}}\)	\(821\)
risch	\(\text {Expression too large to display}\)	\(6775\)

Input:

int(1/(e*x+d)^3/(c*x^2+b*x+a)^2,x,method=_RETURNVERBOSE)
 

Output:

-1/2*e^3/(a*e^2-b*d*e+c*d^2)^2/(e*x+d)^2+2*e^3/(a*e^2-b*d*e+c*d^2)^3*(b*e- 
2*c*d)/(e*x+d)-e^3*(2*a*c*e^2-3*b^2*e^2+10*b*c*d*e-10*c^2*d^2)/(a*e^2-b*d* 
e+c*d^2)^4*ln(e*x+d)+1/(a*e^2-b*d*e+c*d^2)^4*((c*(3*a^2*b*c*e^5-6*a^2*c^2* 
d*e^4-a*b^3*e^5+6*a*b*c^2*d^2*e^3-4*a*c^3*d^3*e^2+b^4*d*e^4-4*b^3*c*d^2*e^ 
3+6*b^2*c^2*d^3*e^2-5*b*c^3*d^4*e+2*c^4*d^5)/(4*a*c-b^2)*x-(2*a^3*c^2*e^5- 
4*a^2*b^2*c*e^5+7*a^2*b*c^2*d*e^4-4*a^2*c^3*d^2*e^3+a*b^4*e^5+a*b^3*c*d*e^ 
4-10*a*b^2*c^2*d^2*e^3+14*a*b*c^3*d^3*e^2-6*a*c^4*d^4*e-b^5*d*e^4+4*b^4*c* 
d^2*e^3-6*b^3*c^2*d^3*e^2+4*b^2*c^3*d^4*e-b*c^4*d^5)/(4*a*c-b^2))/(c*x^2+b 
*x+a)+1/(4*a*c-b^2)*(1/2*(8*a^2*c^3*e^5-14*a*b^2*c^2*e^5+40*a*b*c^3*d*e^4- 
40*a*c^4*d^2*e^3+3*b^4*c*e^5-10*b^3*c^2*d*e^4+10*b^2*c^3*d^2*e^3)/c*ln(c*x 
^2+b*x+a)+2*(19*a^2*b*c^2*e^5-30*a^2*c^3*d*e^4-17*a*b^3*c*e^5+50*a*b^2*c^2 
*d*e^4-50*a*b*c^3*d^2*e^3+20*a*c^4*d^3*e^2+3*b^5*e^5-10*b^4*c*d*e^4+10*b^3 
*d^2*e^3*c^2-5*b*c^4*d^4*e+2*d^5*c^5-1/2*(8*a^2*c^3*e^5-14*a*b^2*c^2*e^5+4 
0*a*b*c^3*d*e^4-40*a*c^4*d^2*e^3+3*b^4*c*e^5-10*b^3*c^2*d*e^4+10*b^2*c^3*d 
^2*e^3)*b/c)/(4*a*c-b^2)^(1/2)*arctan((2*c*x+b)/(4*a*c-b^2)^(1/2))))
 

Fricas [F(-1)]

Timed out. \[ \int \frac {1}{(d+e x)^3 \left (a+b x+c x^2\right )^2} \, dx=\text {Timed out} \] Input:

integrate(1/(e*x+d)^3/(c*x^2+b*x+a)^2,x, algorithm="fricas")
 

Output:

Timed out
 

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{(d+e x)^3 \left (a+b x+c x^2\right )^2} \, dx=\text {Timed out} \] Input:

integrate(1/(e*x+d)**3/(c*x**2+b*x+a)**2,x)
 

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{(d+e x)^3 \left (a+b x+c x^2\right )^2} \, dx=\text {Exception raised: ValueError} \] Input:

integrate(1/(e*x+d)^3/(c*x^2+b*x+a)^2,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 [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1720 vs. \(2 (485) = 970\).

Time = 0.36 (sec) , antiderivative size = 1720, normalized size of antiderivative = 3.49 \[ \int \frac {1}{(d+e x)^3 \left (a+b x+c x^2\right )^2} \, dx=\text {Too large to display} \] Input:

integrate(1/(e*x+d)^3/(c*x^2+b*x+a)^2,x, algorithm="giac")
 

Output:

-1/2*(10*c^2*d^2*e^3 - 10*b*c*d*e^4 + 3*b^2*e^5 - 2*a*c*e^5)*log(c*x^2 + b 
*x + a)/(c^4*d^8 - 4*b*c^3*d^7*e + 6*b^2*c^2*d^6*e^2 + 4*a*c^3*d^6*e^2 - 4 
*b^3*c*d^5*e^3 - 12*a*b*c^2*d^5*e^3 + b^4*d^4*e^4 + 12*a*b^2*c*d^4*e^4 + 6 
*a^2*c^2*d^4*e^4 - 4*a*b^3*d^3*e^5 - 12*a^2*b*c*d^3*e^5 + 6*a^2*b^2*d^2*e^ 
6 + 4*a^3*c*d^2*e^6 - 4*a^3*b*d*e^7 + a^4*e^8) + (10*c^2*d^2*e^4 - 10*b*c* 
d*e^5 + 3*b^2*e^6 - 2*a*c*e^6)*log(abs(e*x + d))/(c^4*d^8*e - 4*b*c^3*d^7* 
e^2 + 6*b^2*c^2*d^6*e^3 + 4*a*c^3*d^6*e^3 - 4*b^3*c*d^5*e^4 - 12*a*b*c^2*d 
^5*e^4 + b^4*d^4*e^5 + 12*a*b^2*c*d^4*e^5 + 6*a^2*c^2*d^4*e^5 - 4*a*b^3*d^ 
3*e^6 - 12*a^2*b*c*d^3*e^6 + 6*a^2*b^2*d^2*e^7 + 4*a^3*c*d^2*e^7 - 4*a^3*b 
*d*e^8 + a^4*e^9) - (4*c^5*d^5 - 10*b*c^4*d^4*e + 40*a*c^4*d^3*e^2 + 10*b^ 
3*c^2*d^2*e^3 - 60*a*b*c^3*d^2*e^3 - 10*b^4*c*d*e^4 + 60*a*b^2*c^2*d*e^4 - 
 60*a^2*c^3*d*e^4 + 3*b^5*e^5 - 20*a*b^3*c*e^5 + 30*a^2*b*c^2*e^5)*arctan( 
(2*c*x + b)/sqrt(-b^2 + 4*a*c))/((b^2*c^4*d^8 - 4*a*c^5*d^8 - 4*b^3*c^3*d^ 
7*e + 16*a*b*c^4*d^7*e + 6*b^4*c^2*d^6*e^2 - 20*a*b^2*c^3*d^6*e^2 - 16*a^2 
*c^4*d^6*e^2 - 4*b^5*c*d^5*e^3 + 4*a*b^3*c^2*d^5*e^3 + 48*a^2*b*c^3*d^5*e^ 
3 + b^6*d^4*e^4 + 8*a*b^4*c*d^4*e^4 - 42*a^2*b^2*c^2*d^4*e^4 - 24*a^3*c^3* 
d^4*e^4 - 4*a*b^5*d^3*e^5 + 4*a^2*b^3*c*d^3*e^5 + 48*a^3*b*c^2*d^3*e^5 + 6 
*a^2*b^4*d^2*e^6 - 20*a^3*b^2*c*d^2*e^6 - 16*a^4*c^2*d^2*e^6 - 4*a^3*b^3*d 
*e^7 + 16*a^4*b*c*d*e^7 + a^4*b^2*e^8 - 4*a^5*c*e^8)*sqrt(-b^2 + 4*a*c)) - 
 1/2*(2*b*c^4*d^7 - 8*b^2*c^3*d^6*e + 12*a*c^4*d^6*e + 12*b^3*c^2*d^5*e...
 

Mupad [B] (verification not implemented)

Time = 10.65 (sec) , antiderivative size = 3748, normalized size of antiderivative = 7.60 \[ \int \frac {1}{(d+e x)^3 \left (a+b x+c x^2\right )^2} \, dx=\text {Too large to display} \] Input:

int(1/((d + e*x)^3*(a + b*x + c*x^2)^2),x)
 

Output:

(log(b^3 + (-(4*a*c - b^2)^3)^(1/2) - 4*a*b*c - 8*a*c^2*x + 2*b^2*c*x)*((3 
*b^8*e^5)/2 + 64*a^4*c^4*e^5 - (3*b^5*e^5*(-(4*a*c - b^2)^3)^(1/2))/2 - 2* 
c^5*d^5*(-(4*a*c - b^2)^3)^(1/2) + 84*a^2*b^4*c^2*e^5 - 144*a^3*b^2*c^3*e^ 
5 - 320*a^3*c^5*d^2*e^3 + 5*b^6*c^2*d^2*e^3 - 19*a*b^6*c*e^5 - 5*b^7*c*d*e 
^4 + 240*a^2*b^2*c^4*d^2*e^3 - 5*b^3*c^2*d^2*e^3*(-(4*a*c - b^2)^3)^(1/2) 
+ 10*a*b^3*c*e^5*(-(4*a*c - b^2)^3)^(1/2) + 60*a*b^5*c^2*d*e^4 + 320*a^3*b 
*c^4*d*e^4 + 5*b*c^4*d^4*e*(-(4*a*c - b^2)^3)^(1/2) + 5*b^4*c*d*e^4*(-(4*a 
*c - b^2)^3)^(1/2) - 15*a^2*b*c^2*e^5*(-(4*a*c - b^2)^3)^(1/2) - 60*a*b^4* 
c^3*d^2*e^3 - 240*a^2*b^3*c^3*d*e^4 - 20*a*c^4*d^3*e^2*(-(4*a*c - b^2)^3)^ 
(1/2) + 30*a^2*c^3*d*e^4*(-(4*a*c - b^2)^3)^(1/2) + 30*a*b*c^3*d^2*e^3*(-( 
4*a*c - b^2)^3)^(1/2) - 30*a*b^2*c^2*d*e^4*(-(4*a*c - b^2)^3)^(1/2)))/(64* 
a^3*c^7*d^8 - a^4*b^6*e^8 + 64*a^7*c^3*e^8 - b^6*c^4*d^8 - b^10*d^4*e^4 + 
12*a*b^4*c^5*d^8 + 12*a^5*b^4*c*e^8 + 4*a*b^9*d^3*e^5 + 4*a^3*b^7*d*e^7 + 
4*b^7*c^3*d^7*e + 4*b^9*c*d^5*e^3 - 48*a^2*b^2*c^6*d^8 - 48*a^6*b^2*c^2*e^ 
8 - 6*a^2*b^8*d^2*e^6 + 256*a^4*c^6*d^6*e^2 + 384*a^5*c^5*d^4*e^4 + 256*a^ 
6*c^4*d^2*e^6 - 6*b^8*c^2*d^6*e^2 - 240*a^2*b^4*c^4*d^6*e^2 + 48*a^2*b^5*c 
^3*d^5*e^3 + 90*a^2*b^6*c^2*d^4*e^4 + 192*a^3*b^2*c^5*d^6*e^2 + 320*a^3*b^ 
3*c^4*d^5*e^3 - 440*a^3*b^4*c^3*d^4*e^4 + 48*a^3*b^5*c^2*d^3*e^5 + 480*a^4 
*b^2*c^4*d^4*e^4 + 320*a^4*b^3*c^3*d^3*e^5 - 240*a^4*b^4*c^2*d^2*e^6 + 192 
*a^5*b^2*c^3*d^2*e^6 - 48*a*b^5*c^4*d^7*e - 256*a^3*b*c^6*d^7*e - 48*a^...
 

Reduce [B] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 13591, normalized size of antiderivative = 27.57 \[ \int \frac {1}{(d+e x)^3 \left (a+b x+c x^2\right )^2} \, dx =\text {Too large to display} \] Input:

int(1/(e*x+d)^3/(c*x^2+b*x+a)^2,x)
 

Output:

(60*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2))*a**3*b**2*c**2 
*d**2*e**6 + 120*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2))*a 
**3*b**2*c**2*d*e**7*x + 60*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a*c 
 - b**2))*a**3*b**2*c**2*e**8*x**2 - 240*sqrt(4*a*c - b**2)*atan((b + 2*c* 
x)/sqrt(4*a*c - b**2))*a**3*c**4*d**4*e**4 - 480*sqrt(4*a*c - b**2)*atan(( 
b + 2*c*x)/sqrt(4*a*c - b**2))*a**3*c**4*d**3*e**5*x - 240*sqrt(4*a*c - b* 
*2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2))*a**3*c**4*d**2*e**6*x**2 - 40*sqr 
t(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2))*a**2*b**4*c*d**2*e**6 
 - 80*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2))*a**2*b**4*c* 
d*e**7*x - 40*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2))*a**2 
*b**4*c*e**8*x**2 + 40*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a*c - b* 
*2))*a**2*b**3*c**2*d**3*e**5 + 140*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sq 
rt(4*a*c - b**2))*a**2*b**3*c**2*d**2*e**6*x + 160*sqrt(4*a*c - b**2)*atan 
((b + 2*c*x)/sqrt(4*a*c - b**2))*a**2*b**3*c**2*d*e**7*x**2 + 60*sqrt(4*a* 
c - b**2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2))*a**2*b**3*c**2*e**8*x**3 + 
120*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2))*a**2*b**2*c**3 
*d**4*e**4 + 240*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2))*a 
**2*b**2*c**3*d**3*e**5*x + 180*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4 
*a*c - b**2))*a**2*b**2*c**3*d**2*e**6*x**2 + 120*sqrt(4*a*c - b**2)*atan( 
(b + 2*c*x)/sqrt(4*a*c - b**2))*a**2*b**2*c**3*d*e**7*x**3 + 60*sqrt(4*...