\(\int \frac {1}{(b d+2 c d x)^3 (a+b x+c x^2)^2} \, dx\) [68]

Optimal result

Integrand size = 24, antiderivative size = 110 \[ \int \frac {1}{(b d+2 c d x)^3 \left (a+b x+c x^2\right )^2} \, dx=-\frac {8 c}{\left (b^2-4 a c\right )^2 d^3 (b+2 c x)^2}-\frac {1}{\left (b^2-4 a c\right ) d^3 (b+2 c x)^2 \left (a+b x+c x^2\right )}+\frac {16 c \log (b+2 c x)}{\left (b^2-4 a c\right )^3 d^3}-\frac {8 c \log \left (a+b x+c x^2\right )}{\left (b^2-4 a c\right )^3 d^3} \] Output:

-8*c/(-4*a*c+b^2)^2/d^3/(2*c*x+b)^2-1/(-4*a*c+b^2)/d^3/(2*c*x+b)^2/(c*x^2+ 
b*x+a)+16*c*ln(2*c*x+b)/(-4*a*c+b^2)^3/d^3-8*c*ln(c*x^2+b*x+a)/(-4*a*c+b^2 
)^3/d^3
 

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.72 \[ \int \frac {1}{(b d+2 c d x)^3 \left (a+b x+c x^2\right )^2} \, dx=\frac {-\frac {4 c \left (b^2-4 a c\right )}{(b+2 c x)^2}+\frac {-b^2+4 a c}{a+x (b+c x)}+16 c \log (b+2 c x)-8 c \log (a+x (b+c x))}{\left (b^2-4 a c\right )^3 d^3} \] Input:

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

Output:

((-4*c*(b^2 - 4*a*c))/(b + 2*c*x)^2 + (-b^2 + 4*a*c)/(a + x*(b + c*x)) + 1 
6*c*Log[b + 2*c*x] - 8*c*Log[a + x*(b + c*x)])/((b^2 - 4*a*c)^3*d^3)
 

Rubi [A] (verified)

Time = 0.29 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.14, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1111, 27, 1117, 1105, 16, 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.

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

\(\Big \downarrow \) 1111

\(\displaystyle -\frac {8 c \int \frac {1}{d^3 (b+2 c x)^3 \left (c x^2+b x+a\right )}dx}{b^2-4 a c}-\frac {1}{d^3 \left (b^2-4 a c\right ) (b+2 c x)^2 \left (a+b x+c x^2\right )}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {8 c \int \frac {1}{(b+2 c x)^3 \left (c x^2+b x+a\right )}dx}{d^3 \left (b^2-4 a c\right )}-\frac {1}{d^3 \left (b^2-4 a c\right ) (b+2 c x)^2 \left (a+b x+c x^2\right )}\)

\(\Big \downarrow \) 1117

\(\displaystyle -\frac {8 c \left (\frac {\int \frac {1}{(b+2 c x) \left (c x^2+b x+a\right )}dx}{b^2-4 a c}+\frac {1}{\left (b^2-4 a c\right ) (b+2 c x)^2}\right )}{d^3 \left (b^2-4 a c\right )}-\frac {1}{d^3 \left (b^2-4 a c\right ) (b+2 c x)^2 \left (a+b x+c x^2\right )}\)

\(\Big \downarrow \) 1105

\(\displaystyle -\frac {8 c \left (\frac {\frac {\int \frac {b+2 c x}{c x^2+b x+a}dx}{b^2-4 a c}-\frac {4 c \int \frac {1}{b+2 c x}dx}{b^2-4 a c}}{b^2-4 a c}+\frac {1}{\left (b^2-4 a c\right ) (b+2 c x)^2}\right )}{d^3 \left (b^2-4 a c\right )}-\frac {1}{d^3 \left (b^2-4 a c\right ) (b+2 c x)^2 \left (a+b x+c x^2\right )}\)

\(\Big \downarrow \) 16

\(\displaystyle -\frac {8 c \left (\frac {\frac {\int \frac {b+2 c x}{c x^2+b x+a}dx}{b^2-4 a c}-\frac {2 \log (b+2 c x)}{b^2-4 a c}}{b^2-4 a c}+\frac {1}{\left (b^2-4 a c\right ) (b+2 c x)^2}\right )}{d^3 \left (b^2-4 a c\right )}-\frac {1}{d^3 \left (b^2-4 a c\right ) (b+2 c x)^2 \left (a+b x+c x^2\right )}\)

\(\Big \downarrow \) 1103

\(\displaystyle -\frac {1}{d^3 \left (b^2-4 a c\right ) (b+2 c x)^2 \left (a+b x+c x^2\right )}-\frac {8 c \left (\frac {\frac {\log \left (a+b x+c x^2\right )}{b^2-4 a c}-\frac {2 \log (b+2 c x)}{b^2-4 a c}}{b^2-4 a c}+\frac {1}{\left (b^2-4 a c\right ) (b+2 c x)^2}\right )}{d^3 \left (b^2-4 a c\right )}\)

Input:

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

Output:

-(1/((b^2 - 4*a*c)*d^3*(b + 2*c*x)^2*(a + b*x + c*x^2))) - (8*c*(1/((b^2 - 
 4*a*c)*(b + 2*c*x)^2) + ((-2*Log[b + 2*c*x])/(b^2 - 4*a*c) + Log[a + b*x 
+ c*x^2]/(b^2 - 4*a*c))/(b^2 - 4*a*c)))/((b^2 - 4*a*c)*d^3)
 

Defintions of rubi rules used

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 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, 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 1105

Int[1/(((d_) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)), x_Symbol] 
:> Simp[-4*b*(c/(d*(b^2 - 4*a*c)))   Int[1/(b + 2*c*x), x], x] + Simp[b^2/( 
d^2*(b^2 - 4*a*c))   Int[(d + e*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b 
, c, d, e}, x] && EqQ[2*c*d - b*e, 0]
 

rule 1111

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_S 
ymbol] :> Simp[2*c*(d + e*x)^(m + 1)*((a + b*x + c*x^2)^(p + 1)/(e*(p + 1)* 
(b^2 - 4*a*c))), x] - Simp[2*c*e*((m + 2*p + 3)/(e*(p + 1)*(b^2 - 4*a*c))) 
  Int[(d + e*x)^m*(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e 
, m}, x] && EqQ[2*c*d - b*e, 0] && NeQ[m + 2*p + 3, 0] && LtQ[p, -1] &&  !G 
tQ[m, 1] && RationalQ[m] && IntegerQ[2*p]
 

rule 1117

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_S 
ymbol] :> Simp[-2*b*d*(d + e*x)^(m + 1)*((a + b*x + c*x^2)^(p + 1)/(d^2*(m 
+ 1)*(b^2 - 4*a*c))), x] + Simp[b^2*((m + 2*p + 3)/(d^2*(m + 1)*(b^2 - 4*a* 
c)))   Int[(d + e*x)^(m + 2)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, 
 d, e, p}, x] && EqQ[2*c*d - b*e, 0] && NeQ[m + 2*p + 3, 0] && LtQ[m, -1] & 
& (IntegerQ[2*p] || (IntegerQ[m] && RationalQ[p]) || IntegerQ[(m + 2*p + 3) 
/2])
 

Maple [A] (verified)

Time = 0.96 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.91

method	result	size
default	\(\frac {-\frac {16 c \ln \left (2 c x +b \right )}{\left (4 a c -b^{2}\right )^{3}}-\frac {4 c}{\left (4 a c -b^{2}\right )^{2} \left (2 c x +b \right )^{2}}+\frac {\frac {-4 a c +b^{2}}{c \,x^{2}+b x +a}+8 c \ln \left (c \,x^{2}+b x +a \right )}{\left (4 a c -b^{2}\right )^{3}}}{d^{3}}\)	\(100\)
risch	\(\frac {-\frac {8 c^{2} x^{2}}{16 a^{2} c^{2}-8 c a \,b^{2}+b^{4}}-\frac {8 b c x}{16 a^{2} c^{2}-8 c a \,b^{2}+b^{4}}-\frac {4 a c +b^{2}}{16 a^{2} c^{2}-8 c a \,b^{2}+b^{4}}}{d^{3} \left (2 c x +b \right )^{2} \left (c \,x^{2}+b x +a \right )}-\frac {16 c \ln \left (2 c x +b \right )}{d^{3} \left (64 a^{3} c^{3}-48 a^{2} b^{2} c^{2}+12 a \,b^{4} c -b^{6}\right )}+\frac {8 c \ln \left (-c \,x^{2}-b x -a \right )}{d^{3} \left (64 a^{3} c^{3}-48 a^{2} b^{2} c^{2}+12 a \,b^{4} c -b^{6}\right )}\)	\(215\)
norman	\(\frac {\frac {-16 a \,c^{4}-4 c^{3} b^{2}}{4 d \,c^{3} \left (16 a^{2} c^{2}-8 c a \,b^{2}+b^{4}\right )}-\frac {8 c^{2} x^{2}}{d \left (16 a^{2} c^{2}-8 c a \,b^{2}+b^{4}\right )}-\frac {8 c b x}{d \left (16 a^{2} c^{2}-8 c a \,b^{2}+b^{4}\right )}}{d^{2} \left (2 c x +b \right )^{2} \left (c \,x^{2}+b x +a \right )}-\frac {16 c \ln \left (2 c x +b \right )}{d^{3} \left (64 a^{3} c^{3}-48 a^{2} b^{2} c^{2}+12 a \,b^{4} c -b^{6}\right )}+\frac {8 c \ln \left (c \,x^{2}+b x +a \right )}{d^{3} \left (64 a^{3} c^{3}-48 a^{2} b^{2} c^{2}+12 a \,b^{4} c -b^{6}\right )}\)	\(230\)
parallelrisch	\(\frac {4 b^{4} c^{3}+32 b^{2} c^{5} x^{2}-128 a b \,c^{5} x -64 a^{2} c^{5}-512 \ln \left (\frac {b}{2}+c x \right ) x^{3} b \,c^{6}+256 \ln \left (c \,x^{2}+b x +a \right ) x^{3} b \,c^{6}-256 \ln \left (\frac {b}{2}+c x \right ) x^{2} a \,c^{6}-320 \ln \left (\frac {b}{2}+c x \right ) x^{2} b^{2} c^{5}+128 \ln \left (c \,x^{2}+b x +a \right ) x^{2} a \,c^{6}+160 \ln \left (c \,x^{2}+b x +a \right ) x^{2} b^{2} c^{5}-64 \ln \left (\frac {b}{2}+c x \right ) x \,b^{3} c^{4}+32 \ln \left (c \,x^{2}+b x +a \right ) x \,b^{3} c^{4}-64 \ln \left (\frac {b}{2}+c x \right ) a \,b^{2} c^{4}+32 \ln \left (c \,x^{2}+b x +a \right ) a \,b^{2} c^{4}-256 \ln \left (\frac {b}{2}+c x \right ) x a b \,c^{5}+128 \ln \left (c \,x^{2}+b x +a \right ) x a b \,c^{5}-128 x^{2} a \,c^{6}+32 x \,b^{3} c^{4}+128 \ln \left (c \,x^{2}+b x +a \right ) x^{4} c^{7}-256 \ln \left (\frac {b}{2}+c x \right ) x^{4} c^{7}}{4 \left (64 a^{3} c^{3}-48 a^{2} b^{2} c^{2}+12 a \,b^{4} c -b^{6}\right ) \left (c \,x^{2}+b x +a \right ) \left (2 c x +b \right )^{2} d^{3} c^{3}}\)	\(376\)

Input:

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

Output:

1/d^3*(-16*c/(4*a*c-b^2)^3*ln(2*c*x+b)-4*c/(4*a*c-b^2)^2/(2*c*x+b)^2+1/(4* 
a*c-b^2)^3*((-4*a*c+b^2)/(c*x^2+b*x+a)+8*c*ln(c*x^2+b*x+a)))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 409 vs. \(2 (110) = 220\).

Time = 0.09 (sec) , antiderivative size = 409, normalized size of antiderivative = 3.72 \[ \int \frac {1}{(b d+2 c d x)^3 \left (a+b x+c x^2\right )^2} \, dx=-\frac {b^{4} - 16 \, a^{2} c^{2} + 8 \, {\left (b^{2} c^{2} - 4 \, a c^{3}\right )} x^{2} + 8 \, {\left (b^{3} c - 4 \, a b c^{2}\right )} x + 8 \, {\left (4 \, c^{4} x^{4} + 8 \, b c^{3} x^{3} + a b^{2} c + {\left (5 \, b^{2} c^{2} + 4 \, a c^{3}\right )} x^{2} + {\left (b^{3} c + 4 \, a b c^{2}\right )} x\right )} \log \left (c x^{2} + b x + a\right ) - 16 \, {\left (4 \, c^{4} x^{4} + 8 \, b c^{3} x^{3} + a b^{2} c + {\left (5 \, b^{2} c^{2} + 4 \, a c^{3}\right )} x^{2} + {\left (b^{3} c + 4 \, a b c^{2}\right )} x\right )} \log \left (2 \, c x + b\right )}{4 \, {\left (b^{6} c^{3} - 12 \, a b^{4} c^{4} + 48 \, a^{2} b^{2} c^{5} - 64 \, a^{3} c^{6}\right )} d^{3} x^{4} + 8 \, {\left (b^{7} c^{2} - 12 \, a b^{5} c^{3} + 48 \, a^{2} b^{3} c^{4} - 64 \, a^{3} b c^{5}\right )} d^{3} x^{3} + {\left (5 \, b^{8} c - 56 \, a b^{6} c^{2} + 192 \, a^{2} b^{4} c^{3} - 128 \, a^{3} b^{2} c^{4} - 256 \, a^{4} c^{5}\right )} d^{3} x^{2} + {\left (b^{9} - 8 \, a b^{7} c + 128 \, a^{3} b^{3} c^{3} - 256 \, a^{4} b c^{4}\right )} d^{3} x + {\left (a b^{8} - 12 \, a^{2} b^{6} c + 48 \, a^{3} b^{4} c^{2} - 64 \, a^{4} b^{2} c^{3}\right )} d^{3}} \] Input:

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

Output:

-(b^4 - 16*a^2*c^2 + 8*(b^2*c^2 - 4*a*c^3)*x^2 + 8*(b^3*c - 4*a*b*c^2)*x + 
 8*(4*c^4*x^4 + 8*b*c^3*x^3 + a*b^2*c + (5*b^2*c^2 + 4*a*c^3)*x^2 + (b^3*c 
 + 4*a*b*c^2)*x)*log(c*x^2 + b*x + a) - 16*(4*c^4*x^4 + 8*b*c^3*x^3 + a*b^ 
2*c + (5*b^2*c^2 + 4*a*c^3)*x^2 + (b^3*c + 4*a*b*c^2)*x)*log(2*c*x + b))/( 
4*(b^6*c^3 - 12*a*b^4*c^4 + 48*a^2*b^2*c^5 - 64*a^3*c^6)*d^3*x^4 + 8*(b^7* 
c^2 - 12*a*b^5*c^3 + 48*a^2*b^3*c^4 - 64*a^3*b*c^5)*d^3*x^3 + (5*b^8*c - 5 
6*a*b^6*c^2 + 192*a^2*b^4*c^3 - 128*a^3*b^2*c^4 - 256*a^4*c^5)*d^3*x^2 + ( 
b^9 - 8*a*b^7*c + 128*a^3*b^3*c^3 - 256*a^4*b*c^4)*d^3*x + (a*b^8 - 12*a^2 
*b^6*c + 48*a^3*b^4*c^2 - 64*a^4*b^2*c^3)*d^3)
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 304 vs. \(2 (110) = 220\).

Time = 1.58 (sec) , antiderivative size = 304, normalized size of antiderivative = 2.76 \[ \int \frac {1}{(b d+2 c d x)^3 \left (a+b x+c x^2\right )^2} \, dx=- \frac {16 c \log {\left (\frac {b}{2 c} + x \right )}}{d^{3} \left (4 a c - b^{2}\right )^{3}} + \frac {8 c \log {\left (\frac {a}{c} + \frac {b x}{c} + x^{2} \right )}}{d^{3} \left (4 a c - b^{2}\right )^{3}} + \frac {- 4 a c - b^{2} - 8 b c x - 8 c^{2} x^{2}}{16 a^{3} b^{2} c^{2} d^{3} - 8 a^{2} b^{4} c d^{3} + a b^{6} d^{3} + x^{4} \cdot \left (64 a^{2} c^{5} d^{3} - 32 a b^{2} c^{4} d^{3} + 4 b^{4} c^{3} d^{3}\right ) + x^{3} \cdot \left (128 a^{2} b c^{4} d^{3} - 64 a b^{3} c^{3} d^{3} + 8 b^{5} c^{2} d^{3}\right ) + x^{2} \cdot \left (64 a^{3} c^{4} d^{3} + 48 a^{2} b^{2} c^{3} d^{3} - 36 a b^{4} c^{2} d^{3} + 5 b^{6} c d^{3}\right ) + x \left (64 a^{3} b c^{3} d^{3} - 16 a^{2} b^{3} c^{2} d^{3} - 4 a b^{5} c d^{3} + b^{7} d^{3}\right )} \] Input:

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

Output:

-16*c*log(b/(2*c) + x)/(d**3*(4*a*c - b**2)**3) + 8*c*log(a/c + b*x/c + x* 
*2)/(d**3*(4*a*c - b**2)**3) + (-4*a*c - b**2 - 8*b*c*x - 8*c**2*x**2)/(16 
*a**3*b**2*c**2*d**3 - 8*a**2*b**4*c*d**3 + a*b**6*d**3 + x**4*(64*a**2*c* 
*5*d**3 - 32*a*b**2*c**4*d**3 + 4*b**4*c**3*d**3) + x**3*(128*a**2*b*c**4* 
d**3 - 64*a*b**3*c**3*d**3 + 8*b**5*c**2*d**3) + x**2*(64*a**3*c**4*d**3 + 
 48*a**2*b**2*c**3*d**3 - 36*a*b**4*c**2*d**3 + 5*b**6*c*d**3) + x*(64*a** 
3*b*c**3*d**3 - 16*a**2*b**3*c**2*d**3 - 4*a*b**5*c*d**3 + b**7*d**3))
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 296 vs. \(2 (110) = 220\).

Time = 0.06 (sec) , antiderivative size = 296, normalized size of antiderivative = 2.69 \[ \int \frac {1}{(b d+2 c d x)^3 \left (a+b x+c x^2\right )^2} \, dx=-\frac {8 \, c^{2} x^{2} + 8 \, b c x + b^{2} + 4 \, a c}{4 \, {\left (b^{4} c^{3} - 8 \, a b^{2} c^{4} + 16 \, a^{2} c^{5}\right )} d^{3} x^{4} + 8 \, {\left (b^{5} c^{2} - 8 \, a b^{3} c^{3} + 16 \, a^{2} b c^{4}\right )} d^{3} x^{3} + {\left (5 \, b^{6} c - 36 \, a b^{4} c^{2} + 48 \, a^{2} b^{2} c^{3} + 64 \, a^{3} c^{4}\right )} d^{3} x^{2} + {\left (b^{7} - 4 \, a b^{5} c - 16 \, a^{2} b^{3} c^{2} + 64 \, a^{3} b c^{3}\right )} d^{3} x + {\left (a b^{6} - 8 \, a^{2} b^{4} c + 16 \, a^{3} b^{2} c^{2}\right )} d^{3}} - \frac {8 \, c \log \left (c x^{2} + b x + a\right )}{{\left (b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}\right )} d^{3}} + \frac {16 \, c \log \left (2 \, c x + b\right )}{{\left (b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}\right )} d^{3}} \] Input:

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

Output:

-(8*c^2*x^2 + 8*b*c*x + b^2 + 4*a*c)/(4*(b^4*c^3 - 8*a*b^2*c^4 + 16*a^2*c^ 
5)*d^3*x^4 + 8*(b^5*c^2 - 8*a*b^3*c^3 + 16*a^2*b*c^4)*d^3*x^3 + (5*b^6*c - 
 36*a*b^4*c^2 + 48*a^2*b^2*c^3 + 64*a^3*c^4)*d^3*x^2 + (b^7 - 4*a*b^5*c - 
16*a^2*b^3*c^2 + 64*a^3*b*c^3)*d^3*x + (a*b^6 - 8*a^2*b^4*c + 16*a^3*b^2*c 
^2)*d^3) - 8*c*log(c*x^2 + b*x + a)/((b^6 - 12*a*b^4*c + 48*a^2*b^2*c^2 - 
64*a^3*c^3)*d^3) + 16*c*log(2*c*x + b)/((b^6 - 12*a*b^4*c + 48*a^2*b^2*c^2 
 - 64*a^3*c^3)*d^3)
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.85 \[ \int \frac {1}{(b d+2 c d x)^3 \left (a+b x+c x^2\right )^2} \, dx=\frac {16 \, c^{2} \log \left ({\left | 2 \, c x + b \right |}\right )}{b^{6} c d^{3} - 12 \, a b^{4} c^{2} d^{3} + 48 \, a^{2} b^{2} c^{3} d^{3} - 64 \, a^{3} c^{4} d^{3}} - \frac {8 \, c \log \left (c x^{2} + b x + a\right )}{b^{6} d^{3} - 12 \, a b^{4} c d^{3} + 48 \, a^{2} b^{2} c^{2} d^{3} - 64 \, a^{3} c^{3} d^{3}} - \frac {b^{4} - 16 \, a^{2} c^{2} + 8 \, {\left (b^{2} c^{2} - 4 \, a c^{3}\right )} x^{2} + 8 \, {\left (b^{3} c - 4 \, a b c^{2}\right )} x}{{\left (c x^{2} + b x + a\right )} {\left (b^{2} - 4 \, a c\right )}^{3} {\left (2 \, c x + b\right )}^{2} d^{3}} \] Input:

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

Output:

16*c^2*log(abs(2*c*x + b))/(b^6*c*d^3 - 12*a*b^4*c^2*d^3 + 48*a^2*b^2*c^3* 
d^3 - 64*a^3*c^4*d^3) - 8*c*log(c*x^2 + b*x + a)/(b^6*d^3 - 12*a*b^4*c*d^3 
 + 48*a^2*b^2*c^2*d^3 - 64*a^3*c^3*d^3) - (b^4 - 16*a^2*c^2 + 8*(b^2*c^2 - 
 4*a*c^3)*x^2 + 8*(b^3*c - 4*a*b*c^2)*x)/((c*x^2 + b*x + a)*(b^2 - 4*a*c)^ 
3*(2*c*x + b)^2*d^3)
 

Mupad [B] (verification not implemented)

Time = 5.85 (sec) , antiderivative size = 278, normalized size of antiderivative = 2.53 \[ \int \frac {1}{(b d+2 c d x)^3 \left (a+b x+c x^2\right )^2} \, dx=\frac {16\,c\,\ln \left (b+2\,c\,x\right )}{-64\,a^3\,c^3\,d^3+48\,a^2\,b^2\,c^2\,d^3-12\,a\,b^4\,c\,d^3+b^6\,d^3}-\frac {8\,c\,\ln \left (c\,x^2+b\,x+a\right )}{-64\,a^3\,c^3\,d^3+48\,a^2\,b^2\,c^2\,d^3-12\,a\,b^4\,c\,d^3+b^6\,d^3}-\frac {\frac {b^2+4\,a\,c}{16\,a^2\,c^2-8\,a\,b^2\,c+b^4}+\frac {8\,c^2\,x^2}{16\,a^2\,c^2-8\,a\,b^2\,c+b^4}+\frac {8\,b\,c\,x}{16\,a^2\,c^2-8\,a\,b^2\,c+b^4}}{x^2\,\left (5\,b^2\,c\,d^3+4\,a\,c^2\,d^3\right )+x\,\left (b^3\,d^3+4\,a\,c\,b\,d^3\right )+a\,b^2\,d^3+4\,c^3\,d^3\,x^4+8\,b\,c^2\,d^3\,x^3} \] Input:

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

Output:

(16*c*log(b + 2*c*x))/(b^6*d^3 - 64*a^3*c^3*d^3 + 48*a^2*b^2*c^2*d^3 - 12* 
a*b^4*c*d^3) - (8*c*log(a + b*x + c*x^2))/(b^6*d^3 - 64*a^3*c^3*d^3 + 48*a 
^2*b^2*c^2*d^3 - 12*a*b^4*c*d^3) - ((4*a*c + b^2)/(b^4 + 16*a^2*c^2 - 8*a* 
b^2*c) + (8*c^2*x^2)/(b^4 + 16*a^2*c^2 - 8*a*b^2*c) + (8*b*c*x)/(b^4 + 16* 
a^2*c^2 - 8*a*b^2*c))/(x^2*(4*a*c^2*d^3 + 5*b^2*c*d^3) + x*(b^3*d^3 + 4*a* 
b*c*d^3) + a*b^2*d^3 + 4*c^3*d^3*x^4 + 8*b*c^2*d^3*x^3)
 

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 528, normalized size of antiderivative = 4.80 \[ \int \frac {1}{(b d+2 c d x)^3 \left (a+b x+c x^2\right )^2} \, dx=\frac {8 \,\mathrm {log}\left (c \,x^{2}+b x +a \right ) a \,b^{2} c +32 \,\mathrm {log}\left (c \,x^{2}+b x +a \right ) a b \,c^{2} x +32 \,\mathrm {log}\left (c \,x^{2}+b x +a \right ) a \,c^{3} x^{2}+8 \,\mathrm {log}\left (c \,x^{2}+b x +a \right ) b^{3} c x +40 \,\mathrm {log}\left (c \,x^{2}+b x +a \right ) b^{2} c^{2} x^{2}+64 \,\mathrm {log}\left (c \,x^{2}+b x +a \right ) b \,c^{3} x^{3}+32 \,\mathrm {log}\left (c \,x^{2}+b x +a \right ) c^{4} x^{4}-16 \,\mathrm {log}\left (2 c x +b \right ) a \,b^{2} c -64 \,\mathrm {log}\left (2 c x +b \right ) a b \,c^{2} x -64 \,\mathrm {log}\left (2 c x +b \right ) a \,c^{3} x^{2}-16 \,\mathrm {log}\left (2 c x +b \right ) b^{3} c x -80 \,\mathrm {log}\left (2 c x +b \right ) b^{2} c^{2} x^{2}-128 \,\mathrm {log}\left (2 c x +b \right ) b \,c^{3} x^{3}-64 \,\mathrm {log}\left (2 c x +b \right ) c^{4} x^{4}-16 a^{2} c^{2}-32 a b \,c^{2} x -32 a \,c^{3} x^{2}+b^{4}+8 b^{3} c x +8 b^{2} c^{2} x^{2}}{d^{3} \left (256 a^{3} c^{6} x^{4}-192 a^{2} b^{2} c^{5} x^{4}+48 a \,b^{4} c^{4} x^{4}-4 b^{6} c^{3} x^{4}+512 a^{3} b \,c^{5} x^{3}-384 a^{2} b^{3} c^{4} x^{3}+96 a \,b^{5} c^{3} x^{3}-8 b^{7} c^{2} x^{3}+256 a^{4} c^{5} x^{2}+128 a^{3} b^{2} c^{4} x^{2}-192 a^{2} b^{4} c^{3} x^{2}+56 a \,b^{6} c^{2} x^{2}-5 b^{8} c \,x^{2}+256 a^{4} b \,c^{4} x -128 a^{3} b^{3} c^{3} x +8 a \,b^{7} c x -b^{9} x +64 a^{4} b^{2} c^{3}-48 a^{3} b^{4} c^{2}+12 a^{2} b^{6} c -a \,b^{8}\right )} \] Input:

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

Output:

(8*log(a + b*x + c*x**2)*a*b**2*c + 32*log(a + b*x + c*x**2)*a*b*c**2*x + 
32*log(a + b*x + c*x**2)*a*c**3*x**2 + 8*log(a + b*x + c*x**2)*b**3*c*x + 
40*log(a + b*x + c*x**2)*b**2*c**2*x**2 + 64*log(a + b*x + c*x**2)*b*c**3* 
x**3 + 32*log(a + b*x + c*x**2)*c**4*x**4 - 16*log(b + 2*c*x)*a*b**2*c - 6 
4*log(b + 2*c*x)*a*b*c**2*x - 64*log(b + 2*c*x)*a*c**3*x**2 - 16*log(b + 2 
*c*x)*b**3*c*x - 80*log(b + 2*c*x)*b**2*c**2*x**2 - 128*log(b + 2*c*x)*b*c 
**3*x**3 - 64*log(b + 2*c*x)*c**4*x**4 - 16*a**2*c**2 - 32*a*b*c**2*x - 32 
*a*c**3*x**2 + b**4 + 8*b**3*c*x + 8*b**2*c**2*x**2)/(d**3*(64*a**4*b**2*c 
**3 + 256*a**4*b*c**4*x + 256*a**4*c**5*x**2 - 48*a**3*b**4*c**2 - 128*a** 
3*b**3*c**3*x + 128*a**3*b**2*c**4*x**2 + 512*a**3*b*c**5*x**3 + 256*a**3* 
c**6*x**4 + 12*a**2*b**6*c - 192*a**2*b**4*c**3*x**2 - 384*a**2*b**3*c**4* 
x**3 - 192*a**2*b**2*c**5*x**4 - a*b**8 + 8*a*b**7*c*x + 56*a*b**6*c**2*x* 
*2 + 96*a*b**5*c**3*x**3 + 48*a*b**4*c**4*x**4 - b**9*x - 5*b**8*c*x**2 - 
8*b**7*c**2*x**3 - 4*b**6*c**3*x**4))