\(\int \frac {1}{(e+f x) (a c+(b c+a d) x+b d x^2)^3} \, dx\) [370]

Optimal result

Integrand size = 29, antiderivative size = 356 \[ \int \frac {1}{(e+f x) \left (a c+(b c+a d) x+b d x^2\right )^3} \, dx=-\frac {b^3}{2 (b c-a d)^3 (b e-a f) (a+b x)^2}+\frac {b^3 (3 b d e+b c f-4 a d f)}{(b c-a d)^4 (b e-a f)^2 (a+b x)}+\frac {d^3}{2 (b c-a d)^3 (d e-c f) (c+d x)^2}+\frac {d^3 (3 b d e-4 b c f+a d f)}{(b c-a d)^4 (d e-c f)^2 (c+d x)}+\frac {b^3 \left (10 a^2 d^2 f^2-5 a b d f (3 d e+c f)+b^2 \left (6 d^2 e^2+3 c d e f+c^2 f^2\right )\right ) \log (a+b x)}{(b c-a d)^5 (b e-a f)^3}-\frac {d^3 \left (a^2 d^2 f^2+a b d f (3 d e-5 c f)+b^2 \left (6 d^2 e^2-15 c d e f+10 c^2 f^2\right )\right ) \log (c+d x)}{(b c-a d)^5 (d e-c f)^3}+\frac {f^5 \log (e+f x)}{(b e-a f)^3 (d e-c f)^3} \] Output:

-1/2*b^3/(-a*d+b*c)^3/(-a*f+b*e)/(b*x+a)^2+b^3*(-4*a*d*f+b*c*f+3*b*d*e)/(- 
a*d+b*c)^4/(-a*f+b*e)^2/(b*x+a)+1/2*d^3/(-a*d+b*c)^3/(-c*f+d*e)/(d*x+c)^2+ 
d^3*(a*d*f-4*b*c*f+3*b*d*e)/(-a*d+b*c)^4/(-c*f+d*e)^2/(d*x+c)+b^3*(10*a^2* 
d^2*f^2-5*a*b*d*f*(c*f+3*d*e)+b^2*(c^2*f^2+3*c*d*e*f+6*d^2*e^2))*ln(b*x+a) 
/(-a*d+b*c)^5/(-a*f+b*e)^3-d^3*(a^2*d^2*f^2+a*b*d*f*(-5*c*f+3*d*e)+b^2*(10 
*c^2*f^2-15*c*d*e*f+6*d^2*e^2))*ln(d*x+c)/(-a*d+b*c)^5/(-c*f+d*e)^3+f^5*ln 
(f*x+e)/(-a*f+b*e)^3/(-c*f+d*e)^3
 

Mathematica [A] (verified)

Time = 0.32 (sec) , antiderivative size = 355, normalized size of antiderivative = 1.00 \[ \int \frac {1}{(e+f x) \left (a c+(b c+a d) x+b d x^2\right )^3} \, dx=-\frac {b^3}{2 (b c-a d)^3 (b e-a f) (a+b x)^2}+\frac {b^3 (3 b d e+b c f-4 a d f)}{(b c-a d)^4 (b e-a f)^2 (a+b x)}-\frac {d^3}{2 (b c-a d)^3 (-d e+c f) (c+d x)^2}+\frac {d^3 (3 b d e-4 b c f+a d f)}{(b c-a d)^4 (d e-c f)^2 (c+d x)}+\frac {b^3 \left (10 a^2 d^2 f^2-5 a b d f (3 d e+c f)+b^2 \left (6 d^2 e^2+3 c d e f+c^2 f^2\right )\right ) \log (a+b x)}{(b c-a d)^5 (b e-a f)^3}+\frac {d^3 \left (a^2 d^2 f^2+a b d f (3 d e-5 c f)+b^2 \left (6 d^2 e^2-15 c d e f+10 c^2 f^2\right )\right ) \log (c+d x)}{(b c-a d)^5 (-d e+c f)^3}+\frac {f^5 \log (e+f x)}{(b e-a f)^3 (d e-c f)^3} \] Input:

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

Output:

-1/2*b^3/((b*c - a*d)^3*(b*e - a*f)*(a + b*x)^2) + (b^3*(3*b*d*e + b*c*f - 
 4*a*d*f))/((b*c - a*d)^4*(b*e - a*f)^2*(a + b*x)) - d^3/(2*(b*c - a*d)^3* 
(-(d*e) + c*f)*(c + d*x)^2) + (d^3*(3*b*d*e - 4*b*c*f + a*d*f))/((b*c - a* 
d)^4*(d*e - c*f)^2*(c + d*x)) + (b^3*(10*a^2*d^2*f^2 - 5*a*b*d*f*(3*d*e + 
c*f) + b^2*(6*d^2*e^2 + 3*c*d*e*f + c^2*f^2))*Log[a + b*x])/((b*c - a*d)^5 
*(b*e - a*f)^3) + (d^3*(a^2*d^2*f^2 + a*b*d*f*(3*d*e - 5*c*f) + b^2*(6*d^2 
*e^2 - 15*c*d*e*f + 10*c^2*f^2))*Log[c + d*x])/((b*c - a*d)^5*(-(d*e) + c* 
f)^3) + (f^5*Log[e + f*x])/((b*e - a*f)^3*(d*e - c*f)^3)
 

Rubi [A] (verified)

Time = 1.05 (sec) , antiderivative size = 369, normalized size of antiderivative = 1.04, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {1141, 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.

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

\(\Big \downarrow \) 1141

\(\displaystyle b^3 d^3 \int \left (\frac {f^6}{b^3 d^3 (b e-a f)^3 (d e-c f)^3 (e+f x)}+\frac {b \left (\left (6 d^2 e^2+3 c d f e+c^2 f^2\right ) b^2-5 a d f (3 d e+c f) b+10 a^2 d^2 f^2\right )}{d^3 (b c-a d)^5 (b e-a f)^3 (a+b x)}-\frac {d \left (\left (6 d^2 e^2-15 c d f e+10 c^2 f^2\right ) b^2+a d f (3 d e-5 c f) b+a^2 d^2 f^2\right )}{b^3 (b c-a d)^5 (d e-c f)^3 (c+d x)}-\frac {b (3 b d e+b c f-4 a d f)}{d^3 (b c-a d)^4 (b e-a f)^2 (a+b x)^2}-\frac {d (3 b d e-4 b c f+a d f)}{b^3 (b c-a d)^4 (d e-c f)^2 (c+d x)^2}+\frac {b}{d^3 (b c-a d)^3 (b e-a f) (a+b x)^3}-\frac {d}{b^3 (b c-a d)^3 (d e-c f) (c+d x)^3}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle b^3 d^3 \left (\frac {\log (a+b x) \left (10 a^2 d^2 f^2-5 a b d f (c f+3 d e)+b^2 \left (c^2 f^2+3 c d e f+6 d^2 e^2\right )\right )}{d^3 (b c-a d)^5 (b e-a f)^3}-\frac {\log (c+d x) \left (a^2 d^2 f^2+a b d f (3 d e-5 c f)+b^2 \left (10 c^2 f^2-15 c d e f+6 d^2 e^2\right )\right )}{b^3 (b c-a d)^5 (d e-c f)^3}+\frac {f^5 \log (e+f x)}{b^3 d^3 (b e-a f)^3 (d e-c f)^3}+\frac {a d f-4 b c f+3 b d e}{b^3 (c+d x) (b c-a d)^4 (d e-c f)^2}+\frac {1}{2 b^3 (c+d x)^2 (b c-a d)^3 (d e-c f)}+\frac {-4 a d f+b c f+3 b d e}{d^3 (a+b x) (b c-a d)^4 (b e-a f)^2}-\frac {1}{2 d^3 (a+b x)^2 (b c-a d)^3 (b e-a f)}\right )\)

Input:

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

Output:

b^3*d^3*(-1/2*1/(d^3*(b*c - a*d)^3*(b*e - a*f)*(a + b*x)^2) + (3*b*d*e + b 
*c*f - 4*a*d*f)/(d^3*(b*c - a*d)^4*(b*e - a*f)^2*(a + b*x)) + 1/(2*b^3*(b* 
c - a*d)^3*(d*e - c*f)*(c + d*x)^2) + (3*b*d*e - 4*b*c*f + a*d*f)/(b^3*(b* 
c - a*d)^4*(d*e - c*f)^2*(c + d*x)) + ((10*a^2*d^2*f^2 - 5*a*b*d*f*(3*d*e 
+ c*f) + b^2*(6*d^2*e^2 + 3*c*d*e*f + c^2*f^2))*Log[a + b*x])/(d^3*(b*c - 
a*d)^5*(b*e - a*f)^3) - ((a^2*d^2*f^2 + a*b*d*f*(3*d*e - 5*c*f) + b^2*(6*d 
^2*e^2 - 15*c*d*e*f + 10*c^2*f^2))*Log[c + d*x])/(b^3*(b*c - a*d)^5*(d*e - 
 c*f)^3) + (f^5*Log[e + f*x])/(b^3*d^3*(b*e - a*f)^3*(d*e - c*f)^3))
 

Defintions of rubi rules used

rule 1141

Int[((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_ 
Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[1/c^p   Int[ExpandIntegrand[ 
(d + e*x)^m*(b/2 - q/2 + c*x)^p*(b/2 + q/2 + c*x)^p, x], x], x] /; EqQ[p, - 
1] ||  !FractionalPowerFactorQ[q]] /; FreeQ[{a, b, c, d, e}, x] && ILtQ[p, 
0] && IntegerQ[m] && NiceSqrtQ[b^2 - 4*a*c]
 

rule 2009

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

Maple [A] (verified)

Time = 1.47 (sec) , antiderivative size = 371, normalized size of antiderivative = 1.04

method	result	size
default	\(-\frac {b^{3}}{2 \left (a f -b e \right ) \left (a d -b c \right )^{3} \left (b x +a \right )^{2}}-\frac {b^{3} \left (4 a d f -b c f -3 b d e \right )}{\left (a f -b e \right )^{2} \left (a d -b c \right )^{4} \left (b x +a \right )}+\frac {b^{3} \left (10 a^{2} d^{2} f^{2}-5 a b c d \,f^{2}-15 a b \,d^{2} e f +b^{2} c^{2} f^{2}+3 b^{2} c d e f +6 d^{2} e^{2} b^{2}\right ) \ln \left (b x +a \right )}{\left (a f -b e \right )^{3} \left (a d -b c \right )^{5}}+\frac {f^{5} \ln \left (f x +e \right )}{\left (a f -b e \right )^{3} \left (c f -d e \right )^{3}}+\frac {d^{3}}{2 \left (c f -d e \right ) \left (a d -b c \right )^{3} \left (d x +c \right )^{2}}+\frac {d^{3} \left (a d f -4 b c f +3 b d e \right )}{\left (c f -d e \right )^{2} \left (a d -b c \right )^{4} \left (d x +c \right )}-\frac {d^{3} \left (a^{2} d^{2} f^{2}-5 a b c d \,f^{2}+3 a b \,d^{2} e f +10 b^{2} c^{2} f^{2}-15 b^{2} c d e f +6 d^{2} e^{2} b^{2}\right ) \ln \left (d x +c \right )}{\left (c f -d e \right )^{3} \left (a d -b c \right )^{5}}\)	\(371\)
norman	\(\text {Expression too large to display}\)	\(2150\)
risch	\(\text {Expression too large to display}\)	\(7427\)
parallelrisch	\(\text {Expression too large to display}\)	\(7462\)

Input:

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

Output:

-1/2*b^3/(a*f-b*e)/(a*d-b*c)^3/(b*x+a)^2-b^3*(4*a*d*f-b*c*f-3*b*d*e)/(a*f- 
b*e)^2/(a*d-b*c)^4/(b*x+a)+b^3*(10*a^2*d^2*f^2-5*a*b*c*d*f^2-15*a*b*d^2*e* 
f+b^2*c^2*f^2+3*b^2*c*d*e*f+6*b^2*d^2*e^2)/(a*f-b*e)^3/(a*d-b*c)^5*ln(b*x+ 
a)+f^5/(a*f-b*e)^3/(c*f-d*e)^3*ln(f*x+e)+1/2*d^3/(c*f-d*e)/(a*d-b*c)^3/(d* 
x+c)^2+d^3*(a*d*f-4*b*c*f+3*b*d*e)/(c*f-d*e)^2/(a*d-b*c)^4/(d*x+c)-d^3*(a^ 
2*d^2*f^2-5*a*b*c*d*f^2+3*a*b*d^2*e*f+10*b^2*c^2*f^2-15*b^2*c*d*e*f+6*b^2* 
d^2*e^2)/(c*f-d*e)^3/(a*d-b*c)^5*ln(d*x+c)
 

Fricas [F(-1)]

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

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

Output:

Timed out
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 3790 vs. \(2 (352) = 704\).

Time = 0.35 (sec) , antiderivative size = 3790, normalized size of antiderivative = 10.65 \[ \int \frac {1}{(e+f x) \left (a c+(b c+a d) x+b d x^2\right )^3} \, dx=\text {Too large to display} \] Input:

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

Output:

f^5*log(f*x + e)/(b^3*d^3*e^6 + a^3*c^3*f^6 - 3*(b^3*c*d^2 + a*b^2*d^3)*e^ 
5*f + 3*(b^3*c^2*d + 3*a*b^2*c*d^2 + a^2*b*d^3)*e^4*f^2 - (b^3*c^3 + 9*a*b 
^2*c^2*d + 9*a^2*b*c*d^2 + a^3*d^3)*e^3*f^3 + 3*(a*b^2*c^3 + 3*a^2*b*c^2*d 
 + a^3*c*d^2)*e^2*f^4 - 3*(a^2*b*c^3 + a^3*c^2*d)*e*f^5) + (6*b^5*d^2*e^2 
+ 3*(b^5*c*d - 5*a*b^4*d^2)*e*f + (b^5*c^2 - 5*a*b^4*c*d + 10*a^2*b^3*d^2) 
*f^2)*log(b*x + a)/((b^8*c^5 - 5*a*b^7*c^4*d + 10*a^2*b^6*c^3*d^2 - 10*a^3 
*b^5*c^2*d^3 + 5*a^4*b^4*c*d^4 - a^5*b^3*d^5)*e^3 - 3*(a*b^7*c^5 - 5*a^2*b 
^6*c^4*d + 10*a^3*b^5*c^3*d^2 - 10*a^4*b^4*c^2*d^3 + 5*a^5*b^3*c*d^4 - a^6 
*b^2*d^5)*e^2*f + 3*(a^2*b^6*c^5 - 5*a^3*b^5*c^4*d + 10*a^4*b^4*c^3*d^2 - 
10*a^5*b^3*c^2*d^3 + 5*a^6*b^2*c*d^4 - a^7*b*d^5)*e*f^2 - (a^3*b^5*c^5 - 5 
*a^4*b^4*c^4*d + 10*a^5*b^3*c^3*d^2 - 10*a^6*b^2*c^2*d^3 + 5*a^7*b*c*d^4 - 
 a^8*d^5)*f^3) - (6*b^2*d^5*e^2 - 3*(5*b^2*c*d^4 - a*b*d^5)*e*f + (10*b^2* 
c^2*d^3 - 5*a*b*c*d^4 + a^2*d^5)*f^2)*log(d*x + c)/((b^5*c^5*d^3 - 5*a*b^4 
*c^4*d^4 + 10*a^2*b^3*c^3*d^5 - 10*a^3*b^2*c^2*d^6 + 5*a^4*b*c*d^7 - a^5*d 
^8)*e^3 - 3*(b^5*c^6*d^2 - 5*a*b^4*c^5*d^3 + 10*a^2*b^3*c^4*d^4 - 10*a^3*b 
^2*c^3*d^5 + 5*a^4*b*c^2*d^6 - a^5*c*d^7)*e^2*f + 3*(b^5*c^7*d - 5*a*b^4*c 
^6*d^2 + 10*a^2*b^3*c^5*d^3 - 10*a^3*b^2*c^4*d^4 + 5*a^4*b*c^3*d^5 - a^5*c 
^2*d^6)*e*f^2 - (b^5*c^8 - 5*a*b^4*c^7*d + 10*a^2*b^3*c^6*d^2 - 10*a^3*b^2 
*c^5*d^3 + 5*a^4*b*c^4*d^4 - a^5*c^3*d^5)*f^3) - 1/2*((b^5*c^3*d^2 - 7*a*b 
^4*c^2*d^3 - 7*a^2*b^3*c*d^4 + a^3*b^2*d^5)*e^3 - (2*b^5*c^4*d - 11*a*b...
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2839 vs. \(2 (352) = 704\).

Time = 0.38 (sec) , antiderivative size = 2839, normalized size of antiderivative = 7.97 \[ \int \frac {1}{(e+f x) \left (a c+(b c+a d) x+b d x^2\right )^3} \, dx=\text {Too large to display} \] Input:

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

Output:

f^6*log(abs(f*x + e))/(b^3*d^3*e^6*f - 3*b^3*c*d^2*e^5*f^2 - 3*a*b^2*d^3*e 
^5*f^2 + 3*b^3*c^2*d*e^4*f^3 + 9*a*b^2*c*d^2*e^4*f^3 + 3*a^2*b*d^3*e^4*f^3 
 - b^3*c^3*e^3*f^4 - 9*a*b^2*c^2*d*e^3*f^4 - 9*a^2*b*c*d^2*e^3*f^4 - a^3*d 
^3*e^3*f^4 + 3*a*b^2*c^3*e^2*f^5 + 9*a^2*b*c^2*d*e^2*f^5 + 3*a^3*c*d^2*e^2 
*f^5 - 3*a^2*b*c^3*e*f^6 - 3*a^3*c^2*d*e*f^6 + a^3*c^3*f^7) + (6*b^6*d^2*e 
^2 + 3*b^6*c*d*e*f - 15*a*b^5*d^2*e*f + b^6*c^2*f^2 - 5*a*b^5*c*d*f^2 + 10 
*a^2*b^4*d^2*f^2)*log(abs(b*x + a))/(b^9*c^5*e^3 - 5*a*b^8*c^4*d*e^3 + 10* 
a^2*b^7*c^3*d^2*e^3 - 10*a^3*b^6*c^2*d^3*e^3 + 5*a^4*b^5*c*d^4*e^3 - a^5*b 
^4*d^5*e^3 - 3*a*b^8*c^5*e^2*f + 15*a^2*b^7*c^4*d*e^2*f - 30*a^3*b^6*c^3*d 
^2*e^2*f + 30*a^4*b^5*c^2*d^3*e^2*f - 15*a^5*b^4*c*d^4*e^2*f + 3*a^6*b^3*d 
^5*e^2*f + 3*a^2*b^7*c^5*e*f^2 - 15*a^3*b^6*c^4*d*e*f^2 + 30*a^4*b^5*c^3*d 
^2*e*f^2 - 30*a^5*b^4*c^2*d^3*e*f^2 + 15*a^6*b^3*c*d^4*e*f^2 - 3*a^7*b^2*d 
^5*e*f^2 - a^3*b^6*c^5*f^3 + 5*a^4*b^5*c^4*d*f^3 - 10*a^5*b^4*c^3*d^2*f^3 
+ 10*a^6*b^3*c^2*d^3*f^3 - 5*a^7*b^2*c*d^4*f^3 + a^8*b*d^5*f^3) - (6*b^2*d 
^6*e^2 - 15*b^2*c*d^5*e*f + 3*a*b*d^6*e*f + 10*b^2*c^2*d^4*f^2 - 5*a*b*c*d 
^5*f^2 + a^2*d^6*f^2)*log(abs(d*x + c))/(b^5*c^5*d^4*e^3 - 5*a*b^4*c^4*d^5 
*e^3 + 10*a^2*b^3*c^3*d^6*e^3 - 10*a^3*b^2*c^2*d^7*e^3 + 5*a^4*b*c*d^8*e^3 
 - a^5*d^9*e^3 - 3*b^5*c^6*d^3*e^2*f + 15*a*b^4*c^5*d^4*e^2*f - 30*a^2*b^3 
*c^4*d^5*e^2*f + 30*a^3*b^2*c^3*d^6*e^2*f - 15*a^4*b*c^2*d^7*e^2*f + 3*a^5 
*c*d^8*e^2*f + 3*b^5*c^7*d^2*e*f^2 - 15*a*b^4*c^6*d^3*e*f^2 + 30*a^2*b^...
 

Mupad [B] (verification not implemented)

Time = 12.84 (sec) , antiderivative size = 3770, normalized size of antiderivative = 10.59 \[ \int \frac {1}{(e+f x) \left (a c+(b c+a d) x+b d x^2\right )^3} \, dx=\text {Too large to display} \] Input:

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

Output:

(f^5*log(e + f*x))/(a^3*c^3*f^6 + b^3*d^3*e^6 - a^3*d^3*e^3*f^3 - b^3*c^3* 
e^3*f^3 - 3*a^2*b*c^3*e*f^5 - 3*a*b^2*d^3*e^5*f - 3*a^3*c^2*d*e*f^5 - 3*b^ 
3*c*d^2*e^5*f + 3*a*b^2*c^3*e^2*f^4 + 3*a^2*b*d^3*e^4*f^2 + 3*a^3*c*d^2*e^ 
2*f^4 + 3*b^3*c^2*d*e^4*f^2 + 9*a*b^2*c*d^2*e^4*f^2 - 9*a*b^2*c^2*d*e^3*f^ 
3 - 9*a^2*b*c*d^2*e^3*f^3 + 9*a^2*b*c^2*d*e^2*f^4) - ((3*a*b^4*c^5*f^3 + 3 
*a^5*c*d^4*f^3 - a^5*d^5*e*f^2 - b^5*c^5*e*f^2 - a^3*b^2*d^5*e^3 - b^5*c^3 
*d^2*e^3 + 2*a^4*b*d^5*e^2*f + 2*b^5*c^4*d*e^2*f + 7*a*b^4*c^2*d^3*e^3 + 7 
*a^2*b^3*c*d^4*e^3 - 9*a^2*b^3*c^4*d*f^3 - 9*a^4*b*c^2*d^3*f^3 - 11*a*b^4* 
c^3*d^2*e^2*f - 11*a^3*b^2*c*d^4*e^2*f - 18*a^2*b^3*c^2*d^3*e^2*f + 18*a^2 
*b^3*c^3*d^2*e*f^2 + 18*a^3*b^2*c^2*d^3*e*f^2 + a*b^4*c^4*d*e*f^2 + a^4*b* 
c*d^4*e*f^2)/(2*(2*a*b^5*c^6*e*f^3 - a^4*b^2*d^6*e^4 - a^6*c^2*d^4*f^4 - b 
^6*c^4*d^2*e^4 - a^6*d^6*e^2*f^2 - b^6*c^6*e^2*f^2 - 6*a^2*b^4*c^2*d^4*e^4 
 - 6*a^4*b^2*c^4*d^2*f^4 - a^2*b^4*c^6*f^4 + 2*a^5*b*d^6*e^3*f + 2*a^6*c*d 
^5*e*f^3 + 2*b^6*c^5*d*e^3*f + 4*a*b^5*c^3*d^3*e^4 + 4*a^3*b^3*c*d^5*e^4 + 
 4*a^3*b^3*c^5*d*f^4 + 4*a^5*b*c^3*d^3*f^4 - 6*a*b^5*c^4*d^2*e^3*f - 6*a^2 
*b^4*c^5*d*e*f^3 - 6*a^4*b^2*c*d^5*e^3*f - 6*a^5*b*c^2*d^4*e*f^3 + 4*a^2*b 
^4*c^3*d^3*e^3*f + 4*a^3*b^3*c^2*d^4*e^3*f + 4*a^3*b^3*c^4*d^2*e*f^3 + 4*a 
^4*b^2*c^3*d^3*e*f^3 + 9*a^2*b^4*c^4*d^2*e^2*f^2 - 16*a^3*b^3*c^3*d^3*e^2* 
f^2 + 9*a^4*b^2*c^2*d^4*e^2*f^2)) + (x*(a^5*d^5*f^3 + b^5*c^5*f^3 + 2*a^2* 
b^3*d^5*e^3 + 2*b^5*c^2*d^3*e^3 - 9*a^2*b^3*c^3*d^2*f^3 - 9*a^3*b^2*c^2...
 

Reduce [B] (verification not implemented)

Time = 0.31 (sec) , antiderivative size = 14564, normalized size of antiderivative = 40.91 \[ \int \frac {1}{(e+f x) \left (a c+(b c+a d) x+b d x^2\right )^3} \, dx =\text {Too large to display} \] Input:

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

Output:

(20*log(a + b*x)*a**5*b**3*c**5*d**3*f**5 - 60*log(a + b*x)*a**5*b**3*c**4 
*d**4*e*f**4 + 40*log(a + b*x)*a**5*b**3*c**4*d**4*f**5*x + 60*log(a + b*x 
)*a**5*b**3*c**3*d**5*e**2*f**3 - 120*log(a + b*x)*a**5*b**3*c**3*d**5*e*f 
**4*x + 20*log(a + b*x)*a**5*b**3*c**3*d**5*f**5*x**2 - 20*log(a + b*x)*a* 
*5*b**3*c**2*d**6*e**3*f**2 + 120*log(a + b*x)*a**5*b**3*c**2*d**6*e**2*f* 
*3*x - 60*log(a + b*x)*a**5*b**3*c**2*d**6*e*f**4*x**2 - 40*log(a + b*x)*a 
**5*b**3*c*d**7*e**3*f**2*x + 60*log(a + b*x)*a**5*b**3*c*d**7*e**2*f**3*x 
**2 - 20*log(a + b*x)*a**5*b**3*d**8*e**3*f**2*x**2 + 10*log(a + b*x)*a**4 
*b**4*c**6*d**2*f**5 - 60*log(a + b*x)*a**4*b**4*c**5*d**3*e*f**4 + 60*log 
(a + b*x)*a**4*b**4*c**5*d**3*f**5*x + 120*log(a + b*x)*a**4*b**4*c**4*d** 
4*e**2*f**3 - 240*log(a + b*x)*a**4*b**4*c**4*d**4*e*f**4*x + 90*log(a + b 
*x)*a**4*b**4*c**4*d**4*f**5*x**2 - 100*log(a + b*x)*a**4*b**4*c**3*d**5*e 
**3*f**2 + 360*log(a + b*x)*a**4*b**4*c**3*d**5*e**2*f**3*x - 300*log(a + 
b*x)*a**4*b**4*c**3*d**5*e*f**4*x**2 + 40*log(a + b*x)*a**4*b**4*c**3*d**5 
*f**5*x**3 + 30*log(a + b*x)*a**4*b**4*c**2*d**6*e**4*f - 240*log(a + b*x) 
*a**4*b**4*c**2*d**6*e**3*f**2*x + 360*log(a + b*x)*a**4*b**4*c**2*d**6*e* 
*2*f**3*x**2 - 120*log(a + b*x)*a**4*b**4*c**2*d**6*e*f**4*x**3 + 60*log(a 
 + b*x)*a**4*b**4*c*d**7*e**4*f*x - 180*log(a + b*x)*a**4*b**4*c*d**7*e**3 
*f**2*x**2 + 120*log(a + b*x)*a**4*b**4*c*d**7*e**2*f**3*x**3 + 30*log(a + 
 b*x)*a**4*b**4*d**8*e**4*f*x**2 - 40*log(a + b*x)*a**4*b**4*d**8*e**3*...