\(\int \frac {A+B x}{(d+e x)^3 (b x+c x^2)} \, dx\) [36]

Optimal result

Integrand size = 24, antiderivative size = 171 \[ \int \frac {A+B x}{(d+e x)^3 \left (b x+c x^2\right )} \, dx=\frac {B d-A e}{2 d (c d-b e) (d+e x)^2}+\frac {B c d^2-A e (2 c d-b e)}{d^2 (c d-b e)^2 (d+e x)}+\frac {A \log (x)}{b d^3}+\frac {c^2 (b B-A c) \log (b+c x)}{b (c d-b e)^3}-\frac {\left (B c^2 d^3-A e \left (3 c^2 d^2-3 b c d e+b^2 e^2\right )\right ) \log (d+e x)}{d^3 (c d-b e)^3} \] Output:

1/2*(-A*e+B*d)/d/(-b*e+c*d)/(e*x+d)^2+(B*c*d^2-A*e*(-b*e+2*c*d))/d^2/(-b*e 
+c*d)^2/(e*x+d)+A*ln(x)/b/d^3+c^2*(-A*c+B*b)*ln(c*x+b)/b/(-b*e+c*d)^3-(B*c 
^2*d^3-A*e*(b^2*e^2-3*b*c*d*e+3*c^2*d^2))*ln(e*x+d)/d^3/(-b*e+c*d)^3
 

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 169, normalized size of antiderivative = 0.99 \[ \int \frac {A+B x}{(d+e x)^3 \left (b x+c x^2\right )} \, dx=\frac {B d-A e}{2 d (c d-b e) (d+e x)^2}+\frac {B c d^2+A e (-2 c d+b e)}{d^2 (c d-b e)^2 (d+e x)}+\frac {A \log (x)}{b d^3}+\frac {c^2 (-b B+A c) \log (b+c x)}{b (-c d+b e)^3}-\frac {\left (B c^2 d^3-A e \left (3 c^2 d^2-3 b c d e+b^2 e^2\right )\right ) \log (d+e x)}{d^3 (c d-b e)^3} \] Input:

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

Output:

(B*d - A*e)/(2*d*(c*d - b*e)*(d + e*x)^2) + (B*c*d^2 + A*e*(-2*c*d + b*e)) 
/(d^2*(c*d - b*e)^2*(d + e*x)) + (A*Log[x])/(b*d^3) + (c^2*(-(b*B) + A*c)* 
Log[b + c*x])/(b*(-(c*d) + b*e)^3) - ((B*c^2*d^3 - A*e*(3*c^2*d^2 - 3*b*c* 
d*e + b^2*e^2))*Log[d + e*x])/(d^3*(c*d - b*e)^3)
 

Rubi [A] (verified)

Time = 0.72 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {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.

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

\(\Big \downarrow \) 1200

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

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {\log (d+e x) \left (B c^2 d^3-A e \left (b^2 e^2-3 b c d e+3 c^2 d^2\right )\right )}{d^3 (c d-b e)^3}+\frac {c^2 (b B-A c) \log (b+c x)}{b (c d-b e)^3}+\frac {B c d^2-A e (2 c d-b e)}{d^2 (d+e x) (c d-b e)^2}+\frac {B d-A e}{2 d (d+e x)^2 (c d-b e)}+\frac {A \log (x)}{b d^3}\)

Input:

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

Output:

(B*d - A*e)/(2*d*(c*d - b*e)*(d + e*x)^2) + (B*c*d^2 - A*e*(2*c*d - b*e))/ 
(d^2*(c*d - b*e)^2*(d + e*x)) + (A*Log[x])/(b*d^3) + (c^2*(b*B - A*c)*Log[ 
b + c*x])/(b*(c*d - b*e)^3) - ((B*c^2*d^3 - A*e*(3*c^2*d^2 - 3*b*c*d*e + b 
^2*e^2))*Log[d + e*x])/(d^3*(c*d - b*e)^3)
 

Defintions of rubi rules used

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 = 1.06 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.00

method	result	size
default	\(\frac {\left (A c -B b \right ) c^{2} \ln \left (c x +b \right )}{b \left (b e -c d \right )^{3}}+\frac {A b \,e^{2}-2 A c d e +B c \,d^{2}}{d^{2} \left (b e -c d \right )^{2} \left (e x +d \right )}-\frac {\left (A \,b^{2} e^{3}-3 A b c d \,e^{2}+3 A \,c^{2} d^{2} e -B \,c^{2} d^{3}\right ) \ln \left (e x +d \right )}{d^{3} \left (b e -c d \right )^{3}}+\frac {A e -B d}{2 d \left (b e -c d \right ) \left (e x +d \right )^{2}}+\frac {A \ln \left (x \right )}{b \,d^{3}}\)	\(171\)
norman	\(\frac {-\frac {e^{2} \left (3 A b \,e^{2}-5 A c d e -B b d e +3 B c \,d^{2}\right ) x^{2}}{2 d^{3} \left (b^{2} e^{2}-2 b c d e +c^{2} d^{2}\right )}-\frac {\left (2 A b \,e^{2}-3 A c d e -B b d e +2 B c \,d^{2}\right ) e x}{d^{2} \left (b^{2} e^{2}-2 b c d e +c^{2} d^{2}\right )}}{\left (e x +d \right )^{2}}+\frac {A \ln \left (x \right )}{b \,d^{3}}+\frac {c^{2} \left (A c -B b \right ) \ln \left (c x +b \right )}{\left (b^{3} e^{3}-3 d \,e^{2} b^{2} c +3 d^{2} e b \,c^{2}-d^{3} c^{3}\right ) b}-\frac {\left (A \,b^{2} e^{3}-3 A b c d \,e^{2}+3 A \,c^{2} d^{2} e -B \,c^{2} d^{3}\right ) \ln \left (e x +d \right )}{d^{3} \left (b^{3} e^{3}-3 d \,e^{2} b^{2} c +3 d^{2} e b \,c^{2}-d^{3} c^{3}\right )}\)	\(284\)
risch	\(\frac {\frac {e \left (A b \,e^{2}-2 A c d e +B c \,d^{2}\right ) x}{d^{2} \left (b^{2} e^{2}-2 b c d e +c^{2} d^{2}\right )}+\frac {3 A b \,e^{2}-5 A c d e -B b d e +3 B c \,d^{2}}{2 d \left (b^{2} e^{2}-2 b c d e +c^{2} d^{2}\right )}}{\left (e x +d \right )^{2}}+\frac {A \ln \left (-x \right )}{d^{3} b}-\frac {\ln \left (-e x -d \right ) A \,b^{2} e^{3}}{d^{3} \left (b^{3} e^{3}-3 d \,e^{2} b^{2} c +3 d^{2} e b \,c^{2}-d^{3} c^{3}\right )}+\frac {3 \ln \left (-e x -d \right ) A b c \,e^{2}}{d^{2} \left (b^{3} e^{3}-3 d \,e^{2} b^{2} c +3 d^{2} e b \,c^{2}-d^{3} c^{3}\right )}-\frac {3 \ln \left (-e x -d \right ) A \,c^{2} e}{d \left (b^{3} e^{3}-3 d \,e^{2} b^{2} c +3 d^{2} e b \,c^{2}-d^{3} c^{3}\right )}+\frac {\ln \left (-e x -d \right ) B \,c^{2}}{b^{3} e^{3}-3 d \,e^{2} b^{2} c +3 d^{2} e b \,c^{2}-d^{3} c^{3}}+\frac {c^{3} \ln \left (c x +b \right ) A}{\left (b^{3} e^{3}-3 d \,e^{2} b^{2} c +3 d^{2} e b \,c^{2}-d^{3} c^{3}\right ) b}-\frac {c^{2} \ln \left (c x +b \right ) B}{b^{3} e^{3}-3 d \,e^{2} b^{2} c +3 d^{2} e b \,c^{2}-d^{3} c^{3}}\)	\(454\)
parallelrisch	\(\frac {2 B \,b^{3} d^{2} e^{3} x -3 A \,b^{3} e^{5} x^{2}-4 A d \,b^{3} e^{4} x -4 B \,b^{2} c \,d^{2} e^{3} x^{2}+3 B b \,c^{2} d^{3} e^{2} x^{2}-5 A b \,c^{2} d^{2} e^{3} x^{2}+12 A \ln \left (x \right ) x b \,c^{2} d^{3} e^{2}+12 A \ln \left (e x +d \right ) x \,b^{2} c \,d^{2} e^{3}-12 A \ln \left (e x +d \right ) x b \,c^{2} d^{3} e^{2}-4 B \ln \left (c x +b \right ) x b \,c^{2} d^{4} e +4 B \ln \left (e x +d \right ) x b \,c^{2} d^{4} e -6 A \ln \left (x \right ) x^{2} b^{2} c d \,e^{4}+6 A \ln \left (x \right ) x^{2} b \,c^{2} d^{2} e^{3}-6 A b \,c^{2} d^{3} e^{2} x -6 B \,b^{2} c \,d^{3} e^{2} x +4 B b \,c^{2} d^{4} e x -12 A \ln \left (x \right ) x \,b^{2} c \,d^{2} e^{3}+6 A \ln \left (e x +d \right ) x^{2} b^{2} c d \,e^{4}-6 A \ln \left (e x +d \right ) x^{2} b \,c^{2} d^{2} e^{3}-2 B \ln \left (c x +b \right ) x^{2} b \,c^{2} d^{3} e^{2}+2 B \ln \left (e x +d \right ) x^{2} b \,c^{2} d^{3} e^{2}+B \,b^{3} d \,e^{4} x^{2}-2 A \ln \left (x \right ) c^{3} d^{5}+2 A \ln \left (c x +b \right ) c^{3} d^{5}+8 A \,b^{2} c d \,e^{4} x^{2}-2 B \ln \left (c x +b \right ) b \,c^{2} d^{5}+2 B \ln \left (e x +d \right ) b \,c^{2} d^{5}+2 A \ln \left (x \right ) x^{2} b^{3} e^{5}-2 A \ln \left (e x +d \right ) x^{2} b^{3} e^{5}+2 A \ln \left (x \right ) b^{3} d^{2} e^{3}-2 A \ln \left (e x +d \right ) b^{3} d^{2} e^{3}+10 A \,b^{2} c \,d^{2} e^{3} x -2 A \ln \left (x \right ) x^{2} c^{3} d^{3} e^{2}+2 A \ln \left (c x +b \right ) x^{2} c^{3} d^{3} e^{2}+4 A \ln \left (x \right ) x \,b^{3} d \,e^{4}-4 A \ln \left (x \right ) x \,c^{3} d^{4} e +4 A \ln \left (c x +b \right ) x \,c^{3} d^{4} e -4 A \ln \left (e x +d \right ) x \,b^{3} d \,e^{4}-6 A \ln \left (x \right ) b^{2} c \,d^{3} e^{2}+6 A \ln \left (x \right ) b \,c^{2} d^{4} e +6 A \ln \left (e x +d \right ) b^{2} c \,d^{3} e^{2}-6 A \ln \left (e x +d \right ) b \,c^{2} d^{4} e}{2 \left (b^{3} e^{3}-3 d \,e^{2} b^{2} c +3 d^{2} e b \,c^{2}-d^{3} c^{3}\right ) b \left (e x +d \right )^{2} d^{3}}\)	\(731\)

Input:

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

Output:

(A*c-B*b)*c^2/b/(b*e-c*d)^3*ln(c*x+b)+(A*b*e^2-2*A*c*d*e+B*c*d^2)/d^2/(b*e 
-c*d)^2/(e*x+d)-(A*b^2*e^3-3*A*b*c*d*e^2+3*A*c^2*d^2*e-B*c^2*d^3)/d^3/(b*e 
-c*d)^3*ln(e*x+d)+1/2*(A*e-B*d)/d/(b*e-c*d)/(e*x+d)^2+A*ln(x)/b/d^3
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 644 vs. \(2 (169) = 338\).

Time = 19.74 (sec) , antiderivative size = 644, normalized size of antiderivative = 3.77 \[ \int \frac {A+B x}{(d+e x)^3 \left (b x+c x^2\right )} \, dx=\frac {3 \, B b c^{2} d^{5} - 3 \, A b^{3} d^{2} e^{3} - {\left (4 \, B b^{2} c + 5 \, A b c^{2}\right )} d^{4} e + {\left (B b^{3} + 8 \, A b^{2} c\right )} d^{3} e^{2} + 2 \, {\left (B b c^{2} d^{4} e + 3 \, A b^{2} c d^{2} e^{3} - A b^{3} d e^{4} - {\left (B b^{2} c + 2 \, A b c^{2}\right )} d^{3} e^{2}\right )} x + 2 \, {\left ({\left (B b c^{2} - A c^{3}\right )} d^{3} e^{2} x^{2} + 2 \, {\left (B b c^{2} - A c^{3}\right )} d^{4} e x + {\left (B b c^{2} - A c^{3}\right )} d^{5}\right )} \log \left (c x + b\right ) - 2 \, {\left (B b c^{2} d^{5} - 3 \, A b c^{2} d^{4} e + 3 \, A b^{2} c d^{3} e^{2} - A b^{3} d^{2} e^{3} + {\left (B b c^{2} d^{3} e^{2} - 3 \, A b c^{2} d^{2} e^{3} + 3 \, A b^{2} c d e^{4} - A b^{3} e^{5}\right )} x^{2} + 2 \, {\left (B b c^{2} d^{4} e - 3 \, A b c^{2} d^{3} e^{2} + 3 \, A b^{2} c d^{2} e^{3} - A b^{3} d e^{4}\right )} x\right )} \log \left (e x + d\right ) + 2 \, {\left (A c^{3} d^{5} - 3 \, A b c^{2} d^{4} e + 3 \, A b^{2} c d^{3} e^{2} - A b^{3} d^{2} e^{3} + {\left (A c^{3} d^{3} e^{2} - 3 \, A b c^{2} d^{2} e^{3} + 3 \, A b^{2} c d e^{4} - A b^{3} e^{5}\right )} x^{2} + 2 \, {\left (A c^{3} d^{4} e - 3 \, A b c^{2} d^{3} e^{2} + 3 \, A b^{2} c d^{2} e^{3} - A b^{3} d e^{4}\right )} x\right )} \log \left (x\right )}{2 \, {\left (b c^{3} d^{8} - 3 \, b^{2} c^{2} d^{7} e + 3 \, b^{3} c d^{6} e^{2} - b^{4} d^{5} e^{3} + {\left (b c^{3} d^{6} e^{2} - 3 \, b^{2} c^{2} d^{5} e^{3} + 3 \, b^{3} c d^{4} e^{4} - b^{4} d^{3} e^{5}\right )} x^{2} + 2 \, {\left (b c^{3} d^{7} e - 3 \, b^{2} c^{2} d^{6} e^{2} + 3 \, b^{3} c d^{5} e^{3} - b^{4} d^{4} e^{4}\right )} x\right )}} \] Input:

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

Output:

1/2*(3*B*b*c^2*d^5 - 3*A*b^3*d^2*e^3 - (4*B*b^2*c + 5*A*b*c^2)*d^4*e + (B* 
b^3 + 8*A*b^2*c)*d^3*e^2 + 2*(B*b*c^2*d^4*e + 3*A*b^2*c*d^2*e^3 - A*b^3*d* 
e^4 - (B*b^2*c + 2*A*b*c^2)*d^3*e^2)*x + 2*((B*b*c^2 - A*c^3)*d^3*e^2*x^2 
+ 2*(B*b*c^2 - A*c^3)*d^4*e*x + (B*b*c^2 - A*c^3)*d^5)*log(c*x + b) - 2*(B 
*b*c^2*d^5 - 3*A*b*c^2*d^4*e + 3*A*b^2*c*d^3*e^2 - A*b^3*d^2*e^3 + (B*b*c^ 
2*d^3*e^2 - 3*A*b*c^2*d^2*e^3 + 3*A*b^2*c*d*e^4 - A*b^3*e^5)*x^2 + 2*(B*b* 
c^2*d^4*e - 3*A*b*c^2*d^3*e^2 + 3*A*b^2*c*d^2*e^3 - A*b^3*d*e^4)*x)*log(e* 
x + d) + 2*(A*c^3*d^5 - 3*A*b*c^2*d^4*e + 3*A*b^2*c*d^3*e^2 - A*b^3*d^2*e^ 
3 + (A*c^3*d^3*e^2 - 3*A*b*c^2*d^2*e^3 + 3*A*b^2*c*d*e^4 - A*b^3*e^5)*x^2 
+ 2*(A*c^3*d^4*e - 3*A*b*c^2*d^3*e^2 + 3*A*b^2*c*d^2*e^3 - A*b^3*d*e^4)*x) 
*log(x))/(b*c^3*d^8 - 3*b^2*c^2*d^7*e + 3*b^3*c*d^6*e^2 - b^4*d^5*e^3 + (b 
*c^3*d^6*e^2 - 3*b^2*c^2*d^5*e^3 + 3*b^3*c*d^4*e^4 - b^4*d^3*e^5)*x^2 + 2* 
(b*c^3*d^7*e - 3*b^2*c^2*d^6*e^2 + 3*b^3*c*d^5*e^3 - b^4*d^4*e^4)*x)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 312, normalized size of antiderivative = 1.82 \[ \int \frac {A+B x}{(d+e x)^3 \left (b x+c x^2\right )} \, dx=\frac {{\left (B b c^{2} - A c^{3}\right )} \log \left (c x + b\right )}{b c^{3} d^{3} - 3 \, b^{2} c^{2} d^{2} e + 3 \, b^{3} c d e^{2} - b^{4} e^{3}} - \frac {{\left (B c^{2} d^{3} - 3 \, A c^{2} d^{2} e + 3 \, A b c d e^{2} - A b^{2} e^{3}\right )} \log \left (e x + d\right )}{c^{3} d^{6} - 3 \, b c^{2} d^{5} e + 3 \, b^{2} c d^{4} e^{2} - b^{3} d^{3} e^{3}} + \frac {3 \, B c d^{3} + 3 \, A b d e^{2} - {\left (B b + 5 \, A c\right )} d^{2} e + 2 \, {\left (B c d^{2} e - 2 \, A c d e^{2} + A b e^{3}\right )} x}{2 \, {\left (c^{2} d^{6} - 2 \, b c d^{5} e + b^{2} d^{4} e^{2} + {\left (c^{2} d^{4} e^{2} - 2 \, b c d^{3} e^{3} + b^{2} d^{2} e^{4}\right )} x^{2} + 2 \, {\left (c^{2} d^{5} e - 2 \, b c d^{4} e^{2} + b^{2} d^{3} e^{3}\right )} x\right )}} + \frac {A \log \left (x\right )}{b d^{3}} \] Input:

integrate((B*x+A)/(e*x+d)^3/(c*x^2+b*x),x, algorithm="maxima")
 

Output:

(B*b*c^2 - A*c^3)*log(c*x + b)/(b*c^3*d^3 - 3*b^2*c^2*d^2*e + 3*b^3*c*d*e^ 
2 - b^4*e^3) - (B*c^2*d^3 - 3*A*c^2*d^2*e + 3*A*b*c*d*e^2 - A*b^2*e^3)*log 
(e*x + d)/(c^3*d^6 - 3*b*c^2*d^5*e + 3*b^2*c*d^4*e^2 - b^3*d^3*e^3) + 1/2* 
(3*B*c*d^3 + 3*A*b*d*e^2 - (B*b + 5*A*c)*d^2*e + 2*(B*c*d^2*e - 2*A*c*d*e^ 
2 + A*b*e^3)*x)/(c^2*d^6 - 2*b*c*d^5*e + b^2*d^4*e^2 + (c^2*d^4*e^2 - 2*b* 
c*d^3*e^3 + b^2*d^2*e^4)*x^2 + 2*(c^2*d^5*e - 2*b*c*d^4*e^2 + b^2*d^3*e^3) 
*x) + A*log(x)/(b*d^3)
 

Giac [A] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 314, normalized size of antiderivative = 1.84 \[ \int \frac {A+B x}{(d+e x)^3 \left (b x+c x^2\right )} \, dx=\frac {{\left (B b c^{3} - A c^{4}\right )} \log \left ({\left | c x + b \right |}\right )}{b c^{4} d^{3} - 3 \, b^{2} c^{3} d^{2} e + 3 \, b^{3} c^{2} d e^{2} - b^{4} c e^{3}} - \frac {{\left (B c^{2} d^{3} e - 3 \, A c^{2} d^{2} e^{2} + 3 \, A b c d e^{3} - A b^{2} e^{4}\right )} \log \left ({\left | e x + d \right |}\right )}{c^{3} d^{6} e - 3 \, b c^{2} d^{5} e^{2} + 3 \, b^{2} c d^{4} e^{3} - b^{3} d^{3} e^{4}} + \frac {A \log \left ({\left | x \right |}\right )}{b d^{3}} + \frac {3 \, B c^{2} d^{5} - 4 \, B b c d^{4} e - 5 \, A c^{2} d^{4} e + B b^{2} d^{3} e^{2} + 8 \, A b c d^{3} e^{2} - 3 \, A b^{2} d^{2} e^{3} + 2 \, {\left (B c^{2} d^{4} e - B b c d^{3} e^{2} - 2 \, A c^{2} d^{3} e^{2} + 3 \, A b c d^{2} e^{3} - A b^{2} d e^{4}\right )} x}{2 \, {\left (c d - b e\right )}^{3} {\left (e x + d\right )}^{2} d^{3}} \] Input:

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

Output:

(B*b*c^3 - A*c^4)*log(abs(c*x + b))/(b*c^4*d^3 - 3*b^2*c^3*d^2*e + 3*b^3*c 
^2*d*e^2 - b^4*c*e^3) - (B*c^2*d^3*e - 3*A*c^2*d^2*e^2 + 3*A*b*c*d*e^3 - A 
*b^2*e^4)*log(abs(e*x + d))/(c^3*d^6*e - 3*b*c^2*d^5*e^2 + 3*b^2*c*d^4*e^3 
 - b^3*d^3*e^4) + A*log(abs(x))/(b*d^3) + 1/2*(3*B*c^2*d^5 - 4*B*b*c*d^4*e 
 - 5*A*c^2*d^4*e + B*b^2*d^3*e^2 + 8*A*b*c*d^3*e^2 - 3*A*b^2*d^2*e^3 + 2*( 
B*c^2*d^4*e - B*b*c*d^3*e^2 - 2*A*c^2*d^3*e^2 + 3*A*b*c*d^2*e^3 - A*b^2*d* 
e^4)*x)/((c*d - b*e)^3*(e*x + d)^2*d^3)
 

Mupad [B] (verification not implemented)

Time = 11.76 (sec) , antiderivative size = 284, normalized size of antiderivative = 1.66 \[ \int \frac {A+B x}{(d+e x)^3 \left (b x+c x^2\right )} \, dx=\frac {\frac {3\,A\,b\,e^2+3\,B\,c\,d^2-5\,A\,c\,d\,e-B\,b\,d\,e}{2\,d\,\left (b^2\,e^2-2\,b\,c\,d\,e+c^2\,d^2\right )}+\frac {x\,\left (B\,c\,d^2\,e-2\,A\,c\,d\,e^2+A\,b\,e^3\right )}{d^2\,\left (b^2\,e^2-2\,b\,c\,d\,e+c^2\,d^2\right )}}{d^2+2\,d\,e\,x+e^2\,x^2}-\frac {\ln \left (d+e\,x\right )\,\left (c^2\,\left (B\,d^3-3\,A\,d^2\,e\right )-A\,b^2\,e^3+3\,A\,b\,c\,d\,e^2\right )}{-b^3\,d^3\,e^3+3\,b^2\,c\,d^4\,e^2-3\,b\,c^2\,d^5\,e+c^3\,d^6}+\frac {\ln \left (b+c\,x\right )\,\left (A\,c^3-B\,b\,c^2\right )}{b^4\,e^3-3\,b^3\,c\,d\,e^2+3\,b^2\,c^2\,d^2\,e-b\,c^3\,d^3}+\frac {A\,\ln \left (x\right )}{b\,d^3} \] Input:

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

Output:

((3*A*b*e^2 + 3*B*c*d^2 - 5*A*c*d*e - B*b*d*e)/(2*d*(b^2*e^2 + c^2*d^2 - 2 
*b*c*d*e)) + (x*(A*b*e^3 - 2*A*c*d*e^2 + B*c*d^2*e))/(d^2*(b^2*e^2 + c^2*d 
^2 - 2*b*c*d*e)))/(d^2 + e^2*x^2 + 2*d*e*x) - (log(d + e*x)*(c^2*(B*d^3 - 
3*A*d^2*e) - A*b^2*e^3 + 3*A*b*c*d*e^2))/(c^3*d^6 - b^3*d^3*e^3 + 3*b^2*c* 
d^4*e^2 - 3*b*c^2*d^5*e) + (log(b + c*x)*(A*c^3 - B*b*c^2))/(b^4*e^3 - b*c 
^3*d^3 + 3*b^2*c^2*d^2*e - 3*b^3*c*d*e^2) + (A*log(x))/(b*d^3)
 

Reduce [B] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 809, normalized size of antiderivative = 4.73 \[ \int \frac {A+B x}{(d+e x)^3 \left (b x+c x^2\right )} \, dx =\text {Too large to display} \] Input:

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

Output:

(2*log(b + c*x)*a*c**3*d**5 + 4*log(b + c*x)*a*c**3*d**4*e*x + 2*log(b + c 
*x)*a*c**3*d**3*e**2*x**2 - 2*log(b + c*x)*b**2*c**2*d**5 - 4*log(b + c*x) 
*b**2*c**2*d**4*e*x - 2*log(b + c*x)*b**2*c**2*d**3*e**2*x**2 - 2*log(d + 
e*x)*a*b**3*d**2*e**3 - 4*log(d + e*x)*a*b**3*d*e**4*x - 2*log(d + e*x)*a* 
b**3*e**5*x**2 + 6*log(d + e*x)*a*b**2*c*d**3*e**2 + 12*log(d + e*x)*a*b** 
2*c*d**2*e**3*x + 6*log(d + e*x)*a*b**2*c*d*e**4*x**2 - 6*log(d + e*x)*a*b 
*c**2*d**4*e - 12*log(d + e*x)*a*b*c**2*d**3*e**2*x - 6*log(d + e*x)*a*b*c 
**2*d**2*e**3*x**2 + 2*log(d + e*x)*b**2*c**2*d**5 + 4*log(d + e*x)*b**2*c 
**2*d**4*e*x + 2*log(d + e*x)*b**2*c**2*d**3*e**2*x**2 + 2*log(x)*a*b**3*d 
**2*e**3 + 4*log(x)*a*b**3*d*e**4*x + 2*log(x)*a*b**3*e**5*x**2 - 6*log(x) 
*a*b**2*c*d**3*e**2 - 12*log(x)*a*b**2*c*d**2*e**3*x - 6*log(x)*a*b**2*c*d 
*e**4*x**2 + 6*log(x)*a*b*c**2*d**4*e + 12*log(x)*a*b*c**2*d**3*e**2*x + 6 
*log(x)*a*b*c**2*d**2*e**3*x**2 - 2*log(x)*a*c**3*d**5 - 4*log(x)*a*c**3*d 
**4*e*x - 2*log(x)*a*c**3*d**3*e**2*x**2 + 2*a*b**3*d**2*e**3 - a*b**3*e** 
5*x**2 - 5*a*b**2*c*d**3*e**2 + 3*a*b**2*c*d*e**4*x**2 + 3*a*b*c**2*d**4*e 
 - 2*a*b*c**2*d**2*e**3*x**2 - b**4*d**3*e**2 + 3*b**3*c*d**4*e - b**3*c*d 
**2*e**3*x**2 - 2*b**2*c**2*d**5 + b**2*c**2*d**3*e**2*x**2)/(2*b*d**3*(b* 
*3*d**2*e**3 + 2*b**3*d*e**4*x + b**3*e**5*x**2 - 3*b**2*c*d**3*e**2 - 6*b 
**2*c*d**2*e**3*x - 3*b**2*c*d*e**4*x**2 + 3*b*c**2*d**4*e + 6*b*c**2*d**3 
*e**2*x + 3*b*c**2*d**2*e**3*x**2 - c**3*d**5 - 2*c**3*d**4*e*x - c**3*...