3.11.45 \(\int \frac {(a+b x)^3 (A+B x)}{(d+e x)^5} \, dx\) [1045]

3.11.45.1 Optimal result
3.11.45.2 Mathematica [A] (verified)
3.11.45.3 Rubi [A] (verified)
3.11.45.4 Maple [B] (verified)
3.11.45.5 Fricas [B] (verification not implemented)
3.11.45.6 Sympy [B] (verification not implemented)
3.11.45.7 Maxima [B] (verification not implemented)
3.11.45.8 Giac [B] (verification not implemented)
3.11.45.9 Mupad [B] (verification not implemented)

3.11.45.1 Optimal result

Integrand size = 20, antiderivative size = 129 \[ \int \frac {(a+b x)^3 (A+B x)}{(d+e x)^5} \, dx=-\frac {(B d-A e) (a+b x)^4}{4 e (b d-a e) (d+e x)^4}+\frac {B (b d-a e)^3}{3 e^5 (d+e x)^3}-\frac {3 b B (b d-a e)^2}{2 e^5 (d+e x)^2}+\frac {3 b^2 B (b d-a e)}{e^5 (d+e x)}+\frac {b^3 B \log (d+e x)}{e^5} \]

output
-1/4*(-A*e+B*d)*(b*x+a)^4/e/(-a*e+b*d)/(e*x+d)^4+1/3*B*(-a*e+b*d)^3/e^5/(e 
*x+d)^3-3/2*b*B*(-a*e+b*d)^2/e^5/(e*x+d)^2+3*b^2*B*(-a*e+b*d)/e^5/(e*x+d)+ 
b^3*B*ln(e*x+d)/e^5
 
3.11.45.2 Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 222, normalized size of antiderivative = 1.72 \[ \int \frac {(a+b x)^3 (A+B x)}{(d+e x)^5} \, dx=\frac {-a^3 e^3 (3 A e+B (d+4 e x))-3 a^2 b e^2 \left (A e (d+4 e x)+B \left (d^2+4 d e x+6 e^2 x^2\right )\right )-3 a b^2 e \left (A e \left (d^2+4 d e x+6 e^2 x^2\right )+3 B \left (d^3+4 d^2 e x+6 d e^2 x^2+4 e^3 x^3\right )\right )+b^3 \left (-3 A e \left (d^3+4 d^2 e x+6 d e^2 x^2+4 e^3 x^3\right )+B d \left (25 d^3+88 d^2 e x+108 d e^2 x^2+48 e^3 x^3\right )\right )+12 b^3 B (d+e x)^4 \log (d+e x)}{12 e^5 (d+e x)^4} \]

input
Integrate[((a + b*x)^3*(A + B*x))/(d + e*x)^5,x]
 
output
(-(a^3*e^3*(3*A*e + B*(d + 4*e*x))) - 3*a^2*b*e^2*(A*e*(d + 4*e*x) + B*(d^ 
2 + 4*d*e*x + 6*e^2*x^2)) - 3*a*b^2*e*(A*e*(d^2 + 4*d*e*x + 6*e^2*x^2) + 3 
*B*(d^3 + 4*d^2*e*x + 6*d*e^2*x^2 + 4*e^3*x^3)) + b^3*(-3*A*e*(d^3 + 4*d^2 
*e*x + 6*d*e^2*x^2 + 4*e^3*x^3) + B*d*(25*d^3 + 88*d^2*e*x + 108*d*e^2*x^2 
 + 48*e^3*x^3)) + 12*b^3*B*(d + e*x)^4*Log[d + e*x])/(12*e^5*(d + e*x)^4)
 
3.11.45.3 Rubi [A] (verified)

Time = 0.29 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.02, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {87, 49, 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.

\(\displaystyle \int \frac {(a+b x)^3 (A+B x)}{(d+e x)^5} \, dx\)

\(\Big \downarrow \) 87

\(\displaystyle \frac {B \int \frac {(a+b x)^3}{(d+e x)^4}dx}{e}-\frac {(a+b x)^4 (B d-A e)}{4 e (d+e x)^4 (b d-a e)}\)

\(\Big \downarrow \) 49

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {B \left (\frac {3 b^2 (b d-a e)}{e^4 (d+e x)}-\frac {3 b (b d-a e)^2}{2 e^4 (d+e x)^2}+\frac {(b d-a e)^3}{3 e^4 (d+e x)^3}+\frac {b^3 \log (d+e x)}{e^4}\right )}{e}-\frac {(a+b x)^4 (B d-A e)}{4 e (d+e x)^4 (b d-a e)}\)

input
Int[((a + b*x)^3*(A + B*x))/(d + e*x)^5,x]
 
output
-1/4*((B*d - A*e)*(a + b*x)^4)/(e*(b*d - a*e)*(d + e*x)^4) + (B*((b*d - a* 
e)^3/(3*e^4*(d + e*x)^3) - (3*b*(b*d - a*e)^2)/(2*e^4*(d + e*x)^2) + (3*b^ 
2*(b*d - a*e))/(e^4*(d + e*x)) + (b^3*Log[d + e*x])/e^4))/e
 

3.11.45.3.1 Defintions of rubi rules used

rule 49
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int 
[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] 
&& IGtQ[m, 0] && IGtQ[m + n + 2, 0]
 

rule 87
Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p 
_.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p 
+ 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p 
+ 1)))/(f*(p + 1)*(c*f - d*e))   Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] 
/; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege 
rQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ[p, n]))))
 

rule 2009
Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 
3.11.45.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(266\) vs. \(2(123)=246\).

Time = 0.69 (sec) , antiderivative size = 267, normalized size of antiderivative = 2.07

method result size
risch \(\frac {-\frac {b^{2} \left (A b e +3 B a e -4 B b d \right ) x^{3}}{e^{2}}-\frac {3 b \left (A a b \,e^{2}+A \,b^{2} d e +B \,a^{2} e^{2}+3 B a b d e -6 b^{2} B \,d^{2}\right ) x^{2}}{2 e^{3}}-\frac {\left (3 A \,a^{2} b \,e^{3}+3 A a \,b^{2} d \,e^{2}+3 A \,b^{3} d^{2} e +B \,a^{3} e^{3}+3 B \,a^{2} b d \,e^{2}+9 B a \,b^{2} d^{2} e -22 b^{3} B \,d^{3}\right ) x}{3 e^{4}}-\frac {3 a^{3} A \,e^{4}+3 A \,a^{2} b d \,e^{3}+3 A a \,b^{2} d^{2} e^{2}+3 A \,b^{3} d^{3} e +B \,a^{3} d \,e^{3}+3 B \,a^{2} b \,d^{2} e^{2}+9 B a \,b^{2} d^{3} e -25 b^{3} B \,d^{4}}{12 e^{5}}}{\left (e x +d \right )^{4}}+\frac {b^{3} B \ln \left (e x +d \right )}{e^{5}}\) \(267\)
norman \(\frac {-\frac {3 a^{3} A \,e^{4}+3 A \,a^{2} b d \,e^{3}+3 A a \,b^{2} d^{2} e^{2}+3 A \,b^{3} d^{3} e +B \,a^{3} d \,e^{3}+3 B \,a^{2} b \,d^{2} e^{2}+9 B a \,b^{2} d^{3} e -25 b^{3} B \,d^{4}}{12 e^{5}}-\frac {\left (A \,b^{3} e +3 B a \,b^{2} e -4 b^{3} B d \right ) x^{3}}{e^{2}}-\frac {3 \left (A a \,b^{2} e^{2}+A \,b^{3} d e +B \,a^{2} b \,e^{2}+3 B a \,b^{2} d e -6 b^{3} B \,d^{2}\right ) x^{2}}{2 e^{3}}-\frac {\left (3 A \,a^{2} b \,e^{3}+3 A a \,b^{2} d \,e^{2}+3 A \,b^{3} d^{2} e +B \,a^{3} e^{3}+3 B \,a^{2} b d \,e^{2}+9 B a \,b^{2} d^{2} e -22 b^{3} B \,d^{3}\right ) x}{3 e^{4}}}{\left (e x +d \right )^{4}}+\frac {b^{3} B \ln \left (e x +d \right )}{e^{5}}\) \(275\)
default \(-\frac {b^{2} \left (A b e +3 B a e -4 B b d \right )}{e^{5} \left (e x +d \right )}-\frac {3 A \,a^{2} b \,e^{3}-6 A a \,b^{2} d \,e^{2}+3 A \,b^{3} d^{2} e +B \,a^{3} e^{3}-6 B \,a^{2} b d \,e^{2}+9 B a \,b^{2} d^{2} e -4 b^{3} B \,d^{3}}{3 e^{5} \left (e x +d \right )^{3}}-\frac {3 b \left (A a b \,e^{2}-A \,b^{2} d e +B \,a^{2} e^{2}-3 B a b d e +2 b^{2} B \,d^{2}\right )}{2 e^{5} \left (e x +d \right )^{2}}+\frac {b^{3} B \ln \left (e x +d \right )}{e^{5}}-\frac {a^{3} A \,e^{4}-3 A \,a^{2} b d \,e^{3}+3 A a \,b^{2} d^{2} e^{2}-A \,b^{3} d^{3} e -B \,a^{3} d \,e^{3}+3 B \,a^{2} b \,d^{2} e^{2}-3 B a \,b^{2} d^{3} e +b^{3} B \,d^{4}}{4 e^{5} \left (e x +d \right )^{4}}\) \(279\)
parallelrisch \(-\frac {3 A a \,b^{2} d^{2} e^{2}-48 B \ln \left (e x +d \right ) x^{3} b^{3} d \,e^{3}+54 B \,x^{2} a \,b^{2} d \,e^{3}+12 A x a \,b^{2} d \,e^{3}+12 B x \,a^{2} b d \,e^{3}+36 B x a \,b^{2} d^{2} e^{2}-25 b^{3} B \,d^{4}+3 a^{3} A \,e^{4}+B \,a^{3} d \,e^{3}+3 A \,b^{3} d^{3} e +3 A \,a^{2} b d \,e^{3}+3 B \,a^{2} b \,d^{2} e^{2}+9 B a \,b^{2} d^{3} e +36 B \,x^{3} a \,b^{2} e^{4}-48 B \,x^{3} b^{3} d \,e^{3}+18 A \,x^{2} a \,b^{2} e^{4}+18 A \,x^{2} b^{3} d \,e^{3}+18 B \,x^{2} a^{2} b \,e^{4}-108 B \,x^{2} b^{3} d^{2} e^{2}+12 A x \,a^{2} b \,e^{4}+12 A x \,b^{3} d^{2} e^{2}-88 B x \,b^{3} d^{3} e +12 A \,x^{3} b^{3} e^{4}-12 B \ln \left (e x +d \right ) x^{4} b^{3} e^{4}+4 B x \,a^{3} e^{4}-12 B \ln \left (e x +d \right ) b^{3} d^{4}-72 B \ln \left (e x +d \right ) x^{2} b^{3} d^{2} e^{2}-48 B \ln \left (e x +d \right ) x \,b^{3} d^{3} e}{12 e^{5} \left (e x +d \right )^{4}}\) \(378\)

input
int((b*x+a)^3*(B*x+A)/(e*x+d)^5,x,method=_RETURNVERBOSE)
 
output
(-b^2*(A*b*e+3*B*a*e-4*B*b*d)/e^2*x^3-3/2*b*(A*a*b*e^2+A*b^2*d*e+B*a^2*e^2 
+3*B*a*b*d*e-6*B*b^2*d^2)/e^3*x^2-1/3*(3*A*a^2*b*e^3+3*A*a*b^2*d*e^2+3*A*b 
^3*d^2*e+B*a^3*e^3+3*B*a^2*b*d*e^2+9*B*a*b^2*d^2*e-22*B*b^3*d^3)/e^4*x-1/1 
2*(3*A*a^3*e^4+3*A*a^2*b*d*e^3+3*A*a*b^2*d^2*e^2+3*A*b^3*d^3*e+B*a^3*d*e^3 
+3*B*a^2*b*d^2*e^2+9*B*a*b^2*d^3*e-25*B*b^3*d^4)/e^5)/(e*x+d)^4+b^3*B*ln(e 
*x+d)/e^5
 
3.11.45.5 Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 354 vs. \(2 (123) = 246\).

Time = 0.23 (sec) , antiderivative size = 354, normalized size of antiderivative = 2.74 \[ \int \frac {(a+b x)^3 (A+B x)}{(d+e x)^5} \, dx=\frac {25 \, B b^{3} d^{4} - 3 \, A a^{3} e^{4} - 3 \, {\left (3 \, B a b^{2} + A b^{3}\right )} d^{3} e - 3 \, {\left (B a^{2} b + A a b^{2}\right )} d^{2} e^{2} - {\left (B a^{3} + 3 \, A a^{2} b\right )} d e^{3} + 12 \, {\left (4 \, B b^{3} d e^{3} - {\left (3 \, B a b^{2} + A b^{3}\right )} e^{4}\right )} x^{3} + 18 \, {\left (6 \, B b^{3} d^{2} e^{2} - {\left (3 \, B a b^{2} + A b^{3}\right )} d e^{3} - {\left (B a^{2} b + A a b^{2}\right )} e^{4}\right )} x^{2} + 4 \, {\left (22 \, B b^{3} d^{3} e - 3 \, {\left (3 \, B a b^{2} + A b^{3}\right )} d^{2} e^{2} - 3 \, {\left (B a^{2} b + A a b^{2}\right )} d e^{3} - {\left (B a^{3} + 3 \, A a^{2} b\right )} e^{4}\right )} x + 12 \, {\left (B b^{3} e^{4} x^{4} + 4 \, B b^{3} d e^{3} x^{3} + 6 \, B b^{3} d^{2} e^{2} x^{2} + 4 \, B b^{3} d^{3} e x + B b^{3} d^{4}\right )} \log \left (e x + d\right )}{12 \, {\left (e^{9} x^{4} + 4 \, d e^{8} x^{3} + 6 \, d^{2} e^{7} x^{2} + 4 \, d^{3} e^{6} x + d^{4} e^{5}\right )}} \]

input
integrate((b*x+a)^3*(B*x+A)/(e*x+d)^5,x, algorithm="fricas")
 
output
1/12*(25*B*b^3*d^4 - 3*A*a^3*e^4 - 3*(3*B*a*b^2 + A*b^3)*d^3*e - 3*(B*a^2* 
b + A*a*b^2)*d^2*e^2 - (B*a^3 + 3*A*a^2*b)*d*e^3 + 12*(4*B*b^3*d*e^3 - (3* 
B*a*b^2 + A*b^3)*e^4)*x^3 + 18*(6*B*b^3*d^2*e^2 - (3*B*a*b^2 + A*b^3)*d*e^ 
3 - (B*a^2*b + A*a*b^2)*e^4)*x^2 + 4*(22*B*b^3*d^3*e - 3*(3*B*a*b^2 + A*b^ 
3)*d^2*e^2 - 3*(B*a^2*b + A*a*b^2)*d*e^3 - (B*a^3 + 3*A*a^2*b)*e^4)*x + 12 
*(B*b^3*e^4*x^4 + 4*B*b^3*d*e^3*x^3 + 6*B*b^3*d^2*e^2*x^2 + 4*B*b^3*d^3*e* 
x + B*b^3*d^4)*log(e*x + d))/(e^9*x^4 + 4*d*e^8*x^3 + 6*d^2*e^7*x^2 + 4*d^ 
3*e^6*x + d^4*e^5)
 
3.11.45.6 Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 359 vs. \(2 (116) = 232\).

Time = 28.96 (sec) , antiderivative size = 359, normalized size of antiderivative = 2.78 \[ \int \frac {(a+b x)^3 (A+B x)}{(d+e x)^5} \, dx=\frac {B b^{3} \log {\left (d + e x \right )}}{e^{5}} + \frac {- 3 A a^{3} e^{4} - 3 A a^{2} b d e^{3} - 3 A a b^{2} d^{2} e^{2} - 3 A b^{3} d^{3} e - B a^{3} d e^{3} - 3 B a^{2} b d^{2} e^{2} - 9 B a b^{2} d^{3} e + 25 B b^{3} d^{4} + x^{3} \left (- 12 A b^{3} e^{4} - 36 B a b^{2} e^{4} + 48 B b^{3} d e^{3}\right ) + x^{2} \left (- 18 A a b^{2} e^{4} - 18 A b^{3} d e^{3} - 18 B a^{2} b e^{4} - 54 B a b^{2} d e^{3} + 108 B b^{3} d^{2} e^{2}\right ) + x \left (- 12 A a^{2} b e^{4} - 12 A a b^{2} d e^{3} - 12 A b^{3} d^{2} e^{2} - 4 B a^{3} e^{4} - 12 B a^{2} b d e^{3} - 36 B a b^{2} d^{2} e^{2} + 88 B b^{3} d^{3} e\right )}{12 d^{4} e^{5} + 48 d^{3} e^{6} x + 72 d^{2} e^{7} x^{2} + 48 d e^{8} x^{3} + 12 e^{9} x^{4}} \]

input
integrate((b*x+a)**3*(B*x+A)/(e*x+d)**5,x)
 
output
B*b**3*log(d + e*x)/e**5 + (-3*A*a**3*e**4 - 3*A*a**2*b*d*e**3 - 3*A*a*b** 
2*d**2*e**2 - 3*A*b**3*d**3*e - B*a**3*d*e**3 - 3*B*a**2*b*d**2*e**2 - 9*B 
*a*b**2*d**3*e + 25*B*b**3*d**4 + x**3*(-12*A*b**3*e**4 - 36*B*a*b**2*e**4 
 + 48*B*b**3*d*e**3) + x**2*(-18*A*a*b**2*e**4 - 18*A*b**3*d*e**3 - 18*B*a 
**2*b*e**4 - 54*B*a*b**2*d*e**3 + 108*B*b**3*d**2*e**2) + x*(-12*A*a**2*b* 
e**4 - 12*A*a*b**2*d*e**3 - 12*A*b**3*d**2*e**2 - 4*B*a**3*e**4 - 12*B*a** 
2*b*d*e**3 - 36*B*a*b**2*d**2*e**2 + 88*B*b**3*d**3*e))/(12*d**4*e**5 + 48 
*d**3*e**6*x + 72*d**2*e**7*x**2 + 48*d*e**8*x**3 + 12*e**9*x**4)
 
3.11.45.7 Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 302 vs. \(2 (123) = 246\).

Time = 0.23 (sec) , antiderivative size = 302, normalized size of antiderivative = 2.34 \[ \int \frac {(a+b x)^3 (A+B x)}{(d+e x)^5} \, dx=\frac {25 \, B b^{3} d^{4} - 3 \, A a^{3} e^{4} - 3 \, {\left (3 \, B a b^{2} + A b^{3}\right )} d^{3} e - 3 \, {\left (B a^{2} b + A a b^{2}\right )} d^{2} e^{2} - {\left (B a^{3} + 3 \, A a^{2} b\right )} d e^{3} + 12 \, {\left (4 \, B b^{3} d e^{3} - {\left (3 \, B a b^{2} + A b^{3}\right )} e^{4}\right )} x^{3} + 18 \, {\left (6 \, B b^{3} d^{2} e^{2} - {\left (3 \, B a b^{2} + A b^{3}\right )} d e^{3} - {\left (B a^{2} b + A a b^{2}\right )} e^{4}\right )} x^{2} + 4 \, {\left (22 \, B b^{3} d^{3} e - 3 \, {\left (3 \, B a b^{2} + A b^{3}\right )} d^{2} e^{2} - 3 \, {\left (B a^{2} b + A a b^{2}\right )} d e^{3} - {\left (B a^{3} + 3 \, A a^{2} b\right )} e^{4}\right )} x}{12 \, {\left (e^{9} x^{4} + 4 \, d e^{8} x^{3} + 6 \, d^{2} e^{7} x^{2} + 4 \, d^{3} e^{6} x + d^{4} e^{5}\right )}} + \frac {B b^{3} \log \left (e x + d\right )}{e^{5}} \]

input
integrate((b*x+a)^3*(B*x+A)/(e*x+d)^5,x, algorithm="maxima")
 
output
1/12*(25*B*b^3*d^4 - 3*A*a^3*e^4 - 3*(3*B*a*b^2 + A*b^3)*d^3*e - 3*(B*a^2* 
b + A*a*b^2)*d^2*e^2 - (B*a^3 + 3*A*a^2*b)*d*e^3 + 12*(4*B*b^3*d*e^3 - (3* 
B*a*b^2 + A*b^3)*e^4)*x^3 + 18*(6*B*b^3*d^2*e^2 - (3*B*a*b^2 + A*b^3)*d*e^ 
3 - (B*a^2*b + A*a*b^2)*e^4)*x^2 + 4*(22*B*b^3*d^3*e - 3*(3*B*a*b^2 + A*b^ 
3)*d^2*e^2 - 3*(B*a^2*b + A*a*b^2)*d*e^3 - (B*a^3 + 3*A*a^2*b)*e^4)*x)/(e^ 
9*x^4 + 4*d*e^8*x^3 + 6*d^2*e^7*x^2 + 4*d^3*e^6*x + d^4*e^5) + B*b^3*log(e 
*x + d)/e^5
 
3.11.45.8 Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 449 vs. \(2 (123) = 246\).

Time = 0.30 (sec) , antiderivative size = 449, normalized size of antiderivative = 3.48 \[ \int \frac {(a+b x)^3 (A+B x)}{(d+e x)^5} \, dx=-\frac {B b^{3} \log \left (\frac {{\left | e x + d \right |}}{{\left (e x + d\right )}^{2} {\left | e \right |}}\right )}{e^{5}} + \frac {\frac {48 \, B b^{3} d e^{15}}{e x + d} - \frac {36 \, B b^{3} d^{2} e^{15}}{{\left (e x + d\right )}^{2}} + \frac {16 \, B b^{3} d^{3} e^{15}}{{\left (e x + d\right )}^{3}} - \frac {3 \, B b^{3} d^{4} e^{15}}{{\left (e x + d\right )}^{4}} - \frac {36 \, B a b^{2} e^{16}}{e x + d} - \frac {12 \, A b^{3} e^{16}}{e x + d} + \frac {54 \, B a b^{2} d e^{16}}{{\left (e x + d\right )}^{2}} + \frac {18 \, A b^{3} d e^{16}}{{\left (e x + d\right )}^{2}} - \frac {36 \, B a b^{2} d^{2} e^{16}}{{\left (e x + d\right )}^{3}} - \frac {12 \, A b^{3} d^{2} e^{16}}{{\left (e x + d\right )}^{3}} + \frac {9 \, B a b^{2} d^{3} e^{16}}{{\left (e x + d\right )}^{4}} + \frac {3 \, A b^{3} d^{3} e^{16}}{{\left (e x + d\right )}^{4}} - \frac {18 \, B a^{2} b e^{17}}{{\left (e x + d\right )}^{2}} - \frac {18 \, A a b^{2} e^{17}}{{\left (e x + d\right )}^{2}} + \frac {24 \, B a^{2} b d e^{17}}{{\left (e x + d\right )}^{3}} + \frac {24 \, A a b^{2} d e^{17}}{{\left (e x + d\right )}^{3}} - \frac {9 \, B a^{2} b d^{2} e^{17}}{{\left (e x + d\right )}^{4}} - \frac {9 \, A a b^{2} d^{2} e^{17}}{{\left (e x + d\right )}^{4}} - \frac {4 \, B a^{3} e^{18}}{{\left (e x + d\right )}^{3}} - \frac {12 \, A a^{2} b e^{18}}{{\left (e x + d\right )}^{3}} + \frac {3 \, B a^{3} d e^{18}}{{\left (e x + d\right )}^{4}} + \frac {9 \, A a^{2} b d e^{18}}{{\left (e x + d\right )}^{4}} - \frac {3 \, A a^{3} e^{19}}{{\left (e x + d\right )}^{4}}}{12 \, e^{20}} \]

input
integrate((b*x+a)^3*(B*x+A)/(e*x+d)^5,x, algorithm="giac")
 
output
-B*b^3*log(abs(e*x + d)/((e*x + d)^2*abs(e)))/e^5 + 1/12*(48*B*b^3*d*e^15/ 
(e*x + d) - 36*B*b^3*d^2*e^15/(e*x + d)^2 + 16*B*b^3*d^3*e^15/(e*x + d)^3 
- 3*B*b^3*d^4*e^15/(e*x + d)^4 - 36*B*a*b^2*e^16/(e*x + d) - 12*A*b^3*e^16 
/(e*x + d) + 54*B*a*b^2*d*e^16/(e*x + d)^2 + 18*A*b^3*d*e^16/(e*x + d)^2 - 
 36*B*a*b^2*d^2*e^16/(e*x + d)^3 - 12*A*b^3*d^2*e^16/(e*x + d)^3 + 9*B*a*b 
^2*d^3*e^16/(e*x + d)^4 + 3*A*b^3*d^3*e^16/(e*x + d)^4 - 18*B*a^2*b*e^17/( 
e*x + d)^2 - 18*A*a*b^2*e^17/(e*x + d)^2 + 24*B*a^2*b*d*e^17/(e*x + d)^3 + 
 24*A*a*b^2*d*e^17/(e*x + d)^3 - 9*B*a^2*b*d^2*e^17/(e*x + d)^4 - 9*A*a*b^ 
2*d^2*e^17/(e*x + d)^4 - 4*B*a^3*e^18/(e*x + d)^3 - 12*A*a^2*b*e^18/(e*x + 
 d)^3 + 3*B*a^3*d*e^18/(e*x + d)^4 + 9*A*a^2*b*d*e^18/(e*x + d)^4 - 3*A*a^ 
3*e^19/(e*x + d)^4)/e^20
 
3.11.45.9 Mupad [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 303, normalized size of antiderivative = 2.35 \[ \int \frac {(a+b x)^3 (A+B x)}{(d+e x)^5} \, dx=\frac {B\,b^3\,\ln \left (d+e\,x\right )}{e^5}-\frac {\frac {B\,a^3\,d\,e^3+3\,A\,a^3\,e^4+3\,B\,a^2\,b\,d^2\,e^2+3\,A\,a^2\,b\,d\,e^3+9\,B\,a\,b^2\,d^3\,e+3\,A\,a\,b^2\,d^2\,e^2-25\,B\,b^3\,d^4+3\,A\,b^3\,d^3\,e}{12\,e^5}+\frac {x\,\left (B\,a^3\,e^3+3\,B\,a^2\,b\,d\,e^2+3\,A\,a^2\,b\,e^3+9\,B\,a\,b^2\,d^2\,e+3\,A\,a\,b^2\,d\,e^2-22\,B\,b^3\,d^3+3\,A\,b^3\,d^2\,e\right )}{3\,e^4}+\frac {3\,x^2\,\left (B\,a^2\,b\,e^2+3\,B\,a\,b^2\,d\,e+A\,a\,b^2\,e^2-6\,B\,b^3\,d^2+A\,b^3\,d\,e\right )}{2\,e^3}+\frac {b^2\,x^3\,\left (A\,b\,e+3\,B\,a\,e-4\,B\,b\,d\right )}{e^2}}{d^4+4\,d^3\,e\,x+6\,d^2\,e^2\,x^2+4\,d\,e^3\,x^3+e^4\,x^4} \]

input
int(((A + B*x)*(a + b*x)^3)/(d + e*x)^5,x)
 
output
(B*b^3*log(d + e*x))/e^5 - ((3*A*a^3*e^4 - 25*B*b^3*d^4 + 3*A*b^3*d^3*e + 
B*a^3*d*e^3 + 3*A*a*b^2*d^2*e^2 + 3*B*a^2*b*d^2*e^2 + 3*A*a^2*b*d*e^3 + 9* 
B*a*b^2*d^3*e)/(12*e^5) + (x*(B*a^3*e^3 - 22*B*b^3*d^3 + 3*A*a^2*b*e^3 + 3 
*A*b^3*d^2*e + 3*A*a*b^2*d*e^2 + 9*B*a*b^2*d^2*e + 3*B*a^2*b*d*e^2))/(3*e^ 
4) + (3*x^2*(A*b^3*d*e - 6*B*b^3*d^2 + A*a*b^2*e^2 + B*a^2*b*e^2 + 3*B*a*b 
^2*d*e))/(2*e^3) + (b^2*x^3*(A*b*e + 3*B*a*e - 4*B*b*d))/e^2)/(d^4 + e^4*x 
^4 + 4*d*e^3*x^3 + 6*d^2*e^2*x^2 + 4*d^3*e*x)