\(\int \frac {(a+b x)^3 (A+B x)}{(d+e x)^9} \, dx\) [1049]

Optimal result

Integrand size = 20, antiderivative size = 163 \[ \int \frac {(a+b x)^3 (A+B x)}{(d+e x)^9} \, dx=-\frac {(b d-a e)^3 (B d-A e)}{8 e^5 (d+e x)^8}+\frac {(b d-a e)^2 (4 b B d-3 A b e-a B e)}{7 e^5 (d+e x)^7}-\frac {b (b d-a e) (2 b B d-A b e-a B e)}{2 e^5 (d+e x)^6}+\frac {b^2 (4 b B d-A b e-3 a B e)}{5 e^5 (d+e x)^5}-\frac {b^3 B}{4 e^5 (d+e x)^4} \]

[Out]

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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {78} \[ \int \frac {(a+b x)^3 (A+B x)}{(d+e x)^9} \, dx=\frac {b^2 (-3 a B e-A b e+4 b B d)}{5 e^5 (d+e x)^5}-\frac {b (b d-a e) (-a B e-A b e+2 b B d)}{2 e^5 (d+e x)^6}+\frac {(b d-a e)^2 (-a B e-3 A b e+4 b B d)}{7 e^5 (d+e x)^7}-\frac {(b d-a e)^3 (B d-A e)}{8 e^5 (d+e x)^8}-\frac {b^3 B}{4 e^5 (d+e x)^4} \]

[In]

Int[((a + b*x)^3*(A + B*x))/(d + e*x)^9,x]

[Out]

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

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {(-b d+a e)^3 (-B d+A e)}{e^4 (d+e x)^9}+\frac {(-b d+a e)^2 (-4 b B d+3 A b e+a B e)}{e^4 (d+e x)^8}-\frac {3 b (b d-a e) (-2 b B d+A b e+a B e)}{e^4 (d+e x)^7}+\frac {b^2 (-4 b B d+A b e+3 a B e)}{e^4 (d+e x)^6}+\frac {b^3 B}{e^4 (d+e x)^5}\right ) \, dx \\ & = -\frac {(b d-a e)^3 (B d-A e)}{8 e^5 (d+e x)^8}+\frac {(b d-a e)^2 (4 b B d-3 A b e-a B e)}{7 e^5 (d+e x)^7}-\frac {b (b d-a e) (2 b B d-A b e-a B e)}{2 e^5 (d+e x)^6}+\frac {b^2 (4 b B d-A b e-3 a B e)}{5 e^5 (d+e x)^5}-\frac {b^3 B}{4 e^5 (d+e x)^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.29 \[ \int \frac {(a+b x)^3 (A+B x)}{(d+e x)^9} \, dx=-\frac {5 a^3 e^3 (7 A e+B (d+8 e x))+5 a^2 b e^2 \left (3 A e (d+8 e x)+B \left (d^2+8 d e x+28 e^2 x^2\right )\right )+a b^2 e \left (5 A e \left (d^2+8 d e x+28 e^2 x^2\right )+3 B \left (d^3+8 d^2 e x+28 d e^2 x^2+56 e^3 x^3\right )\right )+b^3 \left (A e \left (d^3+8 d^2 e x+28 d e^2 x^2+56 e^3 x^3\right )+B \left (d^4+8 d^3 e x+28 d^2 e^2 x^2+56 d e^3 x^3+70 e^4 x^4\right )\right )}{280 e^5 (d+e x)^8} \]

[In]

Integrate[((a + b*x)^3*(A + B*x))/(d + e*x)^9,x]

[Out]

-1/280*(5*a^3*e^3*(7*A*e + B*(d + 8*e*x)) + 5*a^2*b*e^2*(3*A*e*(d + 8*e*x) + B*(d^2 + 8*d*e*x + 28*e^2*x^2)) +
a*b^2*e*(5*A*e*(d^2 + 8*d*e*x + 28*e^2*x^2) + 3*B*(d^3 + 8*d^2*e*x + 28*d*e^2*x^2 + 56*e^3*x^3)) + b^3*(A*e*(
d^3 + 8*d^2*e*x + 28*d*e^2*x^2 + 56*e^3*x^3) + B*(d^4 + 8*d^3*e*x + 28*d^2*e^2*x^2 + 56*d*e^3*x^3 + 70*e^4*x^4
)))/(e^5*(d + e*x)^8)

Maple [A] (verified)

Time = 0.70 (sec) , antiderivative size = 262, normalized size of antiderivative = 1.61

method	result	size
risch	\(\frac {-\frac {b^{3} B \,x^{4}}{4 e}-\frac {b^{2} \left (A b e +3 B a e +B b d \right ) x^{3}}{5 e^{2}}-\frac {b \left (5 A a b \,e^{2}+A \,b^{2} d e +5 B \,a^{2} e^{2}+3 B a b d e +b^{2} B \,d^{2}\right ) x^{2}}{10 e^{3}}-\frac {\left (15 A \,a^{2} b \,e^{3}+5 A a \,b^{2} d \,e^{2}+A \,b^{3} d^{2} e +5 B \,a^{3} e^{3}+5 B \,a^{2} b d \,e^{2}+3 B a \,b^{2} d^{2} e +b^{3} B \,d^{3}\right ) x}{35 e^{4}}-\frac {35 a^{3} A \,e^{4}+15 A \,a^{2} b d \,e^{3}+5 A a \,b^{2} d^{2} e^{2}+A \,b^{3} d^{3} e +5 B \,a^{3} d \,e^{3}+5 B \,a^{2} b \,d^{2} e^{2}+3 B a \,b^{2} d^{3} e +b^{3} B \,d^{4}}{280 e^{5}}}{\left (e x +d \right )^{8}}\)	\(262\)
default	\(-\frac {a^{3} A \,e^{4}-3 A \,a^{2} b d \,e^{3}+3 A a \,b^{2} d^{2} e^{2}-A \,b^{3} d^{3} e -B \,a^{3} d \,e^{3}+3 B \,a^{2} b \,d^{2} e^{2}-3 B a \,b^{2} d^{3} e +b^{3} B \,d^{4}}{8 e^{5} \left (e x +d \right )^{8}}-\frac {3 A \,a^{2} b \,e^{3}-6 A a \,b^{2} d \,e^{2}+3 A \,b^{3} d^{2} e +B \,a^{3} e^{3}-6 B \,a^{2} b d \,e^{2}+9 B a \,b^{2} d^{2} e -4 b^{3} B \,d^{3}}{7 e^{5} \left (e x +d \right )^{7}}-\frac {b^{2} \left (A b e +3 B a e -4 B b d \right )}{5 e^{5} \left (e x +d \right )^{5}}-\frac {b \left (A a b \,e^{2}-A \,b^{2} d e +B \,a^{2} e^{2}-3 B a b d e +2 b^{2} B \,d^{2}\right )}{2 e^{5} \left (e x +d \right )^{6}}-\frac {b^{3} B}{4 e^{5} \left (e x +d \right )^{4}}\)	\(281\)
norman	\(\frac {-\frac {b^{3} B \,x^{4}}{4 e}-\frac {\left (A \,b^{3} e^{4}+3 B a \,b^{2} e^{4}+b^{3} B d \,e^{3}\right ) x^{3}}{5 e^{5}}-\frac {\left (5 A a \,b^{2} e^{5}+A \,b^{3} d \,e^{4}+5 B \,a^{2} b \,e^{5}+3 B a \,b^{2} d \,e^{4}+b^{3} B \,d^{2} e^{3}\right ) x^{2}}{10 e^{6}}-\frac {\left (15 a^{2} b A \,e^{6}+5 A a \,b^{2} d \,e^{5}+A \,b^{3} d^{2} e^{4}+5 a^{3} B \,e^{6}+5 B \,a^{2} b d \,e^{5}+3 B a \,b^{2} d^{2} e^{4}+B \,b^{3} d^{3} e^{3}\right ) x}{35 e^{7}}-\frac {35 a^{3} A \,e^{7}+15 A \,a^{2} b d \,e^{6}+5 A a \,b^{2} d^{2} e^{5}+A \,b^{3} d^{3} e^{4}+5 B \,a^{3} d \,e^{6}+5 B \,a^{2} b \,d^{2} e^{5}+3 B a \,b^{2} d^{3} e^{4}+B \,b^{3} d^{4} e^{3}}{280 e^{8}}}{\left (e x +d \right )^{8}}\)	\(298\)
gosper	\(-\frac {70 B \,x^{4} b^{3} e^{4}+56 A \,x^{3} b^{3} e^{4}+168 B \,x^{3} a \,b^{2} e^{4}+56 B \,x^{3} b^{3} d \,e^{3}+140 A \,x^{2} a \,b^{2} e^{4}+28 A \,x^{2} b^{3} d \,e^{3}+140 B \,x^{2} a^{2} b \,e^{4}+84 B \,x^{2} a \,b^{2} d \,e^{3}+28 B \,x^{2} b^{3} d^{2} e^{2}+120 A x \,a^{2} b \,e^{4}+40 A x a \,b^{2} d \,e^{3}+8 A x \,b^{3} d^{2} e^{2}+40 B x \,a^{3} e^{4}+40 B x \,a^{2} b d \,e^{3}+24 B x a \,b^{2} d^{2} e^{2}+8 B x \,b^{3} d^{3} e +35 a^{3} A \,e^{4}+15 A \,a^{2} b d \,e^{3}+5 A a \,b^{2} d^{2} e^{2}+A \,b^{3} d^{3} e +5 B \,a^{3} d \,e^{3}+5 B \,a^{2} b \,d^{2} e^{2}+3 B a \,b^{2} d^{3} e +b^{3} B \,d^{4}}{280 e^{5} \left (e x +d \right )^{8}}\)	\(299\)
parallelrisch	\(-\frac {70 b^{3} B \,x^{4} e^{7}+56 A \,b^{3} e^{7} x^{3}+168 B a \,b^{2} e^{7} x^{3}+56 B \,b^{3} d \,e^{6} x^{3}+140 A a \,b^{2} e^{7} x^{2}+28 A \,b^{3} d \,e^{6} x^{2}+140 B \,a^{2} b \,e^{7} x^{2}+84 B a \,b^{2} d \,e^{6} x^{2}+28 B \,b^{3} d^{2} e^{5} x^{2}+120 A \,a^{2} b \,e^{7} x +40 A a \,b^{2} d \,e^{6} x +8 A \,b^{3} d^{2} e^{5} x +40 B \,a^{3} e^{7} x +40 B \,a^{2} b d \,e^{6} x +24 B a \,b^{2} d^{2} e^{5} x +8 B \,b^{3} d^{3} e^{4} x +35 a^{3} A \,e^{7}+15 A \,a^{2} b d \,e^{6}+5 A a \,b^{2} d^{2} e^{5}+A \,b^{3} d^{3} e^{4}+5 B \,a^{3} d \,e^{6}+5 B \,a^{2} b \,d^{2} e^{5}+3 B a \,b^{2} d^{3} e^{4}+B \,b^{3} d^{4} e^{3}}{280 e^{8} \left (e x +d \right )^{8}}\)	\(308\)

[In]

int((b*x+a)^3*(B*x+A)/(e*x+d)^9,x,method=_RETURNVERBOSE)

[Out]

(-1/4*b^3*B/e*x^4-1/5*b^2/e^2*(A*b*e+3*B*a*e+B*b*d)*x^3-1/10*b/e^3*(5*A*a*b*e^2+A*b^2*d*e+5*B*a^2*e^2+3*B*a*b*
d*e+B*b^2*d^2)*x^2-1/35/e^4*(15*A*a^2*b*e^3+5*A*a*b^2*d*e^2+A*b^3*d^2*e+5*B*a^3*e^3+5*B*a^2*b*d*e^2+3*B*a*b^2*
d^2*e+B*b^3*d^3)*x-1/280/e^5*(35*A*a^3*e^4+15*A*a^2*b*d*e^3+5*A*a*b^2*d^2*e^2+A*b^3*d^3*e+5*B*a^3*d*e^3+5*B*a^
2*b*d^2*e^2+3*B*a*b^2*d^3*e+B*b^3*d^4))/(e*x+d)^8

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 335 vs. \(2 (153) = 306\).

Time = 0.23 (sec) , antiderivative size = 335, normalized size of antiderivative = 2.06 \[ \int \frac {(a+b x)^3 (A+B x)}{(d+e x)^9} \, dx=-\frac {70 \, B b^{3} e^{4} x^{4} + B b^{3} d^{4} + 35 \, A a^{3} e^{4} + {\left (3 \, B a b^{2} + A b^{3}\right )} d^{3} e + 5 \, {\left (B a^{2} b + A a b^{2}\right )} d^{2} e^{2} + 5 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} d e^{3} + 56 \, {\left (B b^{3} d e^{3} + {\left (3 \, B a b^{2} + A b^{3}\right )} e^{4}\right )} x^{3} + 28 \, {\left (B b^{3} d^{2} e^{2} + {\left (3 \, B a b^{2} + A b^{3}\right )} d e^{3} + 5 \, {\left (B a^{2} b + A a b^{2}\right )} e^{4}\right )} x^{2} + 8 \, {\left (B b^{3} d^{3} e + {\left (3 \, B a b^{2} + A b^{3}\right )} d^{2} e^{2} + 5 \, {\left (B a^{2} b + A a b^{2}\right )} d e^{3} + 5 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} e^{4}\right )} x}{280 \, {\left (e^{13} x^{8} + 8 \, d e^{12} x^{7} + 28 \, d^{2} e^{11} x^{6} + 56 \, d^{3} e^{10} x^{5} + 70 \, d^{4} e^{9} x^{4} + 56 \, d^{5} e^{8} x^{3} + 28 \, d^{6} e^{7} x^{2} + 8 \, d^{7} e^{6} x + d^{8} e^{5}\right )}} \]

[In]

integrate((b*x+a)^3*(B*x+A)/(e*x+d)^9,x, algorithm="fricas")

[Out]

-1/280*(70*B*b^3*e^4*x^4 + B*b^3*d^4 + 35*A*a^3*e^4 + (3*B*a*b^2 + A*b^3)*d^3*e + 5*(B*a^2*b + A*a*b^2)*d^2*e^
2 + 5*(B*a^3 + 3*A*a^2*b)*d*e^3 + 56*(B*b^3*d*e^3 + (3*B*a*b^2 + A*b^3)*e^4)*x^3 + 28*(B*b^3*d^2*e^2 + (3*B*a*
b^2 + A*b^3)*d*e^3 + 5*(B*a^2*b + A*a*b^2)*e^4)*x^2 + 8*(B*b^3*d^3*e + (3*B*a*b^2 + A*b^3)*d^2*e^2 + 5*(B*a^2*
b + A*a*b^2)*d*e^3 + 5*(B*a^3 + 3*A*a^2*b)*e^4)*x)/(e^13*x^8 + 8*d*e^12*x^7 + 28*d^2*e^11*x^6 + 56*d^3*e^10*x^
5 + 70*d^4*e^9*x^4 + 56*d^5*e^8*x^3 + 28*d^6*e^7*x^2 + 8*d^7*e^6*x + d^8*e^5)

Sympy [F(-1)]

Timed out. \[ \int \frac {(a+b x)^3 (A+B x)}{(d+e x)^9} \, dx=\text {Timed out} \]

[In]

integrate((b*x+a)**3*(B*x+A)/(e*x+d)**9,x)

[Out]

Timed out

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 335 vs. \(2 (153) = 306\).

Time = 0.22 (sec) , antiderivative size = 335, normalized size of antiderivative = 2.06 \[ \int \frac {(a+b x)^3 (A+B x)}{(d+e x)^9} \, dx=-\frac {70 \, B b^{3} e^{4} x^{4} + B b^{3} d^{4} + 35 \, A a^{3} e^{4} + {\left (3 \, B a b^{2} + A b^{3}\right )} d^{3} e + 5 \, {\left (B a^{2} b + A a b^{2}\right )} d^{2} e^{2} + 5 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} d e^{3} + 56 \, {\left (B b^{3} d e^{3} + {\left (3 \, B a b^{2} + A b^{3}\right )} e^{4}\right )} x^{3} + 28 \, {\left (B b^{3} d^{2} e^{2} + {\left (3 \, B a b^{2} + A b^{3}\right )} d e^{3} + 5 \, {\left (B a^{2} b + A a b^{2}\right )} e^{4}\right )} x^{2} + 8 \, {\left (B b^{3} d^{3} e + {\left (3 \, B a b^{2} + A b^{3}\right )} d^{2} e^{2} + 5 \, {\left (B a^{2} b + A a b^{2}\right )} d e^{3} + 5 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} e^{4}\right )} x}{280 \, {\left (e^{13} x^{8} + 8 \, d e^{12} x^{7} + 28 \, d^{2} e^{11} x^{6} + 56 \, d^{3} e^{10} x^{5} + 70 \, d^{4} e^{9} x^{4} + 56 \, d^{5} e^{8} x^{3} + 28 \, d^{6} e^{7} x^{2} + 8 \, d^{7} e^{6} x + d^{8} e^{5}\right )}} \]

[In]

integrate((b*x+a)^3*(B*x+A)/(e*x+d)^9,x, algorithm="maxima")

[Out]

-1/280*(70*B*b^3*e^4*x^4 + B*b^3*d^4 + 35*A*a^3*e^4 + (3*B*a*b^2 + A*b^3)*d^3*e + 5*(B*a^2*b + A*a*b^2)*d^2*e^
2 + 5*(B*a^3 + 3*A*a^2*b)*d*e^3 + 56*(B*b^3*d*e^3 + (3*B*a*b^2 + A*b^3)*e^4)*x^3 + 28*(B*b^3*d^2*e^2 + (3*B*a*
b^2 + A*b^3)*d*e^3 + 5*(B*a^2*b + A*a*b^2)*e^4)*x^2 + 8*(B*b^3*d^3*e + (3*B*a*b^2 + A*b^3)*d^2*e^2 + 5*(B*a^2*
b + A*a*b^2)*d*e^3 + 5*(B*a^3 + 3*A*a^2*b)*e^4)*x)/(e^13*x^8 + 8*d*e^12*x^7 + 28*d^2*e^11*x^6 + 56*d^3*e^10*x^
5 + 70*d^4*e^9*x^4 + 56*d^5*e^8*x^3 + 28*d^6*e^7*x^2 + 8*d^7*e^6*x + d^8*e^5)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 298, normalized size of antiderivative = 1.83 \[ \int \frac {(a+b x)^3 (A+B x)}{(d+e x)^9} \, dx=-\frac {70 \, B b^{3} e^{4} x^{4} + 56 \, B b^{3} d e^{3} x^{3} + 168 \, B a b^{2} e^{4} x^{3} + 56 \, A b^{3} e^{4} x^{3} + 28 \, B b^{3} d^{2} e^{2} x^{2} + 84 \, B a b^{2} d e^{3} x^{2} + 28 \, A b^{3} d e^{3} x^{2} + 140 \, B a^{2} b e^{4} x^{2} + 140 \, A a b^{2} e^{4} x^{2} + 8 \, B b^{3} d^{3} e x + 24 \, B a b^{2} d^{2} e^{2} x + 8 \, A b^{3} d^{2} e^{2} x + 40 \, B a^{2} b d e^{3} x + 40 \, A a b^{2} d e^{3} x + 40 \, B a^{3} e^{4} x + 120 \, A a^{2} b e^{4} x + B b^{3} d^{4} + 3 \, B a b^{2} d^{3} e + A b^{3} d^{3} e + 5 \, B a^{2} b d^{2} e^{2} + 5 \, A a b^{2} d^{2} e^{2} + 5 \, B a^{3} d e^{3} + 15 \, A a^{2} b d e^{3} + 35 \, A a^{3} e^{4}}{280 \, {\left (e x + d\right )}^{8} e^{5}} \]

[In]

integrate((b*x+a)^3*(B*x+A)/(e*x+d)^9,x, algorithm="giac")

[Out]

-1/280*(70*B*b^3*e^4*x^4 + 56*B*b^3*d*e^3*x^3 + 168*B*a*b^2*e^4*x^3 + 56*A*b^3*e^4*x^3 + 28*B*b^3*d^2*e^2*x^2
+ 84*B*a*b^2*d*e^3*x^2 + 28*A*b^3*d*e^3*x^2 + 140*B*a^2*b*e^4*x^2 + 140*A*a*b^2*e^4*x^2 + 8*B*b^3*d^3*e*x + 24
*B*a*b^2*d^2*e^2*x + 8*A*b^3*d^2*e^2*x + 40*B*a^2*b*d*e^3*x + 40*A*a*b^2*d*e^3*x + 40*B*a^3*e^4*x + 120*A*a^2*
b*e^4*x + B*b^3*d^4 + 3*B*a*b^2*d^3*e + A*b^3*d^3*e + 5*B*a^2*b*d^2*e^2 + 5*A*a*b^2*d^2*e^2 + 5*B*a^3*d*e^3 +
15*A*a^2*b*d*e^3 + 35*A*a^3*e^4)/((e*x + d)^8*e^5)

Mupad [B] (verification not implemented)

Time = 1.46 (sec) , antiderivative size = 339, normalized size of antiderivative = 2.08 \[ \int \frac {(a+b x)^3 (A+B x)}{(d+e x)^9} \, dx=-\frac {\frac {5\,B\,a^3\,d\,e^3+35\,A\,a^3\,e^4+5\,B\,a^2\,b\,d^2\,e^2+15\,A\,a^2\,b\,d\,e^3+3\,B\,a\,b^2\,d^3\,e+5\,A\,a\,b^2\,d^2\,e^2+B\,b^3\,d^4+A\,b^3\,d^3\,e}{280\,e^5}+\frac {x\,\left (5\,B\,a^3\,e^3+5\,B\,a^2\,b\,d\,e^2+15\,A\,a^2\,b\,e^3+3\,B\,a\,b^2\,d^2\,e+5\,A\,a\,b^2\,d\,e^2+B\,b^3\,d^3+A\,b^3\,d^2\,e\right )}{35\,e^4}+\frac {b^2\,x^3\,\left (A\,b\,e+3\,B\,a\,e+B\,b\,d\right )}{5\,e^2}+\frac {b\,x^2\,\left (5\,B\,a^2\,e^2+3\,B\,a\,b\,d\,e+5\,A\,a\,b\,e^2+B\,b^2\,d^2+A\,b^2\,d\,e\right )}{10\,e^3}+\frac {B\,b^3\,x^4}{4\,e}}{d^8+8\,d^7\,e\,x+28\,d^6\,e^2\,x^2+56\,d^5\,e^3\,x^3+70\,d^4\,e^4\,x^4+56\,d^3\,e^5\,x^5+28\,d^2\,e^6\,x^6+8\,d\,e^7\,x^7+e^8\,x^8} \]

[In]

int(((A + B*x)*(a + b*x)^3)/(d + e*x)^9,x)

[Out]

-((35*A*a^3*e^4 + B*b^3*d^4 + A*b^3*d^3*e + 5*B*a^3*d*e^3 + 5*A*a*b^2*d^2*e^2 + 5*B*a^2*b*d^2*e^2 + 15*A*a^2*b
*d*e^3 + 3*B*a*b^2*d^3*e)/(280*e^5) + (x*(5*B*a^3*e^3 + B*b^3*d^3 + 15*A*a^2*b*e^3 + A*b^3*d^2*e + 5*A*a*b^2*d
*e^2 + 3*B*a*b^2*d^2*e + 5*B*a^2*b*d*e^2))/(35*e^4) + (b^2*x^3*(A*b*e + 3*B*a*e + B*b*d))/(5*e^2) + (b*x^2*(5*
B*a^2*e^2 + B*b^2*d^2 + 5*A*a*b*e^2 + A*b^2*d*e + 3*B*a*b*d*e))/(10*e^3) + (B*b^3*x^4)/(4*e))/(d^8 + e^8*x^8 +
8*d*e^7*x^7 + 28*d^6*e^2*x^2 + 56*d^5*e^3*x^3 + 70*d^4*e^4*x^4 + 56*d^3*e^5*x^5 + 28*d^2*e^6*x^6 + 8*d^7*e*x)