\(\int \frac {(a+b x)^3 (A+B x)}{x^9} \, dx\) [117]

Optimal result

Integrand size = 16, antiderivative size = 75 \[ \int \frac {(a+b x)^3 (A+B x)}{x^9} \, dx=-\frac {a^3 A}{8 x^8}-\frac {a^2 (3 A b+a B)}{7 x^7}-\frac {a b (A b+a B)}{2 x^6}-\frac {b^2 (A b+3 a B)}{5 x^5}-\frac {b^3 B}{4 x^4} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {77} \[ \int \frac {(a+b x)^3 (A+B x)}{x^9} \, dx=-\frac {a^3 A}{8 x^8}-\frac {a^2 (a B+3 A b)}{7 x^7}-\frac {b^2 (3 a B+A b)}{5 x^5}-\frac {a b (a B+A b)}{2 x^6}-\frac {b^3 B}{4 x^4} \]

[In]

Int[((a + b*x)^3*(A + B*x))/x^9,x]

[Out]

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

Rule 77

Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*
x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && (NeQ[n, -1] || EqQ[p, 1]) && N
eQ[b*e + a*f, 0] && ( !IntegerQ[n] || LtQ[9*p + 5*n, 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && Rational
Q[a, b, d, e, f])) && (NeQ[n + p + 3, 0] || EqQ[p, 1])

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.92 \[ \int \frac {(a+b x)^3 (A+B x)}{x^9} \, dx=-\frac {14 b^3 x^3 (4 A+5 B x)+28 a b^2 x^2 (5 A+6 B x)+20 a^2 b x (6 A+7 B x)+5 a^3 (7 A+8 B x)}{280 x^8} \]

[In]

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

[Out]

-1/280*(14*b^3*x^3*(4*A + 5*B*x) + 28*a*b^2*x^2*(5*A + 6*B*x) + 20*a^2*b*x*(6*A + 7*B*x) + 5*a^3*(7*A + 8*B*x)
)/x^8

Maple [A] (verified)

Time = 0.39 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.88

method	result	size
default	\(-\frac {a^{3} A}{8 x^{8}}-\frac {a^{2} \left (3 A b +B a \right )}{7 x^{7}}-\frac {a b \left (A b +B a \right )}{2 x^{6}}-\frac {b^{2} \left (A b +3 B a \right )}{5 x^{5}}-\frac {b^{3} B}{4 x^{4}}\)	\(66\)
norman	\(\frac {-\frac {b^{3} B \,x^{4}}{4}+\left (-\frac {1}{5} b^{3} A -\frac {3}{5} a \,b^{2} B \right ) x^{3}+\left (-\frac {1}{2} a \,b^{2} A -\frac {1}{2} a^{2} b B \right ) x^{2}+\left (-\frac {3}{7} a^{2} b A -\frac {1}{7} a^{3} B \right ) x -\frac {a^{3} A}{8}}{x^{8}}\)	\(74\)
risch	\(\frac {-\frac {b^{3} B \,x^{4}}{4}+\left (-\frac {1}{5} b^{3} A -\frac {3}{5} a \,b^{2} B \right ) x^{3}+\left (-\frac {1}{2} a \,b^{2} A -\frac {1}{2} a^{2} b B \right ) x^{2}+\left (-\frac {3}{7} a^{2} b A -\frac {1}{7} a^{3} B \right ) x -\frac {a^{3} A}{8}}{x^{8}}\)	\(74\)
gosper	\(-\frac {70 b^{3} B \,x^{4}+56 A \,b^{3} x^{3}+168 B a \,b^{2} x^{3}+140 a A \,b^{2} x^{2}+140 B \,a^{2} b \,x^{2}+120 a^{2} A b x +40 a^{3} B x +35 a^{3} A}{280 x^{8}}\)	\(76\)
parallelrisch	\(-\frac {70 b^{3} B \,x^{4}+56 A \,b^{3} x^{3}+168 B a \,b^{2} x^{3}+140 a A \,b^{2} x^{2}+140 B \,a^{2} b \,x^{2}+120 a^{2} A b x +40 a^{3} B x +35 a^{3} A}{280 x^{8}}\)	\(76\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.97 \[ \int \frac {(a+b x)^3 (A+B x)}{x^9} \, dx=-\frac {70 \, B b^{3} x^{4} + 35 \, A a^{3} + 56 \, {\left (3 \, B a b^{2} + A b^{3}\right )} x^{3} + 140 \, {\left (B a^{2} b + A a b^{2}\right )} x^{2} + 40 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} x}{280 \, x^{8}} \]

[In]

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

[Out]

-1/280*(70*B*b^3*x^4 + 35*A*a^3 + 56*(3*B*a*b^2 + A*b^3)*x^3 + 140*(B*a^2*b + A*a*b^2)*x^2 + 40*(B*a^3 + 3*A*a
^2*b)*x)/x^8

Sympy [A] (verification not implemented)

Time = 1.42 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.09 \[ \int \frac {(a+b x)^3 (A+B x)}{x^9} \, dx=\frac {- 35 A a^{3} - 70 B b^{3} x^{4} + x^{3} \left (- 56 A b^{3} - 168 B a b^{2}\right ) + x^{2} \left (- 140 A a b^{2} - 140 B a^{2} b\right ) + x \left (- 120 A a^{2} b - 40 B a^{3}\right )}{280 x^{8}} \]

[In]

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

[Out]

(-35*A*a**3 - 70*B*b**3*x**4 + x**3*(-56*A*b**3 - 168*B*a*b**2) + x**2*(-140*A*a*b**2 - 140*B*a**2*b) + x*(-12
0*A*a**2*b - 40*B*a**3))/(280*x**8)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.97 \[ \int \frac {(a+b x)^3 (A+B x)}{x^9} \, dx=-\frac {70 \, B b^{3} x^{4} + 35 \, A a^{3} + 56 \, {\left (3 \, B a b^{2} + A b^{3}\right )} x^{3} + 140 \, {\left (B a^{2} b + A a b^{2}\right )} x^{2} + 40 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} x}{280 \, x^{8}} \]

[In]

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

[Out]

-1/280*(70*B*b^3*x^4 + 35*A*a^3 + 56*(3*B*a*b^2 + A*b^3)*x^3 + 140*(B*a^2*b + A*a*b^2)*x^2 + 40*(B*a^3 + 3*A*a
^2*b)*x)/x^8

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.00 \[ \int \frac {(a+b x)^3 (A+B x)}{x^9} \, dx=-\frac {70 \, B b^{3} x^{4} + 168 \, B a b^{2} x^{3} + 56 \, A b^{3} x^{3} + 140 \, B a^{2} b x^{2} + 140 \, A a b^{2} x^{2} + 40 \, B a^{3} x + 120 \, A a^{2} b x + 35 \, A a^{3}}{280 \, x^{8}} \]

[In]

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

[Out]

-1/280*(70*B*b^3*x^4 + 168*B*a*b^2*x^3 + 56*A*b^3*x^3 + 140*B*a^2*b*x^2 + 140*A*a*b^2*x^2 + 40*B*a^3*x + 120*A
*a^2*b*x + 35*A*a^3)/x^8

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.99 \[ \int \frac {(a+b x)^3 (A+B x)}{x^9} \, dx=-\frac {x^2\,\left (\frac {B\,a^2\,b}{2}+\frac {A\,a\,b^2}{2}\right )+x\,\left (\frac {B\,a^3}{7}+\frac {3\,A\,b\,a^2}{7}\right )+\frac {A\,a^3}{8}+x^3\,\left (\frac {A\,b^3}{5}+\frac {3\,B\,a\,b^2}{5}\right )+\frac {B\,b^3\,x^4}{4}}{x^8} \]

[In]

int(((A + B*x)*(a + b*x)^3)/x^9,x)

[Out]

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