\(\int \frac {(a+b x)^3}{x^8} \, dx\) [76]

Optimal result

Integrand size = 11, antiderivative size = 43 \[ \int \frac {(a+b x)^3}{x^8} \, dx=-\frac {a^3}{7 x^7}-\frac {a^2 b}{2 x^6}-\frac {3 a b^2}{5 x^5}-\frac {b^3}{4 x^4} \]

[Out]

-1/7*a^3/x^7-1/2*a^2*b/x^6-3/5*a*b^2/x^5-1/4*b^3/x^4

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {45} \[ \int \frac {(a+b x)^3}{x^8} \, dx=-\frac {a^3}{7 x^7}-\frac {a^2 b}{2 x^6}-\frac {3 a b^2}{5 x^5}-\frac {b^3}{4 x^4} \]

[In]

Int[(a + b*x)^3/x^8,x]

[Out]

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

Rule 45

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, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

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

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.00 \[ \int \frac {(a+b x)^3}{x^8} \, dx=-\frac {a^3}{7 x^7}-\frac {a^2 b}{2 x^6}-\frac {3 a b^2}{5 x^5}-\frac {b^3}{4 x^4} \]

[In]

Integrate[(a + b*x)^3/x^8,x]

[Out]

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

Maple [A] (verified)

Time = 0.17 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.81

method	result	size
norman	\(\frac {-\frac {1}{4} b^{3} x^{3}-\frac {3}{5} a \,b^{2} x^{2}-\frac {1}{2} a^{2} b x -\frac {1}{7} a^{3}}{x^{7}}\)	\(35\)
risch	\(\frac {-\frac {1}{4} b^{3} x^{3}-\frac {3}{5} a \,b^{2} x^{2}-\frac {1}{2} a^{2} b x -\frac {1}{7} a^{3}}{x^{7}}\)	\(35\)
gosper	\(-\frac {35 b^{3} x^{3}+84 a \,b^{2} x^{2}+70 a^{2} b x +20 a^{3}}{140 x^{7}}\)	\(36\)
default	\(-\frac {a^{3}}{7 x^{7}}-\frac {a^{2} b}{2 x^{6}}-\frac {3 a \,b^{2}}{5 x^{5}}-\frac {b^{3}}{4 x^{4}}\)	\(36\)
parallelrisch	\(\frac {-35 b^{3} x^{3}-84 a \,b^{2} x^{2}-70 a^{2} b x -20 a^{3}}{140 x^{7}}\)	\(36\)

[In]

int((b*x+a)^3/x^8,x,method=_RETURNVERBOSE)

[Out]

1/x^7*(-1/4*b^3*x^3-3/5*a*b^2*x^2-1/2*a^2*b*x-1/7*a^3)

Fricas [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.81 \[ \int \frac {(a+b x)^3}{x^8} \, dx=-\frac {35 \, b^{3} x^{3} + 84 \, a b^{2} x^{2} + 70 \, a^{2} b x + 20 \, a^{3}}{140 \, x^{7}} \]

[In]

integrate((b*x+a)^3/x^8,x, algorithm="fricas")

[Out]

-1/140*(35*b^3*x^3 + 84*a*b^2*x^2 + 70*a^2*b*x + 20*a^3)/x^7

Sympy [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.86 \[ \int \frac {(a+b x)^3}{x^8} \, dx=\frac {- 20 a^{3} - 70 a^{2} b x - 84 a b^{2} x^{2} - 35 b^{3} x^{3}}{140 x^{7}} \]

[In]

integrate((b*x+a)**3/x**8,x)

[Out]

(-20*a**3 - 70*a**2*b*x - 84*a*b**2*x**2 - 35*b**3*x**3)/(140*x**7)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.81 \[ \int \frac {(a+b x)^3}{x^8} \, dx=-\frac {35 \, b^{3} x^{3} + 84 \, a b^{2} x^{2} + 70 \, a^{2} b x + 20 \, a^{3}}{140 \, x^{7}} \]

[In]

integrate((b*x+a)^3/x^8,x, algorithm="maxima")

[Out]

-1/140*(35*b^3*x^3 + 84*a*b^2*x^2 + 70*a^2*b*x + 20*a^3)/x^7

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.81 \[ \int \frac {(a+b x)^3}{x^8} \, dx=-\frac {35 \, b^{3} x^{3} + 84 \, a b^{2} x^{2} + 70 \, a^{2} b x + 20 \, a^{3}}{140 \, x^{7}} \]

[In]

integrate((b*x+a)^3/x^8,x, algorithm="giac")

[Out]

-1/140*(35*b^3*x^3 + 84*a*b^2*x^2 + 70*a^2*b*x + 20*a^3)/x^7

Mupad [B] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.81 \[ \int \frac {(a+b x)^3}{x^8} \, dx=-\frac {\frac {a^3}{7}+\frac {a^2\,b\,x}{2}+\frac {3\,a\,b^2\,x^2}{5}+\frac {b^3\,x^3}{4}}{x^7} \]

[In]

int((a + b*x)^3/x^8,x)

[Out]

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