\(\int \frac {(a+b x)^2}{x^6} \, dx\) [61]

Optimal result

Integrand size = 11, antiderivative size = 30 \[ \int \frac {(a+b x)^2}{x^6} \, dx=-\frac {a^2}{5 x^5}-\frac {a b}{2 x^4}-\frac {b^2}{3 x^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 30, 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} \[ \int \frac {(a+b x)^2}{x^6} \, dx=-\frac {a^2}{5 x^5}-\frac {a b}{2 x^4}-\frac {b^2}{3 x^3} \]

[In]

Int[(a + b*x)^2/x^6,x]

[Out]

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

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^2}{x^6}+\frac {2 a b}{x^5}+\frac {b^2}{x^4}\right ) \, dx \\ & = -\frac {a^2}{5 x^5}-\frac {a b}{2 x^4}-\frac {b^2}{3 x^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00 \[ \int \frac {(a+b x)^2}{x^6} \, dx=-\frac {a^2}{5 x^5}-\frac {a b}{2 x^4}-\frac {b^2}{3 x^3} \]

[In]

Integrate[(a + b*x)^2/x^6,x]

[Out]

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

Maple [A] (verified)

Time = 0.16 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.80

method	result	size
norman	\(\frac {-\frac {1}{3} b^{2} x^{2}-\frac {1}{2} a b x -\frac {1}{5} a^{2}}{x^{5}}\)	\(24\)
risch	\(\frac {-\frac {1}{3} b^{2} x^{2}-\frac {1}{2} a b x -\frac {1}{5} a^{2}}{x^{5}}\)	\(24\)
gosper	\(-\frac {10 b^{2} x^{2}+15 a b x +6 a^{2}}{30 x^{5}}\)	\(25\)
default	\(-\frac {a^{2}}{5 x^{5}}-\frac {a b}{2 x^{4}}-\frac {b^{2}}{3 x^{3}}\)	\(25\)
parallelrisch	\(\frac {-10 b^{2} x^{2}-15 a b x -6 a^{2}}{30 x^{5}}\)	\(25\)

[In]

int((b*x+a)^2/x^6,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.80 \[ \int \frac {(a+b x)^2}{x^6} \, dx=-\frac {10 \, b^{2} x^{2} + 15 \, a b x + 6 \, a^{2}}{30 \, x^{5}} \]

[In]

integrate((b*x+a)^2/x^6,x, algorithm="fricas")

[Out]

-1/30*(10*b^2*x^2 + 15*a*b*x + 6*a^2)/x^5

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.87 \[ \int \frac {(a+b x)^2}{x^6} \, dx=\frac {- 6 a^{2} - 15 a b x - 10 b^{2} x^{2}}{30 x^{5}} \]

[In]

integrate((b*x+a)**2/x**6,x)

[Out]

(-6*a**2 - 15*a*b*x - 10*b**2*x**2)/(30*x**5)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.80 \[ \int \frac {(a+b x)^2}{x^6} \, dx=-\frac {10 \, b^{2} x^{2} + 15 \, a b x + 6 \, a^{2}}{30 \, x^{5}} \]

[In]

integrate((b*x+a)^2/x^6,x, algorithm="maxima")

[Out]

-1/30*(10*b^2*x^2 + 15*a*b*x + 6*a^2)/x^5

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.80 \[ \int \frac {(a+b x)^2}{x^6} \, dx=-\frac {10 \, b^{2} x^{2} + 15 \, a b x + 6 \, a^{2}}{30 \, x^{5}} \]

[In]

integrate((b*x+a)^2/x^6,x, algorithm="giac")

[Out]

-1/30*(10*b^2*x^2 + 15*a*b*x + 6*a^2)/x^5

Mupad [B] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.80 \[ \int \frac {(a+b x)^2}{x^6} \, dx=-\frac {\frac {a^2}{5}+\frac {a\,b\,x}{2}+\frac {b^2\,x^2}{3}}{x^5} \]

[In]

int((a + b*x)^2/x^6,x)

[Out]

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