\(\int \frac {(a+b x)^2}{x^7} \, dx\) [62]

Optimal result

Integrand size = 11, antiderivative size = 30 \[ \int \frac {(a+b x)^2}{x^7} \, dx=-\frac {a^2}{6 x^6}-\frac {2 a b}{5 x^5}-\frac {b^2}{4 x^4} \]

[Out]

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

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} \[ \int \frac {(a+b x)^2}{x^7} \, dx=-\frac {a^2}{6 x^6}-\frac {2 a b}{5 x^5}-\frac {b^2}{4 x^4} \]

[In]

Int[(a + b*x)^2/x^7,x]

[Out]

-1/6*a^2/x^6 - (2*a*b)/(5*x^5) - b^2/(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^2}{x^7}+\frac {2 a b}{x^6}+\frac {b^2}{x^5}\right ) \, dx \\ & = -\frac {a^2}{6 x^6}-\frac {2 a b}{5 x^5}-\frac {b^2}{4 x^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00 \[ \int \frac {(a+b x)^2}{x^7} \, dx=-\frac {a^2}{6 x^6}-\frac {2 a b}{5 x^5}-\frac {b^2}{4 x^4} \]

[In]

Integrate[(a + b*x)^2/x^7,x]

[Out]

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

Maple [A] (verified)

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

method	result	size
norman	\(\frac {-\frac {1}{4} b^{2} x^{2}-\frac {2}{5} a b x -\frac {1}{6} a^{2}}{x^{6}}\)	\(24\)
risch	\(\frac {-\frac {1}{4} b^{2} x^{2}-\frac {2}{5} a b x -\frac {1}{6} a^{2}}{x^{6}}\)	\(24\)
gosper	\(-\frac {15 b^{2} x^{2}+24 a b x +10 a^{2}}{60 x^{6}}\)	\(25\)
default	\(-\frac {a^{2}}{6 x^{6}}-\frac {2 a b}{5 x^{5}}-\frac {b^{2}}{4 x^{4}}\)	\(25\)
parallelrisch	\(\frac {-15 b^{2} x^{2}-24 a b x -10 a^{2}}{60 x^{6}}\)	\(25\)

[In]

int((b*x+a)^2/x^7,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.80 \[ \int \frac {(a+b x)^2}{x^7} \, dx=-\frac {15 \, b^{2} x^{2} + 24 \, a b x + 10 \, a^{2}}{60 \, x^{6}} \]

[In]

integrate((b*x+a)^2/x^7,x, algorithm="fricas")

[Out]

-1/60*(15*b^2*x^2 + 24*a*b*x + 10*a^2)/x^6

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.87 \[ \int \frac {(a+b x)^2}{x^7} \, dx=\frac {- 10 a^{2} - 24 a b x - 15 b^{2} x^{2}}{60 x^{6}} \]

[In]

integrate((b*x+a)**2/x**7,x)

[Out]

(-10*a**2 - 24*a*b*x - 15*b**2*x**2)/(60*x**6)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.80 \[ \int \frac {(a+b x)^2}{x^7} \, dx=-\frac {15 \, b^{2} x^{2} + 24 \, a b x + 10 \, a^{2}}{60 \, x^{6}} \]

[In]

integrate((b*x+a)^2/x^7,x, algorithm="maxima")

[Out]

-1/60*(15*b^2*x^2 + 24*a*b*x + 10*a^2)/x^6

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.80 \[ \int \frac {(a+b x)^2}{x^7} \, dx=-\frac {15 \, b^{2} x^{2} + 24 \, a b x + 10 \, a^{2}}{60 \, x^{6}} \]

[In]

integrate((b*x+a)^2/x^7,x, algorithm="giac")

[Out]

-1/60*(15*b^2*x^2 + 24*a*b*x + 10*a^2)/x^6

Mupad [B] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.80 \[ \int \frac {(a+b x)^2}{x^7} \, dx=-\frac {\frac {a^2}{6}+\frac {2\,a\,b\,x}{5}+\frac {b^2\,x^2}{4}}{x^6} \]

[In]

int((a + b*x)^2/x^7,x)

[Out]

-(a^2/6 + (b^2*x^2)/4 + (2*a*b*x)/5)/x^6