\(\int \frac {(a+b x)^2}{x^5} \, dx\) [60]

Optimal result

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

[Out]

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

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} \[ \int \frac {(a+b x)^2}{x^5} \, dx=-\frac {a^2}{4 x^4}-\frac {2 a b}{3 x^3}-\frac {b^2}{2 x^2} \]

[In]

Int[(a + b*x)^2/x^5,x]

[Out]

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

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

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00 \[ \int \frac {(a+b x)^2}{x^5} \, dx=-\frac {a^2}{4 x^4}-\frac {2 a b}{3 x^3}-\frac {b^2}{2 x^2} \]

[In]

Integrate[(a + b*x)^2/x^5,x]

[Out]

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

Maple [A] (verified)

Time = 0.15 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.80

method	result	size
norman	\(\frac {-\frac {1}{2} b^{2} x^{2}-\frac {2}{3} a b x -\frac {1}{4} a^{2}}{x^{4}}\)	\(24\)
risch	\(\frac {-\frac {1}{2} b^{2} x^{2}-\frac {2}{3} a b x -\frac {1}{4} a^{2}}{x^{4}}\)	\(24\)
gosper	\(-\frac {6 b^{2} x^{2}+8 a b x +3 a^{2}}{12 x^{4}}\)	\(25\)
default	\(-\frac {a^{2}}{4 x^{4}}-\frac {2 a b}{3 x^{3}}-\frac {b^{2}}{2 x^{2}}\)	\(25\)
parallelrisch	\(\frac {-6 b^{2} x^{2}-8 a b x -3 a^{2}}{12 x^{4}}\)	\(25\)

[In]

int((b*x+a)^2/x^5,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.80 \[ \int \frac {(a+b x)^2}{x^5} \, dx=-\frac {6 \, b^{2} x^{2} + 8 \, a b x + 3 \, a^{2}}{12 \, x^{4}} \]

[In]

integrate((b*x+a)^2/x^5,x, algorithm="fricas")

[Out]

-1/12*(6*b^2*x^2 + 8*a*b*x + 3*a^2)/x^4

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.87 \[ \int \frac {(a+b x)^2}{x^5} \, dx=\frac {- 3 a^{2} - 8 a b x - 6 b^{2} x^{2}}{12 x^{4}} \]

[In]

integrate((b*x+a)**2/x**5,x)

[Out]

(-3*a**2 - 8*a*b*x - 6*b**2*x**2)/(12*x**4)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.80 \[ \int \frac {(a+b x)^2}{x^5} \, dx=-\frac {6 \, b^{2} x^{2} + 8 \, a b x + 3 \, a^{2}}{12 \, x^{4}} \]

[In]

integrate((b*x+a)^2/x^5,x, algorithm="maxima")

[Out]

-1/12*(6*b^2*x^2 + 8*a*b*x + 3*a^2)/x^4

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.80 \[ \int \frac {(a+b x)^2}{x^5} \, dx=-\frac {6 \, b^{2} x^{2} + 8 \, a b x + 3 \, a^{2}}{12 \, x^{4}} \]

[In]

integrate((b*x+a)^2/x^5,x, algorithm="giac")

[Out]

-1/12*(6*b^2*x^2 + 8*a*b*x + 3*a^2)/x^4

Mupad [B] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.80 \[ \int \frac {(a+b x)^2}{x^5} \, dx=-\frac {\frac {a^2}{4}+\frac {2\,a\,b\,x}{3}+\frac {b^2\,x^2}{2}}{x^4} \]

[In]

int((a + b*x)^2/x^5,x)

[Out]

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