\(\int \frac {(a+b x)^2}{x^4} \, dx\) [59]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [A] (verification not implemented)
   Maxima [A] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 11, antiderivative size = 17 \[ \int \frac {(a+b x)^2}{x^4} \, dx=-\frac {(a+b x)^3}{3 a x^3} \]

[Out]

-1/3*(b*x+a)^3/a/x^3

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {37} \[ \int \frac {(a+b x)^2}{x^4} \, dx=-\frac {(a+b x)^3}{3 a x^3} \]

[In]

Int[(a + b*x)^2/x^4,x]

[Out]

-1/3*(a + b*x)^3/(a*x^3)

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n +
1)/((b*c - a*d)*(m + 1))), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rubi steps \begin{align*} \text {integral}& = -\frac {(a+b x)^3}{3 a x^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.53 \[ \int \frac {(a+b x)^2}{x^4} \, dx=-\frac {a^2}{3 x^3}-\frac {a b}{x^2}-\frac {b^2}{x} \]

[In]

Integrate[(a + b*x)^2/x^4,x]

[Out]

-1/3*a^2/x^3 - (a*b)/x^2 - b^2/x

Maple [A] (verified)

Time = 0.17 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.35

method result size
gosper \(-\frac {3 b^{2} x^{2}+3 a b x +a^{2}}{3 x^{3}}\) \(23\)
norman \(\frac {-b^{2} x^{2}-a b x -\frac {1}{3} a^{2}}{x^{3}}\) \(24\)
risch \(\frac {-b^{2} x^{2}-a b x -\frac {1}{3} a^{2}}{x^{3}}\) \(24\)
default \(-\frac {a^{2}}{3 x^{3}}-\frac {b^{2}}{x}-\frac {a b}{x^{2}}\) \(25\)
parallelrisch \(\frac {-3 b^{2} x^{2}-3 a b x -a^{2}}{3 x^{3}}\) \(25\)

[In]

int((b*x+a)^2/x^4,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.29 \[ \int \frac {(a+b x)^2}{x^4} \, dx=-\frac {3 \, b^{2} x^{2} + 3 \, a b x + a^{2}}{3 \, x^{3}} \]

[In]

integrate((b*x+a)^2/x^4,x, algorithm="fricas")

[Out]

-1/3*(3*b^2*x^2 + 3*a*b*x + a^2)/x^3

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.41 \[ \int \frac {(a+b x)^2}{x^4} \, dx=\frac {- a^{2} - 3 a b x - 3 b^{2} x^{2}}{3 x^{3}} \]

[In]

integrate((b*x+a)**2/x**4,x)

[Out]

(-a**2 - 3*a*b*x - 3*b**2*x**2)/(3*x**3)

Maxima [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.29 \[ \int \frac {(a+b x)^2}{x^4} \, dx=-\frac {3 \, b^{2} x^{2} + 3 \, a b x + a^{2}}{3 \, x^{3}} \]

[In]

integrate((b*x+a)^2/x^4,x, algorithm="maxima")

[Out]

-1/3*(3*b^2*x^2 + 3*a*b*x + a^2)/x^3

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.29 \[ \int \frac {(a+b x)^2}{x^4} \, dx=-\frac {3 \, b^{2} x^{2} + 3 \, a b x + a^{2}}{3 \, x^{3}} \]

[In]

integrate((b*x+a)^2/x^4,x, algorithm="giac")

[Out]

-1/3*(3*b^2*x^2 + 3*a*b*x + a^2)/x^3

Mupad [B] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.29 \[ \int \frac {(a+b x)^2}{x^4} \, dx=-\frac {\frac {a^2}{3}+a\,b\,x+b^2\,x^2}{x^3} \]

[In]

int((a + b*x)^2/x^4,x)

[Out]

-(a^2/3 + b^2*x^2 + a*b*x)/x^3