\(\int \frac {(a+b x)^2}{x^3} \, dx\) [58]

Optimal result

Integrand size = 11, antiderivative size = 24 \[ \int \frac {(a+b x)^2}{x^3} \, dx=-\frac {a^2}{2 x^2}-\frac {2 a b}{x}+b^2 \log (x) \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 24, 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} \[ \int \frac {(a+b x)^2}{x^3} \, dx=-\frac {a^2}{2 x^2}-\frac {2 a b}{x}+b^2 \log (x) \]

[In]

Int[(a + b*x)^2/x^3,x]

[Out]

-1/2*a^2/x^2 - (2*a*b)/x + b^2*Log[x]

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

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {(a+b x)^2}{x^3} \, dx=-\frac {a^2}{2 x^2}-\frac {2 a b}{x}+b^2 \log (x) \]

[In]

Integrate[(a + b*x)^2/x^3,x]

[Out]

-1/2*a^2/x^2 - (2*a*b)/x + b^2*Log[x]

Maple [A] (verified)

Time = 0.16 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96

method	result	size
default	\(-\frac {a^{2}}{2 x^{2}}-\frac {2 a b}{x}+b^{2} \ln \left (x \right )\)	\(23\)
norman	\(\frac {-\frac {1}{2} a^{2}-2 a b x}{x^{2}}+b^{2} \ln \left (x \right )\)	\(23\)
risch	\(\frac {-\frac {1}{2} a^{2}-2 a b x}{x^{2}}+b^{2} \ln \left (x \right )\)	\(23\)
parallelrisch	\(\frac {2 b^{2} \ln \left (x \right ) x^{2}-4 a b x -a^{2}}{2 x^{2}}\)	\(27\)

[In]

int((b*x+a)^2/x^3,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08 \[ \int \frac {(a+b x)^2}{x^3} \, dx=\frac {2 \, b^{2} x^{2} \log \left (x\right ) - 4 \, a b x - a^{2}}{2 \, x^{2}} \]

[In]

integrate((b*x+a)^2/x^3,x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int \frac {(a+b x)^2}{x^3} \, dx=b^{2} \log {\left (x \right )} + \frac {- a^{2} - 4 a b x}{2 x^{2}} \]

[In]

integrate((b*x+a)**2/x**3,x)

[Out]

b**2*log(x) + (-a**2 - 4*a*b*x)/(2*x**2)

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.88 \[ \int \frac {(a+b x)^2}{x^3} \, dx=b^{2} \log \left (x\right ) - \frac {4 \, a b x + a^{2}}{2 \, x^{2}} \]

[In]

integrate((b*x+a)^2/x^3,x, algorithm="maxima")

[Out]

b^2*log(x) - 1/2*(4*a*b*x + a^2)/x^2

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int \frac {(a+b x)^2}{x^3} \, dx=b^{2} \log \left ({\left | x \right |}\right ) - \frac {4 \, a b x + a^{2}}{2 \, x^{2}} \]

[In]

integrate((b*x+a)^2/x^3,x, algorithm="giac")

[Out]

b^2*log(abs(x)) - 1/2*(4*a*b*x + a^2)/x^2

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96 \[ \int \frac {(a+b x)^2}{x^3} \, dx=b^2\,\ln \left (x\right )-\frac {\frac {a^2}{2}+2\,b\,x\,a}{x^2} \]

[In]

int((a + b*x)^2/x^3,x)

[Out]

b^2*log(x) - (a^2/2 + 2*a*b*x)/x^2