\(\int \frac {-8+9 x-49 x^2}{9 x^2} \, dx\) [3840]

Optimal result

Integrand size = 17, antiderivative size = 25 \[ \int \frac {-8+9 x-49 x^2}{9 x^2} \, dx=\log \left (e^{-6+\frac {40}{9} \left (\frac {1}{5 x}-x\right )-x} x\right ) \]

[Out]

ln(x/exp(49/9*x+6-8/9/x))

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.60, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {12, 14} \[ \int \frac {-8+9 x-49 x^2}{9 x^2} \, dx=-\frac {49 x}{9}+\frac {8}{9 x}+\log (x) \]

[In]

Int[(-8 + 9*x - 49*x^2)/(9*x^2),x]

[Out]

8/(9*x) - (49*x)/9 + Log[x]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{9} \int \frac {-8+9 x-49 x^2}{x^2} \, dx \\ & = \frac {1}{9} \int \left (-49-\frac {8}{x^2}+\frac {9}{x}\right ) \, dx \\ & = \frac {8}{9 x}-\frac {49 x}{9}+\log (x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.60 \[ \int \frac {-8+9 x-49 x^2}{9 x^2} \, dx=\frac {8}{9 x}-\frac {49 x}{9}+\log (x) \]

[In]

Integrate[(-8 + 9*x - 49*x^2)/(9*x^2),x]

[Out]

8/(9*x) - (49*x)/9 + Log[x]

Maple [A] (verified)

Time = 0.33 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.48

method	result	size
default	\(-\frac {49 x}{9}+\ln \left (x \right )+\frac {8}{9 x}\)	\(12\)
risch	\(-\frac {49 x}{9}+\ln \left (x \right )+\frac {8}{9 x}\)	\(12\)
norman	\(\frac {\frac {8}{9}-\frac {49 x^{2}}{9}}{x}+\ln \left (x \right )\)	\(15\)
parallelrisch	\(\frac {9 x \ln \left (x \right )-49 x^{2}+8}{9 x}\)	\(18\)

[In]

int(1/9*(-49*x^2+9*x-8)/x^2,x,method=_RETURNVERBOSE)

[Out]

-49/9*x+ln(x)+8/9/x

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.68 \[ \int \frac {-8+9 x-49 x^2}{9 x^2} \, dx=-\frac {49 \, x^{2} - 9 \, x \log \left (x\right ) - 8}{9 \, x} \]

[In]

integrate(1/9*(-49*x^2+9*x-8)/x^2,x, algorithm="fricas")

[Out]

-1/9*(49*x^2 - 9*x*log(x) - 8)/x

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.48 \[ \int \frac {-8+9 x-49 x^2}{9 x^2} \, dx=- \frac {49 x}{9} + \log {\left (x \right )} + \frac {8}{9 x} \]

[In]

integrate(1/9*(-49*x**2+9*x-8)/x**2,x)

[Out]

-49*x/9 + log(x) + 8/(9*x)

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.44 \[ \int \frac {-8+9 x-49 x^2}{9 x^2} \, dx=-\frac {49}{9} \, x + \frac {8}{9 \, x} + \log \left (x\right ) \]

[In]

integrate(1/9*(-49*x^2+9*x-8)/x^2,x, algorithm="maxima")

[Out]

-49/9*x + 8/9/x + log(x)

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.48 \[ \int \frac {-8+9 x-49 x^2}{9 x^2} \, dx=-\frac {49}{9} \, x + \frac {8}{9 \, x} + \log \left ({\left | x \right |}\right ) \]

[In]

integrate(1/9*(-49*x^2+9*x-8)/x^2,x, algorithm="giac")

[Out]

-49/9*x + 8/9/x + log(abs(x))

Mupad [B] (verification not implemented)

Time = 9.28 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.44 \[ \int \frac {-8+9 x-49 x^2}{9 x^2} \, dx=\ln \left (x\right )-\frac {49\,x}{9}+\frac {8}{9\,x} \]

[In]

int(-((49*x^2)/9 - x + 8/9)/x^2,x)

[Out]

log(x) - (49*x)/9 + 8/(9*x)