\(\int \frac {-118-125 x-123 x^2}{25 x^2} \, dx\) [8682]

Optimal result

Integrand size = 17, antiderivative size = 29 \[ \int \frac {-118-125 x-123 x^2}{25 x^2} \, dx=-5 x+\frac {2 \left (2+x+\frac {1}{25} (3+x)^2\right )}{x}+5 (4-\log (x)) \]

[Out]

2*(2+1/5*(3+x)*(3/5+1/5*x)+x)/x-5*ln(x)+20-5*x

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.59, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {12, 14} \[ \int \frac {-118-125 x-123 x^2}{25 x^2} \, dx=-\frac {123 x}{25}+\frac {118}{25 x}-5 \log (x) \]

[In]

Int[(-118 - 125*x - 123*x^2)/(25*x^2),x]

[Out]

118/(25*x) - (123*x)/25 - 5*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}{25} \int \frac {-118-125 x-123 x^2}{x^2} \, dx \\ & = \frac {1}{25} \int \left (-123-\frac {118}{x^2}-\frac {125}{x}\right ) \, dx \\ & = \frac {118}{25 x}-\frac {123 x}{25}-5 \log (x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.59 \[ \int \frac {-118-125 x-123 x^2}{25 x^2} \, dx=\frac {1}{25} \left (\frac {118}{x}-123 x-125 \log (x)\right ) \]

[In]

Integrate[(-118 - 125*x - 123*x^2)/(25*x^2),x]

[Out]

(118/x - 123*x - 125*Log[x])/25

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.48

method	result	size
default	\(-\frac {123 x}{25}+\frac {118}{25 x}-5 \ln \left (x \right )\)	\(14\)
risch	\(-\frac {123 x}{25}+\frac {118}{25 x}-5 \ln \left (x \right )\)	\(14\)
norman	\(\frac {\frac {118}{25}-\frac {123 x^{2}}{25}}{x}-5 \ln \left (x \right )\)	\(17\)
parallelrisch	\(-\frac {125 x \ln \left (x \right )+123 x^{2}-118}{25 x}\)	\(18\)

[In]

int(1/25*(-123*x^2-125*x-118)/x^2,x,method=_RETURNVERBOSE)

[Out]

-123/25*x+118/25/x-5*ln(x)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.59 \[ \int \frac {-118-125 x-123 x^2}{25 x^2} \, dx=-\frac {123 \, x^{2} + 125 \, x \log \left (x\right ) - 118}{25 \, x} \]

[In]

integrate(1/25*(-123*x^2-125*x-118)/x^2,x, algorithm="fricas")

[Out]

-1/25*(123*x^2 + 125*x*log(x) - 118)/x

Sympy [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.48 \[ \int \frac {-118-125 x-123 x^2}{25 x^2} \, dx=- \frac {123 x}{25} - 5 \log {\left (x \right )} + \frac {118}{25 x} \]

[In]

integrate(1/25*(-123*x**2-125*x-118)/x**2,x)

[Out]

-123*x/25 - 5*log(x) + 118/(25*x)

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.45 \[ \int \frac {-118-125 x-123 x^2}{25 x^2} \, dx=-\frac {123}{25} \, x + \frac {118}{25 \, x} - 5 \, \log \left (x\right ) \]

[In]

integrate(1/25*(-123*x^2-125*x-118)/x^2,x, algorithm="maxima")

[Out]

-123/25*x + 118/25/x - 5*log(x)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.48 \[ \int \frac {-118-125 x-123 x^2}{25 x^2} \, dx=-\frac {123}{25} \, x + \frac {118}{25 \, x} - 5 \, \log \left ({\left | x \right |}\right ) \]

[In]

integrate(1/25*(-123*x^2-125*x-118)/x^2,x, algorithm="giac")

[Out]

-123/25*x + 118/25/x - 5*log(abs(x))

Mupad [B] (verification not implemented)

Time = 13.35 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.45 \[ \int \frac {-118-125 x-123 x^2}{25 x^2} \, dx=\frac {118}{25\,x}-5\,\ln \left (x\right )-\frac {123\,x}{25} \]

[In]

int(-(5*x + (123*x^2)/25 + 118/25)/x^2,x)

[Out]

118/(25*x) - 5*log(x) - (123*x)/25