\(\int \frac {-20+5 x^2-174 x^3}{25 x^3} \, dx\) [2495]

Optimal result

Integrand size = 19, antiderivative size = 17 \[ \int \frac {-20+5 x^2-174 x^3}{25 x^3} \, dx=\frac {1}{5} \left (\frac {2}{x^2}-\frac {174 x}{5}+\log (x)\right ) \]

[Out]

-174/25*x+2/5/x^2+1/5*ln(x)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.12, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {12, 14} \[ \int \frac {-20+5 x^2-174 x^3}{25 x^3} \, dx=\frac {2}{5 x^2}-\frac {174 x}{25}+\frac {\log (x)}{5} \]

[In]

Int[(-20 + 5*x^2 - 174*x^3)/(25*x^3),x]

[Out]

2/(5*x^2) - (174*x)/25 + Log[x]/5

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 {-20+5 x^2-174 x^3}{x^3} \, dx \\ & = \frac {1}{25} \int \left (-174-\frac {20}{x^3}+\frac {5}{x}\right ) \, dx \\ & = \frac {2}{5 x^2}-\frac {174 x}{25}+\frac {\log (x)}{5} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int \frac {-20+5 x^2-174 x^3}{25 x^3} \, dx=\frac {1}{25} \left (\frac {10}{x^2}-174 x+5 \log (x)\right ) \]

[In]

Integrate[(-20 + 5*x^2 - 174*x^3)/(25*x^3),x]

[Out]

(10/x^2 - 174*x + 5*Log[x])/25

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82

method	result	size
default	\(-\frac {174 x}{25}+\frac {2}{5 x^{2}}+\frac {\ln \left (x \right )}{5}\)	\(14\)
risch	\(-\frac {174 x}{25}+\frac {2}{5 x^{2}}+\frac {\ln \left (x \right )}{5}\)	\(14\)
norman	\(\frac {\frac {2}{5}-\frac {174 x^{3}}{25}}{x^{2}}+\frac {\ln \left (x \right )}{5}\)	\(17\)
parallelrisch	\(\frac {5 x^{2} \ln \left (x \right )-174 x^{3}+10}{25 x^{2}}\)	\(20\)

[In]

int(1/25*(-174*x^3+5*x^2-20)/x^3,x,method=_RETURNVERBOSE)

[Out]

-174/25*x+2/5/x^2+1/5*ln(x)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.12 \[ \int \frac {-20+5 x^2-174 x^3}{25 x^3} \, dx=-\frac {174 \, x^{3} - 5 \, x^{2} \log \left (x\right ) - 10}{25 \, x^{2}} \]

[In]

integrate(1/25*(-174*x^3+5*x^2-20)/x^3,x, algorithm="fricas")

[Out]

-1/25*(174*x^3 - 5*x^2*log(x) - 10)/x^2

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88 \[ \int \frac {-20+5 x^2-174 x^3}{25 x^3} \, dx=- \frac {174 x}{25} + \frac {\log {\left (x \right )}}{5} + \frac {2}{5 x^{2}} \]

[In]

integrate(1/25*(-174*x**3+5*x**2-20)/x**3,x)

[Out]

-174*x/25 + log(x)/5 + 2/(5*x**2)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.76 \[ \int \frac {-20+5 x^2-174 x^3}{25 x^3} \, dx=-\frac {174}{25} \, x + \frac {2}{5 \, x^{2}} + \frac {1}{5} \, \log \left (x\right ) \]

[In]

integrate(1/25*(-174*x^3+5*x^2-20)/x^3,x, algorithm="maxima")

[Out]

-174/25*x + 2/5/x^2 + 1/5*log(x)

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82 \[ \int \frac {-20+5 x^2-174 x^3}{25 x^3} \, dx=-\frac {174}{25} \, x + \frac {2}{5 \, x^{2}} + \frac {1}{5} \, \log \left ({\left | x \right |}\right ) \]

[In]

integrate(1/25*(-174*x^3+5*x^2-20)/x^3,x, algorithm="giac")

[Out]

-174/25*x + 2/5/x^2 + 1/5*log(abs(x))

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.76 \[ \int \frac {-20+5 x^2-174 x^3}{25 x^3} \, dx=\frac {\ln \left (x\right )}{5}-\frac {174\,x}{25}+\frac {2}{5\,x^2} \]

[In]

int(-((174*x^3)/25 - x^2/5 + 4/5)/x^3,x)

[Out]

log(x)/5 - (174*x)/25 + 2/(5*x^2)