\(\int \frac {-5-8 x^3}{8 x^2} \, dx\) [6076]

Optimal result

Integrand size = 14, antiderivative size = 19 \[ \int \frac {-5-8 x^3}{8 x^2} \, dx=\frac {1}{2} \left (\frac {5}{4 x}-x^2+\log (2)\right ) \]

[Out]

5/8/x-1/2*x^2+1/2*ln(2)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {12, 14} \[ \int \frac {-5-8 x^3}{8 x^2} \, dx=\frac {5}{8 x}-\frac {x^2}{2} \]

[In]

Int[(-5 - 8*x^3)/(8*x^2),x]

[Out]

5/(8*x) - x^2/2

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}{8} \int \frac {-5-8 x^3}{x^2} \, dx \\ & = \frac {1}{8} \int \left (-\frac {5}{x^2}-8 x\right ) \, dx \\ & = \frac {5}{8 x}-\frac {x^2}{2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79 \[ \int \frac {-5-8 x^3}{8 x^2} \, dx=\frac {5}{8 x}-\frac {x^2}{2} \]

[In]

Integrate[(-5 - 8*x^3)/(8*x^2),x]

[Out]

5/(8*x) - x^2/2

Maple [A] (verified)

Time = 0.02 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.63

method	result	size
default	\(-\frac {x^{2}}{2}+\frac {5}{8 x}\)	\(12\)
norman	\(\frac {\frac {5}{8}-\frac {x^{3}}{2}}{x}\)	\(12\)
risch	\(-\frac {x^{2}}{2}+\frac {5}{8 x}\)	\(12\)
gosper	\(-\frac {4 x^{3}-5}{8 x}\)	\(13\)
parallelrisch	\(-\frac {4 x^{3}-5}{8 x}\)	\(13\)

[In]

int(1/8*(-8*x^3-5)/x^2,x,method=_RETURNVERBOSE)

[Out]

-1/2*x^2+5/8/x

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.63 \[ \int \frac {-5-8 x^3}{8 x^2} \, dx=-\frac {4 \, x^{3} - 5}{8 \, x} \]

[In]

integrate(1/8*(-8*x^3-5)/x^2,x, algorithm="fricas")

[Out]

-1/8*(4*x^3 - 5)/x

Sympy [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.42 \[ \int \frac {-5-8 x^3}{8 x^2} \, dx=- \frac {x^{2}}{2} + \frac {5}{8 x} \]

[In]

integrate(1/8*(-8*x**3-5)/x**2,x)

[Out]

-x**2/2 + 5/(8*x)

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.58 \[ \int \frac {-5-8 x^3}{8 x^2} \, dx=-\frac {1}{2} \, x^{2} + \frac {5}{8 \, x} \]

[In]

integrate(1/8*(-8*x^3-5)/x^2,x, algorithm="maxima")

[Out]

-1/2*x^2 + 5/8/x

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.58 \[ \int \frac {-5-8 x^3}{8 x^2} \, dx=-\frac {1}{2} \, x^{2} + \frac {5}{8 \, x} \]

[In]

integrate(1/8*(-8*x^3-5)/x^2,x, algorithm="giac")

[Out]

-1/2*x^2 + 5/8/x

Mupad [B] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.63 \[ \int \frac {-5-8 x^3}{8 x^2} \, dx=-\frac {4\,x^3-5}{8\,x} \]

[In]

int(-(x^3 + 5/8)/x^2,x)

[Out]

-(4*x^3 - 5)/(8*x)