\(\int \frac {-30-3 x^3+e (10+x^3)}{x^3} \, dx\) [647]

Optimal result

Integrand size = 18, antiderivative size = 12 \[ \int \frac {-30-3 x^3+e \left (10+x^3\right )}{x^3} \, dx=(-3+e) \left (-5-\frac {5}{x^2}+x\right ) \]

[Out]

(-5-5/x^2+x)*(exp(1)-3)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.58, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {14} \[ \int \frac {-30-3 x^3+e \left (10+x^3\right )}{x^3} \, dx=\frac {5 (3-e)}{x^2}-(3-e) x \]

[In]

Int[(-30 - 3*x^3 + E*(10 + x^3))/x^3,x]

[Out]

(5*(3 - E))/x^2 - (3 - E)*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}& = \int \left (-3 \left (1-\frac {e}{3}\right )+\frac {10 (-3+e)}{x^3}\right ) \, dx \\ & = \frac {5 (3-e)}{x^2}-(3-e) x \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.92 \[ \int \frac {-30-3 x^3+e \left (10+x^3\right )}{x^3} \, dx=(-3+e) \left (-\frac {5}{x^2}+x\right ) \]

[In]

Integrate[(-30 - 3*x^3 + E*(10 + x^3))/x^3,x]

[Out]

(-3 + E)*(-5/x^2 + x)

Maple [A] (verified)

Time = 0.02 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.08

method	result	size
default	\(\left ({\mathrm e}-3\right ) \left (x -\frac {5}{x^{2}}\right )\)	\(13\)
gosper	\(\frac {\left ({\mathrm e}-3\right ) \left (x^{3}-5\right )}{x^{2}}\)	\(14\)
norman	\(\frac {\left ({\mathrm e}-3\right ) x^{3}-5 \,{\mathrm e}+15}{x^{2}}\)	\(19\)
risch	\(x \,{\mathrm e}-3 x -\frac {5 \,{\mathrm e}}{x^{2}}+\frac {15}{x^{2}}\)	\(21\)
parallelrisch	\(\frac {x^{3} {\mathrm e}-3 x^{3}-5 \,{\mathrm e}+15}{x^{2}}\)	\(22\)

[In]

int(((x^3+10)*exp(1)-3*x^3-30)/x^3,x,method=_RETURNVERBOSE)

[Out]

(exp(1)-3)*(x-5/x^2)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.75 \[ \int \frac {-30-3 x^3+e \left (10+x^3\right )}{x^3} \, dx=-\frac {3 \, x^{3} - {\left (x^{3} - 5\right )} e - 15}{x^{2}} \]

[In]

integrate(((x^3+10)*exp(1)-3*x^3-30)/x^3,x, algorithm="fricas")

[Out]

-(3*x^3 - (x^3 - 5)*e - 15)/x^2

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.42 \[ \int \frac {-30-3 x^3+e \left (10+x^3\right )}{x^3} \, dx=- x \left (3 - e\right ) - \frac {-15 + 5 e}{x^{2}} \]

[In]

integrate(((x**3+10)*exp(1)-3*x**3-30)/x**3,x)

[Out]

-x*(3 - E) - (-15 + 5*E)/x**2

Maxima [A] (verification not implemented)

none

Time = 0.17 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.33 \[ \int \frac {-30-3 x^3+e \left (10+x^3\right )}{x^3} \, dx=x {\left (e - 3\right )} - \frac {5 \, {\left (e - 3\right )}}{x^{2}} \]

[In]

integrate(((x^3+10)*exp(1)-3*x^3-30)/x^3,x, algorithm="maxima")

[Out]

x*(e - 3) - 5*(e - 3)/x^2

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.42 \[ \int \frac {-30-3 x^3+e \left (10+x^3\right )}{x^3} \, dx=x e - 3 \, x - \frac {5 \, {\left (e - 3\right )}}{x^{2}} \]

[In]

integrate(((x^3+10)*exp(1)-3*x^3-30)/x^3,x, algorithm="giac")

[Out]

x*e - 3*x - 5*(e - 3)/x^2

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.08 \[ \int \frac {-30-3 x^3+e \left (10+x^3\right )}{x^3} \, dx=\frac {\left (x^3-5\right )\,\left (\mathrm {e}-3\right )}{x^2} \]

[In]

int(-(3*x^3 - exp(1)*(x^3 + 10) + 30)/x^3,x)

[Out]

((x^3 - 5)*(exp(1) - 3))/x^2