\(\int \frac {-5+x^3+e^x x^3+2 x^4}{x^3} \, dx\) [8077]

Optimal result

Integrand size = 21, antiderivative size = 16 \[ \int \frac {-5+x^3+e^x x^3+2 x^4}{x^3} \, dx=2+e^x+\frac {5}{2 x^2}+x+x^2 \]

[Out]

exp(x)+2+x+5/2/x^2+x^2

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.94, number of steps used = 5, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {14, 2225} \[ \int \frac {-5+x^3+e^x x^3+2 x^4}{x^3} \, dx=x^2+\frac {5}{2 x^2}+x+e^x \]

[In]

Int[(-5 + x^3 + E^x*x^3 + 2*x^4)/x^3,x]

[Out]

E^x + 5/(2*x^2) + x + x^2

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]]

Rule 2225

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rubi steps \begin{align*} \text {integral}& = \int \left (e^x+\frac {-5+x^3+2 x^4}{x^3}\right ) \, dx \\ & = \int e^x \, dx+\int \frac {-5+x^3+2 x^4}{x^3} \, dx \\ & = e^x+\int \left (1-\frac {5}{x^3}+2 x\right ) \, dx \\ & = e^x+\frac {5}{2 x^2}+x+x^2 \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.94 \[ \int \frac {-5+x^3+e^x x^3+2 x^4}{x^3} \, dx=e^x+\frac {5}{2 x^2}+x+x^2 \]

[In]

Integrate[(-5 + x^3 + E^x*x^3 + 2*x^4)/x^3,x]

[Out]

E^x + 5/(2*x^2) + x + x^2

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.81

method	result	size
default	\(x^{2}+x +\frac {5}{2 x^{2}}+{\mathrm e}^{x}\)	\(13\)
risch	\(x^{2}+x +\frac {5}{2 x^{2}}+{\mathrm e}^{x}\)	\(13\)
parts	\(x^{2}+x +\frac {5}{2 x^{2}}+{\mathrm e}^{x}\)	\(13\)
norman	\(\frac {\frac {5}{2}+{\mathrm e}^{x} x^{2}+x^{4}+x^{3}}{x^{2}}\)	\(19\)
parallelrisch	\(\frac {2 x^{4}+2 \,{\mathrm e}^{x} x^{2}+2 x^{3}+5}{2 x^{2}}\)	\(25\)

[In]

int((exp(x)*x^3+2*x^4+x^3-5)/x^3,x,method=_RETURNVERBOSE)

[Out]

x^2+x+5/2/x^2+exp(x)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.50 \[ \int \frac {-5+x^3+e^x x^3+2 x^4}{x^3} \, dx=\frac {2 \, x^{4} + 2 \, x^{3} + 2 \, x^{2} e^{x} + 5}{2 \, x^{2}} \]

[In]

integrate((exp(x)*x^3+2*x^4+x^3-5)/x^3,x, algorithm="fricas")

[Out]

1/2*(2*x^4 + 2*x^3 + 2*x^2*e^x + 5)/x^2

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.88 \[ \int \frac {-5+x^3+e^x x^3+2 x^4}{x^3} \, dx=x^{2} + x + e^{x} + \frac {5}{2 x^{2}} \]

[In]

integrate((exp(x)*x**3+2*x**4+x**3-5)/x**3,x)

[Out]

x**2 + x + exp(x) + 5/(2*x**2)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.75 \[ \int \frac {-5+x^3+e^x x^3+2 x^4}{x^3} \, dx=x^{2} + x + \frac {5}{2 \, x^{2}} + e^{x} \]

[In]

integrate((exp(x)*x^3+2*x^4+x^3-5)/x^3,x, algorithm="maxima")

[Out]

x^2 + x + 5/2/x^2 + e^x

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.50 \[ \int \frac {-5+x^3+e^x x^3+2 x^4}{x^3} \, dx=\frac {2 \, x^{4} + 2 \, x^{3} + 2 \, x^{2} e^{x} + 5}{2 \, x^{2}} \]

[In]

integrate((exp(x)*x^3+2*x^4+x^3-5)/x^3,x, algorithm="giac")

[Out]

1/2*(2*x^4 + 2*x^3 + 2*x^2*e^x + 5)/x^2

Mupad [B] (verification not implemented)

Time = 13.04 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.75 \[ \int \frac {-5+x^3+e^x x^3+2 x^4}{x^3} \, dx=x+{\mathrm {e}}^x+\frac {5}{2\,x^2}+x^2 \]

[In]

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

[Out]

x + exp(x) + 5/(2*x^2) + x^2