\(\int \frac {-1-5 e^x x^2-10 x^3}{5 x^2} \, dx\) [7404]

Optimal result

Integrand size = 22, antiderivative size = 19 \[ \int \frac {-1-5 e^x x^2-10 x^3}{5 x^2} \, dx=3-e^x+\frac {1}{5 x}-x^2 \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.95, number of steps used = 6, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {12, 14, 2225} \[ \int \frac {-1-5 e^x x^2-10 x^3}{5 x^2} \, dx=-x^2-e^x+\frac {1}{5 x} \]

[In]

Int[(-1 - 5*E^x*x^2 - 10*x^3)/(5*x^2),x]

[Out]

-E^x + 1/(5*x) - x^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]]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.95 \[ \int \frac {-1-5 e^x x^2-10 x^3}{5 x^2} \, dx=-e^x+\frac {1}{5 x}-x^2 \]

[In]

Integrate[(-1 - 5*E^x*x^2 - 10*x^3)/(5*x^2),x]

[Out]

-E^x + 1/(5*x) - x^2

Maple [A] (verified)

Time = 0.02 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.84

method	result	size
default	\(-x^{2}+\frac {1}{5 x}-{\mathrm e}^{x}\)	\(16\)
risch	\(-x^{2}+\frac {1}{5 x}-{\mathrm e}^{x}\)	\(16\)
parts	\(-x^{2}+\frac {1}{5 x}-{\mathrm e}^{x}\)	\(16\)
norman	\(\frac {\frac {1}{5}-x^{3}-{\mathrm e}^{x} x}{x}\)	\(17\)
parallelrisch	\(-\frac {5 x^{3}+5 \,{\mathrm e}^{x} x -1}{5 x}\)	\(18\)

[In]

int(1/5*(-5*exp(x)*x^2-10*x^3-1)/x^2,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89 \[ \int \frac {-1-5 e^x x^2-10 x^3}{5 x^2} \, dx=-\frac {5 \, x^{3} + 5 \, x e^{x} - 1}{5 \, x} \]

[In]

integrate(1/5*(-5*exp(x)*x^2-10*x^3-1)/x^2,x, algorithm="fricas")

[Out]

-1/5*(5*x^3 + 5*x*e^x - 1)/x

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.53 \[ \int \frac {-1-5 e^x x^2-10 x^3}{5 x^2} \, dx=- x^{2} - e^{x} + \frac {1}{5 x} \]

[In]

integrate(1/5*(-5*exp(x)*x**2-10*x**3-1)/x**2,x)

[Out]

-x**2 - exp(x) + 1/(5*x)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79 \[ \int \frac {-1-5 e^x x^2-10 x^3}{5 x^2} \, dx=-x^{2} + \frac {1}{5 \, x} - e^{x} \]

[In]

integrate(1/5*(-5*exp(x)*x^2-10*x^3-1)/x^2,x, algorithm="maxima")

[Out]

-x^2 + 1/5/x - e^x

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89 \[ \int \frac {-1-5 e^x x^2-10 x^3}{5 x^2} \, dx=-\frac {5 \, x^{3} + 5 \, x e^{x} - 1}{5 \, x} \]

[In]

integrate(1/5*(-5*exp(x)*x^2-10*x^3-1)/x^2,x, algorithm="giac")

[Out]

-1/5*(5*x^3 + 5*x*e^x - 1)/x

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79 \[ \int \frac {-1-5 e^x x^2-10 x^3}{5 x^2} \, dx=\frac {1}{5\,x}-{\mathrm {e}}^x-x^2 \]

[In]

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

[Out]

1/(5*x) - exp(x) - x^2