\(\int \frac {e^{\frac {1}{x}} (112-112 x)-336 x^3}{x} \, dx\) [4805]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [A] (verification not implemented)
   Maxima [C] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 21, antiderivative size = 21 \[ \int \frac {e^{\frac {1}{x}} (112-112 x)-336 x^3}{x} \, dx=112 x \left (-e^{\frac {1}{x}}+\frac {25}{x}-x^2\right ) \]

[Out]

112*(25/x-x^2-exp(1/x))*x

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.67, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {14, 2326} \[ \int \frac {e^{\frac {1}{x}} (112-112 x)-336 x^3}{x} \, dx=-112 x^3-112 e^{\frac {1}{x}} x \]

[In]

Int[(E^x^(-1)*(112 - 112*x) - 336*x^3)/x,x]

[Out]

-112*E^x^(-1)*x - 112*x^3

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 2326

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = v*(y/(Log[F]*D[u, x]))}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {112 e^{\frac {1}{x}} (-1+x)}{x}-336 x^2\right ) \, dx \\ & = -112 x^3-112 \int \frac {e^{\frac {1}{x}} (-1+x)}{x} \, dx \\ & = -112 e^{\frac {1}{x}} x-112 x^3 \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.62 \[ \int \frac {e^{\frac {1}{x}} (112-112 x)-336 x^3}{x} \, dx=-112 \left (e^{\frac {1}{x}} x+x^3\right ) \]

[In]

Integrate[(E^x^(-1)*(112 - 112*x) - 336*x^3)/x,x]

[Out]

-112*(E^x^(-1)*x + x^3)

Maple [A] (verified)

Time = 0.11 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.67

method result size
derivativedivides \(-112 x^{3}-112 x \,{\mathrm e}^{\frac {1}{x}}\) \(14\)
default \(-112 x^{3}-112 x \,{\mathrm e}^{\frac {1}{x}}\) \(14\)
norman \(-112 x^{3}-112 x \,{\mathrm e}^{\frac {1}{x}}\) \(14\)
risch \(-112 x^{3}-112 x \,{\mathrm e}^{\frac {1}{x}}\) \(14\)
parallelrisch \(-112 x^{3}-112 x \,{\mathrm e}^{\frac {1}{x}}\) \(14\)
parts \(-112 x^{3}-112 x \,{\mathrm e}^{\frac {1}{x}}\) \(14\)

[In]

int(((-112*x+112)*exp(1/x)-336*x^3)/x,x,method=_RETURNVERBOSE)

[Out]

-112*x^3-112*x*exp(1/x)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.62 \[ \int \frac {e^{\frac {1}{x}} (112-112 x)-336 x^3}{x} \, dx=-112 \, x^{3} - 112 \, x e^{\frac {1}{x}} \]

[In]

integrate(((-112*x+112)*exp(1/x)-336*x^3)/x,x, algorithm="fricas")

[Out]

-112*x^3 - 112*x*e^(1/x)

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.67 \[ \int \frac {e^{\frac {1}{x}} (112-112 x)-336 x^3}{x} \, dx=- 112 x^{3} - 112 x e^{\frac {1}{x}} \]

[In]

integrate(((-112*x+112)*exp(1/x)-336*x**3)/x,x)

[Out]

-112*x**3 - 112*x*exp(1/x)

Maxima [C] (verification not implemented)

Result contains higher order function than in optimal. Order 4 vs. order 3.

Time = 0.20 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {1}{x}} (112-112 x)-336 x^3}{x} \, dx=-112 \, x^{3} - 112 \, {\rm Ei}\left (\frac {1}{x}\right ) + 112 \, \Gamma \left (-1, -\frac {1}{x}\right ) \]

[In]

integrate(((-112*x+112)*exp(1/x)-336*x^3)/x,x, algorithm="maxima")

[Out]

-112*x^3 - 112*Ei(1/x) + 112*gamma(-1, -1/x)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.71 \[ \int \frac {e^{\frac {1}{x}} (112-112 x)-336 x^3}{x} \, dx=-112 \, x^{3} {\left (\frac {e^{\frac {1}{x}}}{x^{2}} + 1\right )} \]

[In]

integrate(((-112*x+112)*exp(1/x)-336*x^3)/x,x, algorithm="giac")

[Out]

-112*x^3*(e^(1/x)/x^2 + 1)

Mupad [B] (verification not implemented)

Time = 9.79 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.52 \[ \int \frac {e^{\frac {1}{x}} (112-112 x)-336 x^3}{x} \, dx=-112\,x\,\left ({\mathrm {e}}^{1/x}+x^2\right ) \]

[In]

int(-(exp(1/x)*(112*x - 112) + 336*x^3)/x,x)

[Out]

-112*x*(exp(1/x) + x^2)