\(\int \frac {-13 x-6 x^2+e^{\frac {1}{x}} (-4+4 x)}{3 x} \, dx\) [5875]

Optimal result

Integrand size = 27, antiderivative size = 22 \[ \int \frac {-13 x-6 x^2+e^{\frac {1}{x}} (-4+4 x)}{3 x} \, dx=7+x+\left (\frac {4}{3} \left (-4+e^{\frac {1}{x}}\right )-x\right ) x+\log (5) \]

[Out]

x+7+(4/3*exp(1/x)-16/3-x)*x+ln(5)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {12, 14, 2326} \[ \int \frac {-13 x-6 x^2+e^{\frac {1}{x}} (-4+4 x)}{3 x} \, dx=-x^2+\frac {4}{3} e^{\frac {1}{x}} x-\frac {13 x}{3} \]

[In]

Int[(-13*x - 6*x^2 + E^x^(-1)*(-4 + 4*x))/(3*x),x]

[Out]

(-13*x)/3 + (4*E^x^(-1)*x)/3 - 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 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}& = \frac {1}{3} \int \frac {-13 x-6 x^2+e^{\frac {1}{x}} (-4+4 x)}{x} \, dx \\ & = \frac {1}{3} \int \left (-13+\frac {4 e^{\frac {1}{x}} (-1+x)}{x}-6 x\right ) \, dx \\ & = -\frac {13 x}{3}-x^2+\frac {4}{3} \int \frac {e^{\frac {1}{x}} (-1+x)}{x} \, dx \\ & = -\frac {13 x}{3}+\frac {4}{3} e^{\frac {1}{x}} x-x^2 \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95 \[ \int \frac {-13 x-6 x^2+e^{\frac {1}{x}} (-4+4 x)}{3 x} \, dx=\frac {1}{3} \left (-13 x+4 e^{\frac {1}{x}} x-3 x^2\right ) \]

[In]

Integrate[(-13*x - 6*x^2 + E^x^(-1)*(-4 + 4*x))/(3*x),x]

[Out]

(-13*x + 4*E^x^(-1)*x - 3*x^2)/3

Maple [A] (verified)

Time = 0.21 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.77

method	result	size
derivativedivides	\(-x^{2}-\frac {13 x}{3}+\frac {4 x \,{\mathrm e}^{\frac {1}{x}}}{3}\)	\(17\)
default	\(-x^{2}-\frac {13 x}{3}+\frac {4 x \,{\mathrm e}^{\frac {1}{x}}}{3}\)	\(17\)
norman	\(-x^{2}-\frac {13 x}{3}+\frac {4 x \,{\mathrm e}^{\frac {1}{x}}}{3}\)	\(17\)
risch	\(-x^{2}-\frac {13 x}{3}+\frac {4 x \,{\mathrm e}^{\frac {1}{x}}}{3}\)	\(17\)
parallelrisch	\(-x^{2}-\frac {13 x}{3}+\frac {4 x \,{\mathrm e}^{\frac {1}{x}}}{3}\)	\(17\)
parts	\(-x^{2}-\frac {13 x}{3}+\frac {4 x \,{\mathrm e}^{\frac {1}{x}}}{3}\)	\(17\)

[In]

int(1/3*((-4+4*x)*exp(1/x)-6*x^2-13*x)/x,x,method=_RETURNVERBOSE)

[Out]

-x^2-13/3*x+4/3*x*exp(1/x)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.73 \[ \int \frac {-13 x-6 x^2+e^{\frac {1}{x}} (-4+4 x)}{3 x} \, dx=-x^{2} + \frac {4}{3} \, x e^{\frac {1}{x}} - \frac {13}{3} \, x \]

[In]

integrate(1/3*((-4+4*x)*exp(1/x)-6*x^2-13*x)/x,x, algorithm="fricas")

[Out]

-x^2 + 4/3*x*e^(1/x) - 13/3*x

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.77 \[ \int \frac {-13 x-6 x^2+e^{\frac {1}{x}} (-4+4 x)}{3 x} \, dx=- x^{2} + \frac {4 x e^{\frac {1}{x}}}{3} - \frac {13 x}{3} \]

[In]

integrate(1/3*((-4+4*x)*exp(1/x)-6*x**2-13*x)/x,x)

[Out]

-x**2 + 4*x*exp(1/x)/3 - 13*x/3

Maxima [C] (verification not implemented)

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

Time = 0.22 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.09 \[ \int \frac {-13 x-6 x^2+e^{\frac {1}{x}} (-4+4 x)}{3 x} \, dx=-x^{2} - \frac {13}{3} \, x + \frac {4}{3} \, {\rm Ei}\left (\frac {1}{x}\right ) - \frac {4}{3} \, \Gamma \left (-1, -\frac {1}{x}\right ) \]

[In]

integrate(1/3*((-4+4*x)*exp(1/x)-6*x^2-13*x)/x,x, algorithm="maxima")

[Out]

-x^2 - 13/3*x + 4/3*Ei(1/x) - 4/3*gamma(-1, -1/x)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95 \[ \int \frac {-13 x-6 x^2+e^{\frac {1}{x}} (-4+4 x)}{3 x} \, dx=\frac {1}{3} \, x^{2} {\left (\frac {4 \, e^{\frac {1}{x}}}{x} - \frac {13}{x} - 3\right )} \]

[In]

integrate(1/3*((-4+4*x)*exp(1/x)-6*x^2-13*x)/x,x, algorithm="giac")

[Out]

1/3*x^2*(4*e^(1/x)/x - 13/x - 3)

Mupad [B] (verification not implemented)

Time = 12.44 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.64 \[ \int \frac {-13 x-6 x^2+e^{\frac {1}{x}} (-4+4 x)}{3 x} \, dx=-\frac {x\,\left (3\,x-4\,{\mathrm {e}}^{1/x}+13\right )}{3} \]

[In]

int(-((13*x)/3 - (exp(1/x)*(4*x - 4))/3 + 2*x^2)/x,x)

[Out]

-(x*(3*x - 4*exp(1/x) + 13))/3