\(\int \frac {2 x^3-6 x^4+e^{\frac {16-8 x-9 x^2}{x^2}} (-96+24 x)}{x^3} \, dx\) [1010]

Optimal result

Integrand size = 37, antiderivative size = 26 \[ \int \frac {2 x^3-6 x^4+e^{\frac {16-8 x-9 x^2}{x^2}} (-96+24 x)}{x^3} \, dx=-x+3 \left (e^{-10+\left (1-\frac {4}{x}\right )^2}+x-x^2\right ) \]

[Out]

2*x+3*exp((1-4/x)^2-10)-3*x^2

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.08, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.054, Rules used = {14, 6838} \[ \int \frac {2 x^3-6 x^4+e^{\frac {16-8 x-9 x^2}{x^2}} (-96+24 x)}{x^3} \, dx=3 e^{\frac {16}{x^2}-\frac {8}{x}-9}-\frac {1}{3} (1-3 x)^2 \]

[In]

Int[(2*x^3 - 6*x^4 + E^((16 - 8*x - 9*x^2)/x^2)*(-96 + 24*x))/x^3,x]

[Out]

3*E^(-9 + 16/x^2 - 8/x) - (1 - 3*x)^2/3

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 6838

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[q*(F^v/Log[F]), x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {24 e^{-9+\frac {16}{x^2}-\frac {8}{x}} (-4+x)}{x^3}-2 (-1+3 x)\right ) \, dx \\ & = -\frac {1}{3} (1-3 x)^2+24 \int \frac {e^{-9+\frac {16}{x^2}-\frac {8}{x}} (-4+x)}{x^3} \, dx \\ & = 3 e^{-9+\frac {16}{x^2}-\frac {8}{x}}-\frac {1}{3} (1-3 x)^2 \\ \end{align*}

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96 \[ \int \frac {2 x^3-6 x^4+e^{\frac {16-8 x-9 x^2}{x^2}} (-96+24 x)}{x^3} \, dx=3 e^{-9+\frac {16}{x^2}-\frac {8}{x}}+2 x-3 x^2 \]

[In]

Integrate[(2*x^3 - 6*x^4 + E^((16 - 8*x - 9*x^2)/x^2)*(-96 + 24*x))/x^3,x]

[Out]

3*E^(-9 + 16/x^2 - 8/x) + 2*x - 3*x^2

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.04

method	result	size
parts	\(-3 x^{2}+2 x +3 \,{\mathrm e}^{\frac {-9 x^{2}-8 x +16}{x^{2}}}\)	\(27\)
risch	\(-3 x^{2}+2 x +3 \,{\mathrm e}^{-\frac {9 x^{2}+8 x -16}{x^{2}}}\)	\(28\)
parallelrisch	\(-3 x^{2}+2 x +3 \,{\mathrm e}^{-\frac {9 x^{2}+8 x -16}{x^{2}}}\)	\(28\)
norman	\(\frac {2 x^{3}-3 x^{4}+3 x^{2} {\mathrm e}^{\frac {-9 x^{2}-8 x +16}{x^{2}}}}{x^{2}}\)	\(36\)
derivativedivides	\(-3 x^{2}+2 x -3 i {\mathrm e}^{-9} \sqrt {\pi }\, {\mathrm e}^{-1} \operatorname {erf}\left (-\frac {4 i}{x}+i\right )+96 \,{\mathrm e}^{-9} \left (\frac {{\mathrm e}^{-\frac {8}{x}+\frac {16}{x^{2}}}}{32}+\frac {i \sqrt {\pi }\, {\mathrm e}^{-1} \operatorname {erf}\left (-\frac {4 i}{x}+i\right )}{32}\right )\)	\(67\)
default	\(-3 x^{2}+2 x -3 i {\mathrm e}^{-9} \sqrt {\pi }\, {\mathrm e}^{-1} \operatorname {erf}\left (-\frac {4 i}{x}+i\right )+96 \,{\mathrm e}^{-9} \left (\frac {{\mathrm e}^{-\frac {8}{x}+\frac {16}{x^{2}}}}{32}+\frac {i \sqrt {\pi }\, {\mathrm e}^{-1} \operatorname {erf}\left (-\frac {4 i}{x}+i\right )}{32}\right )\)	\(67\)

[In]

int(((24*x-96)*exp((-9*x^2-8*x+16)/x^2)-6*x^4+2*x^3)/x^3,x,method=_RETURNVERBOSE)

[Out]

-3*x^2+2*x+3*exp((-9*x^2-8*x+16)/x^2)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.04 \[ \int \frac {2 x^3-6 x^4+e^{\frac {16-8 x-9 x^2}{x^2}} (-96+24 x)}{x^3} \, dx=-3 \, x^{2} + 2 \, x + 3 \, e^{\left (-\frac {9 \, x^{2} + 8 \, x - 16}{x^{2}}\right )} \]

[In]

integrate(((24*x-96)*exp((-9*x^2-8*x+16)/x^2)-6*x^4+2*x^3)/x^3,x, algorithm="fricas")

[Out]

-3*x^2 + 2*x + 3*e^(-(9*x^2 + 8*x - 16)/x^2)

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \[ \int \frac {2 x^3-6 x^4+e^{\frac {16-8 x-9 x^2}{x^2}} (-96+24 x)}{x^3} \, dx=- 3 x^{2} + 2 x + 3 e^{\frac {- 9 x^{2} - 8 x + 16}{x^{2}}} \]

[In]

integrate(((24*x-96)*exp((-9*x**2-8*x+16)/x**2)-6*x**4+2*x**3)/x**3,x)

[Out]

-3*x**2 + 2*x + 3*exp((-9*x**2 - 8*x + 16)/x**2)

Maxima [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \[ \int \frac {2 x^3-6 x^4+e^{\frac {16-8 x-9 x^2}{x^2}} (-96+24 x)}{x^3} \, dx=-3 \, x^{2} + 2 \, x + 3 \, e^{\left (-\frac {8}{x} + \frac {16}{x^{2}} - 9\right )} \]

[In]

integrate(((24*x-96)*exp((-9*x^2-8*x+16)/x^2)-6*x^4+2*x^3)/x^3,x, algorithm="maxima")

[Out]

-3*x^2 + 2*x + 3*e^(-8/x + 16/x^2 - 9)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \[ \int \frac {2 x^3-6 x^4+e^{\frac {16-8 x-9 x^2}{x^2}} (-96+24 x)}{x^3} \, dx=-3 \, x^{2} + 2 \, x + 3 \, e^{\left (-\frac {8}{x} + \frac {16}{x^{2}} - 9\right )} \]

[In]

integrate(((24*x-96)*exp((-9*x^2-8*x+16)/x^2)-6*x^4+2*x^3)/x^3,x, algorithm="giac")

[Out]

-3*x^2 + 2*x + 3*e^(-8/x + 16/x^2 - 9)

Mupad [B] (verification not implemented)

Time = 9.39 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96 \[ \int \frac {2 x^3-6 x^4+e^{\frac {16-8 x-9 x^2}{x^2}} (-96+24 x)}{x^3} \, dx=2\,x-3\,x^2+3\,{\mathrm {e}}^{-9}\,{\mathrm {e}}^{-\frac {8}{x}}\,{\mathrm {e}}^{\frac {16}{x^2}} \]

[In]

int((exp(-(8*x + 9*x^2 - 16)/x^2)*(24*x - 96) + 2*x^3 - 6*x^4)/x^3,x)

[Out]

2*x - 3*x^2 + 3*exp(-9)*exp(-8/x)*exp(16/x^2)