\(\int \frac {8 x^4+e^{\frac {-1-51 x+4 x^3}{4 x}} (4-17 x+32 x^3-8 x^4)}{4 x^3} \, dx\) [3196]

Optimal result

Integrand size = 48, antiderivative size = 31 \[ \int \frac {8 x^4+e^{\frac {-1-51 x+4 x^3}{4 x}} \left (4-17 x+32 x^3-8 x^4\right )}{4 x^3} \, dx=3+\frac {e^{-13+\frac {-1+x}{4 x}+x^2} (4-x)}{x}+x^2 \]

[Out]

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

Rubi [F]

\[ \int \frac {8 x^4+e^{\frac {-1-51 x+4 x^3}{4 x}} \left (4-17 x+32 x^3-8 x^4\right )}{4 x^3} \, dx=\int \frac {8 x^4+e^{\frac {-1-51 x+4 x^3}{4 x}} \left (4-17 x+32 x^3-8 x^4\right )}{4 x^3} \, dx \]

[In]

Int[(8*x^4 + E^((-1 - 51*x + 4*x^3)/(4*x))*(4 - 17*x + 32*x^3 - 8*x^4))/(4*x^3),x]

[Out]

x^2 + 8*Defer[Int][E^(-51/4 - 1/(4*x) + x^2), x] + Defer[Int][E^(-51/4 - 1/(4*x) + x^2)/x^3, x] - (17*Defer[In
t][E^(-51/4 - 1/(4*x) + x^2)/x^2, x])/4 - 2*Defer[Int][E^(-51/4 - 1/(4*x) + x^2)*x, x]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{4} \int \frac {8 x^4+e^{\frac {-1-51 x+4 x^3}{4 x}} \left (4-17 x+32 x^3-8 x^4\right )}{x^3} \, dx \\ & = \frac {1}{4} \int \left (8 x-\frac {e^{-\frac {51}{4}-\frac {1}{4 x}+x^2} \left (-4+17 x-32 x^3+8 x^4\right )}{x^3}\right ) \, dx \\ & = x^2-\frac {1}{4} \int \frac {e^{-\frac {51}{4}-\frac {1}{4 x}+x^2} \left (-4+17 x-32 x^3+8 x^4\right )}{x^3} \, dx \\ & = x^2-\frac {1}{4} \int \left (-32 e^{-\frac {51}{4}-\frac {1}{4 x}+x^2}-\frac {4 e^{-\frac {51}{4}-\frac {1}{4 x}+x^2}}{x^3}+\frac {17 e^{-\frac {51}{4}-\frac {1}{4 x}+x^2}}{x^2}+8 e^{-\frac {51}{4}-\frac {1}{4 x}+x^2} x\right ) \, dx \\ & = x^2-2 \int e^{-\frac {51}{4}-\frac {1}{4 x}+x^2} x \, dx-\frac {17}{4} \int \frac {e^{-\frac {51}{4}-\frac {1}{4 x}+x^2}}{x^2} \, dx+8 \int e^{-\frac {51}{4}-\frac {1}{4 x}+x^2} \, dx+\int \frac {e^{-\frac {51}{4}-\frac {1}{4 x}+x^2}}{x^3} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.26 \[ \int \frac {8 x^4+e^{\frac {-1-51 x+4 x^3}{4 x}} \left (4-17 x+32 x^3-8 x^4\right )}{4 x^3} \, dx=-\frac {1}{4} e^{-\frac {1}{4 x}+x^2} \left (\frac {4}{e^{51/4}}-\frac {16}{e^{51/4} x}\right )+x^2 \]

[In]

Integrate[(8*x^4 + E^((-1 - 51*x + 4*x^3)/(4*x))*(4 - 17*x + 32*x^3 - 8*x^4))/(4*x^3),x]

[Out]

-1/4*(E^(-1/4*1/x + x^2)*(4/E^(51/4) - 16/(E^(51/4)*x))) + x^2

Maple [A] (verified)

Time = 0.91 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.94

method	result	size
risch	\(x^{2}-\frac {\left (x -4\right ) {\mathrm e}^{\frac {4 x^{3}-51 x -1}{4 x}}}{x}\)	\(29\)
norman	\(\frac {x^{4}+4 \,{\mathrm e}^{\frac {4 x^{3}-51 x -1}{4 x}} x -{\mathrm e}^{\frac {4 x^{3}-51 x -1}{4 x}} x^{2}}{x^{2}}\)	\(49\)
parallelrisch	\(\frac {4 x^{3}-4 \,{\mathrm e}^{\frac {4 x^{3}-51 x -1}{4 x}} x +16 \,{\mathrm e}^{\frac {4 x^{3}-51 x -1}{4 x}}}{4 x}\)	\(49\)
parts	\(x^{2}+\frac {4 \,{\mathrm e}^{\frac {4 x^{3}-51 x -1}{4 x}} x -{\mathrm e}^{\frac {4 x^{3}-51 x -1}{4 x}} x^{2}}{x^{2}}\)	\(50\)

[In]

int(1/4*((-8*x^4+32*x^3-17*x+4)*exp(1/4*(4*x^3-51*x-1)/x)+8*x^4)/x^3,x,method=_RETURNVERBOSE)

[Out]

x^2-(x-4)/x*exp(1/4*(4*x^3-51*x-1)/x)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.94 \[ \int \frac {8 x^4+e^{\frac {-1-51 x+4 x^3}{4 x}} \left (4-17 x+32 x^3-8 x^4\right )}{4 x^3} \, dx=\frac {x^{3} - {\left (x - 4\right )} e^{\left (\frac {4 \, x^{3} - 51 \, x - 1}{4 \, x}\right )}}{x} \]

[In]

integrate(1/4*((-8*x^4+32*x^3-17*x+4)*exp(1/4*(4*x^3-51*x-1)/x)+8*x^4)/x^3,x, algorithm="fricas")

[Out]

(x^3 - (x - 4)*e^(1/4*(4*x^3 - 51*x - 1)/x))/x

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.71 \[ \int \frac {8 x^4+e^{\frac {-1-51 x+4 x^3}{4 x}} \left (4-17 x+32 x^3-8 x^4\right )}{4 x^3} \, dx=x^{2} + \frac {\left (4 - x\right ) e^{\frac {x^{3} - \frac {51 x}{4} - \frac {1}{4}}{x}}}{x} \]

[In]

integrate(1/4*((-8*x**4+32*x**3-17*x+4)*exp(1/4*(4*x**3-51*x-1)/x)+8*x**4)/x**3,x)

[Out]

x**2 + (4 - x)*exp((x**3 - 51*x/4 - 1/4)/x)/x

Maxima [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.94 \[ \int \frac {8 x^4+e^{\frac {-1-51 x+4 x^3}{4 x}} \left (4-17 x+32 x^3-8 x^4\right )}{4 x^3} \, dx=x^{2} - \frac {{\left (x e^{\frac {1}{4}} - 4 \, e^{\frac {1}{4}}\right )} e^{\left (x^{2} - \frac {1}{4 \, x} - 13\right )}}{x} \]

[In]

integrate(1/4*((-8*x^4+32*x^3-17*x+4)*exp(1/4*(4*x^3-51*x-1)/x)+8*x^4)/x^3,x, algorithm="maxima")

[Out]

x^2 - (x*e^(1/4) - 4*e^(1/4))*e^(x^2 - 1/4/x - 13)/x

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.45 \[ \int \frac {8 x^4+e^{\frac {-1-51 x+4 x^3}{4 x}} \left (4-17 x+32 x^3-8 x^4\right )}{4 x^3} \, dx=\frac {x^{3} - x e^{\left (\frac {4 \, x^{3} - 51 \, x - 1}{4 \, x}\right )} + 4 \, e^{\left (\frac {4 \, x^{3} - 51 \, x - 1}{4 \, x}\right )}}{x} \]

[In]

integrate(1/4*((-8*x^4+32*x^3-17*x+4)*exp(1/4*(4*x^3-51*x-1)/x)+8*x^4)/x^3,x, algorithm="giac")

[Out]

(x^3 - x*e^(1/4*(4*x^3 - 51*x - 1)/x) + 4*e^(1/4*(4*x^3 - 51*x - 1)/x))/x

Mupad [B] (verification not implemented)

Time = 9.24 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.06 \[ \int \frac {8 x^4+e^{\frac {-1-51 x+4 x^3}{4 x}} \left (4-17 x+32 x^3-8 x^4\right )}{4 x^3} \, dx=\frac {4\,{\mathrm {e}}^{x^2-\frac {1}{4\,x}-\frac {51}{4}}}{x}-{\mathrm {e}}^{x^2-\frac {1}{4\,x}-\frac {51}{4}}+x^2 \]

[In]

int(-((exp(-((51*x)/4 - x^3 + 1/4)/x)*(17*x - 32*x^3 + 8*x^4 - 4))/4 - 2*x^4)/x^3,x)

[Out]

(4*exp(x^2 - 1/(4*x) - 51/4))/x - exp(x^2 - 1/(4*x) - 51/4) + x^2