\(\int \frac {128 x^2+64 x^3-8 x^6+e^3 (-256-192 x-32 x^2)}{x^5} \, dx\) [2871]

Optimal result

Integrand size = 34, antiderivative size = 26 \[ \int \frac {128 x^2+64 x^3-8 x^6+e^3 \left (-256-192 x-32 x^2\right )}{x^5} \, dx=4 \left (\frac {e^3}{x}-x\right ) \left (x+\frac {(4+2 x)^2}{x^3}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.50, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.029, Rules used = {14} \[ \int \frac {128 x^2+64 x^3-8 x^6+e^3 \left (-256-192 x-32 x^2\right )}{x^5} \, dx=\frac {64 e^3}{x^4}+\frac {64 e^3}{x^3}-4 x^2-\frac {16 \left (4-e^3\right )}{x^2}-\frac {64}{x} \]

[In]

Int[(128*x^2 + 64*x^3 - 8*x^6 + E^3*(-256 - 192*x - 32*x^2))/x^5,x]

[Out]

(64*E^3)/x^4 + (64*E^3)/x^3 - (16*(4 - E^3))/x^2 - 64/x - 4*x^2

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]]

Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {256 e^3}{x^5}-\frac {192 e^3}{x^4}-\frac {32 \left (-4+e^3\right )}{x^3}+\frac {64}{x^2}-8 x\right ) \, dx \\ & = \frac {64 e^3}{x^4}+\frac {64 e^3}{x^3}-\frac {16 \left (4-e^3\right )}{x^2}-\frac {64}{x}-4 x^2 \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.08 \[ \int \frac {128 x^2+64 x^3-8 x^6+e^3 \left (-256-192 x-32 x^2\right )}{x^5} \, dx=-\frac {4 \left (-4 e^3 (2+x)^2+x^2 \left (16+16 x+x^4\right )\right )}{x^4} \]

[In]

Integrate[(128*x^2 + 64*x^3 - 8*x^6 + E^3*(-256 - 192*x - 32*x^2))/x^5,x]

[Out]

(-4*(-4*E^3*(2 + x)^2 + x^2*(16 + 16*x + x^4)))/x^4

Maple [A] (verified)

Time = 0.14 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.35

method	result	size
norman	\(\frac {\left (16 \,{\mathrm e}^{3}-64\right ) x^{2}-64 x^{3}-4 x^{6}+64 x \,{\mathrm e}^{3}+64 \,{\mathrm e}^{3}}{x^{4}}\)	\(35\)
risch	\(-4 x^{2}+\frac {-64 x^{3}+\left (16 \,{\mathrm e}^{3}-64\right ) x^{2}+64 x \,{\mathrm e}^{3}+64 \,{\mathrm e}^{3}}{x^{4}}\)	\(36\)
default	\(-4 x^{2}+\frac {64 \,{\mathrm e}^{3}}{x^{4}}-\frac {64}{x}-\frac {4 \left (-4 \,{\mathrm e}^{3}+16\right )}{x^{2}}+\frac {64 \,{\mathrm e}^{3}}{x^{3}}\)	\(37\)
parallelrisch	\(\frac {-4 x^{6}+16 x^{2} {\mathrm e}^{3}-64 x^{3}+64 x \,{\mathrm e}^{3}-64 x^{2}+64 \,{\mathrm e}^{3}}{x^{4}}\)	\(37\)
gosper	\(\frac {-4 x^{6}+16 x^{2} {\mathrm e}^{3}-64 x^{3}+64 x \,{\mathrm e}^{3}-64 x^{2}+64 \,{\mathrm e}^{3}}{x^{4}}\)	\(38\)

[In]

int(((-32*x^2-192*x-256)*exp(3)-8*x^6+64*x^3+128*x^2)/x^5,x,method=_RETURNVERBOSE)

[Out]

((16*exp(3)-64)*x^2-64*x^3-4*x^6+64*x*exp(3)+64*exp(3))/x^4

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.19 \[ \int \frac {128 x^2+64 x^3-8 x^6+e^3 \left (-256-192 x-32 x^2\right )}{x^5} \, dx=-\frac {4 \, {\left (x^{6} + 16 \, x^{3} + 16 \, x^{2} - 4 \, {\left (x^{2} + 4 \, x + 4\right )} e^{3}\right )}}{x^{4}} \]

[In]

integrate(((-32*x^2-192*x-256)*exp(3)-8*x^6+64*x^3+128*x^2)/x^5,x, algorithm="fricas")

[Out]

-4*(x^6 + 16*x^3 + 16*x^2 - 4*(x^2 + 4*x + 4)*e^3)/x^4

Sympy [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.38 \[ \int \frac {128 x^2+64 x^3-8 x^6+e^3 \left (-256-192 x-32 x^2\right )}{x^5} \, dx=- 4 x^{2} - \frac {64 x^{3} + x^{2} \cdot \left (64 - 16 e^{3}\right ) - 64 x e^{3} - 64 e^{3}}{x^{4}} \]

[In]

integrate(((-32*x**2-192*x-256)*exp(3)-8*x**6+64*x**3+128*x**2)/x**5,x)

[Out]

-4*x**2 - (64*x**3 + x**2*(64 - 16*exp(3)) - 64*x*exp(3) - 64*exp(3))/x**4

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.35 \[ \int \frac {128 x^2+64 x^3-8 x^6+e^3 \left (-256-192 x-32 x^2\right )}{x^5} \, dx=-4 \, x^{2} - \frac {16 \, {\left (4 \, x^{3} - x^{2} {\left (e^{3} - 4\right )} - 4 \, x e^{3} - 4 \, e^{3}\right )}}{x^{4}} \]

[In]

integrate(((-32*x^2-192*x-256)*exp(3)-8*x^6+64*x^3+128*x^2)/x^5,x, algorithm="maxima")

[Out]

-4*x^2 - 16*(4*x^3 - x^2*(e^3 - 4) - 4*x*e^3 - 4*e^3)/x^4

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.46 \[ \int \frac {128 x^2+64 x^3-8 x^6+e^3 \left (-256-192 x-32 x^2\right )}{x^5} \, dx=-4 \, x^{2} - \frac {16 \, {\left (4 \, x^{3} - x^{2} e^{3} + 4 \, x^{2} - 4 \, x e^{3} - 4 \, e^{3}\right )}}{x^{4}} \]

[In]

integrate(((-32*x^2-192*x-256)*exp(3)-8*x^6+64*x^3+128*x^2)/x^5,x, algorithm="giac")

[Out]

-4*x^2 - 16*(4*x^3 - x^2*e^3 + 4*x^2 - 4*x*e^3 - 4*e^3)/x^4

Mupad [B] (verification not implemented)

Time = 9.39 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.35 \[ \int \frac {128 x^2+64 x^3-8 x^6+e^3 \left (-256-192 x-32 x^2\right )}{x^5} \, dx=\frac {-64\,x^3+\left (16\,{\mathrm {e}}^3-64\right )\,x^2+64\,{\mathrm {e}}^3\,x+64\,{\mathrm {e}}^3}{x^4}-4\,x^2 \]

[In]

int(-(exp(3)*(192*x + 32*x^2 + 256) - 128*x^2 - 64*x^3 + 8*x^6)/x^5,x)

[Out]

(64*exp(3) + 64*x*exp(3) + x^2*(16*exp(3) - 64) - 64*x^3)/x^4 - 4*x^2