∫ −5x5+e2x(−16+8x−20x2+20x3)_______________________

Optimal. Leaf size=23 \[ \frac {2 e^{2 x} \left (5+\frac {2}{x^2}\right )}{5 x^2}-x \]

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Rubi [A] time = 0.22, antiderivative size = 26, normalized size of antiderivative = 1.13, number of steps used = 19, number of rules used = 5, integrand size = 34, $\frac {\text {number of rules}}{\text {integrand size}}$ = 0.147, Rules used = {12, 14, 2199, 2177, 2178} \begin {gather*} \frac {4 e^{2 x}}{5 x^4}+\frac {2 e^{2 x}}{x^2}-x \end {gather*}

Int[(-5*x^5 + E^(2*x)*(-16 + 8*x - 20*x^2 + 20*x^3))/(5*x^5),x]

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] && !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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]]

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[((c + d*x)^(m

+ 1)*(b*F^(g*(e + f*x)))^n)/(d*(m + 1)), x] - Dist[(f*g*n*Log[F])/(d*(m + 1)), Int[(c + d*x)^(m + 1)*(b*F^(g*

(e + f*x)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && LtQ[m, -1] && IntegerQ[2*m] && !$UseGamma ===

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - (c*f)/d))*ExpIntegral

Ei[(f*g*(c + d*x)*Log[F])/d])/d, x] /; FreeQ[{F, c, d, e, f, g}, x] && !$UseGamma === True

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> Int[ExpandIntegrand[F^(c*ExpandToSum[v, x]), w*NormalizePo

werOfLinear[u, x]^m, x], x] /; FreeQ[{F, c}, x] && PolynomialQ[w, x] && LinearQ[v, x] && PowerOfLinearQ[u, x]

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{5} \int \frac {-5 x^5+e^{2 x} \left (-16+8 x-20 x^2+20 x^3\right )}{x^5} \, dx\\ &=\frac {1}{5} \int \left (-5+\frac {4 e^{2 x} \left (-4+2 x-5 x^2+5 x^3\right )}{x^5}\right ) \, dx\\ &=-x+\frac {4}{5} \int \frac {e^{2 x} \left (-4+2 x-5 x^2+5 x^3\right )}{x^5} \, dx\\ &=-x+\frac {4}{5} \int \left (-\frac {4 e^{2 x}}{x^5}+\frac {2 e^{2 x}}{x^4}-\frac {5 e^{2 x}}{x^3}+\frac {5 e^{2 x}}{x^2}\right ) \, dx\\ &=-x+\frac {8}{5} \int \frac {e^{2 x}}{x^4} \, dx-\frac {16}{5} \int \frac {e^{2 x}}{x^5} \, dx-4 \int \frac {e^{2 x}}{x^3} \, dx+4 \int \frac {e^{2 x}}{x^2} \, dx\\ &=\frac {4 e^{2 x}}{5 x^4}-\frac {8 e^{2 x}}{15 x^3}+\frac {2 e^{2 x}}{x^2}-\frac {4 e^{2 x}}{x}-x+\frac {16}{15} \int \frac {e^{2 x}}{x^3} \, dx-\frac {8}{5} \int \frac {e^{2 x}}{x^4} \, dx-4 \int \frac {e^{2 x}}{x^2} \, dx+8 \int \frac {e^{2 x}}{x} \, dx\\ &=\frac {4 e^{2 x}}{5 x^4}+\frac {22 e^{2 x}}{15 x^2}-x+8 \text {Ei}(2 x)-\frac {16}{15} \int \frac {e^{2 x}}{x^3} \, dx+\frac {16}{15} \int \frac {e^{2 x}}{x^2} \, dx-8 \int \frac {e^{2 x}}{x} \, dx\\ &=\frac {4 e^{2 x}}{5 x^4}+\frac {2 e^{2 x}}{x^2}-\frac {16 e^{2 x}}{15 x}-x-\frac {16}{15} \int \frac {e^{2 x}}{x^2} \, dx+\frac {32}{15} \int \frac {e^{2 x}}{x} \, dx\\ &=\frac {4 e^{2 x}}{5 x^4}+\frac {2 e^{2 x}}{x^2}-x+\frac {32 \text {Ei}(2 x)}{15}-\frac {32}{15} \int \frac {e^{2 x}}{x} \, dx\\ &=\frac {4 e^{2 x}}{5 x^4}+\frac {2 e^{2 x}}{x^2}-x\\ \end {aligned} \end {gather*}

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Mathematica [A] time = 0.02, size = 25, normalized size = 1.09 \begin {gather*} -x+\frac {2 e^{2 x} \left (2 x+5 x^3\right )}{5 x^5} \end {gather*}

Integrate[(-5*x^5 + E^(2*x)*(-16 + 8*x - 20*x^2 + 20*x^3))/(5*x^5),x]

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fricas [A] time = 0.84, size = 24, normalized size = 1.04 \begin {gather*} -\frac {5 \, x^{5} - 2 \, {\left (5 \, x^{2} + 2\right )} e^{\left (2 \, x\right )}}{5 \, x^{4}} \end {gather*}

integrate(1/5*((20*x^3-20*x^2+8*x-16)*exp(2*x)-5*x^5)/x^5,x, algorithm="fricas")

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giac [A] time = 0.16, size = 26, normalized size = 1.13 \begin {gather*} -\frac {5 \, x^{5} - 10 \, x^{2} e^{\left (2 \, x\right )} - 4 \, e^{\left (2 \, x\right )}}{5 \, x^{4}} \end {gather*}

integrate(1/5*((20*x^3-20*x^2+8*x-16)*exp(2*x)-5*x^5)/x^5,x, algorithm="giac")

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method	result	size

risch	$-x +\frac {2 \left (5 x^{2}+2\right ) {\mathrm e}^{2 x}}{5 x^{4}}$	$21$
derivativedivides	$-x +\frac {4 \,{\mathrm e}^{2 x}}{5 x^{4}}+\frac {2 \,{\mathrm e}^{2 x}}{x^{2}}$	$23$
default	$-x +\frac {4 \,{\mathrm e}^{2 x}}{5 x^{4}}+\frac {2 \,{\mathrm e}^{2 x}}{x^{2}}$	$23$

int(1/5*((20*x^3-20*x^2+8*x-16)*exp(2*x)-5*x^5)/x^5,x,method=_RETURNVERBOSE)

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maxima [C] time = 0.42, size = 32, normalized size = 1.39 \begin {gather*} -x + 8 \, \Gamma \left (-1, -2 \, x\right ) + 16 \, \Gamma \left (-2, -2 \, x\right ) + \frac {64}{5} \, \Gamma \left (-3, -2 \, x\right ) + \frac {256}{5} \, \Gamma \left (-4, -2 \, x\right ) \end {gather*}

integrate(1/5*((20*x^3-20*x^2+8*x-16)*exp(2*x)-5*x^5)/x^5,x, algorithm="maxima")

-x + 8*gamma(-1, -2*x) + 16*gamma(-2, -2*x) + 64/5*gamma(-3, -2*x) + 256/5*gamma(-4, -2*x)

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mupad [B] time = 0.07, size = 24, normalized size = 1.04 \begin {gather*} \frac {\frac {4\,{\mathrm {e}}^{2\,x}}{5}+2\,x^2\,{\mathrm {e}}^{2\,x}}{x^4}-x \end {gather*}

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sympy [A] time = 0.11, size = 17, normalized size = 0.74 \begin {gather*} - x + \frac {\left (10 x^{2} + 4\right ) e^{2 x}}{5 x^{4}} \end {gather*}

integrate(1/5*((20*x**3-20*x**2+8*x-16)*exp(2*x)-5*x**5)/x**5,x)

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3.38.46 \(\int \frac {-5 x^5+e^{2 x} (-16+8 x-20 x^2+20 x^3)}{5 x^5} \, dx\)