\(\int \frac {-5 x^5+e^{2 x} (-16+8 x-20 x^2+20 x^3)}{5 x^5} \, dx\) [3746]

Optimal result

Integrand size = 34, antiderivative size = 23 \[ \int \frac {-5 x^5+e^{2 x} \left (-16+8 x-20 x^2+20 x^3\right )}{5 x^5} \, dx=\frac {2 e^{2 x} \left (5+\frac {2}{x^2}\right )}{5 x^2}-x \]

[Out]

2/5*exp(2*x)*(5+2/x^2)/x^2-x

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.13, number of steps used = 19, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.147, Rules used = {12, 14, 2230, 2208, 2209} \[ \int \frac {-5 x^5+e^{2 x} \left (-16+8 x-20 x^2+20 x^3\right )}{5 x^5} \, dx=\frac {4 e^{2 x}}{5 x^4}+\frac {2 e^{2 x}}{x^2}-x \]

[In]

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

[Out]

(4*E^(2*x))/(5*x^4) + (2*E^(2*x))/x^2 - x

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 2208

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] &&  !TrueQ[$UseGamm
a]

Rule 2209

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - c*(f/d)))/d)*ExpInteg
ralEi[f*g*(c + d*x)*(Log[F]/d)], x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !TrueQ[$UseGamma]

Rule 2230

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]
&& IntegerQ[m] &&  !TrueQ[$UseGamma]

Rubi steps \begin{align*} \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{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.09 \[ \int \frac {-5 x^5+e^{2 x} \left (-16+8 x-20 x^2+20 x^3\right )}{5 x^5} \, dx=-x+\frac {2 e^{2 x} \left (2 x+5 x^3\right )}{5 x^5} \]

[In]

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

[Out]

-x + (2*E^(2*x)*(2*x + 5*x^3))/(5*x^5)

Maple [A] (verified)

Time = 1.93 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91

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	\(-x +\frac {4 \,{\mathrm e}^{2 x}}{5 x^{4}}+\frac {2 \,{\mathrm e}^{2 x}}{x^{2}}\)	\(23\)
parts	\(-x +\frac {4 \,{\mathrm e}^{2 x}}{5 x^{4}}+\frac {2 \,{\mathrm e}^{2 x}}{x^{2}}\)	\(23\)
parallelrisch	\(-\frac {5 x^{5}-10 \,{\mathrm e}^{2 x} x^{2}-4 \,{\mathrm e}^{2 x}}{5 x^{4}}\)	\(27\)

[In]

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

[Out]

-x+2/5*(5*x^2+2)/x^4*exp(2*x)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int \frac {-5 x^5+e^{2 x} \left (-16+8 x-20 x^2+20 x^3\right )}{5 x^5} \, dx=-\frac {5 \, x^{5} - 2 \, {\left (5 \, x^{2} + 2\right )} e^{\left (2 \, x\right )}}{5 \, x^{4}} \]

[In]

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

[Out]

-1/5*(5*x^5 - 2*(5*x^2 + 2)*e^(2*x))/x^4

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.74 \[ \int \frac {-5 x^5+e^{2 x} \left (-16+8 x-20 x^2+20 x^3\right )}{5 x^5} \, dx=- x + \frac {\left (10 x^{2} + 4\right ) e^{2 x}}{5 x^{4}} \]

[In]

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

[Out]

-x + (10*x**2 + 4)*exp(2*x)/(5*x**4)

Maxima [C] (verification not implemented)

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

Time = 0.21 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.39 \[ \int \frac {-5 x^5+e^{2 x} \left (-16+8 x-20 x^2+20 x^3\right )}{5 x^5} \, dx=-x + 8 \, \Gamma \left (-1, -2 \, x\right ) + 16 \, \Gamma \left (-2, -2 \, x\right ) + \frac {64}{5} \, \Gamma \left (-3, -2 \, x\right ) + \frac {256}{5} \, \Gamma \left (-4, -2 \, x\right ) \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.13 \[ \int \frac {-5 x^5+e^{2 x} \left (-16+8 x-20 x^2+20 x^3\right )}{5 x^5} \, dx=-\frac {5 \, x^{5} - 10 \, x^{2} e^{\left (2 \, x\right )} - 4 \, e^{\left (2 \, x\right )}}{5 \, x^{4}} \]

[In]

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

[Out]

-1/5*(5*x^5 - 10*x^2*e^(2*x) - 4*e^(2*x))/x^4

Mupad [B] (verification not implemented)

Time = 8.74 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int \frac {-5 x^5+e^{2 x} \left (-16+8 x-20 x^2+20 x^3\right )}{5 x^5} \, dx=\frac {\frac {4\,{\mathrm {e}}^{2\,x}}{5}+2\,x^2\,{\mathrm {e}}^{2\,x}}{x^4}-x \]

[In]

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

[Out]

((4*exp(2*x))/5 + 2*x^2*exp(2*x))/x^4 - x