3.63.56 \(\int \frac {e^{-\frac {5 x}{-1+5 x}} (3-45 x+75 x^2+e^{\frac {5 x}{-1+5 x}} (-x^2+10 x^3-25 x^4))}{x^2-10 x^3+25 x^4} \, dx\)

Optimal. Leaf size=22 \[ 6-\frac {3 e^{-\frac {x}{-\frac {1}{5}+x}}}{x}-x \]

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Rubi [A]  time = 1.36, antiderivative size = 21, normalized size of antiderivative = 0.95, number of steps used = 27, number of rules used = 11, integrand size = 68, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.162, Rules used = {1594, 27, 6688, 6742, 2232, 2231, 2230, 2210, 2233, 2178, 2209} \begin {gather*} -x-\frac {3 e^{\frac {5 x}{1-5 x}}}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(3 - 45*x + 75*x^2 + E^((5*x)/(-1 + 5*x))*(-x^2 + 10*x^3 - 25*x^4))/(E^((5*x)/(-1 + 5*x))*(x^2 - 10*x^3 +
25*x^4)),x]

[Out]

(-3*E^((5*x)/(1 - 5*x)))/x - x

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 1594

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^
(q - p) + c*x^(r - p))^n, x] /; FreeQ[{a, b, c, p, q, r}, x] && IntegerQ[n] && PosQ[q - p] && PosQ[r - p]

Rule 2178

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

Rule 2209

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[((e + f*x)^n*
F^(a + b*(c + d*x)^n))/(b*f*n*(c + d*x)^n*Log[F]), x] /; FreeQ[{F, a, b, c, d, e, f, n}, x] && EqQ[m, n - 1] &
& EqQ[d*e - c*f, 0]

Rule 2210

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> Simp[(F^a*ExpIntegralEi[
b*(c + d*x)^n*Log[F]])/(f*n), x] /; FreeQ[{F, a, b, c, d, e, f, n}, x] && EqQ[d*e - c*f, 0]

Rule 2230

Int[(F_)^((e_.) + ((f_.)*((a_.) + (b_.)*(x_)))/((c_.) + (d_.)*(x_)))*((g_.) + (h_.)*(x_))^(m_.), x_Symbol] :>
Int[(g + h*x)^m*F^((d*e + b*f)/d - (f*(b*c - a*d))/(d*(c + d*x))), x] /; FreeQ[{F, a, b, c, d, e, f, g, h, m},
x] && NeQ[b*c - a*d, 0] && EqQ[d*g - c*h, 0]

Rule 2231

Int[(F_)^((e_.) + ((f_.)*((a_.) + (b_.)*(x_)))/((c_.) + (d_.)*(x_)))/((g_.) + (h_.)*(x_)), x_Symbol] :> Dist[d
/h, Int[F^(e + (f*(a + b*x))/(c + d*x))/(c + d*x), x], x] - Dist[(d*g - c*h)/h, Int[F^(e + (f*(a + b*x))/(c +
d*x))/((c + d*x)*(g + h*x)), x], x] /; FreeQ[{F, a, b, c, d, e, f, g, h}, x] && NeQ[b*c - a*d, 0] && NeQ[d*g -
c*h, 0]

Rule 2232

Int[(F_)^((e_.) + ((f_.)*((a_.) + (b_.)*(x_)))/((c_.) + (d_.)*(x_)))*((g_.) + (h_.)*(x_))^(m_), x_Symbol] :> S
imp[((g + h*x)^(m + 1)*F^(e + (f*(a + b*x))/(c + d*x)))/(h*(m + 1)), x] - Dist[(f*(b*c - a*d)*Log[F])/(h*(m +
1)), Int[((g + h*x)^(m + 1)*F^(e + (f*(a + b*x))/(c + d*x)))/(c + d*x)^2, x], x] /; FreeQ[{F, a, b, c, d, e, f
, g, h}, x] && NeQ[b*c - a*d, 0] && NeQ[d*g - c*h, 0] && ILtQ[m, -1]

Rule 2233

Int[(F_)^((e_.) + ((f_.)*((a_.) + (b_.)*(x_)))/((c_.) + (d_.)*(x_)))/(((g_.) + (h_.)*(x_))*((i_.) + (j_.)*(x_)
)), x_Symbol] :> -Dist[d/(h*(d*i - c*j)), Subst[Int[F^(e + (f*(b*i - a*j))/(d*i - c*j) - ((b*c - a*d)*f*x)/(d*
i - c*j))/x, x], x, (i + j*x)/(c + d*x)], x] /; FreeQ[{F, a, b, c, d, e, f, g, h}, x] && EqQ[d*g - c*h, 0]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{-\frac {5 x}{-1+5 x}} \left (3-45 x+75 x^2+e^{\frac {5 x}{-1+5 x}} \left (-x^2+10 x^3-25 x^4\right )\right )}{x^2 \left (1-10 x+25 x^2\right )} \, dx\\ &=\int \frac {e^{-\frac {5 x}{-1+5 x}} \left (3-45 x+75 x^2+e^{\frac {5 x}{-1+5 x}} \left (-x^2+10 x^3-25 x^4\right )\right )}{x^2 (-1+5 x)^2} \, dx\\ &=\int \left (-1+\frac {e^{\frac {5 x}{1-5 x}} \left (3-45 x+75 x^2\right )}{(1-5 x)^2 x^2}\right ) \, dx\\ &=-x+\int \frac {e^{\frac {5 x}{1-5 x}} \left (3-45 x+75 x^2\right )}{(1-5 x)^2 x^2} \, dx\\ &=-x+\int \left (\frac {3 e^{\frac {5 x}{1-5 x}}}{x^2}-\frac {15 e^{\frac {5 x}{1-5 x}}}{x}-\frac {75 e^{\frac {5 x}{1-5 x}}}{(-1+5 x)^2}+\frac {75 e^{\frac {5 x}{1-5 x}}}{-1+5 x}\right ) \, dx\\ &=-x+3 \int \frac {e^{\frac {5 x}{1-5 x}}}{x^2} \, dx-15 \int \frac {e^{\frac {5 x}{1-5 x}}}{x} \, dx-75 \int \frac {e^{\frac {5 x}{1-5 x}}}{(-1+5 x)^2} \, dx+75 \int \frac {e^{\frac {5 x}{1-5 x}}}{-1+5 x} \, dx\\ &=-\frac {3 e^{\frac {5 x}{1-5 x}}}{x}-x+15 \int \frac {e^{\frac {5 x}{1-5 x}}}{(1-5 x)^2 x} \, dx-15 \int \frac {e^{\frac {5 x}{1-5 x}}}{(1-5 x) x} \, dx+75 \int \frac {e^{\frac {5 x}{1-5 x}}}{1-5 x} \, dx-75 \int \frac {e^{-1+\frac {1}{1-5 x}}}{(-1+5 x)^2} \, dx+75 \int \frac {e^{-1+\frac {1}{1-5 x}}}{-1+5 x} \, dx\\ &=-15 e^{-1+\frac {1}{1-5 x}}-\frac {3 e^{\frac {5 x}{1-5 x}}}{x}-x-\frac {15 \text {Ei}\left (\frac {1}{1-5 x}\right )}{e}+15 \int \left (\frac {e^{\frac {5 x}{1-5 x}}}{x}+\frac {5 e^{\frac {5 x}{1-5 x}}}{(-1+5 x)^2}-\frac {5 e^{\frac {5 x}{1-5 x}}}{-1+5 x}\right ) \, dx-15 \operatorname {Subst}\left (\int \frac {e^{5 x}}{x} \, dx,x,\frac {x}{1-5 x}\right )+75 \int \frac {e^{-1+\frac {1}{1-5 x}}}{1-5 x} \, dx\\ &=-15 e^{-1+\frac {1}{1-5 x}}-\frac {3 e^{\frac {5 x}{1-5 x}}}{x}-x-15 \text {Ei}\left (\frac {5 x}{1-5 x}\right )+15 \int \frac {e^{\frac {5 x}{1-5 x}}}{x} \, dx+75 \int \frac {e^{\frac {5 x}{1-5 x}}}{(-1+5 x)^2} \, dx-75 \int \frac {e^{\frac {5 x}{1-5 x}}}{-1+5 x} \, dx\\ &=-15 e^{-1+\frac {1}{1-5 x}}-\frac {3 e^{\frac {5 x}{1-5 x}}}{x}-x-15 \text {Ei}\left (\frac {5 x}{1-5 x}\right )+15 \int \frac {e^{\frac {5 x}{1-5 x}}}{(1-5 x) x} \, dx-75 \int \frac {e^{\frac {5 x}{1-5 x}}}{1-5 x} \, dx+75 \int \frac {e^{-1+\frac {1}{1-5 x}}}{(-1+5 x)^2} \, dx-75 \int \frac {e^{-1+\frac {1}{1-5 x}}}{-1+5 x} \, dx\\ &=-\frac {3 e^{\frac {5 x}{1-5 x}}}{x}-x+\frac {15 \text {Ei}\left (\frac {1}{1-5 x}\right )}{e}-15 \text {Ei}\left (\frac {5 x}{1-5 x}\right )+15 \operatorname {Subst}\left (\int \frac {e^{5 x}}{x} \, dx,x,\frac {x}{1-5 x}\right )-75 \int \frac {e^{-1+\frac {1}{1-5 x}}}{1-5 x} \, dx\\ &=-\frac {3 e^{\frac {5 x}{1-5 x}}}{x}-x\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.12, size = 21, normalized size = 0.95 \begin {gather*} -\frac {3 e^{\frac {5 x}{1-5 x}}}{x}-x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(3 - 45*x + 75*x^2 + E^((5*x)/(-1 + 5*x))*(-x^2 + 10*x^3 - 25*x^4))/(E^((5*x)/(-1 + 5*x))*(x^2 - 10*
x^3 + 25*x^4)),x]

[Out]

(-3*E^((5*x)/(1 - 5*x)))/x - x

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fricas [A]  time = 0.53, size = 33, normalized size = 1.50 \begin {gather*} -\frac {{\left (x^{2} e^{\left (\frac {5 \, x}{5 \, x - 1}\right )} + 3\right )} e^{\left (-\frac {5 \, x}{5 \, x - 1}\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-25*x^4+10*x^3-x^2)*exp(5*x/(5*x-1))+75*x^2-45*x+3)/(25*x^4-10*x^3+x^2)/exp(5*x/(5*x-1)),x, algori
thm="fricas")

[Out]

-(x^2*e^(5*x/(5*x - 1)) + 3)*e^(-5*x/(5*x - 1))/x

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giac [B]  time = 0.67, size = 94, normalized size = 4.27 \begin {gather*} -\frac {\frac {150 \, x e^{\left (-\frac {5 \, x}{5 \, x - 1}\right )}}{5 \, x - 1} - \frac {375 \, x^{2} e^{\left (-\frac {5 \, x}{5 \, x - 1}\right )}}{{\left (5 \, x - 1\right )}^{2}} - \frac {x}{5 \, x - 1} - 15 \, e^{\left (-\frac {5 \, x}{5 \, x - 1}\right )}}{5 \, {\left (\frac {x}{5 \, x - 1} - \frac {5 \, x^{2}}{{\left (5 \, x - 1\right )}^{2}}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-25*x^4+10*x^3-x^2)*exp(5*x/(5*x-1))+75*x^2-45*x+3)/(25*x^4-10*x^3+x^2)/exp(5*x/(5*x-1)),x, algori
thm="giac")

[Out]

-1/5*(150*x*e^(-5*x/(5*x - 1))/(5*x - 1) - 375*x^2*e^(-5*x/(5*x - 1))/(5*x - 1)^2 - x/(5*x - 1) - 15*e^(-5*x/(
5*x - 1)))/(x/(5*x - 1) - 5*x^2/(5*x - 1)^2)

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maple [A]  time = 0.17, size = 21, normalized size = 0.95

method	result	size
risch	\(-x -\frac {3 \,{\mathrm e}^{-\frac {5 x}{5 x -1}}}{x}\)	\(21\)
derivativedivides	\(-x +\frac {1}{5}+\frac {15 \,{\mathrm e}^{-1-\frac {1}{5 x -1}}}{1+\frac {1}{5 x -1}}-15 \expIntegralEi \left (1, 1+\frac {1}{5 x -1}\right )\)	\(44\)
default	\(-x +\frac {1}{5}+\frac {15 \,{\mathrm e}^{-1-\frac {1}{5 x -1}}}{1+\frac {1}{5 x -1}}-15 \expIntegralEi \left (1, 1+\frac {1}{5 x -1}\right )\)	\(44\)
norman	\(\frac {\left (3+x^{2} {\mathrm e}^{\frac {5 x}{5 x -1}}-15 x -5 \,{\mathrm e}^{\frac {5 x}{5 x -1}} x^{3}\right ) {\mathrm e}^{-\frac {5 x}{5 x -1}}}{x \left (5 x -1\right )}\)	\(61\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-25*x^4+10*x^3-x^2)*exp(5*x/(5*x-1))+75*x^2-45*x+3)/(25*x^4-10*x^3+x^2)/exp(5*x/(5*x-1)),x,method=_RETUR
NVERBOSE)

[Out]

-x-3/x*exp(-5*x/(5*x-1))

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maxima [A]  time = 0.42, size = 26, normalized size = 1.18 \begin {gather*} -\frac {{\left (x^{2} e + 3 \, e^{\left (-\frac {1}{5 \, x - 1}\right )}\right )} e^{\left (-1\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-25*x^4+10*x^3-x^2)*exp(5*x/(5*x-1))+75*x^2-45*x+3)/(25*x^4-10*x^3+x^2)/exp(5*x/(5*x-1)),x, algori
thm="maxima")

[Out]

-(x^2*e + 3*e^(-1/(5*x - 1)))*e^(-1)/x

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mupad [B]  time = 0.19, size = 20, normalized size = 0.91 \begin {gather*} -x-\frac {3\,{\mathrm {e}}^{-\frac {5\,x}{5\,x-1}}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-(5*x)/(5*x - 1))*(45*x + exp((5*x)/(5*x - 1))*(x^2 - 10*x^3 + 25*x^4) - 75*x^2 - 3))/(x^2 - 10*x^3
+ 25*x^4),x)

[Out]

- x - (3*exp(-(5*x)/(5*x - 1)))/x

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sympy [A]  time = 0.16, size = 15, normalized size = 0.68 \begin {gather*} - x - \frac {3 e^{- \frac {5 x}{5 x - 1}}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-25*x**4+10*x**3-x**2)*exp(5*x/(5*x-1))+75*x**2-45*x+3)/(25*x**4-10*x**3+x**2)/exp(5*x/(5*x-1)),x)

[Out]

-x - 3*exp(-5*x/(5*x - 1))/x

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