3.37.18 \(\int \frac {145 x+50 x^2+5 x^3+e^{-x} x (-25+15 x+9 x^2+x^3)}{100 x+40 x^2+4 x^3} \, dx\)

Optimal. Leaf size=25 \[ -10+x+\frac {x}{5+x}+\frac {1}{4} \left (-2+x-e^{-x} x\right ) \]

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Rubi [A]  time = 0.26, antiderivative size = 36, normalized size of antiderivative = 1.44, number of steps used = 9, number of rules used = 7, integrand size = 51, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.137, Rules used = {1594, 27, 12, 6688, 2176, 2194, 683} \begin {gather*} \frac {1}{4} e^{-x} (1-x)-\frac {e^{-x}}{4}+\frac {5 x}{4}-\frac {5}{x+5} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(145*x + 50*x^2 + 5*x^3 + (x*(-25 + 15*x + 9*x^2 + x^3))/E^x)/(100*x + 40*x^2 + 4*x^3),x]

[Out]

-1/4*1/E^x + (1 - x)/(4*E^x) + (5*x)/4 - 5/(5 + 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 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 683

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d +
e*x)^m*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[2*c*d - b*e,
0] && IGtQ[p, 0] &&  !(EqQ[m, 3] && NeQ[p, 1])

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 2176

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^m
*(b*F^(g*(e + f*x)))^n)/(f*g*n*Log[F]), x] - Dist[(d*m)/(f*g*n*Log[F]), 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] && GtQ[m, 0] && IntegerQ[2*m] &&  !$UseGamma === True

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 6688

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {145 x+50 x^2+5 x^3+e^{-x} x \left (-25+15 x+9 x^2+x^3\right )}{x \left (100+40 x+4 x^2\right )} \, dx\\ &=\int \frac {145 x+50 x^2+5 x^3+e^{-x} x \left (-25+15 x+9 x^2+x^3\right )}{4 x (5+x)^2} \, dx\\ &=\frac {1}{4} \int \frac {145 x+50 x^2+5 x^3+e^{-x} x \left (-25+15 x+9 x^2+x^3\right )}{x (5+x)^2} \, dx\\ &=\frac {1}{4} \int \left (e^{-x} (-1+x)+\frac {5 \left (29+10 x+x^2\right )}{(5+x)^2}\right ) \, dx\\ &=\frac {1}{4} \int e^{-x} (-1+x) \, dx+\frac {5}{4} \int \frac {29+10 x+x^2}{(5+x)^2} \, dx\\ &=\frac {1}{4} e^{-x} (1-x)+\frac {1}{4} \int e^{-x} \, dx+\frac {5}{4} \int \left (1+\frac {4}{(5+x)^2}\right ) \, dx\\ &=-\frac {e^{-x}}{4}+\frac {1}{4} e^{-x} (1-x)+\frac {5 x}{4}-\frac {5}{5+x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 23, normalized size = 0.92 \begin {gather*} \frac {1}{4} \left (5 x-e^{-x} x-\frac {20}{5+x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(145*x + 50*x^2 + 5*x^3 + (x*(-25 + 15*x + 9*x^2 + x^3))/E^x)/(100*x + 40*x^2 + 4*x^3),x]

[Out]

(5*x - x/E^x - 20/(5 + x))/4

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fricas [A]  time = 0.54, size = 29, normalized size = 1.16 \begin {gather*} \frac {5 \, x^{2} - {\left (x + 5\right )} e^{\left (-x + \log \relax (x)\right )} + 25 \, x - 20}{4 \, {\left (x + 5\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^3+9*x^2+15*x-25)*exp(log(x)-x)+5*x^3+50*x^2+145*x)/(4*x^3+40*x^2+100*x),x, algorithm="fricas")

[Out]

1/4*(5*x^2 - (x + 5)*e^(-x + log(x)) + 25*x - 20)/(x + 5)

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giac [A]  time = 0.14, size = 32, normalized size = 1.28 \begin {gather*} -\frac {x^{2} e^{\left (-x\right )} - 5 \, x^{2} + 5 \, x e^{\left (-x\right )} - 25 \, x + 20}{4 \, {\left (x + 5\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^3+9*x^2+15*x-25)*exp(log(x)-x)+5*x^3+50*x^2+145*x)/(4*x^3+40*x^2+100*x),x, algorithm="giac")

[Out]

-1/4*(x^2*e^(-x) - 5*x^2 + 5*x*e^(-x) - 25*x + 20)/(x + 5)

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maple [A]  time = 0.03, size = 19, normalized size = 0.76

method	result	size
default	\(-\frac {x \,{\mathrm e}^{-x}}{4}+\frac {5 x}{4}-\frac {5}{5+x}\)	\(19\)
risch	\(-\frac {x \,{\mathrm e}^{-x}}{4}+\frac {5 x}{4}-\frac {5}{5+x}\)	\(19\)
norman	\(\frac {\frac {5 x^{2}}{4}-\frac {145}{4}-\frac {{\mathrm e}^{\ln \relax (x )-x} x}{4}-\frac {5 \,{\mathrm e}^{\ln \relax (x )-x}}{4}}{5+x}\)	\(33\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x^3+9*x^2+15*x-25)*exp(ln(x)-x)+5*x^3+50*x^2+145*x)/(4*x^3+40*x^2+100*x),x,method=_RETURNVERBOSE)

[Out]

-1/4*x*exp(-x)+5/4*x-5/(5+x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -\frac {1}{4} \, x e^{\left (-x\right )} + \frac {5}{4} \, x + \frac {25 \, e^{5} E_{2}\left (x + 5\right )}{4 \, {\left (x + 5\right )}} - \frac {5}{x + 5} + \frac {25}{4} \, \int \frac {e^{\left (-x\right )}}{x^{2} + 10 \, x + 25}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^3+9*x^2+15*x-25)*exp(log(x)-x)+5*x^3+50*x^2+145*x)/(4*x^3+40*x^2+100*x),x, algorithm="maxima")

[Out]

-1/4*x*e^(-x) + 5/4*x + 25/4*e^5*exp_integral_e(2, x + 5)/(x + 5) - 5/(x + 5) + 25/4*integrate(e^(-x)/(x^2 + 1
0*x + 25), x)

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mupad [B]  time = 2.16, size = 29, normalized size = 1.16 \begin {gather*} \frac {5\,x}{4}-\frac {x\,{\mathrm {e}}^{-x}}{4}-\frac {20\,x^2}{4\,x^3+20\,x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((145*x + exp(log(x) - x)*(15*x + 9*x^2 + x^3 - 25) + 50*x^2 + 5*x^3)/(100*x + 40*x^2 + 4*x^3),x)

[Out]

(5*x)/4 - (x*exp(-x))/4 - (20*x^2)/(20*x^2 + 4*x^3)

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sympy [A]  time = 0.15, size = 15, normalized size = 0.60 \begin {gather*} \frac {5 x}{4} - \frac {x e^{- x}}{4} - \frac {5}{x + 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x**3+9*x**2+15*x-25)*exp(ln(x)-x)+5*x**3+50*x**2+145*x)/(4*x**3+40*x**2+100*x),x)

[Out]

5*x/4 - x*exp(-x)/4 - 5/(x + 5)

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