3.19.56 \(\int \frac {e^{-x} (40+10 x+(-60 x-10 x^2) \log (x))}{64 x+48 x^2+12 x^3+x^4} \, dx\)

Optimal. Leaf size=18 \[ -e+\frac {10 e^{-x} \log (x)}{(4+x)^2} \]

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Rubi [A]  time = 0.81, antiderivative size = 14, normalized size of antiderivative = 0.78, number of steps used = 17, number of rules used = 7, integrand size = 42, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {6688, 12, 6742, 2178, 2177, 2197, 2554} \begin {gather*} \frac {10 e^{-x} \log (x)}{(x+4)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(40 + 10*x + (-60*x - 10*x^2)*Log[x])/(E^x*(64*x + 48*x^2 + 12*x^3 + x^4)),x]

[Out]

(10*Log[x])/(E^x*(4 + x)^2)

Rule 12

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

Rule 2177

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 ===
True

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 2197

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> With[{b = Coefficient[v, x, 1], d = Coefficient[u, x, 0],
e = Coefficient[u, x, 1], f = Coefficient[w, x, 0], g = Coefficient[w, x, 1]}, Simp[(g*u^(m + 1)*F^(c*v))/(b*c
*e*Log[F]), x] /; EqQ[e*g*(m + 1) - b*c*(e*f - d*g)*Log[F], 0]] /; FreeQ[{F, c, m}, x] && LinearQ[{u, v, w}, x
]

Rule 2554

Int[Log[u_]*(v_), x_Symbol] :> With[{w = IntHide[v, x]}, Dist[Log[u], w, x] - Int[SimplifyIntegrand[(w*D[u, x]
)/u, x], x] /; InverseFunctionFreeQ[w, x]] /; InverseFunctionFreeQ[u, x]

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 {10 e^{-x} (4+x-x (6+x) \log (x))}{x (4+x)^3} \, dx\\ &=10 \int \frac {e^{-x} (4+x-x (6+x) \log (x))}{x (4+x)^3} \, dx\\ &=10 \int \left (\frac {e^{-x}}{x (4+x)^2}-\frac {e^{-x} (6+x) \log (x)}{(4+x)^3}\right ) \, dx\\ &=10 \int \frac {e^{-x}}{x (4+x)^2} \, dx-10 \int \frac {e^{-x} (6+x) \log (x)}{(4+x)^3} \, dx\\ &=\frac {10 e^{-x} \log (x)}{(4+x)^2}-10 \int \frac {e^{-x}}{x (4+x)^2} \, dx+10 \int \left (\frac {e^{-x}}{16 x}-\frac {e^{-x}}{4 (4+x)^2}-\frac {e^{-x}}{16 (4+x)}\right ) \, dx\\ &=\frac {10 e^{-x} \log (x)}{(4+x)^2}+\frac {5}{8} \int \frac {e^{-x}}{x} \, dx-\frac {5}{8} \int \frac {e^{-x}}{4+x} \, dx-\frac {5}{2} \int \frac {e^{-x}}{(4+x)^2} \, dx-10 \int \left (\frac {e^{-x}}{16 x}-\frac {e^{-x}}{4 (4+x)^2}-\frac {e^{-x}}{16 (4+x)}\right ) \, dx\\ &=\frac {5 e^{-x}}{2 (4+x)}-\frac {5}{8} e^4 \text {Ei}(-4-x)+\frac {5 \text {Ei}(-x)}{8}+\frac {10 e^{-x} \log (x)}{(4+x)^2}-\frac {5}{8} \int \frac {e^{-x}}{x} \, dx+\frac {5}{8} \int \frac {e^{-x}}{4+x} \, dx+\frac {5}{2} \int \frac {e^{-x}}{(4+x)^2} \, dx+\frac {5}{2} \int \frac {e^{-x}}{4+x} \, dx\\ &=\frac {5}{2} e^4 \text {Ei}(-4-x)+\frac {10 e^{-x} \log (x)}{(4+x)^2}-\frac {5}{2} \int \frac {e^{-x}}{4+x} \, dx\\ &=\frac {10 e^{-x} \log (x)}{(4+x)^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.12, size = 14, normalized size = 0.78 \begin {gather*} \frac {10 e^{-x} \log (x)}{(4+x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(40 + 10*x + (-60*x - 10*x^2)*Log[x])/(E^x*(64*x + 48*x^2 + 12*x^3 + x^4)),x]

[Out]

(10*Log[x])/(E^x*(4 + x)^2)

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fricas [A]  time = 0.84, size = 18, normalized size = 1.00 \begin {gather*} \frac {10 \, e^{\left (-x\right )} \log \relax (x)}{x^{2} + 8 \, x + 16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-10*x^2-60*x)*log(x)+10*x+40)/(x^4+12*x^3+48*x^2+64*x)/exp(x),x, algorithm="fricas")

[Out]

10*e^(-x)*log(x)/(x^2 + 8*x + 16)

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giac [A]  time = 0.33, size = 18, normalized size = 1.00 \begin {gather*} \frac {10 \, e^{\left (-x\right )} \log \relax (x)}{x^{2} + 8 \, x + 16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-10*x^2-60*x)*log(x)+10*x+40)/(x^4+12*x^3+48*x^2+64*x)/exp(x),x, algorithm="giac")

[Out]

10*e^(-x)*log(x)/(x^2 + 8*x + 16)

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maple [A]  time = 0.04, size = 14, normalized size = 0.78

method	result	size
norman	\(\frac {10 \ln \relax (x ) {\mathrm e}^{-x}}{\left (4+x \right )^{2}}\)	\(14\)
risch	\(\frac {10 \ln \relax (x ) {\mathrm e}^{-x}}{\left (4+x \right )^{2}}\)	\(14\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-10*x^2-60*x)*ln(x)+10*x+40)/(x^4+12*x^3+48*x^2+64*x)/exp(x),x,method=_RETURNVERBOSE)

[Out]

10*ln(x)/(4+x)^2/exp(x)

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maxima [A]  time = 0.68, size = 18, normalized size = 1.00 \begin {gather*} \frac {10 \, e^{\left (-x\right )} \log \relax (x)}{x^{2} + 8 \, x + 16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-10*x^2-60*x)*log(x)+10*x+40)/(x^4+12*x^3+48*x^2+64*x)/exp(x),x, algorithm="maxima")

[Out]

10*e^(-x)*log(x)/(x^2 + 8*x + 16)

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mupad [B]  time = 1.42, size = 13, normalized size = 0.72 \begin {gather*} \frac {10\,{\mathrm {e}}^{-x}\,\ln \relax (x)}{{\left (x+4\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-x)*(10*x - log(x)*(60*x + 10*x^2) + 40))/(64*x + 48*x^2 + 12*x^3 + x^4),x)

[Out]

(10*exp(-x)*log(x))/(x + 4)^2

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sympy [A]  time = 0.29, size = 15, normalized size = 0.83 \begin {gather*} \frac {10 e^{- x} \log {\relax (x )}}{x^{2} + 8 x + 16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-10*x**2-60*x)*ln(x)+10*x+40)/(x**4+12*x**3+48*x**2+64*x)/exp(x),x)

[Out]

10*exp(-x)*log(x)/(x**2 + 8*x + 16)

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