3.37.52 \(\int \frac {e^{-4 x} (-25-10 x-x^2+(-25-100 x-39 x^2-4 x^3) \log (\frac {5}{x}))}{x^2} \, dx\)

Optimal. Leaf size=20 \[ \frac {e^{-4 x} (5+x)^2 \log \left (\frac {5}{x}\right )}{x} \]

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Rubi [B]  time = 0.60, antiderivative size = 43, normalized size of antiderivative = 2.15, number of steps used = 15, number of rules used = 7, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {6742, 2199, 2194, 2177, 2178, 2176, 2554} \begin {gather*} 10 e^{-4 x} \log \left (\frac {5}{x}\right )+e^{-4 x} x \log \left (\frac {5}{x}\right )+\frac {25 e^{-4 x} \log \left (\frac {5}{x}\right )}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-25 - 10*x - x^2 + (-25 - 100*x - 39*x^2 - 4*x^3)*Log[5/x])/(E^(4*x)*x^2),x]

[Out]

(10*Log[5/x])/E^(4*x) + (25*Log[5/x])/(E^(4*x)*x) + (x*Log[5/x])/E^(4*x)

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 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 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 2199

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] &&  !$UseGamma === True

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 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-\frac {e^{-4 x} (5+x)^2}{x^2}-\frac {e^{-4 x} (5+x) \left (5+19 x+4 x^2\right ) \log \left (\frac {5}{x}\right )}{x^2}\right ) \, dx\\ &=-\int \frac {e^{-4 x} (5+x)^2}{x^2} \, dx-\int \frac {e^{-4 x} (5+x) \left (5+19 x+4 x^2\right ) \log \left (\frac {5}{x}\right )}{x^2} \, dx\\ &=10 e^{-4 x} \log \left (\frac {5}{x}\right )+\frac {25 e^{-4 x} \log \left (\frac {5}{x}\right )}{x}+e^{-4 x} x \log \left (\frac {5}{x}\right )-\int \left (e^{-4 x}+\frac {25 e^{-4 x}}{x^2}+\frac {10 e^{-4 x}}{x}\right ) \, dx+\int \frac {e^{-4 x} (5+x)^2}{x^2} \, dx\\ &=10 e^{-4 x} \log \left (\frac {5}{x}\right )+\frac {25 e^{-4 x} \log \left (\frac {5}{x}\right )}{x}+e^{-4 x} x \log \left (\frac {5}{x}\right )-10 \int \frac {e^{-4 x}}{x} \, dx-25 \int \frac {e^{-4 x}}{x^2} \, dx-\int e^{-4 x} \, dx+\int \left (e^{-4 x}+\frac {25 e^{-4 x}}{x^2}+\frac {10 e^{-4 x}}{x}\right ) \, dx\\ &=\frac {e^{-4 x}}{4}+\frac {25 e^{-4 x}}{x}-10 \text {Ei}(-4 x)+10 e^{-4 x} \log \left (\frac {5}{x}\right )+\frac {25 e^{-4 x} \log \left (\frac {5}{x}\right )}{x}+e^{-4 x} x \log \left (\frac {5}{x}\right )+10 \int \frac {e^{-4 x}}{x} \, dx+25 \int \frac {e^{-4 x}}{x^2} \, dx+100 \int \frac {e^{-4 x}}{x} \, dx+\int e^{-4 x} \, dx\\ &=100 \text {Ei}(-4 x)+10 e^{-4 x} \log \left (\frac {5}{x}\right )+\frac {25 e^{-4 x} \log \left (\frac {5}{x}\right )}{x}+e^{-4 x} x \log \left (\frac {5}{x}\right )-100 \int \frac {e^{-4 x}}{x} \, dx\\ &=10 e^{-4 x} \log \left (\frac {5}{x}\right )+\frac {25 e^{-4 x} \log \left (\frac {5}{x}\right )}{x}+e^{-4 x} x \log \left (\frac {5}{x}\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.15, size = 20, normalized size = 1.00 \begin {gather*} \frac {e^{-4 x} (5+x)^2 \log \left (\frac {5}{x}\right )}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-25 - 10*x - x^2 + (-25 - 100*x - 39*x^2 - 4*x^3)*Log[5/x])/(E^(4*x)*x^2),x]

[Out]

((5 + x)^2*Log[5/x])/(E^(4*x)*x)

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fricas [A]  time = 0.74, size = 22, normalized size = 1.10 \begin {gather*} \frac {{\left (x^{2} + 10 \, x + 25\right )} e^{\left (-4 \, x\right )} \log \left (\frac {5}{x}\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x^3-39*x^2-100*x-25)*log(5/x)-x^2-10*x-25)/x^2/exp(2*x)^2,x, algorithm="fricas")

[Out]

(x^2 + 10*x + 25)*e^(-4*x)*log(5/x)/x

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giac [B]  time = 0.15, size = 44, normalized size = 2.20 \begin {gather*} \frac {x^{2} e^{\left (-4 \, x\right )} \log \left (\frac {5}{x}\right ) + 10 \, x e^{\left (-4 \, x\right )} \log \left (\frac {5}{x}\right ) + 25 \, e^{\left (-4 \, x\right )} \log \left (\frac {5}{x}\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x^3-39*x^2-100*x-25)*log(5/x)-x^2-10*x-25)/x^2/exp(2*x)^2,x, algorithm="giac")

[Out]

(x^2*e^(-4*x)*log(5/x) + 10*x*e^(-4*x)*log(5/x) + 25*e^(-4*x)*log(5/x))/x

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maple [B]  time = 0.12, size = 47, normalized size = 2.35

method	result	size
risch	\(-\frac {\left (x^{2}+10 x +25\right ) {\mathrm e}^{-4 x} \ln \relax (x )}{x}+\frac {\left (2 x^{2} \ln \relax (5)+20 x \ln \relax (5)+50 \ln \relax (5)\right ) {\mathrm e}^{-4 x}}{2 x}\)	\(47\)
default	\(\frac {\left (25 \ln \relax (5)+25 \ln \left (\frac {1}{x}\right )+25 \ln \relax (x )\right ) {\mathrm e}^{-4 x}+\left (\ln \relax (5)+\ln \left (\frac {1}{x}\right )+\ln \relax (x )\right ) x^{2} {\mathrm e}^{-4 x}+\left (10 \ln \relax (5)+10 \ln \left (\frac {1}{x}\right )+10 \ln \relax (x )\right ) x \,{\mathrm e}^{-4 x}-25 \ln \relax (x ) {\mathrm e}^{-4 x}-10 x \,{\mathrm e}^{-4 x} \ln \relax (x )-x^{2} {\mathrm e}^{-4 x} \ln \relax (x )}{x}\)	\(92\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-4*x^3-39*x^2-100*x-25)*ln(5/x)-x^2-10*x-25)/x^2/exp(2*x)^2,x,method=_RETURNVERBOSE)

[Out]

-(x^2+10*x+25)/x*exp(-4*x)*ln(x)+1/2*(2*x^2*ln(5)+20*x*ln(5)+50*ln(5))/x*exp(-4*x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -\frac {{\left (4 \, x^{2} + x + 100\right )} e^{\left (-4 \, x\right )} \log \relax (x)}{4 \, x} + \frac {39}{4} \, e^{\left (-4 \, x\right )} \log \left (\frac {5}{x}\right ) + \frac {39}{4} \, {\rm Ei}\left (-4 \, x\right ) + \frac {1}{4} \, e^{\left (-4 \, x\right )} + 100 \, \Gamma \left (-1, 4 \, x\right ) - \frac {1}{4} \, \int \frac {{\left (16 \, x^{3} \log \relax (5) - 4 \, x^{2} + x {\left (400 \, \log \relax (5) + 39\right )} + 100 \, \log \relax (5) - 100\right )} e^{\left (-4 \, x\right )}}{x^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x^3-39*x^2-100*x-25)*log(5/x)-x^2-10*x-25)/x^2/exp(2*x)^2,x, algorithm="maxima")

[Out]

-1/4*(4*x^2 + x + 100)*e^(-4*x)*log(x)/x + 39/4*e^(-4*x)*log(5/x) + 39/4*Ei(-4*x) + 1/4*e^(-4*x) + 100*gamma(-
1, 4*x) - 1/4*integrate((16*x^3*log(5) - 4*x^2 + x*(400*log(5) + 39) + 100*log(5) - 100)*e^(-4*x)/x^2, x)

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mupad [B]  time = 2.30, size = 19, normalized size = 0.95 \begin {gather*} \frac {{\mathrm {e}}^{-4\,x}\,\ln \left (\frac {5}{x}\right )\,{\left (x+5\right )}^2}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-4*x)*(10*x + log(5/x)*(100*x + 39*x^2 + 4*x^3 + 25) + x^2 + 25))/x^2,x)

[Out]

(exp(-4*x)*log(5/x)*(x + 5)^2)/x

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sympy [A]  time = 0.32, size = 29, normalized size = 1.45 \begin {gather*} \frac {\left (x^{2} \log {\left (\frac {5}{x} \right )} + 10 x \log {\left (\frac {5}{x} \right )} + 25 \log {\left (\frac {5}{x} \right )}\right ) e^{- 4 x}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x**3-39*x**2-100*x-25)*ln(5/x)-x**2-10*x-25)/x**2/exp(2*x)**2,x)

[Out]

(x**2*log(5/x) + 10*x*log(5/x) + 25*log(5/x))*exp(-4*x)/x

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