3.94.87 \(\int \frac {e^{-e^4} (-e^3 x^3+e^{10} (-1+3 x)+4 e^{10} \log (x))}{x^5} \, dx\)

Optimal. Leaf size=27 \[ \frac {e^{-e^4} \left (e^3-\frac {e^{10} (x+\log (x))}{x^3}\right )}{x} \]

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Rubi [A]  time = 0.05, antiderivative size = 44, normalized size of antiderivative = 1.63, number of steps used = 6, number of rules used = 3, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {12, 14, 2304} \begin {gather*} -\frac {e^{10-e^4} \log (x)}{x^4}-\frac {e^{10-e^4}}{x^3}+\frac {e^{3-e^4}}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-(E^3*x^3) + E^10*(-1 + 3*x) + 4*E^10*Log[x])/(E^E^4*x^5),x]

[Out]

-(E^(10 - E^4)/x^3) + E^(3 - E^4)/x - (E^(10 - E^4)*Log[x])/x^4

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 2304

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Log[c*x^
n]))/(d*(m + 1)), x] - Simp[(b*n*(d*x)^(m + 1))/(d*(m + 1)^2), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1
]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=e^{-e^4} \int \frac {-e^3 x^3+e^{10} (-1+3 x)+4 e^{10} \log (x)}{x^5} \, dx\\ &=e^{-e^4} \int \left (\frac {e^3 \left (-e^7+3 e^7 x-x^3\right )}{x^5}+\frac {4 e^{10} \log (x)}{x^5}\right ) \, dx\\ &=e^{3-e^4} \int \frac {-e^7+3 e^7 x-x^3}{x^5} \, dx+\left (4 e^{10-e^4}\right ) \int \frac {\log (x)}{x^5} \, dx\\ &=-\frac {e^{10-e^4}}{4 x^4}-\frac {e^{10-e^4} \log (x)}{x^4}+e^{3-e^4} \int \left (-\frac {e^7}{x^5}+\frac {3 e^7}{x^4}-\frac {1}{x^2}\right ) \, dx\\ &=-\frac {e^{10-e^4}}{x^3}+\frac {e^{3-e^4}}{x}-\frac {e^{10-e^4} \log (x)}{x^4}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 33, normalized size = 1.22 \begin {gather*} -e^{3-e^4} \left (\frac {e^7}{x^3}-\frac {1}{x}+\frac {e^7 \log (x)}{x^4}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-(E^3*x^3) + E^10*(-1 + 3*x) + 4*E^10*Log[x])/(E^E^4*x^5),x]

[Out]

-(E^(3 - E^4)*(E^7/x^3 - x^(-1) + (E^7*Log[x])/x^4))

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fricas [A]  time = 1.06, size = 27, normalized size = 1.00 \begin {gather*} \frac {{\left (x^{3} e^{3} - x e^{10} - e^{10} \log \relax (x)\right )} e^{\left (-e^{4}\right )}}{x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*exp(5)^2*log(x)+(3*x-1)*exp(5)^2-x^3*exp(3))/x^5/exp(exp(4)),x, algorithm="fricas")

[Out]

(x^3*e^3 - x*e^10 - e^10*log(x))*e^(-e^4)/x^4

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giac [A]  time = 0.23, size = 27, normalized size = 1.00 \begin {gather*} \frac {{\left (x^{3} e^{3} - x e^{10} - e^{10} \log \relax (x)\right )} e^{\left (-e^{4}\right )}}{x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*exp(5)^2*log(x)+(3*x-1)*exp(5)^2-x^3*exp(3))/x^5/exp(exp(4)),x, algorithm="giac")

[Out]

(x^3*e^3 - x*e^10 - e^10*log(x))*e^(-e^4)/x^4

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maple [A]  time = 0.07, size = 36, normalized size = 1.33

method	result	size
risch	\(-\frac {\ln \relax (x ) {\mathrm e}^{-{\mathrm e}^{4}+10}}{x^{4}}-\frac {\left ({\mathrm e}^{7}-x^{2}\right ) {\mathrm e}^{3-{\mathrm e}^{4}}}{x^{3}}\)	\(36\)
norman	\(\frac {{\mathrm e}^{-{\mathrm e}^{4}} {\mathrm e}^{3} x^{3}-{\mathrm e}^{-{\mathrm e}^{4}} {\mathrm e}^{10} x -{\mathrm e}^{-{\mathrm e}^{4}} {\mathrm e}^{10} \ln \relax (x )}{x^{4}}\)	\(42\)
default	\({\mathrm e}^{-{\mathrm e}^{4}} \left (\frac {{\mathrm e}^{3}}{x}+4 \,{\mathrm e}^{10} \left (-\frac {\ln \relax (x )}{4 x^{4}}-\frac {1}{16 x^{4}}\right )-\frac {{\mathrm e}^{10}}{x^{3}}+\frac {{\mathrm e}^{10}}{4 x^{4}}\right )\)	\(51\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*exp(5)^2*ln(x)+(3*x-1)*exp(5)^2-x^3*exp(3))/x^5/exp(exp(4)),x,method=_RETURNVERBOSE)

[Out]

-1/x^4*ln(x)*exp(-exp(4)+10)-1/x^3*(exp(7)-x^2)*exp(3-exp(4))

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maxima [A]  time = 0.35, size = 43, normalized size = 1.59 \begin {gather*} -\frac {1}{4} \, {\left ({\left (\frac {4 \, \log \relax (x)}{x^{4}} + \frac {1}{x^{4}}\right )} e^{10} - \frac {4 \, e^{3}}{x} + \frac {4 \, e^{10}}{x^{3}} - \frac {e^{10}}{x^{4}}\right )} e^{\left (-e^{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*exp(5)^2*log(x)+(3*x-1)*exp(5)^2-x^3*exp(3))/x^5/exp(exp(4)),x, algorithm="maxima")

[Out]

-1/4*((4*log(x)/x^4 + 1/x^4)*e^10 - 4*e^3/x + 4*e^10/x^3 - e^10/x^4)*e^(-e^4)

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mupad [B]  time = 8.11, size = 37, normalized size = 1.37 \begin {gather*} -\frac {x\,{\mathrm {e}}^{10-{\mathrm {e}}^4}-x^3\,{\mathrm {e}}^{3-{\mathrm {e}}^4}+{\mathrm {e}}^{10-{\mathrm {e}}^4}\,\ln \relax (x)}{x^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-exp(4))*(4*exp(10)*log(x) - x^3*exp(3) + exp(10)*(3*x - 1)))/x^5,x)

[Out]

-(x*exp(10 - exp(4)) - x^3*exp(3 - exp(4)) + exp(10 - exp(4))*log(x))/x^4

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sympy [A]  time = 0.20, size = 34, normalized size = 1.26 \begin {gather*} - \frac {- x^{2} e^{3} + e^{10}}{x^{3} e^{e^{4}}} - \frac {e^{10} \log {\relax (x )}}{x^{4} e^{e^{4}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*exp(5)**2*ln(x)+(3*x-1)*exp(5)**2-x**3*exp(3))/x**5/exp(exp(4)),x)

[Out]

-(-x**2*exp(3) + exp(10))*exp(-exp(4))/x**3 - exp(10)*exp(-exp(4))*log(x)/x**4

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