3.15.88 \(\int \frac {e^{e^{\frac {2 e^2+(-4-x) \log (5)}{\log (5)}}} (-3-e^{\frac {2 e^2+(-4-x) \log (5)}{\log (5)}} x)}{x^4} \, dx\)

Optimal. Leaf size=22 \[ \frac {e^{e^{-4-x+\frac {2 e^2}{\log (5)}}}}{x^3} \]

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Rubi [A]  time = 0.15, antiderivative size = 29, normalized size of antiderivative = 1.32, number of steps used = 1, number of rules used = 1, integrand size = 53, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.019, Rules used = {2288} \begin {gather*} \frac {e^{e^{\frac {2 e^2}{\log (5)}} 5^{-\frac {x+4}{\log (5)}}}}{x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

E^(E^((2*E^2)/Log[5])/5^((4 + x)/Log[5]))/x^3

Rule 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {e^{5^{-\frac {4+x}{\log (5)}} e^{\frac {2 e^2}{\log (5)}}}}{x^3}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.41, size = 22, normalized size = 1.00 \begin {gather*} \frac {e^{e^{-4-x+\frac {2 e^2}{\log (5)}}}}{x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

E^E^(-4 - x + (2*E^2)/Log[5])/x^3

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fricas [A]  time = 1.07, size = 23, normalized size = 1.05 \begin {gather*} \frac {e^{\left (e^{\left (-\frac {{\left (x + 4\right )} \log \relax (5) - 2 \, e^{2}}{\log \relax (5)}\right )}\right )}}{x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e^(e^(-((x + 4)*log(5) - 2*e^2)/log(5)))/x^3

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \mathit {undef} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

undef

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maple [A]  time = 0.10, size = 26, normalized size = 1.18

method	result	size
risch	\(\frac {{\mathrm e}^{{\mathrm e}^{\frac {-x \ln \relax (5)+2 \,{\mathrm e}^{2}-4 \ln \relax (5)}{\ln \relax (5)}}}}{x^{3}}\)	\(26\)
norman	\(\frac {{\mathrm e}^{{\mathrm e}^{\frac {\left (-x -4\right ) \ln \relax (5)+2 \,{\mathrm e}^{2}}{\ln \relax (5)}}}}{x^{3}}\)	\(27\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-integrate((x*e^(-((x + 4)*log(5) - 2*e^2)/log(5)) + 3)*e^(e^(-((x + 4)*log(5) - 2*e^2)/log(5)))/x^4, x)

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mupad [B]  time = 0.13, size = 21, normalized size = 0.95 \begin {gather*} \frac {{\mathrm {e}}^{{\mathrm {e}}^{-x}\,{\mathrm {e}}^{-4}\,{\mathrm {e}}^{\frac {2\,{\mathrm {e}}^2}{\ln \relax (5)}}}}{x^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.13, size = 22, normalized size = 1.00 \begin {gather*} \frac {e^{e^{\frac {\left (- x - 4\right ) \log {\relax (5 )} + 2 e^{2}}{\log {\relax (5 )}}}}}{x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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