3.83.87 \(\int \frac {-x^3+e^{\frac {25+x}{x^2}} (50+x)}{(e^{\frac {2 (25+x)}{x^2}} x^3+2 e^{\frac {25+x}{x^2}} x^4+x^5) \log (\log (16))} \, dx\)

Optimal. Leaf size=19 \[ \frac {1}{\left (e^{\frac {25+x}{x^2}}+x\right ) \log (\log (16))} \]

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Rubi [A]  time = 0.34, antiderivative size = 19, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 59, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.051, Rules used = {12, 6688, 6686} \begin {gather*} \frac {1}{\left (e^{\frac {x+25}{x^2}}+x\right ) \log (\log (16))} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-x^3 + E^((25 + x)/x^2)*(50 + x))/((E^((2*(25 + x))/x^2)*x^3 + 2*E^((25 + x)/x^2)*x^4 + x^5)*Log[Log[16]]
),x]

[Out]

1/((E^((25 + x)/x^2) + x)*Log[Log[16]])

Rule 12

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

Rule 6686

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[(q*y^(m + 1))/(m + 1), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]

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} &=\frac {\int \frac {-x^3+e^{\frac {25+x}{x^2}} (50+x)}{e^{\frac {2 (25+x)}{x^2}} x^3+2 e^{\frac {25+x}{x^2}} x^4+x^5} \, dx}{\log (\log (16))}\\ &=\frac {\int \frac {-x^3+e^{\frac {25+x}{x^2}} (50+x)}{x^3 \left (e^{\frac {25+x}{x^2}}+x\right )^2} \, dx}{\log (\log (16))}\\ &=\frac {1}{\left (e^{\frac {25+x}{x^2}}+x\right ) \log (\log (16))}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.05, size = 19, normalized size = 1.00 \begin {gather*} \frac {1}{\left (e^{\frac {25+x}{x^2}}+x\right ) \log (\log (16))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-x^3 + E^((25 + x)/x^2)*(50 + x))/((E^((2*(25 + x))/x^2)*x^3 + 2*E^((25 + x)/x^2)*x^4 + x^5)*Log[Lo
g[16]]),x]

[Out]

1/((E^((25 + x)/x^2) + x)*Log[Log[16]])

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fricas [A]  time = 0.84, size = 20, normalized size = 1.05 \begin {gather*} \frac {1}{{\left (x + e^{\left (\frac {x + 25}{x^{2}}\right )}\right )} \log \left (4 \, \log \relax (2)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((50+x)*exp((x+25)/x^2)-x^3)/(x^3*exp((x+25)/x^2)^2+2*x^4*exp((x+25)/x^2)+x^5)/log(4*log(2)),x, algo
rithm="fricas")

[Out]

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

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giac [A]  time = 0.18, size = 20, normalized size = 1.05 \begin {gather*} \frac {1}{{\left (x + e^{\left (\frac {x + 25}{x^{2}}\right )}\right )} \log \left (4 \, \log \relax (2)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((50+x)*exp((x+25)/x^2)-x^3)/(x^3*exp((x+25)/x^2)^2+2*x^4*exp((x+25)/x^2)+x^5)/log(4*log(2)),x, algo
rithm="giac")

[Out]

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

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maple [A]  time = 0.18, size = 24, normalized size = 1.26

method	result	size
norman	\(\frac {1}{\left (2 \ln \relax (2)+\ln \left (\ln \relax (2)\right )\right ) \left (x +{\mathrm e}^{\frac {x +25}{x^{2}}}\right )}\)	\(24\)
risch	\(\frac {1}{\left (2 \ln \relax (2)+\ln \left (\ln \relax (2)\right )\right ) \left (x +{\mathrm e}^{\frac {x +25}{x^{2}}}\right )}\)	\(24\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((50+x)*exp((x+25)/x^2)-x^3)/(x^3*exp((x+25)/x^2)^2+2*x^4*exp((x+25)/x^2)+x^5)/ln(4*ln(2)),x,method=_RETUR
NVERBOSE)

[Out]

1/(2*ln(2)+ln(ln(2)))/(x+exp((x+25)/x^2))

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maxima [A]  time = 0.41, size = 22, normalized size = 1.16 \begin {gather*} \frac {1}{{\left (x + e^{\left (\frac {1}{x} + \frac {25}{x^{2}}\right )}\right )} \log \left (4 \, \log \relax (2)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((50+x)*exp((x+25)/x^2)-x^3)/(x^3*exp((x+25)/x^2)^2+2*x^4*exp((x+25)/x^2)+x^5)/log(4*log(2)),x, algo
rithm="maxima")

[Out]

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

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mupad [B]  time = 5.36, size = 20, normalized size = 1.05 \begin {gather*} \frac {1}{{\mathrm {e}}^{\frac {x+25}{x^2}}\,\ln \left (\ln \left (16\right )\right )+x\,\ln \left (\ln \left (16\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp((x + 25)/x^2)*(x + 50) - x^3)/(log(4*log(2))*(2*x^4*exp((x + 25)/x^2) + x^3*exp((2*(x + 25))/x^2) + x
^5)),x)

[Out]

1/(exp((x + 25)/x^2)*log(log(16)) + x*log(log(16)))

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sympy [A]  time = 0.15, size = 32, normalized size = 1.68 \begin {gather*} \frac {1}{x \log {\left (\log {\relax (2 )} \right )} + 2 x \log {\relax (2 )} + \left (\log {\left (\log {\relax (2 )} \right )} + 2 \log {\relax (2 )}\right ) e^{\frac {x + 25}{x^{2}}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((50+x)*exp((x+25)/x**2)-x**3)/(x**3*exp((x+25)/x**2)**2+2*x**4*exp((x+25)/x**2)+x**5)/ln(4*ln(2)),x
)

[Out]

1/(x*log(log(2)) + 2*x*log(2) + (log(log(2)) + 2*log(2))*exp((x + 25)/x**2))

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