\(\int \frac {-x^3+e^{\frac {25+x}{x^2}} (50+x)}{(e^{\frac {2 (25+x)}{x^2}} x^3+2 e^{\frac {25+x}{x^2}} x^4+x^5) \log (\log (16))} \, dx\) [8287]

Optimal result

Integrand size = 59, antiderivative size = 19 \[ \int \frac {-x^3+e^{\frac {25+x}{x^2}} (50+x)}{\left (e^{\frac {2 (25+x)}{x^2}} x^3+2 e^{\frac {25+x}{x^2}} x^4+x^5\right ) \log (\log (16))} \, dx=\frac {1}{\left (e^{\frac {25+x}{x^2}}+x\right ) \log (\log (16))} \]

[Out]

1/ln(4*ln(2))/(x+exp((x+25)/x^2))

Rubi [A] (verified)

Time = 0.22 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.051, Rules used = {12, 6820, 6818} \[ \int \frac {-x^3+e^{\frac {25+x}{x^2}} (50+x)}{\left (e^{\frac {2 (25+x)}{x^2}} x^3+2 e^{\frac {25+x}{x^2}} x^4+x^5\right ) \log (\log (16))} \, dx=\frac {1}{\left (e^{\frac {x+25}{x^2}}+x\right ) \log (\log (16))} \]

[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 6818

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 6820

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

Rubi steps \begin{align*} \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{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \frac {-x^3+e^{\frac {25+x}{x^2}} (50+x)}{\left (e^{\frac {2 (25+x)}{x^2}} x^3+2 e^{\frac {25+x}{x^2}} x^4+x^5\right ) \log (\log (16))} \, dx=\frac {1}{\left (e^{\frac {25+x}{x^2}}+x\right ) \log (\log (16))} \]

[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]])

Maple [A] (verified)

Time = 0.26 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.11

method	result	size
parallelrisch	\(\frac {1}{\ln \left (4 \ln \left (2\right )\right ) \left (x +{\mathrm e}^{\frac {x +25}{x^{2}}}\right )}\)	\(21\)
norman	\(\frac {1}{\left (2 \ln \left (2\right )+\ln \left (\ln \left (2\right )\right )\right ) \left (x +{\mathrm e}^{\frac {x +25}{x^{2}}}\right )}\)	\(24\)
risch	\(\frac {1}{\left (2 \ln \left (2\right )+\ln \left (\ln \left (2\right )\right )\right ) \left (x +{\mathrm e}^{\frac {x +25}{x^{2}}}\right )}\)	\(24\)

[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/ln(4*ln(2))/(x+exp((x+25)/x^2))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.05 \[ \int \frac {-x^3+e^{\frac {25+x}{x^2}} (50+x)}{\left (e^{\frac {2 (25+x)}{x^2}} x^3+2 e^{\frac {25+x}{x^2}} x^4+x^5\right ) \log (\log (16))} \, dx=\frac {1}{{\left (x + e^{\left (\frac {x + 25}{x^{2}}\right )}\right )} \log \left (4 \, \log \left (2\right )\right )} \]

[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)))

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.68 \[ \int \frac {-x^3+e^{\frac {25+x}{x^2}} (50+x)}{\left (e^{\frac {2 (25+x)}{x^2}} x^3+2 e^{\frac {25+x}{x^2}} x^4+x^5\right ) \log (\log (16))} \, dx=\frac {1}{x \log {\left (\log {\left (2 \right )} \right )} + 2 x \log {\left (2 \right )} + \left (\log {\left (\log {\left (2 \right )} \right )} + 2 \log {\left (2 \right )}\right ) e^{\frac {x + 25}{x^{2}}}} \]

[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))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.16 \[ \int \frac {-x^3+e^{\frac {25+x}{x^2}} (50+x)}{\left (e^{\frac {2 (25+x)}{x^2}} x^3+2 e^{\frac {25+x}{x^2}} x^4+x^5\right ) \log (\log (16))} \, dx=\frac {1}{{\left (x + e^{\left (\frac {1}{x} + \frac {25}{x^{2}}\right )}\right )} \log \left (4 \, \log \left (2\right )\right )} \]

[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)))

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.05 \[ \int \frac {-x^3+e^{\frac {25+x}{x^2}} (50+x)}{\left (e^{\frac {2 (25+x)}{x^2}} x^3+2 e^{\frac {25+x}{x^2}} x^4+x^5\right ) \log (\log (16))} \, dx=\frac {1}{{\left (x + e^{\left (\frac {x + 25}{x^{2}}\right )}\right )} \log \left (4 \, \log \left (2\right )\right )} \]

[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)))

Mupad [B] (verification not implemented)

Time = 14.12 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.05 \[ \int \frac {-x^3+e^{\frac {25+x}{x^2}} (50+x)}{\left (e^{\frac {2 (25+x)}{x^2}} x^3+2 e^{\frac {25+x}{x^2}} x^4+x^5\right ) \log (\log (16))} \, dx=\frac {1}{{\mathrm {e}}^{\frac {x+25}{x^2}}\,\ln \left (\ln \left (16\right )\right )+x\,\ln \left (\ln \left (16\right )\right )} \]

[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)))