\(\int \frac {e^{-\frac {75}{x^2+x^5-x \log (2)}} (150 x+375 x^4-75 \log (2))}{x^4+x^{10}-2 x^3 \log (2)+x^2 \log ^2(2)+x^4 (2 x^3-2 x^2 \log (2))} \, dx\) [4196]

Optimal result

Integrand size = 73, antiderivative size = 18 \[ \int \frac {e^{-\frac {75}{x^2+x^5-x \log (2)}} \left (150 x+375 x^4-75 \log (2)\right )}{x^4+x^{10}-2 x^3 \log (2)+x^2 \log ^2(2)+x^4 \left (2 x^3-2 x^2 \log (2)\right )} \, dx=e^{-\frac {75}{x \left (x+x^4-\log (2)\right )}} \]

[Out]

exp(-75/x/(x^4-ln(2)+x))

Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.041, Rules used = {6820, 12, 6838} \[ \int \frac {e^{-\frac {75}{x^2+x^5-x \log (2)}} \left (150 x+375 x^4-75 \log (2)\right )}{x^4+x^{10}-2 x^3 \log (2)+x^2 \log ^2(2)+x^4 \left (2 x^3-2 x^2 \log (2)\right )} \, dx=e^{-\frac {75}{x \left (x^4+x-\log (2)\right )}} \]

[In]

Int[(150*x + 375*x^4 - 75*Log[2])/(E^(75/(x^2 + x^5 - x*Log[2]))*(x^4 + x^10 - 2*x^3*Log[2] + x^2*Log[2]^2 + x
^4*(2*x^3 - 2*x^2*Log[2]))),x]

[Out]

E^(-75/(x*(x + x^4 - Log[2])))

Rule 12

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

Rule 6820

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

Rule 6838

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[q*(F^v/Log[F]), x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {75 e^{-\frac {75}{x \left (x+x^4-\log (2)\right )}} \left (2 x+5 x^4-\log (2)\right )}{x^2 \left (x+x^4-\log (2)\right )^2} \, dx \\ & = 75 \int \frac {e^{-\frac {75}{x \left (x+x^4-\log (2)\right )}} \left (2 x+5 x^4-\log (2)\right )}{x^2 \left (x+x^4-\log (2)\right )^2} \, dx \\ & = e^{-\frac {75}{x \left (x+x^4-\log (2)\right )}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.58 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-\frac {75}{x^2+x^5-x \log (2)}} \left (150 x+375 x^4-75 \log (2)\right )}{x^4+x^{10}-2 x^3 \log (2)+x^2 \log ^2(2)+x^4 \left (2 x^3-2 x^2 \log (2)\right )} \, dx=e^{-\frac {75}{x^2+x^5-x \log (2)}} \]

[In]

Integrate[(150*x + 375*x^4 - 75*Log[2])/(E^(75/(x^2 + x^5 - x*Log[2]))*(x^4 + x^10 - 2*x^3*Log[2] + x^2*Log[2]
^2 + x^4*(2*x^3 - 2*x^2*Log[2]))),x]

[Out]

E^(-75/(x^2 + x^5 - x*Log[2]))

Maple [A] (verified)

Time = 5.16 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.11

method	result	size
gosper	\({\mathrm e}^{\frac {75}{x \left (-x^{4}+\ln \left (2\right )-x \right )}}\)	\(20\)
risch	\({\mathrm e}^{\frac {75}{x \left (-x^{4}+\ln \left (2\right )-x \right )}}\)	\(20\)
parallelrisch	\({\mathrm e}^{\frac {75}{x \left (-x^{4}+\ln \left (2\right )-x \right )}}\)	\(20\)
norman	\(\frac {x \ln \left (2\right ) {\mathrm e}^{-\frac {75}{x^{5}-x \ln \left (2\right )+x^{2}}}-x^{2} {\mathrm e}^{-\frac {75}{x^{5}-x \ln \left (2\right )+x^{2}}}-x^{5} {\mathrm e}^{-\frac {75}{x^{5}-x \ln \left (2\right )+x^{2}}}}{x \left (-x^{4}+\ln \left (2\right )-x \right )}\)	\(84\)

[In]

int((375*x^4-75*ln(2)+150*x)*exp(-75/(x^5-x*ln(2)+x^2))/(x^10+(-2*x^2*ln(2)+2*x^3)*x^4+x^2*ln(2)^2-2*x^3*ln(2)
+x^4),x,method=_RETURNVERBOSE)

[Out]

exp(75/x/(-x^4+ln(2)-x))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.94 \[ \int \frac {e^{-\frac {75}{x^2+x^5-x \log (2)}} \left (150 x+375 x^4-75 \log (2)\right )}{x^4+x^{10}-2 x^3 \log (2)+x^2 \log ^2(2)+x^4 \left (2 x^3-2 x^2 \log (2)\right )} \, dx=e^{\left (-\frac {75}{x^{5} + x^{2} - x \log \left (2\right )}\right )} \]

[In]

integrate((375*x^4-75*log(2)+150*x)*exp(-75/(x^5-x*log(2)+x^2))/(x^10+(-2*x^2*log(2)+2*x^3)*x^4+x^2*log(2)^2-2
*x^3*log(2)+x^4),x, algorithm="fricas")

[Out]

e^(-75/(x^5 + x^2 - x*log(2)))

Sympy [A] (verification not implemented)

Time = 0.55 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.83 \[ \int \frac {e^{-\frac {75}{x^2+x^5-x \log (2)}} \left (150 x+375 x^4-75 \log (2)\right )}{x^4+x^{10}-2 x^3 \log (2)+x^2 \log ^2(2)+x^4 \left (2 x^3-2 x^2 \log (2)\right )} \, dx=e^{- \frac {75}{x^{5} + x^{2} - x \log {\left (2 \right )}}} \]

[In]

integrate((375*x**4-75*ln(2)+150*x)*exp(-75/(x**5-x*ln(2)+x**2))/(x**10+(-2*x**2*ln(2)+2*x**3)*x**4+x**2*ln(2)
**2-2*x**3*ln(2)+x**4),x)

[Out]

exp(-75/(x**5 + x**2 - x*log(2)))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 56 vs. \(2 (17) = 34\).

Time = 0.37 (sec) , antiderivative size = 56, normalized size of antiderivative = 3.11 \[ \int \frac {e^{-\frac {75}{x^2+x^5-x \log (2)}} \left (150 x+375 x^4-75 \log (2)\right )}{x^4+x^{10}-2 x^3 \log (2)+x^2 \log ^2(2)+x^4 \left (2 x^3-2 x^2 \log (2)\right )} \, dx=e^{\left (-\frac {75 \, x^{3}}{x^{4} \log \left (2\right ) + x \log \left (2\right ) - \log \left (2\right )^{2}} - \frac {75}{x^{4} \log \left (2\right ) + x \log \left (2\right ) - \log \left (2\right )^{2}} + \frac {75}{x \log \left (2\right )}\right )} \]

[In]

integrate((375*x^4-75*log(2)+150*x)*exp(-75/(x^5-x*log(2)+x^2))/(x^10+(-2*x^2*log(2)+2*x^3)*x^4+x^2*log(2)^2-2
*x^3*log(2)+x^4),x, algorithm="maxima")

[Out]

e^(-75*x^3/(x^4*log(2) + x*log(2) - log(2)^2) - 75/(x^4*log(2) + x*log(2) - log(2)^2) + 75/(x*log(2)))

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.94 \[ \int \frac {e^{-\frac {75}{x^2+x^5-x \log (2)}} \left (150 x+375 x^4-75 \log (2)\right )}{x^4+x^{10}-2 x^3 \log (2)+x^2 \log ^2(2)+x^4 \left (2 x^3-2 x^2 \log (2)\right )} \, dx=e^{\left (-\frac {75}{x^{5} + x^{2} - x \log \left (2\right )}\right )} \]

[In]

integrate((375*x^4-75*log(2)+150*x)*exp(-75/(x^5-x*log(2)+x^2))/(x^10+(-2*x^2*log(2)+2*x^3)*x^4+x^2*log(2)^2-2
*x^3*log(2)+x^4),x, algorithm="giac")

[Out]

e^(-75/(x^5 + x^2 - x*log(2)))

Mupad [B] (verification not implemented)

Time = 10.58 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.94 \[ \int \frac {e^{-\frac {75}{x^2+x^5-x \log (2)}} \left (150 x+375 x^4-75 \log (2)\right )}{x^4+x^{10}-2 x^3 \log (2)+x^2 \log ^2(2)+x^4 \left (2 x^3-2 x^2 \log (2)\right )} \, dx={\mathrm {e}}^{-\frac {75}{x^5+x^2-\ln \left (2\right )\,x}} \]

[In]

int((exp(-75/(x^2 - x*log(2) + x^5))*(150*x - 75*log(2) + 375*x^4))/(x^2*log(2)^2 - x^4*(2*x^2*log(2) - 2*x^3)
- 2*x^3*log(2) + x^4 + x^10),x)

[Out]

exp(-75/(x^2 - x*log(2) + x^5))