\(\int \frac {x^4-2 x^3 \log (2)+x^2 \log ^2(2)+e^{\frac {10}{-x^2+x \log (2)}} (-20 x+10 \log (2))}{x^4-2 x^3 \log (2)+x^2 \log ^2(2)} \, dx\) [1512]

Optimal result

Integrand size = 66, antiderivative size = 23 \[ \int \frac {x^4-2 x^3 \log (2)+x^2 \log ^2(2)+e^{\frac {10}{-x^2+x \log (2)}} (-20 x+10 \log (2))}{x^4-2 x^3 \log (2)+x^2 \log ^2(2)} \, dx=-e^{\frac {10}{x (-x+\log (2))}}+x-\log (5) \]

[Out]

x-ln(5)-exp(25/(5*ln(2)-5*x)/x)^2

Rubi [A] (verified)

Time = 0.58 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.061, Rules used = {1608, 27, 6820, 6838} \[ \int \frac {x^4-2 x^3 \log (2)+x^2 \log ^2(2)+e^{\frac {10}{-x^2+x \log (2)}} (-20 x+10 \log (2))}{x^4-2 x^3 \log (2)+x^2 \log ^2(2)} \, dx=x-e^{-\frac {10}{x (x-\log (2))}} \]

[In]

Int[(x^4 - 2*x^3*Log[2] + x^2*Log[2]^2 + E^(10/(-x^2 + x*Log[2]))*(-20*x + 10*Log[2]))/(x^4 - 2*x^3*Log[2] + x
^2*Log[2]^2),x]

[Out]

-E^(-10/(x*(x - Log[2]))) + x

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 1608

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^
(q - p) + c*x^(r - p))^n, x] /; FreeQ[{a, b, c, p, q, r}, x] && IntegerQ[n] && PosQ[q - p] && PosQ[r - p]

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 {x^4-2 x^3 \log (2)+x^2 \log ^2(2)+e^{\frac {10}{-x^2+x \log (2)}} (-20 x+10 \log (2))}{x^2 \left (x^2-2 x \log (2)+\log ^2(2)\right )} \, dx \\ & = \int \frac {x^4-2 x^3 \log (2)+x^2 \log ^2(2)+e^{\frac {10}{-x^2+x \log (2)}} (-20 x+10 \log (2))}{x^2 (x-\log (2))^2} \, dx \\ & = \int \left (1+\frac {10 e^{-\frac {10}{x (x-\log (2))}} (-2 x+\log (2))}{x^2 (x-\log (2))^2}\right ) \, dx \\ & = x+10 \int \frac {e^{-\frac {10}{x (x-\log (2))}} (-2 x+\log (2))}{x^2 (x-\log (2))^2} \, dx \\ & = -e^{-\frac {10}{x (x-\log (2))}}+x \\ \end{align*}

Mathematica [A] (verified)

Time = 0.56 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.30 \[ \int \frac {x^4-2 x^3 \log (2)+x^2 \log ^2(2)+e^{\frac {10}{-x^2+x \log (2)}} (-20 x+10 \log (2))}{x^4-2 x^3 \log (2)+x^2 \log ^2(2)} \, dx=-e^{\frac {10}{x \log (2)}-\frac {10}{(x-\log (2)) \log (2)}}+x \]

[In]

Integrate[(x^4 - 2*x^3*Log[2] + x^2*Log[2]^2 + E^(10/(-x^2 + x*Log[2]))*(-20*x + 10*Log[2]))/(x^4 - 2*x^3*Log[
2] + x^2*Log[2]^2),x]

[Out]

-E^(10/(x*Log[2]) - 10/((x - Log[2])*Log[2])) + x

Maple [A] (verified)

Time = 0.44 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83

method	result	size
risch	\(-{\mathrm e}^{\frac {10}{x \left (\ln \left (2\right )-x \right )}}+x\)	\(19\)
parts	\(x +\frac {x^{2} {\mathrm e}^{\frac {10}{x \ln \left (2\right )-x^{2}}}-x \ln \left (2\right ) {\mathrm e}^{\frac {10}{x \ln \left (2\right )-x^{2}}}}{x \left (\ln \left (2\right )-x \right )}\)	\(59\)
norman	\(\frac {x^{2} {\mathrm e}^{\frac {10}{x \ln \left (2\right )-x^{2}}}+x \ln \left (2\right )^{2}-x^{3}-x \ln \left (2\right ) {\mathrm e}^{\frac {10}{x \ln \left (2\right )-x^{2}}}}{\left (\ln \left (2\right )-x \right ) x}\)	\(68\)

[In]

int(((10*ln(2)-20*x)*exp(5/(x*ln(2)-x^2))^2+x^2*ln(2)^2-2*x^3*ln(2)+x^4)/(x^2*ln(2)^2-2*x^3*ln(2)+x^4),x,metho
d=_RETURNVERBOSE)

[Out]

-exp(10/x/(ln(2)-x))+x

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.78 \[ \int \frac {x^4-2 x^3 \log (2)+x^2 \log ^2(2)+e^{\frac {10}{-x^2+x \log (2)}} (-20 x+10 \log (2))}{x^4-2 x^3 \log (2)+x^2 \log ^2(2)} \, dx=x - e^{\left (-\frac {10}{x^{2} - x \log \left (2\right )}\right )} \]

[In]

integrate(((10*log(2)-20*x)*exp(5/(x*log(2)-x^2))^2+x^2*log(2)^2-2*x^3*log(2)+x^4)/(x^2*log(2)^2-2*x^3*log(2)+
x^4),x, algorithm="fricas")

[Out]

x - e^(-10/(x^2 - x*log(2)))

Sympy [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.52 \[ \int \frac {x^4-2 x^3 \log (2)+x^2 \log ^2(2)+e^{\frac {10}{-x^2+x \log (2)}} (-20 x+10 \log (2))}{x^4-2 x^3 \log (2)+x^2 \log ^2(2)} \, dx=x - e^{\frac {10}{- x^{2} + x \log {\left (2 \right )}}} \]

[In]

integrate(((10*ln(2)-20*x)*exp(5/(x*ln(2)-x**2))**2+x**2*ln(2)**2-2*x**3*ln(2)+x**4)/(x**2*ln(2)**2-2*x**3*ln(
2)+x**4),x)

[Out]

x - exp(10/(-x**2 + x*log(2)))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 80 vs. \(2 (22) = 44\).

Time = 0.34 (sec) , antiderivative size = 80, normalized size of antiderivative = 3.48 \[ \int \frac {x^4-2 x^3 \log (2)+x^2 \log ^2(2)+e^{\frac {10}{-x^2+x \log (2)}} (-20 x+10 \log (2))}{x^4-2 x^3 \log (2)+x^2 \log ^2(2)} \, dx=2 \, {\left (\frac {\log \left (2\right )}{x - \log \left (2\right )} - \log \left (x - \log \left (2\right )\right )\right )} \log \left (2\right ) + 2 \, \log \left (2\right ) \log \left (x - \log \left (2\right )\right ) + x - \frac {2 \, \log \left (2\right )^{2}}{x - \log \left (2\right )} - e^{\left (-\frac {10}{x \log \left (2\right ) - \log \left (2\right )^{2}} + \frac {10}{x \log \left (2\right )}\right )} \]

[In]

integrate(((10*log(2)-20*x)*exp(5/(x*log(2)-x^2))^2+x^2*log(2)^2-2*x^3*log(2)+x^4)/(x^2*log(2)^2-2*x^3*log(2)+
x^4),x, algorithm="maxima")

[Out]

2*(log(2)/(x - log(2)) - log(x - log(2)))*log(2) + 2*log(2)*log(x - log(2)) + x - 2*log(2)^2/(x - log(2)) - e^
(-10/(x*log(2) - log(2)^2) + 10/(x*log(2)))

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.78 \[ \int \frac {x^4-2 x^3 \log (2)+x^2 \log ^2(2)+e^{\frac {10}{-x^2+x \log (2)}} (-20 x+10 \log (2))}{x^4-2 x^3 \log (2)+x^2 \log ^2(2)} \, dx=x - e^{\left (-\frac {10}{x^{2} - x \log \left (2\right )}\right )} \]

[In]

integrate(((10*log(2)-20*x)*exp(5/(x*log(2)-x^2))^2+x^2*log(2)^2-2*x^3*log(2)+x^4)/(x^2*log(2)^2-2*x^3*log(2)+
x^4),x, algorithm="giac")

[Out]

x - e^(-10/(x^2 - x*log(2)))

Mupad [B] (verification not implemented)

Time = 9.24 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {x^4-2 x^3 \log (2)+x^2 \log ^2(2)+e^{\frac {10}{-x^2+x \log (2)}} (-20 x+10 \log (2))}{x^4-2 x^3 \log (2)+x^2 \log ^2(2)} \, dx=x-{\mathrm {e}}^{\frac {10}{x\,\ln \left (2\right )-x^2}} \]

[In]

int((x^2*log(2)^2 - exp(10/(x*log(2) - x^2))*(20*x - 10*log(2)) - 2*x^3*log(2) + x^4)/(x^2*log(2)^2 - 2*x^3*lo
g(2) + x^4),x)

[Out]

x - exp(10/(x*log(2) - x^2))