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

Optimal. Leaf size=23 \[ -e^{\frac {10}{x (-x+\log (2))}}+x-\log (5) \]

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Rubi [A]  time = 0.93, antiderivative size = 19, normalized size of antiderivative = 0.83, number of steps used = 5, number of rules used = 4, integrand size = 66, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.061, Rules used = {1594, 27, 6688, 6706} \begin {gather*} x-e^{-\frac {10}{x (x-\log (2))}} \end {gather*}

Antiderivative was successfully verified.

[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 1594

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 6688

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

Rule 6706

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 {gather*} \begin {aligned} \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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.18, size = 30, normalized size = 1.30 \begin {gather*} -e^{\frac {10}{x \log (2)}-\frac {10}{(x-\log (2)) \log (2)}}+x \end {gather*}

Antiderivative was successfully verified.

[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

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fricas [A]  time = 0.54, size = 18, normalized size = 0.78 \begin {gather*} x - e^{\left (-\frac {10}{x^{2} - x \log \relax (2)}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)))

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giac [A]  time = 0.38, size = 18, normalized size = 0.78 \begin {gather*} x - e^{\left (-\frac {10}{x^{2} - x \log \relax (2)}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)))

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maple [A]  time = 0.27, size = 19, normalized size = 0.83

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

Verification of antiderivative is not currently implemented for this CAS.

[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

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maxima [B]  time = 0.67, size = 80, normalized size = 3.48 \begin {gather*} 2 \, {\left (\frac {\log \relax (2)}{x - \log \relax (2)} - \log \left (x - \log \relax (2)\right )\right )} \log \relax (2) + 2 \, \log \relax (2) \log \left (x - \log \relax (2)\right ) + x - \frac {2 \, \log \relax (2)^{2}}{x - \log \relax (2)} - e^{\left (-\frac {10}{x \log \relax (2) - \log \relax (2)^{2}} + \frac {10}{x \log \relax (2)}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)))

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mupad [B]  time = 1.27, size = 19, normalized size = 0.83 \begin {gather*} x-{\mathrm {e}}^{\frac {10}{x\,\ln \relax (2)-x^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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sympy [A]  time = 0.26, size = 12, normalized size = 0.52 \begin {gather*} x - e^{\frac {10}{- x^{2} + x \log {\relax (2 )}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)))

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