3.93.58 \(\int \frac {-e^{12}+2 e^6 x^2-x^4+(e^{12}-2 e^6 x^2+x^4) \log (x)+(e^{12} (-1-2 x)+x^2-x^4-2 x^5+e^6 (1+2 x^2+4 x^3)) \log ^2(x)}{(e^{12}-2 e^6 x^2+x^4) \log ^2(x)} \, dx\)

Optimal. Leaf size=36 \[ -1+x \left (-x+\frac {5-x+\frac {x}{e^6-x^2}}{x}\right )+\frac {x}{\log (x)} \]

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Rubi [A]  time = 0.26, antiderivative size = 28, normalized size of antiderivative = 0.78, number of steps used = 9, number of rules used = 6, integrand size = 103, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.058, Rules used = {28, 6688, 1814, 1586, 2297, 2298} \begin {gather*} -x^2+\frac {x}{e^6-x^2}-x+\frac {x}{\log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-E^12 + 2*E^6*x^2 - x^4 + (E^12 - 2*E^6*x^2 + x^4)*Log[x] + (E^12*(-1 - 2*x) + x^2 - x^4 - 2*x^5 + E^6*(1
+ 2*x^2 + 4*x^3))*Log[x]^2)/((E^12 - 2*E^6*x^2 + x^4)*Log[x]^2),x]

[Out]

-x - x^2 + x/(E^6 - x^2) + x/Log[x]

Rule 28

Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/c^p, Int[u*(b/2 + c*x^n)^(2*
p), x], x] /; FreeQ[{a, b, c, n}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 1586

Int[(u_.)*(Px_)^(p_.)*(Qx_)^(q_.), x_Symbol] :> Int[u*PolynomialQuotient[Px, Qx, x]^p*Qx^(p + q), x] /; FreeQ[
q, x] && PolyQ[Px, x] && PolyQ[Qx, x] && EqQ[PolynomialRemainder[Px, Qx, x], 0] && IntegerQ[p] && LtQ[p*q, 0]

Rule 1814

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, a + b*x^2, x], f = Coeff[P
olynomialRemainder[Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 1]}, Simp[((a
*g - b*f*x)*(a + b*x^2)^(p + 1))/(2*a*b*(p + 1)), x] + Dist[1/(2*a*(p + 1)), Int[(a + b*x^2)^(p + 1)*ExpandToS
um[2*a*(p + 1)*Q + f*(2*p + 3), x], x], x]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && LtQ[p, -1]

Rule 2297

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_), x_Symbol] :> Simp[(x*(a + b*Log[c*x^n])^(p + 1))/(b*n*(p + 1))
, x] - Dist[1/(b*n*(p + 1)), Int[(a + b*Log[c*x^n])^(p + 1), x], x] /; FreeQ[{a, b, c, n}, x] && LtQ[p, -1] &&
IntegerQ[2*p]

Rule 2298

Int[Log[(c_.)*(x_)]^(-1), x_Symbol] :> Simp[LogIntegral[c*x]/c, x] /; FreeQ[c, x]

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} &=\int \frac {-e^{12}+2 e^6 x^2-x^4+\left (e^{12}-2 e^6 x^2+x^4\right ) \log (x)+\left (e^{12} (-1-2 x)+x^2-x^4-2 x^5+e^6 \left (1+2 x^2+4 x^3\right )\right ) \log ^2(x)}{\left (-e^6+x^2\right )^2 \log ^2(x)} \, dx\\ &=\int \left (\frac {x^2-x^4-2 x^5-e^{12} (1+2 x)+e^6 \left (1+2 x^2+4 x^3\right )}{\left (e^6-x^2\right )^2}-\frac {1}{\log ^2(x)}+\frac {1}{\log (x)}\right ) \, dx\\ &=\int \frac {x^2-x^4-2 x^5-e^{12} (1+2 x)+e^6 \left (1+2 x^2+4 x^3\right )}{\left (e^6-x^2\right )^2} \, dx-\int \frac {1}{\log ^2(x)} \, dx+\int \frac {1}{\log (x)} \, dx\\ &=\frac {x}{e^6-x^2}+\frac {x}{\log (x)}+\text {li}(x)-\frac {\int \frac {2 e^{12}+4 e^{12} x-2 e^6 x^2-4 e^6 x^3}{e^6-x^2} \, dx}{2 e^6}-\int \frac {1}{\log (x)} \, dx\\ &=\frac {x}{e^6-x^2}+\frac {x}{\log (x)}-\frac {\int \left (2 e^6+4 e^6 x\right ) \, dx}{2 e^6}\\ &=-x-x^2+\frac {x}{e^6-x^2}+\frac {x}{\log (x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.04, size = 29, normalized size = 0.81 \begin {gather*} -x-x^2-\frac {x}{-e^6+x^2}+\frac {x}{\log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-E^12 + 2*E^6*x^2 - x^4 + (E^12 - 2*E^6*x^2 + x^4)*Log[x] + (E^12*(-1 - 2*x) + x^2 - x^4 - 2*x^5 +
E^6*(1 + 2*x^2 + 4*x^3))*Log[x]^2)/((E^12 - 2*E^6*x^2 + x^4)*Log[x]^2),x]

[Out]

-x - x^2 - x/(-E^6 + x^2) + x/Log[x]

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fricas [A]  time = 0.88, size = 45, normalized size = 1.25 \begin {gather*} \frac {x^{3} - x e^{6} - {\left (x^{4} + x^{3} - {\left (x^{2} + x\right )} e^{6} + x\right )} \log \relax (x)}{{\left (x^{2} - e^{6}\right )} \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-2*x-1)*exp(6)^2+(4*x^3+2*x^2+1)*exp(6)-2*x^5-x^4+x^2)*log(x)^2+(exp(6)^2-2*x^2*exp(6)+x^4)*log(x
)-exp(6)^2+2*x^2*exp(6)-x^4)/(exp(6)^2-2*x^2*exp(6)+x^4)/log(x)^2,x, algorithm="fricas")

[Out]

(x^3 - x*e^6 - (x^4 + x^3 - (x^2 + x)*e^6 + x)*log(x))/((x^2 - e^6)*log(x))

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giac [A]  time = 0.16, size = 60, normalized size = 1.67 \begin {gather*} -\frac {x^{4} \log \relax (x) + x^{3} \log \relax (x) - x^{2} e^{6} \log \relax (x) - x^{3} - x e^{6} \log \relax (x) + x e^{6} + 2 \, x \log \relax (x)}{x^{2} \log \relax (x) - e^{6} \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-2*x-1)*exp(6)^2+(4*x^3+2*x^2+1)*exp(6)-2*x^5-x^4+x^2)*log(x)^2+(exp(6)^2-2*x^2*exp(6)+x^4)*log(x
)-exp(6)^2+2*x^2*exp(6)-x^4)/(exp(6)^2-2*x^2*exp(6)+x^4)/log(x)^2,x, algorithm="giac")

[Out]

-(x^4*log(x) + x^3*log(x) - x^2*e^6*log(x) - x^3 - x*e^6*log(x) + x*e^6 + 2*x*log(x))/(x^2*log(x) - e^6*log(x)
)

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maple [A]  time = 0.14, size = 28, normalized size = 0.78

method	result	size
default	\(\frac {x}{\ln \relax (x )}-x^{2}-x +\frac {x}{{\mathrm e}^{6}-x^{2}}\)	\(28\)
risch	\(-\frac {x \left (-x^{3}+x \,{\mathrm e}^{6}-x^{2}+{\mathrm e}^{6}-1\right )}{{\mathrm e}^{6}-x^{2}}+\frac {x}{\ln \relax (x )}\)	\(39\)
norman	\(\frac {x \,{\mathrm e}^{6}+x^{3} \ln \relax (x )+x^{4} \ln \relax (x )-\ln \relax (x ) {\mathrm e}^{12}+\left (-{\mathrm e}^{6}+1\right ) x \ln \relax (x )-x^{3}}{\left ({\mathrm e}^{6}-x^{2}\right ) \ln \relax (x )}\)	\(56\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-2*x-1)*exp(6)^2+(4*x^3+2*x^2+1)*exp(6)-2*x^5-x^4+x^2)*ln(x)^2+(exp(6)^2-2*x^2*exp(6)+x^4)*ln(x)-exp(6)
^2+2*x^2*exp(6)-x^4)/(exp(6)^2-2*x^2*exp(6)+x^4)/ln(x)^2,x,method=_RETURNVERBOSE)

[Out]

x/ln(x)-x^2-x+x/(exp(6)-x^2)

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maxima [A]  time = 0.39, size = 49, normalized size = 1.36 \begin {gather*} \frac {x^{3} - x e^{6} - {\left (x^{4} + x^{3} - x^{2} e^{6} - x {\left (e^{6} - 1\right )}\right )} \log \relax (x)}{{\left (x^{2} - e^{6}\right )} \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-2*x-1)*exp(6)^2+(4*x^3+2*x^2+1)*exp(6)-2*x^5-x^4+x^2)*log(x)^2+(exp(6)^2-2*x^2*exp(6)+x^4)*log(x
)-exp(6)^2+2*x^2*exp(6)-x^4)/(exp(6)^2-2*x^2*exp(6)+x^4)/log(x)^2,x, algorithm="maxima")

[Out]

(x^3 - x*e^6 - (x^4 + x^3 - x^2*e^6 - x*(e^6 - 1))*log(x))/((x^2 - e^6)*log(x))

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mupad [B]  time = 6.25, size = 27, normalized size = 0.75 \begin {gather*} \frac {x}{\ln \relax (x)}-x+\frac {x}{{\mathrm {e}}^6-x^2}-x^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(12) - log(x)*(exp(12) - 2*x^2*exp(6) + x^4) - 2*x^2*exp(6) + log(x)^2*(x^4 - x^2 - exp(6)*(2*x^2 + 4
*x^3 + 1) + 2*x^5 + exp(12)*(2*x + 1)) + x^4)/(log(x)^2*(exp(12) - 2*x^2*exp(6) + x^4)),x)

[Out]

x/log(x) - x + x/(exp(6) - x^2) - x^2

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sympy [A]  time = 0.23, size = 17, normalized size = 0.47 \begin {gather*} - x^{2} - x + \frac {x}{\log {\relax (x )}} - \frac {x}{x^{2} - e^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-2*x-1)*exp(6)**2+(4*x**3+2*x**2+1)*exp(6)-2*x**5-x**4+x**2)*ln(x)**2+(exp(6)**2-2*x**2*exp(6)+x*
*4)*ln(x)-exp(6)**2+2*x**2*exp(6)-x**4)/(exp(6)**2-2*x**2*exp(6)+x**4)/ln(x)**2,x)

[Out]

-x**2 - x + x/log(x) - x/(x**2 - exp(6))

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