3.92.70 \(\int \frac {e^3 (4+8 \log (2+e^5)+4 \log ^2(2+e^5))}{(\frac {e^9}{x^5}-\frac {3 e^6}{x^3}+\frac {3 e^3}{x}-x) x^2} \, dx\)

Optimal. Leaf size=24 \[ 2+\frac {\left (1+\log \left (2+e^5\right )\right )^2}{\left (-1+\frac {e^3}{x^2}\right )^2} \]

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Rubi [A]  time = 0.11, antiderivative size = 25, normalized size of antiderivative = 1.04, number of steps used = 3, number of rules used = 3, integrand size = 56, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.054, Rules used = {12, 6688, 264} \begin {gather*} \frac {x^4 \left (1+\log \left (2+e^5\right )\right )^2}{\left (e^3-x^2\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^3*(4 + 8*Log[2 + E^5] + 4*Log[2 + E^5]^2))/((E^9/x^5 - (3*E^6)/x^3 + (3*E^3)/x - x)*x^2),x]

[Out]

(x^4*(1 + Log[2 + E^5])^2)/(E^3 - x^2)^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 264

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a
*c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[(m + 1)/n + p + 1, 0] && NeQ[m, -1]

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} &=\left (4 e^3 \left (1+\log \left (2+e^5\right )\right )^2\right ) \int \frac {1}{\left (\frac {e^9}{x^5}-\frac {3 e^6}{x^3}+\frac {3 e^3}{x}-x\right ) x^2} \, dx\\ &=\left (4 e^3 \left (1+\log \left (2+e^5\right )\right )^2\right ) \int \frac {x^3}{\left (e^3-x^2\right )^3} \, dx\\ &=\frac {x^4 \left (1+\log \left (2+e^5\right )\right )^2}{\left (e^3-x^2\right )^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 35, normalized size = 1.46 \begin {gather*} -\frac {e^3 \left (e^3-2 x^2\right ) \left (1+\log \left (2+e^5\right )\right )^2}{\left (e^3-x^2\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^3*(4 + 8*Log[2 + E^5] + 4*Log[2 + E^5]^2))/((E^9/x^5 - (3*E^6)/x^3 + (3*E^3)/x - x)*x^2),x]

[Out]

-((E^3*(E^3 - 2*x^2)*(1 + Log[2 + E^5])^2)/(E^3 - x^2)^2)

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fricas [B]  time = 0.49, size = 67, normalized size = 2.79 \begin {gather*} \frac {2 \, x^{2} e^{3} + {\left (2 \, x^{2} e^{3} - e^{6}\right )} \log \left (e^{5} + 2\right )^{2} + 2 \, {\left (2 \, x^{2} e^{3} - e^{6}\right )} \log \left (e^{5} + 2\right ) - e^{6}}{x^{4} - 2 \, x^{2} e^{3} + e^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*log(exp(5)+2)^2+8*log(exp(5)+2)+4)*exp(3-log(x^2))/(x*exp(3-log(x^2))^3-3*x*exp(3-log(x^2))^2+3*x
*exp(3-log(x^2))-x),x, algorithm="fricas")

[Out]

(2*x^2*e^3 + (2*x^2*e^3 - e^6)*log(e^5 + 2)^2 + 2*(2*x^2*e^3 - e^6)*log(e^5 + 2) - e^6)/(x^4 - 2*x^2*e^3 + e^6
)

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giac [A]  time = 0.13, size = 1, normalized size = 0.04 \begin {gather*} 0 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*log(exp(5)+2)^2+8*log(exp(5)+2)+4)*exp(3-log(x^2))/(x*exp(3-log(x^2))^3-3*x*exp(3-log(x^2))^2+3*x
*exp(3-log(x^2))-x),x, algorithm="giac")

[Out]

0

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maple [A]  time = 0.10, size = 47, normalized size = 1.96

method	result	size
risch	\(\frac {\left (4 \ln \left ({\mathrm e}^{5}+2\right )^{2}+8 \ln \left ({\mathrm e}^{5}+2\right )+4\right ) {\mathrm e}^{3} \left (-\frac {{\mathrm e}^{3}}{4}+\frac {x^{2}}{2}\right )}{{\mathrm e}^{6}-2 x^{2} {\mathrm e}^{3}+x^{4}}\)	\(47\)
norman	\(\frac {-{\mathrm e}^{6} \left (\ln \left ({\mathrm e}^{5}+2\right )^{2}+2 \ln \left ({\mathrm e}^{5}+2\right )+1\right ) x +\left (2 \,{\mathrm e}^{3} \ln \left ({\mathrm e}^{5}+2\right )^{2}+4 \,{\mathrm e}^{3} \ln \left ({\mathrm e}^{5}+2\right )+2 \,{\mathrm e}^{3}\right ) x^{3}}{x \left (-x^{2}+{\mathrm e}^{3}\right )^{2}}\)	\(68\)
default	\(\frac {\left (4 \ln \left ({\mathrm e}^{5}+2\right )^{2}+8 \ln \left ({\mathrm e}^{5}+2\right )+4\right ) {\mathrm e}^{3-\ln \left (x^{2}\right )+2 \ln \relax (x )} \left (\munderset {\textit {\_R} =\RootOf \left (-3 \,{\mathrm e}^{3-\ln \left (x^{2}\right )+2 \ln \relax (x )} \textit {\_Z}^{2}+\textit {\_Z}^{3}+3 \,{\mathrm e}^{6-2 \ln \left (x^{2}\right )+4 \ln \relax (x )} \textit {\_Z} -{\mathrm e}^{9-3 \ln \left (x^{2}\right )+6 \ln \relax (x )}\right )}{\sum }\frac {\textit {\_R} \ln \left (x^{2}-\textit {\_R} \right )}{-\textit {\_R}^{2}+2 \,{\mathrm e}^{3-\ln \left (x^{2}\right )+2 \ln \relax (x )} \textit {\_R} -{\mathrm e}^{6-2 \ln \left (x^{2}\right )+4 \ln \relax (x )}}\right )}{6}\)	\(140\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*ln(exp(5)+2)^2+8*ln(exp(5)+2)+4)*exp(3-ln(x^2))/(x*exp(3-ln(x^2))^3-3*x*exp(3-ln(x^2))^2+3*x*exp(3-ln(x
^2))-x),x,method=_RETURNVERBOSE)

[Out]

(4*ln(exp(5)+2)^2+8*ln(exp(5)+2)+4)*exp(3)*(-1/4*exp(3)+1/2*x^2)/(exp(6)-2*x^2*exp(3)+x^4)

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maxima [A]  time = 0.35, size = 44, normalized size = 1.83 \begin {gather*} \frac {{\left (2 \, x^{2} - e^{3}\right )} {\left (\log \left (e^{5} + 2\right )^{2} + 2 \, \log \left (e^{5} + 2\right ) + 1\right )} e^{3}}{x^{4} - 2 \, x^{2} e^{3} + e^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*log(exp(5)+2)^2+8*log(exp(5)+2)+4)*exp(3-log(x^2))/(x*exp(3-log(x^2))^3-3*x*exp(3-log(x^2))^2+3*x
*exp(3-log(x^2))-x),x, algorithm="maxima")

[Out]

(2*x^2 - e^3)*(log(e^5 + 2)^2 + 2*log(e^5 + 2) + 1)*e^3/(x^4 - 2*x^2*e^3 + e^6)

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mupad [B]  time = 0.21, size = 35, normalized size = 1.46 \begin {gather*} \frac {x^4\,\left (2\,\ln \left ({\mathrm {e}}^5+2\right )+{\ln \left ({\mathrm {e}}^5+2\right )}^2+1\right )}{x^4-2\,{\mathrm {e}}^3\,x^2+{\mathrm {e}}^6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(3 - log(x^2))*(8*log(exp(5) + 2) + 4*log(exp(5) + 2)^2 + 4))/(x - 3*x*exp(3 - log(x^2)) + 3*x*exp(6
- 2*log(x^2)) - x*exp(9 - 3*log(x^2))),x)

[Out]

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

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sympy [B]  time = 0.25, size = 58, normalized size = 2.42 \begin {gather*} \frac {\left (- 2 x^{2} + e^{3}\right ) \left (- 4 e^{3} \log {\left (2 + e^{5} \right )}^{2} - 8 e^{3} \log {\left (2 + e^{5} \right )} - 4 e^{3}\right )}{4 x^{4} - 8 x^{2} e^{3} + 4 e^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*ln(exp(5)+2)**2+8*ln(exp(5)+2)+4)*exp(3-ln(x**2))/(x*exp(3-ln(x**2))**3-3*x*exp(3-ln(x**2))**2+3*
x*exp(3-ln(x**2))-x),x)

[Out]

(-2*x**2 + exp(3))*(-4*exp(3)*log(2 + exp(5))**2 - 8*exp(3)*log(2 + exp(5)) - 4*exp(3))/(4*x**4 - 8*x**2*exp(3
) + 4*exp(6))

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