∫ 5 log4(2)+5e2x log2(3)+ex(−3x log2(2)−10 log2(2) log(3))_________________________________________ 5x log4(2)−10exx log2(2) log(3)+5e2xx log2(3) dx [5910]

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

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Rubi [A]
time = 0.64, antiderivative size = 29, normalized size of antiderivative = 1.21, number of steps used = 11, number of rules used = 8, integrand size = 73, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.110, Rules used = {6873, 12, 6874, 2320, 46, 36, 29, 31} \begin {gather*} \log (x)-\frac {3 \log ^2(2)}{5 \log (3) \left (\log ^2(2)-e^x \log (3)\right )} \end {gather*}

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

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

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -

Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x

)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] && !(IGtQ[n, 0] && Lt

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu

nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] && !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F

reeQ[{a, m, n}, x] && IntegerQ[m*n]] && !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x

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

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

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {5 \log ^4(2)+5 e^{2 x} \log ^2(3)+e^x \left (-3 x \log ^2(2)-10 \log ^2(2) \log (3)\right )}{5 x \left (\log ^2(2)-e^x \log (3)\right )^2} \, dx\\ &=\frac {1}{5} \int \frac {5 \log ^4(2)+5 e^{2 x} \log ^2(3)+e^x \left (-3 x \log ^2(2)-10 \log ^2(2) \log (3)\right )}{x \left (\log ^2(2)-e^x \log (3)\right )^2} \, dx\\ &=\frac {1}{5} \int \left (\frac {5}{x}-\frac {3 \log ^4(2)}{\log (3) \left (\log ^2(2)-e^x \log (3)\right )^2}+\frac {3 \log ^2(2)}{\log (3) \left (\log ^2(2)-e^x \log (3)\right )}\right ) \, dx\\ &=\log (x)+\frac {\left (3 \log ^2(2)\right ) \int \frac {1}{\log ^2(2)-e^x \log (3)} \, dx}{5 \log (3)}-\frac {\left (3 \log ^4(2)\right ) \int \frac {1}{\left (\log ^2(2)-e^x \log (3)\right )^2} \, dx}{5 \log (3)}\\ &=\log (x)+\frac {\left (3 \log ^2(2)\right ) \text {Subst}\left (\int \frac {1}{x \left (\log ^2(2)-x \log (3)\right )} \, dx,x,e^x\right )}{5 \log (3)}-\frac {\left (3 \log ^4(2)\right ) \text {Subst}\left (\int \frac {1}{x \left (\log ^2(2)-x \log (3)\right )^2} \, dx,x,e^x\right )}{5 \log (3)}\\ &=\log (x)+\frac {3}{5} \text {Subst}\left (\int \frac {1}{\log ^2(2)-x \log (3)} \, dx,x,e^x\right )+\frac {3 \text {Subst}\left (\int \frac {1}{x} \, dx,x,e^x\right )}{5 \log (3)}-\frac {\left (3 \log ^4(2)\right ) \text {Subst}\left (\int \left (\frac {1}{x \log ^4(2)}+\frac {\log (3)}{\log ^2(2) \left (\log ^2(2)-x \log (3)\right )^2}+\frac {\log (3)}{\log ^4(2) \left (\log ^2(2)-x \log (3)\right )}\right ) \, dx,x,e^x\right )}{5 \log (3)}\\ &=-\frac {3 \log ^2(2)}{5 \log (3) \left (\log ^2(2)-e^x \log (3)\right )}+\log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]
time = 0.12, size = 34, normalized size = 1.42 \begin {gather*} \frac {1}{5} \left (\frac {3 \log ^2(2)}{\log (3) \left (-\log ^2(2)+e^x \log (3)\right )}+5 \log (x)\right ) \end {gather*}

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

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method	result	size

norman	\(\frac {3 \,{\mathrm e}^{x}}{5 \left (\ln \left (3\right ) {\mathrm e}^{x}-\ln \left (2\right )^{2}\right )}+\ln \left (x \right )\)	\(22\)
risch	\(\ln \left (x \right )+\frac {3 \ln \left (2\right )^{2}}{5 \ln \left (3\right ) \left (\ln \left (3\right ) {\mathrm e}^{x}-\ln \left (2\right )^{2}\right )}\)	\(28\)

int((5*ln(3)^2*exp(x)^2+(-10*ln(2)^2*ln(3)-3*x*ln(2)^2)*exp(x)+5*ln(2)^4)/(5*x*ln(3)^2*exp(x)^2-10*x*ln(2)^2*l

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Maxima [A]
time = 0.49, size = 27, normalized size = 1.12 \begin {gather*} \frac {3 \, \log \left (2\right )^{2}}{5 \, {\left (e^{x} \log \left (3\right )^{2} - \log \left (3\right ) \log \left (2\right )^{2}\right )}} + \log \left (x\right ) \end {gather*}

integrate((5*log(3)^2*exp(x)^2+(-10*log(2)^2*log(3)-3*x*log(2)^2)*exp(x)+5*log(2)^4)/(5*x*log(3)^2*exp(x)^2-10

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 47 vs. \(2 (21) = 42\).
time = 0.43, size = 47, normalized size = 1.96 \begin {gather*} \frac {3 \, \log \left (2\right )^{2} + 5 \, {\left (e^{x} \log \left (3\right )^{2} - \log \left (3\right ) \log \left (2\right )^{2}\right )} \log \left (x\right )}{5 \, {\left (e^{x} \log \left (3\right )^{2} - \log \left (3\right ) \log \left (2\right )^{2}\right )}} \end {gather*}

integrate((5*log(3)^2*exp(x)^2+(-10*log(2)^2*log(3)-3*x*log(2)^2)*exp(x)+5*log(2)^4)/(5*x*log(3)^2*exp(x)^2-10

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

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Sympy [A]
time = 0.04, size = 29, normalized size = 1.21 \begin {gather*} \log {\left (x \right )} + \frac {3 \log {\left (2 \right )}^{2}}{5 e^{x} \log {\left (3 \right )}^{2} - 5 \log {\left (2 \right )}^{2} \log {\left (3 \right )}} \end {gather*}

integrate((5*ln(3)**2*exp(x)**2+(-10*ln(2)**2*ln(3)-3*x*ln(2)**2)*exp(x)+5*ln(2)**4)/(5*x*ln(3)**2*exp(x)**2-1

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 47 vs. \(2 (21) = 42\).
time = 0.40, size = 47, normalized size = 1.96 \begin {gather*} \frac {5 \, e^{x} \log \left (3\right )^{2} \log \left (x\right ) - 5 \, \log \left (3\right ) \log \left (2\right )^{2} \log \left (x\right ) + 3 \, \log \left (2\right )^{2}}{5 \, {\left (e^{x} \log \left (3\right )^{2} - \log \left (3\right ) \log \left (2\right )^{2}\right )}} \end {gather*}

integrate((5*log(3)^2*exp(x)^2+(-10*log(2)^2*log(3)-3*x*log(2)^2)*exp(x)+5*log(2)^4)/(5*x*log(3)^2*exp(x)^2-10

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

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Mupad [B]
time = 4.22, size = 22, normalized size = 0.92 \begin {gather*} \ln \left (x\right )+\frac {3\,{\mathrm {e}}^x}{5\,{\mathrm {e}}^x\,\ln \left (3\right )-5\,{\ln \left (2\right )}^2} \end {gather*}

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

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3.60.10 \(\int \frac {5 \log ^4(2)+5 e^{2 x} \log ^2(3)+e^x (-3 x \log ^2(2)-10 \log ^2(2) \log (3))}{5 x \log ^4(2)-10 e^x x \log ^2(2) \log (3)+5 e^{2 x} x \log ^2(3)} \, dx\) [5910]