∫ (−15−5ex2) log( 4_____ 3x log(2))+log(x)(−15−5ex2+10ex2 log( 4_____ 3x log(2))) ______________________________________________________________________________________

Optimal. Leaf size=27 \[ \frac {5 \left (\frac {3}{e}+x^2\right ) \log \left (\frac {4}{3 x \log (2)}\right )}{\log (x)} \]

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Rubi [A] time = 0.95, antiderivative size = 39, normalized size of antiderivative = 1.44, number of steps used = 23, number of rules used = 11, integrand size = 62, $\frac {\text {number of rules}}{\text {integrand size}}$ = 0.177, Rules used = {12, 6742, 2353, 2302, 29, 2309, 2178, 30, 2306, 2366, 6482} \begin {gather*} \frac {5 x^2 \log \left (\frac {4}{x \log (8)}\right )}{\log (x)}+\frac {15 \log \left (\frac {4}{x \log (8)}\right )}{e \log (x)} \end {gather*}

Int[((-15 - 5*E*x^2)*Log[4/(3*x*Log[2])] + Log[x]*(-15 - 5*E*x^2 + 10*E*x^2*Log[4/(3*x*Log[2])]))/(E*x*Log[x]^

(15*Log[4/(x*Log[8])])/(E*Log[x]) + (5*x^2*Log[4/(x*Log[8])])/Log[x]

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

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - (c*f)/d))*ExpIntegral

Ei[(f*g*(c + d*x)*Log[F])/d])/d, x] /; FreeQ[{F, c, d, e, f, g}, x] && !$UseGamma === True

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/(x_), x_Symbol] :> Dist[1/(b*n), Subst[Int[x^p, x], x, a + b*L

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Log

[c*x^n])^(p + 1))/(b*d*n*(p + 1)), x] - Dist[(m + 1)/(b*n*(p + 1)), Int[(d*x)^m*(a + b*Log[c*x^n])^(p + 1), x]

, x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1] && LtQ[p, -1]

Int[((a_.) + Log[(c_.)*(x_)]*(b_.))^(p_)*(x_)^(m_.), x_Symbol] :> Dist[1/c^(m + 1), Subst[Int[E^((m + 1)*x)*(a

+ b*x)^p, x], x, Log[c*x]], x] /; FreeQ[{a, b, c, p}, x] && IntegerQ[m]

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol]

:> With[{u = ExpandIntegrand[(a + b*Log[c*x^n])^p, (f*x)^m*(d + e*x^r)^q, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[

{a, b, c, d, e, f, m, n, p, q, r}, x] && IntegerQ[q] && (GtQ[q, 0] || (IGtQ[p, 0] && IntegerQ[m] && IntegerQ[r

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_.) + Log[(f_.)*(x_)^(r_.)]*(e_.))*((g_.)*(x_))^(m_.), x_Sy

mbol] :> With[{u = IntHide[(g*x)^m*(a + b*Log[c*x^n])^p, x]}, Dist[d + e*Log[f*x^r], u, x] - Dist[e*r, Int[Sim

plifyIntegrand[u/x, x], x], x]] /; FreeQ[{a, b, c, d, e, f, g, m, n, p, r}, x] && !(EqQ[p, 1] && EqQ[a, 0] &&

Int[ExpIntegralEi[(a_.) + (b_.)*(x_)], x_Symbol] :> Simp[((a + b*x)*ExpIntegralEi[a + b*x])/b, x] - Simp[E^(a

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

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {\left (-15-5 e x^2\right ) \log \left (\frac {4}{3 x \log (2)}\right )+\log (x) \left (-15-5 e x^2+10 e x^2 \log \left (\frac {4}{3 x \log (2)}\right )\right )}{x \log ^2(x)} \, dx}{e}\\ &=\frac {\int \left (-\frac {5 \left (3+e x^2\right )}{x \log (x)}+\frac {5 \left (-3-e x^2+2 e x^2 \log (x)\right ) \log \left (\frac {4}{x \log (8)}\right )}{x \log ^2(x)}\right ) \, dx}{e}\\ &=-\frac {5 \int \frac {3+e x^2}{x \log (x)} \, dx}{e}+\frac {5 \int \frac {\left (-3-e x^2+2 e x^2 \log (x)\right ) \log \left (\frac {4}{x \log (8)}\right )}{x \log ^2(x)} \, dx}{e}\\ &=-\frac {5 \int \left (\frac {3}{x \log (x)}+\frac {e x}{\log (x)}\right ) \, dx}{e}+\frac {5 \int \left (-\frac {3 \log \left (\frac {4}{x \log (8)}\right )}{x \log ^2(x)}-\frac {e x \log \left (\frac {4}{x \log (8)}\right )}{\log ^2(x)}+\frac {2 e x \log \left (\frac {4}{x \log (8)}\right )}{\log (x)}\right ) \, dx}{e}\\ &=-\left (5 \int \frac {x}{\log (x)} \, dx\right )-5 \int \frac {x \log \left (\frac {4}{x \log (8)}\right )}{\log ^2(x)} \, dx+10 \int \frac {x \log \left (\frac {4}{x \log (8)}\right )}{\log (x)} \, dx-\frac {15 \int \frac {1}{x \log (x)} \, dx}{e}-\frac {15 \int \frac {\log \left (\frac {4}{x \log (8)}\right )}{x \log ^2(x)} \, dx}{e}\\ &=\frac {15 \log \left (\frac {4}{x \log (8)}\right )}{e \log (x)}+\frac {5 x^2 \log \left (\frac {4}{x \log (8)}\right )}{\log (x)}-5 \int \left (\frac {2 \text {Ei}(2 \log (x))}{x}-\frac {x}{\log (x)}\right ) \, dx-5 \operatorname {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (x)\right )+10 \int \frac {\text {Ei}(2 \log (x))}{x} \, dx+\frac {15 \int \frac {1}{x \log (x)} \, dx}{e}-\frac {15 \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,\log (x)\right )}{e}\\ &=-5 \text {Ei}(2 \log (x))+\frac {15 \log \left (\frac {4}{x \log (8)}\right )}{e \log (x)}+\frac {5 x^2 \log \left (\frac {4}{x \log (8)}\right )}{\log (x)}-\frac {15 \log (\log (x))}{e}+5 \int \frac {x}{\log (x)} \, dx-10 \int \frac {\text {Ei}(2 \log (x))}{x} \, dx+10 \operatorname {Subst}(\int \text {Ei}(2 x) \, dx,x,\log (x))+\frac {15 \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,\log (x)\right )}{e}\\ &=-5 x^2-5 \text {Ei}(2 \log (x))+10 \text {Ei}(2 \log (x)) \log (x)+\frac {15 \log \left (\frac {4}{x \log (8)}\right )}{e \log (x)}+\frac {5 x^2 \log \left (\frac {4}{x \log (8)}\right )}{\log (x)}+5 \operatorname {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (x)\right )-10 \operatorname {Subst}(\int \text {Ei}(2 x) \, dx,x,\log (x))\\ &=\frac {15 \log \left (\frac {4}{x \log (8)}\right )}{e \log (x)}+\frac {5 x^2 \log \left (\frac {4}{x \log (8)}\right )}{\log (x)}\\ \end {aligned} \end {gather*}

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Mathematica [A] time = 0.15, size = 29, normalized size = 1.07 \begin {gather*} \frac {15+\frac {5 \left (3+e x^2\right ) \log \left (\frac {4}{x \log (8)}\right )}{\log (x)}}{e} \end {gather*}

Integrate[((-15 - 5*E*x^2)*Log[4/(3*x*Log[2])] + Log[x]*(-15 - 5*E*x^2 + 10*E*x^2*Log[4/(3*x*Log[2])]))/(E*x*L

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fricas [B] time = 0.88, size = 55, normalized size = 2.04 \begin {gather*} \frac {5 \, {\left (x^{2} e \log \left (\frac {4}{3 \, x \log \relax (2)}\right ) + 3 \, \log \left (\frac {4}{3 \, \log \relax (2)}\right )\right )}}{e \log \left (\frac {4}{3 \, \log \relax (2)}\right ) - e \log \left (\frac {4}{3 \, x \log \relax (2)}\right )} \end {gather*}

integrate(((10*x^2*exp(1)*log(4/3/x/log(2))-5*x^2*exp(1)-15)*log(x)+(-5*x^2*exp(1)-15)*log(4/3/x/log(2)))/x/ex

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

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giac [B] time = 0.16, size = 56, normalized size = 2.07 \begin {gather*} -\frac {5 \, {\left (x^{2} e \log \relax (3) - 2 \, x^{2} e \log \relax (2) + x^{2} e \log \relax (x) + x^{2} e \log \left (\log \relax (2)\right ) + 3 \, \log \relax (3) - 6 \, \log \relax (2) + 3 \, \log \left (\log \relax (2)\right )\right )} e^{\left (-1\right )}}{\log \relax (x)} \end {gather*}

integrate(((10*x^2*exp(1)*log(4/3/x/log(2))-5*x^2*exp(1)-15)*log(x)+(-5*x^2*exp(1)-15)*log(4/3/x/log(2)))/x/ex

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

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method	result	size

risch	$-5 x^{2}+\frac {5 \,{\mathrm e}^{-1} \left (4 \ln \relax (2) {\mathrm e} x^{2}-2 x^{2} {\mathrm e} \ln \left (\ln \relax (2)\right )-2 x^{2} {\mathrm e} \ln \relax (3)+12 \ln \relax (2)-6 \ln \left (\ln \relax (2)\right )-6 \ln \relax (3)\right )}{2 \ln \relax (x )}$	$57$

int(((10*x^2*exp(1)*ln(4/3/x/ln(2))-5*x^2*exp(1)-15)*ln(x)+(-5*x^2*exp(1)-15)*ln(4/3/x/ln(2)))/x/exp(1)/ln(x)^

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

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maxima [C] time = 0.38, size = 108, normalized size = 4.00 \begin {gather*} 5 \, {\left (2 \, {\rm Ei}\left (2 \, \log \relax (x)\right ) e \log \left (\frac {4}{3 \, x \log \relax (2)}\right ) - 2 \, e \Gamma \left (-1, -2 \, \log \relax (x)\right ) \log \left (\frac {4}{3 \, x \log \relax (2)}\right ) - {\left (x^{2} - 2 \, {\rm Ei}\left (2 \, \log \relax (x)\right ) \log \relax (x)\right )} e - {\left (2 \, \Gamma \left (-1, -2 \, \log \relax (x)\right ) \log \relax (x) - {\rm Ei}\left (2 \, \log \relax (x)\right )\right )} e - {\rm Ei}\left (2 \, \log \relax (x)\right ) e + \frac {3 \, \log \left (\frac {4}{3 \, x \log \relax (2)}\right )}{\log \relax (x)}\right )} e^{\left (-1\right )} \end {gather*}

integrate(((10*x^2*exp(1)*log(4/3/x/log(2))-5*x^2*exp(1)-15)*log(x)+(-5*x^2*exp(1)-15)*log(4/3/x/log(2)))/x/ex

5*(2*Ei(2*log(x))*e*log(4/3/(x*log(2))) - 2*e*gamma(-1, -2*log(x))*log(4/3/(x*log(2))) - (x^2 - 2*Ei(2*log(x))

*log(x))*e - (2*gamma(-1, -2*log(x))*log(x) - Ei(2*log(x)))*e - Ei(2*log(x))*e + 3*log(4/3/(x*log(2)))/log(x))

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mupad [B] time = 3.23, size = 57, normalized size = 2.11 \begin {gather*} \frac {x^2\,\left (5\,\ln \left (\frac {1}{x}\right )+10\,\ln \relax (2)-5\,\ln \relax (3)-5\,\ln \left (\ln \relax (2)\right )\right )}{\ln \relax (x)}+\frac {15\,{\mathrm {e}}^{-1}\,\left (\ln \left (\frac {1}{x}\right )+2\,\ln \relax (2)-\ln \relax (3)-\ln \left (\ln \relax (2)\right )+\ln \relax (x)\right )}{\ln \relax (x)} \end {gather*}

int(-(exp(-1)*(log(x)*(5*x^2*exp(1) - 10*x^2*exp(1)*log(4/(3*x*log(2))) + 15) + log(4/(3*x*log(2)))*(5*x^2*exp

(x^2*(5*log(1/x) + 10*log(2) - 5*log(3) - 5*log(log(2))))/log(x) + (15*exp(-1)*(log(1/x) + 2*log(2) - log(3) -

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sympy [B] time = 0.27, size = 65, normalized size = 2.41 \begin {gather*} - 5 x^{2} + \frac {- 5 e x^{2} \log {\relax (3 )} - 5 e x^{2} \log {\left (\log {\relax (2 )} \right )} + 10 e x^{2} \log {\relax (2 )} - 15 \log {\relax (3 )} - 15 \log {\left (\log {\relax (2 )} \right )} + 30 \log {\relax (2 )}}{e \log {\relax (x )}} \end {gather*}

integrate(((10*x**2*exp(1)*ln(4/3/x/ln(2))-5*x**2*exp(1)-15)*ln(x)+(-5*x**2*exp(1)-15)*ln(4/3/x/ln(2)))/x/exp(

-5*x**2 + (-5*E*x**2*log(3) - 5*E*x**2*log(log(2)) + 10*E*x**2*log(2) - 15*log(3) - 15*log(log(2)) + 30*log(2)

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3.43.48 \(\int \frac {(-15-5 e x^2) \log (\frac {4}{3 x \log (2)})+\log (x) (-15-5 e x^2+10 e x^2 \log (\frac {4}{3 x \log (2)}))}{e x \log ^2(x)} \, dx\)