3.23.50 \(\int \frac {10 x-20 x \log (x)+(e^x (-2-2 x)-3 x-6 x^2+2 \log (3)) \log ^2(x)}{2 \log ^2(x)} \, dx\)

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

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Rubi [A]  time = 0.29, antiderivative size = 40, normalized size of antiderivative = 1.48, number of steps used = 12, number of rules used = 8, integrand size = 44, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {12, 6742, 2176, 2194, 6688, 2306, 2309, 2178} \begin {gather*} -x^3-\frac {3 x^2}{4}-\frac {5 x^2}{\log (x)}+e^x-e^x (x+1)+\frac {1}{2} x \log (9) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(10*x - 20*x*Log[x] + (E^x*(-2 - 2*x) - 3*x - 6*x^2 + 2*Log[3])*Log[x]^2)/(2*Log[x]^2),x]

[Out]

E^x - (3*x^2)/4 - x^3 - E^x*(1 + x) + (x*Log[9])/2 - (5*x^2)/Log[x]

Rule 12

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

Rule 2176

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^m
*(b*F^(g*(e + f*x)))^n)/(f*g*n*Log[F]), x] - Dist[(d*m)/(f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*(b*F^(g*(e + f*x
)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && IntegerQ[2*m] &&  !$UseGamma === True

Rule 2178

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

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 2306

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]

Rule 2309

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]

Rule 6688

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

Rule 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{2} \int \frac {10 x-20 x \log (x)+\left (e^x (-2-2 x)-3 x-6 x^2+2 \log (3)\right ) \log ^2(x)}{\log ^2(x)} \, dx\\ &=\frac {1}{2} \int \left (-2 e^x (1+x)+\frac {10 x-20 x \log (x)-3 x \log ^2(x)-6 x^2 \log ^2(x)+\log (9) \log ^2(x)}{\log ^2(x)}\right ) \, dx\\ &=\frac {1}{2} \int \frac {10 x-20 x \log (x)-3 x \log ^2(x)-6 x^2 \log ^2(x)+\log (9) \log ^2(x)}{\log ^2(x)} \, dx-\int e^x (1+x) \, dx\\ &=-e^x (1+x)+\frac {1}{2} \int \left (-3 x-6 x^2+\log (9)+\frac {10 x}{\log ^2(x)}-\frac {20 x}{\log (x)}\right ) \, dx+\int e^x \, dx\\ &=e^x-\frac {3 x^2}{4}-x^3-e^x (1+x)+\frac {1}{2} x \log (9)+5 \int \frac {x}{\log ^2(x)} \, dx-10 \int \frac {x}{\log (x)} \, dx\\ &=e^x-\frac {3 x^2}{4}-x^3-e^x (1+x)+\frac {1}{2} x \log (9)-\frac {5 x^2}{\log (x)}+10 \int \frac {x}{\log (x)} \, dx-10 \operatorname {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (x)\right )\\ &=e^x-\frac {3 x^2}{4}-x^3-e^x (1+x)-10 \text {Ei}(2 \log (x))+\frac {1}{2} x \log (9)-\frac {5 x^2}{\log (x)}+10 \operatorname {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (x)\right )\\ &=e^x-\frac {3 x^2}{4}-x^3-e^x (1+x)+\frac {1}{2} x \log (9)-\frac {5 x^2}{\log (x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.09, size = 35, normalized size = 1.30 \begin {gather*} -e^x x-\frac {3 x^2}{4}-x^3+\frac {1}{2} x \log (9)-\frac {5 x^2}{\log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(10*x - 20*x*Log[x] + (E^x*(-2 - 2*x) - 3*x - 6*x^2 + 2*Log[3])*Log[x]^2)/(2*Log[x]^2),x]

[Out]

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

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fricas [A]  time = 0.92, size = 36, normalized size = 1.33 \begin {gather*} -\frac {20 \, x^{2} + {\left (4 \, x^{3} + 3 \, x^{2} + 4 \, x e^{x} - 4 \, x \log \relax (3)\right )} \log \relax (x)}{4 \, \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(((-2*x-2)*exp(x)+2*log(3)-6*x^2-3*x)*log(x)^2-20*x*log(x)+10*x)/log(x)^2,x, algorithm="fricas")

[Out]

-1/4*(20*x^2 + (4*x^3 + 3*x^2 + 4*x*e^x - 4*x*log(3))*log(x))/log(x)

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giac [A]  time = 0.23, size = 40, normalized size = 1.48 \begin {gather*} -\frac {4 \, x^{3} \log \relax (x) + 3 \, x^{2} \log \relax (x) + 4 \, x e^{x} \log \relax (x) - 4 \, x \log \relax (3) \log \relax (x) + 20 \, x^{2}}{4 \, \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(((-2*x-2)*exp(x)+2*log(3)-6*x^2-3*x)*log(x)^2-20*x*log(x)+10*x)/log(x)^2,x, algorithm="giac")

[Out]

-1/4*(4*x^3*log(x) + 3*x^2*log(x) + 4*x*e^x*log(x) - 4*x*log(3)*log(x) + 20*x^2)/log(x)

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maple [A]  time = 0.06, size = 30, normalized size = 1.11

method	result	size
default	\(-{\mathrm e}^{x} x -\frac {3 x^{2}}{4}-x^{3}-\frac {5 x^{2}}{\ln \relax (x )}+x \ln \relax (3)\)	\(30\)
risch	\(-{\mathrm e}^{x} x -\frac {3 x^{2}}{4}-x^{3}-\frac {5 x^{2}}{\ln \relax (x )}+x \ln \relax (3)\)	\(30\)
norman	\(\frac {x \ln \relax (3) \ln \relax (x )-5 x^{2}-\frac {3 x^{2} \ln \relax (x )}{4}-x^{3} \ln \relax (x )-x \,{\mathrm e}^{x} \ln \relax (x )}{\ln \relax (x )}\)	\(39\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/2*(((-2*x-2)*exp(x)+2*ln(3)-6*x^2-3*x)*ln(x)^2-20*x*ln(x)+10*x)/ln(x)^2,x,method=_RETURNVERBOSE)

[Out]

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

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maxima [C]  time = 0.47, size = 41, normalized size = 1.52 \begin {gather*} -x^{3} - \frac {3}{4} \, x^{2} - {\left (x - 1\right )} e^{x} + x \log \relax (3) - 10 \, {\rm Ei}\left (2 \, \log \relax (x)\right ) - e^{x} + 10 \, \Gamma \left (-1, -2 \, \log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(((-2*x-2)*exp(x)+2*log(3)-6*x^2-3*x)*log(x)^2-20*x*log(x)+10*x)/log(x)^2,x, algorithm="maxima")

[Out]

-x^3 - 3/4*x^2 - (x - 1)*e^x + x*log(3) - 10*Ei(2*log(x)) - e^x + 10*gamma(-1, -2*log(x))

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mupad [B]  time = 1.37, size = 30, normalized size = 1.11 \begin {gather*} -\frac {x\,\left (3\,x-\ln \left (81\right )+4\,{\mathrm {e}}^x+4\,x^2\right )}{4}-\frac {5\,x^2}{\ln \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-((log(x)^2*(3*x - 2*log(3) + exp(x)*(2*x + 2) + 6*x^2))/2 - 5*x + 10*x*log(x))/log(x)^2,x)

[Out]

- (x*(3*x - log(81) + 4*exp(x) + 4*x^2))/4 - (5*x^2)/log(x)

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sympy [A]  time = 0.30, size = 27, normalized size = 1.00 \begin {gather*} - x^{3} - \frac {3 x^{2}}{4} - \frac {5 x^{2}}{\log {\relax (x )}} - x e^{x} + x \log {\relax (3 )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(((-2*x-2)*exp(x)+2*ln(3)-6*x**2-3*x)*ln(x)**2-20*x*ln(x)+10*x)/ln(x)**2,x)

[Out]

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

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