3.22.79 \(\int \frac {-x^2+3 x^2 \log (x)+(-1-9 x-3 x^2-2 x \log (\frac {1}{x})) \log ^2(x)}{\log ^2(x)} \, dx\)

Optimal. Leaf size=22 \[ -x^2 \left (5+\frac {1}{x}+x+\log \left (\frac {1}{x}\right )-\frac {x}{\log (x)}\right ) \]

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Rubi [A]  time = 0.15, antiderivative size = 31, normalized size of antiderivative = 1.41, number of steps used = 8, number of rules used = 5, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {6742, 2304, 2306, 2309, 2178} \begin {gather*} -x^3+\frac {x^3}{\log (x)}-5 x^2-x^2 \log \left (\frac {1}{x}\right )-x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-x^2 + 3*x^2*Log[x] + (-1 - 9*x - 3*x^2 - 2*x*Log[x^(-1)])*Log[x]^2)/Log[x]^2,x]

[Out]

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

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 2304

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Log[c*x^
n]))/(d*(m + 1)), x] - Simp[(b*n*(d*x)^(m + 1))/(d*(m + 1)^2), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1
]

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 6742

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

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 31, normalized size = 1.41 \begin {gather*} -x-5 x^2-x^3-x^2 \log \left (\frac {1}{x}\right )+\frac {x^3}{\log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-x^2 + 3*x^2*Log[x] + (-1 - 9*x - 3*x^2 - 2*x*Log[x^(-1)])*Log[x]^2)/Log[x]^2,x]

[Out]

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

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fricas [A]  time = 0.59, size = 37, normalized size = 1.68 \begin {gather*} -\frac {x^{2} \log \left (\frac {1}{x}\right )^{2} + x^{3} + {\left (x^{3} + 5 \, x^{2} + x\right )} \log \left (\frac {1}{x}\right )}{\log \left (\frac {1}{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x*log(1/x)-3*x^2-9*x-1)*log(x)^2+3*x^2*log(x)-x^2)/log(x)^2,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.24, size = 28, normalized size = 1.27 \begin {gather*} -x^{3} + x^{2} \log \relax (x) - 5 \, x^{2} + \frac {x^{3}}{\log \relax (x)} - x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x*log(1/x)-3*x^2-9*x-1)*log(x)^2+3*x^2*log(x)-x^2)/log(x)^2,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.13, size = 29, normalized size = 1.32

method	result	size
risch	\(-x^{3}+x^{2} \ln \relax (x )-5 x^{2}-x +\frac {x^{3}}{\ln \relax (x )}\)	\(29\)
derivativedivides	\(-x -5 x^{2}-x^{3}-x^{2} \ln \left (\frac {1}{x}\right )+\frac {x^{3}}{\ln \relax (x )}\)	\(32\)
default	\(-x -5 x^{2}-x^{3}-x^{2} \ln \left (\frac {1}{x}\right )+\frac {x^{3}}{\ln \relax (x )}\)	\(32\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-2*x*ln(1/x)-3*x^2-9*x-1)*ln(x)^2+3*x^2*ln(x)-x^2)/ln(x)^2,x,method=_RETURNVERBOSE)

[Out]

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

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maxima [C]  time = 0.49, size = 35, normalized size = 1.59 \begin {gather*} -x^{3} + x^{2} \log \relax (x) - 5 \, x^{2} - x + 3 \, {\rm Ei}\left (3 \, \log \relax (x)\right ) - 3 \, \Gamma \left (-1, -3 \, \log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x*log(1/x)-3*x^2-9*x-1)*log(x)^2+3*x^2*log(x)-x^2)/log(x)^2,x, algorithm="maxima")

[Out]

-x^3 + x^2*log(x) - 5*x^2 - x + 3*Ei(3*log(x)) - 3*gamma(-1, -3*log(x))

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mupad [B]  time = 1.26, size = 26, normalized size = 1.18 \begin {gather*} \frac {x^3}{\ln \relax (x)}-x\,\left (5\,x+x\,\ln \left (\frac {1}{x}\right )+x^2+1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(x)^2*(9*x + 2*x*log(1/x) + 3*x^2 + 1) - 3*x^2*log(x) + x^2)/log(x)^2,x)

[Out]

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

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sympy [A]  time = 0.24, size = 22, normalized size = 1.00 \begin {gather*} - x^{3} + \frac {x^{3}}{\log {\relax (x )}} + x^{2} \log {\relax (x )} - 5 x^{2} - x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x*ln(1/x)-3*x**2-9*x-1)*ln(x)**2+3*x**2*ln(x)-x**2)/ln(x)**2,x)

[Out]

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

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