3.103.71 \(\int \frac {1+(-e^5 x+2 x^2) \log ^2(x)}{x \log ^2(x)} \, dx\)

Optimal. Leaf size=20 \[ 4+e^3-e^5 x+x^2-\frac {1}{\log (x)} \]

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Rubi [A]  time = 0.09, antiderivative size = 16, normalized size of antiderivative = 0.80, number of steps used = 4, number of rules used = 3, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {6688, 2302, 30} \begin {gather*} x^2-e^5 x-\frac {1}{\log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(E^5*x) + x^2 - Log[x]^(-1)

Rule 30

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

Rule 2302

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

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

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Mathematica [A]  time = 0.00, size = 16, normalized size = 0.80 \begin {gather*} -e^5 x+x^2-\frac {1}{\log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(E^5*x) + x^2 - Log[x]^(-1)

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fricas [A]  time = 0.59, size = 19, normalized size = 0.95 \begin {gather*} \frac {{\left (x^{2} - x e^{5}\right )} \log \relax (x) - 1}{\log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x*exp(5)+2*x^2)*log(x)^2+1)/x/log(x)^2,x, algorithm="fricas")

[Out]

((x^2 - x*e^5)*log(x) - 1)/log(x)

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giac [A]  time = 0.12, size = 20, normalized size = 1.00 \begin {gather*} \frac {x^{2} \log \relax (x) - x e^{5} \log \relax (x) - 1}{\log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x*exp(5)+2*x^2)*log(x)^2+1)/x/log(x)^2,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.03, size = 16, normalized size = 0.80

method	result	size
default	\(-x \,{\mathrm e}^{5}+x^{2}-\frac {1}{\ln \relax (x )}\)	\(16\)
risch	\(-x \,{\mathrm e}^{5}+x^{2}-\frac {1}{\ln \relax (x )}\)	\(16\)
norman	\(\frac {-1+x^{2} \ln \relax (x )-x \,{\mathrm e}^{5} \ln \relax (x )}{\ln \relax (x )}\)	\(21\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-x*exp(5)+2*x^2)*ln(x)^2+1)/x/ln(x)^2,x,method=_RETURNVERBOSE)

[Out]

-x*exp(5)+x^2-1/ln(x)

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maxima [A]  time = 0.37, size = 15, normalized size = 0.75 \begin {gather*} x^{2} - x e^{5} - \frac {1}{\log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x*exp(5)+2*x^2)*log(x)^2+1)/x/log(x)^2,x, algorithm="maxima")

[Out]

x^2 - x*e^5 - 1/log(x)

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mupad [B]  time = 6.79, size = 15, normalized size = 0.75 \begin {gather*} x\,\left (x-{\mathrm {e}}^5\right )-\frac {1}{\ln \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(x)^2*(x*exp(5) - 2*x^2) - 1)/(x*log(x)^2),x)

[Out]

x*(x - exp(5)) - 1/log(x)

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sympy [A]  time = 0.10, size = 12, normalized size = 0.60 \begin {gather*} x^{2} - x e^{5} - \frac {1}{\log {\relax (x )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x*exp(5)+2*x**2)*ln(x)**2+1)/x/ln(x)**2,x)

[Out]

x**2 - x*exp(5) - 1/log(x)

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