\(\int \frac {1+(-e^5 x+2 x^2) \log ^2(x)}{x \log ^2(x)} \, dx\) [10271]

Optimal result

Integrand size = 27, antiderivative size = 20 \[ \int \frac {1+\left (-e^5 x+2 x^2\right ) \log ^2(x)}{x \log ^2(x)} \, dx=4+e^3-e^5 x+x^2-\frac {1}{\log (x)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.80, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {6820, 2339, 30} \[ \int \frac {1+\left (-e^5 x+2 x^2\right ) \log ^2(x)}{x \log ^2(x)} \, dx=x^2-e^5 x-\frac {1}{\log (x)} \]

[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 2339

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 6820

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

Rubi steps \begin{align*} \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+\text {Subst}\left (\int \frac {1}{x^2} \, dx,x,\log (x)\right ) \\ & = -e^5 x+x^2-\frac {1}{\log (x)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.80 \[ \int \frac {1+\left (-e^5 x+2 x^2\right ) \log ^2(x)}{x \log ^2(x)} \, dx=-e^5 x+x^2-\frac {1}{\log (x)} \]

[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)

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.80

method	result	size
default	\(-x \,{\mathrm e}^{5}+x^{2}-\frac {1}{\ln \left (x \right )}\)	\(16\)
risch	\(-x \,{\mathrm e}^{5}+x^{2}-\frac {1}{\ln \left (x \right )}\)	\(16\)
parts	\(-x \,{\mathrm e}^{5}+x^{2}-\frac {1}{\ln \left (x \right )}\)	\(16\)
norman	\(\frac {-1+x^{2} \ln \left (x \right )-x \,{\mathrm e}^{5} \ln \left (x \right )}{\ln \left (x \right )}\)	\(21\)
parallelrisch	\(-\frac {x \,{\mathrm e}^{5} \ln \left (x \right )-x^{2} \ln \left (x \right )+1}{\ln \left (x \right )}\)	\(22\)

[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)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95 \[ \int \frac {1+\left (-e^5 x+2 x^2\right ) \log ^2(x)}{x \log ^2(x)} \, dx=\frac {{\left (x^{2} - x e^{5}\right )} \log \left (x\right ) - 1}{\log \left (x\right )} \]

[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)

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.60 \[ \int \frac {1+\left (-e^5 x+2 x^2\right ) \log ^2(x)}{x \log ^2(x)} \, dx=x^{2} - x e^{5} - \frac {1}{\log {\left (x \right )}} \]

[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)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.75 \[ \int \frac {1+\left (-e^5 x+2 x^2\right ) \log ^2(x)}{x \log ^2(x)} \, dx=x^{2} - x e^{5} - \frac {1}{\log \left (x\right )} \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {1+\left (-e^5 x+2 x^2\right ) \log ^2(x)}{x \log ^2(x)} \, dx=\frac {x^{2} \log \left (x\right ) - x e^{5} \log \left (x\right ) - 1}{\log \left (x\right )} \]

[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)

Mupad [B] (verification not implemented)

Time = 15.60 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.75 \[ \int \frac {1+\left (-e^5 x+2 x^2\right ) \log ^2(x)}{x \log ^2(x)} \, dx=x\,\left (x-{\mathrm {e}}^5\right )-\frac {1}{\ln \left (x\right )} \]

[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)