\(\int \frac {-5+(-1+e^{4+x} x-8 x^2) \log ^2(x)}{x \log ^2(x)} \, dx\) [1004]

Optimal result

Integrand size = 29, antiderivative size = 27 \[ \int \frac {-5+\left (-1+e^{4+x} x-8 x^2\right ) \log ^2(x)}{x \log ^2(x)} \, dx=e^{4+x}-4 x^2-\frac {\log (10)}{3}+\frac {5}{\log (x)}-\log (x) \]

[Out]

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

Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.78, number of steps used = 10, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {6874, 2225, 6820, 14, 2339, 30} \[ \int \frac {-5+\left (-1+e^{4+x} x-8 x^2\right ) \log ^2(x)}{x \log ^2(x)} \, dx=-4 x^2+e^{x+4}-\log (x)+\frac {5}{\log (x)} \]

[In]

Int[(-5 + (-1 + E^(4 + x)*x - 8*x^2)*Log[x]^2)/(x*Log[x]^2),x]

[Out]

E^(4 + x) - 4*x^2 + 5/Log[x] - Log[x]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 30

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

Rule 2225

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 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]]

Rule 6874

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

Rubi steps \begin{align*} \text {integral}& = \int \left (e^{4+x}+\frac {-5-\log ^2(x)-8 x^2 \log ^2(x)}{x \log ^2(x)}\right ) \, dx \\ & = \int e^{4+x} \, dx+\int \frac {-5-\log ^2(x)-8 x^2 \log ^2(x)}{x \log ^2(x)} \, dx \\ & = e^{4+x}+\int \frac {-1-8 x^2-\frac {5}{\log ^2(x)}}{x} \, dx \\ & = e^{4+x}+\int \left (\frac {-1-8 x^2}{x}-\frac {5}{x \log ^2(x)}\right ) \, dx \\ & = e^{4+x}-5 \int \frac {1}{x \log ^2(x)} \, dx+\int \frac {-1-8 x^2}{x} \, dx \\ & = e^{4+x}-5 \text {Subst}\left (\int \frac {1}{x^2} \, dx,x,\log (x)\right )+\int \left (-\frac {1}{x}-8 x\right ) \, dx \\ & = e^{4+x}-4 x^2+\frac {5}{\log (x)}-\log (x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.78 \[ \int \frac {-5+\left (-1+e^{4+x} x-8 x^2\right ) \log ^2(x)}{x \log ^2(x)} \, dx=e^{4+x}-4 x^2+\frac {5}{\log (x)}-\log (x) \]

[In]

Integrate[(-5 + (-1 + E^(4 + x)*x - 8*x^2)*Log[x]^2)/(x*Log[x]^2),x]

[Out]

E^(4 + x) - 4*x^2 + 5/Log[x] - Log[x]

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.78

method	result	size
default	\(\frac {5}{\ln \left (x \right )}-4 x^{2}-\ln \left (x \right )+{\mathrm e}^{4+x}\)	\(21\)
risch	\(\frac {5}{\ln \left (x \right )}-4 x^{2}-\ln \left (x \right )+{\mathrm e}^{4+x}\)	\(21\)
parts	\(\frac {5}{\ln \left (x \right )}-4 x^{2}-\ln \left (x \right )+{\mathrm e}^{4+x}\)	\(21\)
norman	\(\frac {5-\ln \left (x \right )^{2}+{\mathrm e}^{4+x} \ln \left (x \right )-4 x^{2} \ln \left (x \right )}{\ln \left (x \right )}\)	\(28\)
parallelrisch	\(\frac {5-\ln \left (x \right )^{2}+{\mathrm e}^{4+x} \ln \left (x \right )-4 x^{2} \ln \left (x \right )}{\ln \left (x \right )}\)	\(28\)

[In]

int(((x*exp(4+x)-8*x^2-1)*ln(x)^2-5)/x/ln(x)^2,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {-5+\left (-1+e^{4+x} x-8 x^2\right ) \log ^2(x)}{x \log ^2(x)} \, dx=-\frac {{\left (4 \, x^{2} - e^{\left (x + 4\right )}\right )} \log \left (x\right ) + \log \left (x\right )^{2} - 5}{\log \left (x\right )} \]

[In]

integrate(((x*exp(4+x)-8*x^2-1)*log(x)^2-5)/x/log(x)^2,x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.63 \[ \int \frac {-5+\left (-1+e^{4+x} x-8 x^2\right ) \log ^2(x)}{x \log ^2(x)} \, dx=- 4 x^{2} + e^{x + 4} - \log {\left (x \right )} + \frac {5}{\log {\left (x \right )}} \]

[In]

integrate(((x*exp(4+x)-8*x**2-1)*ln(x)**2-5)/x/ln(x)**2,x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.74 \[ \int \frac {-5+\left (-1+e^{4+x} x-8 x^2\right ) \log ^2(x)}{x \log ^2(x)} \, dx=-4 \, x^{2} + \frac {5}{\log \left (x\right )} + e^{\left (x + 4\right )} - \log \left (x\right ) \]

[In]

integrate(((x*exp(4+x)-8*x^2-1)*log(x)^2-5)/x/log(x)^2,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {-5+\left (-1+e^{4+x} x-8 x^2\right ) \log ^2(x)}{x \log ^2(x)} \, dx=-\frac {4 \, x^{2} \log \left (x\right ) - e^{\left (x + 4\right )} \log \left (x\right ) + \log \left (x\right )^{2} - 5}{\log \left (x\right )} \]

[In]

integrate(((x*exp(4+x)-8*x^2-1)*log(x)^2-5)/x/log(x)^2,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 9.17 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.74 \[ \int \frac {-5+\left (-1+e^{4+x} x-8 x^2\right ) \log ^2(x)}{x \log ^2(x)} \, dx={\mathrm {e}}^{x+4}-\ln \left (x\right )+\frac {5}{\ln \left (x\right )}-4\,x^2 \]

[In]

int(-(log(x)^2*(8*x^2 - x*exp(x + 4) + 1) + 5)/(x*log(x)^2),x)

[Out]

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