\(\int \frac {9 x^2 \log ^5(x)+e^{\frac {x-9 \log ^2(2) \log ^4(x)}{9 x \log ^4(x)}} (-4 x+9 \log ^2(2) \log ^5(x))}{9 x^2 \log ^5(x)} \, dx\) [5862]

Optimal result

Integrand size = 61, antiderivative size = 25 \[ \int \frac {9 x^2 \log ^5(x)+e^{\frac {x-9 \log ^2(2) \log ^4(x)}{9 x \log ^4(x)}} \left (-4 x+9 \log ^2(2) \log ^5(x)\right )}{9 x^2 \log ^5(x)} \, dx=1+e^{\frac {-\log ^2(2)+\frac {x}{9 \log ^4(x)}}{x}}+x \]

[Out]

1+exp((1/9*x/ln(x)^4-ln(2)^2)/x)+x

Rubi [A] (verified)

Time = 0.64 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.049, Rules used = {12, 6874, 6838} \[ \int \frac {9 x^2 \log ^5(x)+e^{\frac {x-9 \log ^2(2) \log ^4(x)}{9 x \log ^4(x)}} \left (-4 x+9 \log ^2(2) \log ^5(x)\right )}{9 x^2 \log ^5(x)} \, dx=x+e^{\frac {1}{9 \log ^4(x)}-\frac {\log ^2(2)}{x}} \]

[In]

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

[Out]

E^(-(Log[2]^2/x) + 1/(9*Log[x]^4)) + 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 6838

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[q*(F^v/Log[F]), x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rule 6874

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

Rubi steps \begin{align*} \text {integral}& = \frac {1}{9} \int \frac {9 x^2 \log ^5(x)+e^{\frac {x-9 \log ^2(2) \log ^4(x)}{9 x \log ^4(x)}} \left (-4 x+9 \log ^2(2) \log ^5(x)\right )}{x^2 \log ^5(x)} \, dx \\ & = \frac {1}{9} \int \left (9-\frac {e^{-\frac {\log ^2(2)}{x}+\frac {1}{9 \log ^4(x)}} \left (4 x-9 \log ^2(2) \log ^5(x)\right )}{x^2 \log ^5(x)}\right ) \, dx \\ & = x-\frac {1}{9} \int \frac {e^{-\frac {\log ^2(2)}{x}+\frac {1}{9 \log ^4(x)}} \left (4 x-9 \log ^2(2) \log ^5(x)\right )}{x^2 \log ^5(x)} \, dx \\ & = e^{-\frac {\log ^2(2)}{x}+\frac {1}{9 \log ^4(x)}}+x \\ \end{align*}

Mathematica [A] (verified)

Time = 0.47 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {9 x^2 \log ^5(x)+e^{\frac {x-9 \log ^2(2) \log ^4(x)}{9 x \log ^4(x)}} \left (-4 x+9 \log ^2(2) \log ^5(x)\right )}{9 x^2 \log ^5(x)} \, dx=e^{-\frac {\log ^2(2)}{x}+\frac {1}{9 \log ^4(x)}}+x \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 9.83 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08

method	result	size
risch	\(x +{\mathrm e}^{-\frac {9 \ln \left (2\right )^{2} \ln \left (x \right )^{4}-x}{9 x \ln \left (x \right )^{4}}}\)	\(27\)
parallelrisch	\(x +{\mathrm e}^{-\frac {9 \ln \left (2\right )^{2} \ln \left (x \right )^{4}-x}{9 x \ln \left (x \right )^{4}}}\)	\(27\)

[In]

int(1/9*((9*ln(2)^2*ln(x)^5-4*x)*exp(1/9*(-9*ln(2)^2*ln(x)^4+x)/x/ln(x)^4)+9*x^2*ln(x)^5)/x^2/ln(x)^5,x,method
=_RETURNVERBOSE)

[Out]

x+exp(-1/9*(9*ln(2)^2*ln(x)^4-x)/x/ln(x)^4)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.04 \[ \int \frac {9 x^2 \log ^5(x)+e^{\frac {x-9 \log ^2(2) \log ^4(x)}{9 x \log ^4(x)}} \left (-4 x+9 \log ^2(2) \log ^5(x)\right )}{9 x^2 \log ^5(x)} \, dx=x + e^{\left (-\frac {9 \, \log \left (2\right )^{2} \log \left (x\right )^{4} - x}{9 \, x \log \left (x\right )^{4}}\right )} \]

[In]

integrate(1/9*((9*log(2)^2*log(x)^5-4*x)*exp(1/9*(-9*log(2)^2*log(x)^4+x)/x/log(x)^4)+9*x^2*log(x)^5)/x^2/log(
x)^5,x, algorithm="fricas")

[Out]

x + e^(-1/9*(9*log(2)^2*log(x)^4 - x)/(x*log(x)^4))

Sympy [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {9 x^2 \log ^5(x)+e^{\frac {x-9 \log ^2(2) \log ^4(x)}{9 x \log ^4(x)}} \left (-4 x+9 \log ^2(2) \log ^5(x)\right )}{9 x^2 \log ^5(x)} \, dx=x + e^{\frac {\frac {x}{9} - \log {\left (2 \right )}^{2} \log {\left (x \right )}^{4}}{x \log {\left (x \right )}^{4}}} \]

[In]

integrate(1/9*((9*ln(2)**2*ln(x)**5-4*x)*exp(1/9*(-9*ln(2)**2*ln(x)**4+x)/x/ln(x)**4)+9*x**2*ln(x)**5)/x**2/ln
(x)**5,x)

[Out]

x + exp((x/9 - log(2)**2*log(x)**4)/(x*log(x)**4))

Maxima [A] (verification not implemented)

none

Time = 0.36 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.76 \[ \int \frac {9 x^2 \log ^5(x)+e^{\frac {x-9 \log ^2(2) \log ^4(x)}{9 x \log ^4(x)}} \left (-4 x+9 \log ^2(2) \log ^5(x)\right )}{9 x^2 \log ^5(x)} \, dx=x + e^{\left (-\frac {\log \left (2\right )^{2}}{x} + \frac {1}{9 \, \log \left (x\right )^{4}}\right )} \]

[In]

integrate(1/9*((9*log(2)^2*log(x)^5-4*x)*exp(1/9*(-9*log(2)^2*log(x)^4+x)/x/log(x)^4)+9*x^2*log(x)^5)/x^2/log(
x)^5,x, algorithm="maxima")

[Out]

x + e^(-log(2)^2/x + 1/9/log(x)^4)

Giac [F]

\[ \int \frac {9 x^2 \log ^5(x)+e^{\frac {x-9 \log ^2(2) \log ^4(x)}{9 x \log ^4(x)}} \left (-4 x+9 \log ^2(2) \log ^5(x)\right )}{9 x^2 \log ^5(x)} \, dx=\int { \frac {9 \, x^{2} \log \left (x\right )^{5} + {\left (9 \, \log \left (2\right )^{2} \log \left (x\right )^{5} - 4 \, x\right )} e^{\left (-\frac {9 \, \log \left (2\right )^{2} \log \left (x\right )^{4} - x}{9 \, x \log \left (x\right )^{4}}\right )}}{9 \, x^{2} \log \left (x\right )^{5}} \,d x } \]

[In]

integrate(1/9*((9*log(2)^2*log(x)^5-4*x)*exp(1/9*(-9*log(2)^2*log(x)^4+x)/x/log(x)^4)+9*x^2*log(x)^5)/x^2/log(
x)^5,x, algorithm="giac")

[Out]

undef

Mupad [B] (verification not implemented)

Time = 12.41 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80 \[ \int \frac {9 x^2 \log ^5(x)+e^{\frac {x-9 \log ^2(2) \log ^4(x)}{9 x \log ^4(x)}} \left (-4 x+9 \log ^2(2) \log ^5(x)\right )}{9 x^2 \log ^5(x)} \, dx=x+{\mathrm {e}}^{\frac {1}{9\,{\ln \left (x\right )}^4}}\,{\mathrm {e}}^{-\frac {{\ln \left (2\right )}^2}{x}} \]

[In]

int((x^2*log(x)^5 - (exp((x/9 - log(2)^2*log(x)^4)/(x*log(x)^4))*(4*x - 9*log(2)^2*log(x)^5))/9)/(x^2*log(x)^5
),x)

[Out]

x + exp(1/(9*log(x)^4))*exp(-log(2)^2/x)