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

Optimal. Leaf size=25 \[ 1+e^{\frac {-\log ^2(2)+\frac {x}{9 \log ^4(x)}}{x}}+x \]

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Rubi [A]  time = 0.96, antiderivative size = 22, normalized size of antiderivative = 0.88, number of steps used = 4, number of rules used = 3, integrand size = 61, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.049, Rules used = {12, 6742, 6706} \begin {gather*} x+e^{\frac {1}{9 \log ^4(x)}-\frac {\log ^2(2)}{x}} \end {gather*}

Antiderivative was successfully verified.

[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 6706

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 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.17, size = 22, normalized size = 0.88 \begin {gather*} e^{-\frac {\log ^2(2)}{x}+\frac {1}{9 \log ^4(x)}}+x \end {gather*}

Antiderivative was successfully verified.

[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

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fricas [A]  time = 1.46, size = 26, normalized size = 1.04 \begin {gather*} x + e^{\left (-\frac {9 \, \log \relax (2)^{2} \log \relax (x)^{4} - x}{9 \, x \log \relax (x)^{4}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \mathit {undef} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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maple [A]  time = 0.09, size = 27, normalized size = 1.08

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maxima [A]  time = 0.61, size = 19, normalized size = 0.76 \begin {gather*} x + e^{\left (-\frac {\log \relax (2)^{2}}{x} + \frac {1}{9 \, \log \relax (x)^{4}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 4.11, size = 20, normalized size = 0.80 \begin {gather*} x+{\mathrm {e}}^{\frac {1}{9\,{\ln \relax (x)}^4}}\,{\mathrm {e}}^{-\frac {{\ln \relax (2)}^2}{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 0.36, size = 22, normalized size = 0.88 \begin {gather*} x + e^{\frac {\frac {x}{9} - \log {\relax (2 )}^{2} \log {\relax (x )}^{4}}{x \log {\relax (x )}^{4}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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