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

Optimal. Leaf size=23 \[ 9 \left (8-x+\log (2)-\log (3)+\frac {-1+e^5}{\log (x)}\right ) \]

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Rubi [A]  time = 0.19, antiderivative size = 17, normalized size of antiderivative = 0.74, number of steps used = 6, number of rules used = 5, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {6741, 12, 6742, 2302, 30} \begin {gather*} -9 x-\frac {9 \left (1-e^5\right )}{\log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-9*x - (9*(1 - E^5))/Log[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 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 6741

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

Rule 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {9 \left (1-e^5-x \log ^2(x)\right )}{x \log ^2(x)} \, dx\\ &=9 \int \frac {1-e^5-x \log ^2(x)}{x \log ^2(x)} \, dx\\ &=9 \int \left (-1+\frac {1-e^5}{x \log ^2(x)}\right ) \, dx\\ &=-9 x+\left (9 \left (1-e^5\right )\right ) \int \frac {1}{x \log ^2(x)} \, dx\\ &=-9 x+\left (9 \left (1-e^5\right )\right ) \operatorname {Subst}\left (\int \frac {1}{x^2} \, dx,x,\log (x)\right )\\ &=-9 x-\frac {9 \left (1-e^5\right )}{\log (x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 19, normalized size = 0.83 \begin {gather*} -9 x-\frac {9}{\log (x)}+\frac {9 e^5}{\log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-9*x - 9/Log[x] + (9*E^5)/Log[x]

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fricas [A]  time = 0.79, size = 16, normalized size = 0.70 \begin {gather*} -\frac {9 \, {\left (x \log \relax (x) - e^{5} + 1\right )}}{\log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-9*(x*log(x) - e^5 + 1)/log(x)

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giac [A]  time = 0.18, size = 16, normalized size = 0.70 \begin {gather*} -\frac {9 \, {\left (x \log \relax (x) - e^{5} + 1\right )}}{\log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-9*(x*log(x) - e^5 + 1)/log(x)

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maple [A]  time = 0.03, size = 17, normalized size = 0.74

method	result	size
norman	\(\frac {-9 x \ln \relax (x )-9+9 \,{\mathrm e}^{5}}{\ln \relax (x )}\)	\(17\)
default	\(-9 x +\frac {9 \,{\mathrm e}^{5}}{\ln \relax (x )}-\frac {9}{\ln \relax (x )}\)	\(19\)
risch	\(-9 x +\frac {9 \,{\mathrm e}^{5}}{\ln \relax (x )}-\frac {9}{\ln \relax (x )}\)	\(19\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-9*x*ln(x)-9+9*exp(5))/ln(x)

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maxima [A]  time = 0.35, size = 18, normalized size = 0.78 \begin {gather*} -9 \, x + \frac {9 \, e^{5}}{\log \relax (x)} - \frac {9}{\log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-9*x + 9*e^5/log(x) - 9/log(x)

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mupad [B]  time = 8.47, size = 15, normalized size = 0.65 \begin {gather*} \frac {9\,{\mathrm {e}}^5-9}{\ln \relax (x)}-9\,x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(9*exp(5) - 9)/log(x) - 9*x

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sympy [A]  time = 0.08, size = 12, normalized size = 0.52 \begin {gather*} - 9 x + \frac {-9 + 9 e^{5}}{\log {\relax (x )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-9*x + (-9 + 9*exp(5))/log(x)

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