3.1.100 \(\int \frac {e-9 e^{6562-2 x}+(-e+e^{6562-2 x} (9-18 x)) \log (x)+\log ^2(x)}{\log ^2(x)} \, dx\)

Optimal. Leaf size=22 \[ x+\frac {e \left (-x+9 e^{6561-2 x} x\right )}{\log (x)} \]

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Rubi [A]  time = 0.79, antiderivative size = 24, normalized size of antiderivative = 1.09, number of steps used = 8, number of rules used = 4, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {6742, 2288, 2297, 2298} \begin {gather*} x+\frac {9 e^{6562-2 x} x}{\log (x)}-\frac {e x}{\log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E - 9*E^(6562 - 2*x) + (-E + E^(6562 - 2*x)*(9 - 18*x))*Log[x] + Log[x]^2)/Log[x]^2,x]

[Out]

x - (E*x)/Log[x] + (9*E^(6562 - 2*x)*x)/Log[x]

Rule 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rule 2297

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_), x_Symbol] :> Simp[(x*(a + b*Log[c*x^n])^(p + 1))/(b*n*(p + 1))
, x] - Dist[1/(b*n*(p + 1)), Int[(a + b*Log[c*x^n])^(p + 1), x], x] /; FreeQ[{a, b, c, n}, x] && LtQ[p, -1] &&
IntegerQ[2*p]

Rule 2298

Int[Log[(c_.)*(x_)]^(-1), x_Symbol] :> Simp[LogIntegral[c*x]/c, x] /; FreeQ[c, 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} &=\int \left (-\frac {9 e^{6562-2 x} (1-\log (x)+2 x \log (x))}{\log ^2(x)}+\frac {e-e \log (x)+\log ^2(x)}{\log ^2(x)}\right ) \, dx\\ &=-\left (9 \int \frac {e^{6562-2 x} (1-\log (x)+2 x \log (x))}{\log ^2(x)} \, dx\right )+\int \frac {e-e \log (x)+\log ^2(x)}{\log ^2(x)} \, dx\\ &=\frac {9 e^{6562-2 x} x}{\log (x)}+\int \left (1+\frac {e}{\log ^2(x)}-\frac {e}{\log (x)}\right ) \, dx\\ &=x+\frac {9 e^{6562-2 x} x}{\log (x)}+e \int \frac {1}{\log ^2(x)} \, dx-e \int \frac {1}{\log (x)} \, dx\\ &=x-\frac {e x}{\log (x)}+\frac {9 e^{6562-2 x} x}{\log (x)}-e \text {li}(x)+e \int \frac {1}{\log (x)} \, dx\\ &=x-\frac {e x}{\log (x)}+\frac {9 e^{6562-2 x} x}{\log (x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.65, size = 21, normalized size = 0.95 \begin {gather*} \frac {x \left (-e+9 e^{6562-2 x}+\log (x)\right )}{\log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E - 9*E^(6562 - 2*x) + (-E + E^(6562 - 2*x)*(9 - 18*x))*Log[x] + Log[x]^2)/Log[x]^2,x]

[Out]

(x*(-E + 9*E^(6562 - 2*x) + Log[x]))/Log[x]

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fricas [A]  time = 1.34, size = 25, normalized size = 1.14 \begin {gather*} -\frac {x e - 9 \, x e^{\left (-2 \, x + 6562\right )} - x \log \relax (x)}{\log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((log(x)^2+((-18*x+9)*exp(1)*exp(-2*x+6561)-exp(1))*log(x)-9*exp(1)*exp(-2*x+6561)+exp(1))/log(x)^2,x
, algorithm="fricas")

[Out]

-(x*e - 9*x*e^(-2*x + 6562) - x*log(x))/log(x)

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giac [A]  time = 0.39, size = 25, normalized size = 1.14 \begin {gather*} -\frac {x e - 9 \, x e^{\left (-2 \, x + 6562\right )} - x \log \relax (x)}{\log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((log(x)^2+((-18*x+9)*exp(1)*exp(-2*x+6561)-exp(1))*log(x)-9*exp(1)*exp(-2*x+6561)+exp(1))/log(x)^2,x
, algorithm="giac")

[Out]

-(x*e - 9*x*e^(-2*x + 6562) - x*log(x))/log(x)

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maple [A]  time = 0.06, size = 21, normalized size = 0.95

method	result	size
risch	\(x +\frac {x \,{\mathrm e} \left (9 \,{\mathrm e}^{-2 x +6561}-1\right )}{\ln \relax (x )}\)	\(21\)
default	\(x +\frac {9 \,{\mathrm e} x \,{\mathrm e}^{-2 x +6561}}{\ln \relax (x )}-\frac {x \,{\mathrm e}}{\ln \relax (x )}\)	\(27\)
norman	\(\frac {x \ln \relax (x )-x \,{\mathrm e}+9 x \,{\mathrm e} \,{\mathrm e}^{-2 x +6561}}{\ln \relax (x )}\)	\(27\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((ln(x)^2+((-18*x+9)*exp(1)*exp(-2*x+6561)-exp(1))*ln(x)-9*exp(1)*exp(-2*x+6561)+exp(1))/ln(x)^2,x,method=_
RETURNVERBOSE)

[Out]

x+x*exp(1)*(9*exp(-2*x+6561)-1)/ln(x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} e \Gamma \left (-1, -\log \relax (x)\right ) - e \int \frac {1}{\log \relax (x)}\,{d x} + x + \frac {9 \, x e^{\left (-2 \, x + 6562\right )}}{\log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((log(x)^2+((-18*x+9)*exp(1)*exp(-2*x+6561)-exp(1))*log(x)-9*exp(1)*exp(-2*x+6561)+exp(1))/log(x)^2,x
, algorithm="maxima")

[Out]

e*gamma(-1, -log(x)) - e*integrate(1/log(x), x) + x + 9*x*e^(-2*x + 6562)/log(x)

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mupad [B]  time = 0.31, size = 20, normalized size = 0.91 \begin {gather*} x-\frac {x\,\left (\mathrm {e}-9\,{\mathrm {e}}^{6562-2\,x}\right )}{\ln \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(1) + log(x)^2 - 9*exp(1)*exp(6561 - 2*x) - log(x)*(exp(1) + exp(1)*exp(6561 - 2*x)*(18*x - 9)))/log(x
)^2,x)

[Out]

x - (x*(exp(1) - 9*exp(6562 - 2*x)))/log(x)

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sympy [A]  time = 0.28, size = 26, normalized size = 1.18 \begin {gather*} \frac {9 e x e^{6561 - 2 x}}{\log {\relax (x )}} + x - \frac {e x}{\log {\relax (x )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((ln(x)**2+((-18*x+9)*exp(1)*exp(-2*x+6561)-exp(1))*ln(x)-9*exp(1)*exp(-2*x+6561)+exp(1))/ln(x)**2,x)

[Out]

9*E*x*exp(6561 - 2*x)/log(x) + x - E*x/log(x)

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