3.74.90 \(\int \frac {59049 e^{-18+e^x} (-4+4 e^x x \log (x)+4 e^x x \log ^2(x))}{x \log ^2(x)} \, dx\)

Optimal. Leaf size=26 \[ e^{5 \left (-4+\frac {1}{5} \left (2+e^x\right )+\log (9)\right )} \left (4+\frac {4}{\log (x)}\right ) \]

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Rubi [A]  time = 0.20, antiderivative size = 36, normalized size of antiderivative = 1.38, number of steps used = 2, number of rules used = 2, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {12, 2288} \begin {gather*} \frac {236196 e^{-x+e^x-18} \left (e^x x \log ^2(x)+e^x x \log (x)\right )}{x \log ^2(x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(59049*E^(-18 + E^x)*(-4 + 4*E^x*x*Log[x] + 4*E^x*x*Log[x]^2))/(x*Log[x]^2),x]

[Out]

(236196*E^(-18 + E^x - x)*(E^x*x*Log[x] + E^x*x*Log[x]^2))/(x*Log[x]^2)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=59049 \int \frac {e^{-18+e^x} \left (-4+4 e^x x \log (x)+4 e^x x \log ^2(x)\right )}{x \log ^2(x)} \, dx\\ &=\frac {236196 e^{-18+e^x-x} \left (e^x x \log (x)+e^x x \log ^2(x)\right )}{x \log ^2(x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.04, size = 17, normalized size = 0.65 \begin {gather*} \frac {236196 e^{-18+e^x} (1+\log (x))}{\log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(59049*E^(-18 + E^x)*(-4 + 4*E^x*x*Log[x] + 4*E^x*x*Log[x]^2))/(x*Log[x]^2),x]

[Out]

(236196*E^(-18 + E^x)*(1 + Log[x]))/Log[x]

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fricas [A]  time = 0.60, size = 19, normalized size = 0.73 \begin {gather*} \frac {4 \, {\left (\log \relax (x) + 1\right )} e^{\left (e^{x} + 10 \, \log \relax (3) - 18\right )}}{\log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x*exp(x)*log(x)^2+4*x*exp(x)*log(x)-4)*exp(exp(x)+10*log(3)-18)/x/log(x)^2,x, algorithm="fricas")

[Out]

4*(log(x) + 1)*e^(e^x + 10*log(3) - 18)/log(x)

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giac [A]  time = 0.19, size = 26, normalized size = 1.00 \begin {gather*} \frac {236196 \, {\left (e^{\left (x + e^{x}\right )} \log \relax (x) + e^{\left (x + e^{x}\right )}\right )} e^{\left (-x - 18\right )}}{\log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x*exp(x)*log(x)^2+4*x*exp(x)*log(x)-4)*exp(exp(x)+10*log(3)-18)/x/log(x)^2,x, algorithm="giac")

[Out]

236196*(e^(x + e^x)*log(x) + e^(x + e^x))*e^(-x - 18)/log(x)

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maple [A]  time = 0.04, size = 16, normalized size = 0.62

method	result	size
risch	\(\frac {236196 \left (\ln \relax (x )+1\right ) {\mathrm e}^{{\mathrm e}^{x}-18}}{\ln \relax (x )}\)	\(16\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*x*exp(x)*ln(x)^2+4*x*exp(x)*ln(x)-4)*exp(exp(x)+10*ln(3)-18)/x/ln(x)^2,x,method=_RETURNVERBOSE)

[Out]

236196*(ln(x)+1)/ln(x)*exp(exp(x)-18)

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maxima [A]  time = 0.43, size = 19, normalized size = 0.73 \begin {gather*} \frac {236196 \, e^{\left (e^{x} - 18\right )}}{\log \relax (x)} + 236196 \, e^{\left (e^{x} - 18\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x*exp(x)*log(x)^2+4*x*exp(x)*log(x)-4)*exp(exp(x)+10*log(3)-18)/x/log(x)^2,x, algorithm="maxima")

[Out]

236196*e^(e^x - 18)/log(x) + 236196*e^(e^x - 18)

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mupad [B]  time = 5.06, size = 15, normalized size = 0.58 \begin {gather*} \frac {236196\,{\mathrm {e}}^{{\mathrm {e}}^x}\,{\mathrm {e}}^{-18}\,\left (\ln \relax (x)+1\right )}{\ln \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(10*log(3) + exp(x) - 18)*(4*x*exp(x)*log(x) + 4*x*exp(x)*log(x)^2 - 4))/(x*log(x)^2),x)

[Out]

(236196*exp(exp(x))*exp(-18)*(log(x) + 1))/log(x)

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sympy [A]  time = 0.27, size = 15, normalized size = 0.58 \begin {gather*} \frac {\left (236196 \log {\relax (x )} + 236196\right ) e^{e^{x} - 18}}{\log {\relax (x )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x*exp(x)*ln(x)**2+4*x*exp(x)*ln(x)-4)*exp(exp(x)+10*ln(3)-18)/x/ln(x)**2,x)

[Out]

(236196*log(x) + 236196)*exp(exp(x) - 18)/log(x)

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