3.63.87 \(\int \frac {e^{-x} (12+e^x (-4 x-4 x \log (5))+(-12 x-4 e^x x) \log (x))}{3 x} \, dx\)

Optimal. Leaf size=26 \[ 4 \left (3-\frac {1}{3} x \log (5)+e^{-x} \log (x)-\frac {1}{3} x \log (x)\right ) \]

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Rubi [A]  time = 0.51, antiderivative size = 31, normalized size of antiderivative = 1.19, number of steps used = 7, number of rules used = 5, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {12, 6741, 6742, 2295, 2288} \begin {gather*} \frac {4 x}{3}-\frac {4}{3} x \log (x)-\frac {4}{3} x (1+\log (5))+4 e^{-x} \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(12 + E^x*(-4*x - 4*x*Log[5]) + (-12*x - 4*E^x*x)*Log[x])/(3*E^x*x),x]

[Out]

(4*x)/3 - (4*x*(1 + Log[5]))/3 + (4*Log[x])/E^x - (4*x*Log[x])/3

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]

Rule 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, 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} &=\frac {1}{3} \int \frac {e^{-x} \left (12+e^x (-4 x-4 x \log (5))+\left (-12 x-4 e^x x\right ) \log (x)\right )}{x} \, dx\\ &=\frac {1}{3} \int \frac {4 e^{-x} \left (3-e^x x (1+\log (5))-3 x \log (x)-e^x x \log (x)\right )}{x} \, dx\\ &=\frac {4}{3} \int \frac {e^{-x} \left (3-e^x x (1+\log (5))-3 x \log (x)-e^x x \log (x)\right )}{x} \, dx\\ &=\frac {4}{3} \int \left (-1-\log (5)-\log (x)-\frac {3 e^{-x} (-1+x \log (x))}{x}\right ) \, dx\\ &=-\frac {4}{3} x (1+\log (5))-\frac {4}{3} \int \log (x) \, dx-4 \int \frac {e^{-x} (-1+x \log (x))}{x} \, dx\\ &=\frac {4 x}{3}-\frac {4}{3} x (1+\log (5))+4 e^{-x} \log (x)-\frac {4}{3} x \log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.10, size = 22, normalized size = 0.85 \begin {gather*} -\frac {4}{3} \left (x \log (5)-3 e^{-x} \log (x)+x \log (x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(12 + E^x*(-4*x - 4*x*Log[5]) + (-12*x - 4*E^x*x)*Log[x])/(3*E^x*x),x]

[Out]

(-4*(x*Log[5] - (3*Log[x])/E^x + x*Log[x]))/3

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fricas [A]  time = 0.71, size = 22, normalized size = 0.85 \begin {gather*} -\frac {4}{3} \, {\left (x e^{x} \log \relax (5) + {\left (x e^{x} - 3\right )} \log \relax (x)\right )} e^{\left (-x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((-4*exp(x)*x-12*x)*log(x)+(-4*x*log(5)-4*x)*exp(x)+12)/exp(x)/x,x, algorithm="fricas")

[Out]

-4/3*(x*e^x*log(5) + (x*e^x - 3)*log(x))*e^(-x)

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giac [A]  time = 0.20, size = 19, normalized size = 0.73 \begin {gather*} -\frac {4}{3} \, x \log \relax (5) - \frac {4}{3} \, x \log \relax (x) + 4 \, e^{\left (-x\right )} \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((-4*exp(x)*x-12*x)*log(x)+(-4*x*log(5)-4*x)*exp(x)+12)/exp(x)/x,x, algorithm="giac")

[Out]

-4/3*x*log(5) - 4/3*x*log(x) + 4*e^(-x)*log(x)

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maple [A]  time = 0.09, size = 20, normalized size = 0.77

method	result	size
default	\(4 \ln \relax (x ) {\mathrm e}^{-x}-\frac {4 x \ln \relax (5)}{3}-\frac {4 x \ln \relax (x )}{3}\)	\(20\)
risch	\(-\frac {4 \left ({\mathrm e}^{x} x -3\right ) {\mathrm e}^{-x} \ln \relax (x )}{3}-\frac {4 x \ln \relax (5)}{3}\)	\(21\)
norman	\(\left (-\frac {4 x \,{\mathrm e}^{x} \ln \relax (5)}{3}-\frac {4 x \,{\mathrm e}^{x} \ln \relax (x )}{3}+4 \ln \relax (x )\right ) {\mathrm e}^{-x}\)	\(25\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/3*((-4*exp(x)*x-12*x)*ln(x)+(-4*x*ln(5)-4*x)*exp(x)+12)/exp(x)/x,x,method=_RETURNVERBOSE)

[Out]

4*ln(x)/exp(x)-4/3*x*ln(5)-4/3*x*ln(x)

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maxima [A]  time = 0.39, size = 19, normalized size = 0.73 \begin {gather*} -\frac {4}{3} \, x \log \relax (5) - \frac {4}{3} \, x \log \relax (x) + 4 \, e^{\left (-x\right )} \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((-4*exp(x)*x-12*x)*log(x)+(-4*x*log(5)-4*x)*exp(x)+12)/exp(x)/x,x, algorithm="maxima")

[Out]

-4/3*x*log(5) - 4/3*x*log(x) + 4*e^(-x)*log(x)

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mupad [B]  time = 4.25, size = 19, normalized size = 0.73 \begin {gather*} 4\,{\mathrm {e}}^{-x}\,\ln \relax (x)-\frac {4\,x\,\ln \relax (5)}{3}-\frac {4\,x\,\ln \relax (x)}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-x)*((log(x)*(12*x + 4*x*exp(x)))/3 + (exp(x)*(4*x + 4*x*log(5)))/3 - 4))/x,x)

[Out]

4*exp(-x)*log(x) - (4*x*log(5))/3 - (4*x*log(x))/3

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sympy [A]  time = 0.33, size = 24, normalized size = 0.92 \begin {gather*} - \frac {4 x \log {\relax (x )}}{3} - \frac {4 x \log {\relax (5 )}}{3} + 4 e^{- x} \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((-4*exp(x)*x-12*x)*ln(x)+(-4*x*ln(5)-4*x)*exp(x)+12)/exp(x)/x,x)

[Out]

-4*x*log(x)/3 - 4*x*log(5)/3 + 4*exp(-x)*log(x)

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