3.37.15 \(\int \frac {e^{-x} (-x+e^x (3 x^3-3 x^2 \log (3))+(-x+x^2+(1-x) \log (3)) \log (x-\log (3)))}{-x+\log (3)} \, dx\)

Optimal. Leaf size=21 \[ x \left (-x^2+e^{-x} \log (x-\log (3))\right ) \]

________________________________________________________________________________________

Rubi [A]  time = 0.81, antiderivative size = 38, normalized size of antiderivative = 1.81, number of steps used = 14, number of rules used = 7, integrand size = 58, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {6742, 6688, 2199, 2194, 2178, 2176, 2554} \begin {gather*} -x^3+e^{-x} \log (x-\log (3))-e^{-x} (1-x) \log (x-\log (3)) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-x + E^x*(3*x^3 - 3*x^2*Log[3]) + (-x + x^2 + (1 - x)*Log[3])*Log[x - Log[3]])/(E^x*(-x + Log[3])),x]

[Out]

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

Rule 2176

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^m
*(b*F^(g*(e + f*x)))^n)/(f*g*n*Log[F]), x] - Dist[(d*m)/(f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*(b*F^(g*(e + f*x
)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && IntegerQ[2*m] &&  !$UseGamma === True

Rule 2178

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - (c*f)/d))*ExpIntegral
Ei[(f*g*(c + d*x)*Log[F])/d])/d, x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !$UseGamma === True

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 2199

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> Int[ExpandIntegrand[F^(c*ExpandToSum[v, x]), w*NormalizePo
werOfLinear[u, x]^m, x], x] /; FreeQ[{F, c}, x] && PolynomialQ[w, x] && LinearQ[v, x] && PowerOfLinearQ[u, x]
&& IntegerQ[m] &&  !$UseGamma === True

Rule 2554

Int[Log[u_]*(v_), x_Symbol] :> With[{w = IntHide[v, x]}, Dist[Log[u], w, x] - Int[SimplifyIntegrand[(w*D[u, x]
)/u, x], x] /; InverseFunctionFreeQ[w, x]] /; InverseFunctionFreeQ[u, x]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, 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 (-3 x^2+\frac {e^{-x} \left (x-x^2 \log (x-\log (3))-\log (3) \log (x-\log (3))+x (1+\log (3)) \log (x-\log (3))\right )}{x-\log (3)}\right ) \, dx\\ &=-x^3+\int \frac {e^{-x} \left (x-x^2 \log (x-\log (3))-\log (3) \log (x-\log (3))+x (1+\log (3)) \log (x-\log (3))\right )}{x-\log (3)} \, dx\\ &=-x^3+\int e^{-x} \left (\frac {x}{x-\log (3)}-(-1+x) \log (x-\log (3))\right ) \, dx\\ &=-x^3+\int \left (\frac {e^{-x} x}{x-\log (3)}-e^{-x} (-1+x) \log (x-\log (3))\right ) \, dx\\ &=-x^3+\int \frac {e^{-x} x}{x-\log (3)} \, dx-\int e^{-x} (-1+x) \log (x-\log (3)) \, dx\\ &=-x^3+e^{-x} \log (x-\log (3))-e^{-x} (1-x) \log (x-\log (3))+\int \frac {e^{-x} x}{-x+\log (3)} \, dx+\int \left (e^{-x}+\frac {e^{-x} \log (3)}{x-\log (3)}\right ) \, dx\\ &=-x^3+e^{-x} \log (x-\log (3))-e^{-x} (1-x) \log (x-\log (3))+\log (3) \int \frac {e^{-x}}{x-\log (3)} \, dx+\int e^{-x} \, dx+\int \left (-e^{-x}-\frac {e^{-x} \log (3)}{x-\log (3)}\right ) \, dx\\ &=-e^{-x}-x^3+\frac {1}{3} \text {Ei}(-x+\log (3)) \log (3)+e^{-x} \log (x-\log (3))-e^{-x} (1-x) \log (x-\log (3))-\log (3) \int \frac {e^{-x}}{x-\log (3)} \, dx-\int e^{-x} \, dx\\ &=-x^3+e^{-x} \log (x-\log (3))-e^{-x} (1-x) \log (x-\log (3))\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.90, size = 24, normalized size = 1.14 \begin {gather*} -x^3+\log ^3(3)+e^{-x} x \log (x-\log (3)) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-x + E^x*(3*x^3 - 3*x^2*Log[3]) + (-x + x^2 + (1 - x)*Log[3])*Log[x - Log[3]])/(E^x*(-x + Log[3])),
x]

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.62, size = 23, normalized size = 1.10 \begin {gather*} -{\left (x^{3} e^{x} - x \log \left (x - \log \relax (3)\right )\right )} e^{\left (-x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x+1)*log(3)+x^2-x)*log(-log(3)+x)+(-3*x^2*log(3)+3*x^3)*exp(x)-x)/(log(3)-x)/exp(x),x, algorithm
="fricas")

[Out]

-(x^3*e^x - x*log(x - log(3)))*e^(-x)

________________________________________________________________________________________

giac [A]  time = 0.19, size = 19, normalized size = 0.90 \begin {gather*} -x^{3} + x e^{\left (-x\right )} \log \left (x - \log \relax (3)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x+1)*log(3)+x^2-x)*log(-log(3)+x)+(-3*x^2*log(3)+3*x^3)*exp(x)-x)/(log(3)-x)/exp(x),x, algorithm
="giac")

[Out]

-x^3 + x*e^(-x)*log(x - log(3))

________________________________________________________________________________________

maple [A]  time = 0.16, size = 20, normalized size = 0.95

method	result	size
default	\(\ln \left (-\ln \relax (3)+x \right ) x \,{\mathrm e}^{-x}-x^{3}\)	\(20\)
risch	\(\ln \left (-\ln \relax (3)+x \right ) x \,{\mathrm e}^{-x}-x^{3}\)	\(20\)
norman	\(\left (\ln \left (-\ln \relax (3)+x \right ) x -{\mathrm e}^{x} x^{3}\right ) {\mathrm e}^{-x}\)	\(23\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((1-x)*ln(3)+x^2-x)*ln(-ln(3)+x)+(-3*x^2*ln(3)+3*x^3)*exp(x)-x)/(ln(3)-x)/exp(x),x,method=_RETURNVERBOSE)

[Out]

ln(-ln(3)+x)*x/exp(x)-x^3

________________________________________________________________________________________

maxima [A]  time = 0.87, size = 19, normalized size = 0.90 \begin {gather*} -x^{3} + x e^{\left (-x\right )} \log \left (x - \log \relax (3)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x+1)*log(3)+x^2-x)*log(-log(3)+x)+(-3*x^2*log(3)+3*x^3)*exp(x)-x)/(log(3)-x)/exp(x),x, algorithm
="maxima")

[Out]

-x^3 + x*e^(-x)*log(x - log(3))

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.05 \begin {gather*} \int \frac {{\mathrm {e}}^{-x}\,\left (x+\ln \left (x-\ln \relax (3)\right )\,\left (x+\ln \relax (3)\,\left (x-1\right )-x^2\right )+{\mathrm {e}}^x\,\left (3\,x^2\,\ln \relax (3)-3\,x^3\right )\right )}{x-\ln \relax (3)} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-x)*(x + log(x - log(3))*(x + log(3)*(x - 1) - x^2) + exp(x)*(3*x^2*log(3) - 3*x^3)))/(x - log(3)),x)

[Out]

int((exp(-x)*(x + log(x - log(3))*(x + log(3)*(x - 1) - x^2) + exp(x)*(3*x^2*log(3) - 3*x^3)))/(x - log(3)), x
)

________________________________________________________________________________________

sympy [A]  time = 0.42, size = 14, normalized size = 0.67 \begin {gather*} - x^{3} + x e^{- x} \log {\left (x - \log {\relax (3 )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x+1)*ln(3)+x**2-x)*ln(-ln(3)+x)+(-3*x**2*ln(3)+3*x**3)*exp(x)-x)/(ln(3)-x)/exp(x),x)

[Out]

-x**3 + x*exp(-x)*log(x - log(3))

________________________________________________________________________________________