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

Optimal. Leaf size=18 \[ e^x \left (-x-x^2+\log (3) \log (x)\right ) \]

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Rubi [A]  time = 0.10, antiderivative size = 23, normalized size of antiderivative = 1.28, number of steps used = 11, number of rules used = 5, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.147, Rules used = {14, 2194, 2176, 2178, 2554} \begin {gather*} -e^x x^2-e^x x+e^x \log (3) \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(E^x*x) - E^x*x^2 + E^x*Log[3]*Log[x]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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 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]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-e^x-3 e^x x-e^x x^2+\frac {e^x \log (3)}{x}+e^x \log (3) \log (x)\right ) \, dx\\ &=-\left (3 \int e^x x \, dx\right )+\log (3) \int \frac {e^x}{x} \, dx+\log (3) \int e^x \log (x) \, dx-\int e^x \, dx-\int e^x x^2 \, dx\\ &=-e^x-3 e^x x-e^x x^2+\text {Ei}(x) \log (3)+e^x \log (3) \log (x)+2 \int e^x x \, dx+3 \int e^x \, dx-\log (3) \int \frac {e^x}{x} \, dx\\ &=2 e^x-e^x x-e^x x^2+e^x \log (3) \log (x)-2 \int e^x \, dx\\ &=-e^x x-e^x x^2+e^x \log (3) \log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.07, size = 16, normalized size = 0.89 \begin {gather*} -e^x \left (x+x^2-\log (3) \log (x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(E^x*(x + x^2 - Log[3]*Log[x]))

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fricas [A]  time = 0.47, size = 17, normalized size = 0.94 \begin {gather*} e^{x} \log \relax (3) \log \relax (x) - {\left (x^{2} + x\right )} e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e^x*log(3)*log(x) - (x^2 + x)*e^x

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giac [A]  time = 0.93, size = 20, normalized size = 1.11 \begin {gather*} -x^{2} e^{x} + e^{x} \log \relax (3) \log \relax (x) - x e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x^2*e^x + e^x*log(3)*log(x) - x*e^x

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maple [A]  time = 0.05, size = 17, normalized size = 0.94

method	result	size
risch	\(\ln \relax (3) {\mathrm e}^{x} \ln \relax (x )-x \left (x +1\right ) {\mathrm e}^{x}\)	\(17\)
norman	\(\ln \relax (3) {\mathrm e}^{x} \ln \relax (x )-{\mathrm e}^{x} x -{\mathrm e}^{x} x^{2}\)	\(21\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

ln(3)*exp(x)*ln(x)-x*(x+1)*exp(x)

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maxima [B]  time = 0.53, size = 31, normalized size = 1.72 \begin {gather*} e^{x} \log \relax (3) \log \relax (x) - {\left (x^{2} - 2 \, x + 2\right )} e^{x} - 3 \, {\left (x - 1\right )} e^{x} - e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e^x*log(3)*log(x) - (x^2 - 2*x + 2)*e^x - 3*(x - 1)*e^x - e^x

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mupad [B]  time = 0.70, size = 15, normalized size = 0.83 \begin {gather*} -{\mathrm {e}}^x\,\left (x-\ln \relax (3)\,\ln \relax (x)+x^2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-exp(x)*(x - log(3)*log(x) + x^2)

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sympy [A]  time = 0.34, size = 14, normalized size = 0.78 \begin {gather*} \left (- x^{2} - x + \log {\relax (3 )} \log {\relax (x )}\right ) e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-x**2 - x + log(3)*log(x))*exp(x)

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