3.12.31 \(\int (e^x (8+4 x)+e^x (-2-2 x) \log (x)) \, dx\)

Optimal. Leaf size=25 \[ 2 e^x \left (5-x \left (\frac {2 \left (x-x^2\right )}{x^2}+\log (x)\right )\right ) \]

________________________________________________________________________________________

Rubi [A]  time = 0.05, antiderivative size = 31, normalized size of antiderivative = 1.24, number of steps used = 6, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {2176, 2194, 2554, 12} \begin {gather*} 4 e^x (x+2)-2 e^x+2 e^x \log (x)-2 e^x (x+1) \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^x*(8 + 4*x) + E^x*(-2 - 2*x)*Log[x],x]

[Out]

-2*E^x + 4*E^x*(2 + x) + 2*E^x*Log[x] - 2*E^x*(1 + x)*Log[x]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 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 e^x (8+4 x) \, dx+\int e^x (-2-2 x) \log (x) \, dx\\ &=4 e^x (2+x)+2 e^x \log (x)-2 e^x (1+x) \log (x)-4 \int e^x \, dx-\int -2 e^x \, dx\\ &=-4 e^x+4 e^x (2+x)+2 e^x \log (x)-2 e^x (1+x) \log (x)+2 \int e^x \, dx\\ &=-2 e^x+4 e^x (2+x)+2 e^x \log (x)-2 e^x (1+x) \log (x)\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.03, size = 14, normalized size = 0.56 \begin {gather*} -2 e^x (-3-2 x+x \log (x)) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[E^x*(8 + 4*x) + E^x*(-2 - 2*x)*Log[x],x]

[Out]

-2*E^x*(-3 - 2*x + x*Log[x])

________________________________________________________________________________________

fricas [A]  time = 0.67, size = 17, normalized size = 0.68 \begin {gather*} -2 \, x e^{x} \log \relax (x) + 2 \, {\left (2 \, x + 3\right )} e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x-2)*exp(x)*log(x)+(4*x+8)*exp(x),x, algorithm="fricas")

[Out]

-2*x*e^x*log(x) + 2*(2*x + 3)*e^x

________________________________________________________________________________________

giac [A]  time = 0.23, size = 19, normalized size = 0.76 \begin {gather*} -2 \, x e^{x} \log \relax (x) + 4 \, {\left (x + 1\right )} e^{x} + 2 \, e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x-2)*exp(x)*log(x)+(4*x+8)*exp(x),x, algorithm="giac")

[Out]

-2*x*e^x*log(x) + 4*(x + 1)*e^x + 2*e^x

________________________________________________________________________________________

maple [A]  time = 0.04, size = 18, normalized size = 0.72




method result size



default \(4 \,{\mathrm e}^{x} x +6 \,{\mathrm e}^{x}-2 x \,{\mathrm e}^{x} \ln \relax (x )\) \(18\)
norman \(4 \,{\mathrm e}^{x} x +6 \,{\mathrm e}^{x}-2 x \,{\mathrm e}^{x} \ln \relax (x )\) \(18\)
risch \(4 \,{\mathrm e}^{x} x +6 \,{\mathrm e}^{x}-2 x \,{\mathrm e}^{x} \ln \relax (x )\) \(18\)



Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-2*x-2)*exp(x)*ln(x)+(4*x+8)*exp(x),x,method=_RETURNVERBOSE)

[Out]

4*exp(x)*x+6*exp(x)-2*x*exp(x)*ln(x)

________________________________________________________________________________________

maxima [A]  time = 0.45, size = 19, normalized size = 0.76 \begin {gather*} -2 \, x e^{x} \log \relax (x) + 4 \, {\left (x - 1\right )} e^{x} + 10 \, e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x-2)*exp(x)*log(x)+(4*x+8)*exp(x),x, algorithm="maxima")

[Out]

-2*x*e^x*log(x) + 4*(x - 1)*e^x + 10*e^x

________________________________________________________________________________________

mupad [B]  time = 0.78, size = 14, normalized size = 0.56 \begin {gather*} 2\,{\mathrm {e}}^x\,\left (2\,x-x\,\ln \relax (x)+3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(x)*(4*x + 8) - exp(x)*log(x)*(2*x + 2),x)

[Out]

2*exp(x)*(2*x - x*log(x) + 3)

________________________________________________________________________________________

sympy [A]  time = 0.29, size = 14, normalized size = 0.56 \begin {gather*} \left (- 2 x \log {\relax (x )} + 4 x + 6\right ) e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x-2)*exp(x)*ln(x)+(4*x+8)*exp(x),x)

[Out]

(-2*x*log(x) + 4*x + 6)*exp(x)

________________________________________________________________________________________