3.69.20 $\int e^{-3-2x}(4e^{2+2x}+e(1+4x-4x^2)+e(1-2x)\log (x))dx$

Optimal. Leaf size=23 $$-2+\frac{4x}{e}+e^{-2-2x}x(2x+\log (x))$$

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Rubi [A]  time = 0.55, antiderivative size = 30, normalized size of antiderivative = 1.30, number of steps used = 18, number of rules used = 8, integrand size = 39, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.205, Rules used = {6741, 6742, 2194, 2176, 2554, 12, 2178, 2199} $$\begin{array}{c}2e^{-2x-2}{x}^2+\frac{4x}{e}+e^{-2x-2}x\log (x)\end{array}$$

Antiderivative was successfully verified.

[In]

Int[E^(-3 - 2*x)*(4*E^(2 + 2*x) + E*(1 + 4*x - 4*x^2) + E*(1 - 2*x)*Log[x]),x]

[Out]

(4*x)/E + 2*E^(-2 - 2*x)*x^2 + E^(-2 - 2*x)*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 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 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

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int e^{-2-2x}(1+4e^{1+2x}+4x-4x^2+\log (x)-2x\log (x))dx\\ & =\int (\frac{4}{e}+e^{-2-2x}+4e^{-2-2x}x-4e^{-2-2x}{x}^2+e^{-2-2x}\log (x)-2e^{-2-2x}x\log (x))dx\\ & =\frac{4x}{e}-2\int e^{-2-2x}x\log (x)dx+4\int e^{-2-2x}xdx-4\int e^{-2-2x}{x}^{2}dx+\int e^{-2-2x}dx+\int e^{-2-2x}\log (x)dx\\ & =-\frac{1}{2}{e}^{-2-2x}+\frac{4x}{e}-2e^{-2-2x}x+2e^{-2-2x}{x}^2+e^{-2-2x}x\log (x)+2\int e^{-2-2x}dx+2\int \frac{e^{-2-2x}(-1-2x)}{4x}dx-4\int e^{-2-2x}xdx-\int -\frac{e^{-2-2x}}{2x}dx\\ & =-\frac{3}{2}{e}^{-2-2x}+\frac{4x}{e}+2e^{-2-2x}{x}^2+e^{-2-2x}x\log (x)+\frac{1}{2}\int \frac{e^{-2-2x}}{x}dx+\frac{1}{2}\int \frac{e^{-2-2x}(-1-2x)}{x}dx-2\int e^{-2-2x}dx\\ & =-\frac{1}{2}{e}^{-2-2x}+\frac{4x}{e}+2e^{-2-2x}{x}^2+\frac{\text{Ei}(-2x)}{2e^2}+e^{-2-2x}x\log (x)+\frac{1}{2}\int (-2e^{-2-2x}-\frac{e^{-2-2x}}{x})dx\\ & =-\frac{1}{2}{e}^{-2-2x}+\frac{4x}{e}+2e^{-2-2x}{x}^2+\frac{\text{Ei}(-2x)}{2e^2}+e^{-2-2x}x\log (x)-\frac{1}{2}\int \frac{e^{-2-2x}}{x}dx-\int e^{-2-2x}dx\\ & =\frac{4x}{e}+2e^{-2-2x}{x}^2+e^{-2-2x}x\log (x)\end{array}\end{array}$$

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Mathematica [A]  time = 0.30, size = 24, normalized size = 1.04 $$\begin{array}{c}{e}^{-2(1+x)}x(4e^{1+2x}+2x+\log (x))\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[E^(-3 - 2*x)*(4*E^(2 + 2*x) + E*(1 + 4*x - 4*x^2) + E*(1 - 2*x)*Log[x]),x]

[Out]

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

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fricas [A]  time = 0.45, size = 30, normalized size = 1.30 $$\begin{array}{c}(2x^{2}e+xe\log (x)+4x{e}^{(2x+2)})e^{(-2x-3)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((1-2*x)*exp(1)*log(x)+4*exp(x+1)^2+(-4*x^2+4*x+1)*exp(1))/exp(1)/exp(x+1)^2,x, algorithm="fricas")

[Out]

(2*x^2*e + x*e*log(x) + 4*x*e^(2*x + 2))*e^(-2*x - 3)

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giac [A]  time = 0.13, size = 30, normalized size = 1.30 $$\begin{array}{c}(2x^2{e}^{(-2x+1)}+x{e}^{(-2x+1)}\log (x)+4x{e}^2)e^{(-3)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((1-2*x)*exp(1)*log(x)+4*exp(x+1)^2+(-4*x^2+4*x+1)*exp(1))/exp(1)/exp(x+1)^2,x, algorithm="giac")

[Out]

(2*x^2*e^(-2*x + 1) + x*e^(-2*x + 1)*log(x) + 4*x*e^2)*e^(-3)

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maple [A]  time = 0.14, size = 31, normalized size = 1.35

method	result	size
default	${\mathrm{e}}^{-1}(4x+(x\mathrm{e}\ln (x)+2x^2\mathrm{e}){\mathrm{e}}^{-2x-2})$	$31$
norman	$(x\ln (x)+2x^2+4x{\mathrm{e}}^{-1}{\mathrm{e}}^{2x+2}){\mathrm{e}}^{-2x-2}$	$31$
risch	$x{\mathrm{e}}^{-2x-2}\ln (x)+2(x\mathrm{e}+2{\mathrm{e}}^{2x+2})x{\mathrm{e}}^{-2x-3}$	$34$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((1-2*x)*exp(1)*ln(x)+4*exp(x+1)^2+(-4*x^2+4*x+1)*exp(1))/exp(1)/exp(x+1)^2,x,method=_RETURNVERBOSE)

[Out]

1/exp(1)*(4*x+(x*exp(1)*ln(x)+2*x^2*exp(1))/exp(x+1)^2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 $$\begin{array}{c}\frac{1}{2}(2x+1)e^{(-2x-2)}\log (x)+4x{e}^{(-1)}+\frac{1}{2}\mathrm{E}\mathrm{i}(-2x)e^{(-2)}+(2x^2+2x+1)e^{(-2x-2)}-(2x+1)e^{(-2x-2)}-\frac{1}{2}{e}^{(-2x-2)}\log (x)-\frac{1}{2}{e}^{(-2x-2)}-\frac{1}{2}\int \frac{(2x+1)e^{(-2x-2)}}{x}dx\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((1-2*x)*exp(1)*log(x)+4*exp(x+1)^2+(-4*x^2+4*x+1)*exp(1))/exp(1)/exp(x+1)^2,x, algorithm="maxima")

[Out]

1/2*(2*x + 1)*e^(-2*x - 2)*log(x) + 4*x*e^(-1) + 1/2*Ei(-2*x)*e^(-2) + (2*x^2 + 2*x + 1)*e^(-2*x - 2) - (2*x +
1)*e^(-2*x - 2) - 1/2*e^(-2*x - 2)*log(x) - 1/2*e^(-2*x - 2) - 1/2*integrate((2*x + 1)*e^(-2*x - 2)/x, x)

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mupad [B]  time = 4.14, size = 27, normalized size = 1.17 $$\begin{array}{c}4x{\mathrm{e}}^{-1}+2x^2{\mathrm{e}}^{-2x-2}+x{\mathrm{e}}^{-2x-2}\ln (x)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(-1)*exp(- 2*x - 2)*(4*exp(2*x + 2) + exp(1)*(4*x - 4*x^2 + 1) - exp(1)*log(x)*(2*x - 1)),x)

[Out]

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

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sympy [A]  time = 0.32, size = 24, normalized size = 1.04 $$\begin{array}{c}\frac{4x}{e}+(2x^2+x\log (x))e^{-2x-2}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((1-2*x)*exp(1)*ln(x)+4*exp(x+1)**2+(-4*x**2+4*x+1)*exp(1))/exp(1)/exp(x+1)**2,x)

[Out]

4*x*exp(-1) + (2*x**2 + x*log(x))*exp(-2*x - 2)

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