∫ e 1 ________________________________________________ −9−e6x+e3x(6−2x)+6x−x2+log(x) (−3−18x+18e6xx+6x2+e3x(−48x+18x2)) ________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________324x+4e12x x−432x2+216x3−48x4+4x5+e9x(−48x+16x2)+e6x(216x−144x2+24x3)+e3x(−432x+432x2−144x3+16x4)+(−72x−8e6xx+48x2−8x3+e3x(48x−16x2)) log(x)+4x log2(x) dx

3.65.16 \(\int \frac {e^{\frac {1}{-9-e^{6 x}+e^{3 x} (6-2 x)+6 x-x^2+\log (x)}} (-3-18 x+18 e^{6 x} x+6 x^2+e^{3 x} (-48 x+18 x^2))}{324 x+4 e^{12 x} x-432 x^2+216 x^3-48 x^4+4 x^5+e^{9 x} (-48 x+16 x^2)+e^{6 x} (216 x-144 x^2+24 x^3)+e^{3 x} (-432 x+432 x^2-144 x^3+16 x^4)+(-72 x-8 e^{6 x} x+48 x^2-8 x^3+e^{3 x} (48 x-16 x^2)) \log (x)+4 x \log ^2(x)} \, dx\)

Optimal. Leaf size=27 \[ \frac {3}{4} e^{\frac {1}{-\left (3-e^{3 x}-x\right )^2+\log (x)}} \]

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Rubi [A] time = 5.66, antiderivative size = 25, normalized size of antiderivative = 0.93, number of steps used = 3, number of rules used = 3, integrand size = 209, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.014, Rules used = {6688, 12, 6706} \begin {gather*} \frac {3}{4} e^{-\frac {1}{\left (x+e^{3 x}-3\right )^2-\log (x)}} \end {gather*}

Antiderivative was successfully veriﬁed.

[In]

Int[(E^(-9 - E^(6*x) + E^(3*x)*(6 - 2*x) + 6*x - x^2 + Log[x])^(-1)*(-3 - 18*x + 18*E^(6*x)*x + 6*x^2 + E^(3*x

)*(-48*x + 18*x^2)))/(324*x + 4*E^(12*x)*x - 432*x^2 + 216*x^3 - 48*x^4 + 4*x^5 + E^(9*x)*(-48*x + 16*x^2) + E

^(6*x)*(216*x - 144*x^2 + 24*x^3) + E^(3*x)*(-432*x + 432*x^2 - 144*x^3 + 16*x^4) + (-72*x - 8*E^(6*x)*x + 48*

x^2 - 8*x^3 + E^(3*x)*(48*x - 16*x^2))*Log[x] + 4*x*Log[x]^2),x]

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] && !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rule 6706

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[(q*F^v)/Log[F], x] /; !FalseQ[q]

] /; FreeQ[F, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {3 e^{-\frac {1}{\left (-3+e^{3 x}+x\right )^2-\log (x)}} \left (-1+2 \left (-3-8 e^{3 x}+3 e^{6 x}\right ) x+\left (2+6 e^{3 x}\right ) x^2\right )}{4 x \left (\left (-3+e^{3 x}+x\right )^2-\log (x)\right )^2} \, dx\\ &=\frac {3}{4} \int \frac {e^{-\frac {1}{\left (-3+e^{3 x}+x\right )^2-\log (x)}} \left (-1+2 \left (-3-8 e^{3 x}+3 e^{6 x}\right ) x+\left (2+6 e^{3 x}\right ) x^2\right )}{x \left (\left (-3+e^{3 x}+x\right )^2-\log (x)\right )^2} \, dx\\ &=\frac {3}{4} e^{-\frac {1}{\left (-3+e^{3 x}+x\right )^2-\log (x)}}\\ \end {aligned} \end {gather*}

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Mathematica [A] time = 5.14, size = 25, normalized size = 0.93 \begin {gather*} \frac {3}{4} e^{-\frac {1}{\left (-3+e^{3 x}+x\right )^2-\log (x)}} \end {gather*}

Antiderivative was successfully veriﬁed.

[In]

Integrate[(E^(-9 - E^(6*x) + E^(3*x)*(6 - 2*x) + 6*x - x^2 + Log[x])^(-1)*(-3 - 18*x + 18*E^(6*x)*x + 6*x^2 +

E^(3*x)*(-48*x + 18*x^2)))/(324*x + 4*E^(12*x)*x - 432*x^2 + 216*x^3 - 48*x^4 + 4*x^5 + E^(9*x)*(-48*x + 16*x^

2) + E^(6*x)*(216*x - 144*x^2 + 24*x^3) + E^(3*x)*(-432*x + 432*x^2 - 144*x^3 + 16*x^4) + (-72*x - 8*E^(6*x)*x

+ 48*x^2 - 8*x^3 + E^(3*x)*(48*x - 16*x^2))*Log[x] + 4*x*Log[x]^2),x]

[Out]

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

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fricas [A] time = 0.69, size = 32, normalized size = 1.19 \begin {gather*} \frac {3}{4} \, e^{\left (-\frac {1}{x^{2} + 2 \, {\left (x - 3\right )} e^{\left (3 \, x\right )} - 6 \, x + e^{\left (6 \, x\right )} - \log \relax (x) + 9}\right )} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((18*x*exp(3*x)^2+(18*x^2-48*x)*exp(3*x)+6*x^2-18*x-3)*exp(1/(log(x)-exp(3*x)^2+(6-2*x)*exp(3*x)-x^2+

6*x-9))/(4*x*log(x)^2+(-8*x*exp(3*x)^2+(-16*x^2+48*x)*exp(3*x)-8*x^3+48*x^2-72*x)*log(x)+4*x*exp(3*x)^4+(16*x^

2-48*x)*exp(3*x)^3+(24*x^3-144*x^2+216*x)*exp(3*x)^2+(16*x^4-144*x^3+432*x^2-432*x)*exp(3*x)+4*x^5-48*x^4+216*

x^3-432*x^2+324*x),x, algorithm="fricas")

[Out]

3/4*e^(-1/(x^2 + 2*(x - 3)*e^(3*x) - 6*x + e^(6*x) - log(x) + 9))

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giac [A] time = 30.16, size = 36, normalized size = 1.33 \begin {gather*} \frac {3}{4} \, e^{\left (-\frac {1}{x^{2} + 2 \, x e^{\left (3 \, x\right )} - 6 \, x + e^{\left (6 \, x\right )} - 6 \, e^{\left (3 \, x\right )} - \log \relax (x) + 9}\right )} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((18*x*exp(3*x)^2+(18*x^2-48*x)*exp(3*x)+6*x^2-18*x-3)*exp(1/(log(x)-exp(3*x)^2+(6-2*x)*exp(3*x)-x^2+

6*x-9))/(4*x*log(x)^2+(-8*x*exp(3*x)^2+(-16*x^2+48*x)*exp(3*x)-8*x^3+48*x^2-72*x)*log(x)+4*x*exp(3*x)^4+(16*x^

2-48*x)*exp(3*x)^3+(24*x^3-144*x^2+216*x)*exp(3*x)^2+(16*x^4-144*x^3+432*x^2-432*x)*exp(3*x)+4*x^5-48*x^4+216*

x^3-432*x^2+324*x),x, algorithm="giac")

[Out]

3/4*e^(-1/(x^2 + 2*x*e^(3*x) - 6*x + e^(6*x) - 6*e^(3*x) - log(x) + 9))

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maple [A] time = 0.06, size = 37, normalized size = 1.37


method	result	size

risch	\(\frac {3 \,{\mathrm e}^{\frac {1}{-2 x \,{\mathrm e}^{3 x}-x^{2}+\ln \relax (x )-{\mathrm e}^{6 x}+6 \,{\mathrm e}^{3 x}+6 x -9}}}{4}\)	\(37\)

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((18*x*exp(3*x)^2+(18*x^2-48*x)*exp(3*x)+6*x^2-18*x-3)*exp(1/(ln(x)-exp(3*x)^2+(6-2*x)*exp(3*x)-x^2+6*x-9))

/(4*x*ln(x)^2+(-8*x*exp(3*x)^2+(-16*x^2+48*x)*exp(3*x)-8*x^3+48*x^2-72*x)*ln(x)+4*x*exp(3*x)^4+(16*x^2-48*x)*e

xp(3*x)^3+(24*x^3-144*x^2+216*x)*exp(3*x)^2+(16*x^4-144*x^3+432*x^2-432*x)*exp(3*x)+4*x^5-48*x^4+216*x^3-432*x

^2+324*x),x,method=_RETURNVERBOSE)

[Out]

3/4*exp(1/(-2*x*exp(3*x)-x^2+ln(x)-exp(6*x)+6*exp(3*x)+6*x-9))

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maxima [A] time = 2.17, size = 32, normalized size = 1.19 \begin {gather*} \frac {3}{4} \, e^{\left (-\frac {1}{x^{2} + 2 \, {\left (x - 3\right )} e^{\left (3 \, x\right )} - 6 \, x + e^{\left (6 \, x\right )} - \log \relax (x) + 9}\right )} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((18*x*exp(3*x)^2+(18*x^2-48*x)*exp(3*x)+6*x^2-18*x-3)*exp(1/(log(x)-exp(3*x)^2+(6-2*x)*exp(3*x)-x^2+

6*x-9))/(4*x*log(x)^2+(-8*x*exp(3*x)^2+(-16*x^2+48*x)*exp(3*x)-8*x^3+48*x^2-72*x)*log(x)+4*x*exp(3*x)^4+(16*x^

2-48*x)*exp(3*x)^3+(24*x^3-144*x^2+216*x)*exp(3*x)^2+(16*x^4-144*x^3+432*x^2-432*x)*exp(3*x)+4*x^5-48*x^4+216*

x^3-432*x^2+324*x),x, algorithm="maxima")

[Out]

3/4*e^(-1/(x^2 + 2*(x - 3)*e^(3*x) - 6*x + e^(6*x) - log(x) + 9))

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mupad [B] time = 4.56, size = 36, normalized size = 1.33 \begin {gather*} \frac {3\,{\mathrm {e}}^{-\frac {1}{{\mathrm {e}}^{6\,x}-6\,{\mathrm {e}}^{3\,x}-6\,x-\ln \relax (x)+2\,x\,{\mathrm {e}}^{3\,x}+x^2+9}}}{4} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-1/(exp(6*x) - 6*x - log(x) + exp(3*x)*(2*x - 6) + x^2 + 9))*(18*x + exp(3*x)*(48*x - 18*x^2) - 18*x

*exp(6*x) - 6*x^2 + 3))/(324*x - log(x)*(72*x - exp(3*x)*(48*x - 16*x^2) + 8*x*exp(6*x) - 48*x^2 + 8*x^3) - ex

p(9*x)*(48*x - 16*x^2) + 4*x*exp(12*x) + 4*x*log(x)^2 + exp(6*x)*(216*x - 144*x^2 + 24*x^3) - exp(3*x)*(432*x

- 432*x^2 + 144*x^3 - 16*x^4) - 432*x^2 + 216*x^3 - 48*x^4 + 4*x^5),x)

[Out]

(3*exp(-1/(exp(6*x) - 6*exp(3*x) - 6*x - log(x) + 2*x*exp(3*x) + x^2 + 9)))/4

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sympy [A] time = 9.46, size = 32, normalized size = 1.19 \begin {gather*} \frac {3 e^{\frac {1}{- x^{2} + 6 x + \left (6 - 2 x\right ) e^{3 x} - e^{6 x} + \log {\relax (x )} - 9}}}{4} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((18*x*exp(3*x)**2+(18*x**2-48*x)*exp(3*x)+6*x**2-18*x-3)*exp(1/(ln(x)-exp(3*x)**2+(6-2*x)*exp(3*x)-x

**2+6*x-9))/(4*x*ln(x)**2+(-8*x*exp(3*x)**2+(-16*x**2+48*x)*exp(3*x)-8*x**3+48*x**2-72*x)*ln(x)+4*x*exp(3*x)**

4+(16*x**2-48*x)*exp(3*x)**3+(24*x**3-144*x**2+216*x)*exp(3*x)**2+(16*x**4-144*x**3+432*x**2-432*x)*exp(3*x)+4

*x**5-48*x**4+216*x**3-432*x**2+324*x),x)

[Out]

3*exp(1/(-x**2 + 6*x + (6 - 2*x)*exp(3*x) - exp(6*x) + log(x) - 9))/4

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