3.74.36 \(\int \frac {e^4 (4 x^4-4 x^5+x^6)+e^{\frac {2 x^2}{e^4 (-2+x)}} (-512 x^3+128 x^4+e^4 (-512+384 x^2-128 x^3))+(e^{\frac {2 x^2}{e^4 (-2+x)}} (-512 x^2+128 x^3)+e^4 (12 x^3-12 x^4+3 x^5)) \log (x)+e^4 (12 x^2-12 x^3+3 x^4) \log ^2(x)+e^4 (4 x-4 x^2+x^3) \log ^3(x)}{e^4 (4 x^4-4 x^5+x^6)+e^4 (12 x^3-12 x^4+3 x^5) \log (x)+e^4 (12 x^2-12 x^3+3 x^4) \log ^2(x)+e^4 (4 x-4 x^2+x^3) \log ^3(x)} \, dx\)

Optimal. Leaf size=26 \[ -3+x+\frac {64 e^{\frac {2 x^2}{e^4 (-2+x)}}}{(x+\log (x))^2} \]

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Rubi [B]  time = 4.56, antiderivative size = 90, normalized size of antiderivative = 3.46, number of steps used = 12, number of rules used = 4, integrand size = 245, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.016, Rules used = {6688, 12, 6742, 2288} \begin {gather*} \frac {64 e^{-\frac {2 x^2}{e^4 (2-x)}-4} \left ((4-x) x^3+(4-x) x^2 \log (x)\right )}{(2-x)^2 \left (\frac {x^2}{e^4 (2-x)^2}+\frac {2 x}{e^4 (2-x)}\right ) x (x+\log (x))^3}+x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^4*(4*x^4 - 4*x^5 + x^6) + E^((2*x^2)/(E^4*(-2 + x)))*(-512*x^3 + 128*x^4 + E^4*(-512 + 384*x^2 - 128*x^
3)) + (E^((2*x^2)/(E^4*(-2 + x)))*(-512*x^2 + 128*x^3) + E^4*(12*x^3 - 12*x^4 + 3*x^5))*Log[x] + E^4*(12*x^2 -
12*x^3 + 3*x^4)*Log[x]^2 + E^4*(4*x - 4*x^2 + x^3)*Log[x]^3)/(E^4*(4*x^4 - 4*x^5 + x^6) + E^4*(12*x^3 - 12*x^
4 + 3*x^5)*Log[x] + E^4*(12*x^2 - 12*x^3 + 3*x^4)*Log[x]^2 + E^4*(4*x - 4*x^2 + x^3)*Log[x]^3),x]

[Out]

x + (64*E^(-4 - (2*x^2)/(E^4*(2 - x)))*((4 - x)*x^3 + (4 - x)*x^2*Log[x]))/((2 - x)^2*x*((2*x)/(E^4*(2 - x)) +
x^2/(E^4*(2 - x)^2))*(x + Log[x])^3)

Rule 12

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

Rule 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, 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 \frac {128 e^{\frac {2 x^2}{e^4 (-2+x)}} (-4+x) x^3+e^4 (-2+x)^2 x^4-128 e^{4+\frac {2 x^2}{e^4 (-2+x)}} (-2+x)^2 (1+x)+x^2 \left (128 e^{\frac {2 x^2}{e^4 (-2+x)}} (-4+x)+3 e^4 (-2+x)^2 x\right ) \log (x)+3 e^4 (-2+x)^2 x^2 \log ^2(x)+e^4 (-2+x)^2 x \log ^3(x)}{e^4 (2-x)^2 x (x+\log (x))^3} \, dx\\ &=\frac {\int \frac {128 e^{\frac {2 x^2}{e^4 (-2+x)}} (-4+x) x^3+e^4 (-2+x)^2 x^4-128 e^{4+\frac {2 x^2}{e^4 (-2+x)}} (-2+x)^2 (1+x)+x^2 \left (128 e^{\frac {2 x^2}{e^4 (-2+x)}} (-4+x)+3 e^4 (-2+x)^2 x\right ) \log (x)+3 e^4 (-2+x)^2 x^2 \log ^2(x)+e^4 (-2+x)^2 x \log ^3(x)}{(2-x)^2 x (x+\log (x))^3} \, dx}{e^4}\\ &=\frac {\int \left (\frac {e^4 x^3}{(x+\log (x))^3}+\frac {3 e^4 x^2 \log (x)}{(x+\log (x))^3}+\frac {3 e^4 x \log ^2(x)}{(x+\log (x))^3}+\frac {e^4 \log ^3(x)}{(x+\log (x))^3}+\frac {128 e^{\frac {2 x^2}{e^4 (-2+x)}} \left (-4 e^4+3 e^4 x^2-4 \left (1+\frac {e^4}{4}\right ) x^3+x^4-4 x^2 \log (x)+x^3 \log (x)\right )}{(2-x)^2 x (x+\log (x))^3}\right ) \, dx}{e^4}\\ &=3 \int \frac {x^2 \log (x)}{(x+\log (x))^3} \, dx+3 \int \frac {x \log ^2(x)}{(x+\log (x))^3} \, dx+\frac {128 \int \frac {e^{\frac {2 x^2}{e^4 (-2+x)}} \left (-4 e^4+3 e^4 x^2-4 \left (1+\frac {e^4}{4}\right ) x^3+x^4-4 x^2 \log (x)+x^3 \log (x)\right )}{(2-x)^2 x (x+\log (x))^3} \, dx}{e^4}+\int \frac {x^3}{(x+\log (x))^3} \, dx+\int \frac {\log ^3(x)}{(x+\log (x))^3} \, dx\\ &=3 \int \left (-\frac {x^3}{(x+\log (x))^3}+\frac {x^2}{(x+\log (x))^2}\right ) \, dx+3 \int \left (\frac {x^3}{(x+\log (x))^3}-\frac {2 x^2}{(x+\log (x))^2}+\frac {x}{x+\log (x)}\right ) \, dx+\frac {128 \int \frac {e^{\frac {2 x^2}{e^4 (-2+x)}} \left ((-4+x) x^3-e^4 (-2+x)^2 (1+x)+(-4+x) x^2 \log (x)\right )}{(2-x)^2 x (x+\log (x))^3} \, dx}{e^4}+\int \frac {x^3}{(x+\log (x))^3} \, dx+\int \left (1-\frac {x^3}{(x+\log (x))^3}+\frac {3 x^2}{(x+\log (x))^2}-\frac {3 x}{x+\log (x)}\right ) \, dx\\ &=x+\frac {64 e^{-4-\frac {2 x^2}{e^4 (2-x)}} \left ((4-x) x^3+(4-x) x^2 \log (x)\right )}{(2-x)^2 x \left (\frac {2 x}{e^4 (2-x)}+\frac {x^2}{e^4 (2-x)^2}\right ) (x+\log (x))^3}+2 \left (3 \int \frac {x^2}{(x+\log (x))^2} \, dx\right )-6 \int \frac {x^2}{(x+\log (x))^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.24, size = 25, normalized size = 0.96 \begin {gather*} x+\frac {64 e^{\frac {2 x^2}{e^4 (-2+x)}}}{(x+\log (x))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^4*(4*x^4 - 4*x^5 + x^6) + E^((2*x^2)/(E^4*(-2 + x)))*(-512*x^3 + 128*x^4 + E^4*(-512 + 384*x^2 -
128*x^3)) + (E^((2*x^2)/(E^4*(-2 + x)))*(-512*x^2 + 128*x^3) + E^4*(12*x^3 - 12*x^4 + 3*x^5))*Log[x] + E^4*(12
*x^2 - 12*x^3 + 3*x^4)*Log[x]^2 + E^4*(4*x - 4*x^2 + x^3)*Log[x]^3)/(E^4*(4*x^4 - 4*x^5 + x^6) + E^4*(12*x^3 -
12*x^4 + 3*x^5)*Log[x] + E^4*(12*x^2 - 12*x^3 + 3*x^4)*Log[x]^2 + E^4*(4*x - 4*x^2 + x^3)*Log[x]^3),x]

[Out]

x + (64*E^((2*x^2)/(E^4*(-2 + x))))/(x + Log[x])^2

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fricas [A]  time = 0.63, size = 48, normalized size = 1.85 \begin {gather*} \frac {x^{3} + 2 \, x^{2} \log \relax (x) + x \log \relax (x)^{2} + 64 \, e^{\left (\frac {2 \, x^{2} e^{\left (-4\right )}}{x - 2}\right )}}{x^{2} + 2 \, x \log \relax (x) + \log \relax (x)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^3-4*x^2+4*x)*exp(2)^2*log(x)^3+(3*x^4-12*x^3+12*x^2)*exp(2)^2*log(x)^2+((128*x^3-512*x^2)*exp(x^
2/(x-2)/exp(2)^2)^2+(3*x^5-12*x^4+12*x^3)*exp(2)^2)*log(x)+((-128*x^3+384*x^2-512)*exp(2)^2+128*x^4-512*x^3)*e
xp(x^2/(x-2)/exp(2)^2)^2+(x^6-4*x^5+4*x^4)*exp(2)^2)/((x^3-4*x^2+4*x)*exp(2)^2*log(x)^3+(3*x^4-12*x^3+12*x^2)*
exp(2)^2*log(x)^2+(3*x^5-12*x^4+12*x^3)*exp(2)^2*log(x)+(x^6-4*x^5+4*x^4)*exp(2)^2),x, algorithm="fricas")

[Out]

(x^3 + 2*x^2*log(x) + x*log(x)^2 + 64*e^(2*x^2*e^(-4)/(x - 2)))/(x^2 + 2*x*log(x) + log(x)^2)

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giac [B]  time = 0.40, size = 52, normalized size = 2.00 \begin {gather*} \frac {x^{3} + 2 \, x^{2} \log \relax (x) + x \log \relax (x)^{2} + 64 \, e^{\left (\frac {2 \, x^{2}}{x e^{4} - 2 \, e^{4}}\right )}}{x^{2} + 2 \, x \log \relax (x) + \log \relax (x)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^3-4*x^2+4*x)*exp(2)^2*log(x)^3+(3*x^4-12*x^3+12*x^2)*exp(2)^2*log(x)^2+((128*x^3-512*x^2)*exp(x^
2/(x-2)/exp(2)^2)^2+(3*x^5-12*x^4+12*x^3)*exp(2)^2)*log(x)+((-128*x^3+384*x^2-512)*exp(2)^2+128*x^4-512*x^3)*e
xp(x^2/(x-2)/exp(2)^2)^2+(x^6-4*x^5+4*x^4)*exp(2)^2)/((x^3-4*x^2+4*x)*exp(2)^2*log(x)^3+(3*x^4-12*x^3+12*x^2)*
exp(2)^2*log(x)^2+(3*x^5-12*x^4+12*x^3)*exp(2)^2*log(x)+(x^6-4*x^5+4*x^4)*exp(2)^2),x, algorithm="giac")

[Out]

(x^3 + 2*x^2*log(x) + x*log(x)^2 + 64*e^(2*x^2/(x*e^4 - 2*e^4)))/(x^2 + 2*x*log(x) + log(x)^2)

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maple [A]  time = 0.09, size = 24, normalized size = 0.92

method	result	size
risch	\(x +\frac {64 \,{\mathrm e}^{\frac {2 x^{2} {\mathrm e}^{-4}}{x -2}}}{\left (x +\ln \relax (x )\right )^{2}}\)	\(24\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x^3-4*x^2+4*x)*exp(2)^2*ln(x)^3+(3*x^4-12*x^3+12*x^2)*exp(2)^2*ln(x)^2+((128*x^3-512*x^2)*exp(x^2/(x-2)/
exp(2)^2)^2+(3*x^5-12*x^4+12*x^3)*exp(2)^2)*ln(x)+((-128*x^3+384*x^2-512)*exp(2)^2+128*x^4-512*x^3)*exp(x^2/(x
-2)/exp(2)^2)^2+(x^6-4*x^5+4*x^4)*exp(2)^2)/((x^3-4*x^2+4*x)*exp(2)^2*ln(x)^3+(3*x^4-12*x^3+12*x^2)*exp(2)^2*l
n(x)^2+(3*x^5-12*x^4+12*x^3)*exp(2)^2*ln(x)+(x^6-4*x^5+4*x^4)*exp(2)^2),x,method=_RETURNVERBOSE)

[Out]

x+64*exp(2*x^2/(x-2)*exp(-4))/(x+ln(x))^2

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maxima [B]  time = 0.43, size = 59, normalized size = 2.27 \begin {gather*} \frac {x^{3} + 2 \, x^{2} \log \relax (x) + x \log \relax (x)^{2} + 64 \, e^{\left (2 \, x e^{\left (-4\right )} + \frac {8}{x e^{4} - 2 \, e^{4}} + 4 \, e^{\left (-4\right )}\right )}}{x^{2} + 2 \, x \log \relax (x) + \log \relax (x)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^3-4*x^2+4*x)*exp(2)^2*log(x)^3+(3*x^4-12*x^3+12*x^2)*exp(2)^2*log(x)^2+((128*x^3-512*x^2)*exp(x^
2/(x-2)/exp(2)^2)^2+(3*x^5-12*x^4+12*x^3)*exp(2)^2)*log(x)+((-128*x^3+384*x^2-512)*exp(2)^2+128*x^4-512*x^3)*e
xp(x^2/(x-2)/exp(2)^2)^2+(x^6-4*x^5+4*x^4)*exp(2)^2)/((x^3-4*x^2+4*x)*exp(2)^2*log(x)^3+(3*x^4-12*x^3+12*x^2)*
exp(2)^2*log(x)^2+(3*x^5-12*x^4+12*x^3)*exp(2)^2*log(x)+(x^6-4*x^5+4*x^4)*exp(2)^2),x, algorithm="maxima")

[Out]

(x^3 + 2*x^2*log(x) + x*log(x)^2 + 64*e^(2*x*e^(-4) + 8/(x*e^4 - 2*e^4) + 4*e^(-4)))/(x^2 + 2*x*log(x) + log(x
)^2)

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mupad [B]  time = 5.93, size = 23, normalized size = 0.88 \begin {gather*} x+\frac {64\,{\mathrm {e}}^{\frac {2\,x^2\,{\mathrm {e}}^{-4}}{x-2}}}{{\left (x+\ln \relax (x)\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(4)*(4*x^4 - 4*x^5 + x^6) - log(x)*(exp((2*x^2*exp(-4))/(x - 2))*(512*x^2 - 128*x^3) - exp(4)*(12*x^3
- 12*x^4 + 3*x^5)) - exp((2*x^2*exp(-4))/(x - 2))*(exp(4)*(128*x^3 - 384*x^2 + 512) + 512*x^3 - 128*x^4) + exp
(4)*log(x)^2*(12*x^2 - 12*x^3 + 3*x^4) + exp(4)*log(x)^3*(4*x - 4*x^2 + x^3))/(exp(4)*(4*x^4 - 4*x^5 + x^6) +
exp(4)*log(x)^2*(12*x^2 - 12*x^3 + 3*x^4) + exp(4)*log(x)^3*(4*x - 4*x^2 + x^3) + exp(4)*log(x)*(12*x^3 - 12*x
^4 + 3*x^5)),x)

[Out]

x + (64*exp((2*x^2*exp(-4))/(x - 2)))/(x + log(x))^2

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sympy [A]  time = 0.75, size = 31, normalized size = 1.19 \begin {gather*} x + \frac {64 e^{\frac {2 x^{2}}{\left (x - 2\right ) e^{4}}}}{x^{2} + 2 x \log {\relax (x )} + \log {\relax (x )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x**3-4*x**2+4*x)*exp(2)**2*ln(x)**3+(3*x**4-12*x**3+12*x**2)*exp(2)**2*ln(x)**2+((128*x**3-512*x**
2)*exp(x**2/(x-2)/exp(2)**2)**2+(3*x**5-12*x**4+12*x**3)*exp(2)**2)*ln(x)+((-128*x**3+384*x**2-512)*exp(2)**2+
128*x**4-512*x**3)*exp(x**2/(x-2)/exp(2)**2)**2+(x**6-4*x**5+4*x**4)*exp(2)**2)/((x**3-4*x**2+4*x)*exp(2)**2*l
n(x)**3+(3*x**4-12*x**3+12*x**2)*exp(2)**2*ln(x)**2+(3*x**5-12*x**4+12*x**3)*exp(2)**2*ln(x)+(x**6-4*x**5+4*x*
*4)*exp(2)**2),x)

[Out]

x + 64*exp(2*x**2*exp(-4)/(x - 2))/(x**2 + 2*x*log(x) + log(x)**2)

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