3.58.74 \(\int \frac {e^3 (x^2-2 x^3+x^4) \log ^3(x)+e^{\frac {1}{x^2 \log ^2(x)}} (-6+6 x+(-8+8 x) \log (x)+(-2+2 x) \log ^2(x)+(x^2+2 x^3) \log ^3(x)+x^3 \log ^4(x))}{(x^3-2 x^4+x^5) \log ^3(x)} \, dx\)

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

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Rubi [B]  time = 2.82, antiderivative size = 76, normalized size of antiderivative = 2.81, number of steps used = 5, number of rules used = 4, integrand size = 96, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.042, Rules used = {1594, 27, 6742, 2288} \begin {gather*} \frac {e^{\frac {1}{x^2 \log ^2(x)}} \left (-3 x-x \log ^2(x)+\log ^2(x)-4 x \log (x)+4 \log (x)+3\right )}{(1-x)^2 x^3 \left (\frac {1}{x^3 \log ^3(x)}+\frac {1}{x^3 \log ^2(x)}\right ) \log ^3(x)}+e^3 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^3*(x^2 - 2*x^3 + x^4)*Log[x]^3 + E^(1/(x^2*Log[x]^2))*(-6 + 6*x + (-8 + 8*x)*Log[x] + (-2 + 2*x)*Log[x]
^2 + (x^2 + 2*x^3)*Log[x]^3 + x^3*Log[x]^4))/((x^3 - 2*x^4 + x^5)*Log[x]^3),x]

[Out]

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

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 1594

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^
(q - p) + c*x^(r - p))^n, x] /; FreeQ[{a, b, c, p, q, r}, x] && IntegerQ[n] && PosQ[q - p] && PosQ[r - p]

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

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Mathematica [A]  time = 0.18, size = 42, normalized size = 1.56 \begin {gather*} \frac {-3 e^{\frac {1}{x^2 \log ^2(x)}}-e^{\frac {1}{x^2 \log ^2(x)}} \log (x)+e^3 (-1+x) \log (x)}{-1+x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^3*(x^2 - 2*x^3 + x^4)*Log[x]^3 + E^(1/(x^2*Log[x]^2))*(-6 + 6*x + (-8 + 8*x)*Log[x] + (-2 + 2*x)*
Log[x]^2 + (x^2 + 2*x^3)*Log[x]^3 + x^3*Log[x]^4))/((x^3 - 2*x^4 + x^5)*Log[x]^3),x]

[Out]

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

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fricas [A]  time = 0.79, size = 30, normalized size = 1.11 \begin {gather*} \frac {{\left (x - 1\right )} e^{3} \log \relax (x) - {\left (\log \relax (x) + 3\right )} e^{\left (\frac {1}{x^{2} \log \relax (x)^{2}}\right )}}{x - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^3*log(x)^4+(2*x^3+x^2)*log(x)^3+(2*x-2)*log(x)^2+(8*x-8)*log(x)+6*x-6)*exp(1/x^2/log(x)^2)+(x^4-
2*x^3+x^2)*exp(3)*log(x)^3)/(x^5-2*x^4+x^3)/log(x)^3,x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \mathit {undef} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^3*log(x)^4+(2*x^3+x^2)*log(x)^3+(2*x-2)*log(x)^2+(8*x-8)*log(x)+6*x-6)*exp(1/x^2/log(x)^2)+(x^4-
2*x^3+x^2)*exp(3)*log(x)^3)/(x^5-2*x^4+x^3)/log(x)^3,x, algorithm="giac")

[Out]

undef

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maple [A]  time = 0.08, size = 27, normalized size = 1.00

method	result	size
risch	\(\ln \relax (x ) {\mathrm e}^{3}-\frac {\left (3+\ln \relax (x )\right ) {\mathrm e}^{\frac {1}{x^{2} \ln \relax (x )^{2}}}}{x -1}\)	\(27\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x^3*ln(x)^4+(2*x^3+x^2)*ln(x)^3+(2*x-2)*ln(x)^2+(8*x-8)*ln(x)+6*x-6)*exp(1/x^2/ln(x)^2)+(x^4-2*x^3+x^2)*
exp(3)*ln(x)^3)/(x^5-2*x^4+x^3)/ln(x)^3,x,method=_RETURNVERBOSE)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} e^{3} \log \relax (x) + \int \frac {{\left (x^{3} \log \relax (x)^{4} + {\left (2 \, x^{3} + x^{2}\right )} \log \relax (x)^{3} + 2 \, {\left (x - 1\right )} \log \relax (x)^{2} + 8 \, {\left (x - 1\right )} \log \relax (x) + 6 \, x - 6\right )} e^{\left (\frac {1}{x^{2} \log \relax (x)^{2}}\right )}}{{\left (x^{5} - 2 \, x^{4} + x^{3}\right )} \log \relax (x)^{3}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^3*log(x)^4+(2*x^3+x^2)*log(x)^3+(2*x-2)*log(x)^2+(8*x-8)*log(x)+6*x-6)*exp(1/x^2/log(x)^2)+(x^4-
2*x^3+x^2)*exp(3)*log(x)^3)/(x^5-2*x^4+x^3)/log(x)^3,x, algorithm="maxima")

[Out]

e^3*log(x) + integrate((x^3*log(x)^4 + (2*x^3 + x^2)*log(x)^3 + 2*(x - 1)*log(x)^2 + 8*(x - 1)*log(x) + 6*x -
6)*e^(1/(x^2*log(x)^2))/((x^5 - 2*x^4 + x^3)*log(x)^3), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int \frac {{\mathrm {e}}^{\frac {1}{x^2\,{\ln \relax (x)}^2}}\,\left (6\,x+{\ln \relax (x)}^3\,\left (2\,x^3+x^2\right )+\ln \relax (x)\,\left (8\,x-8\right )+x^3\,{\ln \relax (x)}^4+{\ln \relax (x)}^2\,\left (2\,x-2\right )-6\right )+{\mathrm {e}}^3\,{\ln \relax (x)}^3\,\left (x^4-2\,x^3+x^2\right )}{{\ln \relax (x)}^3\,\left (x^5-2\,x^4+x^3\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(1/(x^2*log(x)^2))*(6*x + log(x)^3*(x^2 + 2*x^3) + log(x)*(8*x - 8) + x^3*log(x)^4 + log(x)^2*(2*x - 2
) - 6) + exp(3)*log(x)^3*(x^2 - 2*x^3 + x^4))/(log(x)^3*(x^3 - 2*x^4 + x^5)),x)

[Out]

int((exp(1/(x^2*log(x)^2))*(6*x + log(x)^3*(x^2 + 2*x^3) + log(x)*(8*x - 8) + x^3*log(x)^4 + log(x)^2*(2*x - 2
) - 6) + exp(3)*log(x)^3*(x^2 - 2*x^3 + x^4))/(log(x)^3*(x^3 - 2*x^4 + x^5)), x)

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sympy [A]  time = 0.36, size = 27, normalized size = 1.00 \begin {gather*} e^{3} \log {\relax (x )} + \frac {\left (- \log {\relax (x )} - 3\right ) e^{\frac {1}{x^{2} \log {\relax (x )}^{2}}}}{x - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x**3*ln(x)**4+(2*x**3+x**2)*ln(x)**3+(2*x-2)*ln(x)**2+(8*x-8)*ln(x)+6*x-6)*exp(1/x**2/ln(x)**2)+(x
**4-2*x**3+x**2)*exp(3)*ln(x)**3)/(x**5-2*x**4+x**3)/ln(x)**3,x)

[Out]

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

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