3.46.27 \(\int \frac {e^{\frac {2+x^3+x \log (\frac {5 \log (\frac {1}{x})-5 \log (x)}{x})}{x}} (2 x+(2+x-2 x^3) \log (\frac {1}{x})+(-2-x+2 x^3) \log (x))}{-x^2 \log (\frac {1}{x})+x^2 \log (x)} \, dx\) [4527]

Optimal. Leaf size=25 \[ \frac {5 e^{\frac {2}{x}+x^2} \left (\log \left (\frac {1}{x}\right )-\log (x)\right )}{x} \]

[Out]

exp(2/x+ln(5*(ln(1/x)-ln(x))/x)+x^2)

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Rubi [B] Leaf count is larger than twice the leaf count of optimal. \(71\) vs. \(2(25)=50\).
time = 1.21, antiderivative size = 71, normalized size of antiderivative = 2.84, number of steps used = 9, number of rules used = 5, integrand size = 78, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.064, Rules used = {6820, 12, 6874, 2326, 2634} \begin {gather*} \frac {5 e^{x^2+\frac {2}{x}} \left (1-x^3\right ) \log \left (\frac {1}{x}\right )}{\left (\frac {1}{x^2}-x\right ) x^3}-\frac {5 e^{x^2+\frac {2}{x}} \left (1-x^3\right ) \log (x)}{\left (\frac {1}{x^2}-x\right ) x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(5*E^(2/x + x^2)*(1 - x^3)*Log[x^(-1)])/((x^(-2) - x)*x^3) - (5*E^(2/x + x^2)*(1 - x^3)*Log[x])/((x^(-2) - x)*
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 2326

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 2634

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 6820

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

Rule 6874

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {5 e^{\frac {2}{x}+x^2} \left (-2 x+\left (-2-x+2 x^3\right ) \log \left (\frac {1}{x}\right )+\left (2+x-2 x^3\right ) \log (x)\right )}{x^3} \, dx\\ &=5 \int \frac {e^{\frac {2}{x}+x^2} \left (-2 x+\left (-2-x+2 x^3\right ) \log \left (\frac {1}{x}\right )+\left (2+x-2 x^3\right ) \log (x)\right )}{x^3} \, dx\\ &=5 \int \left (\frac {e^{\frac {2}{x}+x^2} \left (-2 x-2 \log \left (\frac {1}{x}\right )-x \log \left (\frac {1}{x}\right )+2 x^3 \log \left (\frac {1}{x}\right )\right )}{x^3}-\frac {e^{\frac {2}{x}+x^2} \left (-2-x+2 x^3\right ) \log (x)}{x^3}\right ) \, dx\\ &=5 \int \frac {e^{\frac {2}{x}+x^2} \left (-2 x-2 \log \left (\frac {1}{x}\right )-x \log \left (\frac {1}{x}\right )+2 x^3 \log \left (\frac {1}{x}\right )\right )}{x^3} \, dx-5 \int \frac {e^{\frac {2}{x}+x^2} \left (-2-x+2 x^3\right ) \log (x)}{x^3} \, dx\\ &=-\frac {5 e^{\frac {2}{x}+x^2} \left (1-x^3\right ) \log (x)}{\left (\frac {1}{x^2}-x\right ) x^3}+5 \int \frac {e^{\frac {2}{x}+x^2}}{x^2} \, dx+5 \int \frac {e^{\frac {2}{x}+x^2} \left (-2 x+\left (-2-x+2 x^3\right ) \log \left (\frac {1}{x}\right )\right )}{x^3} \, dx\\ &=-\frac {5 e^{\frac {2}{x}+x^2} \left (1-x^3\right ) \log (x)}{\left (\frac {1}{x^2}-x\right ) x^3}+5 \int \frac {e^{\frac {2}{x}+x^2}}{x^2} \, dx+5 \int \left (-\frac {2 e^{\frac {2}{x}+x^2}}{x^2}+\frac {e^{\frac {2}{x}+x^2} \left (-2-x+2 x^3\right ) \log \left (\frac {1}{x}\right )}{x^3}\right ) \, dx\\ &=-\frac {5 e^{\frac {2}{x}+x^2} \left (1-x^3\right ) \log (x)}{\left (\frac {1}{x^2}-x\right ) x^3}+5 \int \frac {e^{\frac {2}{x}+x^2}}{x^2} \, dx+5 \int \frac {e^{\frac {2}{x}+x^2} \left (-2-x+2 x^3\right ) \log \left (\frac {1}{x}\right )}{x^3} \, dx-10 \int \frac {e^{\frac {2}{x}+x^2}}{x^2} \, dx\\ &=\frac {5 e^{\frac {2}{x}+x^2} \left (1-x^3\right ) \log \left (\frac {1}{x}\right )}{\left (\frac {1}{x^2}-x\right ) x^3}-\frac {5 e^{\frac {2}{x}+x^2} \left (1-x^3\right ) \log (x)}{\left (\frac {1}{x^2}-x\right ) x^3}+2 \left (5 \int \frac {e^{\frac {2}{x}+x^2}}{x^2} \, dx\right )-10 \int \frac {e^{\frac {2}{x}+x^2}}{x^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]
time = 0.07, size = 25, normalized size = 1.00 \begin {gather*} \frac {5 e^{\frac {2}{x}+x^2} \left (\log \left (\frac {1}{x}\right )-\log (x)\right )}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 3.86, size = 141, normalized size = 5.64

method	result	size
risch	\(-{\mathrm e}^{\frac {i x \pi \mathrm {csgn}\left (\frac {i \ln \left (x \right )}{x}\right )^{3}+i x \pi \mathrm {csgn}\left (\frac {i \ln \left (x \right )}{x}\right )^{2} \mathrm {csgn}\left (\frac {i}{x}\right )+i x \pi \mathrm {csgn}\left (\frac {i \ln \left (x \right )}{x}\right )^{2} \mathrm {csgn}\left (i \ln \left (x \right )\right )-i x \pi \,\mathrm {csgn}\left (\frac {i \ln \left (x \right )}{x}\right ) \mathrm {csgn}\left (\frac {i}{x}\right ) \mathrm {csgn}\left (i \ln \left (x \right )\right )-2 i x \pi \mathrm {csgn}\left (\frac {i \ln \left (x \right )}{x}\right )^{2}+2 x^{3}-2 x \ln \left (x \right )+2 x \ln \left (5\right )+2 x \ln \left (\ln \left (x \right )\right )+2 x \ln \left (2\right )+4}{2 x}}\)	\(141\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-exp(1/2*(I*x*Pi*csgn(I/x*ln(x))^3+I*x*Pi*csgn(I/x*ln(x))^2*csgn(I/x)+I*x*Pi*csgn(I/x*ln(x))^2*csgn(I*ln(x))-I
*x*Pi*csgn(I/x*ln(x))*csgn(I/x)*csgn(I*ln(x))-2*I*x*Pi*csgn(I/x*ln(x))^2+2*x^3-2*x*ln(x)+2*x*ln(5)+2*x*ln(ln(x
))+2*x*ln(2)+4)/x)

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Maxima [A]
time = 0.33, size = 17, normalized size = 0.68 \begin {gather*} -\frac {10 \, e^{\left (x^{2} + \frac {2}{x}\right )} \log \left (x\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-10*e^(x^2 + 2/x)*log(x)/x

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Fricas [A]
time = 0.34, size = 22, normalized size = 0.88 \begin {gather*} e^{\left (\frac {x^{3} + x \log \left (\frac {10 \, \log \left (\frac {1}{x}\right )}{x}\right ) + 2}{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e^((x^3 + x*log(10*log(1/x)/x) + 2)/x)

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Sympy [A]
time = 0.25, size = 19, normalized size = 0.76 \begin {gather*} e^{\frac {x^{3} + x \log {\left (- \frac {10 \log {\left (x \right )}}{x} \right )} + 2}{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

exp((x**3 + x*log(-10*log(x)/x) + 2)/x)

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Giac [A]
time = 0.40, size = 18, normalized size = 0.72 \begin {gather*} e^{\left (x^{2} + \frac {2}{x} + \log \left (-\frac {10 \, \log \left (x\right )}{x}\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e^(x^2 + 2/x + log(-10*log(x)/x))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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