3.84.66 \(\int \frac {-3 e^{\frac {1}{9} (45-18 x^2+9 e^3 x^2-5 x^3)}-3 x+(3 x+e^{\frac {1}{9} (45-18 x^2+9 e^3 x^2-5 x^3)} (-12 x^2+6 e^3 x^2-5 x^3)) \log (x)}{3 x \log ^2(x)} \, dx\)

Optimal. Leaf size=32 \[ \frac {e^{5+x \left (x-\left (3-e^3\right ) x-\frac {5 x^2}{9}\right )}+x}{\log (x)} \]

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Rubi [B]  time = 1.55, antiderivative size = 78, normalized size of antiderivative = 2.44, number of steps used = 9, number of rules used = 6, integrand size = 96, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {12, 6742, 2360, 2297, 2298, 2288} \begin {gather*} \frac {e^{-\frac {5 x^3}{9}-\left (2-e^3\right ) x^2+5} \left (5 x^3 \log (x)+6 \left (2-e^3\right ) x^2 \log (x)\right )}{\left (5 x^2+6 \left (2-e^3\right ) x\right ) x \log ^2(x)}+\frac {x}{\log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-3*E^((45 - 18*x^2 + 9*E^3*x^2 - 5*x^3)/9) - 3*x + (3*x + E^((45 - 18*x^2 + 9*E^3*x^2 - 5*x^3)/9)*(-12*x^
2 + 6*E^3*x^2 - 5*x^3))*Log[x])/(3*x*Log[x]^2),x]

[Out]

x/Log[x] + (E^(5 - (2 - E^3)*x^2 - (5*x^3)/9)*(6*(2 - E^3)*x^2*Log[x] + 5*x^3*Log[x]))/(x*(6*(2 - E^3)*x + 5*x
^2)*Log[x]^2)

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 2297

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_), x_Symbol] :> Simp[(x*(a + b*Log[c*x^n])^(p + 1))/(b*n*(p + 1))
, x] - Dist[1/(b*n*(p + 1)), Int[(a + b*Log[c*x^n])^(p + 1), x], x] /; FreeQ[{a, b, c, n}, x] && LtQ[p, -1] &&
IntegerQ[2*p]

Rule 2298

Int[Log[(c_.)*(x_)]^(-1), x_Symbol] :> Simp[LogIntegral[c*x]/c, x] /; FreeQ[c, x]

Rule 2360

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(Log[(c_.)*(x_)^(n_.)]*(e_.) + (d_))^(q_.), x_Symbol] :> Int[E
xpandIntegrand[(a + b*Log[c*x^n])^p*(d + e*Log[c*x^n])^q, x], x] /; FreeQ[{a, b, c, d, e, n}, x] && IntegerQ[p
] && IntegerQ[q]

Rule 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{3} \int \frac {-3 e^{\frac {1}{9} \left (45-18 x^2+9 e^3 x^2-5 x^3\right )}-3 x+\left (3 x+e^{\frac {1}{9} \left (45-18 x^2+9 e^3 x^2-5 x^3\right )} \left (-12 x^2+6 e^3 x^2-5 x^3\right )\right ) \log (x)}{x \log ^2(x)} \, dx\\ &=\frac {1}{3} \int \left (\frac {3 (-1+\log (x))}{\log ^2(x)}+\frac {e^{5-\left (2-e^3\right ) x^2-\frac {5 x^3}{9}} \left (-3-12 \left (1-\frac {e^3}{2}\right ) x^2 \log (x)-5 x^3 \log (x)\right )}{x \log ^2(x)}\right ) \, dx\\ &=\frac {1}{3} \int \frac {e^{5-\left (2-e^3\right ) x^2-\frac {5 x^3}{9}} \left (-3-12 \left (1-\frac {e^3}{2}\right ) x^2 \log (x)-5 x^3 \log (x)\right )}{x \log ^2(x)} \, dx+\int \frac {-1+\log (x)}{\log ^2(x)} \, dx\\ &=\frac {e^{5-\left (2-e^3\right ) x^2-\frac {5 x^3}{9}} \left (6 \left (2-e^3\right ) x^2 \log (x)+5 x^3 \log (x)\right )}{x \left (6 \left (2-e^3\right ) x+5 x^2\right ) \log ^2(x)}+\int \left (-\frac {1}{\log ^2(x)}+\frac {1}{\log (x)}\right ) \, dx\\ &=\frac {e^{5-\left (2-e^3\right ) x^2-\frac {5 x^3}{9}} \left (6 \left (2-e^3\right ) x^2 \log (x)+5 x^3 \log (x)\right )}{x \left (6 \left (2-e^3\right ) x+5 x^2\right ) \log ^2(x)}-\int \frac {1}{\log ^2(x)} \, dx+\int \frac {1}{\log (x)} \, dx\\ &=\frac {x}{\log (x)}+\frac {e^{5-\left (2-e^3\right ) x^2-\frac {5 x^3}{9}} \left (6 \left (2-e^3\right ) x^2 \log (x)+5 x^3 \log (x)\right )}{x \left (6 \left (2-e^3\right ) x+5 x^2\right ) \log ^2(x)}+\text {li}(x)-\int \frac {1}{\log (x)} \, dx\\ &=\frac {x}{\log (x)}+\frac {e^{5-\left (2-e^3\right ) x^2-\frac {5 x^3}{9}} \left (6 \left (2-e^3\right ) x^2 \log (x)+5 x^3 \log (x)\right )}{x \left (6 \left (2-e^3\right ) x+5 x^2\right ) \log ^2(x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.54, size = 27, normalized size = 0.84 \begin {gather*} \frac {e^{5+\left (-2+e^3\right ) x^2-\frac {5 x^3}{9}}+x}{\log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-3*E^((45 - 18*x^2 + 9*E^3*x^2 - 5*x^3)/9) - 3*x + (3*x + E^((45 - 18*x^2 + 9*E^3*x^2 - 5*x^3)/9)*(
-12*x^2 + 6*E^3*x^2 - 5*x^3))*Log[x])/(3*x*Log[x]^2),x]

[Out]

(E^(5 + (-2 + E^3)*x^2 - (5*x^3)/9) + x)/Log[x]

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fricas [A]  time = 0.57, size = 26, normalized size = 0.81 \begin {gather*} \frac {x + e^{\left (-\frac {5}{9} \, x^{3} + x^{2} e^{3} - 2 \, x^{2} + 5\right )}}{\log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(((6*x^2*exp(3)-5*x^3-12*x^2)*exp(x^2*exp(3)-5/9*x^3-2*x^2+5)+3*x)*log(x)-3*exp(x^2*exp(3)-5/9*x
^3-2*x^2+5)-3*x)/x/log(x)^2,x, algorithm="fricas")

[Out]

(x + e^(-5/9*x^3 + x^2*e^3 - 2*x^2 + 5))/log(x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {{\left ({\left (5 \, x^{3} - 6 \, x^{2} e^{3} + 12 \, x^{2}\right )} e^{\left (-\frac {5}{9} \, x^{3} + x^{2} e^{3} - 2 \, x^{2} + 5\right )} - 3 \, x\right )} \log \relax (x) + 3 \, x + 3 \, e^{\left (-\frac {5}{9} \, x^{3} + x^{2} e^{3} - 2 \, x^{2} + 5\right )}}{3 \, x \log \relax (x)^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(((6*x^2*exp(3)-5*x^3-12*x^2)*exp(x^2*exp(3)-5/9*x^3-2*x^2+5)+3*x)*log(x)-3*exp(x^2*exp(3)-5/9*x
^3-2*x^2+5)-3*x)/x/log(x)^2,x, algorithm="giac")

[Out]

integrate(-1/3*(((5*x^3 - 6*x^2*e^3 + 12*x^2)*e^(-5/9*x^3 + x^2*e^3 - 2*x^2 + 5) - 3*x)*log(x) + 3*x + 3*e^(-5
/9*x^3 + x^2*e^3 - 2*x^2 + 5))/(x*log(x)^2), x)

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maple [A]  time = 0.08, size = 27, normalized size = 0.84

method	result	size
risch	\(\frac {{\mathrm e}^{x^{2} {\mathrm e}^{3}-\frac {5 x^{3}}{9}-2 x^{2}+5}+x}{\ln \relax (x )}\)	\(27\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/3*(((6*x^2*exp(3)-5*x^3-12*x^2)*exp(x^2*exp(3)-5/9*x^3-2*x^2+5)+3*x)*ln(x)-3*exp(x^2*exp(3)-5/9*x^3-2*x^
2+5)-3*x)/x/ln(x)^2,x,method=_RETURNVERBOSE)

[Out]

(exp(x^2*exp(3)-5/9*x^3-2*x^2+5)+x)/ln(x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {e^{\left (-\frac {5}{9} \, x^{3} + x^{2} e^{3} - 2 \, x^{2} + 5\right )}}{\log \relax (x)} - \Gamma \left (-1, -\log \relax (x)\right ) + \int \frac {1}{\log \relax (x)}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(((6*x^2*exp(3)-5*x^3-12*x^2)*exp(x^2*exp(3)-5/9*x^3-2*x^2+5)+3*x)*log(x)-3*exp(x^2*exp(3)-5/9*x
^3-2*x^2+5)-3*x)/x/log(x)^2,x, algorithm="maxima")

[Out]

e^(-5/9*x^3 + x^2*e^3 - 2*x^2 + 5)/log(x) - gamma(-1, -log(x)) + integrate(1/log(x), x)

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mupad [B]  time = 5.74, size = 31, normalized size = 0.97 \begin {gather*} \frac {x}{\ln \relax (x)}+\frac {{\mathrm {e}}^{x^2\,{\mathrm {e}}^3-2\,x^2-\frac {5\,x^3}{9}+5}}{\ln \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(x + exp(x^2*exp(3) - 2*x^2 - (5*x^3)/9 + 5) - (log(x)*(3*x - exp(x^2*exp(3) - 2*x^2 - (5*x^3)/9 + 5)*(12
*x^2 - 6*x^2*exp(3) + 5*x^3)))/3)/(x*log(x)^2),x)

[Out]

x/log(x) + exp(x^2*exp(3) - 2*x^2 - (5*x^3)/9 + 5)/log(x)

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sympy [A]  time = 0.34, size = 29, normalized size = 0.91 \begin {gather*} \frac {x}{\log {\relax (x )}} + \frac {e^{- \frac {5 x^{3}}{9} - 2 x^{2} + x^{2} e^{3} + 5}}{\log {\relax (x )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(((6*x**2*exp(3)-5*x**3-12*x**2)*exp(x**2*exp(3)-5/9*x**3-2*x**2+5)+3*x)*ln(x)-3*exp(x**2*exp(3)
-5/9*x**3-2*x**2+5)-3*x)/x/ln(x)**2,x)

[Out]

x/log(x) + exp(-5*x**3/9 - 2*x**2 + x**2*exp(3) + 5)/log(x)

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