3.64.41 \(\int \frac {150+50 \log (x^3)+(x^3-e^2 x^3-2 x^4+x^3 \log (2)) \log ^3(x^3)}{x^3 \log ^3(x^3)} \, dx\)

Optimal. Leaf size=29 \[ 2+(1+x) \left (2-e^2-x+\log (2)\right )-\frac {25}{x^2 \log ^2\left (x^3\right )} \]

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Rubi [A]  time = 0.24, antiderivative size = 28, normalized size of antiderivative = 0.97, number of steps used = 9, number of rules used = 4, integrand size = 48, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {6688, 2306, 2310, 2178} \begin {gather*} -x^2-\frac {25}{x^2 \log ^2\left (x^3\right )}+x \left (1-e^2+\log (2)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(150 + 50*Log[x^3] + (x^3 - E^2*x^3 - 2*x^4 + x^3*Log[2])*Log[x^3]^3)/(x^3*Log[x^3]^3),x]

[Out]

-x^2 + x*(1 - E^2 + Log[2]) - 25/(x^2*Log[x^3]^2)

Rule 2178

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - (c*f)/d))*ExpIntegral
Ei[(f*g*(c + d*x)*Log[F])/d])/d, x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !$UseGamma === True

Rule 2306

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

Rule 2310

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_)*((d_.)*(x_))^(m_.), x_Symbol] :> Dist[(d*x)^(m + 1)/(d*n*(c*x^n
)^((m + 1)/n)), Subst[Int[E^(((m + 1)*x)/n)*(a + b*x)^p, x], x, Log[c*x^n]], x] /; FreeQ[{a, b, c, d, m, n, p}
, x]

Rule 6688

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (1-e^2-2 x+\log (2)+\frac {150}{x^3 \log ^3\left (x^3\right )}+\frac {50}{x^3 \log ^2\left (x^3\right )}\right ) \, dx\\ &=-x^2+x \left (1-e^2+\log (2)\right )+50 \int \frac {1}{x^3 \log ^2\left (x^3\right )} \, dx+150 \int \frac {1}{x^3 \log ^3\left (x^3\right )} \, dx\\ &=-x^2+x \left (1-e^2+\log (2)\right )-\frac {25}{x^2 \log ^2\left (x^3\right )}-\frac {50}{3 x^2 \log \left (x^3\right )}-\frac {100}{3} \int \frac {1}{x^3 \log \left (x^3\right )} \, dx-50 \int \frac {1}{x^3 \log ^2\left (x^3\right )} \, dx\\ &=-x^2+x \left (1-e^2+\log (2)\right )-\frac {25}{x^2 \log ^2\left (x^3\right )}+\frac {100}{3} \int \frac {1}{x^3 \log \left (x^3\right )} \, dx-\frac {\left (100 \left (x^3\right )^{2/3}\right ) \operatorname {Subst}\left (\int \frac {e^{-2 x/3}}{x} \, dx,x,\log \left (x^3\right )\right )}{9 x^2}\\ &=-x^2-\frac {100 \left (x^3\right )^{2/3} \text {Ei}\left (-\frac {2}{3} \log \left (x^3\right )\right )}{9 x^2}+x \left (1-e^2+\log (2)\right )-\frac {25}{x^2 \log ^2\left (x^3\right )}+\frac {\left (100 \left (x^3\right )^{2/3}\right ) \operatorname {Subst}\left (\int \frac {e^{-2 x/3}}{x} \, dx,x,\log \left (x^3\right )\right )}{9 x^2}\\ &=-x^2+x \left (1-e^2+\log (2)\right )-\frac {25}{x^2 \log ^2\left (x^3\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.05, size = 28, normalized size = 0.97 \begin {gather*} x-e^2 x-x^2+x \log (2)-\frac {25}{x^2 \log ^2\left (x^3\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(150 + 50*Log[x^3] + (x^3 - E^2*x^3 - 2*x^4 + x^3*Log[2])*Log[x^3]^3)/(x^3*Log[x^3]^3),x]

[Out]

x - E^2*x - x^2 + x*Log[2] - 25/(x^2*Log[x^3]^2)

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fricas [A]  time = 0.81, size = 42, normalized size = 1.45 \begin {gather*} -\frac {{\left (x^{4} + x^{3} e^{2} - x^{3} \log \relax (2) - x^{3}\right )} \log \left (x^{3}\right )^{2} + 25}{x^{2} \log \left (x^{3}\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^3*log(2)-x^3*exp(2)-2*x^4+x^3)*log(x^3)^3+50*log(x^3)+150)/x^3/log(x^3)^3,x, algorithm="fricas")

[Out]

-((x^4 + x^3*e^2 - x^3*log(2) - x^3)*log(x^3)^2 + 25)/(x^2*log(x^3)^2)

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giac [B]  time = 0.18, size = 59, normalized size = 2.03 \begin {gather*} -\frac {x^{4} \log \left (x^{3}\right )^{2} + x^{3} e^{2} \log \left (x^{3}\right )^{2} - x^{3} \log \relax (2) \log \left (x^{3}\right )^{2} - x^{3} \log \left (x^{3}\right )^{2} + 25}{x^{2} \log \left (x^{3}\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^3*log(2)-x^3*exp(2)-2*x^4+x^3)*log(x^3)^3+50*log(x^3)+150)/x^3/log(x^3)^3,x, algorithm="giac")

[Out]

-(x^4*log(x^3)^2 + x^3*e^2*log(x^3)^2 - x^3*log(2)*log(x^3)^2 - x^3*log(x^3)^2 + 25)/(x^2*log(x^3)^2)

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maple [A]  time = 0.07, size = 28, normalized size = 0.97

method	result	size
risch	\(-{\mathrm e}^{2} x +x \ln \relax (2)-x^{2}+x -\frac {25}{\ln \left (x^{3}\right )^{2} x^{2}}\)	\(28\)
norman	\(\frac {-25+\left (-{\mathrm e}^{2}+\ln \relax (2)+1\right ) x^{3} \ln \left (x^{3}\right )^{2}-x^{4} \ln \left (x^{3}\right )^{2}}{x^{2} \ln \left (x^{3}\right )^{2}}\)	\(42\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x^3*ln(2)-x^3*exp(2)-2*x^4+x^3)*ln(x^3)^3+50*ln(x^3)+150)/x^3/ln(x^3)^3,x,method=_RETURNVERBOSE)

[Out]

-exp(2)*x+x*ln(2)-x^2+x-25/ln(x^3)^2/x^2

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maxima [A]  time = 0.40, size = 25, normalized size = 0.86 \begin {gather*} -x^{2} - x e^{2} + x \log \relax (2) + x - \frac {25}{9 \, x^{2} \log \relax (x)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^3*log(2)-x^3*exp(2)-2*x^4+x^3)*log(x^3)^3+50*log(x^3)+150)/x^3/log(x^3)^3,x, algorithm="maxima")

[Out]

-x^2 - x*e^2 + x*log(2) + x - 25/9/(x^2*log(x)^2)

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mupad [B]  time = 4.34, size = 27, normalized size = 0.93 \begin {gather*} x-x\,{\mathrm {e}}^2+x\,\ln \relax (2)-x^2-\frac {25}{x^2\,{\ln \left (x^3\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((50*log(x^3) - log(x^3)^3*(x^3*exp(2) - x^3*log(2) - x^3 + 2*x^4) + 150)/(x^3*log(x^3)^3),x)

[Out]

x - x*exp(2) + x*log(2) - x^2 - 25/(x^2*log(x^3)^2)

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sympy [A]  time = 0.11, size = 24, normalized size = 0.83 \begin {gather*} - x^{2} + x \left (- e^{2} + \log {\relax (2 )} + 1\right ) - \frac {25}{x^{2} \log {\left (x^{3} \right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x**3*ln(2)-x**3*exp(2)-2*x**4+x**3)*ln(x**3)**3+50*ln(x**3)+150)/x**3/ln(x**3)**3,x)

[Out]

-x**2 + x*(-exp(2) + log(2) + 1) - 25/(x**2*log(x**3)**2)

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