3.14.22 \(\int \frac {-40-10 x-e^8 x+e^4 (-30 x-10 x^2)+(-30 x-2 e^4 x-10 x^2) \log (x)-x \log ^2(x)}{3 e^8 x+e^4 (120 x+30 x^2)+(120 x+6 e^4 x+30 x^2) \log (x)+3 x \log ^2(x)} \, dx\)

Optimal. Leaf size=29 \[ \frac {1}{3} \left (-5-x+\log \left (-2-\frac {4 (5+5 (3+x))}{e^4+\log (x)}\right )\right ) \]

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Rubi [A]  time = 0.77, antiderivative size = 32, normalized size of antiderivative = 1.10, number of steps used = 8, number of rules used = 7, integrand size = 97, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.072, Rules used = {6, 6688, 12, 6742, 2302, 29, 6684} \begin {gather*} -\frac {x}{3}-\frac {1}{3} \log \left (\log (x)+e^4\right )+\frac {1}{3} \log \left (10 x+\log (x)+e^4+40\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-40 - 10*x - E^8*x + E^4*(-30*x - 10*x^2) + (-30*x - 2*E^4*x - 10*x^2)*Log[x] - x*Log[x]^2)/(3*E^8*x + E^
4*(120*x + 30*x^2) + (120*x + 6*E^4*x + 30*x^2)*Log[x] + 3*x*Log[x]^2),x]

[Out]

-1/3*x - Log[E^4 + Log[x]]/3 + Log[40 + E^4 + 10*x + Log[x]]/3

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

Rule 12

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

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 2302

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

Rule 6684

Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*Log[RemoveContent[y, x]], x] /;  !Fa
lseQ[q]]

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 {-40+\left (-10-e^8\right ) x+e^4 \left (-30 x-10 x^2\right )+\left (-30 x-2 e^4 x-10 x^2\right ) \log (x)-x \log ^2(x)}{3 e^8 x+e^4 \left (120 x+30 x^2\right )+\left (120 x+6 e^4 x+30 x^2\right ) \log (x)+3 x \log ^2(x)} \, dx\\ &=\int \frac {-40-\left (10+30 e^4+e^8\right ) x-10 e^4 x^2-2 x \left (e^4+5 (3+x)\right ) \log (x)-x \log ^2(x)}{3 x \left (e^4+\log (x)\right ) \left (e^4+10 (4+x)+\log (x)\right )} \, dx\\ &=\frac {1}{3} \int \frac {-40-\left (10+30 e^4+e^8\right ) x-10 e^4 x^2-2 x \left (e^4+5 (3+x)\right ) \log (x)-x \log ^2(x)}{x \left (e^4+\log (x)\right ) \left (e^4+10 (4+x)+\log (x)\right )} \, dx\\ &=\frac {1}{3} \int \left (-1-\frac {1}{x \left (e^4+\log (x)\right )}+\frac {1+10 x}{x \left (40 \left (1+\frac {e^4}{40}\right )+10 x+\log (x)\right )}\right ) \, dx\\ &=-\frac {x}{3}-\frac {1}{3} \int \frac {1}{x \left (e^4+\log (x)\right )} \, dx+\frac {1}{3} \int \frac {1+10 x}{x \left (40 \left (1+\frac {e^4}{40}\right )+10 x+\log (x)\right )} \, dx\\ &=-\frac {x}{3}+\frac {1}{3} \log \left (40+e^4+10 x+\log (x)\right )-\frac {1}{3} \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,e^4+\log (x)\right )\\ &=-\frac {x}{3}-\frac {1}{3} \log \left (e^4+\log (x)\right )+\frac {1}{3} \log \left (40+e^4+10 x+\log (x)\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.24, size = 28, normalized size = 0.97 \begin {gather*} \frac {1}{3} \left (-x-\log \left (e^4+\log (x)\right )+\log \left (40+e^4+10 x+\log (x)\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-40 - 10*x - E^8*x + E^4*(-30*x - 10*x^2) + (-30*x - 2*E^4*x - 10*x^2)*Log[x] - x*Log[x]^2)/(3*E^8*
x + E^4*(120*x + 30*x^2) + (120*x + 6*E^4*x + 30*x^2)*Log[x] + 3*x*Log[x]^2),x]

[Out]

(-x - Log[E^4 + Log[x]] + Log[40 + E^4 + 10*x + Log[x]])/3

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fricas [A]  time = 0.79, size = 24, normalized size = 0.83 \begin {gather*} -\frac {1}{3} \, x + \frac {1}{3} \, \log \left (10 \, x + e^{4} + \log \relax (x) + 40\right ) - \frac {1}{3} \, \log \left (e^{4} + \log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x*log(x)^2+(-2*x*exp(4)-10*x^2-30*x)*log(x)-x*exp(4)^2+(-10*x^2-30*x)*exp(4)-10*x-40)/(3*x*log(x)^
2+(6*x*exp(4)+30*x^2+120*x)*log(x)+3*x*exp(4)^2+(30*x^2+120*x)*exp(4)),x, algorithm="fricas")

[Out]

-1/3*x + 1/3*log(10*x + e^4 + log(x) + 40) - 1/3*log(e^4 + log(x))

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giac [A]  time = 1.00, size = 28, normalized size = 0.97 \begin {gather*} -\frac {1}{3} \, x + \frac {1}{3} \, \log \left (-10 \, x - e^{4} - \log \relax (x) - 40\right ) - \frac {1}{3} \, \log \left (e^{4} + \log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x*log(x)^2+(-2*x*exp(4)-10*x^2-30*x)*log(x)-x*exp(4)^2+(-10*x^2-30*x)*exp(4)-10*x-40)/(3*x*log(x)^
2+(6*x*exp(4)+30*x^2+120*x)*log(x)+3*x*exp(4)^2+(30*x^2+120*x)*exp(4)),x, algorithm="giac")

[Out]

-1/3*x + 1/3*log(-10*x - e^4 - log(x) - 40) - 1/3*log(e^4 + log(x))

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maple [A]  time = 0.09, size = 25, normalized size = 0.86

method	result	size
norman	\(-\frac {x}{3}-\frac {\ln \left ({\mathrm e}^{4}+\ln \relax (x )\right )}{3}+\frac {\ln \left ({\mathrm e}^{4}+10 x +\ln \relax (x )+40\right )}{3}\)	\(25\)
risch	\(-\frac {x}{3}-\frac {\ln \left ({\mathrm e}^{4}+\ln \relax (x )\right )}{3}+\frac {\ln \left ({\mathrm e}^{4}+10 x +\ln \relax (x )+40\right )}{3}\)	\(25\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-x*ln(x)^2+(-2*x*exp(4)-10*x^2-30*x)*ln(x)-x*exp(4)^2+(-10*x^2-30*x)*exp(4)-10*x-40)/(3*x*ln(x)^2+(6*x*ex
p(4)+30*x^2+120*x)*ln(x)+3*x*exp(4)^2+(30*x^2+120*x)*exp(4)),x,method=_RETURNVERBOSE)

[Out]

-1/3*x-1/3*ln(exp(4)+ln(x))+1/3*ln(exp(4)+10*x+ln(x)+40)

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maxima [A]  time = 0.47, size = 24, normalized size = 0.83 \begin {gather*} -\frac {1}{3} \, x + \frac {1}{3} \, \log \left (10 \, x + e^{4} + \log \relax (x) + 40\right ) - \frac {1}{3} \, \log \left (e^{4} + \log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x*log(x)^2+(-2*x*exp(4)-10*x^2-30*x)*log(x)-x*exp(4)^2+(-10*x^2-30*x)*exp(4)-10*x-40)/(3*x*log(x)^
2+(6*x*exp(4)+30*x^2+120*x)*log(x)+3*x*exp(4)^2+(30*x^2+120*x)*exp(4)),x, algorithm="maxima")

[Out]

-1/3*x + 1/3*log(10*x + e^4 + log(x) + 40) - 1/3*log(e^4 + log(x))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int -\frac {x\,{\ln \relax (x)}^2+\left (30\,x+2\,x\,{\mathrm {e}}^4+10\,x^2\right )\,\ln \relax (x)+10\,x+{\mathrm {e}}^4\,\left (10\,x^2+30\,x\right )+x\,{\mathrm {e}}^8+40}{3\,x\,{\ln \relax (x)}^2+\left (120\,x+6\,x\,{\mathrm {e}}^4+30\,x^2\right )\,\ln \relax (x)+{\mathrm {e}}^4\,\left (30\,x^2+120\,x\right )+3\,x\,{\mathrm {e}}^8} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(10*x + x*log(x)^2 + exp(4)*(30*x + 10*x^2) + x*exp(8) + log(x)*(30*x + 2*x*exp(4) + 10*x^2) + 40)/(3*x*l
og(x)^2 + exp(4)*(120*x + 30*x^2) + 3*x*exp(8) + log(x)*(120*x + 6*x*exp(4) + 30*x^2)),x)

[Out]

int(-(10*x + x*log(x)^2 + exp(4)*(30*x + 10*x^2) + x*exp(8) + log(x)*(30*x + 2*x*exp(4) + 10*x^2) + 40)/(3*x*l
og(x)^2 + exp(4)*(120*x + 30*x^2) + 3*x*exp(8) + log(x)*(120*x + 6*x*exp(4) + 30*x^2)), x)

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sympy [A]  time = 0.32, size = 27, normalized size = 0.93 \begin {gather*} - \frac {x}{3} - \frac {\log {\left (\log {\relax (x )} + e^{4} \right )}}{3} + \frac {\log {\left (10 x + \log {\relax (x )} + 40 + e^{4} \right )}}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x*ln(x)**2+(-2*x*exp(4)-10*x**2-30*x)*ln(x)-x*exp(4)**2+(-10*x**2-30*x)*exp(4)-10*x-40)/(3*x*ln(x)
**2+(6*x*exp(4)+30*x**2+120*x)*ln(x)+3*x*exp(4)**2+(30*x**2+120*x)*exp(4)),x)

[Out]

-x/3 - log(log(x) + exp(4))/3 + log(10*x + log(x) + 40 + exp(4))/3

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