3.33.74 \(\int \frac {-10-10 x+(-10-10 x) \log (x)-10 x \log ^2(x)+e^{3 e^{-e^5}} (-20-20 x+(-20-20 x) \log (x)-20 x \log ^2(x))+e^{6 e^{-e^5}} (-10-10 x+(-10-10 x) \log (x)-10 x \log ^2(x))}{36+12 x^2+x^4+(24 x+4 x^3) \log (x)+(16 x^2+2 x^4) \log ^2(x)+4 x^3 \log ^3(x)+x^4 \log ^4(x)+e^{6 e^{-e^5}} (1+2 x^2+x^4+(4 x+4 x^3) \log (x)+(6 x^2+2 x^4) \log ^2(x)+4 x^3 \log ^3(x)+x^4 \log ^4(x))+e^{3 e^{-e^5}} (12+14 x^2+2 x^4+(28 x+8 x^3) \log (x)+(22 x^2+4 x^4) \log ^2(x)+8 x^3 \log ^3(x)+2 x^4 \log ^4(x))} \, dx\)

Optimal. Leaf size=33 \[ \frac {5}{\frac {5}{1+e^{3 e^{-e^5}}}+x^2+(1+x \log (x))^2} \]

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Rubi [B]  time = 0.71, antiderivative size = 78, normalized size of antiderivative = 2.36, number of steps used = 3, number of rules used = 3, integrand size = 279, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.011, Rules used = {6688, 12, 6686} \begin {gather*} \frac {5 \left (1+e^{3 e^{-e^5}}\right )}{x^2+e^{3 e^{-e^5}} \left (x^2+1\right )+\left (1+e^{3 e^{-e^5}}\right ) x^2 \log ^2(x)+2 \left (1+e^{3 e^{-e^5}}\right ) x \log (x)+6} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-10 - 10*x + (-10 - 10*x)*Log[x] - 10*x*Log[x]^2 + E^(3/E^E^5)*(-20 - 20*x + (-20 - 20*x)*Log[x] - 20*x*L
og[x]^2) + E^(6/E^E^5)*(-10 - 10*x + (-10 - 10*x)*Log[x] - 10*x*Log[x]^2))/(36 + 12*x^2 + x^4 + (24*x + 4*x^3)
*Log[x] + (16*x^2 + 2*x^4)*Log[x]^2 + 4*x^3*Log[x]^3 + x^4*Log[x]^4 + E^(6/E^E^5)*(1 + 2*x^2 + x^4 + (4*x + 4*
x^3)*Log[x] + (6*x^2 + 2*x^4)*Log[x]^2 + 4*x^3*Log[x]^3 + x^4*Log[x]^4) + E^(3/E^E^5)*(12 + 14*x^2 + 2*x^4 + (
28*x + 8*x^3)*Log[x] + (22*x^2 + 4*x^4)*Log[x]^2 + 8*x^3*Log[x]^3 + 2*x^4*Log[x]^4)),x]

[Out]

(5*(1 + E^(3/E^E^5)))/(6 + x^2 + E^(3/E^E^5)*(1 + x^2) + 2*(1 + E^(3/E^E^5))*x*Log[x] + (1 + E^(3/E^E^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 6686

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[(q*y^(m + 1))/(m + 1), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]

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

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Mathematica [B]  time = 0.12, size = 97, normalized size = 2.94 \begin {gather*} \frac {10 \left (1+e^{3 e^{-e^5}}\right )^2}{\left (2+2 e^{3 e^{-e^5}}\right ) \left (6+x^2+e^{3 e^{-e^5}} \left (1+x^2\right )+2 \left (1+e^{3 e^{-e^5}}\right ) x \log (x)+\left (1+e^{3 e^{-e^5}}\right ) x^2 \log ^2(x)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-10 - 10*x + (-10 - 10*x)*Log[x] - 10*x*Log[x]^2 + E^(3/E^E^5)*(-20 - 20*x + (-20 - 20*x)*Log[x] -
20*x*Log[x]^2) + E^(6/E^E^5)*(-10 - 10*x + (-10 - 10*x)*Log[x] - 10*x*Log[x]^2))/(36 + 12*x^2 + x^4 + (24*x +
4*x^3)*Log[x] + (16*x^2 + 2*x^4)*Log[x]^2 + 4*x^3*Log[x]^3 + x^4*Log[x]^4 + E^(6/E^E^5)*(1 + 2*x^2 + x^4 + (4*
x + 4*x^3)*Log[x] + (6*x^2 + 2*x^4)*Log[x]^2 + 4*x^3*Log[x]^3 + x^4*Log[x]^4) + E^(3/E^E^5)*(12 + 14*x^2 + 2*x
^4 + (28*x + 8*x^3)*Log[x] + (22*x^2 + 4*x^4)*Log[x]^2 + 8*x^3*Log[x]^3 + 2*x^4*Log[x]^4)),x]

[Out]

(10*(1 + E^(3/E^E^5))^2)/((2 + 2*E^(3/E^E^5))*(6 + x^2 + E^(3/E^E^5)*(1 + x^2) + 2*(1 + E^(3/E^E^5))*x*Log[x]
+ (1 + E^(3/E^E^5))*x^2*Log[x]^2))

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fricas [A]  time = 0.53, size = 59, normalized size = 1.79 \begin {gather*} \frac {5 \, {\left (e^{\left (3 \, e^{\left (-e^{5}\right )}\right )} + 1\right )}}{x^{2} \log \relax (x)^{2} + x^{2} + {\left (x^{2} \log \relax (x)^{2} + x^{2} + 2 \, x \log \relax (x) + 1\right )} e^{\left (3 \, e^{\left (-e^{5}\right )}\right )} + 2 \, x \log \relax (x) + 6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-10*x*log(x)^2+(-10*x-10)*log(x)-10*x-10)*exp(3/exp(exp(5)))^2+(-20*x*log(x)^2+(-20*x-20)*log(x)-2
0*x-20)*exp(3/exp(exp(5)))-10*x*log(x)^2+(-10*x-10)*log(x)-10*x-10)/((x^4*log(x)^4+4*x^3*log(x)^3+(2*x^4+6*x^2
)*log(x)^2+(4*x^3+4*x)*log(x)+x^4+2*x^2+1)*exp(3/exp(exp(5)))^2+(2*x^4*log(x)^4+8*x^3*log(x)^3+(4*x^4+22*x^2)*
log(x)^2+(8*x^3+28*x)*log(x)+2*x^4+14*x^2+12)*exp(3/exp(exp(5)))+x^4*log(x)^4+4*x^3*log(x)^3+(2*x^4+16*x^2)*lo
g(x)^2+(4*x^3+24*x)*log(x)+x^4+12*x^2+36),x, algorithm="fricas")

[Out]

5*(e^(3*e^(-e^5)) + 1)/(x^2*log(x)^2 + x^2 + (x^2*log(x)^2 + x^2 + 2*x*log(x) + 1)*e^(3*e^(-e^5)) + 2*x*log(x)
+ 6)

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giac [B]  time = 3.08, size = 81, normalized size = 2.45 \begin {gather*} \frac {5 \, {\left (e^{\left (3 \, e^{\left (-e^{5}\right )}\right )} + 1\right )}}{x^{2} e^{\left (3 \, e^{\left (-e^{5}\right )}\right )} \log \relax (x)^{2} + x^{2} \log \relax (x)^{2} + x^{2} e^{\left (3 \, e^{\left (-e^{5}\right )}\right )} + 2 \, x e^{\left (3 \, e^{\left (-e^{5}\right )}\right )} \log \relax (x) + x^{2} + 2 \, x \log \relax (x) + e^{\left (3 \, e^{\left (-e^{5}\right )}\right )} + 6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-10*x*log(x)^2+(-10*x-10)*log(x)-10*x-10)*exp(3/exp(exp(5)))^2+(-20*x*log(x)^2+(-20*x-20)*log(x)-2
0*x-20)*exp(3/exp(exp(5)))-10*x*log(x)^2+(-10*x-10)*log(x)-10*x-10)/((x^4*log(x)^4+4*x^3*log(x)^3+(2*x^4+6*x^2
)*log(x)^2+(4*x^3+4*x)*log(x)+x^4+2*x^2+1)*exp(3/exp(exp(5)))^2+(2*x^4*log(x)^4+8*x^3*log(x)^3+(4*x^4+22*x^2)*
log(x)^2+(8*x^3+28*x)*log(x)+2*x^4+14*x^2+12)*exp(3/exp(exp(5)))+x^4*log(x)^4+4*x^3*log(x)^3+(2*x^4+16*x^2)*lo
g(x)^2+(4*x^3+24*x)*log(x)+x^4+12*x^2+36),x, algorithm="giac")

[Out]

5*(e^(3*e^(-e^5)) + 1)/(x^2*e^(3*e^(-e^5))*log(x)^2 + x^2*log(x)^2 + x^2*e^(3*e^(-e^5)) + 2*x*e^(3*e^(-e^5))*l
og(x) + x^2 + 2*x*log(x) + e^(3*e^(-e^5)) + 6)

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maple [B]  time = 0.72, size = 83, normalized size = 2.52

method	result	size
norman	\(\frac {5 \,{\mathrm e}^{3 \,{\mathrm e}^{-{\mathrm e}^{5}}}+5}{\ln \relax (x )^{2} {\mathrm e}^{3 \,{\mathrm e}^{-{\mathrm e}^{5}}} x^{2}+{\mathrm e}^{3 \,{\mathrm e}^{-{\mathrm e}^{5}}} x^{2}+2 \ln \relax (x ) {\mathrm e}^{3 \,{\mathrm e}^{-{\mathrm e}^{5}}} x +x^{2} \ln \relax (x )^{2}+{\mathrm e}^{3 \,{\mathrm e}^{-{\mathrm e}^{5}}}+x^{2}+2 x \ln \relax (x )+6}\)	\(83\)
risch	\(\frac {5 \,{\mathrm e}^{3 \,{\mathrm e}^{-{\mathrm e}^{5}}}}{\ln \relax (x )^{2} {\mathrm e}^{3 \,{\mathrm e}^{-{\mathrm e}^{5}}} x^{2}+{\mathrm e}^{3 \,{\mathrm e}^{-{\mathrm e}^{5}}} x^{2}+2 \ln \relax (x ) {\mathrm e}^{3 \,{\mathrm e}^{-{\mathrm e}^{5}}} x +x^{2} \ln \relax (x )^{2}+{\mathrm e}^{3 \,{\mathrm e}^{-{\mathrm e}^{5}}}+x^{2}+2 x \ln \relax (x )+6}+\frac {5}{\ln \relax (x )^{2} {\mathrm e}^{3 \,{\mathrm e}^{-{\mathrm e}^{5}}} x^{2}+{\mathrm e}^{3 \,{\mathrm e}^{-{\mathrm e}^{5}}} x^{2}+2 \ln \relax (x ) {\mathrm e}^{3 \,{\mathrm e}^{-{\mathrm e}^{5}}} x +x^{2} \ln \relax (x )^{2}+{\mathrm e}^{3 \,{\mathrm e}^{-{\mathrm e}^{5}}}+x^{2}+2 x \ln \relax (x )+6}\)	\(152\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-10*x*ln(x)^2+(-10*x-10)*ln(x)-10*x-10)*exp(3/exp(exp(5)))^2+(-20*x*ln(x)^2+(-20*x-20)*ln(x)-20*x-20)*ex
p(3/exp(exp(5)))-10*x*ln(x)^2+(-10*x-10)*ln(x)-10*x-10)/((x^4*ln(x)^4+4*x^3*ln(x)^3+(2*x^4+6*x^2)*ln(x)^2+(4*x
^3+4*x)*ln(x)+x^4+2*x^2+1)*exp(3/exp(exp(5)))^2+(2*x^4*ln(x)^4+8*x^3*ln(x)^3+(4*x^4+22*x^2)*ln(x)^2+(8*x^3+28*
x)*ln(x)+2*x^4+14*x^2+12)*exp(3/exp(exp(5)))+x^4*ln(x)^4+4*x^3*ln(x)^3+(2*x^4+16*x^2)*ln(x)^2+(4*x^3+24*x)*ln(
x)+x^4+12*x^2+36),x,method=_RETURNVERBOSE)

[Out]

(5*exp(1/exp(exp(5)))^3+5)/(ln(x)^2*exp(3/exp(exp(5)))*x^2+x^2*ln(x)^2+2*ln(x)*exp(3/exp(exp(5)))*x+exp(3/exp(
exp(5)))*x^2+2*x*ln(x)+x^2+exp(3/exp(exp(5)))+6)

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maxima [B]  time = 0.92, size = 71, normalized size = 2.15 \begin {gather*} \frac {5 \, {\left (e^{\left (3 \, e^{\left (-e^{5}\right )}\right )} + 1\right )}}{x^{2} {\left (e^{\left (3 \, e^{\left (-e^{5}\right )}\right )} + 1\right )} \log \relax (x)^{2} + x^{2} {\left (e^{\left (3 \, e^{\left (-e^{5}\right )}\right )} + 1\right )} + 2 \, x {\left (e^{\left (3 \, e^{\left (-e^{5}\right )}\right )} + 1\right )} \log \relax (x) + e^{\left (3 \, e^{\left (-e^{5}\right )}\right )} + 6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-10*x*log(x)^2+(-10*x-10)*log(x)-10*x-10)*exp(3/exp(exp(5)))^2+(-20*x*log(x)^2+(-20*x-20)*log(x)-2
0*x-20)*exp(3/exp(exp(5)))-10*x*log(x)^2+(-10*x-10)*log(x)-10*x-10)/((x^4*log(x)^4+4*x^3*log(x)^3+(2*x^4+6*x^2
)*log(x)^2+(4*x^3+4*x)*log(x)+x^4+2*x^2+1)*exp(3/exp(exp(5)))^2+(2*x^4*log(x)^4+8*x^3*log(x)^3+(4*x^4+22*x^2)*
log(x)^2+(8*x^3+28*x)*log(x)+2*x^4+14*x^2+12)*exp(3/exp(exp(5)))+x^4*log(x)^4+4*x^3*log(x)^3+(2*x^4+16*x^2)*lo
g(x)^2+(4*x^3+24*x)*log(x)+x^4+12*x^2+36),x, algorithm="maxima")

[Out]

5*(e^(3*e^(-e^5)) + 1)/(x^2*(e^(3*e^(-e^5)) + 1)*log(x)^2 + x^2*(e^(3*e^(-e^5)) + 1) + 2*x*(e^(3*e^(-e^5)) + 1
)*log(x) + e^(3*e^(-e^5)) + 6)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int -\frac {10\,x+10\,x\,{\ln \relax (x)}^2+{\mathrm {e}}^{6\,{\mathrm {e}}^{-{\mathrm {e}}^5}}\,\left (10\,x\,{\ln \relax (x)}^2+\left (10\,x+10\right )\,\ln \relax (x)+10\,x+10\right )+{\mathrm {e}}^{3\,{\mathrm {e}}^{-{\mathrm {e}}^5}}\,\left (20\,x\,{\ln \relax (x)}^2+\left (20\,x+20\right )\,\ln \relax (x)+20\,x+20\right )+\ln \relax (x)\,\left (10\,x+10\right )+10}{{\mathrm {e}}^{3\,{\mathrm {e}}^{-{\mathrm {e}}^5}}\,\left ({\ln \relax (x)}^2\,\left (4\,x^4+22\,x^2\right )+8\,x^3\,{\ln \relax (x)}^3+2\,x^4\,{\ln \relax (x)}^4+\ln \relax (x)\,\left (8\,x^3+28\,x\right )+14\,x^2+2\,x^4+12\right )+{\ln \relax (x)}^2\,\left (2\,x^4+16\,x^2\right )+4\,x^3\,{\ln \relax (x)}^3+x^4\,{\ln \relax (x)}^4+\ln \relax (x)\,\left (4\,x^3+24\,x\right )+{\mathrm {e}}^{6\,{\mathrm {e}}^{-{\mathrm {e}}^5}}\,\left ({\ln \relax (x)}^2\,\left (2\,x^4+6\,x^2\right )+4\,x^3\,{\ln \relax (x)}^3+x^4\,{\ln \relax (x)}^4+\ln \relax (x)\,\left (4\,x^3+4\,x\right )+2\,x^2+x^4+1\right )+12\,x^2+x^4+36} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(10*x + 10*x*log(x)^2 + exp(6*exp(-exp(5)))*(10*x + 10*x*log(x)^2 + log(x)*(10*x + 10) + 10) + exp(3*exp(
-exp(5)))*(20*x + 20*x*log(x)^2 + log(x)*(20*x + 20) + 20) + log(x)*(10*x + 10) + 10)/(exp(3*exp(-exp(5)))*(lo
g(x)^2*(22*x^2 + 4*x^4) + 8*x^3*log(x)^3 + 2*x^4*log(x)^4 + log(x)*(28*x + 8*x^3) + 14*x^2 + 2*x^4 + 12) + log
(x)^2*(16*x^2 + 2*x^4) + 4*x^3*log(x)^3 + x^4*log(x)^4 + log(x)*(24*x + 4*x^3) + exp(6*exp(-exp(5)))*(log(x)^2
*(6*x^2 + 2*x^4) + 4*x^3*log(x)^3 + x^4*log(x)^4 + log(x)*(4*x + 4*x^3) + 2*x^2 + x^4 + 1) + 12*x^2 + x^4 + 36
),x)

[Out]

int(-(10*x + 10*x*log(x)^2 + exp(6*exp(-exp(5)))*(10*x + 10*x*log(x)^2 + log(x)*(10*x + 10) + 10) + exp(3*exp(
-exp(5)))*(20*x + 20*x*log(x)^2 + log(x)*(20*x + 20) + 20) + log(x)*(10*x + 10) + 10)/(exp(3*exp(-exp(5)))*(lo
g(x)^2*(22*x^2 + 4*x^4) + 8*x^3*log(x)^3 + 2*x^4*log(x)^4 + log(x)*(28*x + 8*x^3) + 14*x^2 + 2*x^4 + 12) + log
(x)^2*(16*x^2 + 2*x^4) + 4*x^3*log(x)^3 + x^4*log(x)^4 + log(x)*(24*x + 4*x^3) + exp(6*exp(-exp(5)))*(log(x)^2
*(6*x^2 + 2*x^4) + 4*x^3*log(x)^3 + x^4*log(x)^4 + log(x)*(4*x + 4*x^3) + 2*x^2 + x^4 + 1) + 12*x^2 + x^4 + 36
), x)

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sympy [B]  time = 0.78, size = 75, normalized size = 2.27 \begin {gather*} \frac {5 + 5 e^{\frac {3}{e^{e^{5}}}}}{x^{2} + x^{2} e^{\frac {3}{e^{e^{5}}}} + \left (2 x + 2 x e^{\frac {3}{e^{e^{5}}}}\right ) \log {\relax (x )} + \left (x^{2} + x^{2} e^{\frac {3}{e^{e^{5}}}}\right ) \log {\relax (x )}^{2} + e^{\frac {3}{e^{e^{5}}}} + 6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-10*x*ln(x)**2+(-10*x-10)*ln(x)-10*x-10)*exp(3/exp(exp(5)))**2+(-20*x*ln(x)**2+(-20*x-20)*ln(x)-20
*x-20)*exp(3/exp(exp(5)))-10*x*ln(x)**2+(-10*x-10)*ln(x)-10*x-10)/((x**4*ln(x)**4+4*x**3*ln(x)**3+(2*x**4+6*x*
*2)*ln(x)**2+(4*x**3+4*x)*ln(x)+x**4+2*x**2+1)*exp(3/exp(exp(5)))**2+(2*x**4*ln(x)**4+8*x**3*ln(x)**3+(4*x**4+
22*x**2)*ln(x)**2+(8*x**3+28*x)*ln(x)+2*x**4+14*x**2+12)*exp(3/exp(exp(5)))+x**4*ln(x)**4+4*x**3*ln(x)**3+(2*x
**4+16*x**2)*ln(x)**2+(4*x**3+24*x)*ln(x)+x**4+12*x**2+36),x)

[Out]

(5 + 5*exp(3*exp(-exp(5))))/(x**2 + x**2*exp(3*exp(-exp(5))) + (2*x + 2*x*exp(3*exp(-exp(5))))*log(x) + (x**2
+ x**2*exp(3*exp(-exp(5))))*log(x)**2 + exp(3*exp(-exp(5))) + 6)

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