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

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

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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} $$\begin{array}{c}\frac{5(1+e^{3e^{-e^5}})}{x^2+e^{3e^{-e^5}}(x^2+1)+(1+e^{3e^{-e^5}})x^2{\log }^2(x)+2(1+e^{3e^{-e^5}})x\log (x)+6}\end{array}$$

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

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int \frac{10{(1+e^{3e^{-e^5}})}^2(-1-x-(1+x)\log (x)-x{\log }^2(x))}{{(6+x^2+e^{3e^{-e^5}}(1+x^2)+2(1+e^{3e^{-e^5}})x\log (x)+(1+e^{3e^{-e^5}})x^2{\log }^2(x))}^2}dx\\ & =\left(10{(1+e^{3e^{-e^5}})}^2\right)\int \frac{-1-x-(1+x)\log (x)-x{\log }^2(x)}{{(6+x^2+e^{3e^{-e^5}}(1+x^2)+2(1+e^{3e^{-e^5}})x\log (x)+(1+e^{3e^{-e^5}})x^2{\log }^2(x))}^2}dx\\ & =\frac{5(1+e^{3e^{-e^5}})}{6+x^2+e^{3e^{-e^5}}(1+x^2)+2(1+e^{3e^{-e^5}})x\log (x)+(1+e^{3e^{-e^5}})x^2{\log }^2(x)}\end{array}\end{array}$$

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

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 $$\begin{array}{c}\frac{5(e^{\left(3e^{(-e^5)}\right)}+1)}{x^2\log (x{)}^2+x^2+(x^2\log (x{)}^2+x^2+2x\log (x)+1)e^{\left(3e^{(-e^5)}\right)}+2x\log (x)+6}\end{array}$$

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 $$\begin{array}{c}\frac{5(e^{\left(3e^{(-e^5)}\right)}+1)}{x^2{e}^{\left(3e^{(-e^5)}\right)}\log (x{)}^2+x^2\log (x{)}^2+x^2{e}^{\left(3e^{(-e^5)}\right)}+2x{e}^{\left(3e^{(-e^5)}\right)}\log (x)+x^2+2x\log (x)+e^{\left(3e^{(-e^5)}\right)}+6}\end{array}$$

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	$\frac{5{\mathrm{e}}^{3{\mathrm{e}}^{-{\mathrm{e}}^5}}+5}{\ln (x{)}^2{\mathrm{e}}^{3{\mathrm{e}}^{-{\mathrm{e}}^5}}{x}^2+{\mathrm{e}}^{3{\mathrm{e}}^{-{\mathrm{e}}^5}}{x}^2+2\ln (x){\mathrm{e}}^{3{\mathrm{e}}^{-{\mathrm{e}}^5}}x+x^2\ln (x{)}^2+{\mathrm{e}}^{3{\mathrm{e}}^{-{\mathrm{e}}^5}}+x^2+2x\ln (x)+6}$	$83$
risch	$\frac{5{\mathrm{e}}^{3{\mathrm{e}}^{-{\mathrm{e}}^5}}}{\ln (x{)}^2{\mathrm{e}}^{3{\mathrm{e}}^{-{\mathrm{e}}^5}}{x}^2+{\mathrm{e}}^{3{\mathrm{e}}^{-{\mathrm{e}}^5}}{x}^2+2\ln (x){\mathrm{e}}^{3{\mathrm{e}}^{-{\mathrm{e}}^5}}x+x^2\ln (x{)}^2+{\mathrm{e}}^{3{\mathrm{e}}^{-{\mathrm{e}}^5}}+x^2+2x\ln (x)+6}+\frac{5}{\ln (x{)}^2{\mathrm{e}}^{3{\mathrm{e}}^{-{\mathrm{e}}^5}}{x}^2+{\mathrm{e}}^{3{\mathrm{e}}^{-{\mathrm{e}}^5}}{x}^2+2\ln (x){\mathrm{e}}^{3{\mathrm{e}}^{-{\mathrm{e}}^5}}x+x^2\ln (x{)}^2+{\mathrm{e}}^{3{\mathrm{e}}^{-{\mathrm{e}}^5}}+x^2+2x\ln (x)+6}$	$152$

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 $$\begin{array}{c}\frac{5(e^{\left(3e^{(-e^5)}\right)}+1)}{x^2(e^{\left(3e^{(-e^5)}\right)}+1)\log (x{)}^2+x^2(e^{\left(3e^{(-e^5)}\right)}+1)+2x(e^{\left(3e^{(-e^5)}\right)}+1)\log (x)+e^{\left(3e^{(-e^5)}\right)}+6}\end{array}$$

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 $$\begin{array}{c}\int -\frac{10x+10x{\ln (x)}^2+{\mathrm{e}}^{6{\mathrm{e}}^{-{\mathrm{e}}^5}}(10x{\ln (x)}^2+(10x+10)\ln (x)+10x+10)+{\mathrm{e}}^{3{\mathrm{e}}^{-{\mathrm{e}}^5}}(20x{\ln (x)}^2+(20x+20)\ln (x)+20x+20)+\ln (x)(10x+10)+10}{{\mathrm{e}}^{3{\mathrm{e}}^{-{\mathrm{e}}^5}}({\ln (x)}^2(4x^4+22x^2)+8x^3{\ln (x)}^3+2x^4{\ln (x)}^4+\ln (x)(8x^3+28x)+14x^2+2x^4+12)+{\ln (x)}^2(2x^4+16x^2)+4x^3{\ln (x)}^3+x^4{\ln (x)}^4+\ln (x)(4x^3+24x)+{\mathrm{e}}^{6{\mathrm{e}}^{-{\mathrm{e}}^5}}({\ln (x)}^2(2x^4+6x^2)+4x^3{\ln (x)}^3+x^4{\ln (x)}^4+\ln (x)(4x^3+4x)+2x^2+x^4+1)+12x^2+x^4+36}dx\end{array}$$

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 $$\begin{array}{c}\frac{5+5e^{\frac{3}{e^{e^5}}}}{x^2+x^2{e}^{\frac{3}{e^{e^5}}}+(2x+2x{e}^{\frac{3}{e^{e^5}}})\log (x)+(x^2+x^2{e}^{\frac{3}{e^{e^5}}})\log {(x)}^2+e^{\frac{3}{e^{e^5}}}+6}\end{array}$$

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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