3.88.62 \(\int \frac {e^{15} (10-24 x+6 x^2)+e^{15} (-10+22 x-4 x^2) \log (5)+e^{15} (-10+22 x-4 x^2) \log (5-x)}{-5 x+6 x^2-x^3+(5 x-6 x^2+x^3) \log (5)+(5 x-6 x^2+x^3) \log (5-x)+(e^5 (-15 x+18 x^2-3 x^3)+e^5 (15 x-18 x^2+3 x^3) \log (5)+e^5 (15 x-18 x^2+3 x^3) \log (5-x)) \log (\frac {-x+x^2}{-1+\log (5)+\log (5-x)})+(e^{10} (-15 x+18 x^2-3 x^3)+e^{10} (15 x-18 x^2+3 x^3) \log (5)+e^{10} (15 x-18 x^2+3 x^3) \log (5-x)) \log ^2(\frac {-x+x^2}{-1+\log (5)+\log (5-x)})+(e^{15} (-5 x+6 x^2-x^3)+e^{15} (5 x-6 x^2+x^3) \log (5)+e^{15} (5 x-6 x^2+x^3) \log (5-x)) \log ^3(\frac {-x+x^2}{-1+\log (5)+\log (5-x)})} \, dx\)
Optimal. Leaf size=35 \[ \frac {x^2}{\left (\frac {x}{e^5}+x \log \left (\frac {-x+x^2}{-1+\log (5)+\log (5-x)}\right )\right )^2} \]
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Rubi [A] time = 1.94, antiderivative size = 34, normalized size of antiderivative = 0.97,
number of steps used = 4, number of rules used = 4, integrand size = 357, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.011, Rules used
= {6688, 12, 6708, 32} \begin {gather*} \frac {e^{10}}{\left (e^5 \log \left (\frac {(1-x) x}{1-\log (5 (5-x))}\right )+1\right )^2} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[(E^15*(10 - 24*x + 6*x^2) + E^15*(-10 + 22*x - 4*x^2)*Log[5] + E^15*(-10 + 22*x - 4*x^2)*Log[5 - x])/(-5*x
+ 6*x^2 - x^3 + (5*x - 6*x^2 + x^3)*Log[5] + (5*x - 6*x^2 + x^3)*Log[5 - x] + (E^5*(-15*x + 18*x^2 - 3*x^3) +
E^5*(15*x - 18*x^2 + 3*x^3)*Log[5] + E^5*(15*x - 18*x^2 + 3*x^3)*Log[5 - x])*Log[(-x + x^2)/(-1 + Log[5] + Lo
g[5 - x])] + (E^10*(-15*x + 18*x^2 - 3*x^3) + E^10*(15*x - 18*x^2 + 3*x^3)*Log[5] + E^10*(15*x - 18*x^2 + 3*x^
3)*Log[5 - x])*Log[(-x + x^2)/(-1 + Log[5] + Log[5 - x])]^2 + (E^15*(-5*x + 6*x^2 - x^3) + E^15*(5*x - 6*x^2 +
x^3)*Log[5] + E^15*(5*x - 6*x^2 + x^3)*Log[5 - x])*Log[(-x + x^2)/(-1 + Log[5] + Log[5 - x])]^3),x]
[Out]
E^10/(1 + E^5*Log[((1 - x)*x)/(1 - Log[5*(5 - 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 32
Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]
Rule 6688
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]
Rule 6708
Int[(u_)*((a_) + (b_.)*(v_)^(p_.)*(w_)^(p_.))^(m_.), x_Symbol] :> With[{c = Simplify[u/(w*D[v, x] + v*D[w, x])
]}, Dist[c, Subst[Int[(a + b*x^p)^m, x], x, v*w], x] /; FreeQ[c, x]] /; FreeQ[{a, b, m, p}, x] && IntegerQ[p]
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {2 e^{15} \left (x (12-11 \log (5))+5 (-1+\log (5))+x^2 (-3+\log (25))+\left (5-11 x+2 x^2\right ) \log (5-x)\right )}{x \left (5-6 x+x^2\right ) (1-\log (-5 (-5+x))) \left (1+e^5 \log \left (\frac {(-1+x) x}{-1+\log (-5 (-5+x))}\right )\right )^3} \, dx\\ &=\left (2 e^{15}\right ) \int \frac {x (12-11 \log (5))+5 (-1+\log (5))+x^2 (-3+\log (25))+\left (5-11 x+2 x^2\right ) \log (5-x)}{x \left (5-6 x+x^2\right ) (1-\log (-5 (-5+x))) \left (1+e^5 \log \left (\frac {(-1+x) x}{-1+\log (-5 (-5+x))}\right )\right )^3} \, dx\\ &=-\left (\left (2 e^{10}\right ) \operatorname {Subst}\left (\int \frac {1}{(1+x)^3} \, dx,x,e^5 \log \left (\frac {(-1+x) x}{-1+\log (-5 (-5+x))}\right )\right )\right )\\ &=\frac {e^{10}}{\left (1+e^5 \log \left (\frac {(1-x) x}{1-\log (5 (5-x))}\right )\right )^2}\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.37, size = 28, normalized size = 0.80 \begin {gather*} \frac {e^{10}}{\left (1+e^5 \log \left (\frac {(-1+x) x}{-1+\log (-5 (-5+x))}\right )\right )^2} \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[(E^15*(10 - 24*x + 6*x^2) + E^15*(-10 + 22*x - 4*x^2)*Log[5] + E^15*(-10 + 22*x - 4*x^2)*Log[5 - x])
/(-5*x + 6*x^2 - x^3 + (5*x - 6*x^2 + x^3)*Log[5] + (5*x - 6*x^2 + x^3)*Log[5 - x] + (E^5*(-15*x + 18*x^2 - 3*
x^3) + E^5*(15*x - 18*x^2 + 3*x^3)*Log[5] + E^5*(15*x - 18*x^2 + 3*x^3)*Log[5 - x])*Log[(-x + x^2)/(-1 + Log[5
] + Log[5 - x])] + (E^10*(-15*x + 18*x^2 - 3*x^3) + E^10*(15*x - 18*x^2 + 3*x^3)*Log[5] + E^10*(15*x - 18*x^2
+ 3*x^3)*Log[5 - x])*Log[(-x + x^2)/(-1 + Log[5] + Log[5 - x])]^2 + (E^15*(-5*x + 6*x^2 - x^3) + E^15*(5*x - 6
*x^2 + x^3)*Log[5] + E^15*(5*x - 6*x^2 + x^3)*Log[5 - x])*Log[(-x + x^2)/(-1 + Log[5] + Log[5 - x])]^3),x]
[Out]
E^10/(1 + E^5*Log[((-1 + x)*x)/(-1 + Log[-5*(-5 + x)])])^2
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fricas [A] time = 0.50, size = 58, normalized size = 1.66 \begin {gather*} \frac {e^{10}}{e^{10} \log \left (\frac {x^{2} - x}{\log \relax (5) + \log \left (-x + 5\right ) - 1}\right )^{2} + 2 \, e^{5} \log \left (\frac {x^{2} - x}{\log \relax (5) + \log \left (-x + 5\right ) - 1}\right ) + 1} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(((-4*x^2+22*x-10)*exp(5)^3*log(5-x)+(-4*x^2+22*x-10)*exp(5)^3*log(5)+(6*x^2-24*x+10)*exp(5)^3)/(((x^
3-6*x^2+5*x)*exp(5)^3*log(5-x)+(x^3-6*x^2+5*x)*exp(5)^3*log(5)+(-x^3+6*x^2-5*x)*exp(5)^3)*log((x^2-x)/(log(5-x
)+log(5)-1))^3+((3*x^3-18*x^2+15*x)*exp(5)^2*log(5-x)+(3*x^3-18*x^2+15*x)*exp(5)^2*log(5)+(-3*x^3+18*x^2-15*x)
*exp(5)^2)*log((x^2-x)/(log(5-x)+log(5)-1))^2+((3*x^3-18*x^2+15*x)*exp(5)*log(5-x)+(3*x^3-18*x^2+15*x)*exp(5)*
log(5)+(-3*x^3+18*x^2-15*x)*exp(5))*log((x^2-x)/(log(5-x)+log(5)-1))+(x^3-6*x^2+5*x)*log(5-x)+(x^3-6*x^2+5*x)*
log(5)-x^3+6*x^2-5*x),x, algorithm="fricas")
[Out]
e^10/(e^10*log((x^2 - x)/(log(5) + log(-x + 5) - 1))^2 + 2*e^5*log((x^2 - x)/(log(5) + log(-x + 5) - 1)) + 1)
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giac [B] time = 25.80, size = 86, normalized size = 2.46 \begin {gather*} \frac {e^{10}}{e^{10} \log \left (x^{2} - x\right )^{2} - 2 \, e^{10} \log \left (x^{2} - x\right ) \log \left (\log \relax (5) + \log \left (-x + 5\right ) - 1\right ) + e^{10} \log \left (\log \relax (5) + \log \left (-x + 5\right ) - 1\right )^{2} + 2 \, e^{5} \log \left (x^{2} - x\right ) - 2 \, e^{5} \log \left (\log \relax (5) + \log \left (-x + 5\right ) - 1\right ) + 1} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(((-4*x^2+22*x-10)*exp(5)^3*log(5-x)+(-4*x^2+22*x-10)*exp(5)^3*log(5)+(6*x^2-24*x+10)*exp(5)^3)/(((x^
3-6*x^2+5*x)*exp(5)^3*log(5-x)+(x^3-6*x^2+5*x)*exp(5)^3*log(5)+(-x^3+6*x^2-5*x)*exp(5)^3)*log((x^2-x)/(log(5-x
)+log(5)-1))^3+((3*x^3-18*x^2+15*x)*exp(5)^2*log(5-x)+(3*x^3-18*x^2+15*x)*exp(5)^2*log(5)+(-3*x^3+18*x^2-15*x)
*exp(5)^2)*log((x^2-x)/(log(5-x)+log(5)-1))^2+((3*x^3-18*x^2+15*x)*exp(5)*log(5-x)+(3*x^3-18*x^2+15*x)*exp(5)*
log(5)+(-3*x^3+18*x^2-15*x)*exp(5))*log((x^2-x)/(log(5-x)+log(5)-1))+(x^3-6*x^2+5*x)*log(5-x)+(x^3-6*x^2+5*x)*
log(5)-x^3+6*x^2-5*x),x, algorithm="giac")
[Out]
e^10/(e^10*log(x^2 - x)^2 - 2*e^10*log(x^2 - x)*log(log(5) + log(-x + 5) - 1) + e^10*log(log(5) + log(-x + 5)
- 1)^2 + 2*e^5*log(x^2 - x) - 2*e^5*log(log(5) + log(-x + 5) - 1) + 1)
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maple [C] time = 2.18, size = 340, normalized size = 9.71
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\(-\frac {4 \,{\mathrm e}^{10}}{\left (\pi \,{\mathrm e}^{5} \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (\frac {i \left (x -1\right )}{\ln \left (5-x \right )+\ln \relax (5)-1}\right ) \mathrm {csgn}\left (\frac {i x \left (x -1\right )}{\ln \left (5-x \right )+\ln \relax (5)-1}\right )-\pi \,{\mathrm e}^{5} \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (\frac {i x \left (x -1\right )}{\ln \left (5-x \right )+\ln \relax (5)-1}\right )^{2}-\pi \,{\mathrm e}^{5} \mathrm {csgn}\left (\frac {i}{\ln \left (5-x \right )+\ln \relax (5)-1}\right ) \mathrm {csgn}\left (\frac {i \left (x -1\right )}{\ln \left (5-x \right )+\ln \relax (5)-1}\right )^{2}+\pi \,{\mathrm e}^{5} \mathrm {csgn}\left (\frac {i}{\ln \left (5-x \right )+\ln \relax (5)-1}\right ) \mathrm {csgn}\left (\frac {i \left (x -1\right )}{\ln \left (5-x \right )+\ln \relax (5)-1}\right ) \mathrm {csgn}\left (i \left (x -1\right )\right )+\pi \,{\mathrm e}^{5} \mathrm {csgn}\left (\frac {i \left (x -1\right )}{\ln \left (5-x \right )+\ln \relax (5)-1}\right )^{3}-\pi \,{\mathrm e}^{5} \mathrm {csgn}\left (\frac {i \left (x -1\right )}{\ln \left (5-x \right )+\ln \relax (5)-1}\right )^{2} \mathrm {csgn}\left (i \left (x -1\right )\right )-\pi \,{\mathrm e}^{5} \mathrm {csgn}\left (\frac {i \left (x -1\right )}{\ln \left (5-x \right )+\ln \relax (5)-1}\right ) \mathrm {csgn}\left (\frac {i x \left (x -1\right )}{\ln \left (5-x \right )+\ln \relax (5)-1}\right )^{2}+\pi \,{\mathrm e}^{5} \mathrm {csgn}\left (\frac {i x \left (x -1\right )}{\ln \left (5-x \right )+\ln \relax (5)-1}\right )^{3}+2 i {\mathrm e}^{5} \ln \relax (x )+2 i {\mathrm e}^{5} \ln \left (x -1\right )-2 i {\mathrm e}^{5} \ln \left (\ln \left (5-x \right )+\ln \relax (5)-1\right )+2 i\right )^{2}}\) |
\(340\) |
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Verification of antiderivative is not currently implemented for this CAS.
[In]
int(((-4*x^2+22*x-10)*exp(5)^3*ln(5-x)+(-4*x^2+22*x-10)*exp(5)^3*ln(5)+(6*x^2-24*x+10)*exp(5)^3)/(((x^3-6*x^2+
5*x)*exp(5)^3*ln(5-x)+(x^3-6*x^2+5*x)*exp(5)^3*ln(5)+(-x^3+6*x^2-5*x)*exp(5)^3)*ln((x^2-x)/(ln(5-x)+ln(5)-1))^
3+((3*x^3-18*x^2+15*x)*exp(5)^2*ln(5-x)+(3*x^3-18*x^2+15*x)*exp(5)^2*ln(5)+(-3*x^3+18*x^2-15*x)*exp(5)^2)*ln((
x^2-x)/(ln(5-x)+ln(5)-1))^2+((3*x^3-18*x^2+15*x)*exp(5)*ln(5-x)+(3*x^3-18*x^2+15*x)*exp(5)*ln(5)+(-3*x^3+18*x^
2-15*x)*exp(5))*ln((x^2-x)/(ln(5-x)+ln(5)-1))+(x^3-6*x^2+5*x)*ln(5-x)+(x^3-6*x^2+5*x)*ln(5)-x^3+6*x^2-5*x),x,m
ethod=_RETURNVERBOSE)
[Out]
-4*exp(10)/(Pi*exp(5)*csgn(I*x)*csgn(I*(x-1)/(ln(5-x)+ln(5)-1))*csgn(I*x/(ln(5-x)+ln(5)-1)*(x-1))-Pi*exp(5)*cs
gn(I*x)*csgn(I*x/(ln(5-x)+ln(5)-1)*(x-1))^2-Pi*exp(5)*csgn(I/(ln(5-x)+ln(5)-1))*csgn(I*(x-1)/(ln(5-x)+ln(5)-1)
)^2+Pi*exp(5)*csgn(I/(ln(5-x)+ln(5)-1))*csgn(I*(x-1)/(ln(5-x)+ln(5)-1))*csgn(I*(x-1))+Pi*exp(5)*csgn(I*(x-1)/(
ln(5-x)+ln(5)-1))^3-Pi*exp(5)*csgn(I*(x-1)/(ln(5-x)+ln(5)-1))^2*csgn(I*(x-1))-Pi*exp(5)*csgn(I*(x-1)/(ln(5-x)+
ln(5)-1))*csgn(I*x/(ln(5-x)+ln(5)-1)*(x-1))^2+Pi*exp(5)*csgn(I*x/(ln(5-x)+ln(5)-1)*(x-1))^3+2*I*exp(5)*ln(x)+2
*I*exp(5)*ln(x-1)-2*I*exp(5)*ln(ln(5-x)+ln(5)-1)+2*I)^2
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maxima [B] time = 0.76, size = 87, normalized size = 2.49 \begin {gather*} \frac {e^{10}}{e^{10} \log \left (x - 1\right )^{2} + e^{10} \log \relax (x)^{2} + e^{10} \log \left (\log \relax (5) + \log \left (-x + 5\right ) - 1\right )^{2} + 2 \, {\left (e^{10} \log \relax (x) + e^{5}\right )} \log \left (x - 1\right ) + 2 \, e^{5} \log \relax (x) - 2 \, {\left (e^{10} \log \left (x - 1\right ) + e^{10} \log \relax (x) + e^{5}\right )} \log \left (\log \relax (5) + \log \left (-x + 5\right ) - 1\right ) + 1} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(((-4*x^2+22*x-10)*exp(5)^3*log(5-x)+(-4*x^2+22*x-10)*exp(5)^3*log(5)+(6*x^2-24*x+10)*exp(5)^3)/(((x^
3-6*x^2+5*x)*exp(5)^3*log(5-x)+(x^3-6*x^2+5*x)*exp(5)^3*log(5)+(-x^3+6*x^2-5*x)*exp(5)^3)*log((x^2-x)/(log(5-x
)+log(5)-1))^3+((3*x^3-18*x^2+15*x)*exp(5)^2*log(5-x)+(3*x^3-18*x^2+15*x)*exp(5)^2*log(5)+(-3*x^3+18*x^2-15*x)
*exp(5)^2)*log((x^2-x)/(log(5-x)+log(5)-1))^2+((3*x^3-18*x^2+15*x)*exp(5)*log(5-x)+(3*x^3-18*x^2+15*x)*exp(5)*
log(5)+(-3*x^3+18*x^2-15*x)*exp(5))*log((x^2-x)/(log(5-x)+log(5)-1))+(x^3-6*x^2+5*x)*log(5-x)+(x^3-6*x^2+5*x)*
log(5)-x^3+6*x^2-5*x),x, algorithm="maxima")
[Out]
e^10/(e^10*log(x - 1)^2 + e^10*log(x)^2 + e^10*log(log(5) + log(-x + 5) - 1)^2 + 2*(e^10*log(x) + e^5)*log(x -
1) + 2*e^5*log(x) - 2*(e^10*log(x - 1) + e^10*log(x) + e^5)*log(log(5) + log(-x + 5) - 1) + 1)
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mupad [B] time = 7.37, size = 32, normalized size = 0.91 \begin {gather*} \frac {{\mathrm {e}}^{10}}{{\left ({\mathrm {e}}^5\,\ln \left (-\frac {x-x^2}{\ln \relax (5)+\ln \left (5-x\right )-1}\right )+1\right )}^2} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(-(exp(15)*log(5)*(4*x^2 - 22*x + 10) - exp(15)*(6*x^2 - 24*x + 10) + exp(15)*log(5 - x)*(4*x^2 - 22*x + 10
))/(log(5)*(5*x - 6*x^2 + x^3) - 5*x + log(-(x - x^2)/(log(5) + log(5 - x) - 1))^2*(exp(10)*log(5 - x)*(15*x -
18*x^2 + 3*x^3) - exp(10)*(15*x - 18*x^2 + 3*x^3) + exp(10)*log(5)*(15*x - 18*x^2 + 3*x^3)) + log(5 - x)*(5*x
- 6*x^2 + x^3) + log(-(x - x^2)/(log(5) + log(5 - x) - 1))*(exp(5)*log(5 - x)*(15*x - 18*x^2 + 3*x^3) - exp(5
)*(15*x - 18*x^2 + 3*x^3) + exp(5)*log(5)*(15*x - 18*x^2 + 3*x^3)) + 6*x^2 - x^3 + log(-(x - x^2)/(log(5) + lo
g(5 - x) - 1))^3*(exp(15)*log(5)*(5*x - 6*x^2 + x^3) - exp(15)*(5*x - 6*x^2 + x^3) + exp(15)*log(5 - x)*(5*x -
6*x^2 + x^3))),x)
[Out]
exp(10)/(exp(5)*log(-(x - x^2)/(log(5) + log(5 - x) - 1)) + 1)^2
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sympy [A] time = 1.03, size = 48, normalized size = 1.37 \begin {gather*} \frac {e^{10}}{e^{10} \log {\left (\frac {x^{2} - x}{\log {\left (5 - x \right )} - 1 + \log {\relax (5 )}} \right )}^{2} + 2 e^{5} \log {\left (\frac {x^{2} - x}{\log {\left (5 - x \right )} - 1 + \log {\relax (5 )}} \right )} + 1} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(((-4*x**2+22*x-10)*exp(5)**3*ln(5-x)+(-4*x**2+22*x-10)*exp(5)**3*ln(5)+(6*x**2-24*x+10)*exp(5)**3)/(
((x**3-6*x**2+5*x)*exp(5)**3*ln(5-x)+(x**3-6*x**2+5*x)*exp(5)**3*ln(5)+(-x**3+6*x**2-5*x)*exp(5)**3)*ln((x**2-
x)/(ln(5-x)+ln(5)-1))**3+((3*x**3-18*x**2+15*x)*exp(5)**2*ln(5-x)+(3*x**3-18*x**2+15*x)*exp(5)**2*ln(5)+(-3*x*
*3+18*x**2-15*x)*exp(5)**2)*ln((x**2-x)/(ln(5-x)+ln(5)-1))**2+((3*x**3-18*x**2+15*x)*exp(5)*ln(5-x)+(3*x**3-18
*x**2+15*x)*exp(5)*ln(5)+(-3*x**3+18*x**2-15*x)*exp(5))*ln((x**2-x)/(ln(5-x)+ln(5)-1))+(x**3-6*x**2+5*x)*ln(5-
x)+(x**3-6*x**2+5*x)*ln(5)-x**3+6*x**2-5*x),x)
[Out]
exp(10)/(exp(10)*log((x**2 - x)/(log(5 - x) - 1 + log(5)))**2 + 2*exp(5)*log((x**2 - x)/(log(5 - x) - 1 + log(
5))) + 1)
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