3.84.36 \(\int \frac {-19 x-12 x^2+e^2 (15+10 x)+(-10 e^2+12 x) \log ^2(5)+(-3 x-2 x^2+e^2 (3+2 x)+(-2 e^2+2 x) \log ^2(5)) \log (-e^2+x)+(2+x-\log ^2(5)) \log (2+x-\log ^2(5))}{-20 x^2-10 x^3+e^2 (20 x+10 x^2)+(-10 e^2 x+10 x^2) \log ^2(5)+(e^2 (-10-5 x)+10 x+5 x^2+(5 e^2-5 x) \log ^2(5)) \log (2+x-\log ^2(5))+\log (-e^2+x) (-4 x^2-2 x^3+e^2 (4 x+2 x^2)+(-2 e^2 x+2 x^2) \log ^2(5)+(e^2 (-2-x)+2 x+x^2+(e^2-x) \log ^2(5)) \log (2+x-\log ^2(5)))} \, dx\) [8336]
Optimal. Leaf size=30 \[ \log \left (\left (5+\log \left (-e^2+x\right )\right ) \left (-x+\frac {1}{2} \log \left (2+x-\log ^2(5)\right )\right )\right ) \]
[Out]
ln((ln(x-exp(2))+5)*(1/2*ln(-ln(5)^2+2+x)-x))
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Rubi [A]
time = 3.12, antiderivative size = 29, normalized size of antiderivative = 0.97, number of steps
used = 7, number of rules used = 6, integrand size = 269, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.022, Rules used = {6820, 6874,
2437, 2339, 29, 6816} \begin {gather*} \log \left (2 x-\log \left (x+2-\log ^2(5)\right )\right )+\log \left (\log \left (x-e^2\right )+5\right ) \end {gather*}
Antiderivative was successfully verified.
[In]
Int[(-19*x - 12*x^2 + E^2*(15 + 10*x) + (-10*E^2 + 12*x)*Log[5]^2 + (-3*x - 2*x^2 + E^2*(3 + 2*x) + (-2*E^2 +
2*x)*Log[5]^2)*Log[-E^2 + x] + (2 + x - Log[5]^2)*Log[2 + x - Log[5]^2])/(-20*x^2 - 10*x^3 + E^2*(20*x + 10*x^
2) + (-10*E^2*x + 10*x^2)*Log[5]^2 + (E^2*(-10 - 5*x) + 10*x + 5*x^2 + (5*E^2 - 5*x)*Log[5]^2)*Log[2 + x - Log
[5]^2] + Log[-E^2 + x]*(-4*x^2 - 2*x^3 + E^2*(4*x + 2*x^2) + (-2*E^2*x + 2*x^2)*Log[5]^2 + (E^2*(-2 - x) + 2*x
+ x^2 + (E^2 - x)*Log[5]^2)*Log[2 + x - Log[5]^2])),x]
[Out]
Log[5 + Log[-E^2 + x]] + Log[2*x - Log[2 + x - Log[5]^2]]
Rule 29
Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]
Rule 2339
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 2437
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_) + (g_.)*(x_))^(q_.), x_Symbol] :> Dist[1/
e, Subst[Int[(f*(x/d))^q*(a + b*Log[c*x^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p, q}, x]
&& EqQ[e*f - d*g, 0]
Rule 6816
Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*Log[RemoveContent[y, x]], x] /; !Fa
lseQ[q]]
Rule 6820
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]
Rule 6874
Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-12 x^2+15 e^2 \left (1-\frac {2 \log ^2(5)}{3}\right )-19 x \left (1-\frac {2}{19} \left (5 e^2+6 \log ^2(5)\right )\right )+\left (e^2-x\right ) \left (3+2 x-2 \log ^2(5)\right ) \log \left (-e^2+x\right )+\left (2+x-\log ^2(5)\right ) \log \left (2+x-\log ^2(5)\right )}{\left (e^2-x\right ) \left (2+x-\log ^2(5)\right ) \left (5+\log \left (-e^2+x\right )\right ) \left (2 x-\log \left (2+x-\log ^2(5)\right )\right )} \, dx\\ &=\int \left (-\frac {1}{\left (e^2-x\right ) \left (5+\log \left (-e^2+x\right )\right )}+\frac {3+2 x-2 \log ^2(5)}{\left (2+x-\log ^2(5)\right ) \left (2 x-\log \left (2+x-\log ^2(5)\right )\right )}\right ) \, dx\\ &=-\int \frac {1}{\left (e^2-x\right ) \left (5+\log \left (-e^2+x\right )\right )} \, dx+\int \frac {3+2 x-2 \log ^2(5)}{\left (2+x-\log ^2(5)\right ) \left (2 x-\log \left (2+x-\log ^2(5)\right )\right )} \, dx\\ &=\log \left (2 x-\log \left (2+x-\log ^2(5)\right )\right )+\text {Subst}\left (\int \frac {1}{x (5+\log (x))} \, dx,x,-e^2+x\right )\\ &=\log \left (2 x-\log \left (2+x-\log ^2(5)\right )\right )+\text {Subst}\left (\int \frac {1}{x} \, dx,x,5+\log \left (-e^2+x\right )\right )\\ &=\log \left (5+\log \left (-e^2+x\right )\right )+\log \left (2 x-\log \left (2+x-\log ^2(5)\right )\right )\\ \end {aligned} \end {gather*}
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Mathematica [A]
time = 0.10, size = 29, normalized size = 0.97 \begin {gather*} \log \left (5+\log \left (-e^2+x\right )\right )+\log \left (2 x-\log \left (2+x-\log ^2(5)\right )\right ) \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[(-19*x - 12*x^2 + E^2*(15 + 10*x) + (-10*E^2 + 12*x)*Log[5]^2 + (-3*x - 2*x^2 + E^2*(3 + 2*x) + (-2*
E^2 + 2*x)*Log[5]^2)*Log[-E^2 + x] + (2 + x - Log[5]^2)*Log[2 + x - Log[5]^2])/(-20*x^2 - 10*x^3 + E^2*(20*x +
10*x^2) + (-10*E^2*x + 10*x^2)*Log[5]^2 + (E^2*(-10 - 5*x) + 10*x + 5*x^2 + (5*E^2 - 5*x)*Log[5]^2)*Log[2 + x
- Log[5]^2] + Log[-E^2 + x]*(-4*x^2 - 2*x^3 + E^2*(4*x + 2*x^2) + (-2*E^2*x + 2*x^2)*Log[5]^2 + (E^2*(-2 - x)
+ 2*x + x^2 + (E^2 - x)*Log[5]^2)*Log[2 + x - Log[5]^2])),x]
[Out]
Log[5 + Log[-E^2 + x]] + Log[2*x - Log[2 + x - Log[5]^2]]
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Maple [A]
time = 4.06, size = 29, normalized size = 0.97
| | |
| method |
result |
size |
| | |
| risch |
\(\ln \left (\ln \left (x -{\mathrm e}^{2}\right )+5\right )+\ln \left (-2 x +\ln \left (-\ln \left (5\right )^{2}+2+x \right )\right )\) |
\(27\) |
| default |
\(\ln \left (\ln \left (x -{\mathrm e}^{2}\right )+5\right )+\ln \left (2 x -\ln \left (-\ln \left (5\right )^{2}+2+x \right )\right )\) |
\(29\) |
| | |
|
|
|
|
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((((-2*exp(2)+2*x)*ln(5)^2+(2*x+3)*exp(2)-2*x^2-3*x)*ln(x-exp(2))+(-ln(5)^2+2+x)*ln(-ln(5)^2+2+x)+(-10*exp(
2)+12*x)*ln(5)^2+(10*x+15)*exp(2)-12*x^2-19*x)/((((exp(2)-x)*ln(5)^2+(-x-2)*exp(2)+x^2+2*x)*ln(-ln(5)^2+2+x)+(
-2*exp(2)*x+2*x^2)*ln(5)^2+(2*x^2+4*x)*exp(2)-2*x^3-4*x^2)*ln(x-exp(2))+((5*exp(2)-5*x)*ln(5)^2+(-5*x-10)*exp(
2)+5*x^2+10*x)*ln(-ln(5)^2+2+x)+(-10*exp(2)*x+10*x^2)*ln(5)^2+(10*x^2+20*x)*exp(2)-10*x^3-20*x^2),x,method=_RE
TURNVERBOSE)
[Out]
ln(ln(x-exp(2))+5)+ln(2*x-ln(-ln(5)^2+2+x))
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Maxima [A]
time = 0.53, size = 26, normalized size = 0.87 \begin {gather*} \log \left (-2 \, x + \log \left (-\log \left (5\right )^{2} + x + 2\right )\right ) + \log \left (\log \left (x - e^{2}\right ) + 5\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((((-2*exp(2)+2*x)*log(5)^2+(3+2*x)*exp(2)-2*x^2-3*x)*log(x-exp(2))+(-log(5)^2+2+x)*log(-log(5)^2+2+x
)+(-10*exp(2)+12*x)*log(5)^2+(10*x+15)*exp(2)-12*x^2-19*x)/((((exp(2)-x)*log(5)^2+(-2-x)*exp(2)+x^2+2*x)*log(-
log(5)^2+2+x)+(-2*exp(2)*x+2*x^2)*log(5)^2+(2*x^2+4*x)*exp(2)-2*x^3-4*x^2)*log(x-exp(2))+((5*exp(2)-5*x)*log(5
)^2+(-5*x-10)*exp(2)+5*x^2+10*x)*log(-log(5)^2+2+x)+(-10*exp(2)*x+10*x^2)*log(5)^2+(10*x^2+20*x)*exp(2)-10*x^3
-20*x^2),x, algorithm="maxima")
[Out]
log(-2*x + log(-log(5)^2 + x + 2)) + log(log(x - e^2) + 5)
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Fricas [A]
time = 0.46, size = 26, normalized size = 0.87 \begin {gather*} \log \left (-2 \, x + \log \left (-\log \left (5\right )^{2} + x + 2\right )\right ) + \log \left (\log \left (x - e^{2}\right ) + 5\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((((-2*exp(2)+2*x)*log(5)^2+(3+2*x)*exp(2)-2*x^2-3*x)*log(x-exp(2))+(-log(5)^2+2+x)*log(-log(5)^2+2+x
)+(-10*exp(2)+12*x)*log(5)^2+(10*x+15)*exp(2)-12*x^2-19*x)/((((exp(2)-x)*log(5)^2+(-2-x)*exp(2)+x^2+2*x)*log(-
log(5)^2+2+x)+(-2*exp(2)*x+2*x^2)*log(5)^2+(2*x^2+4*x)*exp(2)-2*x^3-4*x^2)*log(x-exp(2))+((5*exp(2)-5*x)*log(5
)^2+(-5*x-10)*exp(2)+5*x^2+10*x)*log(-log(5)^2+2+x)+(-10*exp(2)*x+10*x^2)*log(5)^2+(10*x^2+20*x)*exp(2)-10*x^3
-20*x^2),x, algorithm="fricas")
[Out]
log(-2*x + log(-log(5)^2 + x + 2)) + log(log(x - e^2) + 5)
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Sympy [A]
time = 0.25, size = 24, normalized size = 0.80 \begin {gather*} \log {\left (- 2 x + \log {\left (x - \log {\left (5 \right )}^{2} + 2 \right )} \right )} + \log {\left (\log {\left (x - e^{2} \right )} + 5 \right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((((-2*exp(2)+2*x)*ln(5)**2+(3+2*x)*exp(2)-2*x**2-3*x)*ln(x-exp(2))+(-ln(5)**2+2+x)*ln(-ln(5)**2+2+x)
+(-10*exp(2)+12*x)*ln(5)**2+(10*x+15)*exp(2)-12*x**2-19*x)/((((exp(2)-x)*ln(5)**2+(-2-x)*exp(2)+x**2+2*x)*ln(-
ln(5)**2+2+x)+(-2*exp(2)*x+2*x**2)*ln(5)**2+(2*x**2+4*x)*exp(2)-2*x**3-4*x**2)*ln(x-exp(2))+((5*exp(2)-5*x)*ln
(5)**2+(-5*x-10)*exp(2)+5*x**2+10*x)*ln(-ln(5)**2+2+x)+(-10*exp(2)*x+10*x**2)*ln(5)**2+(10*x**2+20*x)*exp(2)-1
0*x**3-20*x**2),x)
[Out]
log(-2*x + log(x - log(5)**2 + 2)) + log(log(x - exp(2)) + 5)
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Giac [A]
time = 0.46, size = 28, normalized size = 0.93 \begin {gather*} \log \left (2 \, x - \log \left (-\log \left (5\right )^{2} + x + 2\right )\right ) + \log \left (\log \left (x - e^{2}\right ) + 5\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((((-2*exp(2)+2*x)*log(5)^2+(3+2*x)*exp(2)-2*x^2-3*x)*log(x-exp(2))+(-log(5)^2+2+x)*log(-log(5)^2+2+x
)+(-10*exp(2)+12*x)*log(5)^2+(10*x+15)*exp(2)-12*x^2-19*x)/((((exp(2)-x)*log(5)^2+(-2-x)*exp(2)+x^2+2*x)*log(-
log(5)^2+2+x)+(-2*exp(2)*x+2*x^2)*log(5)^2+(2*x^2+4*x)*exp(2)-2*x^3-4*x^2)*log(x-exp(2))+((5*exp(2)-5*x)*log(5
)^2+(-5*x-10)*exp(2)+5*x^2+10*x)*log(-log(5)^2+2+x)+(-10*exp(2)*x+10*x^2)*log(5)^2+(10*x^2+20*x)*exp(2)-10*x^3
-20*x^2),x, algorithm="giac")
[Out]
log(2*x - log(-log(5)^2 + x + 2)) + log(log(x - e^2) + 5)
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Mupad [B]
time = 0.62, size = 26, normalized size = 0.87 \begin {gather*} \ln \left (\ln \left (x-{\ln \left (5\right )}^2+2\right )-2\,x\right )+\ln \left (\ln \left (x-{\mathrm {e}}^2\right )+5\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((19*x - log(5)^2*(12*x - 10*exp(2)) - log(x - log(5)^2 + 2)*(x - log(5)^2 + 2) + 12*x^2 + log(x - exp(2))*
(3*x - log(5)^2*(2*x - 2*exp(2)) + 2*x^2 - exp(2)*(2*x + 3)) - exp(2)*(10*x + 15))/(log(x - exp(2))*(4*x^2 - l
og(x - log(5)^2 + 2)*(2*x - exp(2)*(x + 2) - log(5)^2*(x - exp(2)) + x^2) - exp(2)*(4*x + 2*x^2) + 2*x^3 + log
(5)^2*(2*x*exp(2) - 2*x^2)) - log(x - log(5)^2 + 2)*(10*x - log(5)^2*(5*x - 5*exp(2)) + 5*x^2 - exp(2)*(5*x +
10)) - exp(2)*(20*x + 10*x^2) + 20*x^2 + 10*x^3 + log(5)^2*(10*x*exp(2) - 10*x^2)),x)
[Out]
log(log(x - log(5)^2 + 2) - 2*x) + log(log(x - exp(2)) + 5)
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