3.33.72 \(\int \frac {12 x^2+e^x (-60 x-48 x^2+12 x^3+e^{4/x} (-240-12 x+12 x^2))+(-60 x+12 x^2+e^{4/x} (-240-12 x+12 x^2)) \log (5-x)+(e^x (60 x-12 x^2)+(60 x-12 x^2) \log (5-x)) \log (e^x+\log (5-x))}{e^x (-5 x+x^2+e^{8/x} (-5 x+x^2)+e^{4/x} (-10 x+2 x^2))+(-5 x+x^2+e^{8/x} (-5 x+x^2)+e^{4/x} (-10 x+2 x^2)) \log (5-x)+(e^x (10 x-2 x^2+e^{4/x} (10 x-2 x^2))+(10 x-2 x^2+e^{4/x} (10 x-2 x^2)) \log (5-x)) \log (e^x+\log (5-x))+(e^x (-5 x+x^2)+(-5 x+x^2) \log (5-x)) \log ^2(e^x+\log (5-x))} \, dx\)
Optimal. Leaf size=27 \[ \frac {12 x}{1+e^{4/x}-\log \left (e^x+\log (5-x)\right )} \]
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Rubi [A] time = 2.99, antiderivative size = 27, normalized size of antiderivative = 1.00,
number of steps used = 3, number of rules used = 3, integrand size = 327, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.009, Rules used
= {6688, 12, 6687} \begin {gather*} \frac {12 x}{e^{4/x}-\log \left (e^x+\log (5-x)\right )+1} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[(12*x^2 + E^x*(-60*x - 48*x^2 + 12*x^3 + E^(4/x)*(-240 - 12*x + 12*x^2)) + (-60*x + 12*x^2 + E^(4/x)*(-240
- 12*x + 12*x^2))*Log[5 - x] + (E^x*(60*x - 12*x^2) + (60*x - 12*x^2)*Log[5 - x])*Log[E^x + Log[5 - x]])/(E^x
*(-5*x + x^2 + E^(8/x)*(-5*x + x^2) + E^(4/x)*(-10*x + 2*x^2)) + (-5*x + x^2 + E^(8/x)*(-5*x + x^2) + E^(4/x)*
(-10*x + 2*x^2))*Log[5 - x] + (E^x*(10*x - 2*x^2 + E^(4/x)*(10*x - 2*x^2)) + (10*x - 2*x^2 + E^(4/x)*(10*x - 2
*x^2))*Log[5 - x])*Log[E^x + Log[5 - x]] + (E^x*(-5*x + x^2) + (-5*x + x^2)*Log[5 - x])*Log[E^x + Log[5 - x]]^
2),x]
[Out]
(12*x)/(1 + E^(4/x) - Log[E^x + Log[5 - 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 6687
Int[(u_)*(y_)^(m_.)*(z_)^(n_.), x_Symbol] :> With[{q = DerivativeDivides[y*z, u*z^(n - m), x]}, Simp[(q*y^(m +
1)*z^(m + 1))/(m + 1), x] /; !FalseQ[q]] /; FreeQ[{m, n}, 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 {12 \left (20 e^{\frac {4}{x}+x}+5 e^x x+e^{\frac {4}{x}+x} x-x^2+4 e^x x^2-e^{\frac {4}{x}+x} x^2-e^x x^3+e^x (-5+x) x \log \left (e^x+\log (5-x)\right )-(-5+x) \log (5-x) \left (x+e^{4/x} (4+x)-x \log \left (e^x+\log (5-x)\right )\right )\right )}{(5-x) x \left (e^x+\log (5-x)\right ) \left (1+e^{4/x}-\log \left (e^x+\log (5-x)\right )\right )^2} \, dx\\ &=12 \int \frac {20 e^{\frac {4}{x}+x}+5 e^x x+e^{\frac {4}{x}+x} x-x^2+4 e^x x^2-e^{\frac {4}{x}+x} x^2-e^x x^3+e^x (-5+x) x \log \left (e^x+\log (5-x)\right )-(-5+x) \log (5-x) \left (x+e^{4/x} (4+x)-x \log \left (e^x+\log (5-x)\right )\right )}{(5-x) x \left (e^x+\log (5-x)\right ) \left (1+e^{4/x}-\log \left (e^x+\log (5-x)\right )\right )^2} \, dx\\ &=\frac {12 x}{1+e^{4/x}-\log \left (e^x+\log (5-x)\right )}\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.21, size = 27, normalized size = 1.00 \begin {gather*} -\frac {12 x}{-1-e^{4/x}+\log \left (e^x+\log (5-x)\right )} \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[(12*x^2 + E^x*(-60*x - 48*x^2 + 12*x^3 + E^(4/x)*(-240 - 12*x + 12*x^2)) + (-60*x + 12*x^2 + E^(4/x)
*(-240 - 12*x + 12*x^2))*Log[5 - x] + (E^x*(60*x - 12*x^2) + (60*x - 12*x^2)*Log[5 - x])*Log[E^x + Log[5 - x]]
)/(E^x*(-5*x + x^2 + E^(8/x)*(-5*x + x^2) + E^(4/x)*(-10*x + 2*x^2)) + (-5*x + x^2 + E^(8/x)*(-5*x + x^2) + E^
(4/x)*(-10*x + 2*x^2))*Log[5 - x] + (E^x*(10*x - 2*x^2 + E^(4/x)*(10*x - 2*x^2)) + (10*x - 2*x^2 + E^(4/x)*(10
*x - 2*x^2))*Log[5 - x])*Log[E^x + Log[5 - x]] + (E^x*(-5*x + x^2) + (-5*x + x^2)*Log[5 - x])*Log[E^x + Log[5
- x]]^2),x]
[Out]
(-12*x)/(-1 - E^(4/x) + Log[E^x + Log[5 - x]])
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fricas [A] time = 0.58, size = 25, normalized size = 0.93 \begin {gather*} \frac {12 \, x}{e^{\frac {4}{x}} - \log \left (e^{x} + \log \left (-x + 5\right )\right ) + 1} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((((-12*x^2+60*x)*log(5-x)+(-12*x^2+60*x)*exp(x))*log(log(5-x)+exp(x))+((12*x^2-12*x-240)*exp(4/x)+12
*x^2-60*x)*log(5-x)+((12*x^2-12*x-240)*exp(4/x)+12*x^3-48*x^2-60*x)*exp(x)+12*x^2)/(((x^2-5*x)*log(5-x)+(x^2-5
*x)*exp(x))*log(log(5-x)+exp(x))^2+(((-2*x^2+10*x)*exp(4/x)-2*x^2+10*x)*log(5-x)+((-2*x^2+10*x)*exp(4/x)-2*x^2
+10*x)*exp(x))*log(log(5-x)+exp(x))+((x^2-5*x)*exp(4/x)^2+(2*x^2-10*x)*exp(4/x)+x^2-5*x)*log(5-x)+((x^2-5*x)*e
xp(4/x)^2+(2*x^2-10*x)*exp(4/x)+x^2-5*x)*exp(x)),x, algorithm="fricas")
[Out]
12*x/(e^(4/x) - log(e^x + log(-x + 5)) + 1)
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giac [A] time = 0.97, size = 25, normalized size = 0.93 \begin {gather*} \frac {12 \, x}{e^{\frac {4}{x}} - \log \left (e^{x} + \log \left (-x + 5\right )\right ) + 1} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((((-12*x^2+60*x)*log(5-x)+(-12*x^2+60*x)*exp(x))*log(log(5-x)+exp(x))+((12*x^2-12*x-240)*exp(4/x)+12
*x^2-60*x)*log(5-x)+((12*x^2-12*x-240)*exp(4/x)+12*x^3-48*x^2-60*x)*exp(x)+12*x^2)/(((x^2-5*x)*log(5-x)+(x^2-5
*x)*exp(x))*log(log(5-x)+exp(x))^2+(((-2*x^2+10*x)*exp(4/x)-2*x^2+10*x)*log(5-x)+((-2*x^2+10*x)*exp(4/x)-2*x^2
+10*x)*exp(x))*log(log(5-x)+exp(x))+((x^2-5*x)*exp(4/x)^2+(2*x^2-10*x)*exp(4/x)+x^2-5*x)*log(5-x)+((x^2-5*x)*e
xp(4/x)^2+(2*x^2-10*x)*exp(4/x)+x^2-5*x)*exp(x)),x, algorithm="giac")
[Out]
12*x/(e^(4/x) - log(e^x + log(-x + 5)) + 1)
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maple [A] time = 0.06, size = 26, normalized size = 0.96
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\(\frac {12 x}{{\mathrm e}^{\frac {4}{x}}-\ln \left (\ln \left (5-x \right )+{\mathrm e}^{x}\right )+1}\) |
\(26\) |
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Verification of antiderivative is not currently implemented for this CAS.
[In]
int((((-12*x^2+60*x)*ln(5-x)+(-12*x^2+60*x)*exp(x))*ln(ln(5-x)+exp(x))+((12*x^2-12*x-240)*exp(4/x)+12*x^2-60*x
)*ln(5-x)+((12*x^2-12*x-240)*exp(4/x)+12*x^3-48*x^2-60*x)*exp(x)+12*x^2)/(((x^2-5*x)*ln(5-x)+(x^2-5*x)*exp(x))
*ln(ln(5-x)+exp(x))^2+(((-2*x^2+10*x)*exp(4/x)-2*x^2+10*x)*ln(5-x)+((-2*x^2+10*x)*exp(4/x)-2*x^2+10*x)*exp(x))
*ln(ln(5-x)+exp(x))+((x^2-5*x)*exp(4/x)^2+(2*x^2-10*x)*exp(4/x)+x^2-5*x)*ln(5-x)+((x^2-5*x)*exp(4/x)^2+(2*x^2-
10*x)*exp(4/x)+x^2-5*x)*exp(x)),x,method=_RETURNVERBOSE)
[Out]
12*x/(exp(4/x)-ln(ln(5-x)+exp(x))+1)
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maxima [A] time = 1.01, size = 25, normalized size = 0.93 \begin {gather*} \frac {12 \, x}{e^{\frac {4}{x}} - \log \left (e^{x} + \log \left (-x + 5\right )\right ) + 1} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((((-12*x^2+60*x)*log(5-x)+(-12*x^2+60*x)*exp(x))*log(log(5-x)+exp(x))+((12*x^2-12*x-240)*exp(4/x)+12
*x^2-60*x)*log(5-x)+((12*x^2-12*x-240)*exp(4/x)+12*x^3-48*x^2-60*x)*exp(x)+12*x^2)/(((x^2-5*x)*log(5-x)+(x^2-5
*x)*exp(x))*log(log(5-x)+exp(x))^2+(((-2*x^2+10*x)*exp(4/x)-2*x^2+10*x)*log(5-x)+((-2*x^2+10*x)*exp(4/x)-2*x^2
+10*x)*exp(x))*log(log(5-x)+exp(x))+((x^2-5*x)*exp(4/x)^2+(2*x^2-10*x)*exp(4/x)+x^2-5*x)*log(5-x)+((x^2-5*x)*e
xp(4/x)^2+(2*x^2-10*x)*exp(4/x)+x^2-5*x)*exp(x)),x, algorithm="maxima")
[Out]
12*x/(e^(4/x) - log(e^x + log(-x + 5)) + 1)
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mupad [F] time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int -\frac {\ln \left (\ln \left (5-x\right )+{\mathrm {e}}^x\right )\,\left (\ln \left (5-x\right )\,\left (60\,x-12\,x^2\right )+{\mathrm {e}}^x\,\left (60\,x-12\,x^2\right )\right )-{\mathrm {e}}^x\,\left (60\,x+{\mathrm {e}}^{4/x}\,\left (-12\,x^2+12\,x+240\right )+48\,x^2-12\,x^3\right )+12\,x^2-\ln \left (5-x\right )\,\left (60\,x+{\mathrm {e}}^{4/x}\,\left (-12\,x^2+12\,x+240\right )-12\,x^2\right )}{\left (\ln \left (5-x\right )\,\left (5\,x-x^2\right )+{\mathrm {e}}^x\,\left (5\,x-x^2\right )\right )\,{\ln \left (\ln \left (5-x\right )+{\mathrm {e}}^x\right )}^2+\left (-{\mathrm {e}}^x\,\left (10\,x+{\mathrm {e}}^{4/x}\,\left (10\,x-2\,x^2\right )-2\,x^2\right )-\ln \left (5-x\right )\,\left (10\,x+{\mathrm {e}}^{4/x}\,\left (10\,x-2\,x^2\right )-2\,x^2\right )\right )\,\ln \left (\ln \left (5-x\right )+{\mathrm {e}}^x\right )+\ln \left (5-x\right )\,\left (5\,x+{\mathrm {e}}^{8/x}\,\left (5\,x-x^2\right )+{\mathrm {e}}^{4/x}\,\left (10\,x-2\,x^2\right )-x^2\right )+{\mathrm {e}}^x\,\left (5\,x+{\mathrm {e}}^{8/x}\,\left (5\,x-x^2\right )+{\mathrm {e}}^{4/x}\,\left (10\,x-2\,x^2\right )-x^2\right )} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(-(log(log(5 - x) + exp(x))*(log(5 - x)*(60*x - 12*x^2) + exp(x)*(60*x - 12*x^2)) - exp(x)*(60*x + exp(4/x)
*(12*x - 12*x^2 + 240) + 48*x^2 - 12*x^3) + 12*x^2 - log(5 - x)*(60*x + exp(4/x)*(12*x - 12*x^2 + 240) - 12*x^
2))/(log(5 - x)*(5*x + exp(8/x)*(5*x - x^2) + exp(4/x)*(10*x - 2*x^2) - x^2) + exp(x)*(5*x + exp(8/x)*(5*x - x
^2) + exp(4/x)*(10*x - 2*x^2) - x^2) + log(log(5 - x) + exp(x))^2*(log(5 - x)*(5*x - x^2) + exp(x)*(5*x - x^2)
) - log(log(5 - x) + exp(x))*(exp(x)*(10*x + exp(4/x)*(10*x - 2*x^2) - 2*x^2) + log(5 - x)*(10*x + exp(4/x)*(1
0*x - 2*x^2) - 2*x^2))),x)
[Out]
int(-(log(log(5 - x) + exp(x))*(log(5 - x)*(60*x - 12*x^2) + exp(x)*(60*x - 12*x^2)) - exp(x)*(60*x + exp(4/x)
*(12*x - 12*x^2 + 240) + 48*x^2 - 12*x^3) + 12*x^2 - log(5 - x)*(60*x + exp(4/x)*(12*x - 12*x^2 + 240) - 12*x^
2))/(log(5 - x)*(5*x + exp(8/x)*(5*x - x^2) + exp(4/x)*(10*x - 2*x^2) - x^2) + exp(x)*(5*x + exp(8/x)*(5*x - x
^2) + exp(4/x)*(10*x - 2*x^2) - x^2) + log(log(5 - x) + exp(x))^2*(log(5 - x)*(5*x - x^2) + exp(x)*(5*x - x^2)
) - log(log(5 - x) + exp(x))*(exp(x)*(10*x + exp(4/x)*(10*x - 2*x^2) - 2*x^2) + log(5 - x)*(10*x + exp(4/x)*(1
0*x - 2*x^2) - 2*x^2))), x)
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sympy [A] time = 3.67, size = 20, normalized size = 0.74 \begin {gather*} - \frac {12 x}{- e^{\frac {4}{x}} + \log {\left (e^{x} + \log {\left (5 - x \right )} \right )} - 1} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((((-12*x**2+60*x)*ln(5-x)+(-12*x**2+60*x)*exp(x))*ln(ln(5-x)+exp(x))+((12*x**2-12*x-240)*exp(4/x)+12
*x**2-60*x)*ln(5-x)+((12*x**2-12*x-240)*exp(4/x)+12*x**3-48*x**2-60*x)*exp(x)+12*x**2)/(((x**2-5*x)*ln(5-x)+(x
**2-5*x)*exp(x))*ln(ln(5-x)+exp(x))**2+(((-2*x**2+10*x)*exp(4/x)-2*x**2+10*x)*ln(5-x)+((-2*x**2+10*x)*exp(4/x)
-2*x**2+10*x)*exp(x))*ln(ln(5-x)+exp(x))+((x**2-5*x)*exp(4/x)**2+(2*x**2-10*x)*exp(4/x)+x**2-5*x)*ln(5-x)+((x*
*2-5*x)*exp(4/x)**2+(2*x**2-10*x)*exp(4/x)+x**2-5*x)*exp(x)),x)
[Out]
-12*x/(-exp(4/x) + log(exp(x) + log(5 - x)) - 1)
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