3.2.24 \(\int \frac {-900+e^x (18000-900 x)}{-2500+e^x (2500-125 x)+125 x+(2500-125 x) \log (20-x)+(1500-75 x+e^x (-1500+75 x)+(-1500+75 x) \log (20-x)) \log (1-2 e^x+e^{2 x}+(-2+2 e^x) \log (20-x)+\log ^2(20-x))+(-300+e^x (300-15 x)+15 x+(300-15 x) \log (20-x)) \log ^2(1-2 e^x+e^{2 x}+(-2+2 e^x) \log (20-x)+\log ^2(20-x))+(20+e^x (-20+x)-x+(-20+x) \log (20-x)) \log ^3(1-2 e^x+e^{2 x}+(-2+2 e^x) \log (20-x)+\log ^2(20-x))} \, dx\)
Optimal. Leaf size=30 \[ \frac {25}{\left (-2+\frac {1}{3} \left (1+\log \left (\left (1-e^x-\log (20-x)\right )^2\right )\right )\right )^2} \]
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Rubi [A] time = 0.43, antiderivative size = 22, normalized size of antiderivative = 0.73,
number of steps used = 3, number of rules used = 3, integrand size = 226, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.013, Rules used
= {6688, 12, 6686} \begin {gather*} \frac {225}{\left (5-\log \left (\left (e^x+\log (20-x)-1\right )^2\right )\right )^2} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[(-900 + E^x*(18000 - 900*x))/(-2500 + E^x*(2500 - 125*x) + 125*x + (2500 - 125*x)*Log[20 - x] + (1500 - 75
*x + E^x*(-1500 + 75*x) + (-1500 + 75*x)*Log[20 - x])*Log[1 - 2*E^x + E^(2*x) + (-2 + 2*E^x)*Log[20 - x] + Log
[20 - x]^2] + (-300 + E^x*(300 - 15*x) + 15*x + (300 - 15*x)*Log[20 - x])*Log[1 - 2*E^x + E^(2*x) + (-2 + 2*E^
x)*Log[20 - x] + Log[20 - x]^2]^2 + (20 + E^x*(-20 + x) - x + (-20 + x)*Log[20 - x])*Log[1 - 2*E^x + E^(2*x) +
(-2 + 2*E^x)*Log[20 - x] + Log[20 - x]^2]^3),x]
[Out]
225/(5 - Log[(-1 + E^x + Log[20 - x])^2])^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 {900 \left (1+e^x (-20+x)\right )}{(20-x) \left (1-e^x-\log (20-x)\right ) \left (5-\log \left (\left (-1+e^x+\log (20-x)\right )^2\right )\right )^3} \, dx\\ &=900 \int \frac {1+e^x (-20+x)}{(20-x) \left (1-e^x-\log (20-x)\right ) \left (5-\log \left (\left (-1+e^x+\log (20-x)\right )^2\right )\right )^3} \, dx\\ &=\frac {225}{\left (5-\log \left (\left (-1+e^x+\log (20-x)\right )^2\right )\right )^2}\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.04, size = 20, normalized size = 0.67 \begin {gather*} \frac {225}{\left (-5+\log \left (\left (-1+e^x+\log (20-x)\right )^2\right )\right )^2} \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[(-900 + E^x*(18000 - 900*x))/(-2500 + E^x*(2500 - 125*x) + 125*x + (2500 - 125*x)*Log[20 - x] + (150
0 - 75*x + E^x*(-1500 + 75*x) + (-1500 + 75*x)*Log[20 - x])*Log[1 - 2*E^x + E^(2*x) + (-2 + 2*E^x)*Log[20 - x]
+ Log[20 - x]^2] + (-300 + E^x*(300 - 15*x) + 15*x + (300 - 15*x)*Log[20 - x])*Log[1 - 2*E^x + E^(2*x) + (-2
+ 2*E^x)*Log[20 - x] + Log[20 - x]^2]^2 + (20 + E^x*(-20 + x) - x + (-20 + x)*Log[20 - x])*Log[1 - 2*E^x + E^(
2*x) + (-2 + 2*E^x)*Log[20 - x] + Log[20 - x]^2]^3),x]
[Out]
225/(-5 + Log[(-1 + E^x + Log[20 - x])^2])^2
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fricas [B] time = 0.57, size = 72, normalized size = 2.40 \begin {gather*} \frac {225}{\log \left (2 \, {\left (e^{x} - 1\right )} \log \left (-x + 20\right ) + \log \left (-x + 20\right )^{2} + e^{\left (2 \, x\right )} - 2 \, e^{x} + 1\right )^{2} - 10 \, \log \left (2 \, {\left (e^{x} - 1\right )} \log \left (-x + 20\right ) + \log \left (-x + 20\right )^{2} + e^{\left (2 \, x\right )} - 2 \, e^{x} + 1\right ) + 25} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(((-900*x+18000)*exp(x)-900)/(((x-20)*log(-x+20)+(x-20)*exp(x)-x+20)*log(log(-x+20)^2+(2*exp(x)-2)*lo
g(-x+20)+exp(x)^2-2*exp(x)+1)^3+((-15*x+300)*log(-x+20)+(-15*x+300)*exp(x)+15*x-300)*log(log(-x+20)^2+(2*exp(x
)-2)*log(-x+20)+exp(x)^2-2*exp(x)+1)^2+((75*x-1500)*log(-x+20)+(75*x-1500)*exp(x)-75*x+1500)*log(log(-x+20)^2+
(2*exp(x)-2)*log(-x+20)+exp(x)^2-2*exp(x)+1)+(-125*x+2500)*log(-x+20)+(-125*x+2500)*exp(x)+125*x-2500),x, algo
rithm="fricas")
[Out]
225/(log(2*(e^x - 1)*log(-x + 20) + log(-x + 20)^2 + e^(2*x) - 2*e^x + 1)^2 - 10*log(2*(e^x - 1)*log(-x + 20)
+ log(-x + 20)^2 + e^(2*x) - 2*e^x + 1) + 25)
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giac [B] time = 1.04, size = 84, normalized size = 2.80 \begin {gather*} \frac {225}{\log \left (2 \, e^{x} \log \left (-x + 20\right ) + \log \left (-x + 20\right )^{2} + e^{\left (2 \, x\right )} - 2 \, e^{x} - 2 \, \log \left (-x + 20\right ) + 1\right )^{2} - 10 \, \log \left (2 \, e^{x} \log \left (-x + 20\right ) + \log \left (-x + 20\right )^{2} + e^{\left (2 \, x\right )} - 2 \, e^{x} - 2 \, \log \left (-x + 20\right ) + 1\right ) + 25} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(((-900*x+18000)*exp(x)-900)/(((x-20)*log(-x+20)+(x-20)*exp(x)-x+20)*log(log(-x+20)^2+(2*exp(x)-2)*lo
g(-x+20)+exp(x)^2-2*exp(x)+1)^3+((-15*x+300)*log(-x+20)+(-15*x+300)*exp(x)+15*x-300)*log(log(-x+20)^2+(2*exp(x
)-2)*log(-x+20)+exp(x)^2-2*exp(x)+1)^2+((75*x-1500)*log(-x+20)+(75*x-1500)*exp(x)-75*x+1500)*log(log(-x+20)^2+
(2*exp(x)-2)*log(-x+20)+exp(x)^2-2*exp(x)+1)+(-125*x+2500)*log(-x+20)+(-125*x+2500)*exp(x)+125*x-2500),x, algo
rithm="giac")
[Out]
225/(log(2*e^x*log(-x + 20) + log(-x + 20)^2 + e^(2*x) - 2*e^x - 2*log(-x + 20) + 1)^2 - 10*log(2*e^x*log(-x +
20) + log(-x + 20)^2 + e^(2*x) - 2*e^x - 2*log(-x + 20) + 1) + 25)
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maple [C] time = 0.24, size = 111, normalized size = 3.70
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\(-\frac {900}{\left (\pi \mathrm {csgn}\left (i \left (-1+{\mathrm e}^{x}+\ln \left (-x +20\right )\right )\right )^{2} \mathrm {csgn}\left (i \left (-1+{\mathrm e}^{x}+\ln \left (-x +20\right )\right )^{2}\right )-2 \pi \,\mathrm {csgn}\left (i \left (-1+{\mathrm e}^{x}+\ln \left (-x +20\right )\right )\right ) \mathrm {csgn}\left (i \left (-1+{\mathrm e}^{x}+\ln \left (-x +20\right )\right )^{2}\right )^{2}+\pi \mathrm {csgn}\left (i \left (-1+{\mathrm e}^{x}+\ln \left (-x +20\right )\right )^{2}\right )^{3}+4 i \ln \left (-1+{\mathrm e}^{x}+\ln \left (-x +20\right )\right )-10 i\right )^{2}}\) |
\(111\) |
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Verification of antiderivative is not currently implemented for this CAS.
[In]
int(((-900*x+18000)*exp(x)-900)/(((x-20)*ln(-x+20)+(x-20)*exp(x)-x+20)*ln(ln(-x+20)^2+(2*exp(x)-2)*ln(-x+20)+e
xp(x)^2-2*exp(x)+1)^3+((-15*x+300)*ln(-x+20)+(-15*x+300)*exp(x)+15*x-300)*ln(ln(-x+20)^2+(2*exp(x)-2)*ln(-x+20
)+exp(x)^2-2*exp(x)+1)^2+((75*x-1500)*ln(-x+20)+(75*x-1500)*exp(x)-75*x+1500)*ln(ln(-x+20)^2+(2*exp(x)-2)*ln(-
x+20)+exp(x)^2-2*exp(x)+1)+(-125*x+2500)*ln(-x+20)+(-125*x+2500)*exp(x)+125*x-2500),x,method=_RETURNVERBOSE)
[Out]
-900/(Pi*csgn(I*(-1+exp(x)+ln(-x+20)))^2*csgn(I*(-1+exp(x)+ln(-x+20))^2)-2*Pi*csgn(I*(-1+exp(x)+ln(-x+20)))*cs
gn(I*(-1+exp(x)+ln(-x+20))^2)^2+Pi*csgn(I*(-1+exp(x)+ln(-x+20))^2)^3+4*I*ln(-1+exp(x)+ln(-x+20))-10*I)^2
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maxima [A] time = 1.84, size = 34, normalized size = 1.13 \begin {gather*} \frac {225}{4 \, \log \left (e^{x} + \log \left (-x + 20\right ) - 1\right )^{2} - 20 \, \log \left (e^{x} + \log \left (-x + 20\right ) - 1\right ) + 25} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(((-900*x+18000)*exp(x)-900)/(((x-20)*log(-x+20)+(x-20)*exp(x)-x+20)*log(log(-x+20)^2+(2*exp(x)-2)*lo
g(-x+20)+exp(x)^2-2*exp(x)+1)^3+((-15*x+300)*log(-x+20)+(-15*x+300)*exp(x)+15*x-300)*log(log(-x+20)^2+(2*exp(x
)-2)*log(-x+20)+exp(x)^2-2*exp(x)+1)^2+((75*x-1500)*log(-x+20)+(75*x-1500)*exp(x)-75*x+1500)*log(log(-x+20)^2+
(2*exp(x)-2)*log(-x+20)+exp(x)^2-2*exp(x)+1)+(-125*x+2500)*log(-x+20)+(-125*x+2500)*exp(x)+125*x-2500),x, algo
rithm="maxima")
[Out]
225/(4*log(e^x + log(-x + 20) - 1)^2 - 20*log(e^x + log(-x + 20) - 1) + 25)
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mupad [B] time = 0.37, size = 38, normalized size = 1.27 \begin {gather*} \frac {225}{{\left (\ln \left ({\ln \left (20-x\right )}^2+\left (2\,{\mathrm {e}}^x-2\right )\,\ln \left (20-x\right )+{\mathrm {e}}^{2\,x}-2\,{\mathrm {e}}^x+1\right )-5\right )}^2} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((exp(x)*(900*x - 18000) + 900)/(log(20 - x)*(125*x - 2500) - log(exp(2*x) - 2*exp(x) + log(20 - x)^2 + log
(20 - x)*(2*exp(x) - 2) + 1)*(log(20 - x)*(75*x - 1500) - 75*x + exp(x)*(75*x - 1500) + 1500) - 125*x + exp(x)
*(125*x - 2500) + log(exp(2*x) - 2*exp(x) + log(20 - x)^2 + log(20 - x)*(2*exp(x) - 2) + 1)^2*(log(20 - x)*(15
*x - 300) - 15*x + exp(x)*(15*x - 300) + 300) - log(exp(2*x) - 2*exp(x) + log(20 - x)^2 + log(20 - x)*(2*exp(x
) - 2) + 1)^3*(exp(x)*(x - 20) - x + log(20 - x)*(x - 20) + 20) + 2500),x)
[Out]
225/(log(exp(2*x) - 2*exp(x) + log(20 - x)^2 + log(20 - x)*(2*exp(x) - 2) + 1) - 5)^2
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sympy [B] time = 1.21, size = 70, normalized size = 2.33 \begin {gather*} \frac {225}{\log {\left (\left (2 e^{x} - 2\right ) \log {\left (20 - x \right )} + e^{2 x} - 2 e^{x} + \log {\left (20 - x \right )}^{2} + 1 \right )}^{2} - 10 \log {\left (\left (2 e^{x} - 2\right ) \log {\left (20 - x \right )} + e^{2 x} - 2 e^{x} + \log {\left (20 - x \right )}^{2} + 1 \right )} + 25} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(((-900*x+18000)*exp(x)-900)/(((x-20)*ln(-x+20)+(x-20)*exp(x)-x+20)*ln(ln(-x+20)**2+(2*exp(x)-2)*ln(-
x+20)+exp(x)**2-2*exp(x)+1)**3+((-15*x+300)*ln(-x+20)+(-15*x+300)*exp(x)+15*x-300)*ln(ln(-x+20)**2+(2*exp(x)-2
)*ln(-x+20)+exp(x)**2-2*exp(x)+1)**2+((75*x-1500)*ln(-x+20)+(75*x-1500)*exp(x)-75*x+1500)*ln(ln(-x+20)**2+(2*e
xp(x)-2)*ln(-x+20)+exp(x)**2-2*exp(x)+1)+(-125*x+2500)*ln(-x+20)+(-125*x+2500)*exp(x)+125*x-2500),x)
[Out]
225/(log((2*exp(x) - 2)*log(20 - x) + exp(2*x) - 2*exp(x) + log(20 - x)**2 + 1)**2 - 10*log((2*exp(x) - 2)*log
(20 - x) + exp(2*x) - 2*exp(x) + log(20 - x)**2 + 1) + 25)
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