3.100.98 \(\int \frac {e^x (-16+32 x-18 x^2)+e^x (32 x-32 x^2) \log (\frac {x}{5})+(e^x x^3+e^x (16 x-32 x^2+16 x^3) \log (\frac {x}{5})) \log (\frac {1}{16} (x^2+(16-32 x+16 x^2) \log (\frac {x}{5})))}{(x^3+(16 x-32 x^2+16 x^3) \log (\frac {x}{5})) \log ^2(\frac {1}{16} (x^2+(16-32 x+16 x^2) \log (\frac {x}{5})))} \, dx\)
Optimal. Leaf size=27 \[ \frac {e^x}{\log \left (\frac {x^2}{16}+(-1+x)^2 \log \left (\frac {x}{5}\right )\right )} \]
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Rubi [F] time = 14.78, antiderivative size = 0, normalized size of antiderivative = 0.00,
number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used =
{} \begin {gather*} \int \frac {e^x \left (-16+32 x-18 x^2\right )+e^x \left (32 x-32 x^2\right ) \log \left (\frac {x}{5}\right )+\left (e^x x^3+e^x \left (16 x-32 x^2+16 x^3\right ) \log \left (\frac {x}{5}\right )\right ) \log \left (\frac {1}{16} \left (x^2+\left (16-32 x+16 x^2\right ) \log \left (\frac {x}{5}\right )\right )\right )}{\left (x^3+\left (16 x-32 x^2+16 x^3\right ) \log \left (\frac {x}{5}\right )\right ) \log ^2\left (\frac {1}{16} \left (x^2+\left (16-32 x+16 x^2\right ) \log \left (\frac {x}{5}\right )\right )\right )} \, dx \end {gather*}
Verification is not applicable to the result.
[In]
Int[(E^x*(-16 + 32*x - 18*x^2) + E^x*(32*x - 32*x^2)*Log[x/5] + (E^x*x^3 + E^x*(16*x - 32*x^2 + 16*x^3)*Log[x/
5])*Log[(x^2 + (16 - 32*x + 16*x^2)*Log[x/5])/16])/((x^3 + (16*x - 32*x^2 + 16*x^3)*Log[x/5])*Log[(x^2 + (16 -
32*x + 16*x^2)*Log[x/5])/16]^2),x]
[Out]
-16*Defer[Int][E^x/(x*(x^2 + 16*(-1 + x)^2*Log[x/5])*Log[(x^2 + 16*(-1 + x)^2*Log[x/5])/16]^2), x] - 18*Defer[
Int][(E^x*x)/((x^2 + 16*(-1 + x)^2*Log[x/5])*Log[(x^2 + 16*(-1 + x)^2*Log[x/5])/16]^2), x] - 32*Defer[Int][(E^
x*x*Log[x/5])/((x^2 + 16*(-1 + x)^2*Log[x/5])*Log[(x^2 + 16*(-1 + x)^2*Log[x/5])/16]^2), x] + Defer[Int][(E^x*
x^2)/((x^2 + 16*(-1 + x)^2*Log[x/5])*Log[(x^2 + 16*(-1 + x)^2*Log[x/5])/16]), x] - 32*Defer[Int][(E^x*x*Log[x/
5])/((x^2 + 16*(-1 + x)^2*Log[x/5])*Log[(x^2 + 16*(-1 + x)^2*Log[x/5])/16]), x] + 16*Defer[Int][(E^x*x^2*Log[x
/5])/((x^2 + 16*(-1 + x)^2*Log[x/5])*Log[(x^2 + 16*(-1 + x)^2*Log[x/5])/16]), x] + 160*Defer[Subst][Defer[Int]
[E^(5*x)/((25*x^2 + 16*(-1 + 5*x)^2*Log[x])*Log[(25*x^2 + 16*(-1 + 5*x)^2*Log[x])/16]^2), x], x, x/5] + 160*De
fer[Subst][Defer[Int][(E^(5*x)*Log[x])/((25*x^2 + 16*(-1 + 5*x)^2*Log[x])*Log[(25*x^2 + 16*(-1 + 5*x)^2*Log[x]
)/16]^2), x], x, x/5] + 80*Defer[Subst][Defer[Int][(E^(5*x)*Log[x])/((25*x^2 + 16*(-1 + 5*x)^2*Log[x])*Log[(25
*x^2 + 16*(-1 + 5*x)^2*Log[x])/16]), x], x, x/5]
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}
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Mathematica [A] time = 0.12, size = 28, normalized size = 1.04 \begin {gather*} \frac {e^x}{\log \left (\frac {1}{16} \left (x^2+16 (-1+x)^2 \log \left (\frac {x}{5}\right )\right )\right )} \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[(E^x*(-16 + 32*x - 18*x^2) + E^x*(32*x - 32*x^2)*Log[x/5] + (E^x*x^3 + E^x*(16*x - 32*x^2 + 16*x^3)*
Log[x/5])*Log[(x^2 + (16 - 32*x + 16*x^2)*Log[x/5])/16])/((x^3 + (16*x - 32*x^2 + 16*x^3)*Log[x/5])*Log[(x^2 +
(16 - 32*x + 16*x^2)*Log[x/5])/16]^2),x]
[Out]
E^x/Log[(x^2 + 16*(-1 + x)^2*Log[x/5])/16]
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fricas [A] time = 1.45, size = 25, normalized size = 0.93 \begin {gather*} \frac {e^{x}}{\log \left (\frac {1}{16} \, x^{2} + {\left (x^{2} - 2 \, x + 1\right )} \log \left (\frac {1}{5} \, x\right )\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((((16*x^3-32*x^2+16*x)*exp(x)*log(1/5*x)+exp(x)*x^3)*log(1/16*(16*x^2-32*x+16)*log(1/5*x)+1/16*x^2)+
(-32*x^2+32*x)*exp(x)*log(1/5*x)+(-18*x^2+32*x-16)*exp(x))/((16*x^3-32*x^2+16*x)*log(1/5*x)+x^3)/log(1/16*(16*
x^2-32*x+16)*log(1/5*x)+1/16*x^2)^2,x, algorithm="fricas")
[Out]
e^x/log(1/16*x^2 + (x^2 - 2*x + 1)*log(1/5*x))
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giac [B] time = 1.21, size = 1234, normalized size = 45.70 result too large to
display
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((((16*x^3-32*x^2+16*x)*exp(x)*log(1/5*x)+exp(x)*x^3)*log(1/16*(16*x^2-32*x+16)*log(1/5*x)+1/16*x^2)+
(-32*x^2+32*x)*exp(x)*log(1/5*x)+(-18*x^2+32*x-16)*exp(x))/((16*x^3-32*x^2+16*x)*log(1/5*x)+x^3)/log(1/16*(16*
x^2-32*x+16)*log(1/5*x)+1/16*x^2)^2,x, algorithm="giac")
[Out]
-(256*x^4*e^x*log(5)*log(1/5*x) - 256*x^4*e^x*log(1/5*x)*log(x) + 16*x^4*e^x*log(5) - 144*x^4*e^x*log(1/5*x) -
768*x^3*e^x*log(5)*log(1/5*x) - 16*x^4*e^x*log(x) + 768*x^3*e^x*log(1/5*x)*log(x) - 9*x^4*e^x - 16*x^3*e^x*lo
g(5) + 544*x^3*e^x*log(1/5*x) + 768*x^2*e^x*log(5)*log(1/5*x) + 16*x^3*e^x*log(x) - 768*x^2*e^x*log(1/5*x)*log
(x) + 16*x^3*e^x - 784*x^2*e^x*log(1/5*x) - 256*x*e^x*log(5)*log(1/5*x) + 256*x*e^x*log(1/5*x)*log(x) - 8*x^2*
e^x + 512*x*e^x*log(1/5*x) - 128*e^x*log(1/5*x))/(1024*x^4*log(5)*log(2)*log(1/5*x) - 256*x^4*log(5)*log(16*x^
2*log(1/5*x) + x^2 - 32*x*log(1/5*x) + 16*log(1/5*x))*log(1/5*x) - 1024*x^4*log(2)*log(1/5*x)*log(x) + 256*x^4
*log(16*x^2*log(1/5*x) + x^2 - 32*x*log(1/5*x) + 16*log(1/5*x))*log(1/5*x)*log(x) + 576*x^4*log(5)*log(2) - 14
4*x^4*log(5)*log(16*x^2*log(1/5*x) + x^2 - 32*x*log(1/5*x) + 16*log(1/5*x)) - 64*x^4*log(2)*log(1/5*x) - 3072*
x^3*log(5)*log(2)*log(1/5*x) + 16*x^4*log(16*x^2*log(1/5*x) + x^2 - 32*x*log(1/5*x) + 16*log(1/5*x))*log(1/5*x
) + 768*x^3*log(5)*log(16*x^2*log(1/5*x) + x^2 - 32*x*log(1/5*x) + 16*log(1/5*x))*log(1/5*x) - 576*x^4*log(2)*
log(x) + 144*x^4*log(16*x^2*log(1/5*x) + x^2 - 32*x*log(1/5*x) + 16*log(1/5*x))*log(x) + 3072*x^3*log(2)*log(1
/5*x)*log(x) - 768*x^3*log(16*x^2*log(1/5*x) + x^2 - 32*x*log(1/5*x) + 16*log(1/5*x))*log(1/5*x)*log(x) - 36*x
^4*log(2) - 2176*x^3*log(5)*log(2) + 9*x^4*log(16*x^2*log(1/5*x) + x^2 - 32*x*log(1/5*x) + 16*log(1/5*x)) + 54
4*x^3*log(5)*log(16*x^2*log(1/5*x) + x^2 - 32*x*log(1/5*x) + 16*log(1/5*x)) + 64*x^3*log(2)*log(1/5*x) + 3072*
x^2*log(5)*log(2)*log(1/5*x) - 16*x^3*log(16*x^2*log(1/5*x) + x^2 - 32*x*log(1/5*x) + 16*log(1/5*x))*log(1/5*x
) - 768*x^2*log(5)*log(16*x^2*log(1/5*x) + x^2 - 32*x*log(1/5*x) + 16*log(1/5*x))*log(1/5*x) + 2176*x^3*log(2)
*log(x) - 544*x^3*log(16*x^2*log(1/5*x) + x^2 - 32*x*log(1/5*x) + 16*log(1/5*x))*log(x) - 3072*x^2*log(2)*log(
1/5*x)*log(x) + 768*x^2*log(16*x^2*log(1/5*x) + x^2 - 32*x*log(1/5*x) + 16*log(1/5*x))*log(1/5*x)*log(x) + 64*
x^3*log(2) + 3136*x^2*log(5)*log(2) - 16*x^3*log(16*x^2*log(1/5*x) + x^2 - 32*x*log(1/5*x) + 16*log(1/5*x)) -
784*x^2*log(5)*log(16*x^2*log(1/5*x) + x^2 - 32*x*log(1/5*x) + 16*log(1/5*x)) - 1024*x*log(5)*log(2)*log(1/5*x
) + 256*x*log(5)*log(16*x^2*log(1/5*x) + x^2 - 32*x*log(1/5*x) + 16*log(1/5*x))*log(1/5*x) - 3136*x^2*log(2)*l
og(x) + 784*x^2*log(16*x^2*log(1/5*x) + x^2 - 32*x*log(1/5*x) + 16*log(1/5*x))*log(x) + 1024*x*log(2)*log(1/5*
x)*log(x) - 256*x*log(16*x^2*log(1/5*x) + x^2 - 32*x*log(1/5*x) + 16*log(1/5*x))*log(1/5*x)*log(x) - 32*x^2*lo
g(2) - 2048*x*log(5)*log(2) + 8*x^2*log(16*x^2*log(1/5*x) + x^2 - 32*x*log(1/5*x) + 16*log(1/5*x)) + 512*x*log
(5)*log(16*x^2*log(1/5*x) + x^2 - 32*x*log(1/5*x) + 16*log(1/5*x)) + 2048*x*log(2)*log(x) - 512*x*log(16*x^2*l
og(1/5*x) + x^2 - 32*x*log(1/5*x) + 16*log(1/5*x))*log(x) + 512*log(5)*log(2) - 128*log(5)*log(16*x^2*log(1/5*
x) + x^2 - 32*x*log(1/5*x) + 16*log(1/5*x)) - 512*log(2)*log(x) + 128*log(16*x^2*log(1/5*x) + x^2 - 32*x*log(1
/5*x) + 16*log(1/5*x))*log(x))
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maple [A] time = 0.04, size = 29, normalized size = 1.07
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size |
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\(\frac {{\mathrm e}^{x}}{\ln \left (\frac {\left (16 x^{2}-32 x +16\right ) \ln \left (\frac {x}{5}\right )}{16}+\frac {x^{2}}{16}\right )}\) |
\(29\) |
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Verification of antiderivative is not currently implemented for this CAS.
[In]
int((((16*x^3-32*x^2+16*x)*exp(x)*ln(1/5*x)+exp(x)*x^3)*ln(1/16*(16*x^2-32*x+16)*ln(1/5*x)+1/16*x^2)+(-32*x^2+
32*x)*exp(x)*ln(1/5*x)+(-18*x^2+32*x-16)*exp(x))/((16*x^3-32*x^2+16*x)*ln(1/5*x)+x^3)/ln(1/16*(16*x^2-32*x+16)
*ln(1/5*x)+1/16*x^2)^2,x,method=_RETURNVERBOSE)
[Out]
exp(x)/ln(1/16*(16*x^2-32*x+16)*ln(1/5*x)+1/16*x^2)
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maxima [B] time = 0.51, size = 47, normalized size = 1.74 \begin {gather*} -\frac {e^{x}}{4 \, \log \relax (2) - \log \left (-x^{2} {\left (16 \, \log \relax (5) - 1\right )} + 32 \, x \log \relax (5) + 16 \, {\left (x^{2} - 2 \, x + 1\right )} \log \relax (x) - 16 \, \log \relax (5)\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((((16*x^3-32*x^2+16*x)*exp(x)*log(1/5*x)+exp(x)*x^3)*log(1/16*(16*x^2-32*x+16)*log(1/5*x)+1/16*x^2)+
(-32*x^2+32*x)*exp(x)*log(1/5*x)+(-18*x^2+32*x-16)*exp(x))/((16*x^3-32*x^2+16*x)*log(1/5*x)+x^3)/log(1/16*(16*
x^2-32*x+16)*log(1/5*x)+1/16*x^2)^2,x, algorithm="maxima")
[Out]
-e^x/(4*log(2) - log(-x^2*(16*log(5) - 1) + 32*x*log(5) + 16*(x^2 - 2*x + 1)*log(x) - 16*log(5)))
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mupad [F] time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int \frac {\ln \left (\frac {\ln \left (\frac {x}{5}\right )\,\left (16\,x^2-32\,x+16\right )}{16}+\frac {x^2}{16}\right )\,\left (x^3\,{\mathrm {e}}^x+\ln \left (\frac {x}{5}\right )\,{\mathrm {e}}^x\,\left (16\,x^3-32\,x^2+16\,x\right )\right )-{\mathrm {e}}^x\,\left (18\,x^2-32\,x+16\right )+\ln \left (\frac {x}{5}\right )\,{\mathrm {e}}^x\,\left (32\,x-32\,x^2\right )}{{\ln \left (\frac {\ln \left (\frac {x}{5}\right )\,\left (16\,x^2-32\,x+16\right )}{16}+\frac {x^2}{16}\right )}^2\,\left (\ln \left (\frac {x}{5}\right )\,\left (16\,x^3-32\,x^2+16\,x\right )+x^3\right )} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((log((log(x/5)*(16*x^2 - 32*x + 16))/16 + x^2/16)*(x^3*exp(x) + log(x/5)*exp(x)*(16*x - 32*x^2 + 16*x^3))
- exp(x)*(18*x^2 - 32*x + 16) + log(x/5)*exp(x)*(32*x - 32*x^2))/(log((log(x/5)*(16*x^2 - 32*x + 16))/16 + x^2
/16)^2*(log(x/5)*(16*x - 32*x^2 + 16*x^3) + x^3)),x)
[Out]
int((log((log(x/5)*(16*x^2 - 32*x + 16))/16 + x^2/16)*(x^3*exp(x) + log(x/5)*exp(x)*(16*x - 32*x^2 + 16*x^3))
- exp(x)*(18*x^2 - 32*x + 16) + log(x/5)*exp(x)*(32*x - 32*x^2))/(log((log(x/5)*(16*x^2 - 32*x + 16))/16 + x^2
/16)^2*(log(x/5)*(16*x - 32*x^2 + 16*x^3) + x^3)), x)
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sympy [A] time = 0.92, size = 22, normalized size = 0.81 \begin {gather*} \frac {e^{x}}{\log {\left (\frac {x^{2}}{16} + \left (x^{2} - 2 x + 1\right ) \log {\left (\frac {x}{5} \right )} \right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((((16*x**3-32*x**2+16*x)*exp(x)*ln(1/5*x)+exp(x)*x**3)*ln(1/16*(16*x**2-32*x+16)*ln(1/5*x)+1/16*x**2
)+(-32*x**2+32*x)*exp(x)*ln(1/5*x)+(-18*x**2+32*x-16)*exp(x))/((16*x**3-32*x**2+16*x)*ln(1/5*x)+x**3)/ln(1/16*
(16*x**2-32*x+16)*ln(1/5*x)+1/16*x**2)**2,x)
[Out]
exp(x)/log(x**2/16 + (x**2 - 2*x + 1)*log(x/5))
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