3.8.97 \(\int \frac {e^{\frac {4 x^4}{4 e^{\frac {6 x}{10+2 \log (x)}} x^2-4 e^{\frac {3 x}{10+2 \log (x)}} x \log (\frac {1}{x})+\log ^2(\frac {1}{x})}} (-200 x^3-400 x^3 \log (\frac {1}{x})+(-80 x^3-160 x^3 \log (\frac {1}{x})) \log (x)+(-8 x^3-16 x^3 \log (\frac {1}{x})) \log ^2(x)+e^{\frac {3 x}{10+2 \log (x)}} (400 x^4-96 x^5+(160 x^4-24 x^5) \log (x)+16 x^4 \log ^2(x)))}{-25 \log ^3(\frac {1}{x})-10 \log ^3(\frac {1}{x}) \log (x)-\log ^3(\frac {1}{x}) \log ^2(x)+e^{\frac {9 x}{10+2 \log (x)}} (200 x^3+80 x^3 \log (x)+8 x^3 \log ^2(x))+e^{\frac {6 x}{10+2 \log (x)}} (-300 x^2 \log (\frac {1}{x})-120 x^2 \log (\frac {1}{x}) \log (x)-12 x^2 \log (\frac {1}{x}) \log ^2(x))+e^{\frac {3 x}{10+2 \log (x)}} (150 x \log ^2(\frac {1}{x})+60 x \log ^2(\frac {1}{x}) \log (x)+6 x \log ^2(\frac {1}{x}) \log ^2(x))} \, dx\)
Optimal. Leaf size=33 \[ e^{\frac {x^2}{\left (e^{\frac {3 x}{2 (5+\log (x))}}-\frac {\log \left (\frac {1}{x}\right )}{2 x}\right )^2}} \]
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Rubi [A] time = 34.25, antiderivative size = 32, normalized size of antiderivative = 0.97,
number of steps used = 3, number of rules used = 3, integrand size = 321, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.009, Rules used
= {6688, 12, 6706} \begin {gather*} \exp \left (\frac {4 x^4}{\left (2 x e^{\frac {3 x}{2 (\log (x)+5)}}-\log \left (\frac {1}{x}\right )\right )^2}\right ) \end {gather*}
Antiderivative was successfully verified.
[In]
Int[(E^((4*x^4)/(4*E^((6*x)/(10 + 2*Log[x]))*x^2 - 4*E^((3*x)/(10 + 2*Log[x]))*x*Log[x^(-1)] + Log[x^(-1)]^2))
*(-200*x^3 - 400*x^3*Log[x^(-1)] + (-80*x^3 - 160*x^3*Log[x^(-1)])*Log[x] + (-8*x^3 - 16*x^3*Log[x^(-1)])*Log[
x]^2 + E^((3*x)/(10 + 2*Log[x]))*(400*x^4 - 96*x^5 + (160*x^4 - 24*x^5)*Log[x] + 16*x^4*Log[x]^2)))/(-25*Log[x
^(-1)]^3 - 10*Log[x^(-1)]^3*Log[x] - Log[x^(-1)]^3*Log[x]^2 + E^((9*x)/(10 + 2*Log[x]))*(200*x^3 + 80*x^3*Log[
x] + 8*x^3*Log[x]^2) + E^((6*x)/(10 + 2*Log[x]))*(-300*x^2*Log[x^(-1)] - 120*x^2*Log[x^(-1)]*Log[x] - 12*x^2*L
og[x^(-1)]*Log[x]^2) + E^((3*x)/(10 + 2*Log[x]))*(150*x*Log[x^(-1)]^2 + 60*x*Log[x^(-1)]^2*Log[x] + 6*x*Log[x^
(-1)]^2*Log[x]^2)),x]
[Out]
E^((4*x^4)/(2*E^((3*x)/(2*(5 + Log[x])))*x - Log[x^(-1)])^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 6688
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]
Rule 6706
Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[(q*F^v)/Log[F], x] /; !FalseQ[q]
] /; FreeQ[F, x]
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {8 \exp \left (\frac {4 x^4}{\left (-2 e^{\frac {3 x}{2 (5+\log (x))}} x+\log \left (\frac {1}{x}\right )\right )^2}\right ) x^3 \left (-25+50 e^{\frac {3 x}{2 (5+\log (x))}} x-12 e^{\frac {3 x}{2 (5+\log (x))}} x^2-\left (10+e^{\frac {3 x}{2 (5+\log (x))}} x (-20+3 x)\right ) \log (x)-\left (1-2 e^{\frac {3 x}{2 (5+\log (x))}} x\right ) \log ^2(x)-2 \log \left (\frac {1}{x}\right ) (5+\log (x))^2\right )}{\left (2 e^{\frac {3 x}{2 (5+\log (x))}} x-\log \left (\frac {1}{x}\right )\right )^3 (5+\log (x))^2} \, dx\\ &=8 \int \frac {\exp \left (\frac {4 x^4}{\left (-2 e^{\frac {3 x}{2 (5+\log (x))}} x+\log \left (\frac {1}{x}\right )\right )^2}\right ) x^3 \left (-25+50 e^{\frac {3 x}{2 (5+\log (x))}} x-12 e^{\frac {3 x}{2 (5+\log (x))}} x^2-\left (10+e^{\frac {3 x}{2 (5+\log (x))}} x (-20+3 x)\right ) \log (x)-\left (1-2 e^{\frac {3 x}{2 (5+\log (x))}} x\right ) \log ^2(x)-2 \log \left (\frac {1}{x}\right ) (5+\log (x))^2\right )}{\left (2 e^{\frac {3 x}{2 (5+\log (x))}} x-\log \left (\frac {1}{x}\right )\right )^3 (5+\log (x))^2} \, dx\\ &=\exp \left (\frac {4 x^4}{\left (2 e^{\frac {3 x}{2 (5+\log (x))}} x-\log \left (\frac {1}{x}\right )\right )^2}\right )\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.49, size = 32, normalized size = 0.97 \begin {gather*} e^{\frac {4 x^4}{\left (2 e^{\frac {3 x}{2 (5+\log (x))}} x-\log \left (\frac {1}{x}\right )\right )^2}} \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[(E^((4*x^4)/(4*E^((6*x)/(10 + 2*Log[x]))*x^2 - 4*E^((3*x)/(10 + 2*Log[x]))*x*Log[x^(-1)] + Log[x^(-1
)]^2))*(-200*x^3 - 400*x^3*Log[x^(-1)] + (-80*x^3 - 160*x^3*Log[x^(-1)])*Log[x] + (-8*x^3 - 16*x^3*Log[x^(-1)]
)*Log[x]^2 + E^((3*x)/(10 + 2*Log[x]))*(400*x^4 - 96*x^5 + (160*x^4 - 24*x^5)*Log[x] + 16*x^4*Log[x]^2)))/(-25
*Log[x^(-1)]^3 - 10*Log[x^(-1)]^3*Log[x] - Log[x^(-1)]^3*Log[x]^2 + E^((9*x)/(10 + 2*Log[x]))*(200*x^3 + 80*x^
3*Log[x] + 8*x^3*Log[x]^2) + E^((6*x)/(10 + 2*Log[x]))*(-300*x^2*Log[x^(-1)] - 120*x^2*Log[x^(-1)]*Log[x] - 12
*x^2*Log[x^(-1)]*Log[x]^2) + E^((3*x)/(10 + 2*Log[x]))*(150*x*Log[x^(-1)]^2 + 60*x*Log[x^(-1)]^2*Log[x] + 6*x*
Log[x^(-1)]^2*Log[x]^2)),x]
[Out]
E^((4*x^4)/(2*E^((3*x)/(2*(5 + Log[x])))*x - Log[x^(-1)])^2)
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fricas [A] time = 1.06, size = 51, normalized size = 1.55 \begin {gather*} e^{\left (\frac {4 \, x^{4}}{4 \, x^{2} e^{\left (-\frac {3 \, x}{\log \left (\frac {1}{x}\right ) - 5}\right )} - 4 \, x e^{\left (-\frac {3 \, x}{2 \, {\left (\log \left (\frac {1}{x}\right ) - 5\right )}}\right )} \log \left (\frac {1}{x}\right ) + \log \left (\frac {1}{x}\right )^{2}}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(((16*x^4*log(x)^2+(-24*x^5+160*x^4)*log(x)-96*x^5+400*x^4)*exp(3*x/(2*log(x)+10))+(-16*x^3*log(1/x)-
8*x^3)*log(x)^2+(-160*x^3*log(1/x)-80*x^3)*log(x)-400*x^3*log(1/x)-200*x^3)*exp(4*x^4/(4*x^2*exp(3*x/(2*log(x)
+10))^2-4*x*log(1/x)*exp(3*x/(2*log(x)+10))+log(1/x)^2))/((8*x^3*log(x)^2+80*x^3*log(x)+200*x^3)*exp(3*x/(2*lo
g(x)+10))^3+(-12*x^2*log(1/x)*log(x)^2-120*x^2*log(1/x)*log(x)-300*x^2*log(1/x))*exp(3*x/(2*log(x)+10))^2+(6*x
*log(1/x)^2*log(x)^2+60*x*log(1/x)^2*log(x)+150*x*log(1/x)^2)*exp(3*x/(2*log(x)+10))-log(1/x)^3*log(x)^2-10*lo
g(1/x)^3*log(x)-25*log(1/x)^3),x, algorithm="fricas")
[Out]
e^(4*x^4/(4*x^2*e^(-3*x/(log(1/x) - 5)) - 4*x*e^(-3/2*x/(log(1/x) - 5))*log(1/x) + log(1/x)^2))
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giac [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(((16*x^4*log(x)^2+(-24*x^5+160*x^4)*log(x)-96*x^5+400*x^4)*exp(3*x/(2*log(x)+10))+(-16*x^3*log(1/x)-
8*x^3)*log(x)^2+(-160*x^3*log(1/x)-80*x^3)*log(x)-400*x^3*log(1/x)-200*x^3)*exp(4*x^4/(4*x^2*exp(3*x/(2*log(x)
+10))^2-4*x*log(1/x)*exp(3*x/(2*log(x)+10))+log(1/x)^2))/((8*x^3*log(x)^2+80*x^3*log(x)+200*x^3)*exp(3*x/(2*lo
g(x)+10))^3+(-12*x^2*log(1/x)*log(x)^2-120*x^2*log(1/x)*log(x)-300*x^2*log(1/x))*exp(3*x/(2*log(x)+10))^2+(6*x
*log(1/x)^2*log(x)^2+60*x*log(1/x)^2*log(x)+150*x*log(1/x)^2)*exp(3*x/(2*log(x)+10))-log(1/x)^3*log(x)^2-10*lo
g(1/x)^3*log(x)-25*log(1/x)^3),x, algorithm="giac")
[Out]
Timed out
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maple [A] time = 0.77, size = 44, normalized size = 1.33
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\({\mathrm e}^{\frac {4 x^{4}}{4 x^{2} {\mathrm e}^{\frac {3 x}{5+\ln \relax (x )}}+4 \ln \relax (x ) {\mathrm e}^{\frac {3 x}{2 \left (5+\ln \relax (x )\right )}} x +\ln \relax (x )^{2}}}\) |
\(44\) |
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Verification of antiderivative is not currently implemented for this CAS.
[In]
int(((16*x^4*ln(x)^2+(-24*x^5+160*x^4)*ln(x)-96*x^5+400*x^4)*exp(3*x/(2*ln(x)+10))+(-16*x^3*ln(1/x)-8*x^3)*ln(
x)^2+(-160*x^3*ln(1/x)-80*x^3)*ln(x)-400*x^3*ln(1/x)-200*x^3)*exp(4*x^4/(4*x^2*exp(3*x/(2*ln(x)+10))^2-4*x*ln(
1/x)*exp(3*x/(2*ln(x)+10))+ln(1/x)^2))/((8*x^3*ln(x)^2+80*x^3*ln(x)+200*x^3)*exp(3*x/(2*ln(x)+10))^3+(-12*x^2*
ln(1/x)*ln(x)^2-120*x^2*ln(1/x)*ln(x)-300*x^2*ln(1/x))*exp(3*x/(2*ln(x)+10))^2+(6*x*ln(1/x)^2*ln(x)^2+60*x*ln(
1/x)^2*ln(x)+150*x*ln(1/x)^2)*exp(3*x/(2*ln(x)+10))-ln(1/x)^3*ln(x)^2-10*ln(1/x)^3*ln(x)-25*ln(1/x)^3),x,metho
d=_RETURNVERBOSE)
[Out]
exp(4*x^4/(4*x^2*exp(3*x/(5+ln(x)))+4*ln(x)*exp(3/2*x/(5+ln(x)))*x+ln(x)^2))
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maxima [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(((16*x^4*log(x)^2+(-24*x^5+160*x^4)*log(x)-96*x^5+400*x^4)*exp(3*x/(2*log(x)+10))+(-16*x^3*log(1/x)-
8*x^3)*log(x)^2+(-160*x^3*log(1/x)-80*x^3)*log(x)-400*x^3*log(1/x)-200*x^3)*exp(4*x^4/(4*x^2*exp(3*x/(2*log(x)
+10))^2-4*x*log(1/x)*exp(3*x/(2*log(x)+10))+log(1/x)^2))/((8*x^3*log(x)^2+80*x^3*log(x)+200*x^3)*exp(3*x/(2*lo
g(x)+10))^3+(-12*x^2*log(1/x)*log(x)^2-120*x^2*log(1/x)*log(x)-300*x^2*log(1/x))*exp(3*x/(2*log(x)+10))^2+(6*x
*log(1/x)^2*log(x)^2+60*x*log(1/x)^2*log(x)+150*x*log(1/x)^2)*exp(3*x/(2*log(x)+10))-log(1/x)^3*log(x)^2-10*lo
g(1/x)^3*log(x)-25*log(1/x)^3),x, algorithm="maxima")
[Out]
Exception raised: RuntimeError >> ECL says: In function CAR, the value of the first argument is 0which is not
of the expected type LIST
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mupad [B] time = 1.74, size = 49, normalized size = 1.48 \begin {gather*} {\mathrm {e}}^{\frac {4\,x^4}{4\,x^2\,{\mathrm {e}}^{\frac {3\,x}{\ln \relax (x)+5}}+{\ln \left (\frac {1}{x}\right )}^2-4\,x\,\ln \left (\frac {1}{x}\right )\,{\mathrm {e}}^{\frac {3\,x}{2\,\ln \relax (x)+10}}}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((exp((4*x^4)/(4*x^2*exp((6*x)/(2*log(x) + 10)) + log(1/x)^2 - 4*x*log(1/x)*exp((3*x)/(2*log(x) + 10))))*(l
og(x)*(160*x^3*log(1/x) + 80*x^3) + 400*x^3*log(1/x) + log(x)^2*(16*x^3*log(1/x) + 8*x^3) + 200*x^3 - exp((3*x
)/(2*log(x) + 10))*(log(x)*(160*x^4 - 24*x^5) + 16*x^4*log(x)^2 + 400*x^4 - 96*x^5)))/(log(1/x)^3*log(x)^2 - e
xp((3*x)/(2*log(x) + 10))*(150*x*log(1/x)^2 + 60*x*log(1/x)^2*log(x) + 6*x*log(1/x)^2*log(x)^2) + exp((6*x)/(2
*log(x) + 10))*(300*x^2*log(1/x) + 120*x^2*log(1/x)*log(x) + 12*x^2*log(1/x)*log(x)^2) + 25*log(1/x)^3 - exp((
9*x)/(2*log(x) + 10))*(80*x^3*log(x) + 8*x^3*log(x)^2 + 200*x^3) + 10*log(1/x)^3*log(x)),x)
[Out]
exp((4*x^4)/(4*x^2*exp((3*x)/(log(x) + 5)) + log(1/x)^2 - 4*x*log(1/x)*exp((3*x)/(2*log(x) + 10))))
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sympy [A] time = 6.08, size = 46, normalized size = 1.39 \begin {gather*} e^{\frac {4 x^{4}}{4 x^{2} e^{\frac {6 x}{2 \log {\relax (x )} + 10}} + 4 x e^{\frac {3 x}{2 \log {\relax (x )} + 10}} \log {\relax (x )} + \log {\relax (x )}^{2}}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(((16*x**4*ln(x)**2+(-24*x**5+160*x**4)*ln(x)-96*x**5+400*x**4)*exp(3*x/(2*ln(x)+10))+(-16*x**3*ln(1/
x)-8*x**3)*ln(x)**2+(-160*x**3*ln(1/x)-80*x**3)*ln(x)-400*x**3*ln(1/x)-200*x**3)*exp(4*x**4/(4*x**2*exp(3*x/(2
*ln(x)+10))**2-4*x*ln(1/x)*exp(3*x/(2*ln(x)+10))+ln(1/x)**2))/((8*x**3*ln(x)**2+80*x**3*ln(x)+200*x**3)*exp(3*
x/(2*ln(x)+10))**3+(-12*x**2*ln(1/x)*ln(x)**2-120*x**2*ln(1/x)*ln(x)-300*x**2*ln(1/x))*exp(3*x/(2*ln(x)+10))**
2+(6*x*ln(1/x)**2*ln(x)**2+60*x*ln(1/x)**2*ln(x)+150*x*ln(1/x)**2)*exp(3*x/(2*ln(x)+10))-ln(1/x)**3*ln(x)**2-1
0*ln(1/x)**3*ln(x)-25*ln(1/x)**3),x)
[Out]
exp(4*x**4/(4*x**2*exp(6*x/(2*log(x) + 10)) + 4*x*exp(3*x/(2*log(x) + 10))*log(x) + log(x)**2))
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