3.73.14 \(\int \frac {e^{\frac {x^2+3 x \log (5)+3 \log (5) \log (x)}{x \log (5)}} (-72 x^2-216 \log (5)+216 \log (5) \log (x))}{(4 x^2 \log (5)+4 e^{\frac {x^2+3 x \log (5)+3 \log (5) \log (x)}{x \log (5)}} x^2 \log (5)+(x^2 \log (5)+e^{\frac {x^2+3 x \log (5)+3 \log (5) \log (x)}{x \log (5)}} x^2 \log (5)) \log (1+e^{\frac {x^2+3 x \log (5)+3 \log (5) \log (x)}{x \log (5)}})) \log ^3(4+\log (1+e^{\frac {x^2+3 x \log (5)+3 \log (5) \log (x)}{x \log (5)}}))} \, dx\)

Optimal. Leaf size=28 \[ \frac {36}{\log ^2\left (4+\log \left (1+e^{\frac {x}{\log (5)}+\frac {3 (x+\log (x))}{x}}\right )\right )} \]

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Rubi [A]  time = 1.21, antiderivative size = 34, normalized size of antiderivative = 1.21, number of steps used = 1, number of rules used = 1, integrand size = 184, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.005, Rules used = {6686} \begin {gather*} \frac {36}{\log ^2\left (\log \left (x^{3/x} 5^{\frac {3}{\log (5)}} e^{\frac {x}{\log (5)}}+1\right )+4\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^((x^2 + 3*x*Log[5] + 3*Log[5]*Log[x])/(x*Log[5]))*(-72*x^2 - 216*Log[5] + 216*Log[5]*Log[x]))/((4*x^2*L
og[5] + 4*E^((x^2 + 3*x*Log[5] + 3*Log[5]*Log[x])/(x*Log[5]))*x^2*Log[5] + (x^2*Log[5] + E^((x^2 + 3*x*Log[5]
+ 3*Log[5]*Log[x])/(x*Log[5]))*x^2*Log[5])*Log[1 + E^((x^2 + 3*x*Log[5] + 3*Log[5]*Log[x])/(x*Log[5]))])*Log[4
+ Log[1 + E^((x^2 + 3*x*Log[5] + 3*Log[5]*Log[x])/(x*Log[5]))]]^3),x]

[Out]

36/Log[4 + Log[1 + 5^(3/Log[5])*E^(x/Log[5])*x^(3/x)]]^2

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]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {36}{\log ^2\left (4+\log \left (1+5^{\frac {3}{\log (5)}} e^{\frac {x}{\log (5)}} x^{3/x}\right )\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.24, size = 28, normalized size = 1.00 \begin {gather*} \frac {36}{\log ^2\left (4+\log \left (1+e^{3+\frac {x}{\log (5)}} x^{3/x}\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((x^2 + 3*x*Log[5] + 3*Log[5]*Log[x])/(x*Log[5]))*(-72*x^2 - 216*Log[5] + 216*Log[5]*Log[x]))/((4
*x^2*Log[5] + 4*E^((x^2 + 3*x*Log[5] + 3*Log[5]*Log[x])/(x*Log[5]))*x^2*Log[5] + (x^2*Log[5] + E^((x^2 + 3*x*L
og[5] + 3*Log[5]*Log[x])/(x*Log[5]))*x^2*Log[5])*Log[1 + E^((x^2 + 3*x*Log[5] + 3*Log[5]*Log[x])/(x*Log[5]))])
*Log[4 + Log[1 + E^((x^2 + 3*x*Log[5] + 3*Log[5]*Log[x])/(x*Log[5]))]]^3),x]

[Out]

36/Log[4 + Log[1 + E^(3 + x/Log[5])*x^(3/x)]]^2

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fricas [A]  time = 0.88, size = 34, normalized size = 1.21 \begin {gather*} \frac {36}{\log \left (\log \left (e^{\left (\frac {x^{2} + 3 \, x \log \relax (5) + 3 \, \log \relax (5) \log \relax (x)}{x \log \relax (5)}\right )} + 1\right ) + 4\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((216*log(5)*log(x)-216*log(5)-72*x^2)*exp((3*log(5)*log(x)+3*x*log(5)+x^2)/x/log(5))/((x^2*log(5)*ex
p((3*log(5)*log(x)+3*x*log(5)+x^2)/x/log(5))+x^2*log(5))*log(exp((3*log(5)*log(x)+3*x*log(5)+x^2)/x/log(5))+1)
+4*x^2*log(5)*exp((3*log(5)*log(x)+3*x*log(5)+x^2)/x/log(5))+4*x^2*log(5))/log(log(exp((3*log(5)*log(x)+3*x*lo
g(5)+x^2)/x/log(5))+1)+4)^3,x, algorithm="fricas")

[Out]

36/log(log(e^((x^2 + 3*x*log(5) + 3*log(5)*log(x))/(x*log(5))) + 1) + 4)^2

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giac [A]  time = 1.87, size = 34, normalized size = 1.21 \begin {gather*} \frac {36}{\log \left (\log \left (e^{\left (\frac {x^{2} + 3 \, x \log \relax (5) + 3 \, \log \relax (5) \log \relax (x)}{x \log \relax (5)}\right )} + 1\right ) + 4\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((216*log(5)*log(x)-216*log(5)-72*x^2)*exp((3*log(5)*log(x)+3*x*log(5)+x^2)/x/log(5))/((x^2*log(5)*ex
p((3*log(5)*log(x)+3*x*log(5)+x^2)/x/log(5))+x^2*log(5))*log(exp((3*log(5)*log(x)+3*x*log(5)+x^2)/x/log(5))+1)
+4*x^2*log(5)*exp((3*log(5)*log(x)+3*x*log(5)+x^2)/x/log(5))+4*x^2*log(5))/log(log(exp((3*log(5)*log(x)+3*x*lo
g(5)+x^2)/x/log(5))+1)+4)^3,x, algorithm="giac")

[Out]

36/log(log(e^((x^2 + 3*x*log(5) + 3*log(5)*log(x))/(x*log(5))) + 1) + 4)^2

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maple [A]  time = 0.05, size = 31, normalized size = 1.11

method	result	size
risch	\(\frac {36}{\ln \left (\ln \left (x^{\frac {3}{x}} {\mathrm e}^{\frac {3 \ln \relax (5)+x}{\ln \relax (5)}}+1\right )+4\right )^{2}}\)	\(31\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((216*ln(5)*ln(x)-216*ln(5)-72*x^2)*exp((3*ln(5)*ln(x)+3*x*ln(5)+x^2)/x/ln(5))/((x^2*ln(5)*exp((3*ln(5)*ln(
x)+3*x*ln(5)+x^2)/x/ln(5))+x^2*ln(5))*ln(exp((3*ln(5)*ln(x)+3*x*ln(5)+x^2)/x/ln(5))+1)+4*x^2*ln(5)*exp((3*ln(5
)*ln(x)+3*x*ln(5)+x^2)/x/ln(5))+4*x^2*ln(5))/ln(ln(exp((3*ln(5)*ln(x)+3*x*ln(5)+x^2)/x/ln(5))+1)+4)^3,x,method
=_RETURNVERBOSE)

[Out]

36/ln(ln(x^(3/x)*exp((3*ln(5)+x)/ln(5))+1)+4)^2

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maxima [A]  time = 0.65, size = 26, normalized size = 0.93 \begin {gather*} \frac {36}{\log \left (\log \left (e^{\left (\frac {x}{\log \relax (5)} + \frac {3 \, \log \relax (x)}{x} + 3\right )} + 1\right ) + 4\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((216*log(5)*log(x)-216*log(5)-72*x^2)*exp((3*log(5)*log(x)+3*x*log(5)+x^2)/x/log(5))/((x^2*log(5)*ex
p((3*log(5)*log(x)+3*x*log(5)+x^2)/x/log(5))+x^2*log(5))*log(exp((3*log(5)*log(x)+3*x*log(5)+x^2)/x/log(5))+1)
+4*x^2*log(5)*exp((3*log(5)*log(x)+3*x*log(5)+x^2)/x/log(5))+4*x^2*log(5))/log(log(exp((3*log(5)*log(x)+3*x*lo
g(5)+x^2)/x/log(5))+1)+4)^3,x, algorithm="maxima")

[Out]

36/log(log(e^(x/log(5) + 3*log(x)/x + 3) + 1) + 4)^2

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mupad [B]  time = 5.56, size = 27, normalized size = 0.96 \begin {gather*} \frac {36}{{\ln \left (\ln \left (x^{3/x}\,{\mathrm {e}}^{\frac {x}{\ln \relax (5)}}\,{\mathrm {e}}^3+1\right )+4\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp((3*x*log(5) + 3*log(5)*log(x) + x^2)/(x*log(5)))*(216*log(5) - 216*log(5)*log(x) + 72*x^2))/(log(log
(exp((3*x*log(5) + 3*log(5)*log(x) + x^2)/(x*log(5))) + 1) + 4)^3*(4*x^2*log(5) + log(exp((3*x*log(5) + 3*log(
5)*log(x) + x^2)/(x*log(5))) + 1)*(x^2*log(5) + x^2*exp((3*x*log(5) + 3*log(5)*log(x) + x^2)/(x*log(5)))*log(5
)) + 4*x^2*exp((3*x*log(5) + 3*log(5)*log(x) + x^2)/(x*log(5)))*log(5))),x)

[Out]

36/log(log(x^(3/x)*exp(x/log(5))*exp(3) + 1) + 4)^2

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sympy [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((216*ln(5)*ln(x)-216*ln(5)-72*x**2)*exp((3*ln(5)*ln(x)+3*x*ln(5)+x**2)/x/ln(5))/((x**2*ln(5)*exp((3*
ln(5)*ln(x)+3*x*ln(5)+x**2)/x/ln(5))+x**2*ln(5))*ln(exp((3*ln(5)*ln(x)+3*x*ln(5)+x**2)/x/ln(5))+1)+4*x**2*ln(5
)*exp((3*ln(5)*ln(x)+3*x*ln(5)+x**2)/x/ln(5))+4*x**2*ln(5))/ln(ln(exp((3*ln(5)*ln(x)+3*x*ln(5)+x**2)/x/ln(5))+
1)+4)**3,x)

[Out]

Timed out

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