3.17.34 \(\int \frac {160 e^{1+\frac {5 e}{4 x \log (3)}}}{-125 x^2 \log (3)+75 e^{\frac {5 e}{4 x \log (3)}} x^2 \log (3)-15 e^{\frac {5 e}{2 x \log (3)}} x^2 \log (3)+e^{\frac {15 e}{4 x \log (3)}} x^2 \log (3)} \, dx\)

Optimal. Leaf size=20 \[ \frac {64}{\left (-5+e^{\frac {5 e}{4 x \log (3)}}\right )^2} \]

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Rubi [A]  time = 0.35, antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 90, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.033, Rules used = {12, 6688, 6686} \begin {gather*} \frac {64}{\left (5-e^{\frac {5 e}{x \log (81)}}\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(160*E^(1 + (5*E)/(4*x*Log[3])))/(-125*x^2*Log[3] + 75*E^((5*E)/(4*x*Log[3]))*x^2*Log[3] - 15*E^((5*E)/(2*
x*Log[3]))*x^2*Log[3] + E^((15*E)/(4*x*Log[3]))*x^2*Log[3]),x]

[Out]

64/(5 - E^((5*E)/(x*Log[81])))^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} &=160 \int \frac {e^{1+\frac {5 e}{4 x \log (3)}}}{-125 x^2 \log (3)+75 e^{\frac {5 e}{4 x \log (3)}} x^2 \log (3)-15 e^{\frac {5 e}{2 x \log (3)}} x^2 \log (3)+e^{\frac {15 e}{4 x \log (3)}} x^2 \log (3)} \, dx\\ &=160 \int \frac {e^{1+\frac {5 e}{x \log (81)}}}{\left (-5+e^{\frac {5 e}{x \log (81)}}\right )^3 x^2 \log (3)} \, dx\\ &=\frac {160 \int \frac {e^{1+\frac {5 e}{x \log (81)}}}{\left (-5+e^{\frac {5 e}{x \log (81)}}\right )^3 x^2} \, dx}{\log (3)}\\ &=\frac {64}{\left (5-e^{\frac {5 e}{x \log (81)}}\right )^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.04, size = 18, normalized size = 0.90 \begin {gather*} \frac {64}{\left (-5+e^{\frac {5 e}{x \log (81)}}\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(160*E^(1 + (5*E)/(4*x*Log[3])))/(-125*x^2*Log[3] + 75*E^((5*E)/(4*x*Log[3]))*x^2*Log[3] - 15*E^((5*
E)/(2*x*Log[3]))*x^2*Log[3] + E^((15*E)/(4*x*Log[3]))*x^2*Log[3]),x]

[Out]

64/(-5 + E^((5*E)/(x*Log[81])))^2

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fricas [B]  time = 0.77, size = 85, normalized size = 4.25 \begin {gather*} -\frac {64 \, e^{\left (2 \, \log \relax (5) + 2\right )}}{10 \, e^{\left (\frac {4 \, x \log \relax (5) \log \relax (3) + 4 \, x \log \relax (3) + e^{\left (\log \relax (5) + 1\right )}}{4 \, x \log \relax (3)} + \log \relax (5) + 1\right )} - e^{\left (\frac {4 \, x \log \relax (5) \log \relax (3) + 4 \, x \log \relax (3) + e^{\left (\log \relax (5) + 1\right )}}{2 \, x \log \relax (3)}\right )} - 25 \, e^{\left (2 \, \log \relax (5) + 2\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(32*exp(log(5)+1)*exp(1/4*exp(log(5)+1)/x/log(3))/(x^2*log(3)*exp(1/4*exp(log(5)+1)/x/log(3))^3-15*x^
2*log(3)*exp(1/4*exp(log(5)+1)/x/log(3))^2+75*x^2*log(3)*exp(1/4*exp(log(5)+1)/x/log(3))-125*x^2*log(3)),x, al
gorithm="fricas")

[Out]

-64*e^(2*log(5) + 2)/(10*e^(1/4*(4*x*log(5)*log(3) + 4*x*log(3) + e^(log(5) + 1))/(x*log(3)) + log(5) + 1) - e
^(1/2*(4*x*log(5)*log(3) + 4*x*log(3) + e^(log(5) + 1))/(x*log(3))) - 25*e^(2*log(5) + 2))

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giac [A]  time = 0.55, size = 41, normalized size = 2.05 \begin {gather*} \frac {64 \, e}{25 \, e + e^{\left (\frac {5 \, e}{2 \, x \log \relax (3)} + 1\right )} - 10 \, e^{\left (\frac {5 \, e}{4 \, x \log \relax (3)} + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(32*exp(log(5)+1)*exp(1/4*exp(log(5)+1)/x/log(3))/(x^2*log(3)*exp(1/4*exp(log(5)+1)/x/log(3))^3-15*x^
2*log(3)*exp(1/4*exp(log(5)+1)/x/log(3))^2+75*x^2*log(3)*exp(1/4*exp(log(5)+1)/x/log(3))-125*x^2*log(3)),x, al
gorithm="giac")

[Out]

64*e/(25*e + e^(5/2*e/(x*log(3)) + 1) - 10*e^(5/4*e/(x*log(3)) + 1))

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maple [A]  time = 0.18, size = 19, normalized size = 0.95

method	result	size
risch	\(\frac {64}{\left ({\mathrm e}^{\frac {5 \,{\mathrm e}}{4 x \ln \relax (3)}}-5\right )^{2}}\)	\(19\)
norman	\(\frac {64}{\left ({\mathrm e}^{\frac {{\mathrm e}^{\ln \relax (5)+1}}{4 x \ln \relax (3)}}-5\right )^{2}}\)	\(22\)
derivativedivides	\(\frac {64 \,{\mathrm e}^{2} {\mathrm e}^{-2}}{\left ({\mathrm e}^{\frac {{\mathrm e}^{\ln \relax (5)+1}}{4 x \ln \relax (3)}}-5\right )^{2}}\)	\(33\)
default	\(\frac {64 \,{\mathrm e}^{2} {\mathrm e}^{-2}}{\left ({\mathrm e}^{\frac {{\mathrm e}^{\ln \relax (5)+1}}{4 x \ln \relax (3)}}-5\right )^{2}}\)	\(33\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(32*exp(ln(5)+1)*exp(1/4*exp(ln(5)+1)/x/ln(3))/(x^2*ln(3)*exp(1/4*exp(ln(5)+1)/x/ln(3))^3-15*x^2*ln(3)*exp(
1/4*exp(ln(5)+1)/x/ln(3))^2+75*x^2*ln(3)*exp(1/4*exp(ln(5)+1)/x/ln(3))-125*x^2*ln(3)),x,method=_RETURNVERBOSE)

[Out]

64/(exp(5/4*exp(1)/x/ln(3))-5)^2

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maxima [B]  time = 0.54, size = 59, normalized size = 2.95 \begin {gather*} -\frac {64 \, {\left (e^{\left (\frac {5 \, e}{2 \, x \log \relax (3)}\right )} - 10 \, e^{\left (\frac {5 \, e}{4 \, x \log \relax (3)}\right )}\right )}}{25 \, {\left (e^{\left (\frac {5 \, e}{2 \, x \log \relax (3)}\right )} - 10 \, e^{\left (\frac {5 \, e}{4 \, x \log \relax (3)}\right )} + 25\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(32*exp(log(5)+1)*exp(1/4*exp(log(5)+1)/x/log(3))/(x^2*log(3)*exp(1/4*exp(log(5)+1)/x/log(3))^3-15*x^
2*log(3)*exp(1/4*exp(log(5)+1)/x/log(3))^2+75*x^2*log(3)*exp(1/4*exp(log(5)+1)/x/log(3))-125*x^2*log(3)),x, al
gorithm="maxima")

[Out]

-64/25*(e^(5/2*e/(x*log(3))) - 10*e^(5/4*e/(x*log(3))))/(e^(5/2*e/(x*log(3))) - 10*e^(5/4*e/(x*log(3))) + 25)

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mupad [B]  time = 1.28, size = 18, normalized size = 0.90 \begin {gather*} \frac {64}{{\left ({\mathrm {e}}^{\frac {5\,\mathrm {e}}{4\,x\,\ln \relax (3)}}-5\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(32*exp(log(5) + 1)*exp(exp(log(5) + 1)/(4*x*log(3))))/(125*x^2*log(3) + 15*x^2*exp(exp(log(5) + 1)/(2*x*
log(3)))*log(3) - 75*x^2*exp(exp(log(5) + 1)/(4*x*log(3)))*log(3) - x^2*exp((3*exp(log(5) + 1))/(4*x*log(3)))*
log(3)),x)

[Out]

64/(exp((5*exp(1))/(4*x*log(3))) - 5)^2

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sympy [A]  time = 0.14, size = 31, normalized size = 1.55 \begin {gather*} \frac {64}{- 10 e^{\frac {5 e}{4 x \log {\relax (3 )}}} + e^{\frac {5 e}{2 x \log {\relax (3 )}}} + 25} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(32*exp(ln(5)+1)*exp(1/4*exp(ln(5)+1)/x/ln(3))/(x**2*ln(3)*exp(1/4*exp(ln(5)+1)/x/ln(3))**3-15*x**2*l
n(3)*exp(1/4*exp(ln(5)+1)/x/ln(3))**2+75*x**2*ln(3)*exp(1/4*exp(ln(5)+1)/x/ln(3))-125*x**2*ln(3)),x)

[Out]

64/(-10*exp(5*E/(4*x*log(3))) + exp(5*E/(2*x*log(3))) + 25)

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