3.68.86 \(\int \frac {2^{\frac {10 x}{6 \log (x)+2 e^{\frac {x^2}{e^5}} \log (x)}} (-15 \log (2)+15 \log (2) \log (x)+e^{\frac {x^2}{e^5}} (-5 \log (2)+5 \log (2) \log (x)-\frac {10 x^2 \log (2) \log (x)}{e^5}))}{9 \log ^2(x)+6 e^{\frac {x^2}{e^5}} \log ^2(x)+e^{\frac {2 x^2}{e^5}} \log ^2(x)} \, dx\)

Optimal. Leaf size=22 \[ 2^{\frac {5 x}{\left (3+e^{\frac {x^2}{e^5}}\right ) \log (x)}} \]

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Rubi [F]  time = 4.98, 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 {2^{\frac {10 x}{6 \log (x)+2 e^{\frac {x^2}{e^5}} \log (x)}} \left (-15 \log (2)+15 \log (2) \log (x)+e^{\frac {x^2}{e^5}} \left (-5 \log (2)+5 \log (2) \log (x)-\frac {10 x^2 \log (2) \log (x)}{e^5}\right )\right )}{9 \log ^2(x)+6 e^{\frac {x^2}{e^5}} \log ^2(x)+e^{\frac {2 x^2}{e^5}} \log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(2^((10*x)/(6*Log[x] + 2*E^(x^2/E^5)*Log[x]))*(-15*Log[2] + 15*Log[2]*Log[x] + E^(x^2/E^5)*(-5*Log[2] + 5*
Log[2]*Log[x] - (10*x^2*Log[2]*Log[x])/E^5)))/(9*Log[x]^2 + 6*E^(x^2/E^5)*Log[x]^2 + E^((2*x^2)/E^5)*Log[x]^2)
,x]

[Out]

-5*Log[2]*Defer[Int][2^((5*x)/((3 + E^(x^2/E^5))*Log[x]))/((3 + E^(x^2/E^5))*Log[x]^2), x] + 5*Log[2]*Defer[In
t][2^((5*x)/((3 + E^(x^2/E^5))*Log[x]))/((3 + E^(x^2/E^5))*Log[x]), x] + (15*Log[2]*Defer[Int][(2^(1 + (5*x)/(
(3 + E^(x^2/E^5))*Log[x]))*x^2)/((3 + E^(x^2/E^5))^2*Log[x]), x])/E^5 - (5*Log[2]*Defer[Int][(2^(1 + (5*x)/((3
+ E^(x^2/E^5))*Log[x]))*x^2)/((3 + E^(x^2/E^5))*Log[x]), x])/E^5

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {2^{\frac {5 x}{\left (3+e^{\frac {x^2}{e^5}}\right ) \log (x)}} \left (-15 \log (2)+15 \log (2) \log (x)+e^{\frac {x^2}{e^5}} \left (-5 \log (2)+5 \log (2) \log (x)-\frac {10 x^2 \log (2) \log (x)}{e^5}\right )\right )}{\left (3+e^{\frac {x^2}{e^5}}\right )^2 \log ^2(x)} \, dx\\ &=\int \left (\frac {15\ 2^{1+\frac {5 x}{\left (3+e^{\frac {x^2}{e^5}}\right ) \log (x)}} x^2 \log (2)}{e^5 \left (3+e^{\frac {x^2}{e^5}}\right )^2 \log (x)}+\frac {5\ 2^{\frac {5 x}{\left (3+e^{\frac {x^2}{e^5}}\right ) \log (x)}} \log (2) \left (-e^5+e^5 \log (x)-2 x^2 \log (x)\right )}{e^5 \left (3+e^{\frac {x^2}{e^5}}\right ) \log ^2(x)}\right ) \, dx\\ &=\frac {(5 \log (2)) \int \frac {2^{\frac {5 x}{\left (3+e^{\frac {x^2}{e^5}}\right ) \log (x)}} \left (-e^5+e^5 \log (x)-2 x^2 \log (x)\right )}{\left (3+e^{\frac {x^2}{e^5}}\right ) \log ^2(x)} \, dx}{e^5}+\frac {(15 \log (2)) \int \frac {2^{1+\frac {5 x}{\left (3+e^{\frac {x^2}{e^5}}\right ) \log (x)}} x^2}{\left (3+e^{\frac {x^2}{e^5}}\right )^2 \log (x)} \, dx}{e^5}\\ &=\frac {(5 \log (2)) \int \left (-\frac {2^{\frac {5 x}{\left (3+e^{\frac {x^2}{e^5}}\right ) \log (x)}} e^5}{\left (3+e^{\frac {x^2}{e^5}}\right ) \log ^2(x)}+\frac {2^{\frac {5 x}{\left (3+e^{\frac {x^2}{e^5}}\right ) \log (x)}} e^5}{\left (3+e^{\frac {x^2}{e^5}}\right ) \log (x)}-\frac {2^{1+\frac {5 x}{\left (3+e^{\frac {x^2}{e^5}}\right ) \log (x)}} x^2}{\left (3+e^{\frac {x^2}{e^5}}\right ) \log (x)}\right ) \, dx}{e^5}+\frac {(15 \log (2)) \int \frac {2^{1+\frac {5 x}{\left (3+e^{\frac {x^2}{e^5}}\right ) \log (x)}} x^2}{\left (3+e^{\frac {x^2}{e^5}}\right )^2 \log (x)} \, dx}{e^5}\\ &=-\left ((5 \log (2)) \int \frac {2^{\frac {5 x}{\left (3+e^{\frac {x^2}{e^5}}\right ) \log (x)}}}{\left (3+e^{\frac {x^2}{e^5}}\right ) \log ^2(x)} \, dx\right )+(5 \log (2)) \int \frac {2^{\frac {5 x}{\left (3+e^{\frac {x^2}{e^5}}\right ) \log (x)}}}{\left (3+e^{\frac {x^2}{e^5}}\right ) \log (x)} \, dx-\frac {(5 \log (2)) \int \frac {2^{1+\frac {5 x}{\left (3+e^{\frac {x^2}{e^5}}\right ) \log (x)}} x^2}{\left (3+e^{\frac {x^2}{e^5}}\right ) \log (x)} \, dx}{e^5}+\frac {(15 \log (2)) \int \frac {2^{1+\frac {5 x}{\left (3+e^{\frac {x^2}{e^5}}\right ) \log (x)}} x^2}{\left (3+e^{\frac {x^2}{e^5}}\right )^2 \log (x)} \, dx}{e^5}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.11, size = 30, normalized size = 1.36 \begin {gather*} \frac {5\ 2^{\frac {5 x}{\left (3+e^{\frac {x^2}{e^5}}\right ) \log (x)}} \log (2)}{\log (32)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2^((10*x)/(6*Log[x] + 2*E^(x^2/E^5)*Log[x]))*(-15*Log[2] + 15*Log[2]*Log[x] + E^(x^2/E^5)*(-5*Log[2
] + 5*Log[2]*Log[x] - (10*x^2*Log[2]*Log[x])/E^5)))/(9*Log[x]^2 + 6*E^(x^2/E^5)*Log[x]^2 + E^((2*x^2)/E^5)*Log
[x]^2),x]

[Out]

(5*2^((5*x)/((3 + E^(x^2/E^5))*Log[x]))*Log[2])/Log[32]

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fricas [A]  time = 0.62, size = 20, normalized size = 0.91 \begin {gather*} 2^{\frac {5 \, x}{{\left (e^{\left (x^{2} e^{\left (-5\right )}\right )} + 3\right )} \log \relax (x)}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-10*log(2)*log(x)*exp(log(x^2)-5)+5*log(2)*log(x)-5*log(2))*exp(exp(log(x^2)-5))+15*log(2)*log(x)-
15*log(2))*exp(5*x*log(2)/(2*log(x)*exp(exp(log(x^2)-5))+6*log(x)))^2/(log(x)^2*exp(exp(log(x^2)-5))^2+6*log(x
)^2*exp(exp(log(x^2)-5))+9*log(x)^2),x, algorithm="fricas")

[Out]

2^(5*x/((e^(x^2*e^(-5)) + 3)*log(x)))

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giac [A]  time = 0.16, size = 22, normalized size = 1.00 \begin {gather*} 2^{\frac {5 \, x}{e^{\left (x^{2} e^{\left (-5\right )}\right )} \log \relax (x) + 3 \, \log \relax (x)}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-10*log(2)*log(x)*exp(log(x^2)-5)+5*log(2)*log(x)-5*log(2))*exp(exp(log(x^2)-5))+15*log(2)*log(x)-
15*log(2))*exp(5*x*log(2)/(2*log(x)*exp(exp(log(x^2)-5))+6*log(x)))^2/(log(x)^2*exp(exp(log(x^2)-5))^2+6*log(x
)^2*exp(exp(log(x^2)-5))+9*log(x)^2),x, algorithm="giac")

[Out]

2^(5*x/(e^(x^2*e^(-5))*log(x) + 3*log(x)))

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maple [A]  time = 0.23, size = 23, normalized size = 1.05

method	result	size
risch	\(2^{\frac {5 x}{\ln \relax (x ) \left ({\mathrm e}^{x^{2} {\mathrm e}^{-5}}+3\right )}}\)	\(23\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-10*ln(2)*ln(x)*exp(ln(x^2)-5)+5*ln(2)*ln(x)-5*ln(2))*exp(exp(ln(x^2)-5))+15*ln(2)*ln(x)-15*ln(2))*exp(5
*x*ln(2)/(2*ln(x)*exp(exp(ln(x^2)-5))+6*ln(x)))^2/(ln(x)^2*exp(exp(ln(x^2)-5))^2+6*ln(x)^2*exp(exp(ln(x^2)-5))
+9*ln(x)^2),x,method=_RETURNVERBOSE)

[Out]

(2^(5/2*x/ln(x)/(exp(x^2*exp(-5))+3)))^2

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maxima [A]  time = 0.54, size = 20, normalized size = 0.91 \begin {gather*} 2^{\frac {5 \, x}{{\left (e^{\left (x^{2} e^{\left (-5\right )}\right )} + 3\right )} \log \relax (x)}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-10*log(2)*log(x)*exp(log(x^2)-5)+5*log(2)*log(x)-5*log(2))*exp(exp(log(x^2)-5))+15*log(2)*log(x)-
15*log(2))*exp(5*x*log(2)/(2*log(x)*exp(exp(log(x^2)-5))+6*log(x)))^2/(log(x)^2*exp(exp(log(x^2)-5))^2+6*log(x
)^2*exp(exp(log(x^2)-5))+9*log(x)^2),x, algorithm="maxima")

[Out]

2^(5*x/((e^(x^2*e^(-5)) + 3)*log(x)))

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mupad [B]  time = 4.64, size = 23, normalized size = 1.05 \begin {gather*} {\mathrm {e}}^{\frac {5\,x\,\ln \relax (2)}{3\,\ln \relax (x)+{\mathrm {e}}^{x^2\,{\mathrm {e}}^{-5}}\,\ln \relax (x)}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp((10*x*log(2))/(6*log(x) + 2*exp(exp(log(x^2) - 5))*log(x)))*(15*log(2) + exp(exp(log(x^2) - 5))*(5*l
og(2) - 5*log(2)*log(x) + 10*exp(log(x^2) - 5)*log(2)*log(x)) - 15*log(2)*log(x)))/(6*exp(exp(log(x^2) - 5))*l
og(x)^2 + 9*log(x)^2 + exp(2*exp(log(x^2) - 5))*log(x)^2),x)

[Out]

exp((5*x*log(2))/(3*log(x) + exp(x^2*exp(-5))*log(x)))

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sympy [A]  time = 1.13, size = 26, normalized size = 1.18 \begin {gather*} e^{\frac {10 x \log {\relax (2 )}}{2 e^{\frac {x^{2}}{e^{5}}} \log {\relax (x )} + 6 \log {\relax (x )}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-10*ln(2)*ln(x)*exp(ln(x**2)-5)+5*ln(2)*ln(x)-5*ln(2))*exp(exp(ln(x**2)-5))+15*ln(2)*ln(x)-15*ln(2
))*exp(5*x*ln(2)/(2*ln(x)*exp(exp(ln(x**2)-5))+6*ln(x)))**2/(ln(x)**2*exp(exp(ln(x**2)-5))**2+6*ln(x)**2*exp(e
xp(ln(x**2)-5))+9*ln(x)**2),x)

[Out]

exp(10*x*log(2)/(2*exp(x**2*exp(-5))*log(x) + 6*log(x)))

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