3.36.87 \(\int \frac {e^{\frac {25+(10+10 x) \log (2 x)+(1+2 x+x^2) \log ^2(2 x)}{25+10 x \log (2 x)+x^2 \log ^2(2 x)}} (50+(10+10 x) \log (2 x)-10 x \log ^2(2 x)+(-2 x-2 x^2) \log ^3(2 x))}{125 x+75 x^2 \log (2 x)+15 x^3 \log ^2(2 x)+x^4 \log ^3(2 x)} \, dx\)

Optimal. Leaf size=20 \[ 1+e^{\left (1+\frac {1}{x+\frac {5}{\log (2 x)}}\right )^2} \]

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Rubi [F]  time = 8.42, 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 {\exp \left (\frac {25+(10+10 x) \log (2 x)+\left (1+2 x+x^2\right ) \log ^2(2 x)}{25+10 x \log (2 x)+x^2 \log ^2(2 x)}\right ) \left (50+(10+10 x) \log (2 x)-10 x \log ^2(2 x)+\left (-2 x-2 x^2\right ) \log ^3(2 x)\right )}{125 x+75 x^2 \log (2 x)+15 x^3 \log ^2(2 x)+x^4 \log ^3(2 x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^((25 + (10 + 10*x)*Log[2*x] + (1 + 2*x + x^2)*Log[2*x]^2)/(25 + 10*x*Log[2*x] + x^2*Log[2*x]^2))*(50 +
(10 + 10*x)*Log[2*x] - 10*x*Log[2*x]^2 + (-2*x - 2*x^2)*Log[2*x]^3))/(125*x + 75*x^2*Log[2*x] + 15*x^3*Log[2*x
]^2 + x^4*Log[2*x]^3),x]

[Out]

-2*Defer[Int][E^((5 + Log[2*x] + x*Log[2*x])^2/(5 + x*Log[2*x])^2)/x^3, x] - 2*Defer[Int][E^((5 + Log[2*x] + x
*Log[2*x])^2/(5 + x*Log[2*x])^2)/x^2, x] + 250*Defer[Int][E^((5 + Log[2*x] + x*Log[2*x])^2/(5 + x*Log[2*x])^2)
/(x^3*(5 + x*Log[2*x])^3), x] - 50*Defer[Int][E^((5 + Log[2*x] + x*Log[2*x])^2/(5 + x*Log[2*x])^2)/(x^2*(5 + x
*Log[2*x])^3), x] - 150*Defer[Int][E^((5 + Log[2*x] + x*Log[2*x])^2/(5 + x*Log[2*x])^2)/(x^3*(5 + x*Log[2*x])^
2), x] - 40*Defer[Int][E^((5 + Log[2*x] + x*Log[2*x])^2/(5 + x*Log[2*x])^2)/(x^2*(5 + x*Log[2*x])^2), x] + 10*
Defer[Int][E^((5 + Log[2*x] + x*Log[2*x])^2/(5 + x*Log[2*x])^2)/(x*(5 + x*Log[2*x])^2), x] + 30*Defer[Int][E^(
(5 + Log[2*x] + x*Log[2*x])^2/(5 + x*Log[2*x])^2)/(x^3*(5 + x*Log[2*x])), x] + 20*Defer[Int][E^((5 + Log[2*x]
+ x*Log[2*x])^2/(5 + x*Log[2*x])^2)/(x^2*(5 + x*Log[2*x])), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\exp \left (\frac {(5+\log (2 x)+x \log (2 x))^2}{(5+x \log (2 x))^2}\right ) \left (50+(10+10 x) \log (2 x)-10 x \log ^2(2 x)+\left (-2 x-2 x^2\right ) \log ^3(2 x)\right )}{x (5+x \log (2 x))^3} \, dx\\ &=\int \left (-\frac {2 \exp \left (\frac {(5+\log (2 x)+x \log (2 x))^2}{(5+x \log (2 x))^2}\right ) (1+x)}{x^3}-\frac {50 \exp \left (\frac {(5+\log (2 x)+x \log (2 x))^2}{(5+x \log (2 x))^2}\right ) (-5+x)}{x^3 (5+x \log (2 x))^3}+\frac {10 \exp \left (\frac {(5+\log (2 x)+x \log (2 x))^2}{(5+x \log (2 x))^2}\right ) \left (-15-4 x+x^2\right )}{x^3 (5+x \log (2 x))^2}+\frac {10 \exp \left (\frac {(5+\log (2 x)+x \log (2 x))^2}{(5+x \log (2 x))^2}\right ) (3+2 x)}{x^3 (5+x \log (2 x))}\right ) \, dx\\ &=-\left (2 \int \frac {\exp \left (\frac {(5+\log (2 x)+x \log (2 x))^2}{(5+x \log (2 x))^2}\right ) (1+x)}{x^3} \, dx\right )+10 \int \frac {\exp \left (\frac {(5+\log (2 x)+x \log (2 x))^2}{(5+x \log (2 x))^2}\right ) \left (-15-4 x+x^2\right )}{x^3 (5+x \log (2 x))^2} \, dx+10 \int \frac {\exp \left (\frac {(5+\log (2 x)+x \log (2 x))^2}{(5+x \log (2 x))^2}\right ) (3+2 x)}{x^3 (5+x \log (2 x))} \, dx-50 \int \frac {\exp \left (\frac {(5+\log (2 x)+x \log (2 x))^2}{(5+x \log (2 x))^2}\right ) (-5+x)}{x^3 (5+x \log (2 x))^3} \, dx\\ &=-\left (2 \int \left (\frac {\exp \left (\frac {(5+\log (2 x)+x \log (2 x))^2}{(5+x \log (2 x))^2}\right )}{x^3}+\frac {\exp \left (\frac {(5+\log (2 x)+x \log (2 x))^2}{(5+x \log (2 x))^2}\right )}{x^2}\right ) \, dx\right )+10 \int \left (-\frac {15 \exp \left (\frac {(5+\log (2 x)+x \log (2 x))^2}{(5+x \log (2 x))^2}\right )}{x^3 (5+x \log (2 x))^2}-\frac {4 \exp \left (\frac {(5+\log (2 x)+x \log (2 x))^2}{(5+x \log (2 x))^2}\right )}{x^2 (5+x \log (2 x))^2}+\frac {\exp \left (\frac {(5+\log (2 x)+x \log (2 x))^2}{(5+x \log (2 x))^2}\right )}{x (5+x \log (2 x))^2}\right ) \, dx+10 \int \left (\frac {3 \exp \left (\frac {(5+\log (2 x)+x \log (2 x))^2}{(5+x \log (2 x))^2}\right )}{x^3 (5+x \log (2 x))}+\frac {2 \exp \left (\frac {(5+\log (2 x)+x \log (2 x))^2}{(5+x \log (2 x))^2}\right )}{x^2 (5+x \log (2 x))}\right ) \, dx-50 \int \left (-\frac {5 \exp \left (\frac {(5+\log (2 x)+x \log (2 x))^2}{(5+x \log (2 x))^2}\right )}{x^3 (5+x \log (2 x))^3}+\frac {\exp \left (\frac {(5+\log (2 x)+x \log (2 x))^2}{(5+x \log (2 x))^2}\right )}{x^2 (5+x \log (2 x))^3}\right ) \, dx\\ &=-\left (2 \int \frac {\exp \left (\frac {(5+\log (2 x)+x \log (2 x))^2}{(5+x \log (2 x))^2}\right )}{x^3} \, dx\right )-2 \int \frac {\exp \left (\frac {(5+\log (2 x)+x \log (2 x))^2}{(5+x \log (2 x))^2}\right )}{x^2} \, dx+10 \int \frac {\exp \left (\frac {(5+\log (2 x)+x \log (2 x))^2}{(5+x \log (2 x))^2}\right )}{x (5+x \log (2 x))^2} \, dx+20 \int \frac {\exp \left (\frac {(5+\log (2 x)+x \log (2 x))^2}{(5+x \log (2 x))^2}\right )}{x^2 (5+x \log (2 x))} \, dx+30 \int \frac {\exp \left (\frac {(5+\log (2 x)+x \log (2 x))^2}{(5+x \log (2 x))^2}\right )}{x^3 (5+x \log (2 x))} \, dx-40 \int \frac {\exp \left (\frac {(5+\log (2 x)+x \log (2 x))^2}{(5+x \log (2 x))^2}\right )}{x^2 (5+x \log (2 x))^2} \, dx-50 \int \frac {\exp \left (\frac {(5+\log (2 x)+x \log (2 x))^2}{(5+x \log (2 x))^2}\right )}{x^2 (5+x \log (2 x))^3} \, dx-150 \int \frac {\exp \left (\frac {(5+\log (2 x)+x \log (2 x))^2}{(5+x \log (2 x))^2}\right )}{x^3 (5+x \log (2 x))^2} \, dx+250 \int \frac {\exp \left (\frac {(5+\log (2 x)+x \log (2 x))^2}{(5+x \log (2 x))^2}\right )}{x^3 (5+x \log (2 x))^3} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [F]  time = 5.36, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {e^{\frac {25+(10+10 x) \log (2 x)+\left (1+2 x+x^2\right ) \log ^2(2 x)}{25+10 x \log (2 x)+x^2 \log ^2(2 x)}} \left (50+(10+10 x) \log (2 x)-10 x \log ^2(2 x)+\left (-2 x-2 x^2\right ) \log ^3(2 x)\right )}{125 x+75 x^2 \log (2 x)+15 x^3 \log ^2(2 x)+x^4 \log ^3(2 x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[(E^((25 + (10 + 10*x)*Log[2*x] + (1 + 2*x + x^2)*Log[2*x]^2)/(25 + 10*x*Log[2*x] + x^2*Log[2*x]^2))*
(50 + (10 + 10*x)*Log[2*x] - 10*x*Log[2*x]^2 + (-2*x - 2*x^2)*Log[2*x]^3))/(125*x + 75*x^2*Log[2*x] + 15*x^3*L
og[2*x]^2 + x^4*Log[2*x]^3),x]

[Out]

Integrate[(E^((25 + (10 + 10*x)*Log[2*x] + (1 + 2*x + x^2)*Log[2*x]^2)/(25 + 10*x*Log[2*x] + x^2*Log[2*x]^2))*
(50 + (10 + 10*x)*Log[2*x] - 10*x*Log[2*x]^2 + (-2*x - 2*x^2)*Log[2*x]^3))/(125*x + 75*x^2*Log[2*x] + 15*x^3*L
og[2*x]^2 + x^4*Log[2*x]^3), x]

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fricas [B]  time = 0.87, size = 49, normalized size = 2.45 \begin {gather*} e^{\left (\frac {{\left (x^{2} + 2 \, x + 1\right )} \log \left (2 \, x\right )^{2} + 10 \, {\left (x + 1\right )} \log \left (2 \, x\right ) + 25}{x^{2} \log \left (2 \, x\right )^{2} + 10 \, x \log \left (2 \, x\right ) + 25}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x^2-2*x)*log(2*x)^3-10*x*log(2*x)^2+(10*x+10)*log(2*x)+50)*exp(((x^2+2*x+1)*log(2*x)^2+(10*x+10
)*log(2*x)+25)/(x^2*log(2*x)^2+10*x*log(2*x)+25))/(x^4*log(2*x)^3+15*x^3*log(2*x)^2+75*x^2*log(2*x)+125*x),x,
algorithm="fricas")

[Out]

e^(((x^2 + 2*x + 1)*log(2*x)^2 + 10*(x + 1)*log(2*x) + 25)/(x^2*log(2*x)^2 + 10*x*log(2*x) + 25))

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giac [B]  time = 48.24, size = 169, normalized size = 8.45 \begin {gather*} e^{\left (\frac {x^{2} \log \left (2 \, x\right )^{2}}{x^{2} \log \left (2 \, x\right )^{2} + 10 \, x \log \left (2 \, x\right ) + 25} + \frac {2 \, x \log \left (2 \, x\right )^{2}}{x^{2} \log \left (2 \, x\right )^{2} + 10 \, x \log \left (2 \, x\right ) + 25} + \frac {10 \, x \log \left (2 \, x\right )}{x^{2} \log \left (2 \, x\right )^{2} + 10 \, x \log \left (2 \, x\right ) + 25} + \frac {\log \left (2 \, x\right )^{2}}{x^{2} \log \left (2 \, x\right )^{2} + 10 \, x \log \left (2 \, x\right ) + 25} + \frac {10 \, \log \left (2 \, x\right )}{x^{2} \log \left (2 \, x\right )^{2} + 10 \, x \log \left (2 \, x\right ) + 25} + \frac {25}{x^{2} \log \left (2 \, x\right )^{2} + 10 \, x \log \left (2 \, x\right ) + 25}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x^2-2*x)*log(2*x)^3-10*x*log(2*x)^2+(10*x+10)*log(2*x)+50)*exp(((x^2+2*x+1)*log(2*x)^2+(10*x+10
)*log(2*x)+25)/(x^2*log(2*x)^2+10*x*log(2*x)+25))/(x^4*log(2*x)^3+15*x^3*log(2*x)^2+75*x^2*log(2*x)+125*x),x,
algorithm="giac")

[Out]

e^(x^2*log(2*x)^2/(x^2*log(2*x)^2 + 10*x*log(2*x) + 25) + 2*x*log(2*x)^2/(x^2*log(2*x)^2 + 10*x*log(2*x) + 25)
+ 10*x*log(2*x)/(x^2*log(2*x)^2 + 10*x*log(2*x) + 25) + log(2*x)^2/(x^2*log(2*x)^2 + 10*x*log(2*x) + 25) + 10
*log(2*x)/(x^2*log(2*x)^2 + 10*x*log(2*x) + 25) + 25/(x^2*log(2*x)^2 + 10*x*log(2*x) + 25))

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maple [A]  time = 0.07, size = 27, normalized size = 1.35

method	result	size
risch	\({\mathrm e}^{\frac {\left (x \ln \left (2 x \right )+\ln \left (2 x \right )+5\right )^{2}}{\left (x \ln \left (2 x \right )+5\right )^{2}}}\)	\(27\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-2*x^2-2*x)*ln(2*x)^3-10*x*ln(2*x)^2+(10*x+10)*ln(2*x)+50)*exp(((x^2+2*x+1)*ln(2*x)^2+(10*x+10)*ln(2*x)+
25)/(x^2*ln(2*x)^2+10*x*ln(2*x)+25))/(x^4*ln(2*x)^3+15*x^3*ln(2*x)^2+75*x^2*ln(2*x)+125*x),x,method=_RETURNVER
BOSE)

[Out]

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

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maxima [B]  time = 1.38, size = 168, normalized size = 8.40 \begin {gather*} e^{\left (\frac {\log \relax (2)^{2}}{x^{2} \log \relax (2)^{2} + x^{2} \log \relax (x)^{2} + 10 \, x \log \relax (2) + 2 \, {\left (x^{2} \log \relax (2) + 5 \, x\right )} \log \relax (x) + 25} + \frac {2 \, \log \relax (2) \log \relax (x)}{x^{2} \log \relax (2)^{2} + x^{2} \log \relax (x)^{2} + 10 \, x \log \relax (2) + 2 \, {\left (x^{2} \log \relax (2) + 5 \, x\right )} \log \relax (x) + 25} + \frac {\log \relax (x)^{2}}{x^{2} \log \relax (2)^{2} + x^{2} \log \relax (x)^{2} + 10 \, x \log \relax (2) + 2 \, {\left (x^{2} \log \relax (2) + 5 \, x\right )} \log \relax (x) + 25} + \frac {2 \, \log \relax (2)}{x \log \relax (2) + x \log \relax (x) + 5} + \frac {2 \, \log \relax (x)}{x \log \relax (2) + x \log \relax (x) + 5} + 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x^2-2*x)*log(2*x)^3-10*x*log(2*x)^2+(10*x+10)*log(2*x)+50)*exp(((x^2+2*x+1)*log(2*x)^2+(10*x+10
)*log(2*x)+25)/(x^2*log(2*x)^2+10*x*log(2*x)+25))/(x^4*log(2*x)^3+15*x^3*log(2*x)^2+75*x^2*log(2*x)+125*x),x,
algorithm="maxima")

[Out]

e^(log(2)^2/(x^2*log(2)^2 + x^2*log(x)^2 + 10*x*log(2) + 2*(x^2*log(2) + 5*x)*log(x) + 25) + 2*log(2)*log(x)/(
x^2*log(2)^2 + x^2*log(x)^2 + 10*x*log(2) + 2*(x^2*log(2) + 5*x)*log(x) + 25) + log(x)^2/(x^2*log(2)^2 + x^2*l
og(x)^2 + 10*x*log(2) + 2*(x^2*log(2) + 5*x)*log(x) + 25) + 2*log(2)/(x*log(2) + x*log(x) + 5) + 2*log(x)/(x*l
og(2) + x*log(x) + 5) + 1)

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mupad [B]  time = 2.60, size = 429, normalized size = 21.45 \begin {gather*} {1024}^{\frac {x+1}{x^2\,{\ln \relax (x)}^2+2\,\ln \relax (2)\,x^2\,\ln \relax (x)+{\ln \relax (2)}^2\,x^2+10\,x\,\ln \relax (x)+10\,\ln \relax (2)\,x+25}}\,x^{\frac {2\,\left (5\,x+\ln \relax (2)+2\,x\,\ln \relax (2)+x^2\,\ln \relax (2)+5\right )}{x^2\,{\ln \relax (x)}^2+2\,\ln \relax (2)\,x^2\,\ln \relax (x)+{\ln \relax (2)}^2\,x^2+10\,x\,\ln \relax (x)+10\,\ln \relax (2)\,x+25}}\,{\mathrm {e}}^{\frac {x^2\,{\ln \relax (2)}^2}{x^2\,{\ln \relax (x)}^2+2\,\ln \relax (2)\,x^2\,\ln \relax (x)+{\ln \relax (2)}^2\,x^2+10\,x\,\ln \relax (x)+10\,\ln \relax (2)\,x+25}}\,{\mathrm {e}}^{\frac {2\,x\,{\ln \relax (x)}^2}{x^2\,{\ln \relax (x)}^2+2\,\ln \relax (2)\,x^2\,\ln \relax (x)+{\ln \relax (2)}^2\,x^2+10\,x\,\ln \relax (x)+10\,\ln \relax (2)\,x+25}}\,{\mathrm {e}}^{\frac {{\ln \relax (2)}^2}{x^2\,{\ln \relax (x)}^2+2\,\ln \relax (2)\,x^2\,\ln \relax (x)+{\ln \relax (2)}^2\,x^2+10\,x\,\ln \relax (x)+10\,\ln \relax (2)\,x+25}}\,{\mathrm {e}}^{\frac {25}{x^2\,{\ln \relax (x)}^2+2\,\ln \relax (2)\,x^2\,\ln \relax (x)+{\ln \relax (2)}^2\,x^2+10\,x\,\ln \relax (x)+10\,\ln \relax (2)\,x+25}}\,{\mathrm {e}}^{\frac {x^2\,{\ln \relax (x)}^2}{x^2\,{\ln \relax (x)}^2+2\,\ln \relax (2)\,x^2\,\ln \relax (x)+{\ln \relax (2)}^2\,x^2+10\,x\,\ln \relax (x)+10\,\ln \relax (2)\,x+25}}\,{\mathrm {e}}^{\frac {{\ln \relax (x)}^2}{x^2\,{\ln \relax (x)}^2+2\,\ln \relax (2)\,x^2\,\ln \relax (x)+{\ln \relax (2)}^2\,x^2+10\,x\,\ln \relax (x)+10\,\ln \relax (2)\,x+25}}\,{\mathrm {e}}^{\frac {2\,x\,{\ln \relax (2)}^2}{x^2\,{\ln \relax (x)}^2+2\,\ln \relax (2)\,x^2\,\ln \relax (x)+{\ln \relax (2)}^2\,x^2+10\,x\,\ln \relax (x)+10\,\ln \relax (2)\,x+25}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp((log(2*x)^2*(2*x + x^2 + 1) + log(2*x)*(10*x + 10) + 25)/(10*x*log(2*x) + x^2*log(2*x)^2 + 25))*(log
(2*x)^3*(2*x + 2*x^2) + 10*x*log(2*x)^2 - log(2*x)*(10*x + 10) - 50))/(125*x + 75*x^2*log(2*x) + 15*x^3*log(2*
x)^2 + x^4*log(2*x)^3),x)

[Out]

1024^((x + 1)/(x^2*log(2)^2 + 10*x*log(2) + x^2*log(x)^2 + 10*x*log(x) + 2*x^2*log(2)*log(x) + 25))*x^((2*(5*x
+ log(2) + 2*x*log(2) + x^2*log(2) + 5))/(x^2*log(2)^2 + 10*x*log(2) + x^2*log(x)^2 + 10*x*log(x) + 2*x^2*log
(2)*log(x) + 25))*exp((x^2*log(2)^2)/(x^2*log(2)^2 + 10*x*log(2) + x^2*log(x)^2 + 10*x*log(x) + 2*x^2*log(2)*l
og(x) + 25))*exp((2*x*log(x)^2)/(x^2*log(2)^2 + 10*x*log(2) + x^2*log(x)^2 + 10*x*log(x) + 2*x^2*log(2)*log(x)
+ 25))*exp(log(2)^2/(x^2*log(2)^2 + 10*x*log(2) + x^2*log(x)^2 + 10*x*log(x) + 2*x^2*log(2)*log(x) + 25))*exp
(25/(x^2*log(2)^2 + 10*x*log(2) + x^2*log(x)^2 + 10*x*log(x) + 2*x^2*log(2)*log(x) + 25))*exp((x^2*log(x)^2)/(
x^2*log(2)^2 + 10*x*log(2) + x^2*log(x)^2 + 10*x*log(x) + 2*x^2*log(2)*log(x) + 25))*exp(log(x)^2/(x^2*log(2)^
2 + 10*x*log(2) + x^2*log(x)^2 + 10*x*log(x) + 2*x^2*log(2)*log(x) + 25))*exp((2*x*log(2)^2)/(x^2*log(2)^2 + 1
0*x*log(2) + x^2*log(x)^2 + 10*x*log(x) + 2*x^2*log(2)*log(x) + 25))

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sympy [B]  time = 0.86, size = 48, normalized size = 2.40 \begin {gather*} e^{\frac {\left (10 x + 10\right ) \log {\left (2 x \right )} + \left (x^{2} + 2 x + 1\right ) \log {\left (2 x \right )}^{2} + 25}{x^{2} \log {\left (2 x \right )}^{2} + 10 x \log {\left (2 x \right )} + 25}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x**2-2*x)*ln(2*x)**3-10*x*ln(2*x)**2+(10*x+10)*ln(2*x)+50)*exp(((x**2+2*x+1)*ln(2*x)**2+(10*x+1
0)*ln(2*x)+25)/(x**2*ln(2*x)**2+10*x*ln(2*x)+25))/(x**4*ln(2*x)**3+15*x**3*ln(2*x)**2+75*x**2*ln(2*x)+125*x),x
)

[Out]

exp(((10*x + 10)*log(2*x) + (x**2 + 2*x + 1)*log(2*x)**2 + 25)/(x**2*log(2*x)**2 + 10*x*log(2*x) + 25))

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