3.12.70 \(\int \frac {x \log ^3(2 x)+e^{\frac {1-10 x+4 \log ^2(2 x)}{2 \log ^2(2 x)}} (1-10 x+5 x \log (2 x))}{x \log ^3(2 x)} \, dx\)

Optimal. Leaf size=23 \[ 4-e^{2+\frac {\frac {1}{2}-5 x}{\log ^2(2 x)}}+x \]

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Rubi [F]  time = 2.16, 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 {x \log ^3(2 x)+e^{\frac {1-10 x+4 \log ^2(2 x)}{2 \log ^2(2 x)}} (1-10 x+5 x \log (2 x))}{x \log ^3(2 x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(x*Log[2*x]^3 + E^((1 - 10*x + 4*Log[2*x]^2)/(2*Log[2*x]^2))*(1 - 10*x + 5*x*Log[2*x]))/(x*Log[2*x]^3),x]

[Out]

x - 10*Defer[Int][E^(2 + (1/2 - 5*x)/Log[2*x]^2)/Log[2*x]^3, x] + Defer[Int][E^(2 + (1/2 - 5*x)/Log[2*x]^2)/(x
*Log[2*x]^3), x] + 5*Defer[Int][E^(2 + (1/2 - 5*x)/Log[2*x]^2)/Log[2*x]^2, x]

Rubi steps

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

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Mathematica [A]  time = 0.54, size = 23, normalized size = 1.00 \begin {gather*} -e^{2+\frac {1-10 x}{2 \log ^2(2 x)}}+x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x*Log[2*x]^3 + E^((1 - 10*x + 4*Log[2*x]^2)/(2*Log[2*x]^2))*(1 - 10*x + 5*x*Log[2*x]))/(x*Log[2*x]^
3),x]

[Out]

-E^(2 + (1 - 10*x)/(2*Log[2*x]^2)) + x

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fricas [A]  time = 0.62, size = 26, normalized size = 1.13 \begin {gather*} x - e^{\left (\frac {4 \, \log \left (2 \, x\right )^{2} - 10 \, x + 1}{2 \, \log \left (2 \, x\right )^{2}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*x*log(2*x)-10*x+1)*exp(1/2*(4*log(2*x)^2-10*x+1)/log(2*x)^2)+x*log(2*x)^3)/x/log(2*x)^3,x, algor
ithm="fricas")

[Out]

x - e^(1/2*(4*log(2*x)^2 - 10*x + 1)/log(2*x)^2)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \mathit {undef} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*x*log(2*x)-10*x+1)*exp(1/2*(4*log(2*x)^2-10*x+1)/log(2*x)^2)+x*log(2*x)^3)/x/log(2*x)^3,x, algor
ithm="giac")

[Out]

undef

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maple [A]  time = 0.04, size = 27, normalized size = 1.17

method	result	size
default	\(x -{\mathrm e}^{\frac {4 \ln \left (2 x \right )^{2}-10 x +1}{2 \ln \left (2 x \right )^{2}}}\)	\(27\)
risch	\(x -{\mathrm e}^{\frac {4 \ln \left (2 x \right )^{2}-10 x +1}{2 \ln \left (2 x \right )^{2}}}\)	\(27\)
norman	\(\frac {x \ln \left (2 x \right )^{2}-\ln \left (2 x \right )^{2} {\mathrm e}^{\frac {4 \ln \left (2 x \right )^{2}-10 x +1}{2 \ln \left (2 x \right )^{2}}}}{\ln \left (2 x \right )^{2}}\)	\(47\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((5*x*ln(2*x)-10*x+1)*exp(1/2*(4*ln(2*x)^2-10*x+1)/ln(2*x)^2)+x*ln(2*x)^3)/x/ln(2*x)^3,x,method=_RETURNVER
BOSE)

[Out]

x-exp(1/2*(4*ln(2*x)^2-10*x+1)/ln(2*x)^2)

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maxima [B]  time = 0.61, size = 46, normalized size = 2.00 \begin {gather*} x - e^{\left (-\frac {5 \, x}{\log \relax (2)^{2} + 2 \, \log \relax (2) \log \relax (x) + \log \relax (x)^{2}} + \frac {1}{2 \, {\left (\log \relax (2)^{2} + 2 \, \log \relax (2) \log \relax (x) + \log \relax (x)^{2}\right )}} + 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*x*log(2*x)-10*x+1)*exp(1/2*(4*log(2*x)^2-10*x+1)/log(2*x)^2)+x*log(2*x)^3)/x/log(2*x)^3,x, algor
ithm="maxima")

[Out]

x - e^(-5*x/(log(2)^2 + 2*log(2)*log(x) + log(x)^2) + 1/2/(log(2)^2 + 2*log(2)*log(x) + log(x)^2) + 2)

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mupad [B]  time = 0.94, size = 118, normalized size = 5.13 \begin {gather*} x-x^{\frac {4\,\ln \relax (2)}{{\ln \relax (x)}^2+2\,\ln \relax (2)\,\ln \relax (x)+{\ln \relax (2)}^2}}\,{\mathrm {e}}^{\frac {2\,{\ln \relax (2)}^2}{{\ln \relax (x)}^2+2\,\ln \relax (2)\,\ln \relax (x)+{\ln \relax (2)}^2}}\,{\mathrm {e}}^{\frac {1}{2\,\left ({\ln \relax (x)}^2+2\,\ln \relax (2)\,\ln \relax (x)+{\ln \relax (2)}^2\right )}}\,{\mathrm {e}}^{\frac {2\,{\ln \relax (x)}^2}{{\ln \relax (x)}^2+2\,\ln \relax (2)\,\ln \relax (x)+{\ln \relax (2)}^2}}\,{\mathrm {e}}^{-\frac {5\,x}{{\ln \relax (x)}^2+2\,\ln \relax (2)\,\ln \relax (x)+{\ln \relax (2)}^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp((2*log(2*x)^2 - 5*x + 1/2)/log(2*x)^2)*(5*x*log(2*x) - 10*x + 1) + x*log(2*x)^3)/(x*log(2*x)^3),x)

[Out]

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

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sympy [A]  time = 0.37, size = 24, normalized size = 1.04 \begin {gather*} x - e^{\frac {- 5 x + 2 \log {\left (2 x \right )}^{2} + \frac {1}{2}}{\log {\left (2 x \right )}^{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*x*ln(2*x)-10*x+1)*exp(1/2*(4*ln(2*x)**2-10*x+1)/ln(2*x)**2)+x*ln(2*x)**3)/x/ln(2*x)**3,x)

[Out]

x - exp((-5*x + 2*log(2*x)**2 + 1/2)/log(2*x)**2)

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