3.56.74 \(\int \frac {-32 e^9+32 e^{9+x} x+160 e^9 x \log (x)+80 e^{9+x} x^2 \log ^2(x)}{2 x+5 x^2 \log ^2(x)} \, dx\)

Optimal. Leaf size=22 \[ 16 e^9 \left (e^x+\log \left (\frac {2}{5 x}+\log ^2(x)\right )\right ) \]

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Rubi [F]  time = 1.17, 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 {-32 e^9+32 e^{9+x} x+160 e^9 x \log (x)+80 e^{9+x} x^2 \log ^2(x)}{2 x+5 x^2 \log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-32*E^9 + 32*E^(9 + x)*x + 160*E^9*x*Log[x] + 80*E^(9 + x)*x^2*Log[x]^2)/(2*x + 5*x^2*Log[x]^2),x]

[Out]

16*E^(9 + x) - 32*E^9*Defer[Int][1/(x*(2 + 5*x*Log[x]^2)), x] + 160*E^9*Defer[Int][Log[x]/(2 + 5*x*Log[x]^2),
x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-32 e^9+32 e^{9+x} x+160 e^9 x \log (x)+80 e^{9+x} x^2 \log ^2(x)}{x^2 \left (\frac {2}{x}+5 \log ^2(x)\right )} \, dx\\ &=\int \frac {-32 e^9+32 e^{9+x} x+160 e^9 x \log (x)+80 e^{9+x} x^2 \log ^2(x)}{x \left (2+5 x \log ^2(x)\right )} \, dx\\ &=\int \frac {16 e^9 \left (-2+2 e^x x+10 x \log (x)+5 e^x x^2 \log ^2(x)\right )}{x \left (2+5 x \log ^2(x)\right )} \, dx\\ &=\left (16 e^9\right ) \int \frac {-2+2 e^x x+10 x \log (x)+5 e^x x^2 \log ^2(x)}{x \left (2+5 x \log ^2(x)\right )} \, dx\\ &=\left (16 e^9\right ) \int \left (e^x+\frac {2 (-1+5 x \log (x))}{x \left (2+5 x \log ^2(x)\right )}\right ) \, dx\\ &=\left (16 e^9\right ) \int e^x \, dx+\left (32 e^9\right ) \int \frac {-1+5 x \log (x)}{x \left (2+5 x \log ^2(x)\right )} \, dx\\ &=16 e^{9+x}+\left (32 e^9\right ) \int \left (-\frac {1}{x \left (2+5 x \log ^2(x)\right )}+\frac {5 \log (x)}{2+5 x \log ^2(x)}\right ) \, dx\\ &=16 e^{9+x}-\left (32 e^9\right ) \int \frac {1}{x \left (2+5 x \log ^2(x)\right )} \, dx+\left (160 e^9\right ) \int \frac {\log (x)}{2+5 x \log ^2(x)} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.12, size = 23, normalized size = 1.05 \begin {gather*} 16 e^9 \left (e^x-\log (x)+\log \left (2+5 x \log ^2(x)\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-32*E^9 + 32*E^(9 + x)*x + 160*E^9*x*Log[x] + 80*E^(9 + x)*x^2*Log[x]^2)/(2*x + 5*x^2*Log[x]^2),x]

[Out]

16*E^9*(E^x - Log[x] + Log[2 + 5*x*Log[x]^2])

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fricas [A]  time = 0.68, size = 25, normalized size = 1.14 \begin {gather*} 16 \, e^{9} \log \left (\frac {5 \, x \log \relax (x)^{2} + 2}{x}\right ) + 16 \, e^{\left (x + 9\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((80*x^2*exp(9)*exp(x)*log(x)^2+160*x*exp(9)*log(x)+32*x*exp(9)*exp(x)-32*exp(9))/(5*x^2*log(x)^2+2*x
),x, algorithm="fricas")

[Out]

16*e^9*log((5*x*log(x)^2 + 2)/x) + 16*e^(x + 9)

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giac [A]  time = 0.30, size = 27, normalized size = 1.23 \begin {gather*} 16 \, e^{9} \log \left (5 \, x \log \relax (x)^{2} + 2\right ) - 16 \, e^{9} \log \relax (x) + 16 \, e^{\left (x + 9\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((80*x^2*exp(9)*exp(x)*log(x)^2+160*x*exp(9)*log(x)+32*x*exp(9)*exp(x)-32*exp(9))/(5*x^2*log(x)^2+2*x
),x, algorithm="giac")

[Out]

16*e^9*log(5*x*log(x)^2 + 2) - 16*e^9*log(x) + 16*e^(x + 9)

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maple [A]  time = 0.16, size = 23, normalized size = 1.05

method	result	size
risch	\(16 \,{\mathrm e}^{x +9}+16 \,{\mathrm e}^{9} \ln \left (\ln \relax (x )^{2}+\frac {2}{5 x}\right )\)	\(23\)
default	\(-16 \,{\mathrm e}^{9} \ln \relax (x )+16 \,{\mathrm e}^{9} \ln \left (5 x \ln \relax (x )^{2}+2\right )+16 \,{\mathrm e}^{9} {\mathrm e}^{x}\)	\(28\)
norman	\(-16 \,{\mathrm e}^{9} \ln \relax (x )+16 \,{\mathrm e}^{9} \ln \left (5 x \ln \relax (x )^{2}+2\right )+16 \,{\mathrm e}^{9} {\mathrm e}^{x}\)	\(28\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((80*x^2*exp(9)*exp(x)*ln(x)^2+160*x*exp(9)*ln(x)+32*x*exp(9)*exp(x)-32*exp(9))/(5*x^2*ln(x)^2+2*x),x,metho
d=_RETURNVERBOSE)

[Out]

16*exp(x+9)+16*exp(9)*ln(ln(x)^2+2/5/x)

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maxima [A]  time = 0.39, size = 26, normalized size = 1.18 \begin {gather*} 16 \, e^{9} \log \left (\frac {5 \, x \log \relax (x)^{2} + 2}{5 \, x}\right ) + 16 \, e^{\left (x + 9\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((80*x^2*exp(9)*exp(x)*log(x)^2+160*x*exp(9)*log(x)+32*x*exp(9)*exp(x)-32*exp(9))/(5*x^2*log(x)^2+2*x
),x, algorithm="maxima")

[Out]

16*e^9*log(1/5*(5*x*log(x)^2 + 2)/x) + 16*e^(x + 9)

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mupad [B]  time = 3.63, size = 25, normalized size = 1.14 \begin {gather*} 16\,{\mathrm {e}}^9\,{\mathrm {e}}^x+16\,\ln \left (\frac {5\,x\,{\ln \relax (x)}^2+2}{x}\right )\,{\mathrm {e}}^9 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((32*x*exp(9)*exp(x) - 32*exp(9) + 160*x*exp(9)*log(x) + 80*x^2*exp(9)*exp(x)*log(x)^2)/(2*x + 5*x^2*log(x)
^2),x)

[Out]

16*exp(9)*exp(x) + 16*log((5*x*log(x)^2 + 2)/x)*exp(9)

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sympy [A]  time = 0.35, size = 24, normalized size = 1.09 \begin {gather*} 16 e^{9} e^{x} + 16 e^{9} \log {\left (\log {\relax (x )}^{2} + \frac {2}{5 x} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((80*x**2*exp(9)*exp(x)*ln(x)**2+160*x*exp(9)*ln(x)+32*x*exp(9)*exp(x)-32*exp(9))/(5*x**2*ln(x)**2+2*
x),x)

[Out]

16*exp(9)*exp(x) + 16*exp(9)*log(log(x)**2 + 2/(5*x))

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