3.27.13 \(\int \frac {10 e^x x+e^x (20 x+10 x^2) \log (x)}{4+5 e^x x^2 \log (x)} \, dx\)

Optimal. Leaf size=19 \[ -5+\log \left (\left (-1-\frac {5}{4} e^x x^2 \log (x)\right )^2\right ) \]

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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 {10 e^x x+e^x \left (20 x+10 x^2\right ) \log (x)}{4+5 e^x x^2 \log (x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(10*E^x*x + E^x*(20*x + 10*x^2)*Log[x])/(4 + 5*E^x*x^2*Log[x]),x]

[Out]

2*x - 8*Defer[Int][(4 + 5*E^x*x^2*Log[x])^(-1), x] + 10*Defer[Int][(E^x*x)/(4 + 5*E^x*x^2*Log[x]), x] + 20*Def
er[Int][(E^x*x*Log[x])/(4 + 5*E^x*x^2*Log[x]), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {10 e^x x (1+(2+x) \log (x))}{4+5 e^x x^2 \log (x)} \, dx\\ &=10 \int \frac {e^x x (1+(2+x) \log (x))}{4+5 e^x x^2 \log (x)} \, dx\\ &=10 \int \left (\frac {e^x x}{4+5 e^x x^2 \log (x)}+\frac {2 e^x x \log (x)}{4+5 e^x x^2 \log (x)}+\frac {e^x x^2 \log (x)}{4+5 e^x x^2 \log (x)}\right ) \, dx\\ &=10 \int \frac {e^x x}{4+5 e^x x^2 \log (x)} \, dx+10 \int \frac {e^x x^2 \log (x)}{4+5 e^x x^2 \log (x)} \, dx+20 \int \frac {e^x x \log (x)}{4+5 e^x x^2 \log (x)} \, dx\\ &=10 \int \frac {e^x x}{4+5 e^x x^2 \log (x)} \, dx+10 \int \left (\frac {1}{5}-\frac {4}{5 \left (4+5 e^x x^2 \log (x)\right )}\right ) \, dx+20 \int \frac {e^x x \log (x)}{4+5 e^x x^2 \log (x)} \, dx\\ &=2 x-8 \int \frac {1}{4+5 e^x x^2 \log (x)} \, dx+10 \int \frac {e^x x}{4+5 e^x x^2 \log (x)} \, dx+20 \int \frac {e^x x \log (x)}{4+5 e^x x^2 \log (x)} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.15, size = 15, normalized size = 0.79 \begin {gather*} 2 \log \left (4+5 e^x x^2 \log (x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(10*E^x*x + E^x*(20*x + 10*x^2)*Log[x])/(4 + 5*E^x*x^2*Log[x]),x]

[Out]

2*Log[4 + 5*E^x*x^2*Log[x]]

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fricas [A]  time = 0.61, size = 30, normalized size = 1.58 \begin {gather*} 2 \, x + 4 \, \log \relax (x) + 2 \, \log \left (\frac {{\left (5 \, x^{2} e^{x} \log \relax (x) + 4\right )} e^{\left (-x\right )}}{x^{2}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((10*x^2+20*x)*exp(x)*log(x)+10*exp(x)*x)/(5*x^2*exp(x)*log(x)+4),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.15, size = 14, normalized size = 0.74 \begin {gather*} 2 \, \log \left (5 \, x^{2} e^{x} \log \relax (x) + 4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((10*x^2+20*x)*exp(x)*log(x)+10*exp(x)*x)/(5*x^2*exp(x)*log(x)+4),x, algorithm="giac")

[Out]

2*log(5*x^2*e^x*log(x) + 4)

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maple [A]  time = 0.02, size = 15, normalized size = 0.79

method	result	size
norman	\(2 \ln \left (5 x^{2} {\mathrm e}^{x} \ln \relax (x )+4\right )\)	\(15\)
risch	\(2 x +4 \ln \relax (x )+2 \ln \left (\ln \relax (x )+\frac {4 \,{\mathrm e}^{-x}}{5 x^{2}}\right )\)	\(24\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((10*x^2+20*x)*exp(x)*ln(x)+10*exp(x)*x)/(5*x^2*exp(x)*ln(x)+4),x,method=_RETURNVERBOSE)

[Out]

2*ln(5*x^2*exp(x)*ln(x)+4)

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maxima [A]  time = 0.49, size = 33, normalized size = 1.74 \begin {gather*} 4 \, \log \relax (x) + 2 \, \log \left (\frac {5 \, x^{2} e^{x} \log \relax (x) + 4}{5 \, x^{2} \log \relax (x)}\right ) + 2 \, \log \left (\log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((10*x^2+20*x)*exp(x)*log(x)+10*exp(x)*x)/(5*x^2*exp(x)*log(x)+4),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.66, size = 14, normalized size = 0.74 \begin {gather*} 2\,\ln \left (5\,x^2\,{\mathrm {e}}^x\,\ln \relax (x)+4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((10*x*exp(x) + exp(x)*log(x)*(20*x + 10*x^2))/(5*x^2*exp(x)*log(x) + 4),x)

[Out]

2*log(5*x^2*exp(x)*log(x) + 4)

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sympy [A]  time = 0.39, size = 27, normalized size = 1.42 \begin {gather*} 4 \log {\relax (x )} + 2 \log {\left (e^{x} + \frac {4}{5 x^{2} \log {\relax (x )}} \right )} + 2 \log {\left (\log {\relax (x )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((10*x**2+20*x)*exp(x)*ln(x)+10*exp(x)*x)/(5*x**2*exp(x)*ln(x)+4),x)

[Out]

4*log(x) + 2*log(exp(x) + 4/(5*x**2*log(x))) + 2*log(log(x))

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