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

Optimal. Leaf size=22 \[ \frac {(-e+x) (-2 (2+2 x)+\log (x+\log (x)))}{x} \]

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Rubi [F]  time = 0.96, 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 {e (-1-5 x)+x+x^2-4 x^3+\left (-4 e-4 x^2\right ) \log (x)+(e x+e \log (x)) \log (x+\log (x))}{x^3+x^2 \log (x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e (-1-5 x)+x+x^2-4 x^3+\left (-4 e-4 x^2\right ) \log (x)+(e x+e \log (x)) \log (x+\log (x))}{x^2 (x+\log (x))} \, dx\\ &=\int \left (\frac {-e+(1-5 e) x+x^2-4 x^3-4 e \log (x)-4 x^2 \log (x)}{x^2 (x+\log (x))}+\frac {e \log (x+\log (x))}{x^2}\right ) \, dx\\ &=e \int \frac {\log (x+\log (x))}{x^2} \, dx+\int \frac {-e+(1-5 e) x+x^2-4 x^3-4 e \log (x)-4 x^2 \log (x)}{x^2 (x+\log (x))} \, dx\\ &=e \int \frac {\log (x+\log (x))}{x^2} \, dx+\int \left (-\frac {4 \left (e+x^2\right )}{x^2}-\frac {(e-x) (1+x)}{x^2 (x+\log (x))}\right ) \, dx\\ &=-\left (4 \int \frac {e+x^2}{x^2} \, dx\right )+e \int \frac {\log (x+\log (x))}{x^2} \, dx-\int \frac {(e-x) (1+x)}{x^2 (x+\log (x))} \, dx\\ &=-\left (4 \int \left (1+\frac {e}{x^2}\right ) \, dx\right )+e \int \frac {\log (x+\log (x))}{x^2} \, dx-\int \left (\frac {1}{-x-\log (x)}+\frac {e}{x^2 (x+\log (x))}+\frac {-1+e}{x (x+\log (x))}\right ) \, dx\\ &=\frac {4 e}{x}-4 x-(-1+e) \int \frac {1}{x (x+\log (x))} \, dx-e \int \frac {1}{x^2 (x+\log (x))} \, dx+e \int \frac {\log (x+\log (x))}{x^2} \, dx-\int \frac {1}{-x-\log (x)} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.19, size = 26, normalized size = 1.18 \begin {gather*} \frac {4 e}{x}-4 x+\log (x+\log (x))-\frac {e \log (x+\log (x))}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(4*E)/x - 4*x + Log[x + Log[x]] - (E*Log[x + Log[x]])/x

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fricas [A]  time = 0.65, size = 28, normalized size = 1.27 \begin {gather*} -\frac {4 \, x^{2} - {\left (x - e\right )} \log \left (x + \log \relax (x)\right ) - 4 \, e}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(1)*log(x)+x*exp(1))*log(x+log(x))+(-4*exp(1)-4*x^2)*log(x)+(-5*x-1)*exp(1)-4*x^3+x^2+x)/(x^2*l
og(x)+x^3),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.24, size = 35, normalized size = 1.59 \begin {gather*} -\frac {4 \, x^{2} + e \log \left (x + \log \relax (x)\right ) - x \log \left (-x - \log \relax (x)\right ) - 4 \, e}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(1)*log(x)+x*exp(1))*log(x+log(x))+(-4*exp(1)-4*x^2)*log(x)+(-5*x-1)*exp(1)-4*x^3+x^2+x)/(x^2*l
og(x)+x^3),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.05, size = 35, normalized size = 1.59

method	result	size
risch	\(-\frac {{\mathrm e} \ln \left (x +\ln \relax (x )\right )}{x}+\frac {\ln \left (x +\ln \relax (x )\right ) x -4 x^{2}+4 \,{\mathrm e}}{x}\)	\(35\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.49, size = 28, normalized size = 1.27 \begin {gather*} -\frac {4 \, x^{2} - {\left (x - e\right )} \log \left (x + \log \relax (x)\right ) - 4 \, e}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(1)*log(x)+x*exp(1))*log(x+log(x))+(-4*exp(1)-4*x^2)*log(x)+(-5*x-1)*exp(1)-4*x^3+x^2+x)/(x^2*l
og(x)+x^3),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.16, size = 28, normalized size = 1.27 \begin {gather*} \ln \left (x+\ln \relax (x)\right )-4\,x+\frac {4\,\mathrm {e}}{x}-\frac {\ln \left (x+\ln \relax (x)\right )\,\mathrm {e}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x - log(x)*(4*exp(1) + 4*x^2) + log(x + log(x))*(x*exp(1) + exp(1)*log(x)) + x^2 - 4*x^3 - exp(1)*(5*x +
1))/(x^2*log(x) + x^3),x)

[Out]

log(x + log(x)) - 4*x + (4*exp(1))/x - (log(x + log(x))*exp(1))/x

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sympy [A]  time = 0.43, size = 27, normalized size = 1.23 \begin {gather*} - 4 x + \log {\left (x + \log {\relax (x )} \right )} - \frac {e \log {\left (x + \log {\relax (x )} \right )}}{x} + \frac {4 e}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4*x + log(x + log(x)) - E*log(x + log(x))/x + 4*E/x

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