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

Optimal. Leaf size=22 \[ -1+\frac {-1+5 \left (e^4+x+\log (x)\right )^x}{2 x} \]

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Rubi [F]  time = 1.56, 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^4+x+\log (x)+\left (e^4+x+\log (x)\right )^x \left (-5 e^4+5 x^2-5 \log (x)+\left (5 e^4 x+5 x^2+5 x \log (x)\right ) \log \left (e^4+x+\log (x)\right )\right )}{2 e^4 x^2+2 x^3+2 x^2 \log (x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 0.33, size = 20, normalized size = 0.91 \begin {gather*} \frac {-1+5 \left (e^4+x+\log (x)\right )^x}{2 x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-1 + 5*(E^4 + x + Log[x])^x)/(2*x)

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fricas [A]  time = 0.86, size = 17, normalized size = 0.77 \begin {gather*} \frac {5 \, {\left (x + e^{4} + \log \relax (x)\right )}^{x} - 1}{2 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(5*(x + e^4 + log(x))^x - 1)/x

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {5 \, {\left (x^{2} + {\left (x^{2} + x e^{4} + x \log \relax (x)\right )} \log \left (x + e^{4} + \log \relax (x)\right ) - e^{4} - \log \relax (x)\right )} {\left (x + e^{4} + \log \relax (x)\right )}^{x} + x + e^{4} + \log \relax (x)}{2 \, {\left (x^{3} + x^{2} e^{4} + x^{2} \log \relax (x)\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/2*(5*(x^2 + (x^2 + x*e^4 + x*log(x))*log(x + e^4 + log(x)) - e^4 - log(x))*(x + e^4 + log(x))^x +
x + e^4 + log(x))/(x^3 + x^2*e^4 + x^2*log(x)), x)

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maple [A]  time = 0.15, size = 20, normalized size = 0.91

method	result	size
risch	\(-\frac {1}{2 x}+\frac {5 \left (\ln \relax (x )+x +{\mathrm e}^{4}\right )^{x}}{2 x}\)	\(20\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.39, size = 17, normalized size = 0.77 \begin {gather*} \frac {5 \, {\left (x + e^{4} + \log \relax (x)\right )}^{x} - 1}{2 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(5*(x + e^4 + log(x))^x - 1)/x

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mupad [B]  time = 5.81, size = 17, normalized size = 0.77 \begin {gather*} \frac {5\,{\left (x+{\mathrm {e}}^4+\ln \relax (x)\right )}^x-1}{2\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(5*(x + exp(4) + log(x))^x - 1)/(2*x)

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sympy [A]  time = 0.60, size = 22, normalized size = 1.00 \begin {gather*} \frac {5 e^{x \log {\left (x + \log {\relax (x )} + e^{4} \right )}}}{2 x} - \frac {1}{2 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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