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

Optimal. Leaf size=22 \[ 4-\log (5+x) \left (-e^{2+e^x} x+\log (\log (3))\right ) \]

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Rubi [F]  time = 1.39, 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^{2+e^x} x \left (x+\left (5+x+e^x \left (5 x+x^2\right )\right ) \log (5+x)\right )-x \log (\log (3))}{5 x+x^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

E^2*ExpIntegralEi[E^x] + E^2*ExpIntegralEi[E^x]*Log[5 + x] - Log[5 + x]*Log[Log[3]] + Log[5 + x]*Defer[Int][E^
(2 + E^x + x)*x, x] - 5*Defer[Int][E^(2 + E^x)/(5 + x), x] - E^2*Defer[Int][ExpIntegralEi[E^x]/(5 + x), x] - D
efer[Int][Defer[Int][E^(2 + E^x + x)*x, x]/(5 + x), x]

Rubi steps

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

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Mathematica [A]  time = 0.25, size = 20, normalized size = 0.91 \begin {gather*} \log (5+x) \left (e^{2+e^x} x-\log (\log (3))\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

Log[5 + x]*(E^(2 + E^x)*x - Log[Log[3]])

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fricas [A]  time = 0.57, size = 22, normalized size = 1.00 \begin {gather*} e^{\left (e^{x} + \log \relax (x) + 2\right )} \log \left (x + 5\right ) - \log \left (x + 5\right ) \log \left (\log \relax (3)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e^(e^x + log(x) + 2)*log(x + 5) - log(x + 5)*log(log(3))

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giac [A]  time = 0.20, size = 29, normalized size = 1.32 \begin {gather*} {\left (x e^{\left (x + e^{x} + 2\right )} \log \left (x + 5\right ) - e^{x} \log \left (x + 5\right ) \log \left (\log \relax (3)\right )\right )} e^{\left (-x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(x*e^(x + e^x + 2)*log(x + 5) - e^x*log(x + 5)*log(log(3)))*e^(-x)

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maple [A]  time = 0.29, size = 22, normalized size = 1.00

method	result	size
risch	\(-\ln \left (\ln \relax (3)\right ) \ln \left (5+x \right )+\ln \left (5+x \right ) {\mathrm e}^{{\mathrm e}^{x}+2} x\)	\(22\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-ln(ln(3))*ln(5+x)+ln(5+x)*exp(exp(x)+2)*x

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maxima [A]  time = 0.42, size = 21, normalized size = 0.95 \begin {gather*} x e^{\left (e^{x} + 2\right )} \log \left (x + 5\right ) - \log \left (x + 5\right ) \log \left (\log \relax (3)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*e^(e^x + 2)*log(x + 5) - log(x + 5)*log(log(3))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.05 \begin {gather*} -\int -\frac {{\mathrm {e}}^{{\mathrm {e}}^x+\ln \relax (x)+2}\,\left (x+\ln \left (x+5\right )\,\left (x+{\mathrm {e}}^x\,\left (x^2+5\,x\right )+5\right )\right )-x\,\ln \left (\ln \relax (3)\right )}{x^2+5\,x} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 11.45, size = 22, normalized size = 1.00 \begin {gather*} x e^{e^{x} + 2} \log {\left (x + 5 \right )} - \log {\left (x + 5 \right )} \log {\left (\log {\relax (3 )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*exp(exp(x) + 2)*log(x + 5) - log(x + 5)*log(log(3))

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