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

Optimal. Leaf size=30 \[ \frac {-3+e^{-5+x} \left (e^{1-x}-x^2\right ) \log (x)}{\log (e+x)} \]

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Rubi [F]  time = 5.81, 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 {3 x+e^{-5+x} \left (-e^{1-x} x+x^3\right ) \log (x)+\left (e^{-5+x} \left (-e x^2-x^3+e^{1-x} (e+x)\right )+e^{-5+x} \left (-2 x^3-x^4+e \left (-2 x^2-x^3\right )\right ) \log (x)\right ) \log (e+x)}{\left (e x+x^2\right ) \log ^2(e+x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 0.72, size = 31, normalized size = 1.03 \begin {gather*} -\frac {3 e^5-e \log (x)+e^x x^2 \log (x)}{e^5 \log (e+x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.77, size = 29, normalized size = 0.97 \begin {gather*} -\frac {{\left ({\left (x^{2} e^{\left (x - 1\right )} - 1\right )} \log \relax (x) + 3 \, e^{4}\right )} e^{\left (-4\right )}}{\log \left (x + e\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.22, size = 30, normalized size = 1.00 \begin {gather*} -\frac {{\left (x^{2} e^{x} \log \relax (x) - e \log \relax (x) + 3 \, e^{5}\right )} e^{\left (-5\right )}}{\log \left (x + e\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.30, size = 41, normalized size = 1.37

method	result	size
risch	\(-\frac {\left ({\mathrm e}^{-4} x^{2} \ln \relax (x )-\ln \relax (x ) {\mathrm e}^{-3-x}+3 \,{\mathrm e}^{1-x}\right ) {\mathrm e}^{x -1}}{\ln \left (x +{\mathrm e}\right )}\)	\(41\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(exp(-4)*x^2*ln(x)-ln(x)*exp(-3-x)+3*exp(1-x))*exp(x-1)/ln(x+exp(1))

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maxima [A]  time = 0.52, size = 36, normalized size = 1.20 \begin {gather*} -\frac {{\left (x^{2} e^{x} \log \relax (x) - e \log \relax (x)\right )} e^{\left (-5\right )}}{\log \left (x + e\right )} - \frac {3}{\log \left (x + e\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int -\frac {\ln \left (x+\mathrm {e}\right )\,\left ({\mathrm {e}}^{x-5}\,\left (x^2\,\mathrm {e}-{\mathrm {e}}^{1-x}\,\left (x+\mathrm {e}\right )+x^3\right )+{\mathrm {e}}^{x-5}\,\ln \relax (x)\,\left (\mathrm {e}\,\left (x^3+2\,x^2\right )+2\,x^3+x^4\right )\right )-3\,x+{\mathrm {e}}^{x-5}\,\ln \relax (x)\,\left (x\,{\mathrm {e}}^{1-x}-x^3\right )}{{\ln \left (x+\mathrm {e}\right )}^2\,\left (x^2+\mathrm {e}\,x\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.53, size = 39, normalized size = 1.30 \begin {gather*} - \frac {x^{2} e^{x - 1} \log {\relax (x )}}{e^{4} \log {\left (x + e \right )}} + \frac {\log {\relax (x )} - 3 e^{4}}{e^{4} \log {\left (x + e \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x**2*exp(-4)*exp(x - 1)*log(x)/log(x + E) + (log(x) - 3*exp(4))*exp(-4)/log(x + E)

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