3.24.44 \(\int \frac {x^2+2 x \log (x)+\log ^2(x)+e^{\frac {2 (25+e x^2+e x \log (x))}{x^2+x \log (x)}} (-25-50 x+x^3+(-25+2 x^2) \log (x)+x \log ^2(x))}{x^2+2 x \log (x)+\log ^2(x)} \, dx\)

Optimal. Leaf size=27 \[ 4+x+\frac {1}{2} e^{2 e+\frac {50}{x (x+\log (x))}} x^2 \]

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Rubi [F]  time = 5.29, 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 {x^2+2 x \log (x)+\log ^2(x)+e^{\frac {2 \left (25+e x^2+e x \log (x)\right )}{x^2+x \log (x)}} \left (-25-50 x+x^3+\left (-25+2 x^2\right ) \log (x)+x \log ^2(x)\right )}{x^2+2 x \log (x)+\log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(x^2 + 2*x*Log[x] + Log[x]^2 + E^((2*(25 + E*x^2 + E*x*Log[x]))/(x^2 + x*Log[x]))*(-25 - 50*x + x^3 + (-25
+ 2*x^2)*Log[x] + x*Log[x]^2))/(x^2 + 2*x*Log[x] + Log[x]^2),x]

[Out]

x + Defer[Int][E^((2*(25 + E*x^2))/(x*(x + Log[x])))*x^(1 + (2*E)/(x + Log[x])), x] - 25*Defer[Int][(E^((2*(25
+ E*x^2))/(x*(x + Log[x])))*x^((2*E)/(x + Log[x])))/(x + Log[x])^2, x] - 25*Defer[Int][(E^((2*(25 + E*x^2))/(
x*(x + Log[x])))*x^(1 + (2*E)/(x + Log[x])))/(x + Log[x])^2, x] - 25*Defer[Int][(E^((2*(25 + E*x^2))/(x*(x + L
og[x])))*x^((2*E)/(x + Log[x])))/(x + Log[x]), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {x^2+2 x \log (x)+\log ^2(x)+e^{\frac {2 \left (25+e x^2+e x \log (x)\right )}{x^2+x \log (x)}} \left (-25-50 x+x^3+\left (-25+2 x^2\right ) \log (x)+x \log ^2(x)\right )}{(x+\log (x))^2} \, dx\\ &=\int \left (1+\frac {e^{\frac {2 \left (25+e x^2\right )}{x (x+\log (x))}} x^{\frac {2 e}{x+\log (x)}} \left (-25-50 x+x^3-25 \log (x)+2 x^2 \log (x)+x \log ^2(x)\right )}{(x+\log (x))^2}\right ) \, dx\\ &=x+\int \frac {e^{\frac {2 \left (25+e x^2\right )}{x (x+\log (x))}} x^{\frac {2 e}{x+\log (x)}} \left (-25-50 x+x^3-25 \log (x)+2 x^2 \log (x)+x \log ^2(x)\right )}{(x+\log (x))^2} \, dx\\ &=x+\int \left (e^{\frac {2 \left (25+e x^2\right )}{x (x+\log (x))}} x^{1+\frac {2 e}{x+\log (x)}}-\frac {25 e^{\frac {2 \left (25+e x^2\right )}{x (x+\log (x))}} x^{\frac {2 e}{x+\log (x)}} (1+x)}{(x+\log (x))^2}-\frac {25 e^{\frac {2 \left (25+e x^2\right )}{x (x+\log (x))}} x^{\frac {2 e}{x+\log (x)}}}{x+\log (x)}\right ) \, dx\\ &=x-25 \int \frac {e^{\frac {2 \left (25+e x^2\right )}{x (x+\log (x))}} x^{\frac {2 e}{x+\log (x)}} (1+x)}{(x+\log (x))^2} \, dx-25 \int \frac {e^{\frac {2 \left (25+e x^2\right )}{x (x+\log (x))}} x^{\frac {2 e}{x+\log (x)}}}{x+\log (x)} \, dx+\int e^{\frac {2 \left (25+e x^2\right )}{x (x+\log (x))}} x^{1+\frac {2 e}{x+\log (x)}} \, dx\\ &=x-25 \int \frac {e^{\frac {2 \left (25+e x^2\right )}{x (x+\log (x))}} x^{\frac {2 e}{x+\log (x)}}}{x+\log (x)} \, dx-25 \int \left (\frac {e^{\frac {2 \left (25+e x^2\right )}{x (x+\log (x))}} x^{\frac {2 e}{x+\log (x)}}}{(x+\log (x))^2}+\frac {e^{\frac {2 \left (25+e x^2\right )}{x (x+\log (x))}} x^{1+\frac {2 e}{x+\log (x)}}}{(x+\log (x))^2}\right ) \, dx+\int e^{\frac {2 \left (25+e x^2\right )}{x (x+\log (x))}} x^{1+\frac {2 e}{x+\log (x)}} \, dx\\ &=x-25 \int \frac {e^{\frac {2 \left (25+e x^2\right )}{x (x+\log (x))}} x^{\frac {2 e}{x+\log (x)}}}{(x+\log (x))^2} \, dx-25 \int \frac {e^{\frac {2 \left (25+e x^2\right )}{x (x+\log (x))}} x^{1+\frac {2 e}{x+\log (x)}}}{(x+\log (x))^2} \, dx-25 \int \frac {e^{\frac {2 \left (25+e x^2\right )}{x (x+\log (x))}} x^{\frac {2 e}{x+\log (x)}}}{x+\log (x)} \, dx+\int e^{\frac {2 \left (25+e x^2\right )}{x (x+\log (x))}} x^{1+\frac {2 e}{x+\log (x)}} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 3.51, size = 26, normalized size = 0.96 \begin {gather*} x+\frac {1}{2} e^{2 e+\frac {50}{x (x+\log (x))}} x^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^2 + 2*x*Log[x] + Log[x]^2 + E^((2*(25 + E*x^2 + E*x*Log[x]))/(x^2 + x*Log[x]))*(-25 - 50*x + x^3
+ (-25 + 2*x^2)*Log[x] + x*Log[x]^2))/(x^2 + 2*x*Log[x] + Log[x]^2),x]

[Out]

x + (E^(2*E + 50/(x*(x + Log[x])))*x^2)/2

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fricas [A]  time = 0.69, size = 34, normalized size = 1.26 \begin {gather*} \frac {1}{2} \, x^{2} e^{\left (\frac {2 \, {\left (x^{2} e + x e \log \relax (x) + 25\right )}}{x^{2} + x \log \relax (x)}\right )} + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*log(x)^2+(2*x^2-25)*log(x)+x^3-50*x-25)*exp((x*exp(1)*log(x)+x^2*exp(1)+25)/(x*log(x)+x^2))^2+lo
g(x)^2+2*x*log(x)+x^2)/(log(x)^2+2*x*log(x)+x^2),x, algorithm="fricas")

[Out]

1/2*x^2*e^(2*(x^2*e + x*e*log(x) + 25)/(x^2 + x*log(x))) + x

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giac [A]  time = 0.73, size = 34, normalized size = 1.26 \begin {gather*} \frac {1}{2} \, x^{2} e^{\left (\frac {2 \, {\left (x^{2} e + x e \log \relax (x) + 25\right )}}{x^{2} + x \log \relax (x)}\right )} + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*log(x)^2+(2*x^2-25)*log(x)+x^3-50*x-25)*exp((x*exp(1)*log(x)+x^2*exp(1)+25)/(x*log(x)+x^2))^2+lo
g(x)^2+2*x*log(x)+x^2)/(log(x)^2+2*x*log(x)+x^2),x, algorithm="giac")

[Out]

1/2*x^2*e^(2*(x^2*e + x*e*log(x) + 25)/(x^2 + x*log(x))) + x

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maple [A]  time = 0.13, size = 34, normalized size = 1.26

method	result	size
risch	\(x +\frac {x^{2} {\mathrm e}^{\frac {2 x \,{\mathrm e} \ln \relax (x )+2 x^{2} {\mathrm e}+50}{\left (x +\ln \relax (x )\right ) x}}}{2}\)	\(34\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x*ln(x)^2+(2*x^2-25)*ln(x)+x^3-50*x-25)*exp((x*exp(1)*ln(x)+x^2*exp(1)+25)/(x*ln(x)+x^2))^2+ln(x)^2+2*x*
ln(x)+x^2)/(ln(x)^2+2*x*ln(x)+x^2),x,method=_RETURNVERBOSE)

[Out]

x+1/2*x^2*exp(2*(x*exp(1)*ln(x)+x^2*exp(1)+25)/x/(x+ln(x)))

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*log(x)^2+(2*x^2-25)*log(x)+x^3-50*x-25)*exp((x*exp(1)*log(x)+x^2*exp(1)+25)/(x*log(x)+x^2))^2+lo
g(x)^2+2*x*log(x)+x^2)/(log(x)^2+2*x*log(x)+x^2),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: In function CAR, the value of the first argument is  0which is not
of the expected type LIST

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mupad [B]  time = 1.76, size = 55, normalized size = 2.04 \begin {gather*} x+\frac {x^{\frac {2\,x\,\mathrm {e}}{x\,\ln \relax (x)+x^2}}\,x^2\,{\mathrm {e}}^{\frac {50}{x\,\ln \relax (x)+x^2}}\,{\mathrm {e}}^{\frac {2\,x^2\,\mathrm {e}}{x\,\ln \relax (x)+x^2}}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(x)^2 + exp((2*(x^2*exp(1) + x*exp(1)*log(x) + 25))/(x*log(x) + x^2))*(x*log(x)^2 - 50*x + x^3 + log(x
)*(2*x^2 - 25) - 25) + 2*x*log(x) + x^2)/(log(x)^2 + 2*x*log(x) + x^2),x)

[Out]

x + (x^((2*x*exp(1))/(x*log(x) + x^2))*x^2*exp(50/(x*log(x) + x^2))*exp((2*x^2*exp(1))/(x*log(x) + x^2)))/2

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sympy [A]  time = 2.68, size = 36, normalized size = 1.33 \begin {gather*} \frac {x^{2} e^{\frac {2 \left (e x^{2} + e x \log {\relax (x )} + 25\right )}{x^{2} + x \log {\relax (x )}}}}{2} + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*ln(x)**2+(2*x**2-25)*ln(x)+x**3-50*x-25)*exp((x*exp(1)*ln(x)+x**2*exp(1)+25)/(x*ln(x)+x**2))**2+
ln(x)**2+2*x*ln(x)+x**2)/(ln(x)**2+2*x*ln(x)+x**2),x)

[Out]

x**2*exp(2*(E*x**2 + E*x*log(x) + 25)/(x**2 + x*log(x)))/2 + x

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