3.62.14 \(\int \frac {e^{\frac {5+e^{\frac {5+6 x+2 x^2}{1+x}}+x-\log (x)}{x}} (-6-12 x-6 x^2+e^{\frac {5+6 x+2 x^2}{1+x}} (-1-x+3 x^2+2 x^3)+(1+2 x+x^2) \log (x))}{x^2+2 x^3+x^4} \, dx\)

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

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Rubi [F]  time = 25.81, 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 {e^{\frac {5+e^{\frac {5+6 x+2 x^2}{1+x}}+x-\log (x)}{x}} \left (-6-12 x-6 x^2+e^{\frac {5+6 x+2 x^2}{1+x}} \left (-1-x+3 x^2+2 x^3\right )+\left (1+2 x+x^2\right ) \log (x)\right )}{x^2+2 x^3+x^4} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^((5 + E^((5 + 6*x + 2*x^2)/(1 + x)) + x - Log[x])/x)*(-6 - 12*x - 6*x^2 + E^((5 + 6*x + 2*x^2)/(1 + x))
*(-1 - x + 3*x^2 + 2*x^3) + (1 + 2*x + x^2)*Log[x]))/(x^2 + 2*x^3 + x^4),x]

[Out]

-6*Defer[Int][E^(1 + 5/x + E^((5 + 6*x + 2*x^2)/(1 + x))/x - Log[x]/x)/x^2, x] - Defer[Int][E^(1 + 5/x + E^((5
+ 6*x + 2*x^2)/(1 + x))/x + (5 + 6*x + 2*x^2)/(1 + x) - Log[x]/x)/x^2, x] + Defer[Int][E^(1 + 5/x + E^((5 + 6
*x + 2*x^2)/(1 + x))/x + (5 + 6*x + 2*x^2)/(1 + x) - Log[x]/x)/x, x] + Defer[Int][E^(1 + 5/x + E^((5 + 6*x + 2
*x^2)/(1 + x))/x + (5 + 6*x + 2*x^2)/(1 + x) - Log[x]/x)/(1 + x)^2, x] + Defer[Int][E^(1 + 5/x + E^((5 + 6*x +
2*x^2)/(1 + x))/x + (5 + 6*x + 2*x^2)/(1 + x) - Log[x]/x)/(1 + x), x] + Defer[Int][(E^(1 + 5/x + E^((5 + 6*x
+ 2*x^2)/(1 + x))/x - Log[x]/x)*Log[x])/x^2, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{\frac {5+e^{\frac {5+6 x+2 x^2}{1+x}}+x-\log (x)}{x}} \left (-6-12 x-6 x^2+e^{\frac {5+6 x+2 x^2}{1+x}} \left (-1-x+3 x^2+2 x^3\right )+\left (1+2 x+x^2\right ) \log (x)\right )}{x^2 \left (1+2 x+x^2\right )} \, dx\\ &=\int \frac {e^{\frac {5+e^{\frac {5+6 x+2 x^2}{1+x}}+x-\log (x)}{x}} \left (-6-12 x-6 x^2+e^{\frac {5+6 x+2 x^2}{1+x}} \left (-1-x+3 x^2+2 x^3\right )+\left (1+2 x+x^2\right ) \log (x)\right )}{x^2 (1+x)^2} \, dx\\ &=\int \frac {e^{1+\frac {5}{x}+\frac {e^{\frac {5+6 x+2 x^2}{1+x}}}{x}-\frac {\log (x)}{x}} \left (-6-12 x-6 x^2+e^{\frac {5+6 x+2 x^2}{1+x}} \left (-1-x+3 x^2+2 x^3\right )+\left (1+2 x+x^2\right ) \log (x)\right )}{x^2 (1+x)^2} \, dx\\ &=\int \left (\frac {\exp \left (1+\frac {5}{x}+\frac {e^{\frac {5+6 x+2 x^2}{1+x}}}{x}+\frac {5+6 x+2 x^2}{1+x}-\frac {\log (x)}{x}\right ) (1+2 x) \left (-1+x+x^2\right )}{x^2 (1+x)^2}+\frac {e^{1+\frac {5}{x}+\frac {e^{\frac {5+6 x+2 x^2}{1+x}}}{x}-\frac {\log (x)}{x}} (-6+\log (x))}{x^2}\right ) \, dx\\ &=\int \frac {\exp \left (1+\frac {5}{x}+\frac {e^{\frac {5+6 x+2 x^2}{1+x}}}{x}+\frac {5+6 x+2 x^2}{1+x}-\frac {\log (x)}{x}\right ) (1+2 x) \left (-1+x+x^2\right )}{x^2 (1+x)^2} \, dx+\int \frac {e^{1+\frac {5}{x}+\frac {e^{\frac {5+6 x+2 x^2}{1+x}}}{x}-\frac {\log (x)}{x}} (-6+\log (x))}{x^2} \, dx\\ &=\int \left (-\frac {\exp \left (1+\frac {5}{x}+\frac {e^{\frac {5+6 x+2 x^2}{1+x}}}{x}+\frac {5+6 x+2 x^2}{1+x}-\frac {\log (x)}{x}\right )}{x^2}+\frac {\exp \left (1+\frac {5}{x}+\frac {e^{\frac {5+6 x+2 x^2}{1+x}}}{x}+\frac {5+6 x+2 x^2}{1+x}-\frac {\log (x)}{x}\right )}{x}+\frac {\exp \left (1+\frac {5}{x}+\frac {e^{\frac {5+6 x+2 x^2}{1+x}}}{x}+\frac {5+6 x+2 x^2}{1+x}-\frac {\log (x)}{x}\right )}{(1+x)^2}+\frac {\exp \left (1+\frac {5}{x}+\frac {e^{\frac {5+6 x+2 x^2}{1+x}}}{x}+\frac {5+6 x+2 x^2}{1+x}-\frac {\log (x)}{x}\right )}{1+x}\right ) \, dx+\int \left (-\frac {6 e^{1+\frac {5}{x}+\frac {e^{\frac {5+6 x+2 x^2}{1+x}}}{x}-\frac {\log (x)}{x}}}{x^2}+\frac {e^{1+\frac {5}{x}+\frac {e^{\frac {5+6 x+2 x^2}{1+x}}}{x}-\frac {\log (x)}{x}} \log (x)}{x^2}\right ) \, dx\\ &=-\left (6 \int \frac {e^{1+\frac {5}{x}+\frac {e^{\frac {5+6 x+2 x^2}{1+x}}}{x}-\frac {\log (x)}{x}}}{x^2} \, dx\right )-\int \frac {\exp \left (1+\frac {5}{x}+\frac {e^{\frac {5+6 x+2 x^2}{1+x}}}{x}+\frac {5+6 x+2 x^2}{1+x}-\frac {\log (x)}{x}\right )}{x^2} \, dx+\int \frac {\exp \left (1+\frac {5}{x}+\frac {e^{\frac {5+6 x+2 x^2}{1+x}}}{x}+\frac {5+6 x+2 x^2}{1+x}-\frac {\log (x)}{x}\right )}{x} \, dx+\int \frac {\exp \left (1+\frac {5}{x}+\frac {e^{\frac {5+6 x+2 x^2}{1+x}}}{x}+\frac {5+6 x+2 x^2}{1+x}-\frac {\log (x)}{x}\right )}{(1+x)^2} \, dx+\int \frac {\exp \left (1+\frac {5}{x}+\frac {e^{\frac {5+6 x+2 x^2}{1+x}}}{x}+\frac {5+6 x+2 x^2}{1+x}-\frac {\log (x)}{x}\right )}{1+x} \, dx+\int \frac {e^{1+\frac {5}{x}+\frac {e^{\frac {5+6 x+2 x^2}{1+x}}}{x}-\frac {\log (x)}{x}} \log (x)}{x^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.24, size = 29, normalized size = 0.97 \begin {gather*} e^{\frac {5+e^{4+2 x+\frac {1}{1+x}}+x}{x}} x^{-1/x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((5 + E^((5 + 6*x + 2*x^2)/(1 + x)) + x - Log[x])/x)*(-6 - 12*x - 6*x^2 + E^((5 + 6*x + 2*x^2)/(1
+ x))*(-1 - x + 3*x^2 + 2*x^3) + (1 + 2*x + x^2)*Log[x]))/(x^2 + 2*x^3 + x^4),x]

[Out]

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

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fricas [A]  time = 0.49, size = 29, normalized size = 0.97 \begin {gather*} e^{\left (\frac {x + e^{\left (\frac {2 \, x^{2} + 6 \, x + 5}{x + 1}\right )} - \log \relax (x) + 5}{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^2+2*x+1)*log(x)+(2*x^3+3*x^2-x-1)*exp((2*x^2+6*x+5)/(x+1))-6*x^2-12*x-6)*exp((-log(x)+exp((2*x^2
+6*x+5)/(x+1))+5+x)/x)/(x^4+2*x^3+x^2),x, algorithm="fricas")

[Out]

e^((x + e^((2*x^2 + 6*x + 5)/(x + 1)) - log(x) + 5)/x)

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giac [A]  time = 0.45, size = 46, normalized size = 1.53 \begin {gather*} e^{\left (\frac {e^{\left (\frac {2 \, x^{2}}{x + 1} + \frac {6 \, x}{x + 1} + \frac {5}{x + 1}\right )}}{x} - \frac {\log \relax (x)}{x} + \frac {5}{x} + 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^2+2*x+1)*log(x)+(2*x^3+3*x^2-x-1)*exp((2*x^2+6*x+5)/(x+1))-6*x^2-12*x-6)*exp((-log(x)+exp((2*x^2
+6*x+5)/(x+1))+5+x)/x)/(x^4+2*x^3+x^2),x, algorithm="giac")

[Out]

e^(e^(2*x^2/(x + 1) + 6*x/(x + 1) + 5/(x + 1))/x - log(x)/x + 5/x + 1)

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maple [A]  time = 0.03, size = 33, normalized size = 1.10

method	result	size
risch	\({\mathrm e}^{-\frac {\ln \relax (x )-{\mathrm e}^{\frac {2 x^{2}+6 x +5}{x +1}}-5-x}{x}}\)	\(33\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x^2+2*x+1)*ln(x)+(2*x^3+3*x^2-x-1)*exp((2*x^2+6*x+5)/(x+1))-6*x^2-12*x-6)*exp((-ln(x)+exp((2*x^2+6*x+5)/
(x+1))+5+x)/x)/(x^4+2*x^3+x^2),x,method=_RETURNVERBOSE)

[Out]

exp(-(ln(x)-exp((2*x^2+6*x+5)/(x+1))-5-x)/x)

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maxima [A]  time = 0.59, size = 30, normalized size = 1.00 \begin {gather*} e^{\left (\frac {e^{\left (2 \, x + \frac {1}{x + 1} + 4\right )}}{x} - \frac {\log \relax (x)}{x} + \frac {5}{x} + 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^2+2*x+1)*log(x)+(2*x^3+3*x^2-x-1)*exp((2*x^2+6*x+5)/(x+1))-6*x^2-12*x-6)*exp((-log(x)+exp((2*x^2
+6*x+5)/(x+1))+5+x)/x)/(x^4+2*x^3+x^2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 4.91, size = 49, normalized size = 1.63 \begin {gather*} \frac {\mathrm {e}\,{\mathrm {e}}^{5/x}\,{\mathrm {e}}^{\frac {{\mathrm {e}}^{\frac {6\,x}{x+1}}\,{\mathrm {e}}^{\frac {2\,x^2}{x+1}}\,{\mathrm {e}}^{\frac {5}{x+1}}}{x}}}{x^{1/x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp((x + exp((6*x + 2*x^2 + 5)/(x + 1)) - log(x) + 5)/x)*(12*x + exp((6*x + 2*x^2 + 5)/(x + 1))*(x - 3*x
^2 - 2*x^3 + 1) - log(x)*(2*x + x^2 + 1) + 6*x^2 + 6))/(x^2 + 2*x^3 + x^4),x)

[Out]

(exp(1)*exp(5/x)*exp((exp((6*x)/(x + 1))*exp((2*x^2)/(x + 1))*exp(5/(x + 1)))/x))/x^(1/x)

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sympy [A]  time = 1.13, size = 24, normalized size = 0.80 \begin {gather*} e^{\frac {x + e^{\frac {2 x^{2} + 6 x + 5}{x + 1}} - \log {\relax (x )} + 5}{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x**2+2*x+1)*ln(x)+(2*x**3+3*x**2-x-1)*exp((2*x**2+6*x+5)/(x+1))-6*x**2-12*x-6)*exp((-ln(x)+exp((2*
x**2+6*x+5)/(x+1))+5+x)/x)/(x**4+2*x**3+x**2),x)

[Out]

exp((x + exp((2*x**2 + 6*x + 5)/(x + 1)) - log(x) + 5)/x)

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