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

Optimal. Leaf size=21 \[ 4+\frac {2}{3} e^{5+x+\frac {1}{e^x+x}} x \log (3) \]

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Rubi [F]  time = 6.44, 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 {\exp \left (\frac {1+5 x+x^2+e^x (5+x)+\left (e^x+x\right ) \log (x)}{e^x+x}\right ) \left (e^{2 x} (2+2 x) \log (3)+e^x \left (2 x+4 x^2\right ) \log (3)+\left (-2 x+2 x^2+2 x^3\right ) \log (3)\right )}{3 e^{2 x} x+6 e^x x^2+3 x^3} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 0.67, size = 19, normalized size = 0.90 \begin {gather*} \frac {2}{3} e^{5+x+\frac {1}{e^x+x}} x \log (3) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.60, size = 33, normalized size = 1.57 \begin {gather*} \frac {2}{3} \, e^{\left (\frac {x^{2} + {\left (x + 5\right )} e^{x} + {\left (x + e^{x}\right )} \log \relax (x) + 5 \, x + 1}{x + e^{x}}\right )} \log \relax (3) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*e^((x^2 + (x + 5)*e^x + (x + e^x)*log(x) + 5*x + 1)/(x + e^x))*log(3)

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giac [A]  time = 0.48, size = 29, normalized size = 1.38 \begin {gather*} \frac {2}{3} \, x e^{\left (\frac {x^{2} + x e^{x} + 5 \, x + 5 \, e^{x} + 1}{x + e^{x}}\right )} \log \relax (3) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*x*e^((x^2 + x*e^x + 5*x + 5*e^x + 1)/(x + e^x))*log(3)

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maple [A]  time = 0.05, size = 38, normalized size = 1.81

method	result	size
risch	\(\frac {2 \ln \relax (3) {\mathrm e}^{\frac {{\mathrm e}^{x} \ln \relax (x )+x \ln \relax (x )+{\mathrm e}^{x} x +x^{2}+5 \,{\mathrm e}^{x}+5 x +1}{{\mathrm e}^{x}+x}}}{3}\)	\(38\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*ln(3)*exp((exp(x)*ln(x)+x*ln(x)+exp(x)*x+x^2+5*exp(x)+5*x+1)/(exp(x)+x))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {2}{3} \, \int \frac {{\left ({\left (x + 1\right )} e^{\left (2 \, x\right )} \log \relax (3) + {\left (2 \, x^{2} + x\right )} e^{x} \log \relax (3) + {\left (x^{3} + x^{2} - x\right )} \log \relax (3)\right )} e^{\left (\frac {x^{2} + {\left (x + 5\right )} e^{x} + {\left (x + e^{x}\right )} \log \relax (x) + 5 \, x + 1}{x + e^{x}}\right )}}{x^{3} + 2 \, x^{2} e^{x} + x e^{\left (2 \, x\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.05 \begin {gather*} \int \frac {{\mathrm {e}}^{\frac {5\,x+{\mathrm {e}}^x\,\left (x+5\right )+x^2+\ln \relax (x)\,\left (x+{\mathrm {e}}^x\right )+1}{x+{\mathrm {e}}^x}}\,\left (\ln \relax (3)\,\left (2\,x^3+2\,x^2-2\,x\right )+{\mathrm {e}}^x\,\ln \relax (3)\,\left (4\,x^2+2\,x\right )+{\mathrm {e}}^{2\,x}\,\ln \relax (3)\,\left (2\,x+2\right )\right )}{3\,x\,{\mathrm {e}}^{2\,x}+6\,x^2\,{\mathrm {e}}^x+3\,x^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.70, size = 36, normalized size = 1.71 \begin {gather*} \frac {2 e^{\frac {x^{2} + 5 x + \left (x + 5\right ) e^{x} + \left (x + e^{x}\right ) \log {\relax (x )} + 1}{x + e^{x}}} \log {\relax (3 )}}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*exp((x**2 + 5*x + (x + 5)*exp(x) + (x + exp(x))*log(x) + 1)/(x + exp(x)))*log(3)/3

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