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

Optimal. Leaf size=24 \[ \log \left (e^3+2 e^{4+\frac {x^2}{e^2}} x \left (3+x^2\right )\right ) \]

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Rubi [F]  time = 3.20, 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 {4 e^2+x^2+e^2 \log (x)}{e^2}} \left (12 x^2+4 x^4+e^2 \left (6+6 x^2\right )\right )}{e^5 x+e^{2+\frac {4 e^2+x^2+e^2 \log (x)}{e^2}} \left (6 x+2 x^3\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

x^2/E^2 + Log[x] + Log[3 + x^2] + Defer[Int][1/((I*Sqrt[3] - x)*(1 + 2*E^(1 + x^2/E^2)*x*(3 + x^2))), x] - Def
er[Int][1/(x*(1 + 2*E^(1 + x^2/E^2)*x*(3 + x^2))), x] - (2*Defer[Int][x/(1 + 2*E^(1 + x^2/E^2)*x*(3 + x^2)), x
])/E^2 - Defer[Int][1/((I*Sqrt[3] + x)*(1 + 2*E^(1 + x^2/E^2)*x*(3 + x^2))), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{\frac {4 e^2+x^2}{e^2}} x \left (12 x^2+4 x^4+e^2 \left (6+6 x^2\right )\right )}{e^5 x+e^{2+\frac {4 e^2+x^2+e^2 \log (x)}{e^2}} \left (6 x+2 x^3\right )} \, dx\\ &=\int \frac {6 e^2+6 \left (2+e^2\right ) x^2+4 x^4}{e \left (e^{-\frac {x^2}{e^2}}+2 e x \left (3+x^2\right )\right )} \, dx\\ &=\frac {\int \frac {6 e^2+6 \left (2+e^2\right ) x^2+4 x^4}{e^{-\frac {x^2}{e^2}}+2 e x \left (3+x^2\right )} \, dx}{e}\\ &=\frac {\int \left (\frac {-3 e^2-3 \left (2+e^2\right ) x^2-2 x^4}{e x \left (3+x^2\right ) \left (1+6 e^{1+\frac {x^2}{e^2}} x+2 e^{1+\frac {x^2}{e^2}} x^3\right )}+\frac {3 e^2+3 \left (2+e^2\right ) x^2+2 x^4}{e x \left (3+x^2\right )}\right ) \, dx}{e}\\ &=\frac {\int \frac {-3 e^2-3 \left (2+e^2\right ) x^2-2 x^4}{x \left (3+x^2\right ) \left (1+6 e^{1+\frac {x^2}{e^2}} x+2 e^{1+\frac {x^2}{e^2}} x^3\right )} \, dx}{e^2}+\frac {\int \frac {3 e^2+3 \left (2+e^2\right ) x^2+2 x^4}{x \left (3+x^2\right )} \, dx}{e^2}\\ &=\frac {\operatorname {Subst}\left (\int \frac {3 e^2+3 \left (2+e^2\right ) x+2 x^2}{x (3+x)} \, dx,x,x^2\right )}{2 e^2}+\frac {\int \frac {-3 e^2-3 \left (2+e^2\right ) x^2-2 x^4}{x \left (3+x^2\right ) \left (1+2 e^{1+\frac {x^2}{e^2}} x \left (3+x^2\right )\right )} \, dx}{e^2}\\ &=\frac {\operatorname {Subst}\left (\int \left (2+\frac {e^2}{x}+\frac {2 e^2}{3+x}\right ) \, dx,x,x^2\right )}{2 e^2}+\frac {\int \left (-\frac {e^2}{x \left (1+6 e^{1+\frac {x^2}{e^2}} x+2 e^{1+\frac {x^2}{e^2}} x^3\right )}-\frac {2 x}{1+6 e^{1+\frac {x^2}{e^2}} x+2 e^{1+\frac {x^2}{e^2}} x^3}-\frac {2 e^2 x}{\left (3+x^2\right ) \left (1+6 e^{1+\frac {x^2}{e^2}} x+2 e^{1+\frac {x^2}{e^2}} x^3\right )}\right ) \, dx}{e^2}\\ &=\frac {x^2}{e^2}+\log (x)+\log \left (3+x^2\right )-2 \int \frac {x}{\left (3+x^2\right ) \left (1+6 e^{1+\frac {x^2}{e^2}} x+2 e^{1+\frac {x^2}{e^2}} x^3\right )} \, dx-\frac {2 \int \frac {x}{1+6 e^{1+\frac {x^2}{e^2}} x+2 e^{1+\frac {x^2}{e^2}} x^3} \, dx}{e^2}-\int \frac {1}{x \left (1+6 e^{1+\frac {x^2}{e^2}} x+2 e^{1+\frac {x^2}{e^2}} x^3\right )} \, dx\\ &=\frac {x^2}{e^2}+\log (x)+\log \left (3+x^2\right )-2 \int \frac {x}{\left (3+x^2\right ) \left (1+2 e^{1+\frac {x^2}{e^2}} x \left (3+x^2\right )\right )} \, dx-\frac {2 \int \frac {x}{1+2 e^{1+\frac {x^2}{e^2}} x \left (3+x^2\right )} \, dx}{e^2}-\int \frac {1}{x \left (1+2 e^{1+\frac {x^2}{e^2}} x \left (3+x^2\right )\right )} \, dx\\ &=\frac {x^2}{e^2}+\log (x)+\log \left (3+x^2\right )-2 \int \left (-\frac {1}{2 \left (i \sqrt {3}-x\right ) \left (1+2 e^{1+\frac {x^2}{e^2}} x \left (3+x^2\right )\right )}+\frac {1}{2 \left (i \sqrt {3}+x\right ) \left (1+2 e^{1+\frac {x^2}{e^2}} x \left (3+x^2\right )\right )}\right ) \, dx-\frac {2 \int \frac {x}{1+2 e^{1+\frac {x^2}{e^2}} x \left (3+x^2\right )} \, dx}{e^2}-\int \frac {1}{x \left (1+2 e^{1+\frac {x^2}{e^2}} x \left (3+x^2\right )\right )} \, dx\\ &=\frac {x^2}{e^2}+\log (x)+\log \left (3+x^2\right )-\frac {2 \int \frac {x}{1+2 e^{1+\frac {x^2}{e^2}} x \left (3+x^2\right )} \, dx}{e^2}+\int \frac {1}{\left (i \sqrt {3}-x\right ) \left (1+2 e^{1+\frac {x^2}{e^2}} x \left (3+x^2\right )\right )} \, dx-\int \frac {1}{x \left (1+2 e^{1+\frac {x^2}{e^2}} x \left (3+x^2\right )\right )} \, dx-\int \frac {1}{\left (i \sqrt {3}+x\right ) \left (1+2 e^{1+\frac {x^2}{e^2}} x \left (3+x^2\right )\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.46, size = 36, normalized size = 1.50 \begin {gather*} \frac {\frac {x^2}{e}+e \log \left (e^{-\frac {x^2}{e^2}}+6 e x+2 e x^3\right )}{e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x^2/E + E*Log[E^(-(x^2/E^2)) + 6*E*x + 2*E*x^3])/E

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fricas [A]  time = 0.59, size = 43, normalized size = 1.79 \begin {gather*} \log \left (x^{2} + 3\right ) + \log \left (\frac {2 \, {\left (x^{2} + 3\right )} e^{\left ({\left (x^{2} + e^{2} \log \relax (x) + 6 \, e^{2}\right )} e^{\left (-2\right )}\right )} + e^{5}}{x^{2} + 3}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(x^2 + 3) + log((2*(x^2 + 3)*e^((x^2 + e^2*log(x) + 6*e^2)*e^(-2)) + e^5)/(x^2 + 3))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {2 \, {\left (2 \, x^{4} + 6 \, x^{2} + 3 \, {\left (x^{2} + 1\right )} e^{2}\right )} e^{\left ({\left (x^{2} + e^{2} \log \relax (x) + 4 \, e^{2}\right )} e^{\left (-2\right )}\right )}}{x e^{5} + 2 \, {\left (x^{3} + 3 \, x\right )} e^{\left ({\left (x^{2} + e^{2} \log \relax (x) + 4 \, e^{2}\right )} e^{\left (-2\right )} + 2\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.16, size = 48, normalized size = 2.00

method	result	size
risch	\(\ln \left (x^{3}+3 x \right )-{\mathrm e}^{2} \ln \relax (x ) {\mathrm e}^{-2}-4 \,{\mathrm e}^{2} {\mathrm e}^{-2}+\ln \left (x \,{\mathrm e}^{{\mathrm e}^{-2} x^{2}+4}+\frac {{\mathrm e}^{3}}{2 x^{2}+6}\right )\)	\(48\)
norman	\(\ln \left (2 \,{\mathrm e}^{\left ({\mathrm e}^{2} \ln \relax (x )+4 \,{\mathrm e}^{2}+x^{2}\right ) {\mathrm e}^{-2}} x^{2}+{\mathrm e}^{3}+6 \,{\mathrm e}^{\left ({\mathrm e}^{2} \ln \relax (x )+4 \,{\mathrm e}^{2}+x^{2}\right ) {\mathrm e}^{-2}}\right )\)	\(50\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.44, size = 49, normalized size = 2.04 \begin {gather*} \log \left (x^{2} + 3\right ) + \log \relax (x) + \log \left (\frac {2 \, {\left (x^{3} e + 3 \, x e\right )} e^{\left (x^{2} e^{\left (-2\right )}\right )} + 1}{2 \, {\left (x^{3} e + 3 \, x e\right )}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 7.00, size = 30, normalized size = 1.25 \begin {gather*} \ln \left ({\mathrm {e}}^3+6\,x\,{\mathrm {e}}^{{\mathrm {e}}^{-2}\,x^2+4}+2\,x^3\,{\mathrm {e}}^{{\mathrm {e}}^{-2}\,x^2+4}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(exp(3) + 6*x*exp(x^2*exp(-2) + 4) + 2*x^3*exp(x^2*exp(-2) + 4))

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sympy [A]  time = 0.57, size = 37, normalized size = 1.54 \begin {gather*} \log {\left (x^{2} + 3 \right )} + \log {\left (e^{\frac {x^{2} + e^{2} \log {\relax (x )} + 4 e^{2}}{e^{2}}} + \frac {e^{3}}{2 x^{2} + 6} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(x**2 + 3) + log(exp((x**2 + exp(2)*log(x) + 4*exp(2))*exp(-2)) + exp(3)/(2*x**2 + 6))

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