3.44.3 \(\int \frac {e^{\frac {x^4+x^2 \log (5)}{2+x^3+x \log (5)}} (8 x^3+x^6+(4 x+2 x^4) \log (5)+x^2 \log ^2(5))}{20+20 x^3+5 x^6+(20 x+10 x^4) \log (5)+5 x^2 \log ^2(5)} \, dx\)

Optimal. Leaf size=24 \[ \frac {1}{5} e^{\frac {x^2}{x+\frac {2}{x^2+\log (5)}}} \]

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Rubi [F]  time = 1.38, 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 {x^4+x^2 \log (5)}{2+x^3+x \log (5)}} \left (8 x^3+x^6+\left (4 x+2 x^4\right ) \log (5)+x^2 \log ^2(5)\right )}{20+20 x^3+5 x^6+\left (20 x+10 x^4\right ) \log (5)+5 x^2 \log ^2(5)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^((x^4 + x^2*Log[5])/(2 + x^3 + x*Log[5]))*(8*x^3 + x^6 + (4*x + 2*x^4)*Log[5] + x^2*Log[5]^2))/(20 + 20
*x^3 + 5*x^6 + (20*x + 10*x^4)*Log[5] + 5*x^2*Log[5]^2),x]

[Out]

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{\frac {x^2 \left (x^2+\log (5)\right )}{2+x^3+x \log (5)}} \left (8 x^3+x^6+\left (4 x+2 x^4\right ) \log (5)+x^2 \log ^2(5)\right )}{5 \left (2+x^3+x \log (5)\right )^2} \, dx\\ &=\frac {1}{5} \int \frac {e^{\frac {x^2 \left (x^2+\log (5)\right )}{2+x^3+x \log (5)}} \left (8 x^3+x^6+\left (4 x+2 x^4\right ) \log (5)+x^2 \log ^2(5)\right )}{\left (2+x^3+x \log (5)\right )^2} \, dx\\ &=\frac {1}{5} \int \left (e^{\frac {x^2 \left (x^2+\log (5)\right )}{2+x^3+x \log (5)}}-\frac {4 e^{\frac {x^2 \left (x^2+\log (5)\right )}{2+x^3+x \log (5)}} (3+x \log (5))}{\left (2+x^3+x \log (5)\right )^2}+\frac {4 e^{\frac {x^2 \left (x^2+\log (5)\right )}{2+x^3+x \log (5)}}}{2+x^3+x \log (5)}\right ) \, dx\\ &=\frac {1}{5} \int e^{\frac {x^2 \left (x^2+\log (5)\right )}{2+x^3+x \log (5)}} \, dx-\frac {4}{5} \int \frac {e^{\frac {x^2 \left (x^2+\log (5)\right )}{2+x^3+x \log (5)}} (3+x \log (5))}{\left (2+x^3+x \log (5)\right )^2} \, dx+\frac {4}{5} \int \frac {e^{\frac {x^2 \left (x^2+\log (5)\right )}{2+x^3+x \log (5)}}}{2+x^3+x \log (5)} \, dx\\ &=\frac {1}{5} \int e^{\frac {x^2 \left (x^2+\log (5)\right )}{2+x^3+x \log (5)}} \, dx+\frac {4}{5} \int \frac {e^{\frac {x^2 \left (x^2+\log (5)\right )}{2+x^3+x \log (5)}}}{2+x^3+x \log (5)} \, dx-\frac {4}{5} \int \left (\frac {3 e^{\frac {x^2 \left (x^2+\log (5)\right )}{2+x^3+x \log (5)}}}{\left (2+x^3+x \log (5)\right )^2}+\frac {e^{\frac {x^2 \left (x^2+\log (5)\right )}{2+x^3+x \log (5)}} x \log (5)}{\left (2+x^3+x \log (5)\right )^2}\right ) \, dx\\ &=\frac {1}{5} \int e^{\frac {x^2 \left (x^2+\log (5)\right )}{2+x^3+x \log (5)}} \, dx+\frac {4}{5} \int \frac {e^{\frac {x^2 \left (x^2+\log (5)\right )}{2+x^3+x \log (5)}}}{2+x^3+x \log (5)} \, dx-\frac {12}{5} \int \frac {e^{\frac {x^2 \left (x^2+\log (5)\right )}{2+x^3+x \log (5)}}}{\left (2+x^3+x \log (5)\right )^2} \, dx-\frac {1}{5} (4 \log (5)) \int \frac {e^{\frac {x^2 \left (x^2+\log (5)\right )}{2+x^3+x \log (5)}} x}{\left (2+x^3+x \log (5)\right )^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [F]  time = 2.73, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {e^{\frac {x^4+x^2 \log (5)}{2+x^3+x \log (5)}} \left (8 x^3+x^6+\left (4 x+2 x^4\right ) \log (5)+x^2 \log ^2(5)\right )}{20+20 x^3+5 x^6+\left (20 x+10 x^4\right ) \log (5)+5 x^2 \log ^2(5)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[(E^((x^4 + x^2*Log[5])/(2 + x^3 + x*Log[5]))*(8*x^3 + x^6 + (4*x + 2*x^4)*Log[5] + x^2*Log[5]^2))/(2
0 + 20*x^3 + 5*x^6 + (20*x + 10*x^4)*Log[5] + 5*x^2*Log[5]^2),x]

[Out]

Integrate[(E^((x^4 + x^2*Log[5])/(2 + x^3 + x*Log[5]))*(8*x^3 + x^6 + (4*x + 2*x^4)*Log[5] + x^2*Log[5]^2))/(2
0 + 20*x^3 + 5*x^6 + (20*x + 10*x^4)*Log[5] + 5*x^2*Log[5]^2), x]

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fricas [A]  time = 0.55, size = 25, normalized size = 1.04 \begin {gather*} \frac {1}{5} \, e^{\left (\frac {x^{4} + x^{2} \log \relax (5)}{x^{3} + x \log \relax (5) + 2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2*log(5)^2+(2*x^4+4*x)*log(5)+x^6+8*x^3)*exp((x^2*log(5)+x^4)/(x*log(5)+x^3+2))/(5*x^2*log(5)^2+(
10*x^4+20*x)*log(5)+5*x^6+20*x^3+20),x, algorithm="fricas")

[Out]

1/5*e^((x^4 + x^2*log(5))/(x^3 + x*log(5) + 2))

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giac [A]  time = 0.12, size = 36, normalized size = 1.50 \begin {gather*} \frac {1}{5} \, e^{\left (\frac {x^{4}}{x^{3} + x \log \relax (5) + 2} + \frac {x^{2} \log \relax (5)}{x^{3} + x \log \relax (5) + 2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2*log(5)^2+(2*x^4+4*x)*log(5)+x^6+8*x^3)*exp((x^2*log(5)+x^4)/(x*log(5)+x^3+2))/(5*x^2*log(5)^2+(
10*x^4+20*x)*log(5)+5*x^6+20*x^3+20),x, algorithm="giac")

[Out]

1/5*e^(x^4/(x^3 + x*log(5) + 2) + x^2*log(5)/(x^3 + x*log(5) + 2))

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maple [A]  time = 0.26, size = 25, normalized size = 1.04

method	result	size
gosper	\(\frac {{\mathrm e}^{\frac {x^{2} \left (x^{2}+\ln \relax (5)\right )}{x \ln \relax (5)+x^{3}+2}}}{5}\)	\(25\)
risch	\(\frac {{\mathrm e}^{\frac {x^{2} \left (x^{2}+\ln \relax (5)\right )}{x \ln \relax (5)+x^{3}+2}}}{5}\)	\(25\)
norman	\(\frac {\frac {x^{3} {\mathrm e}^{\frac {x^{2} \ln \relax (5)+x^{4}}{x \ln \relax (5)+x^{3}+2}}}{5}+\frac {x \ln \relax (5) {\mathrm e}^{\frac {x^{2} \ln \relax (5)+x^{4}}{x \ln \relax (5)+x^{3}+2}}}{5}+\frac {2 \,{\mathrm e}^{\frac {x^{2} \ln \relax (5)+x^{4}}{x \ln \relax (5)+x^{3}+2}}}{5}}{x \ln \relax (5)+x^{3}+2}\)	\(95\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2*ln(5)^2+(2*x^4+4*x)*ln(5)+x^6+8*x^3)*exp((x^2*ln(5)+x^4)/(x*ln(5)+x^3+2))/(5*x^2*ln(5)^2+(10*x^4+20*x
)*ln(5)+5*x^6+20*x^3+20),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.58, size = 19, normalized size = 0.79 \begin {gather*} \frac {1}{5} \, e^{\left (x - \frac {2 \, x}{x^{3} + x \log \relax (5) + 2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2*log(5)^2+(2*x^4+4*x)*log(5)+x^6+8*x^3)*exp((x^2*log(5)+x^4)/(x*log(5)+x^3+2))/(5*x^2*log(5)^2+(
10*x^4+20*x)*log(5)+5*x^6+20*x^3+20),x, algorithm="maxima")

[Out]

1/5*e^(x - 2*x/(x^3 + x*log(5) + 2))

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mupad [B]  time = 4.30, size = 36, normalized size = 1.50 \begin {gather*} \frac {{\mathrm {e}}^{\frac {x^4}{x^3+\ln \relax (5)\,x+2}}\,{\mathrm {e}}^{\frac {x^2\,\ln \relax (5)}{x^3+\ln \relax (5)\,x+2}}}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp((x^2*log(5) + x^4)/(x*log(5) + x^3 + 2))*(x^2*log(5)^2 + log(5)*(4*x + 2*x^4) + 8*x^3 + x^6))/(5*x^2*
log(5)^2 + log(5)*(20*x + 10*x^4) + 20*x^3 + 5*x^6 + 20),x)

[Out]

(exp(x^4/(x*log(5) + x^3 + 2))*exp((x^2*log(5))/(x*log(5) + x^3 + 2)))/5

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sympy [A]  time = 0.50, size = 22, normalized size = 0.92 \begin {gather*} \frac {e^{\frac {x^{4} + x^{2} \log {\relax (5 )}}{x^{3} + x \log {\relax (5 )} + 2}}}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**2*ln(5)**2+(2*x**4+4*x)*ln(5)+x**6+8*x**3)*exp((x**2*ln(5)+x**4)/(x*ln(5)+x**3+2))/(5*x**2*ln(5)
**2+(10*x**4+20*x)*ln(5)+5*x**6+20*x**3+20),x)

[Out]

exp((x**4 + x**2*log(5))/(x**3 + x*log(5) + 2))/5

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