3.1.24 \(\int \frac {e^{\frac {2 x^2-x^6+x^2 \log (x)}{2+5 x+2 x^2}} (10 x+15 x^2+2 x^3-12 x^5-25 x^6-8 x^7+(4 x+5 x^2) \log (x))}{4+20 x+33 x^2+20 x^3+4 x^4} \, dx\)

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

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Rubi [F]  time = 13.94, 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 {2 x^2-x^6+x^2 \log (x)}{2+5 x+2 x^2}} \left (10 x+15 x^2+2 x^3-12 x^5-25 x^6-8 x^7+\left (4 x+5 x^2\right ) \log (x)\right )}{4+20 x+33 x^2+20 x^3+4 x^4} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^((2*x^2 - x^6 + x^2*Log[x])/(2 + 5*x + 2*x^2))*(10*x + 15*x^2 + 2*x^3 - 12*x^5 - 25*x^6 - 8*x^7 + (4*x
+ 5*x^2)*Log[x]))/(4 + 20*x + 33*x^2 + 20*x^3 + 4*x^4),x]

[Out]

(85*Defer[Int][E^((2*x^2 - x^6 + x^2*Log[x])/(2 + 5*x + 2*x^2)), x])/16 - (21*Defer[Int][E^((2*x^2 - x^6 + x^2
*Log[x])/(2 + 5*x + 2*x^2))*x, x])/4 + (15*Defer[Int][E^((2*x^2 - x^6 + x^2*Log[x])/(2 + 5*x + 2*x^2))*x^2, x]
)/4 - 2*Defer[Int][E^((2*x^2 - x^6 + x^2*Log[x])/(2 + 5*x + 2*x^2))*x^3, x] - (56*Defer[Int][E^((2*x^2 - x^6 +
x^2*Log[x])/(2 + 5*x + 2*x^2))/(2 + x)^2, x])/3 + (2*Defer[Int][E^((2*x^2 - x^6 + x^2*Log[x])/(2 + 5*x + 2*x^
2))/(2 + x), x])/3 - (31*Defer[Int][E^((2*x^2 - x^6 + x^2*Log[x])/(2 + 5*x + 2*x^2))/(1 + 2*x)^2, x])/48 - Def
er[Int][E^((2*x^2 - x^6 + x^2*Log[x])/(2 + 5*x + 2*x^2))/(1 + 2*x), x]/3 + (4*Defer[Int][(E^((2*x^2 - x^6 + x^
2*Log[x])/(2 + 5*x + 2*x^2))*Log[x])/(2 + x)^2, x])/3 - Defer[Int][(E^((2*x^2 - x^6 + x^2*Log[x])/(2 + 5*x + 2
*x^2))*Log[x])/(1 + 2*x)^2, x]/3

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {10 e^{\frac {2 x^2-x^6+x^2 \log (x)}{2+5 x+2 x^2}} x}{(2+x)^2 (1+2 x)^2}+\frac {15 e^{\frac {2 x^2-x^6+x^2 \log (x)}{2+5 x+2 x^2}} x^2}{(2+x)^2 (1+2 x)^2}+\frac {2 e^{\frac {2 x^2-x^6+x^2 \log (x)}{2+5 x+2 x^2}} x^3}{(2+x)^2 (1+2 x)^2}-\frac {12 e^{\frac {2 x^2-x^6+x^2 \log (x)}{2+5 x+2 x^2}} x^5}{(2+x)^2 (1+2 x)^2}-\frac {25 e^{\frac {2 x^2-x^6+x^2 \log (x)}{2+5 x+2 x^2}} x^6}{(2+x)^2 (1+2 x)^2}-\frac {8 e^{\frac {2 x^2-x^6+x^2 \log (x)}{2+5 x+2 x^2}} x^7}{(2+x)^2 (1+2 x)^2}+\frac {e^{\frac {2 x^2-x^6+x^2 \log (x)}{2+5 x+2 x^2}} x (4+5 x) \log (x)}{(2+x)^2 (1+2 x)^2}\right ) \, dx\\ &=2 \int \frac {e^{\frac {2 x^2-x^6+x^2 \log (x)}{2+5 x+2 x^2}} x^3}{(2+x)^2 (1+2 x)^2} \, dx-8 \int \frac {e^{\frac {2 x^2-x^6+x^2 \log (x)}{2+5 x+2 x^2}} x^7}{(2+x)^2 (1+2 x)^2} \, dx+10 \int \frac {e^{\frac {2 x^2-x^6+x^2 \log (x)}{2+5 x+2 x^2}} x}{(2+x)^2 (1+2 x)^2} \, dx-12 \int \frac {e^{\frac {2 x^2-x^6+x^2 \log (x)}{2+5 x+2 x^2}} x^5}{(2+x)^2 (1+2 x)^2} \, dx+15 \int \frac {e^{\frac {2 x^2-x^6+x^2 \log (x)}{2+5 x+2 x^2}} x^2}{(2+x)^2 (1+2 x)^2} \, dx-25 \int \frac {e^{\frac {2 x^2-x^6+x^2 \log (x)}{2+5 x+2 x^2}} x^6}{(2+x)^2 (1+2 x)^2} \, dx+\int \frac {e^{\frac {2 x^2-x^6+x^2 \log (x)}{2+5 x+2 x^2}} x (4+5 x) \log (x)}{(2+x)^2 (1+2 x)^2} \, dx\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.10, size = 43, normalized size = 1.59 \begin {gather*} e^{-\frac {x^2 \left (-2+x^4\right )}{2+5 x+2 x^2}} x^{\frac {x^2}{2+5 x+2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((2*x^2 - x^6 + x^2*Log[x])/(2 + 5*x + 2*x^2))*(10*x + 15*x^2 + 2*x^3 - 12*x^5 - 25*x^6 - 8*x^7 +
(4*x + 5*x^2)*Log[x]))/(4 + 20*x + 33*x^2 + 20*x^3 + 4*x^4),x]

[Out]

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

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fricas [A]  time = 0.52, size = 31, normalized size = 1.15 \begin {gather*} e^{\left (-\frac {x^{6} - x^{2} \log \relax (x) - 2 \, x^{2}}{2 \, x^{2} + 5 \, x + 2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*x^2+4*x)*log(x)-8*x^7-25*x^6-12*x^5+2*x^3+15*x^2+10*x)*exp((x^2*log(x)-x^6+2*x^2)/(2*x^2+5*x+2))
/(4*x^4+20*x^3+33*x^2+20*x+4),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.52, size = 54, normalized size = 2.00 \begin {gather*} e^{\left (-\frac {x^{6}}{2 \, x^{2} + 5 \, x + 2} + \frac {x^{2} \log \relax (x)}{2 \, x^{2} + 5 \, x + 2} + \frac {2 \, x^{2}}{2 \, x^{2} + 5 \, x + 2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*x^2+4*x)*log(x)-8*x^7-25*x^6-12*x^5+2*x^3+15*x^2+10*x)*exp((x^2*log(x)-x^6+2*x^2)/(2*x^2+5*x+2))
/(4*x^4+20*x^3+33*x^2+20*x+4),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.02, size = 27, normalized size = 1.00

method	result	size
risch	\({\mathrm e}^{\frac {x^{2} \left (2-x^{4}+\ln \relax (x )\right )}{\left (2+x \right ) \left (2 x +1\right )}}\)	\(27\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((5*x^2+4*x)*ln(x)-8*x^7-25*x^6-12*x^5+2*x^3+15*x^2+10*x)*exp((x^2*ln(x)-x^6+2*x^2)/(2*x^2+5*x+2))/(4*x^4+
20*x^3+33*x^2+20*x+4),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -\int \frac {{\left (8 \, x^{7} + 25 \, x^{6} + 12 \, x^{5} - 2 \, x^{3} - 15 \, x^{2} - {\left (5 \, x^{2} + 4 \, x\right )} \log \relax (x) - 10 \, x\right )} e^{\left (-\frac {x^{6} - x^{2} \log \relax (x) - 2 \, x^{2}}{2 \, x^{2} + 5 \, x + 2}\right )}}{4 \, x^{4} + 20 \, x^{3} + 33 \, x^{2} + 20 \, x + 4}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*x^2+4*x)*log(x)-8*x^7-25*x^6-12*x^5+2*x^3+15*x^2+10*x)*exp((x^2*log(x)-x^6+2*x^2)/(2*x^2+5*x+2))
/(4*x^4+20*x^3+33*x^2+20*x+4),x, algorithm="maxima")

[Out]

-integrate((8*x^7 + 25*x^6 + 12*x^5 - 2*x^3 - 15*x^2 - (5*x^2 + 4*x)*log(x) - 10*x)*e^(-(x^6 - x^2*log(x) - 2*
x^2)/(2*x^2 + 5*x + 2))/(4*x^4 + 20*x^3 + 33*x^2 + 20*x + 4), x)

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mupad [B]  time = 0.43, size = 44, normalized size = 1.63 \begin {gather*} x^{\frac {x^2}{2\,x^2+5\,x+2}}\,{\mathrm {e}}^{\frac {2\,x^2-x^6}{2\,x^2+5\,x+2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp((x^2*log(x) + 2*x^2 - x^6)/(5*x + 2*x^2 + 2))*(10*x + log(x)*(4*x + 5*x^2) + 15*x^2 + 2*x^3 - 12*x^5
- 25*x^6 - 8*x^7))/(20*x + 33*x^2 + 20*x^3 + 4*x^4 + 4),x)

[Out]

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

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sympy [A]  time = 0.55, size = 26, normalized size = 0.96 \begin {gather*} e^{\frac {- x^{6} + x^{2} \log {\relax (x )} + 2 x^{2}}{2 x^{2} + 5 x + 2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*x**2+4*x)*ln(x)-8*x**7-25*x**6-12*x**5+2*x**3+15*x**2+10*x)*exp((x**2*ln(x)-x**6+2*x**2)/(2*x**2
+5*x+2))/(4*x**4+20*x**3+33*x**2+20*x+4),x)

[Out]

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

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