3.90.66 \(\int \frac {5 e^{8+18 x} x^2+5 x^4+e^{.\frac {2}{5}/x} (-9 x+15 x^2)+e^{4+9 x} (10 x^3+e^{.\frac {2}{5}/x} (6+135 x^2))+(-6 e^{.\frac {2}{5}/x}-10 e^{4+9 x} x^2-10 x^3) \log (x)+5 x^2 \log ^2(x)}{60 e^{8+18 x} x^2+120 e^{4+9 x} x^3+60 x^4+(-120 e^{4+9 x} x^2-120 x^3) \log (x)+60 x^2 \log ^2(x)} \, dx\)

Optimal. Leaf size=37 \[ \frac {1}{4} \left (\frac {x}{3}+\frac {e^{\left .\frac {2}{5}\right /x}}{-e^{4+9 x}-x+\log (x)}\right ) \]

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Rubi [F]  time = 4.04, 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 {5 e^{8+18 x} x^2+5 x^4+e^{\left .\frac {2}{5}\right /x} \left (-9 x+15 x^2\right )+e^{4+9 x} \left (10 x^3+e^{\left .\frac {2}{5}\right /x} \left (6+135 x^2\right )\right )+\left (-6 e^{\left .\frac {2}{5}\right /x}-10 e^{4+9 x} x^2-10 x^3\right ) \log (x)+5 x^2 \log ^2(x)}{60 e^{8+18 x} x^2+120 e^{4+9 x} x^3+60 x^4+\left (-120 e^{4+9 x} x^2-120 x^3\right ) \log (x)+60 x^2 \log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(5*E^(8 + 18*x)*x^2 + 5*x^4 + E^(2/(5*x))*(-9*x + 15*x^2) + E^(4 + 9*x)*(10*x^3 + E^(2/(5*x))*(6 + 135*x^2
)) + (-6*E^(2/(5*x)) - 10*E^(4 + 9*x)*x^2 - 10*x^3)*Log[x] + 5*x^2*Log[x]^2)/(60*E^(8 + 18*x)*x^2 + 120*E^(4 +
9*x)*x^3 + 60*x^4 + (-120*E^(4 + 9*x)*x^2 - 120*x^3)*Log[x] + 60*x^2*Log[x]^2),x]

[Out]

x/12 + Defer[Int][E^(2/(5*x))/(E^(4 + 9*x) + x - Log[x])^2, x]/4 - Defer[Int][E^(2/(5*x))/(x*(E^(4 + 9*x) + x
- Log[x])^2), x]/4 - (9*Defer[Int][(E^(2/(5*x))*x)/(E^(4 + 9*x) + x - Log[x])^2, x])/4 + (9*Defer[Int][E^(2/(5
*x))/(E^(4 + 9*x) + x - Log[x]), x])/4 + Defer[Int][E^(2/(5*x))/(x^2*(E^(4 + 9*x) + x - Log[x])), x]/10 + (9*D
efer[Int][(E^(2/(5*x))*Log[x])/(E^(4 + 9*x) + x - Log[x])^2, x])/4

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {5 e^{8+18 x} x^2+5 x^4+e^{\left .\frac {2}{5}\right /x} \left (-9 x+15 x^2\right )+e^{4+9 x} \left (10 x^3+e^{\left .\frac {2}{5}\right /x} \left (6+135 x^2\right )\right )+\left (-6 e^{\left .\frac {2}{5}\right /x}-10 e^{4+9 x} x^2-10 x^3\right ) \log (x)+5 x^2 \log ^2(x)}{60 x^2 \left (e^{4+9 x}+x-\log (x)\right )^2} \, dx\\ &=\frac {1}{60} \int \frac {5 e^{8+18 x} x^2+5 x^4+e^{\left .\frac {2}{5}\right /x} \left (-9 x+15 x^2\right )+e^{4+9 x} \left (10 x^3+e^{\left .\frac {2}{5}\right /x} \left (6+135 x^2\right )\right )+\left (-6 e^{\left .\frac {2}{5}\right /x}-10 e^{4+9 x} x^2-10 x^3\right ) \log (x)+5 x^2 \log ^2(x)}{x^2 \left (e^{4+9 x}+x-\log (x)\right )^2} \, dx\\ &=\frac {1}{60} \int \left (5+\frac {3 e^{\left .\frac {2}{5}\right /x} \left (2+45 x^2\right )}{x^2 \left (e^{4+9 x}+x-\log (x)\right )}-\frac {15 e^{\left .\frac {2}{5}\right /x} \left (1-x+9 x^2-9 x \log (x)\right )}{x \left (e^{4+9 x}+x-\log (x)\right )^2}\right ) \, dx\\ &=\frac {x}{12}+\frac {1}{20} \int \frac {e^{\left .\frac {2}{5}\right /x} \left (2+45 x^2\right )}{x^2 \left (e^{4+9 x}+x-\log (x)\right )} \, dx-\frac {1}{4} \int \frac {e^{\left .\frac {2}{5}\right /x} \left (1-x+9 x^2-9 x \log (x)\right )}{x \left (e^{4+9 x}+x-\log (x)\right )^2} \, dx\\ &=\frac {x}{12}+\frac {1}{20} \int \left (\frac {45 e^{\left .\frac {2}{5}\right /x}}{e^{4+9 x}+x-\log (x)}+\frac {2 e^{\left .\frac {2}{5}\right /x}}{x^2 \left (e^{4+9 x}+x-\log (x)\right )}\right ) \, dx-\frac {1}{4} \int \left (-\frac {e^{\left .\frac {2}{5}\right /x}}{\left (e^{4+9 x}+x-\log (x)\right )^2}+\frac {e^{\left .\frac {2}{5}\right /x}}{x \left (e^{4+9 x}+x-\log (x)\right )^2}+\frac {9 e^{\left .\frac {2}{5}\right /x} x}{\left (e^{4+9 x}+x-\log (x)\right )^2}-\frac {9 e^{\left .\frac {2}{5}\right /x} \log (x)}{\left (e^{4+9 x}+x-\log (x)\right )^2}\right ) \, dx\\ &=\frac {x}{12}+\frac {1}{10} \int \frac {e^{\left .\frac {2}{5}\right /x}}{x^2 \left (e^{4+9 x}+x-\log (x)\right )} \, dx+\frac {1}{4} \int \frac {e^{\left .\frac {2}{5}\right /x}}{\left (e^{4+9 x}+x-\log (x)\right )^2} \, dx-\frac {1}{4} \int \frac {e^{\left .\frac {2}{5}\right /x}}{x \left (e^{4+9 x}+x-\log (x)\right )^2} \, dx-\frac {9}{4} \int \frac {e^{\left .\frac {2}{5}\right /x} x}{\left (e^{4+9 x}+x-\log (x)\right )^2} \, dx+\frac {9}{4} \int \frac {e^{\left .\frac {2}{5}\right /x}}{e^{4+9 x}+x-\log (x)} \, dx+\frac {9}{4} \int \frac {e^{\left .\frac {2}{5}\right /x} \log (x)}{\left (e^{4+9 x}+x-\log (x)\right )^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.12, size = 48, normalized size = 1.30 \begin {gather*} \frac {-3 e^{\left .\frac {2}{5}\right /x}+e^{4+9 x} x+x^2-x \log (x)}{12 \left (e^{4+9 x}+x-\log (x)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(5*E^(8 + 18*x)*x^2 + 5*x^4 + E^(2/(5*x))*(-9*x + 15*x^2) + E^(4 + 9*x)*(10*x^3 + E^(2/(5*x))*(6 + 1
35*x^2)) + (-6*E^(2/(5*x)) - 10*E^(4 + 9*x)*x^2 - 10*x^3)*Log[x] + 5*x^2*Log[x]^2)/(60*E^(8 + 18*x)*x^2 + 120*
E^(4 + 9*x)*x^3 + 60*x^4 + (-120*E^(4 + 9*x)*x^2 - 120*x^3)*Log[x] + 60*x^2*Log[x]^2),x]

[Out]

(-3*E^(2/(5*x)) + E^(4 + 9*x)*x + x^2 - x*Log[x])/(12*(E^(4 + 9*x) + x - Log[x]))

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fricas [A]  time = 0.60, size = 41, normalized size = 1.11 \begin {gather*} \frac {x^{2} + x e^{\left (9 \, x + 4\right )} - x \log \relax (x) - 3 \, e^{\left (\frac {2}{5 \, x}\right )}}{12 \, {\left (x + e^{\left (9 \, x + 4\right )} - \log \relax (x)\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x^2*log(x)^2+(-10*x^2*exp(9*x+4)-6*exp(2/5/x)-10*x^3)*log(x)+5*x^2*exp(9*x+4)^2+((135*x^2+6)*exp(
2/5/x)+10*x^3)*exp(9*x+4)+(15*x^2-9*x)*exp(2/5/x)+5*x^4)/(60*x^2*log(x)^2+(-120*x^2*exp(9*x+4)-120*x^3)*log(x)
+60*x^2*exp(9*x+4)^2+120*x^3*exp(9*x+4)+60*x^4),x, algorithm="fricas")

[Out]

1/12*(x^2 + x*e^(9*x + 4) - x*log(x) - 3*e^(2/5/x))/(x + e^(9*x + 4) - log(x))

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x^2*log(x)^2+(-10*x^2*exp(9*x+4)-6*exp(2/5/x)-10*x^3)*log(x)+5*x^2*exp(9*x+4)^2+((135*x^2+6)*exp(
2/5/x)+10*x^3)*exp(9*x+4)+(15*x^2-9*x)*exp(2/5/x)+5*x^4)/(60*x^2*log(x)^2+(-120*x^2*exp(9*x+4)-120*x^3)*log(x)
+60*x^2*exp(9*x+4)^2+120*x^3*exp(9*x+4)+60*x^4),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Unable to divide, perhaps due to rounding error%%%{-109350,[0,3,6,0]%%%}+%%%{-4860,[0,3,4,0]%%%}+%%%{273375
,[0,2,8,0]%

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maple [A]  time = 0.09, size = 27, normalized size = 0.73

method	result	size
risch	\(\frac {x}{12}-\frac {{\mathrm e}^{\frac {2}{5 x}}}{4 \left (x +{\mathrm e}^{9 x +4}-\ln \relax (x )\right )}\)	\(27\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((5*x^2*ln(x)^2+(-10*x^2*exp(9*x+4)-6*exp(2/5/x)-10*x^3)*ln(x)+5*x^2*exp(9*x+4)^2+((135*x^2+6)*exp(2/5/x)+1
0*x^3)*exp(9*x+4)+(15*x^2-9*x)*exp(2/5/x)+5*x^4)/(60*x^2*ln(x)^2+(-120*x^2*exp(9*x+4)-120*x^3)*ln(x)+60*x^2*ex
p(9*x+4)^2+120*x^3*exp(9*x+4)+60*x^4),x,method=_RETURNVERBOSE)

[Out]

1/12*x-1/4*exp(2/5/x)/(x+exp(9*x+4)-ln(x))

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maxima [A]  time = 0.40, size = 41, normalized size = 1.11 \begin {gather*} \frac {x^{2} + x e^{\left (9 \, x + 4\right )} - x \log \relax (x) - 3 \, e^{\left (\frac {2}{5 \, x}\right )}}{12 \, {\left (x + e^{\left (9 \, x + 4\right )} - \log \relax (x)\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x^2*log(x)^2+(-10*x^2*exp(9*x+4)-6*exp(2/5/x)-10*x^3)*log(x)+5*x^2*exp(9*x+4)^2+((135*x^2+6)*exp(
2/5/x)+10*x^3)*exp(9*x+4)+(15*x^2-9*x)*exp(2/5/x)+5*x^4)/(60*x^2*log(x)^2+(-120*x^2*exp(9*x+4)-120*x^3)*log(x)
+60*x^2*exp(9*x+4)^2+120*x^3*exp(9*x+4)+60*x^4),x, algorithm="maxima")

[Out]

1/12*(x^2 + x*e^(9*x + 4) - x*log(x) - 3*e^(2/5/x))/(x + e^(9*x + 4) - log(x))

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mupad [B]  time = 7.75, size = 30, normalized size = 0.81 \begin {gather*} \frac {x}{12}-\frac {{\mathrm {e}}^{\frac {2}{5\,x}}}{4\,\left (x+{\mathrm {e}}^{9\,x+4}-\ln \relax (x)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(9*x + 4)*(exp(2/(5*x))*(135*x^2 + 6) + 10*x^3) - exp(2/(5*x))*(9*x - 15*x^2) + 5*x^2*log(x)^2 - log(x
)*(6*exp(2/(5*x)) + 10*x^2*exp(9*x + 4) + 10*x^3) + 5*x^2*exp(18*x + 8) + 5*x^4)/(60*x^2*log(x)^2 - log(x)*(12
0*x^2*exp(9*x + 4) + 120*x^3) + 120*x^3*exp(9*x + 4) + 60*x^2*exp(18*x + 8) + 60*x^4),x)

[Out]

x/12 - exp(2/(5*x))/(4*(x + exp(9*x + 4) - log(x)))

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sympy [A]  time = 0.34, size = 26, normalized size = 0.70 \begin {gather*} \frac {x}{12} - \frac {e^{\frac {2}{5 x}}}{4 x + 4 e^{9 x + 4} - 4 \log {\relax (x )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x**2*ln(x)**2+(-10*x**2*exp(9*x+4)-6*exp(2/5/x)-10*x**3)*ln(x)+5*x**2*exp(9*x+4)**2+((135*x**2+6)
*exp(2/5/x)+10*x**3)*exp(9*x+4)+(15*x**2-9*x)*exp(2/5/x)+5*x**4)/(60*x**2*ln(x)**2+(-120*x**2*exp(9*x+4)-120*x
**3)*ln(x)+60*x**2*exp(9*x+4)**2+120*x**3*exp(9*x+4)+60*x**4),x)

[Out]

x/12 - exp(2/(5*x))/(4*x + 4*exp(9*x + 4) - 4*log(x))

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