3.23.39 \(\int \frac {-3 x-x^2+e^{5+x} (-9 x-3 x^2)+(3 x+2 x^2+e^{5+x} (9+6 x)) \log (\frac {\log (5)}{3 e^{5+x}+x})}{45 e^{5+x}+15 x} \, dx\)

Optimal. Leaf size=30 \[ 2+\frac {1}{15} x (3+x) \log \left (\frac {\log (5)}{3 \left (e^{5+x}+\frac {x}{3}\right )}\right ) \]

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Rubi [F]  time = 1.31, 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 {-3 x-x^2+e^{5+x} \left (-9 x-3 x^2\right )+\left (3 x+2 x^2+e^{5+x} (9+6 x)\right ) \log \left (\frac {\log (5)}{3 e^{5+x}+x}\right )}{45 e^{5+x}+15 x} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-3*x - x^2 + E^(5 + x)*(-9*x - 3*x^2) + (3*x + 2*x^2 + E^(5 + x)*(9 + 6*x))*Log[Log[5]/(3*E^(5 + x) + x)]
)/(45*E^(5 + x) + 15*x),x]

[Out]

-1/10*x^2 - x^3/45 + (3 + 2*x)^3/360 + ((3 + 2*x)^2*Log[Log[5]/(3*E^(5 + x) + x)])/60 + (3*Defer[Int][(3*E^(5
+ x) + x)^(-1), x])/20 - (3*Defer[Int][x/(3*E^(5 + x) + x), x])/20

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {1}{15} \left (-\frac {\left (1+3 e^{5+x}\right ) x (3+x)}{3 e^{5+x}+x}+(3+2 x) \log \left (\frac {\log (5)}{3 e^{5+x}+x}\right )\right ) \, dx\\ &=\frac {1}{15} \int \left (-\frac {\left (1+3 e^{5+x}\right ) x (3+x)}{3 e^{5+x}+x}+(3+2 x) \log \left (\frac {\log (5)}{3 e^{5+x}+x}\right )\right ) \, dx\\ &=-\left (\frac {1}{15} \int \frac {\left (1+3 e^{5+x}\right ) x (3+x)}{3 e^{5+x}+x} \, dx\right )+\frac {1}{15} \int (3+2 x) \log \left (\frac {\log (5)}{3 e^{5+x}+x}\right ) \, dx\\ &=\frac {1}{60} (3+2 x)^2 \log \left (\frac {\log (5)}{3 e^{5+x}+x}\right )+\frac {1}{60} \int \frac {\left (1+3 e^{5+x}\right ) (3+2 x)^2}{3 e^{5+x}+x} \, dx-\frac {1}{15} \int \left (x (3+x)-\frac {x \left (-3+2 x+x^2\right )}{3 e^{5+x}+x}\right ) \, dx\\ &=\frac {1}{60} (3+2 x)^2 \log \left (\frac {\log (5)}{3 e^{5+x}+x}\right )+\frac {1}{60} \int \left ((3+2 x)^2-\frac {(-1+x) (3+2 x)^2}{3 e^{5+x}+x}\right ) \, dx-\frac {1}{15} \int x (3+x) \, dx+\frac {1}{15} \int \frac {x \left (-3+2 x+x^2\right )}{3 e^{5+x}+x} \, dx\\ &=\frac {1}{360} (3+2 x)^3+\frac {1}{60} (3+2 x)^2 \log \left (\frac {\log (5)}{3 e^{5+x}+x}\right )-\frac {1}{60} \int \frac {(-1+x) (3+2 x)^2}{3 e^{5+x}+x} \, dx-\frac {1}{15} \int \left (3 x+x^2\right ) \, dx+\frac {1}{15} \int \left (-\frac {3 x}{3 e^{5+x}+x}+\frac {2 x^2}{3 e^{5+x}+x}+\frac {x^3}{3 e^{5+x}+x}\right ) \, dx\\ &=-\frac {x^2}{10}-\frac {x^3}{45}+\frac {1}{360} (3+2 x)^3+\frac {1}{60} (3+2 x)^2 \log \left (\frac {\log (5)}{3 e^{5+x}+x}\right )-\frac {1}{60} \int \left (-\frac {9}{3 e^{5+x}+x}-\frac {3 x}{3 e^{5+x}+x}+\frac {8 x^2}{3 e^{5+x}+x}+\frac {4 x^3}{3 e^{5+x}+x}\right ) \, dx+\frac {1}{15} \int \frac {x^3}{3 e^{5+x}+x} \, dx+\frac {2}{15} \int \frac {x^2}{3 e^{5+x}+x} \, dx-\frac {1}{5} \int \frac {x}{3 e^{5+x}+x} \, dx\\ &=-\frac {x^2}{10}-\frac {x^3}{45}+\frac {1}{360} (3+2 x)^3+\frac {1}{60} (3+2 x)^2 \log \left (\frac {\log (5)}{3 e^{5+x}+x}\right )+\frac {1}{20} \int \frac {x}{3 e^{5+x}+x} \, dx+\frac {3}{20} \int \frac {1}{3 e^{5+x}+x} \, dx-\frac {1}{5} \int \frac {x}{3 e^{5+x}+x} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 7.46, size = 23, normalized size = 0.77 \begin {gather*} \frac {1}{15} x (3+x) \log \left (\frac {\log (5)}{3 e^{5+x}+x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-3*x - x^2 + E^(5 + x)*(-9*x - 3*x^2) + (3*x + 2*x^2 + E^(5 + x)*(9 + 6*x))*Log[Log[5]/(3*E^(5 + x)
+ x)])/(45*E^(5 + x) + 15*x),x]

[Out]

(x*(3 + x)*Log[Log[5]/(3*E^(5 + x) + x)])/15

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fricas [A]  time = 0.94, size = 23, normalized size = 0.77 \begin {gather*} \frac {1}{15} \, {\left (x^{2} + 3 \, x\right )} \log \left (\frac {\log \relax (5)}{x + 3 \, e^{\left (x + 5\right )}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((6*x+9)*exp(5+x)+2*x^2+3*x)*log(log(5)/(3*exp(5+x)+x))+(-3*x^2-9*x)*exp(5+x)-x^2-3*x)/(45*exp(5+x)
+15*x),x, algorithm="fricas")

[Out]

1/15*(x^2 + 3*x)*log(log(5)/(x + 3*e^(x + 5)))

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giac [A]  time = 0.33, size = 41, normalized size = 1.37 \begin {gather*} -\frac {1}{15} \, x^{2} \log \left (x + 3 \, e^{\left (x + 5\right )}\right ) + \frac {1}{15} \, x^{2} \log \left (\log \relax (5)\right ) - \frac {1}{5} \, x \log \left (x + 3 \, e^{\left (x + 5\right )}\right ) + \frac {1}{5} \, x \log \left (\log \relax (5)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((6*x+9)*exp(5+x)+2*x^2+3*x)*log(log(5)/(3*exp(5+x)+x))+(-3*x^2-9*x)*exp(5+x)-x^2-3*x)/(45*exp(5+x)
+15*x),x, algorithm="giac")

[Out]

-1/15*x^2*log(x + 3*e^(x + 5)) + 1/15*x^2*log(log(5)) - 1/5*x*log(x + 3*e^(x + 5)) + 1/5*x*log(log(5))

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maple [A]  time = 0.12, size = 35, normalized size = 1.17

method	result	size
risch	\(\left (-\frac {1}{15} x^{2}-\frac {1}{5} x \right ) \ln \left (3 \,{\mathrm e}^{5+x}+x \right )+\frac {\ln \left (\ln \relax (5)\right ) x^{2}}{15}+\frac {x \ln \left (\ln \relax (5)\right )}{5}\)	\(35\)
norman	\(\frac {\ln \left (\frac {\ln \relax (5)}{3 \,{\mathrm e}^{5+x}+x}\right ) x}{5}+\frac {\ln \left (\frac {\ln \relax (5)}{3 \,{\mathrm e}^{5+x}+x}\right ) x^{2}}{15}\)	\(38\)
default	\(\frac {x^{2} \left (\ln \left (\frac {\ln \relax (5)}{3 \,{\mathrm e}^{5+x}+x}\right )+\ln \left (3 \,{\mathrm e}^{5+x}+x \right )\right )}{15}+\frac {x \left (\ln \left (\frac {\ln \relax (5)}{3 \,{\mathrm e}^{5+x}+x}\right )+\ln \left (3 \,{\mathrm e}^{5+x}+x \right )\right )}{5}-\frac {\ln \left (3 \,{\mathrm e}^{5+x}+x \right ) x}{5}-\frac {\ln \left (3 \,{\mathrm e}^{5+x}+x \right ) x^{2}}{15}\)	\(84\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((6*x+9)*exp(5+x)+2*x^2+3*x)*ln(ln(5)/(3*exp(5+x)+x))+(-3*x^2-9*x)*exp(5+x)-x^2-3*x)/(45*exp(5+x)+15*x),x
,method=_RETURNVERBOSE)

[Out]

(-1/15*x^2-1/5*x)*ln(3*exp(5+x)+x)+1/15*ln(ln(5))*x^2+1/5*x*ln(ln(5))

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maxima [A]  time = 0.60, size = 33, normalized size = 1.10 \begin {gather*} \frac {1}{15} \, x^{2} \log \left (\log \relax (5)\right ) - \frac {1}{15} \, {\left (x^{2} + 3 \, x\right )} \log \left (x + 3 \, e^{\left (x + 5\right )}\right ) + \frac {1}{5} \, x \log \left (\log \relax (5)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((6*x+9)*exp(5+x)+2*x^2+3*x)*log(log(5)/(3*exp(5+x)+x))+(-3*x^2-9*x)*exp(5+x)-x^2-3*x)/(45*exp(5+x)
+15*x),x, algorithm="maxima")

[Out]

1/15*x^2*log(log(5)) - 1/15*(x^2 + 3*x)*log(x + 3*e^(x + 5)) + 1/5*x*log(log(5))

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mupad [B]  time = 1.50, size = 25, normalized size = 0.83 \begin {gather*} \left (\frac {x^2}{15}+\frac {x}{5}\right )\,\left (\ln \left (\frac {1}{x+3\,{\mathrm {e}}^5\,{\mathrm {e}}^x}\right )+\ln \left (\ln \relax (5)\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(3*x + exp(x + 5)*(9*x + 3*x^2) - log(log(5)/(x + 3*exp(x + 5)))*(3*x + exp(x + 5)*(6*x + 9) + 2*x^2) + x
^2)/(15*x + 45*exp(x + 5)),x)

[Out]

(x/5 + x^2/15)*(log(1/(x + 3*exp(5)*exp(x))) + log(log(5)))

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sympy [A]  time = 0.34, size = 20, normalized size = 0.67 \begin {gather*} \left (\frac {x^{2}}{15} + \frac {x}{5}\right ) \log {\left (\frac {\log {\relax (5 )}}{x + 3 e^{x + 5}} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((6*x+9)*exp(5+x)+2*x**2+3*x)*ln(ln(5)/(3*exp(5+x)+x))+(-3*x**2-9*x)*exp(5+x)-x**2-3*x)/(45*exp(5+x
)+15*x),x)

[Out]

(x**2/15 + x/5)*log(log(5)/(x + 3*exp(x + 5)))

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