3.88.67 \(\int \frac {5 x^3+2 x^4+e^5 (-2 x^2-x^3)+(15 x^2-3 e^5 x^2+6 x^3) \log (\frac {4}{x})+(15 x+6 x^2) \log ^2(\frac {4}{x})+(5+2 x) \log ^3(\frac {4}{x})}{x^3+3 x^2 \log (\frac {4}{x})+3 x \log ^2(\frac {4}{x})+\log ^3(\frac {4}{x})} \, dx\)

Optimal. Leaf size=23 \[ x \left (5+x-\frac {e^5 x^2}{\left (x+\log \left (\frac {4}{x}\right )\right )^2}\right ) \]

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Rubi [F]  time = 0.45, 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 x^3+2 x^4+e^5 \left (-2 x^2-x^3\right )+\left (15 x^2-3 e^5 x^2+6 x^3\right ) \log \left (\frac {4}{x}\right )+\left (15 x+6 x^2\right ) \log ^2\left (\frac {4}{x}\right )+(5+2 x) \log ^3\left (\frac {4}{x}\right )}{x^3+3 x^2 \log \left (\frac {4}{x}\right )+3 x \log ^2\left (\frac {4}{x}\right )+\log ^3\left (\frac {4}{x}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(5*x^3 + 2*x^4 + E^5*(-2*x^2 - x^3) + (15*x^2 - 3*E^5*x^2 + 6*x^3)*Log[4/x] + (15*x + 6*x^2)*Log[4/x]^2 +
(5 + 2*x)*Log[4/x]^3)/(x^3 + 3*x^2*Log[4/x] + 3*x*Log[4/x]^2 + Log[4/x]^3),x]

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 0.04, size = 25, normalized size = 1.09 \begin {gather*} 5 x+x^2-\frac {e^5 x^3}{\left (x+\log \left (\frac {4}{x}\right )\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(5*x^3 + 2*x^4 + E^5*(-2*x^2 - x^3) + (15*x^2 - 3*E^5*x^2 + 6*x^3)*Log[4/x] + (15*x + 6*x^2)*Log[4/x
]^2 + (5 + 2*x)*Log[4/x]^3)/(x^3 + 3*x^2*Log[4/x] + 3*x*Log[4/x]^2 + Log[4/x]^3),x]

[Out]

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

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fricas [B]  time = 0.53, size = 73, normalized size = 3.17 \begin {gather*} \frac {x^{4} - x^{3} e^{5} + 5 \, x^{3} + {\left (x^{2} + 5 \, x\right )} \log \left (\frac {4}{x}\right )^{2} + 2 \, {\left (x^{3} + 5 \, x^{2}\right )} \log \left (\frac {4}{x}\right )}{x^{2} + 2 \, x \log \left (\frac {4}{x}\right ) + \log \left (\frac {4}{x}\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.17, size = 92, normalized size = 4.00 \begin {gather*} -\frac {\frac {e^{5}}{x} - \frac {2 \, \log \left (\frac {4}{x}\right )}{x} - \frac {\log \left (\frac {4}{x}\right )^{2}}{x^{2}} - \frac {5}{x} - \frac {10 \, \log \left (\frac {4}{x}\right )}{x^{2}} - \frac {5 \, \log \left (\frac {4}{x}\right )^{2}}{x^{3}} - 1}{\frac {1}{x^{2}} + \frac {2 \, \log \left (\frac {4}{x}\right )}{x^{3}} + \frac {\log \left (\frac {4}{x}\right )^{2}}{x^{4}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(e^5/x - 2*log(4/x)/x - log(4/x)^2/x^2 - 5/x - 10*log(4/x)/x^2 - 5*log(4/x)^2/x^3 - 1)/(1/x^2 + 2*log(4/x)/x^
3 + log(4/x)^2/x^4)

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maple [A]  time = 0.16, size = 25, normalized size = 1.09

method	result	size
risch	\(x^{2}+5 x -\frac {{\mathrm e}^{5} x^{3}}{\left (\ln \left (\frac {4}{x}\right )+x \right )^{2}}\)	\(25\)
norman	\(\frac {x^{4}+x^{2} \ln \left (\frac {4}{x}\right )^{2}+\left (5-{\mathrm e}^{5}\right ) x^{3}+5 x \ln \left (\frac {4}{x}\right )^{2}+10 x^{2} \ln \left (\frac {4}{x}\right )+2 x^{3} \ln \left (\frac {4}{x}\right )}{\left (\ln \left (\frac {4}{x}\right )+x \right )^{2}}\)	\(71\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((5+2*x)*ln(4/x)^3+(6*x^2+15*x)*ln(4/x)^2+(-3*x^2*exp(5)+6*x^3+15*x^2)*ln(4/x)+(-x^3-2*x^2)*exp(5)+2*x^4+5
*x^3)/(ln(4/x)^3+3*x*ln(4/x)^2+3*x^2*ln(4/x)+x^3),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [B]  time = 0.48, size = 105, normalized size = 4.57 \begin {gather*} \frac {x^{4} - x^{3} {\left (e^{5} - 4 \, \log \relax (2) - 5\right )} + 4 \, {\left (\log \relax (2)^{2} + 5 \, \log \relax (2)\right )} x^{2} + 20 \, x \log \relax (2)^{2} + {\left (x^{2} + 5 \, x\right )} \log \relax (x)^{2} - 2 \, {\left (x^{3} + x^{2} {\left (2 \, \log \relax (2) + 5\right )} + 10 \, x \log \relax (2)\right )} \log \relax (x)}{x^{2} + 4 \, x \log \relax (2) + 4 \, \log \relax (2)^{2} - 2 \, {\left (x + 2 \, \log \relax (2)\right )} \log \relax (x) + \log \relax (x)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(x^4 - x^3*(e^5 - 4*log(2) - 5) + 4*(log(2)^2 + 5*log(2))*x^2 + 20*x*log(2)^2 + (x^2 + 5*x)*log(x)^2 - 2*(x^3
+ x^2*(2*log(2) + 5) + 10*x*log(2))*log(x))/(x^2 + 4*x*log(2) + 4*log(2)^2 - 2*(x + 2*log(2))*log(x) + log(x)^
2)

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mupad [B]  time = 5.86, size = 73, normalized size = 3.17 \begin {gather*} x\,\left (x+5\right )-\frac {x^3\,\left (x+5\right )+\ln \left (\frac {4}{x}\right )\,\left (2\,x^2\,\left (x+5\right )-x\,\left (2\,x^2+10\,x\right )\right )-x\,\left (5\,x^2-x^2\,{\mathrm {e}}^5+x^3\right )}{{\left (x+\ln \left (\frac {4}{x}\right )\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(4/x)*(15*x^2 - 3*x^2*exp(5) + 6*x^3) - exp(5)*(2*x^2 + x^3) + log(4/x)^3*(2*x + 5) + log(4/x)^2*(15*x
+ 6*x^2) + 5*x^3 + 2*x^4)/(log(4/x)^3 + x^3 + 3*x*log(4/x)^2 + 3*x^2*log(4/x)),x)

[Out]

x*(x + 5) - (x^3*(x + 5) + log(4/x)*(2*x^2*(x + 5) - x*(10*x + 2*x^2)) - x*(5*x^2 - x^2*exp(5) + x^3))/(x + lo
g(4/x))^2

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sympy [A]  time = 0.18, size = 31, normalized size = 1.35 \begin {gather*} - \frac {x^{3} e^{5}}{x^{2} + 2 x \log {\left (\frac {4}{x} \right )} + \log {\left (\frac {4}{x} \right )}^{2}} + x^{2} + 5 x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5+2*x)*ln(4/x)**3+(6*x**2+15*x)*ln(4/x)**2+(-3*x**2*exp(5)+6*x**3+15*x**2)*ln(4/x)+(-x**3-2*x**2)*
exp(5)+2*x**4+5*x**3)/(ln(4/x)**3+3*x*ln(4/x)**2+3*x**2*ln(4/x)+x**3),x)

[Out]

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

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