3.9.54 \(\int \frac {5+x+(5 x^2+x^3) \log (4)+(20 x+8 x^2) \log (4) \log ^3(\frac {1}{2} (5 x+x^2))}{(5 x^2+x^3) \log (4)} \, dx\)

Optimal. Leaf size=24 \[ x+\frac {-1+x}{x \log (4)}+\log ^4\left (\frac {1}{2} x (5+x)\right ) \]

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Rubi [F]  time = 0.33, 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+\left (5 x^2+x^3\right ) \log (4)+\left (20 x+8 x^2\right ) \log (4) \log ^3\left (\frac {1}{2} \left (5 x+x^2\right )\right )}{\left (5 x^2+x^3\right ) \log (4)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

x - 1/(x*Log[4]) + 4*Defer[Int][Log[(x*(5 + x))/2]^3/x, x] + 4*Defer[Int][Log[(x*(5 + x))/2]^3/(5 + x), x]

Rubi steps

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

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Mathematica [A]  time = 0.14, size = 22, normalized size = 0.92 \begin {gather*} x-\frac {1}{x \log (4)}+\log ^4\left (\frac {1}{2} x (5+x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

x - 1/(x*Log[4]) + Log[(x*(5 + x))/2]^4

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fricas [A]  time = 0.91, size = 35, normalized size = 1.46 \begin {gather*} \frac {2 \, x \log \relax (2) \log \left (\frac {1}{2} \, x^{2} + \frac {5}{2} \, x\right )^{4} + 2 \, x^{2} \log \relax (2) - 1}{2 \, x \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(2*(8*x^2+20*x)*log(2)*log(1/2*x^2+5/2*x)^3+2*(x^3+5*x^2)*log(2)+5+x)/(x^3+5*x^2)/log(2),x, algo
rithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {8 \, {\left (2 \, x^{2} + 5 \, x\right )} \log \relax (2) \log \left (\frac {1}{2} \, x^{2} + \frac {5}{2} \, x\right )^{3} + 2 \, {\left (x^{3} + 5 \, x^{2}\right )} \log \relax (2) + x + 5}{2 \, {\left (x^{3} + 5 \, x^{2}\right )} \log \relax (2)}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(2*(8*x^2+20*x)*log(2)*log(1/2*x^2+5/2*x)^3+2*(x^3+5*x^2)*log(2)+5+x)/(x^3+5*x^2)/log(2),x, algo
rithm="giac")

[Out]

integrate(1/2*(8*(2*x^2 + 5*x)*log(2)*log(1/2*x^2 + 5/2*x)^3 + 2*(x^3 + 5*x^2)*log(2) + x + 5)/((x^3 + 5*x^2)*
log(2)), x)

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maple [A]  time = 0.21, size = 29, normalized size = 1.21

method	result	size
norman	\(\frac {x^{2}+x \ln \left (\frac {1}{2} x^{2}+\frac {5}{2} x \right )^{4}-\frac {1}{2 \ln \relax (2)}}{x}\)	\(29\)
risch	\(\ln \left (\frac {1}{2} x^{2}+\frac {5}{2} x \right )^{4}+\frac {2 x^{2} \ln \relax (2)-1}{2 \ln \relax (2) x}\)	\(32\)
default	\(\frac {-\frac {1}{x}+2 \ln \relax (2) \ln \left (x^{2}+5 x \right )^{4}+2 x \ln \relax (2)-8 \ln \relax (2)^{4} \ln \left (\left (5+x \right ) x \right )-12 \ln \relax (2)^{3} \ln \relax (x )^{2}+24 \ln \relax (2)^{3} \ln \relax (x ) \ln \left (x^{2}+5 x \right )-24 \ln \relax (2)^{3} \ln \relax (x ) \ln \left (1+\frac {x}{5}\right )+24 \ln \relax (2)^{3} \ln \left (5+x \right ) \ln \left (x^{2}+5 x \right )+24 \ln \relax (2)^{3} \ln \left (-\frac {x}{5}\right ) \ln \left (1+\frac {x}{5}\right )-12 \ln \relax (2)^{3} \ln \left (5+x \right )^{2}-24 \ln \relax (2)^{3} \ln \left (-\frac {x}{5}\right ) \ln \left (5+x \right )-8 \ln \relax (2)^{2} \ln \left (x^{2}+5 x \right )^{3}}{2 \ln \relax (2)}\)	\(160\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/2*(2*(8*x^2+20*x)*ln(2)*ln(1/2*x^2+5/2*x)^3+2*(x^3+5*x^2)*ln(2)+5+x)/(x^3+5*x^2)/ln(2),x,method=_RETURNV
ERBOSE)

[Out]

(x^2+x*ln(1/2*x^2+5/2*x)^4-1/2/ln(2))/x

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maxima [B]  time = 0.65, size = 162, normalized size = 6.75 \begin {gather*} \frac {2 \, \log \relax (2) \log \left (x + 5\right )^{4} - 8 \, \log \relax (2)^{4} \log \relax (x) + 12 \, \log \relax (2)^{3} \log \relax (x)^{2} - 8 \, \log \relax (2)^{2} \log \relax (x)^{3} + 2 \, \log \relax (2) \log \relax (x)^{4} - 8 \, {\left (\log \relax (2)^{2} - \log \relax (2) \log \relax (x)\right )} \log \left (x + 5\right )^{3} + 12 \, {\left (\log \relax (2)^{3} - 2 \, \log \relax (2)^{2} \log \relax (x) + \log \relax (2) \log \relax (x)^{2}\right )} \log \left (x + 5\right )^{2} + 2 \, {\left (x - 5 \, \log \left (x + 5\right )\right )} \log \relax (2) - 8 \, {\left (\log \relax (2)^{4} - 3 \, \log \relax (2)^{3} \log \relax (x) + 3 \, \log \relax (2)^{2} \log \relax (x)^{2} - \log \relax (2) \log \relax (x)^{3}\right )} \log \left (x + 5\right ) + 10 \, \log \relax (2) \log \left (x + 5\right ) - \frac {1}{x}}{2 \, \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(2*(8*x^2+20*x)*log(2)*log(1/2*x^2+5/2*x)^3+2*(x^3+5*x^2)*log(2)+5+x)/(x^3+5*x^2)/log(2),x, algo
rithm="maxima")

[Out]

1/2*(2*log(2)*log(x + 5)^4 - 8*log(2)^4*log(x) + 12*log(2)^3*log(x)^2 - 8*log(2)^2*log(x)^3 + 2*log(2)*log(x)^
4 - 8*(log(2)^2 - log(2)*log(x))*log(x + 5)^3 + 12*(log(2)^3 - 2*log(2)^2*log(x) + log(2)*log(x)^2)*log(x + 5)
^2 + 2*(x - 5*log(x + 5))*log(2) - 8*(log(2)^4 - 3*log(2)^3*log(x) + 3*log(2)^2*log(x)^2 - log(2)*log(x)^3)*lo
g(x + 5) + 10*log(2)*log(x + 5) - 1/x)/log(2)

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mupad [B]  time = 0.91, size = 23, normalized size = 0.96 \begin {gather*} x-\frac {1}{2\,x\,\ln \relax (2)}+{\ln \left (\frac {x^2}{2}+\frac {5\,x}{2}\right )}^4 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x/2 + log(2)*(5*x^2 + x^3) + log(2)*log((5*x)/2 + x^2/2)^3*(20*x + 8*x^2) + 5/2)/(log(2)*(5*x^2 + x^3)),x
)

[Out]

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

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sympy [A]  time = 0.18, size = 27, normalized size = 1.12 \begin {gather*} \frac {2 x \log {\relax (2 )} - \frac {1}{x}}{2 \log {\relax (2 )}} + \log {\left (\frac {x^{2}}{2} + \frac {5 x}{2} \right )}^{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(2*(8*x**2+20*x)*ln(2)*ln(1/2*x**2+5/2*x)**3+2*(x**3+5*x**2)*ln(2)+5+x)/(x**3+5*x**2)/ln(2),x)

[Out]

(2*x*log(2) - 1/x)/(2*log(2)) + log(x**2/2 + 5*x/2)**4

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