3.52.86 \(\int \frac {25-25 x^2-25 \log (-3 x)+\log (\frac {1}{4} (4+e^5)) (-3-x+4 x^2+(3+2 x) \log (-3 x))}{\log (\frac {1}{4} (4+e^5)) (x^2-2 x^3+x^4+(2 x-2 x^2) \log (-3 x)+\log ^2(-3 x))} \, dx\)

Optimal. Leaf size=33 \[ \frac {3+x-\frac {25}{\log \left (1+\frac {e^5}{4}\right )}}{1-x+\frac {\log (-3 x)}{x}} \]

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Rubi [F]  time = 0.78, 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 {25-25 x^2-25 \log (-3 x)+\log \left (\frac {1}{4} \left (4+e^5\right )\right ) \left (-3-x+4 x^2+(3+2 x) \log (-3 x)\right )}{\log \left (\frac {1}{4} \left (4+e^5\right )\right ) \left (x^2-2 x^3+x^4+\left (2 x-2 x^2\right ) \log (-3 x)+\log ^2(-3 x)\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(25 - 25*x^2 - 25*Log[-3*x] + Log[(4 + E^5)/4]*(-3 - x + 4*x^2 + (3 + 2*x)*Log[-3*x]))/(Log[(4 + E^5)/4]*(
x^2 - 2*x^3 + x^4 + (2*x - 2*x^2)*Log[-3*x] + Log[-3*x]^2)),x]

[Out]

((25 + Log[64] - 3*Log[4 + E^5])*Defer[Int][(-x + x^2 - Log[-3*x])^(-2), x])/Log[(4 + E^5)/4] - (4 - 25/Log[(4
+ E^5)/4])*Defer[Int][x/(-x + x^2 - Log[-3*x])^2, x] + 5*(1 - 10/Log[(4 + E^5)/4])*Defer[Int][x^2/(-x + x^2 -
Log[-3*x])^2, x] + 2*Defer[Int][x^3/(-x + x^2 - Log[-3*x])^2, x] + ((25 + Log[64] - 3*Log[4 + E^5])*Defer[Int
][(-x + x^2 - Log[-3*x])^(-1), x])/Log[(4 + E^5)/4] - 2*Defer[Int][x/(-x + x^2 - Log[-3*x]), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {25-25 x^2-25 \log (-3 x)+\log \left (\frac {1}{4} \left (4+e^5\right )\right ) \left (-3-x+4 x^2+(3+2 x) \log (-3 x)\right )}{x^2-2 x^3+x^4+\left (2 x-2 x^2\right ) \log (-3 x)+\log ^2(-3 x)} \, dx}{\log \left (\frac {1}{4} \left (4+e^5\right )\right )}\\ &=\frac {\int \frac {25-25 x^2-25 \log (-3 x)+\log \left (\frac {1}{4} \left (4+e^5\right )\right ) \left (-3-x+4 x^2+(3+2 x) \log (-3 x)\right )}{\left (x-x^2+\log (-3 x)\right )^2} \, dx}{\log \left (\frac {1}{4} \left (4+e^5\right )\right )}\\ &=\frac {\int \left (\frac {\left (-1-x+2 x^2\right ) \left (-25+3 \log \left (\frac {1}{4} \left (4+e^5\right )\right )+x \log \left (\frac {1}{4} \left (4+e^5\right )\right )\right )}{\left (-x+x^2-\log (-3 x)\right )^2}+\frac {25-3 \log \left (\frac {1}{4} \left (4+e^5\right )\right )-2 x \log \left (\frac {1}{4} \left (4+e^5\right )\right )}{-x+x^2-\log (-3 x)}\right ) \, dx}{\log \left (\frac {1}{4} \left (4+e^5\right )\right )}\\ &=\frac {\int \frac {\left (-1-x+2 x^2\right ) \left (-25+3 \log \left (\frac {1}{4} \left (4+e^5\right )\right )+x \log \left (\frac {1}{4} \left (4+e^5\right )\right )\right )}{\left (-x+x^2-\log (-3 x)\right )^2} \, dx}{\log \left (\frac {1}{4} \left (4+e^5\right )\right )}+\frac {\int \frac {25-3 \log \left (\frac {1}{4} \left (4+e^5\right )\right )-2 x \log \left (\frac {1}{4} \left (4+e^5\right )\right )}{-x+x^2-\log (-3 x)} \, dx}{\log \left (\frac {1}{4} \left (4+e^5\right )\right )}\\ &=\frac {\int \left (\frac {5 x^2 \left (-10+\log \left (\frac {1}{4} \left (4+e^5\right )\right )\right )}{\left (-x+x^2-\log (-3 x)\right )^2}+\frac {2 x^3 \log \left (\frac {1}{4} \left (4+e^5\right )\right )}{\left (-x+x^2-\log (-3 x)\right )^2}-\frac {x \left (-25+4 \log \left (\frac {1}{4} \left (4+e^5\right )\right )\right )}{\left (-x+x^2-\log (-3 x)\right )^2}+\frac {25 \left (1+\frac {1}{25} \left (\log (64)-3 \log \left (4+e^5\right )\right )\right )}{\left (-x+x^2-\log (-3 x)\right )^2}\right ) \, dx}{\log \left (\frac {1}{4} \left (4+e^5\right )\right )}+\frac {\int \left (\frac {25 \left (1-\frac {3}{25} \log \left (\frac {1}{4} \left (4+e^5\right )\right )\right )}{-x+x^2-\log (-3 x)}-\frac {2 x \log \left (\frac {1}{4} \left (4+e^5\right )\right )}{-x+x^2-\log (-3 x)}\right ) \, dx}{\log \left (\frac {1}{4} \left (4+e^5\right )\right )}\\ &=2 \int \frac {x^3}{\left (-x+x^2-\log (-3 x)\right )^2} \, dx-2 \int \frac {x}{-x+x^2-\log (-3 x)} \, dx+\left (5 \left (1-\frac {10}{\log \left (\frac {1}{4} \left (4+e^5\right )\right )}\right )\right ) \int \frac {x^2}{\left (-x+x^2-\log (-3 x)\right )^2} \, dx+\left (-4+\frac {25}{\log \left (\frac {1}{4} \left (4+e^5\right )\right )}\right ) \int \frac {x}{\left (-x+x^2-\log (-3 x)\right )^2} \, dx+\frac {\left (25+\log (64)-3 \log \left (4+e^5\right )\right ) \int \frac {1}{\left (-x+x^2-\log (-3 x)\right )^2} \, dx}{\log \left (\frac {1}{4} \left (4+e^5\right )\right )}+\frac {\left (25+\log (64)-3 \log \left (4+e^5\right )\right ) \int \frac {1}{-x+x^2-\log (-3 x)} \, dx}{\log \left (\frac {1}{4} \left (4+e^5\right )\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.33, size = 53, normalized size = 1.61 \begin {gather*} \frac {x \left (-25+3 \log \left (\frac {1}{4} \left (4+e^5\right )\right )+x \log \left (\frac {1}{4} \left (4+e^5\right )\right )\right )}{\log \left (\frac {1}{4} \left (4+e^5\right )\right ) \left (x-x^2+\log (-3 x)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(25 - 25*x^2 - 25*Log[-3*x] + Log[(4 + E^5)/4]*(-3 - x + 4*x^2 + (3 + 2*x)*Log[-3*x]))/(Log[(4 + E^5
)/4]*(x^2 - 2*x^3 + x^4 + (2*x - 2*x^2)*Log[-3*x] + Log[-3*x]^2)),x]

[Out]

(x*(-25 + 3*Log[(4 + E^5)/4] + x*Log[(4 + E^5)/4]))/(Log[(4 + E^5)/4]*(x - x^2 + Log[-3*x]))

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fricas [A]  time = 0.48, size = 45, normalized size = 1.36 \begin {gather*} -\frac {{\left (x^{2} + 3 \, x\right )} \log \left (\frac {1}{4} \, e^{5} + 1\right ) - 25 \, x}{{\left (x^{2} - x - \log \left (-3 \, x\right )\right )} \log \left (\frac {1}{4} \, e^{5} + 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x+3)*log(-3*x)+4*x^2-x-3)*log(1+1/4*exp(5))-25*log(-3*x)-25*x^2+25)/(log(-3*x)^2+(-2*x^2+2*x)*l
og(-3*x)+x^4-2*x^3+x^2)/log(1+1/4*exp(5)),x, algorithm="fricas")

[Out]

-((x^2 + 3*x)*log(1/4*e^5 + 1) - 25*x)/((x^2 - x - log(-3*x))*log(1/4*e^5 + 1))

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giac [A]  time = 0.19, size = 59, normalized size = 1.79 \begin {gather*} \frac {2 \, x^{2} \log \relax (2) - x^{2} \log \left (e^{5} + 4\right ) + 6 \, x \log \relax (2) - 3 \, x \log \left (e^{5} + 4\right ) + 25 \, x}{{\left (x^{2} - x - \log \left (-3 \, x\right )\right )} \log \left (\frac {1}{4} \, e^{5} + 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x+3)*log(-3*x)+4*x^2-x-3)*log(1+1/4*exp(5))-25*log(-3*x)-25*x^2+25)/(log(-3*x)^2+(-2*x^2+2*x)*l
og(-3*x)+x^4-2*x^3+x^2)/log(1+1/4*exp(5)),x, algorithm="giac")

[Out]

(2*x^2*log(2) - x^2*log(e^5 + 4) + 6*x*log(2) - 3*x*log(e^5 + 4) + 25*x)/((x^2 - x - log(-3*x))*log(1/4*e^5 +
1))

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maple [A]  time = 0.33, size = 52, normalized size = 1.58

method	result	size
norman	\(\frac {-\ln \left (-3 x \right )-\frac {\left (4 \ln \left (4+{\mathrm e}^{5}\right )-8 \ln \relax (2)-25\right ) x}{-2 \ln \relax (2)+\ln \left (4+{\mathrm e}^{5}\right )}}{x^{2}-x -\ln \left (-3 x \right )}\)	\(52\)
risch	\(-\frac {\left (x \ln \left (4+{\mathrm e}^{5}\right )-2 x \ln \relax (2)+3 \ln \left (4+{\mathrm e}^{5}\right )-6 \ln \relax (2)-25\right ) x}{\left (-2 \ln \relax (2)+\ln \left (4+{\mathrm e}^{5}\right )\right ) \left (x^{2}-x -\ln \left (-3 x \right )\right )}\)	\(56\)
derivativedivides	\(-\frac {3 \left (-\frac {75 x}{9 x^{2}-9 x -9 \ln \left (-3 x \right )}+\frac {-6 \ln \left (-3 x \right ) \ln \relax (2)-24 x \ln \relax (2)}{9 x^{2}-9 x -9 \ln \left (-3 x \right )}+\frac {3 \ln \left (-3 x \right ) \ln \left (4+{\mathrm e}^{5}\right )+12 x \ln \left (4+{\mathrm e}^{5}\right )}{9 x^{2}-9 x -9 \ln \left (-3 x \right )}\right )}{\ln \left (1+\frac {{\mathrm e}^{5}}{4}\right )}\)	\(103\)
default	\(\frac {\frac {225 x}{9 x^{2}-9 x -9 \ln \left (-3 x \right )}-\frac {3 \left (-6 \ln \left (-3 x \right ) \ln \relax (2)-24 x \ln \relax (2)\right )}{9 x^{2}-9 x -9 \ln \left (-3 x \right )}-\frac {3 \left (3 \ln \left (-3 x \right ) \ln \left (4+{\mathrm e}^{5}\right )+12 x \ln \left (4+{\mathrm e}^{5}\right )\right )}{9 x^{2}-9 x -9 \ln \left (-3 x \right )}}{\ln \left (1+\frac {{\mathrm e}^{5}}{4}\right )}\)	\(104\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((2*x+3)*ln(-3*x)+4*x^2-x-3)*ln(1+1/4*exp(5))-25*ln(-3*x)-25*x^2+25)/(ln(-3*x)^2+(-2*x^2+2*x)*ln(-3*x)+x^
4-2*x^3+x^2)/ln(1+1/4*exp(5)),x,method=_RETURNVERBOSE)

[Out]

(-ln(-3*x)-1/(-2*ln(2)+ln(4+exp(5)))*(4*ln(4+exp(5))-8*ln(2)-25)*x)/(x^2-x-ln(-3*x))

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maxima [B]  time = 0.47, size = 61, normalized size = 1.85 \begin {gather*} \frac {x^{2} {\left (2 \, \log \relax (2) - \log \left (e^{5} + 4\right )\right )} + x {\left (6 \, \log \relax (2) - 3 \, \log \left (e^{5} + 4\right ) + 25\right )}}{{\left (x^{2} - x - \log \relax (3) - \log \left (-x\right )\right )} \log \left (\frac {1}{4} \, e^{5} + 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x+3)*log(-3*x)+4*x^2-x-3)*log(1+1/4*exp(5))-25*log(-3*x)-25*x^2+25)/(log(-3*x)^2+(-2*x^2+2*x)*l
og(-3*x)+x^4-2*x^3+x^2)/log(1+1/4*exp(5)),x, algorithm="maxima")

[Out]

(x^2*(2*log(2) - log(e^5 + 4)) + x*(6*log(2) - 3*log(e^5 + 4) + 25))/((x^2 - x - log(3) - log(-x))*log(1/4*e^5
+ 1))

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mupad [B]  time = 4.14, size = 44, normalized size = 1.33 \begin {gather*} \frac {x\,\left (3\,\ln \left (\frac {{\mathrm {e}}^5}{4}+1\right )+x\,\ln \left (\frac {{\mathrm {e}}^5}{4}+1\right )-25\right )}{\ln \left (\frac {{\mathrm {e}}^5}{4}+1\right )\,\left (x+\ln \left (-3\,x\right )-x^2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(25*log(-3*x) + 25*x^2 + log(exp(5)/4 + 1)*(x - 4*x^2 - log(-3*x)*(2*x + 3) + 3) - 25)/(log(exp(5)/4 + 1)
*(log(-3*x)*(2*x - 2*x^2) + log(-3*x)^2 + x^2 - 2*x^3 + x^4)),x)

[Out]

(x*(3*log(exp(5)/4 + 1) + x*log(exp(5)/4 + 1) - 25))/(log(exp(5)/4 + 1)*(x + log(-3*x) - x^2))

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sympy [B]  time = 0.17, size = 63, normalized size = 1.91 \begin {gather*} \frac {x^{2} \log {\left (1 + \frac {e^{5}}{4} \right )} - 25 x + 3 x \log {\left (1 + \frac {e^{5}}{4} \right )}}{- x^{2} \log {\left (1 + \frac {e^{5}}{4} \right )} + x \log {\left (1 + \frac {e^{5}}{4} \right )} + \log {\left (- 3 x \right )} \log {\left (1 + \frac {e^{5}}{4} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x+3)*ln(-3*x)+4*x**2-x-3)*ln(1+1/4*exp(5))-25*ln(-3*x)-25*x**2+25)/(ln(-3*x)**2+(-2*x**2+2*x)*l
n(-3*x)+x**4-2*x**3+x**2)/ln(1+1/4*exp(5)),x)

[Out]

(x**2*log(1 + exp(5)/4) - 25*x + 3*x*log(1 + exp(5)/4))/(-x**2*log(1 + exp(5)/4) + x*log(1 + exp(5)/4) + log(-
3*x)*log(1 + exp(5)/4))

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