3.78.93 \(\int \frac {-3 x^3+(10 x^3-x^4) \log (x)+(-5+5 x) \log ^2(x)+\log (-\frac {e^x}{4 x}) (-x^3+3 x^3 \log (x))}{x^7-10 x^4 \log (x)+25 x \log ^2(x)} \, dx\)

Optimal. Leaf size=27 \[ \frac {3+\log \left (-\frac {e^x}{4 x}\right )}{5-\frac {x^3}{\log (x)}} \]

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Rubi [F]  time = 1.74, 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^3+\left (10 x^3-x^4\right ) \log (x)+(-5+5 x) \log ^2(x)+\log \left (-\frac {e^x}{4 x}\right ) \left (-x^3+3 x^3 \log (x)\right )}{x^7-10 x^4 \log (x)+25 x \log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-3*x^3 + (10*x^3 - x^4)*Log[x] + (-5 + 5*x)*Log[x]^2 + Log[-1/4*E^x/x]*(-x^3 + 3*x^3*Log[x]))/(x^7 - 10*x
^4*Log[x] + 25*x*Log[x]^2),x]

[Out]

x/5 - Log[x]/5 - 3*Defer[Int][x^2/(x^3 - 5*Log[x])^2, x] + (9*Defer[Int][x^5/(x^3 - 5*Log[x])^2, x])/5 - Defer
[Int][(x^2*Log[-1/4*E^x/x])/(x^3 - 5*Log[x])^2, x] + (3*Defer[Int][(x^5*Log[-1/4*E^x/x])/(x^3 - 5*Log[x])^2, x
])/5 - (8*Defer[Int][x^2/(x^3 - 5*Log[x]), x])/5 - Defer[Int][x^3/(x^3 - 5*Log[x]), x]/5 - (3*Defer[Int][(x^2*
Log[-1/4*E^x/x])/(x^3 - 5*Log[x]), x])/5

Rubi steps

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

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Mathematica [A]  time = 0.28, size = 54, normalized size = 2.00 \begin {gather*} \frac {(-3+x) x^3-x^3 \log \left (-\frac {e^x}{4 x}\right )-x \left (5+x^2\right ) \log (x)+5 \log ^2(x)}{5 \left (x^3-5 \log (x)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-3*x^3 + (10*x^3 - x^4)*Log[x] + (-5 + 5*x)*Log[x]^2 + Log[-1/4*E^x/x]*(-x^3 + 3*x^3*Log[x]))/(x^7
- 10*x^4*Log[x] + 25*x*Log[x]^2),x]

[Out]

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

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fricas [A]  time = 0.57, size = 36, normalized size = 1.33 \begin {gather*} \frac {2 \, x^{3} \log \relax (2) - 3 \, x^{3} - 5 \, x \log \relax (x) + 5 \, \log \relax (x)^{2}}{5 \, {\left (x^{3} - 5 \, \log \relax (x)\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*x^3*log(x)-x^3)*log(-1/4*exp(x)/x)+(5*x-5)*log(x)^2+(-x^4+10*x^3)*log(x)-3*x^3)/(25*x*log(x)^2-1
0*x^4*log(x)+x^7),x, algorithm="fricas")

[Out]

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

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giac [C]  time = 0.21, size = 52, normalized size = 1.93 \begin {gather*} -\frac {1}{25} \, x^{3} + \frac {1}{5} \, x + \frac {x^{6} - 5 i \, \pi x^{3} - 5 \, x^{4} + 10 \, x^{3} \log \relax (2) - 15 \, x^{3}}{25 \, {\left (x^{3} - 5 \, \log \relax (x)\right )}} - \frac {1}{5} \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*x^3*log(x)-x^3)*log(-1/4*exp(x)/x)+(5*x-5)*log(x)^2+(-x^4+10*x^3)*log(x)-3*x^3)/(25*x*log(x)^2-1
0*x^4*log(x)+x^7),x, algorithm="giac")

[Out]

-1/25*x^3 + 1/5*x + 1/25*(x^6 - 5*I*pi*x^3 - 5*x^4 + 10*x^3*log(2) - 15*x^3)/(x^3 - 5*log(x)) - 1/5*log(x)

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maple [A]  time = 0.16, size = 42, normalized size = 1.56

method	result	size
default	\(\frac {\ln \relax (x )^{2}+\left (-\ln \left (-\frac {{\mathrm e}^{x}}{4 x}\right )+x -\ln \relax (x )-3\right ) \ln \relax (x )-x \ln \relax (x )}{x^{3}-5 \ln \relax (x )}\)	\(42\)
risch	\(-\frac {x^{3} \ln \left ({\mathrm e}^{x}\right )}{5 \left (x^{3}-5 \ln \relax (x )\right )}+\frac {-i \pi \,x^{3} \mathrm {csgn}\left (\frac {i {\mathrm e}^{x}}{x}\right )^{2} \mathrm {csgn}\left (\frac {i}{x}\right )-i \pi \,x^{3} \mathrm {csgn}\left (\frac {i {\mathrm e}^{x}}{x}\right )^{2} \mathrm {csgn}\left (i {\mathrm e}^{x}\right )+2 i \pi \,x^{3} \mathrm {csgn}\left (\frac {i {\mathrm e}^{x}}{x}\right )^{2}+i \pi \,x^{3} \mathrm {csgn}\left (\frac {i {\mathrm e}^{x}}{x}\right ) \mathrm {csgn}\left (\frac {i}{x}\right ) \mathrm {csgn}\left (i {\mathrm e}^{x}\right )-i \pi \,x^{3} \mathrm {csgn}\left (\frac {i {\mathrm e}^{x}}{x}\right )^{3}-2 i \pi \,x^{3}+4 x^{3} \ln \relax (2)+2 x^{4}-6 x^{3}+10 \ln \relax (x )^{2}-10 x \ln \relax (x )}{10 x^{3}-50 \ln \relax (x )}\)	\(182\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((3*x^3*ln(x)-x^3)*ln(-1/4*exp(x)/x)+(5*x-5)*ln(x)^2+(-x^4+10*x^3)*ln(x)-3*x^3)/(25*x*ln(x)^2-10*x^4*ln(x)
+x^7),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.50, size = 47, normalized size = 1.74 \begin {gather*} \frac {x^{4} + x^{3} {\left (2 \, \log \relax (2) - 3\right )} - x^{3} \log \left (-e^{x}\right ) - 5 \, x \log \relax (x) + 5 \, \log \relax (x)^{2}}{5 \, {\left (x^{3} - 5 \, \log \relax (x)\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*x^3*log(x)-x^3)*log(-1/4*exp(x)/x)+(5*x-5)*log(x)^2+(-x^4+10*x^3)*log(x)-3*x^3)/(25*x*log(x)^2-1
0*x^4*log(x)+x^7),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 5.27, size = 151, normalized size = 5.59 \begin {gather*} \frac {x^3}{25}-\frac {\ln \relax (x)}{5}-\frac {\frac {x\,\left (25\,x^2\,\left (\ln \left (-\frac {1}{4\,x}\right )+\ln \relax (x)\right )+75\,x^2+25\,x^3-5\,x^5+5\,x^6-3\,x^8\right )}{25\,\left (3\,x^3-5\right )}-\frac {x\,\ln \relax (x)\,\left (15\,x^2\,\left (\ln \left (-\frac {1}{4\,x}\right )+\ln \relax (x)\right )+45\,x^2+20\,x^3-6\,x^5\right )}{5\,\left (3\,x^3-5\right )}}{5\,\ln \relax (x)-x^3}-\frac {20\,x+15\,\ln \left (-\frac {1}{4\,x}\right )+15\,\ln \relax (x)+35}{45\,x^3-75}-\frac {x}{15} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(x)*(10*x^3 - x^4) + log(-exp(x)/(4*x))*(3*x^3*log(x) - x^3) - 3*x^3 + log(x)^2*(5*x - 5))/(25*x*log(x
)^2 - 10*x^4*log(x) + x^7),x)

[Out]

x^3/25 - log(x)/5 - ((x*(25*x^2*(log(-1/(4*x)) + log(x)) + 75*x^2 + 25*x^3 - 5*x^5 + 5*x^6 - 3*x^8))/(25*(3*x^
3 - 5)) - (x*log(x)*(15*x^2*(log(-1/(4*x)) + log(x)) + 45*x^2 + 20*x^3 - 6*x^5))/(5*(3*x^3 - 5)))/(5*log(x) -
x^3) - (20*x + 15*log(-1/(4*x)) + 15*log(x) + 35)/(45*x^3 - 75) - x/15

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sympy [C]  time = 0.47, size = 53, normalized size = 1.96 \begin {gather*} - \frac {x^{3}}{25} + \frac {x}{5} - \frac {\log {\relax (x )}}{5} + \frac {- x^{6} + 5 x^{4} - 10 x^{3} \log {\relax (2 )} + 15 x^{3} + 5 i \pi x^{3}}{- 25 x^{3} + 125 \log {\relax (x )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*x**3*ln(x)-x**3)*ln(-1/4*exp(x)/x)+(5*x-5)*ln(x)**2+(-x**4+10*x**3)*ln(x)-3*x**3)/(25*x*ln(x)**2
-10*x**4*ln(x)+x**7),x)

[Out]

-x**3/25 + x/5 - log(x)/5 + (-x**6 + 5*x**4 - 10*x**3*log(2) + 15*x**3 + 5*I*pi*x**3)/(-25*x**3 + 125*log(x))

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