3.97.9 \(\int \frac {30-75 x+10 x^2-25 x^3+(-55 x-70 x^2+25 x^3) \log (x)+(-10 x+25 x^2+10 x \log (x)) \log (5 x^2)}{(4 x-20 x^2+25 x^3) \log ^2(x)} \, dx\)

Optimal. Leaf size=27 \[ \frac {5 \left (-3+x \left (-x+\log \left (5 x^2\right )\right )\right )}{(2-5 x) \log (x)} \]

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Rubi [F]  time = 1.50, 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 {30-75 x+10 x^2-25 x^3+\left (-55 x-70 x^2+25 x^3\right ) \log (x)+\left (-10 x+25 x^2+10 x \log (x)\right ) \log \left (5 x^2\right )}{\left (4 x-20 x^2+25 x^3\right ) \log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(30 - 75*x + 10*x^2 - 25*x^3 + (-55*x - 70*x^2 + 25*x^3)*Log[x] + (-10*x + 25*x^2 + 10*x*Log[x])*Log[5*x^2
])/((4*x - 20*x^2 + 25*x^3)*Log[x]^2),x]

[Out]

-75*Defer[Int][1/((-2 + 5*x)^2*Log[x]^2), x] + 30*Defer[Int][1/(x*(-2 + 5*x)^2*Log[x]^2), x] + 10*Defer[Int][x
/((-2 + 5*x)^2*Log[x]^2), x] - 25*Defer[Int][x^2/((-2 + 5*x)^2*Log[x]^2), x] + 5*Defer[Int][(-11 - 14*x + 5*x^
2)/((-2 + 5*x)^2*Log[x]), x] + 5*Defer[Int][Log[5*x^2]/((-2 + 5*x)*Log[x]^2), x] + 10*Defer[Int][Log[5*x^2]/((
-2 + 5*x)^2*Log[x]), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {30-75 x+10 x^2-25 x^3+\left (-55 x-70 x^2+25 x^3\right ) \log (x)+\left (-10 x+25 x^2+10 x \log (x)\right ) \log \left (5 x^2\right )}{x \left (4-20 x+25 x^2\right ) \log ^2(x)} \, dx\\ &=\int \frac {30-75 x+10 x^2-25 x^3+\left (-55 x-70 x^2+25 x^3\right ) \log (x)+\left (-10 x+25 x^2+10 x \log (x)\right ) \log \left (5 x^2\right )}{x (-2+5 x)^2 \log ^2(x)} \, dx\\ &=\int \left (-\frac {75}{(-2+5 x)^2 \log ^2(x)}+\frac {30}{x (-2+5 x)^2 \log ^2(x)}+\frac {10 x}{(-2+5 x)^2 \log ^2(x)}-\frac {25 x^2}{(-2+5 x)^2 \log ^2(x)}+\frac {5 \left (-11-14 x+5 x^2\right )}{(-2+5 x)^2 \log (x)}+\frac {5 (-2+5 x+2 \log (x)) \log \left (5 x^2\right )}{(-2+5 x)^2 \log ^2(x)}\right ) \, dx\\ &=5 \int \frac {-11-14 x+5 x^2}{(-2+5 x)^2 \log (x)} \, dx+5 \int \frac {(-2+5 x+2 \log (x)) \log \left (5 x^2\right )}{(-2+5 x)^2 \log ^2(x)} \, dx+10 \int \frac {x}{(-2+5 x)^2 \log ^2(x)} \, dx-25 \int \frac {x^2}{(-2+5 x)^2 \log ^2(x)} \, dx+30 \int \frac {1}{x (-2+5 x)^2 \log ^2(x)} \, dx-75 \int \frac {1}{(-2+5 x)^2 \log ^2(x)} \, dx\\ &=5 \int \frac {-11-14 x+5 x^2}{(-2+5 x)^2 \log (x)} \, dx+5 \int \left (-\frac {2 \log \left (5 x^2\right )}{(-2+5 x)^2 \log ^2(x)}+\frac {5 x \log \left (5 x^2\right )}{(-2+5 x)^2 \log ^2(x)}+\frac {2 \log \left (5 x^2\right )}{(-2+5 x)^2 \log (x)}\right ) \, dx+10 \int \frac {x}{(-2+5 x)^2 \log ^2(x)} \, dx-25 \int \frac {x^2}{(-2+5 x)^2 \log ^2(x)} \, dx+30 \int \frac {1}{x (-2+5 x)^2 \log ^2(x)} \, dx-75 \int \frac {1}{(-2+5 x)^2 \log ^2(x)} \, dx\\ &=5 \int \frac {-11-14 x+5 x^2}{(-2+5 x)^2 \log (x)} \, dx+10 \int \frac {x}{(-2+5 x)^2 \log ^2(x)} \, dx-10 \int \frac {\log \left (5 x^2\right )}{(-2+5 x)^2 \log ^2(x)} \, dx+10 \int \frac {\log \left (5 x^2\right )}{(-2+5 x)^2 \log (x)} \, dx-25 \int \frac {x^2}{(-2+5 x)^2 \log ^2(x)} \, dx+25 \int \frac {x \log \left (5 x^2\right )}{(-2+5 x)^2 \log ^2(x)} \, dx+30 \int \frac {1}{x (-2+5 x)^2 \log ^2(x)} \, dx-75 \int \frac {1}{(-2+5 x)^2 \log ^2(x)} \, dx\\ &=5 \int \frac {-11-14 x+5 x^2}{(-2+5 x)^2 \log (x)} \, dx+10 \int \frac {x}{(-2+5 x)^2 \log ^2(x)} \, dx-10 \int \frac {\log \left (5 x^2\right )}{(-2+5 x)^2 \log ^2(x)} \, dx+10 \int \frac {\log \left (5 x^2\right )}{(-2+5 x)^2 \log (x)} \, dx-25 \int \frac {x^2}{(-2+5 x)^2 \log ^2(x)} \, dx+25 \int \left (\frac {2 \log \left (5 x^2\right )}{5 (-2+5 x)^2 \log ^2(x)}+\frac {\log \left (5 x^2\right )}{5 (-2+5 x) \log ^2(x)}\right ) \, dx+30 \int \frac {1}{x (-2+5 x)^2 \log ^2(x)} \, dx-75 \int \frac {1}{(-2+5 x)^2 \log ^2(x)} \, dx\\ &=5 \int \frac {-11-14 x+5 x^2}{(-2+5 x)^2 \log (x)} \, dx+5 \int \frac {\log \left (5 x^2\right )}{(-2+5 x) \log ^2(x)} \, dx+10 \int \frac {x}{(-2+5 x)^2 \log ^2(x)} \, dx+10 \int \frac {\log \left (5 x^2\right )}{(-2+5 x)^2 \log (x)} \, dx-25 \int \frac {x^2}{(-2+5 x)^2 \log ^2(x)} \, dx+30 \int \frac {1}{x (-2+5 x)^2 \log ^2(x)} \, dx-75 \int \frac {1}{(-2+5 x)^2 \log ^2(x)} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.21, size = 29, normalized size = 1.07 \begin {gather*} 2+\frac {5 \left (3+x^2-x \log \left (5 x^2\right )\right )}{(-2+5 x) \log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(30 - 75*x + 10*x^2 - 25*x^3 + (-55*x - 70*x^2 + 25*x^3)*Log[x] + (-10*x + 25*x^2 + 10*x*Log[x])*Log
[5*x^2])/((4*x - 20*x^2 + 25*x^3)*Log[x]^2),x]

[Out]

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

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fricas [A]  time = 0.79, size = 28, normalized size = 1.04 \begin {gather*} \frac {5 \, x^{2} - 5 \, x \log \relax (5) - 4 \, \log \relax (x) + 15}{{\left (5 \, x - 2\right )} \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((10*x*log(x)+25*x^2-10*x)*log(5*x^2)+(25*x^3-70*x^2-55*x)*log(x)-25*x^3+10*x^2-75*x+30)/(25*x^3-20*
x^2+4*x)/log(x)^2,x, algorithm="fricas")

[Out]

(5*x^2 - 5*x*log(5) - 4*log(x) + 15)/((5*x - 2)*log(x))

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giac [A]  time = 0.19, size = 34, normalized size = 1.26 \begin {gather*} \frac {5 \, {\left (x^{2} - x \log \relax (5) + 3\right )}}{5 \, x \log \relax (x) - 2 \, \log \relax (x)} - \frac {4}{5 \, x - 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((10*x*log(x)+25*x^2-10*x)*log(5*x^2)+(25*x^3-70*x^2-55*x)*log(x)-25*x^3+10*x^2-75*x+30)/(25*x^3-20*
x^2+4*x)/log(x)^2,x, algorithm="giac")

[Out]

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

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maple [C]  time = 0.12, size = 88, normalized size = 3.26

method	result	size
risch	\(-\frac {4}{5 x -2}+\frac {\frac {5 i \pi x \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )}{2}-5 i \pi x \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}+15+\frac {5 i \pi x \mathrm {csgn}\left (i x^{2}\right )^{3}}{2}-5 x \ln \relax (5)+5 x^{2}}{\left (5 x -2\right ) \ln \relax (x )}\)	\(88\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((10*x*ln(x)+25*x^2-10*x)*ln(5*x^2)+(25*x^3-70*x^2-55*x)*ln(x)-25*x^3+10*x^2-75*x+30)/(25*x^3-20*x^2+4*x)/
ln(x)^2,x,method=_RETURNVERBOSE)

[Out]

-4/(5*x-2)+5/2*(I*Pi*x*csgn(I*x)^2*csgn(I*x^2)-2*I*Pi*x*csgn(I*x)*csgn(I*x^2)^2+6+I*Pi*x*csgn(I*x^2)^3-2*x*ln(
5)+2*x^2)/(5*x-2)/ln(x)

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maxima [A]  time = 0.48, size = 28, normalized size = 1.04 \begin {gather*} \frac {5 \, x^{2} - 5 \, x \log \relax (5) - 4 \, \log \relax (x) + 15}{{\left (5 \, x - 2\right )} \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((10*x*log(x)+25*x^2-10*x)*log(5*x^2)+(25*x^3-70*x^2-55*x)*log(x)-25*x^3+10*x^2-75*x+30)/(25*x^3-20*
x^2+4*x)/log(x)^2,x, algorithm="maxima")

[Out]

(5*x^2 - 5*x*log(5) - 4*log(x) + 15)/((5*x - 2)*log(x))

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mupad [B]  time = 5.59, size = 27, normalized size = 1.00 \begin {gather*} \frac {5\,\left (x^2-x\,\ln \left (5\,x^2\right )+3\right )}{\ln \relax (x)\,\left (5\,x-2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(75*x - log(5*x^2)*(10*x*log(x) - 10*x + 25*x^2) - 10*x^2 + 25*x^3 + log(x)*(55*x + 70*x^2 - 25*x^3) - 30
)/(log(x)^2*(4*x - 20*x^2 + 25*x^3)),x)

[Out]

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

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sympy [A]  time = 0.29, size = 27, normalized size = 1.00 \begin {gather*} - \frac {20}{25 x - 10} + \frac {5 x^{2} - 5 x \log {\relax (5 )} + 15}{\left (5 x - 2\right ) \log {\relax (x )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((10*x*ln(x)+25*x**2-10*x)*ln(5*x**2)+(25*x**3-70*x**2-55*x)*ln(x)-25*x**3+10*x**2-75*x+30)/(25*x**3
-20*x**2+4*x)/ln(x)**2,x)

[Out]

-20/(25*x - 10) + (5*x**2 - 5*x*log(5) + 15)/((5*x - 2)*log(x))

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