3.87.40 \(\int \frac {-30 x+100 x^2+400 x^3+(6-2 x) \log (x)+(-19 x-80 x^2) \log ^2(x)+(1+4 x) \log ^4(x)}{100 x^3-20 x^2 \log ^2(x)+x \log ^4(x)} \, dx\)

Optimal. Leaf size=29 \[ 1-x+5 (4+x)+\log (x)-\frac {3-x}{-10 x+\log ^2(x)} \]

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Rubi [F]  time = 0.68, 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 x+100 x^2+400 x^3+(6-2 x) \log (x)+\left (-19 x-80 x^2\right ) \log ^2(x)+(1+4 x) \log ^4(x)}{100 x^3-20 x^2 \log ^2(x)+x \log ^4(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-30*x + 100*x^2 + 400*x^3 + (6 - 2*x)*Log[x] + (-19*x - 80*x^2)*Log[x]^2 + (1 + 4*x)*Log[x]^4)/(100*x^3 -
20*x^2*Log[x]^2 + x*Log[x]^4),x]

[Out]

4*x + Log[x] + 3/(10*x - Log[x]^2) + 10*Defer[Int][x/(10*x - Log[x]^2)^2, x] - 2*Defer[Int][Log[x]/(-10*x + Lo
g[x]^2)^2, x] + Defer[Int][(-10*x + Log[x]^2)^(-1), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-30 x+100 x^2+400 x^3+(6-2 x) \log (x)+\left (-19 x-80 x^2\right ) \log ^2(x)+(1+4 x) \log ^4(x)}{x \left (10 x-\log ^2(x)\right )^2} \, dx\\ &=\int \left (\frac {1+4 x}{x}+\frac {2 (-3+x) (5 x-\log (x))}{x \left (10 x-\log ^2(x)\right )^2}+\frac {1}{-10 x+\log ^2(x)}\right ) \, dx\\ &=2 \int \frac {(-3+x) (5 x-\log (x))}{x \left (10 x-\log ^2(x)\right )^2} \, dx+\int \frac {1+4 x}{x} \, dx+\int \frac {1}{-10 x+\log ^2(x)} \, dx\\ &=2 \int \left (\frac {5 x-\log (x)}{\left (10 x-\log ^2(x)\right )^2}-\frac {3 (5 x-\log (x))}{x \left (10 x-\log ^2(x)\right )^2}\right ) \, dx+\int \left (4+\frac {1}{x}\right ) \, dx+\int \frac {1}{-10 x+\log ^2(x)} \, dx\\ &=4 x+\log (x)+2 \int \frac {5 x-\log (x)}{\left (10 x-\log ^2(x)\right )^2} \, dx-6 \int \frac {5 x-\log (x)}{x \left (10 x-\log ^2(x)\right )^2} \, dx+\int \frac {1}{-10 x+\log ^2(x)} \, dx\\ &=4 x+\log (x)+\frac {3}{10 x-\log ^2(x)}+2 \int \left (\frac {5 x}{\left (10 x-\log ^2(x)\right )^2}-\frac {\log (x)}{\left (-10 x+\log ^2(x)\right )^2}\right ) \, dx+\int \frac {1}{-10 x+\log ^2(x)} \, dx\\ &=4 x+\log (x)+\frac {3}{10 x-\log ^2(x)}-2 \int \frac {\log (x)}{\left (-10 x+\log ^2(x)\right )^2} \, dx+10 \int \frac {x}{\left (10 x-\log ^2(x)\right )^2} \, dx+\int \frac {1}{-10 x+\log ^2(x)} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.18, size = 20, normalized size = 0.69 \begin {gather*} 4 x+\log (x)+\frac {-3+x}{-10 x+\log ^2(x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-30*x + 100*x^2 + 400*x^3 + (6 - 2*x)*Log[x] + (-19*x - 80*x^2)*Log[x]^2 + (1 + 4*x)*Log[x]^4)/(100
*x^3 - 20*x^2*Log[x]^2 + x*Log[x]^4),x]

[Out]

4*x + Log[x] + (-3 + x)/(-10*x + Log[x]^2)

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fricas [A]  time = 2.14, size = 35, normalized size = 1.21 \begin {gather*} \frac {4 \, x \log \relax (x)^{2} + \log \relax (x)^{3} - 40 \, x^{2} - 10 \, x \log \relax (x) + x - 3}{\log \relax (x)^{2} - 10 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x+1)*log(x)^4+(-80*x^2-19*x)*log(x)^2+(6-2*x)*log(x)+400*x^3+100*x^2-30*x)/(x*log(x)^4-20*x^2*lo
g(x)^2+100*x^3),x, algorithm="fricas")

[Out]

(4*x*log(x)^2 + log(x)^3 - 40*x^2 - 10*x*log(x) + x - 3)/(log(x)^2 - 10*x)

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giac [A]  time = 0.20, size = 20, normalized size = 0.69 \begin {gather*} 4 \, x + \frac {x - 3}{\log \relax (x)^{2} - 10 \, x} + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x+1)*log(x)^4+(-80*x^2-19*x)*log(x)^2+(6-2*x)*log(x)+400*x^3+100*x^2-30*x)/(x*log(x)^4-20*x^2*lo
g(x)^2+100*x^3),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.04, size = 24, normalized size = 0.83

method	result	size
risch	\(4 x +\ln \relax (x )-\frac {x -3}{-\ln \relax (x )^{2}+10 x}\)	\(24\)
norman	\(\frac {3-\frac {\ln \relax (x )^{2}}{10}+40 x^{2}-4 x \ln \relax (x )^{2}}{-\ln \relax (x )^{2}+10 x}+\ln \relax (x )\)	\(37\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((4*x+1)*ln(x)^4+(-80*x^2-19*x)*ln(x)^2+(6-2*x)*ln(x)+400*x^3+100*x^2-30*x)/(x*ln(x)^4-20*x^2*ln(x)^2+100*
x^3),x,method=_RETURNVERBOSE)

[Out]

4*x+ln(x)-(x-3)/(-ln(x)^2+10*x)

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maxima [A]  time = 0.39, size = 29, normalized size = 1.00 \begin {gather*} \frac {4 \, x \log \relax (x)^{2} - 40 \, x^{2} + x - 3}{\log \relax (x)^{2} - 10 \, x} + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x+1)*log(x)^4+(-80*x^2-19*x)*log(x)^2+(6-2*x)*log(x)+400*x^3+100*x^2-30*x)/(x*log(x)^4-20*x^2*lo
g(x)^2+100*x^3),x, algorithm="maxima")

[Out]

(4*x*log(x)^2 - 40*x^2 + x - 3)/(log(x)^2 - 10*x) + log(x)

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mupad [B]  time = 5.36, size = 23, normalized size = 0.79 \begin {gather*} 4\,x+\ln \relax (x)-\frac {x-3}{10\,x-{\ln \relax (x)}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(30*x + log(x)^2*(19*x + 80*x^2) + log(x)*(2*x - 6) - 100*x^2 - 400*x^3 - log(x)^4*(4*x + 1))/(x*log(x)^4
- 20*x^2*log(x)^2 + 100*x^3),x)

[Out]

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

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sympy [A]  time = 0.13, size = 17, normalized size = 0.59 \begin {gather*} 4 x + \log {\relax (x )} + \frac {x - 3}{- 10 x + \log {\relax (x )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x+1)*ln(x)**4+(-80*x**2-19*x)*ln(x)**2+(6-2*x)*ln(x)+400*x**3+100*x**2-30*x)/(x*ln(x)**4-20*x**2
*ln(x)**2+100*x**3),x)

[Out]

4*x + log(x) + (x - 3)/(-10*x + log(x)**2)

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