3.49.8 \(\int \frac {-5760+1728 x+180 x^2-36 x^3+(-960+192 x) \log (2)+(2880-576 x) \log (-5 x^2+x^3)}{-45 x^4+9 x^5+(-480 x^2+96 x^3) \log (2)+(-1280+256 x) \log ^2(2)+(1440 x^2-288 x^3+(7680-1536 x) \log (2)) \log (-5 x^2+x^3)+(-11520+2304 x) \log ^2(-5 x^2+x^3)} \, dx\)

Optimal. Leaf size=33 \[ -1+\frac {x}{4 \left (\frac {x^2}{16}+\frac {\log (2)}{3}-\log \left ((-5+x) x^2\right )\right )} \]

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Rubi [F]  time = 0.73, 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 {-5760+1728 x+180 x^2-36 x^3+(-960+192 x) \log (2)+(2880-576 x) \log \left (-5 x^2+x^3\right )}{-45 x^4+9 x^5+\left (-480 x^2+96 x^3\right ) \log (2)+(-1280+256 x) \log ^2(2)+\left (1440 x^2-288 x^3+(7680-1536 x) \log (2)\right ) \log \left (-5 x^2+x^3\right )+(-11520+2304 x) \log ^2\left (-5 x^2+x^3\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-5760 + 1728*x + 180*x^2 - 36*x^3 + (-960 + 192*x)*Log[2] + (2880 - 576*x)*Log[-5*x^2 + x^3])/(-45*x^4 +
9*x^5 + (-480*x^2 + 96*x^3)*Log[2] + (-1280 + 256*x)*Log[2]^2 + (1440*x^2 - 288*x^3 + (7680 - 1536*x)*Log[2])*
Log[-5*x^2 + x^3] + (-11520 + 2304*x)*Log[-5*x^2 + x^3]^2),x]

[Out]

1728*Defer[Int][(3*x^2 + 16*Log[2] - 48*Log[(-5 + x)*x^2])^(-2), x] + 2880*Defer[Int][1/((-5 + x)*(3*x^2 + 16*
Log[2] - 48*Log[(-5 + x)*x^2])^2), x] - 72*Defer[Int][x^2/(3*x^2 + 16*Log[2] - 48*Log[(-5 + x)*x^2])^2, x] + 1
2*Defer[Int][(3*x^2 + 16*Log[2] - 48*Log[(-5 + x)*x^2])^(-1), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {12 \left (-15 x^2+3 x^3+80 (6+\log (2))-16 x (9+\log (2))+48 (-5+x) \log \left ((-5+x) x^2\right )\right )}{(5-x) \left (3 x^2+16 \log (2)-48 \log \left ((-5+x) x^2\right )\right )^2} \, dx\\ &=12 \int \frac {-15 x^2+3 x^3+80 (6+\log (2))-16 x (9+\log (2))+48 (-5+x) \log \left ((-5+x) x^2\right )}{(5-x) \left (3 x^2+16 \log (2)-48 \log \left ((-5+x) x^2\right )\right )^2} \, dx\\ &=12 \int \left (-\frac {6 \left (80-24 x-5 x^2+x^3\right )}{(-5+x) \left (3 x^2+16 \log (2)-48 \log \left ((-5+x) x^2\right )\right )^2}+\frac {1}{3 x^2+16 \log (2)-48 \log \left ((-5+x) x^2\right )}\right ) \, dx\\ &=12 \int \frac {1}{3 x^2+16 \log (2)-48 \log \left ((-5+x) x^2\right )} \, dx-72 \int \frac {80-24 x-5 x^2+x^3}{(-5+x) \left (3 x^2+16 \log (2)-48 \log \left ((-5+x) x^2\right )\right )^2} \, dx\\ &=12 \int \frac {1}{3 x^2+16 \log (2)-48 \log \left ((-5+x) x^2\right )} \, dx-72 \int \left (-\frac {24}{\left (3 x^2+16 \log (2)-48 \log \left ((-5+x) x^2\right )\right )^2}-\frac {40}{(-5+x) \left (3 x^2+16 \log (2)-48 \log \left ((-5+x) x^2\right )\right )^2}+\frac {x^2}{\left (3 x^2+16 \log (2)-48 \log \left ((-5+x) x^2\right )\right )^2}\right ) \, dx\\ &=12 \int \frac {1}{3 x^2+16 \log (2)-48 \log \left ((-5+x) x^2\right )} \, dx-72 \int \frac {x^2}{\left (3 x^2+16 \log (2)-48 \log \left ((-5+x) x^2\right )\right )^2} \, dx+1728 \int \frac {1}{\left (3 x^2+16 \log (2)-48 \log \left ((-5+x) x^2\right )\right )^2} \, dx+2880 \int \frac {1}{(-5+x) \left (3 x^2+16 \log (2)-48 \log \left ((-5+x) x^2\right )\right )^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.69, size = 25, normalized size = 0.76 \begin {gather*} -\frac {12 x}{-3 x^2-16 \log (2)+48 \log \left ((-5+x) x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-5760 + 1728*x + 180*x^2 - 36*x^3 + (-960 + 192*x)*Log[2] + (2880 - 576*x)*Log[-5*x^2 + x^3])/(-45*
x^4 + 9*x^5 + (-480*x^2 + 96*x^3)*Log[2] + (-1280 + 256*x)*Log[2]^2 + (1440*x^2 - 288*x^3 + (7680 - 1536*x)*Lo
g[2])*Log[-5*x^2 + x^3] + (-11520 + 2304*x)*Log[-5*x^2 + x^3]^2),x]

[Out]

(-12*x)/(-3*x^2 - 16*Log[2] + 48*Log[(-5 + x)*x^2])

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fricas [A]  time = 0.72, size = 27, normalized size = 0.82 \begin {gather*} \frac {12 \, x}{3 \, x^{2} + 16 \, \log \relax (2) - 48 \, \log \left (x^{3} - 5 \, x^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-576*x+2880)*log(x^3-5*x^2)+(192*x-960)*log(2)-36*x^3+180*x^2+1728*x-5760)/((2304*x-11520)*log(x^3
-5*x^2)^2+((-1536*x+7680)*log(2)-288*x^3+1440*x^2)*log(x^3-5*x^2)+(256*x-1280)*log(2)^2+(96*x^3-480*x^2)*log(2
)+9*x^5-45*x^4),x, algorithm="fricas")

[Out]

12*x/(3*x^2 + 16*log(2) - 48*log(x^3 - 5*x^2))

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giac [A]  time = 0.21, size = 27, normalized size = 0.82 \begin {gather*} \frac {12 \, x}{3 \, x^{2} + 16 \, \log \relax (2) - 48 \, \log \left (x^{3} - 5 \, x^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-576*x+2880)*log(x^3-5*x^2)+(192*x-960)*log(2)-36*x^3+180*x^2+1728*x-5760)/((2304*x-11520)*log(x^3
-5*x^2)^2+((-1536*x+7680)*log(2)-288*x^3+1440*x^2)*log(x^3-5*x^2)+(256*x-1280)*log(2)^2+(96*x^3-480*x^2)*log(2
)+9*x^5-45*x^4),x, algorithm="giac")

[Out]

12*x/(3*x^2 + 16*log(2) - 48*log(x^3 - 5*x^2))

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maple [A]  time = 0.21, size = 28, normalized size = 0.85

method	result	size
norman	\(\frac {12 x}{3 x^{2}+16 \ln \relax (2)-48 \ln \left (x^{3}-5 x^{2}\right )}\)	\(28\)
risch	\(\frac {12 x}{3 x^{2}+16 \ln \relax (2)-48 \ln \left (x^{3}-5 x^{2}\right )}\)	\(28\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-576*x+2880)*ln(x^3-5*x^2)+(192*x-960)*ln(2)-36*x^3+180*x^2+1728*x-5760)/((2304*x-11520)*ln(x^3-5*x^2)^2
+((-1536*x+7680)*ln(2)-288*x^3+1440*x^2)*ln(x^3-5*x^2)+(256*x-1280)*ln(2)^2+(96*x^3-480*x^2)*ln(2)+9*x^5-45*x^
4),x,method=_RETURNVERBOSE)

[Out]

12*x/(3*x^2+16*ln(2)-48*ln(x^3-5*x^2))

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maxima [A]  time = 0.49, size = 25, normalized size = 0.76 \begin {gather*} \frac {12 \, x}{3 \, x^{2} + 16 \, \log \relax (2) - 48 \, \log \left (x - 5\right ) - 96 \, \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-576*x+2880)*log(x^3-5*x^2)+(192*x-960)*log(2)-36*x^3+180*x^2+1728*x-5760)/((2304*x-11520)*log(x^3
-5*x^2)^2+((-1536*x+7680)*log(2)-288*x^3+1440*x^2)*log(x^3-5*x^2)+(256*x-1280)*log(2)^2+(96*x^3-480*x^2)*log(2
)+9*x^5-45*x^4),x, algorithm="maxima")

[Out]

12*x/(3*x^2 + 16*log(2) - 48*log(x - 5) - 96*log(x))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int -\frac {1728\,x+\ln \relax (2)\,\left (192\,x-960\right )-\ln \left (x^3-5\,x^2\right )\,\left (576\,x-2880\right )+180\,x^2-36\,x^3-5760}{\ln \left (x^3-5\,x^2\right )\,\left (\ln \relax (2)\,\left (1536\,x-7680\right )-1440\,x^2+288\,x^3\right )-{\ln \relax (2)}^2\,\left (256\,x-1280\right )+\ln \relax (2)\,\left (480\,x^2-96\,x^3\right )-{\ln \left (x^3-5\,x^2\right )}^2\,\left (2304\,x-11520\right )+45\,x^4-9\,x^5} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(1728*x + log(2)*(192*x - 960) - log(x^3 - 5*x^2)*(576*x - 2880) + 180*x^2 - 36*x^3 - 5760)/(log(x^3 - 5*
x^2)*(log(2)*(1536*x - 7680) - 1440*x^2 + 288*x^3) - log(2)^2*(256*x - 1280) + log(2)*(480*x^2 - 96*x^3) - log
(x^3 - 5*x^2)^2*(2304*x - 11520) + 45*x^4 - 9*x^5),x)

[Out]

int(-(1728*x + log(2)*(192*x - 960) - log(x^3 - 5*x^2)*(576*x - 2880) + 180*x^2 - 36*x^3 - 5760)/(log(x^3 - 5*
x^2)*(log(2)*(1536*x - 7680) - 1440*x^2 + 288*x^3) - log(2)^2*(256*x - 1280) + log(2)*(480*x^2 - 96*x^3) - log
(x^3 - 5*x^2)^2*(2304*x - 11520) + 45*x^4 - 9*x^5), x)

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sympy [A]  time = 0.22, size = 26, normalized size = 0.79 \begin {gather*} - \frac {12 x}{- 3 x^{2} + 48 \log {\left (x^{3} - 5 x^{2} \right )} - 16 \log {\relax (2 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-576*x+2880)*ln(x**3-5*x**2)+(192*x-960)*ln(2)-36*x**3+180*x**2+1728*x-5760)/((2304*x-11520)*ln(x*
*3-5*x**2)**2+((-1536*x+7680)*ln(2)-288*x**3+1440*x**2)*ln(x**3-5*x**2)+(256*x-1280)*ln(2)**2+(96*x**3-480*x**
2)*ln(2)+9*x**5-45*x**4),x)

[Out]

-12*x/(-3*x**2 + 48*log(x**3 - 5*x**2) - 16*log(2))

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