3.62.70 \(\int \frac {5 x^6+48 x^7+112 x^8+(80-40 x^5-384 x^6-896 x^7) \log (3)+(80 x^4+768 x^5+1792 x^6) \log ^2(3)}{x^2-8 x \log (3)+16 \log ^2(3)} \, dx\)

Optimal. Leaf size=26 \[ -2+x^5 (1+4 x)^2+\frac {20 x}{-x+4 \log (3)} \]

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Rubi [A]  time = 0.08, antiderivative size = 26, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 75, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.027, Rules used = {27, 1850} \begin {gather*} 16 x^7+8 x^6+x^5-\frac {80 \log (3)}{x-4 \log (3)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(5*x^6 + 48*x^7 + 112*x^8 + (80 - 40*x^5 - 384*x^6 - 896*x^7)*Log[3] + (80*x^4 + 768*x^5 + 1792*x^6)*Log[3
]^2)/(x^2 - 8*x*Log[3] + 16*Log[3]^2),x]

[Out]

x^5 + 8*x^6 + 16*x^7 - (80*Log[3])/(x - 4*Log[3])

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 1850

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq*(a + b*x^n)^p, x], x] /; FreeQ[
{a, b, n}, x] && PolyQ[Pq, x] && (IGtQ[p, 0] || EqQ[n, 1])

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {5 x^6+48 x^7+112 x^8+\left (80-40 x^5-384 x^6-896 x^7\right ) \log (3)+\left (80 x^4+768 x^5+1792 x^6\right ) \log ^2(3)}{(x-4 \log (3))^2} \, dx\\ &=\int \left (5 x^4+48 x^5+112 x^6+\frac {80 \log (3)}{(x-4 \log (3))^2}\right ) \, dx\\ &=x^5+8 x^6+16 x^7-\frac {80 \log (3)}{x-4 \log (3)}\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.04, size = 81, normalized size = 3.12 \begin {gather*} \frac {16 x^8+x^7 (8-64 \log (3))+x^6 (1-32 \log (3))-4 x^5 \log (3)-1024 x \log ^5(3) (1+16 \log (3))^2+16 \log (3) \left (-5+256 \log ^5(3)+8192 \log ^6(3)+65536 \log ^7(3)\right )}{x-4 \log (3)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(5*x^6 + 48*x^7 + 112*x^8 + (80 - 40*x^5 - 384*x^6 - 896*x^7)*Log[3] + (80*x^4 + 768*x^5 + 1792*x^6)
*Log[3]^2)/(x^2 - 8*x*Log[3] + 16*Log[3]^2),x]

[Out]

(16*x^8 + x^7*(8 - 64*Log[3]) + x^6*(1 - 32*Log[3]) - 4*x^5*Log[3] - 1024*x*Log[3]^5*(1 + 16*Log[3])^2 + 16*Lo
g[3]*(-5 + 256*Log[3]^5 + 8192*Log[3]^6 + 65536*Log[3]^7))/(x - 4*Log[3])

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fricas [A]  time = 0.57, size = 42, normalized size = 1.62 \begin {gather*} \frac {16 \, x^{8} + 8 \, x^{7} + x^{6} - 4 \, {\left (16 \, x^{7} + 8 \, x^{6} + x^{5} + 20\right )} \log \relax (3)}{x - 4 \, \log \relax (3)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((1792*x^6+768*x^5+80*x^4)*log(3)^2+(-896*x^7-384*x^6-40*x^5+80)*log(3)+112*x^8+48*x^7+5*x^6)/(16*lo
g(3)^2-8*x*log(3)+x^2),x, algorithm="fricas")

[Out]

(16*x^8 + 8*x^7 + x^6 - 4*(16*x^7 + 8*x^6 + x^5 + 20)*log(3))/(x - 4*log(3))

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giac [A]  time = 0.19, size = 26, normalized size = 1.00 \begin {gather*} 16 \, x^{7} + 8 \, x^{6} + x^{5} - \frac {80 \, \log \relax (3)}{x - 4 \, \log \relax (3)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((1792*x^6+768*x^5+80*x^4)*log(3)^2+(-896*x^7-384*x^6-40*x^5+80)*log(3)+112*x^8+48*x^7+5*x^6)/(16*lo
g(3)^2-8*x*log(3)+x^2),x, algorithm="giac")

[Out]

16*x^7 + 8*x^6 + x^5 - 80*log(3)/(x - 4*log(3))

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maple [A]  time = 0.53, size = 27, normalized size = 1.04

method	result	size
default	\(16 x^{7}+8 x^{6}+x^{5}-\frac {80 \ln \relax (3)}{-4 \ln \relax (3)+x}\)	\(27\)
risch	\(16 x^{7}+8 x^{6}+x^{5}+\frac {20 \ln \relax (3)}{\ln \relax (3)-\frac {x}{4}}\)	\(27\)
norman	\(\frac {-16 x^{8}+\left (64 \ln \relax (3)-8\right ) x^{7}+\left (32 \ln \relax (3)-1\right ) x^{6}+4 x^{5} \ln \relax (3)+80 \ln \relax (3)}{4 \ln \relax (3)-x}\)	\(49\)
gosper	\(\frac {64 \ln \relax (3) x^{7}-16 x^{8}+32 x^{6} \ln \relax (3)-8 x^{7}+4 x^{5} \ln \relax (3)-x^{6}+80 \ln \relax (3)}{4 \ln \relax (3)-x}\)	\(53\)
meijerg	\(-1835008 \ln \relax (3)^{7} \left (-\frac {x \left (-\frac {45 x^{7}}{16384 \ln \relax (3)^{7}}-\frac {15 x^{6}}{1024 \ln \relax (3)^{6}}-\frac {21 x^{5}}{256 \ln \relax (3)^{5}}-\frac {63 x^{4}}{128 \ln \relax (3)^{4}}-\frac {105 x^{3}}{32 \ln \relax (3)^{3}}-\frac {105 x^{2}}{4 \ln \relax (3)^{2}}-\frac {315 x}{\ln \relax (3)}+2520\right )}{1260 \ln \relax (3) \left (1-\frac {x}{4 \ln \relax (3)}\right )}-8 \ln \left (1-\frac {x}{4 \ln \relax (3)}\right )\right )+65536 \ln \relax (3)^{6} \left (-56 \ln \relax (3)+3\right ) \left (\frac {x \left (-\frac {5 x^{6}}{1024 \ln \relax (3)^{6}}-\frac {7 x^{5}}{256 \ln \relax (3)^{5}}-\frac {21 x^{4}}{128 \ln \relax (3)^{4}}-\frac {35 x^{3}}{32 \ln \relax (3)^{3}}-\frac {35 x^{2}}{4 \ln \relax (3)^{2}}-\frac {105 x}{\ln \relax (3)}+840\right )}{480 \ln \relax (3) \left (1-\frac {x}{4 \ln \relax (3)}\right )}+7 \ln \left (1-\frac {x}{4 \ln \relax (3)}\right )\right )-16384 \ln \relax (3)^{5} \left (112 \ln \relax (3)^{2}-24 \ln \relax (3)+\frac {5}{16}\right ) \left (-\frac {x \left (-\frac {7 x^{5}}{512 \ln \relax (3)^{5}}-\frac {21 x^{4}}{256 \ln \relax (3)^{4}}-\frac {35 x^{3}}{64 \ln \relax (3)^{3}}-\frac {35 x^{2}}{8 \ln \relax (3)^{2}}-\frac {105 x}{2 \ln \relax (3)}+420\right )}{280 \ln \relax (3) \left (1-\frac {x}{4 \ln \relax (3)}\right )}-6 \ln \left (1-\frac {x}{4 \ln \relax (3)}\right )\right )+4096 \ln \relax (3)^{4} \left (48 \ln \relax (3)^{2}-\frac {5 \ln \relax (3)}{2}\right ) \left (\frac {x \left (-\frac {3 x^{4}}{256 \ln \relax (3)^{4}}-\frac {5 x^{3}}{64 \ln \relax (3)^{3}}-\frac {5 x^{2}}{8 \ln \relax (3)^{2}}-\frac {15 x}{2 \ln \relax (3)}+60\right )}{48 \ln \relax (3) \left (1-\frac {x}{4 \ln \relax (3)}\right )}+5 \ln \left (1-\frac {x}{4 \ln \relax (3)}\right )\right )-5120 \ln \relax (3)^{5} \left (-\frac {x \left (-\frac {5 x^{3}}{64 \ln \relax (3)^{3}}-\frac {5 x^{2}}{8 \ln \relax (3)^{2}}-\frac {15 x}{2 \ln \relax (3)}+60\right )}{60 \ln \relax (3) \left (1-\frac {x}{4 \ln \relax (3)}\right )}-4 \ln \left (1-\frac {x}{4 \ln \relax (3)}\right )\right )+\frac {5 x}{\ln \relax (3) \left (1-\frac {x}{4 \ln \relax (3)}\right )}\)	\(459\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((1792*x^6+768*x^5+80*x^4)*ln(3)^2+(-896*x^7-384*x^6-40*x^5+80)*ln(3)+112*x^8+48*x^7+5*x^6)/(16*ln(3)^2-8*
x*ln(3)+x^2),x,method=_RETURNVERBOSE)

[Out]

16*x^7+8*x^6+x^5-80*ln(3)/(-4*ln(3)+x)

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maxima [A]  time = 0.35, size = 26, normalized size = 1.00 \begin {gather*} 16 \, x^{7} + 8 \, x^{6} + x^{5} - \frac {80 \, \log \relax (3)}{x - 4 \, \log \relax (3)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((1792*x^6+768*x^5+80*x^4)*log(3)^2+(-896*x^7-384*x^6-40*x^5+80)*log(3)+112*x^8+48*x^7+5*x^6)/(16*lo
g(3)^2-8*x*log(3)+x^2),x, algorithm="maxima")

[Out]

16*x^7 + 8*x^6 + x^5 - 80*log(3)/(x - 4*log(3))

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mupad [B]  time = 0.14, size = 26, normalized size = 1.00 \begin {gather*} x^5-\frac {80\,\ln \relax (3)}{x-4\,\ln \relax (3)}+8\,x^6+16\,x^7 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(3)^2*(80*x^4 + 768*x^5 + 1792*x^6) - log(3)*(40*x^5 + 384*x^6 + 896*x^7 - 80) + 5*x^6 + 48*x^7 + 112*
x^8)/(16*log(3)^2 - 8*x*log(3) + x^2),x)

[Out]

x^5 - (80*log(3))/(x - 4*log(3)) + 8*x^6 + 16*x^7

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sympy [A]  time = 0.15, size = 24, normalized size = 0.92 \begin {gather*} 16 x^{7} + 8 x^{6} + x^{5} - \frac {80 \log {\relax (3 )}}{x - 4 \log {\relax (3 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((1792*x**6+768*x**5+80*x**4)*ln(3)**2+(-896*x**7-384*x**6-40*x**5+80)*ln(3)+112*x**8+48*x**7+5*x**6
)/(16*ln(3)**2-8*x*ln(3)+x**2),x)

[Out]

16*x**7 + 8*x**6 + x**5 - 80*log(3)/(x - 4*log(3))

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