3.26.15 \(\int \frac {-800+100 x^3 \log (3)-2 x^6 \log ^2(3)+(-200 x^6-10 x^9 \log (3)) \log ^2(4)+100 x^{12} \log ^4(4)}{25 x^9} \, dx\)

Optimal. Leaf size=25 \[ \left (\frac {2}{x^4}-\frac {\log (3)}{5 x}+x^2 \log ^2(4)\right )^2 \]

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Rubi [B]  time = 0.04, antiderivative size = 52, normalized size of antiderivative = 2.08, number of steps used = 3, number of rules used = 2, integrand size = 52, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.038, Rules used = {12, 14} \begin {gather*} \frac {4}{x^8}-\frac {4 \log (3)}{5 x^5}+x^4 \log ^4(4)+\frac {\log ^2(3)+100 \log ^2(4)}{25 x^2}-\frac {2}{5} x \log (3) \log ^2(4) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-800 + 100*x^3*Log[3] - 2*x^6*Log[3]^2 + (-200*x^6 - 10*x^9*Log[3])*Log[4]^2 + 100*x^12*Log[4]^4)/(25*x^9
),x]

[Out]

4/x^8 - (4*Log[3])/(5*x^5) - (2*x*Log[3]*Log[4]^2)/5 + x^4*Log[4]^4 + (Log[3]^2 + 100*Log[4]^2)/(25*x^2)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{25} \int \frac {-800+100 x^3 \log (3)-2 x^6 \log ^2(3)+\left (-200 x^6-10 x^9 \log (3)\right ) \log ^2(4)+100 x^{12} \log ^4(4)}{x^9} \, dx\\ &=\frac {1}{25} \int \left (-\frac {800}{x^9}+\frac {100 \log (3)}{x^6}-10 \log (3) \log ^2(4)+100 x^3 \log ^4(4)-\frac {2 \left (\log ^2(3)+100 \log ^2(4)\right )}{x^3}\right ) \, dx\\ &=\frac {4}{x^8}-\frac {4 \log (3)}{5 x^5}-\frac {2}{5} x \log (3) \log ^2(4)+x^4 \log ^4(4)+\frac {\log ^2(3)+100 \log ^2(4)}{25 x^2}\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.03, size = 52, normalized size = 2.08 \begin {gather*} \frac {4}{x^8}-\frac {2}{5} x \log (3) \log ^2(4)+x^4 \log ^4(4)+\frac {\log ^2(3)+100 \log ^2(4)}{25 x^2}-\frac {\log (81)}{5 x^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-800 + 100*x^3*Log[3] - 2*x^6*Log[3]^2 + (-200*x^6 - 10*x^9*Log[3])*Log[4]^2 + 100*x^12*Log[4]^4)/(
25*x^9),x]

[Out]

4/x^8 - (2*x*Log[3]*Log[4]^2)/5 + x^4*Log[4]^4 + (Log[3]^2 + 100*Log[4]^2)/(25*x^2) - Log[81]/(5*x^5)

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fricas [A]  time = 0.62, size = 50, normalized size = 2.00 \begin {gather*} \frac {400 \, x^{12} \log \relax (2)^{4} + x^{6} \log \relax (3)^{2} + 400 \, x^{6} \log \relax (2)^{2} - 20 \, {\left (2 \, x^{9} \log \relax (2)^{2} + x^{3}\right )} \log \relax (3) + 100}{25 \, x^{8}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/25*(1600*x^12*log(2)^4+4*(-10*x^9*log(3)-200*x^6)*log(2)^2-2*x^6*log(3)^2+100*x^3*log(3)-800)/x^9,
x, algorithm="fricas")

[Out]

1/25*(400*x^12*log(2)^4 + x^6*log(3)^2 + 400*x^6*log(2)^2 - 20*(2*x^9*log(2)^2 + x^3)*log(3) + 100)/x^8

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giac [A]  time = 0.14, size = 50, normalized size = 2.00 \begin {gather*} 16 \, x^{4} \log \relax (2)^{4} - \frac {8}{5} \, x \log \relax (3) \log \relax (2)^{2} + \frac {x^{6} \log \relax (3)^{2} + 400 \, x^{6} \log \relax (2)^{2} - 20 \, x^{3} \log \relax (3) + 100}{25 \, x^{8}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/25*(1600*x^12*log(2)^4+4*(-10*x^9*log(3)-200*x^6)*log(2)^2-2*x^6*log(3)^2+100*x^3*log(3)-800)/x^9,
x, algorithm="giac")

[Out]

16*x^4*log(2)^4 - 8/5*x*log(3)*log(2)^2 + 1/25*(x^6*log(3)^2 + 400*x^6*log(2)^2 - 20*x^3*log(3) + 100)/x^8

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maple [A]  time = 0.07, size = 49, normalized size = 1.96

method	result	size
risch	\(16 x^{4} \ln \relax (2)^{4}-\frac {8 \ln \relax (2)^{2} \ln \relax (3) x}{5}+\frac {100+\left (400 \ln \relax (2)^{2}+\ln \relax (3)^{2}\right ) x^{6}-20 x^{3} \ln \relax (3)}{25 x^{8}}\)	\(49\)
default	\(16 x^{4} \ln \relax (2)^{4}-\frac {8 \ln \relax (2)^{2} \ln \relax (3) x}{5}-\frac {-400 \ln \relax (2)^{2}-\ln \relax (3)^{2}}{25 x^{2}}+\frac {4}{x^{8}}-\frac {4 \ln \relax (3)}{5 x^{5}}\)	\(50\)
norman	\(\frac {4+\left (16 \ln \relax (2)^{2}+\frac {\ln \relax (3)^{2}}{25}\right ) x^{6}-\frac {4 x^{3} \ln \relax (3)}{5}+16 x^{12} \ln \relax (2)^{4}-\frac {8 \ln \relax (2)^{2} \ln \relax (3) x^{9}}{5}}{x^{8}}\)	\(51\)
gosper	\(\frac {400 x^{12} \ln \relax (2)^{4}-40 \ln \relax (2)^{2} \ln \relax (3) x^{9}+400 x^{6} \ln \relax (2)^{2}+x^{6} \ln \relax (3)^{2}-20 x^{3} \ln \relax (3)+100}{25 x^{8}}\)	\(52\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/25*(1600*x^12*ln(2)^4+4*(-10*x^9*ln(3)-200*x^6)*ln(2)^2-2*x^6*ln(3)^2+100*x^3*ln(3)-800)/x^9,x,method=_R
ETURNVERBOSE)

[Out]

16*x^4*ln(2)^4-8/5*ln(2)^2*ln(3)*x+1/25*(100+(400*ln(2)^2+ln(3)^2)*x^6-20*x^3*ln(3))/x^8

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maxima [A]  time = 0.38, size = 48, normalized size = 1.92 \begin {gather*} 16 \, x^{4} \log \relax (2)^{4} - \frac {8}{5} \, x \log \relax (3) \log \relax (2)^{2} + \frac {{\left (\log \relax (3)^{2} + 400 \, \log \relax (2)^{2}\right )} x^{6} - 20 \, x^{3} \log \relax (3) + 100}{25 \, x^{8}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/25*(1600*x^12*log(2)^4+4*(-10*x^9*log(3)-200*x^6)*log(2)^2-2*x^6*log(3)^2+100*x^3*log(3)-800)/x^9,
x, algorithm="maxima")

[Out]

16*x^4*log(2)^4 - 8/5*x*log(3)*log(2)^2 + 1/25*((log(3)^2 + 400*log(2)^2)*x^6 - 20*x^3*log(3) + 100)/x^8

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mupad [B]  time = 1.50, size = 25, normalized size = 1.00 \begin {gather*} \frac {{\left (20\,{\ln \relax (2)}^2\,x^6-\ln \relax (3)\,x^3+10\right )}^2}{25\,x^8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-((2*x^6*log(3)^2)/25 - 64*x^12*log(2)^4 + (4*log(2)^2*(10*x^9*log(3) + 200*x^6))/25 - 4*x^3*log(3) + 32)/
x^9,x)

[Out]

(20*x^6*log(2)^2 - x^3*log(3) + 10)^2/(25*x^8)

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sympy [B]  time = 0.58, size = 53, normalized size = 2.12 \begin {gather*} 16 x^{4} \log {\relax (2 )}^{4} - \frac {8 x \log {\relax (2 )}^{2} \log {\relax (3 )}}{5} + \frac {x^{6} \left (\log {\relax (3 )}^{2} + 400 \log {\relax (2 )}^{2}\right ) - 20 x^{3} \log {\relax (3 )} + 100}{25 x^{8}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/25*(1600*x**12*ln(2)**4+4*(-10*x**9*ln(3)-200*x**6)*ln(2)**2-2*x**6*ln(3)**2+100*x**3*ln(3)-800)/x
**9,x)

[Out]

16*x**4*log(2)**4 - 8*x*log(2)**2*log(3)/5 + (x**6*(log(3)**2 + 400*log(2)**2) - 20*x**3*log(3) + 100)/(25*x**
8)

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