3.7.20 \(\int \frac {51 x^5+98 x^6+(1920 x^2+2240 x^3+240 x^5+1120 x^6) \log (5)+(-102400+12800 x^3+3200 x^6) \log ^2(5)}{9 x^5} \, dx\)

Optimal. Leaf size=30 \[ x+\left (1-x-\frac {10}{3} \left (-x+4 \left (\frac {4}{x^2}-x\right ) \log (5)\right )\right )^2 \]

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Rubi [B]  time = 0.05, antiderivative size = 62, normalized size of antiderivative = 2.07, number of steps used = 3, number of rules used = 2, integrand size = 59, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.034, Rules used = {12, 14} \begin {gather*} \frac {25600 \log ^2(5)}{9 x^4}+\frac {1}{9} x^2 (7+40 \log (5))^2-\frac {320 \log (5)}{3 x^2}+\frac {1}{3} x (17+80 \log (5))-\frac {320 \log (5) (7+40 \log (5))}{9 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(51*x^5 + 98*x^6 + (1920*x^2 + 2240*x^3 + 240*x^5 + 1120*x^6)*Log[5] + (-102400 + 12800*x^3 + 3200*x^6)*Lo
g[5]^2)/(9*x^5),x]

[Out]

(-320*Log[5])/(3*x^2) + (25600*Log[5]^2)/(9*x^4) - (320*Log[5]*(7 + 40*Log[5]))/(9*x) + (x^2*(7 + 40*Log[5])^2
)/9 + (x*(17 + 80*Log[5]))/3

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}{9} \int \frac {51 x^5+98 x^6+\left (1920 x^2+2240 x^3+240 x^5+1120 x^6\right ) \log (5)+\left (-102400+12800 x^3+3200 x^6\right ) \log ^2(5)}{x^5} \, dx\\ &=\frac {1}{9} \int \left (\frac {1920 \log (5)}{x^3}-\frac {102400 \log ^2(5)}{x^5}+\frac {320 \log (5) (7+40 \log (5))}{x^2}+2 x (7+40 \log (5))^2+3 (17+80 \log (5))\right ) \, dx\\ &=-\frac {320 \log (5)}{3 x^2}+\frac {25600 \log ^2(5)}{9 x^4}-\frac {320 \log (5) (7+40 \log (5))}{9 x}+\frac {1}{9} x^2 (7+40 \log (5))^2+\frac {1}{3} x (17+80 \log (5))\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 55, normalized size = 1.83 \begin {gather*} \frac {1}{9} \left (-\frac {960 \log (5)}{x^2}+\frac {25600 \log ^2(5)}{x^4}-\frac {320 \log (5) (7+40 \log (5))}{x}+x^2 (7+40 \log (5))^2+3 x (17+80 \log (5))\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(51*x^5 + 98*x^6 + (1920*x^2 + 2240*x^3 + 240*x^5 + 1120*x^6)*Log[5] + (-102400 + 12800*x^3 + 3200*x
^6)*Log[5]^2)/(9*x^5),x]

[Out]

((-960*Log[5])/x^2 + (25600*Log[5]^2)/x^4 - (320*Log[5]*(7 + 40*Log[5]))/x + x^2*(7 + 40*Log[5])^2 + 3*x*(17 +
80*Log[5]))/9

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fricas [B]  time = 0.87, size = 57, normalized size = 1.90 \begin {gather*} \frac {49 \, x^{6} + 51 \, x^{5} + 1600 \, {\left (x^{6} - 8 \, x^{3} + 16\right )} \log \relax (5)^{2} + 80 \, {\left (7 \, x^{6} + 3 \, x^{5} - 28 \, x^{3} - 12 \, x^{2}\right )} \log \relax (5)}{9 \, x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/9*((3200*x^6+12800*x^3-102400)*log(5)^2+(1120*x^6+240*x^5+2240*x^3+1920*x^2)*log(5)+98*x^6+51*x^5)
/x^5,x, algorithm="fricas")

[Out]

1/9*(49*x^6 + 51*x^5 + 1600*(x^6 - 8*x^3 + 16)*log(5)^2 + 80*(7*x^6 + 3*x^5 - 28*x^3 - 12*x^2)*log(5))/x^4

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giac [B]  time = 0.35, size = 65, normalized size = 2.17 \begin {gather*} \frac {1600}{9} \, x^{2} \log \relax (5)^{2} + \frac {560}{9} \, x^{2} \log \relax (5) + \frac {49}{9} \, x^{2} + \frac {80}{3} \, x \log \relax (5) + \frac {17}{3} \, x - \frac {320 \, {\left (40 \, x^{3} \log \relax (5)^{2} + 7 \, x^{3} \log \relax (5) + 3 \, x^{2} \log \relax (5) - 80 \, \log \relax (5)^{2}\right )}}{9 \, x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/9*((3200*x^6+12800*x^3-102400)*log(5)^2+(1120*x^6+240*x^5+2240*x^3+1920*x^2)*log(5)+98*x^6+51*x^5)
/x^5,x, algorithm="giac")

[Out]

1600/9*x^2*log(5)^2 + 560/9*x^2*log(5) + 49/9*x^2 + 80/3*x*log(5) + 17/3*x - 320/9*(40*x^3*log(5)^2 + 7*x^3*lo
g(5) + 3*x^2*log(5) - 80*log(5)^2)/x^4

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maple [B]  time = 0.06, size = 60, normalized size = 2.00

method	result	size
default	\(\frac {1600 x^{2} \ln \relax (5)^{2}}{9}+\frac {560 x^{2} \ln \relax (5)}{9}+\frac {80 x \ln \relax (5)}{3}+\frac {49 x^{2}}{9}+\frac {17 x}{3}-\frac {320 \ln \relax (5)}{3 x^{2}}-\frac {320 \ln \relax (5) \left (40 \ln \relax (5)+7\right )}{9 x}+\frac {25600 \ln \relax (5)^{2}}{9 x^{4}}\)	\(60\)
norman	\(\frac {\left (-\frac {12800 \ln \relax (5)^{2}}{9}-\frac {2240 \ln \relax (5)}{9}\right ) x^{3}+\left (\frac {80 \ln \relax (5)}{3}+\frac {17}{3}\right ) x^{5}+\left (\frac {1600 \ln \relax (5)^{2}}{9}+\frac {560 \ln \relax (5)}{9}+\frac {49}{9}\right ) x^{6}+\frac {25600 \ln \relax (5)^{2}}{9}-\frac {320 x^{2} \ln \relax (5)}{3}}{x^{4}}\)	\(60\)
risch	\(\frac {1600 x^{2} \ln \relax (5)^{2}}{9}+\frac {560 x^{2} \ln \relax (5)}{9}+\frac {80 x \ln \relax (5)}{3}+\frac {49 x^{2}}{9}+\frac {17 x}{3}+\frac {\left (-12800 \ln \relax (5)^{2}-2240 \ln \relax (5)\right ) x^{3}-960 x^{2} \ln \relax (5)+25600 \ln \relax (5)^{2}}{9 x^{4}}\)	\(65\)
gosper	\(\frac {1600 x^{6} \ln \relax (5)^{2}+560 x^{6} \ln \relax (5)+240 x^{5} \ln \relax (5)+49 x^{6}-12800 x^{3} \ln \relax (5)^{2}+51 x^{5}-2240 x^{3} \ln \relax (5)-960 x^{2} \ln \relax (5)+25600 \ln \relax (5)^{2}}{9 x^{4}}\)	\(69\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/9*((3200*x^6+12800*x^3-102400)*ln(5)^2+(1120*x^6+240*x^5+2240*x^3+1920*x^2)*ln(5)+98*x^6+51*x^5)/x^5,x,m
ethod=_RETURNVERBOSE)

[Out]

1600/9*x^2*ln(5)^2+560/9*x^2*ln(5)+80/3*x*ln(5)+49/9*x^2+17/3*x-320/3*ln(5)/x^2-320/9*ln(5)*(40*ln(5)+7)/x+256
00/9*ln(5)^2/x^4

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maxima [B]  time = 0.52, size = 61, normalized size = 2.03 \begin {gather*} \frac {1}{9} \, {\left (1600 \, \log \relax (5)^{2} + 560 \, \log \relax (5) + 49\right )} x^{2} + \frac {1}{3} \, x {\left (80 \, \log \relax (5) + 17\right )} - \frac {320 \, {\left ({\left (40 \, \log \relax (5)^{2} + 7 \, \log \relax (5)\right )} x^{3} + 3 \, x^{2} \log \relax (5) - 80 \, \log \relax (5)^{2}\right )}}{9 \, x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/9*((3200*x^6+12800*x^3-102400)*log(5)^2+(1120*x^6+240*x^5+2240*x^3+1920*x^2)*log(5)+98*x^6+51*x^5)
/x^5,x, algorithm="maxima")

[Out]

1/9*(1600*log(5)^2 + 560*log(5) + 49)*x^2 + 1/3*x*(80*log(5) + 17) - 320/9*((40*log(5)^2 + 7*log(5))*x^3 + 3*x
^2*log(5) - 80*log(5)^2)/x^4

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mupad [B]  time = 0.17, size = 56, normalized size = 1.87 \begin {gather*} x\,\left (\frac {80\,\ln \relax (5)}{3}+\frac {17}{3}\right )+\frac {x^2\,{\left (40\,\ln \relax (5)+7\right )}^2}{9}-\frac {\left (2240\,\ln \relax (5)+12800\,{\ln \relax (5)}^2\right )\,x^3+960\,\ln \relax (5)\,x^2-25600\,{\ln \relax (5)}^2}{9\,x^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((log(5)*(1920*x^2 + 2240*x^3 + 240*x^5 + 1120*x^6))/9 + (17*x^5)/3 + (98*x^6)/9 + (log(5)^2*(12800*x^3 +
3200*x^6 - 102400))/9)/x^5,x)

[Out]

x*((80*log(5))/3 + 17/3) + (x^2*(40*log(5) + 7)^2)/9 - (x^3*(2240*log(5) + 12800*log(5)^2) + 960*x^2*log(5) -
25600*log(5)^2)/(9*x^4)

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sympy [B]  time = 0.45, size = 65, normalized size = 2.17 \begin {gather*} \frac {x^{2} \left (49 + 560 \log {\relax (5 )} + 1600 \log {\relax (5 )}^{2}\right )}{9} + \frac {x \left (51 + 240 \log {\relax (5 )}\right )}{9} + \frac {x^{3} \left (- 12800 \log {\relax (5 )}^{2} - 2240 \log {\relax (5 )}\right ) - 960 x^{2} \log {\relax (5 )} + 25600 \log {\relax (5 )}^{2}}{9 x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/9*((3200*x**6+12800*x**3-102400)*ln(5)**2+(1120*x**6+240*x**5+2240*x**3+1920*x**2)*ln(5)+98*x**6+5
1*x**5)/x**5,x)

[Out]

x**2*(49 + 560*log(5) + 1600*log(5)**2)/9 + x*(51 + 240*log(5))/9 + (x**3*(-12800*log(5)**2 - 2240*log(5)) - 9
60*x**2*log(5) + 25600*log(5)**2)/(9*x**4)

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