3.25.93 \(\int \frac {-16+24 x^2-16 x^3+3 x^4+(-32+24 x^2-8 x^3) \log (25)+(-24+6 x^2) \log ^2(25)-8 \log ^3(25)-\log ^4(25)}{x^2} \, dx\)

Optimal. Leaf size=19 \[ \frac {\left (2 x-x^2+x \log (25)\right )^4}{x^5} \]

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Rubi [A]  time = 0.02, antiderivative size = 32, normalized size of antiderivative = 1.68, number of steps used = 2, number of rules used = 1, integrand size = 60, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.017, Rules used = {14} \begin {gather*} x^3-4 x^2 (2+\log (25))+6 x (2+\log (25))^2+\frac {(2+\log (25))^4}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-16 + 24*x^2 - 16*x^3 + 3*x^4 + (-32 + 24*x^2 - 8*x^3)*Log[25] + (-24 + 6*x^2)*Log[25]^2 - 8*Log[25]^3 -
Log[25]^4)/x^2,x]

[Out]

x^3 - 4*x^2*(2 + Log[25]) + 6*x*(2 + Log[25])^2 + (2 + Log[25])^4/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} &=\int \left (3 x^2-8 x (2+\log (25))+6 (2+\log (25))^2-\frac {(2+\log (25))^4}{x^2}\right ) \, dx\\ &=x^3-4 x^2 (2+\log (25))+6 x (2+\log (25))^2+\frac {(2+\log (25))^4}{x}\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.02, size = 40, normalized size = 2.11 \begin {gather*} x^3-4 x^2 (2+\log (25))+6 x (2+\log (25))^2-3 (2+\log (25))^3+\frac {(2+\log (25))^4}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-16 + 24*x^2 - 16*x^3 + 3*x^4 + (-32 + 24*x^2 - 8*x^3)*Log[25] + (-24 + 6*x^2)*Log[25]^2 - 8*Log[25
]^3 - Log[25]^4)/x^2,x]

[Out]

x^3 - 4*x^2*(2 + Log[25]) + 6*x*(2 + Log[25])^2 - 3*(2 + Log[25])^3 + (2 + Log[25])^4/x

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fricas [B]  time = 0.89, size = 56, normalized size = 2.95 \begin {gather*} \frac {x^{4} + 16 \, \log \relax (5)^{4} - 8 \, x^{3} + 24 \, {\left (x^{2} + 4\right )} \log \relax (5)^{2} + 64 \, \log \relax (5)^{3} + 24 \, x^{2} - 8 \, {\left (x^{3} - 6 \, x^{2} - 8\right )} \log \relax (5) + 16}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-16*log(5)^4-64*log(5)^3+4*(6*x^2-24)*log(5)^2+2*(-8*x^3+24*x^2-32)*log(5)+3*x^4-16*x^3+24*x^2-16)/
x^2,x, algorithm="fricas")

[Out]

(x^4 + 16*log(5)^4 - 8*x^3 + 24*(x^2 + 4)*log(5)^2 + 64*log(5)^3 + 24*x^2 - 8*(x^3 - 6*x^2 - 8)*log(5) + 16)/x

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giac [B]  time = 0.15, size = 58, normalized size = 3.05 \begin {gather*} x^{3} - 8 \, x^{2} \log \relax (5) + 24 \, x \log \relax (5)^{2} - 8 \, x^{2} + 48 \, x \log \relax (5) + 24 \, x + \frac {16 \, {\left (\log \relax (5)^{4} + 4 \, \log \relax (5)^{3} + 6 \, \log \relax (5)^{2} + 4 \, \log \relax (5) + 1\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-16*log(5)^4-64*log(5)^3+4*(6*x^2-24)*log(5)^2+2*(-8*x^3+24*x^2-32)*log(5)+3*x^4-16*x^3+24*x^2-16)/
x^2,x, algorithm="giac")

[Out]

x^3 - 8*x^2*log(5) + 24*x*log(5)^2 - 8*x^2 + 48*x*log(5) + 24*x + 16*(log(5)^4 + 4*log(5)^3 + 6*log(5)^2 + 4*l
og(5) + 1)/x

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maple [B]  time = 0.05, size = 58, normalized size = 3.05

method	result	size
norman	\(\frac {x^{4}+\left (-8 \ln \relax (5)-8\right ) x^{3}+\left (24 \ln \relax (5)^{2}+48 \ln \relax (5)+24\right ) x^{2}+16 \ln \relax (5)^{4}+64 \ln \relax (5)^{3}+96 \ln \relax (5)^{2}+64 \ln \relax (5)+16}{x}\)	\(58\)
default	\(24 x \ln \relax (5)^{2}-8 x^{2} \ln \relax (5)+x^{3}+48 x \ln \relax (5)-8 x^{2}+24 x -\frac {-16 \ln \relax (5)^{4}-64 \ln \relax (5)^{3}-96 \ln \relax (5)^{2}-64 \ln \relax (5)-16}{x}\)	\(61\)
gosper	\(\frac {16 \ln \relax (5)^{4}+24 x^{2} \ln \relax (5)^{2}-8 x^{3} \ln \relax (5)+x^{4}+64 \ln \relax (5)^{3}+48 x^{2} \ln \relax (5)-8 x^{3}+96 \ln \relax (5)^{2}+24 x^{2}+64 \ln \relax (5)+16}{x}\)	\(65\)
risch	\(24 x \ln \relax (5)^{2}-8 x^{2} \ln \relax (5)+x^{3}+48 x \ln \relax (5)-8 x^{2}+24 x +\frac {16 \ln \relax (5)^{4}}{x}+\frac {64 \ln \relax (5)^{3}}{x}+\frac {96 \ln \relax (5)^{2}}{x}+\frac {64 \ln \relax (5)}{x}+\frac {16}{x}\)	\(71\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-16*ln(5)^4-64*ln(5)^3+4*(6*x^2-24)*ln(5)^2+2*(-8*x^3+24*x^2-32)*ln(5)+3*x^4-16*x^3+24*x^2-16)/x^2,x,meth
od=_RETURNVERBOSE)

[Out]

(x^4+(-8*ln(5)-8)*x^3+(24*ln(5)^2+48*ln(5)+24)*x^2+16*ln(5)^4+64*ln(5)^3+96*ln(5)^2+64*ln(5)+16)/x

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maxima [B]  time = 0.49, size = 53, normalized size = 2.79 \begin {gather*} x^{3} - 8 \, x^{2} {\left (\log \relax (5) + 1\right )} + 24 \, {\left (\log \relax (5)^{2} + 2 \, \log \relax (5) + 1\right )} x + \frac {16 \, {\left (\log \relax (5)^{4} + 4 \, \log \relax (5)^{3} + 6 \, \log \relax (5)^{2} + 4 \, \log \relax (5) + 1\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-16*log(5)^4-64*log(5)^3+4*(6*x^2-24)*log(5)^2+2*(-8*x^3+24*x^2-32)*log(5)+3*x^4-16*x^3+24*x^2-16)/
x^2,x, algorithm="maxima")

[Out]

x^3 - 8*x^2*(log(5) + 1) + 24*(log(5)^2 + 2*log(5) + 1)*x + 16*(log(5)^4 + 4*log(5)^3 + 6*log(5)^2 + 4*log(5)
+ 1)/x

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mupad [B]  time = 0.11, size = 55, normalized size = 2.89 \begin {gather*} x\,\left (48\,\ln \relax (5)+24\,{\ln \relax (5)}^2+24\right )-x^2\,\left (8\,\ln \relax (5)+8\right )+\frac {32\,\ln \left (25\right )+24\,{\ln \left (25\right )}^2+8\,{\ln \left (25\right )}^3+{\ln \left (25\right )}^4+16}{x}+x^3 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(2*log(5)*(8*x^3 - 24*x^2 + 32) - 4*log(5)^2*(6*x^2 - 24) + 64*log(5)^3 + 16*log(5)^4 - 24*x^2 + 16*x^3 -
3*x^4 + 16)/x^2,x)

[Out]

x*(48*log(5) + 24*log(5)^2 + 24) - x^2*(8*log(5) + 8) + (32*log(25) + 24*log(25)^2 + 8*log(25)^3 + log(25)^4 +
16)/x + x^3

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sympy [B]  time = 0.16, size = 58, normalized size = 3.05 \begin {gather*} x^{3} + x^{2} \left (- 8 \log {\relax (5 )} - 8\right ) + x \left (24 + 24 \log {\relax (5 )}^{2} + 48 \log {\relax (5 )}\right ) + \frac {16 + 64 \log {\relax (5 )} + 16 \log {\relax (5 )}^{4} + 96 \log {\relax (5 )}^{2} + 64 \log {\relax (5 )}^{3}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-16*ln(5)**4-64*ln(5)**3+4*(6*x**2-24)*ln(5)**2+2*(-8*x**3+24*x**2-32)*ln(5)+3*x**4-16*x**3+24*x**2
-16)/x**2,x)

[Out]

x**3 + x**2*(-8*log(5) - 8) + x*(24 + 24*log(5)**2 + 48*log(5)) + (16 + 64*log(5) + 16*log(5)**4 + 96*log(5)**
2 + 64*log(5)**3)/x

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