\(\int \frac {-16+24 x^2-16 x^3+3 x^4+(-32+24 x^2-8 x^3) \log (25)+(-24+6 x^2) \log ^2(25)-8 \log ^3(25)-\log ^4(25)}{x^2} \, dx\) [2493]

   Optimal result
   Rubi [A] (verified)
   Mathematica [B] (verified)
   Maple [B] (verified)
   Fricas [B] (verification not implemented)
   Sympy [B] (verification not implemented)
   Maxima [B] (verification not implemented)
   Giac [B] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 60, antiderivative size = 19 \[ \int \frac {-16+24 x^2-16 x^3+3 x^4+\left (-32+24 x^2-8 x^3\right ) \log (25)+\left (-24+6 x^2\right ) \log ^2(25)-8 \log ^3(25)-\log ^4(25)}{x^2} \, dx=\frac {\left (2 x-x^2+x \log (25)\right )^4}{x^5} \]

[Out]

(2*x*ln(5)+2*x-x^2)^4/x^5

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.68, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.017, Rules used = {14} \[ \int \frac {-16+24 x^2-16 x^3+3 x^4+\left (-32+24 x^2-8 x^3\right ) \log (25)+\left (-24+6 x^2\right ) \log ^2(25)-8 \log ^3(25)-\log ^4(25)}{x^2} \, dx=x^3-4 x^2 (2+\log (25))+6 x (2+\log (25))^2+\frac {(2+\log (25))^4}{x} \]

[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{align*} \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{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(40\) vs. \(2(19)=38\).

Time = 0.02 (sec) , antiderivative size = 40, normalized size of antiderivative = 2.11 \[ \int \frac {-16+24 x^2-16 x^3+3 x^4+\left (-32+24 x^2-8 x^3\right ) \log (25)+\left (-24+6 x^2\right ) \log ^2(25)-8 \log ^3(25)-\log ^4(25)}{x^2} \, dx=x^3-4 x^2 (2+\log (25))+6 x (2+\log (25))^2-3 (2+\log (25))^3+\frac {(2+\log (25))^4}{x} \]

[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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(57\) vs. \(2(20)=40\).

Time = 0.09 (sec) , antiderivative size = 58, normalized size of antiderivative = 3.05

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

[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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 56 vs. \(2 (18) = 36\).

Time = 0.23 (sec) , antiderivative size = 56, normalized size of antiderivative = 2.95 \[ \int \frac {-16+24 x^2-16 x^3+3 x^4+\left (-32+24 x^2-8 x^3\right ) \log (25)+\left (-24+6 x^2\right ) \log ^2(25)-8 \log ^3(25)-\log ^4(25)}{x^2} \, dx=\frac {x^{4} + 16 \, \log \left (5\right )^{4} - 8 \, x^{3} + 24 \, {\left (x^{2} + 4\right )} \log \left (5\right )^{2} + 64 \, \log \left (5\right )^{3} + 24 \, x^{2} - 8 \, {\left (x^{3} - 6 \, x^{2} - 8\right )} \log \left (5\right ) + 16}{x} \]

[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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 58 vs. \(2 (17) = 34\).

Time = 0.09 (sec) , antiderivative size = 58, normalized size of antiderivative = 3.05 \[ \int \frac {-16+24 x^2-16 x^3+3 x^4+\left (-32+24 x^2-8 x^3\right ) \log (25)+\left (-24+6 x^2\right ) \log ^2(25)-8 \log ^3(25)-\log ^4(25)}{x^2} \, dx=x^{3} + x^{2} \left (- 8 \log {\left (5 \right )} - 8\right ) + x \left (24 + 24 \log {\left (5 \right )}^{2} + 48 \log {\left (5 \right )}\right ) + \frac {16 + 64 \log {\left (5 \right )} + 16 \log {\left (5 \right )}^{4} + 96 \log {\left (5 \right )}^{2} + 64 \log {\left (5 \right )}^{3}}{x} \]

[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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 53 vs. \(2 (18) = 36\).

Time = 0.19 (sec) , antiderivative size = 53, normalized size of antiderivative = 2.79 \[ \int \frac {-16+24 x^2-16 x^3+3 x^4+\left (-32+24 x^2-8 x^3\right ) \log (25)+\left (-24+6 x^2\right ) \log ^2(25)-8 \log ^3(25)-\log ^4(25)}{x^2} \, dx=x^{3} - 8 \, x^{2} {\left (\log \left (5\right ) + 1\right )} + 24 \, {\left (\log \left (5\right )^{2} + 2 \, \log \left (5\right ) + 1\right )} x + \frac {16 \, {\left (\log \left (5\right )^{4} + 4 \, \log \left (5\right )^{3} + 6 \, \log \left (5\right )^{2} + 4 \, \log \left (5\right ) + 1\right )}}{x} \]

[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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 58 vs. \(2 (18) = 36\).

Time = 0.26 (sec) , antiderivative size = 58, normalized size of antiderivative = 3.05 \[ \int \frac {-16+24 x^2-16 x^3+3 x^4+\left (-32+24 x^2-8 x^3\right ) \log (25)+\left (-24+6 x^2\right ) \log ^2(25)-8 \log ^3(25)-\log ^4(25)}{x^2} \, dx=x^{3} - 8 \, x^{2} \log \left (5\right ) + 24 \, x \log \left (5\right )^{2} - 8 \, x^{2} + 48 \, x \log \left (5\right ) + 24 \, x + \frac {16 \, {\left (\log \left (5\right )^{4} + 4 \, \log \left (5\right )^{3} + 6 \, \log \left (5\right )^{2} + 4 \, \log \left (5\right ) + 1\right )}}{x} \]

[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

Mupad [B] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 55, normalized size of antiderivative = 2.89 \[ \int \frac {-16+24 x^2-16 x^3+3 x^4+\left (-32+24 x^2-8 x^3\right ) \log (25)+\left (-24+6 x^2\right ) \log ^2(25)-8 \log ^3(25)-\log ^4(25)}{x^2} \, dx=x\,\left (48\,\ln \left (5\right )+24\,{\ln \left (5\right )}^2+24\right )-x^2\,\left (8\,\ln \left (5\right )+8\right )+\frac {32\,\ln \left (25\right )+24\,{\ln \left (25\right )}^2+8\,{\ln \left (25\right )}^3+{\ln \left (25\right )}^4+16}{x}+x^3 \]

[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