Integrand size = 53, antiderivative size = 20 \[ \int \frac {-1875-1496 x-450 x^2-48 x^3-3 x^4+\left (500 x+300 x^2+60 x^3+4 x^4\right ) \log \left (\frac {25}{x^3}\right )}{4 x} \, dx=x+x^3+\frac {1}{4} (5+x)^4 \log \left (\frac {25}{x^3}\right ) \] Output:
1/4*(5+x)^4*ln(25/x^3)+x^3+x
Leaf count is larger than twice the leaf count of optimal. \(57\) vs. \(2(20)=40\).
Time = 0.01 (sec) , antiderivative size = 57, normalized size of antiderivative = 2.85 \[ \int \frac {-1875-1496 x-450 x^2-48 x^3-3 x^4+\left (500 x+300 x^2+60 x^3+4 x^4\right ) \log \left (\frac {25}{x^3}\right )}{4 x} \, dx=x+x^3+125 x \log \left (\frac {25}{x^3}\right )+\frac {75}{2} x^2 \log \left (\frac {25}{x^3}\right )+5 x^3 \log \left (\frac {25}{x^3}\right )+\frac {1}{4} x^4 \log \left (\frac {25}{x^3}\right )-\frac {1875 \log (x)}{4} \] Input:
Integrate[(-1875 - 1496*x - 450*x^2 - 48*x^3 - 3*x^4 + (500*x + 300*x^2 + 60*x^3 + 4*x^4)*Log[25/x^3])/(4*x),x]
Output:
x + x^3 + 125*x*Log[25/x^3] + (75*x^2*Log[25/x^3])/2 + 5*x^3*Log[25/x^3] + (x^4*Log[25/x^3])/4 - (1875*Log[x])/4
Time = 0.22 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.25, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.075, Rules used = {27, 25, 2010, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {-3 x^4-48 x^3-450 x^2+\left (4 x^4+60 x^3+300 x^2+500 x\right ) \log \left (\frac {25}{x^3}\right )-1496 x-1875}{4 x} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{4} \int -\frac {3 x^4+48 x^3+450 x^2+1496 x-4 \left (x^4+15 x^3+75 x^2+125 x\right ) \log \left (\frac {25}{x^3}\right )+1875}{x}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {1}{4} \int \frac {3 x^4+48 x^3+450 x^2+1496 x-4 \left (x^4+15 x^3+75 x^2+125 x\right ) \log \left (\frac {25}{x^3}\right )+1875}{x}dx\) |
\(\Big \downarrow \) 2010 |
\(\displaystyle -\frac {1}{4} \int \left (\frac {3 x^4+48 x^3+450 x^2+1496 x+1875}{x}-4 (x+5)^3 \log \left (\frac {25}{x^3}\right )\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{4} \left (4 x^3+(x+5)^4 \log \left (\frac {25}{x^3}\right )+4 x\right )\) |
Input:
Int[(-1875 - 1496*x - 450*x^2 - 48*x^3 - 3*x^4 + (500*x + 300*x^2 + 60*x^3 + 4*x^4)*Log[25/x^3])/(4*x),x]
Output:
(4*x + 4*x^3 + (5 + x)^4*Log[25/x^3])/4
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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]]
Time = 0.95 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95
method | result | size |
risch | \(\frac {\left (5+x \right )^{4} \ln \left (\frac {25}{x^{3}}\right )}{4}+x^{3}+x\) | \(19\) |
norman | \(x +x^{3}+125 \ln \left (\frac {25}{x^{3}}\right ) x +\frac {75 \ln \left (\frac {25}{x^{3}}\right ) x^{2}}{2}+5 \ln \left (\frac {25}{x^{3}}\right ) x^{3}+\frac {\ln \left (\frac {25}{x^{3}}\right ) x^{4}}{4}-\frac {1875 \ln \left (x \right )}{4}\) | \(52\) |
parallelrisch | \(\frac {\ln \left (\frac {25}{x^{3}}\right ) x^{4}}{4}+5 \ln \left (\frac {25}{x^{3}}\right ) x^{3}+x^{3}+\frac {75 \ln \left (\frac {25}{x^{3}}\right ) x^{2}}{2}+125 \ln \left (\frac {25}{x^{3}}\right ) x +x +\frac {625 \ln \left (\frac {25}{x^{3}}\right )}{4}\) | \(56\) |
parts | \(125 x \ln \left (\frac {1}{x^{3}}\right )+x +\frac {75 x^{2} \ln \left (\frac {1}{x^{3}}\right )}{2}+5 x^{3} \ln \left (\frac {1}{x^{3}}\right )+x^{3}+\frac {x^{4} \ln \left (\frac {1}{x^{3}}\right )}{4}-2 \ln \left (5\right ) \left (-\frac {75}{2} x^{2}-125 x -5 x^{3}-\frac {1}{4} x^{4}\right )-\frac {1875 \ln \left (x \right )}{4}\) | \(67\) |
derivativedivides | \(x +x^{3}+\frac {1875 \ln \left (\frac {1}{x}\right )}{4}-2 \ln \left (5\right ) \left (-\frac {75}{2} x^{2}-125 x -5 x^{3}-\frac {1}{4} x^{4}\right )+\frac {75 x^{2} \ln \left (\frac {1}{x^{3}}\right )}{2}+125 x \ln \left (\frac {1}{x^{3}}\right )+5 x^{3} \ln \left (\frac {1}{x^{3}}\right )+\frac {x^{4} \ln \left (\frac {1}{x^{3}}\right )}{4}\) | \(69\) |
default | \(x +x^{3}+\frac {1875 \ln \left (\frac {1}{x}\right )}{4}-2 \ln \left (5\right ) \left (-\frac {75}{2} x^{2}-125 x -5 x^{3}-\frac {1}{4} x^{4}\right )+\frac {75 x^{2} \ln \left (\frac {1}{x^{3}}\right )}{2}+125 x \ln \left (\frac {1}{x^{3}}\right )+5 x^{3} \ln \left (\frac {1}{x^{3}}\right )+\frac {x^{4} \ln \left (\frac {1}{x^{3}}\right )}{4}\) | \(69\) |
orering | \(\frac {\left (7 x^{6}+178 x^{5}+1545 x^{4}+7316 x^{3}-111785 x^{2}+6250 x -15625\right ) \left (\left (4 x^{4}+60 x^{3}+300 x^{2}+500 x \right ) \ln \left (\frac {25}{x^{3}}\right )-3 x^{4}-48 x^{3}-450 x^{2}-1496 x -1875\right )}{16 \left (4 x^{5}+79 x^{4}+460 x^{3}+1254 x^{2}-3125\right ) x}-\frac {x \left (3 x^{6}+94 x^{5}+1085 x^{4}+7212 x^{3}-99245 x^{2}-580900 x +46875\right ) \left (\frac {\left (16 x^{3}+180 x^{2}+600 x +500\right ) \ln \left (\frac {25}{x^{3}}\right )-\frac {3 \left (4 x^{4}+60 x^{3}+300 x^{2}+500 x \right )}{x}-12 x^{3}-144 x^{2}-900 x -1496}{4 x}-\frac {\left (4 x^{4}+60 x^{3}+300 x^{2}+500 x \right ) \ln \left (\frac {25}{x^{3}}\right )-3 x^{4}-48 x^{3}-450 x^{2}-1496 x -1875}{4 x^{2}}\right )}{12 \left (4 x^{5}+79 x^{4}+460 x^{3}+1254 x^{2}-3125\right )}\) | \(282\) |
Input:
int(1/4*((4*x^4+60*x^3+300*x^2+500*x)*ln(25/x^3)-3*x^4-48*x^3-450*x^2-1496 *x-1875)/x,x,method=_RETURNVERBOSE)
Output:
1/4*(5+x)^4*ln(25/x^3)+x^3+x
Time = 0.09 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.55 \[ \int \frac {-1875-1496 x-450 x^2-48 x^3-3 x^4+\left (500 x+300 x^2+60 x^3+4 x^4\right ) \log \left (\frac {25}{x^3}\right )}{4 x} \, dx=x^{3} + \frac {1}{4} \, {\left (x^{4} + 20 \, x^{3} + 150 \, x^{2} + 500 \, x + 625\right )} \log \left (\frac {25}{x^{3}}\right ) + x \] Input:
integrate(1/4*((4*x^4+60*x^3+300*x^2+500*x)*log(25/x^3)-3*x^4-48*x^3-450*x ^2-1496*x-1875)/x,x, algorithm="fricas")
Output:
x^3 + 1/4*(x^4 + 20*x^3 + 150*x^2 + 500*x + 625)*log(25/x^3) + x
Leaf count of result is larger than twice the leaf count of optimal. 37 vs. \(2 (17) = 34\).
Time = 0.08 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.85 \[ \int \frac {-1875-1496 x-450 x^2-48 x^3-3 x^4+\left (500 x+300 x^2+60 x^3+4 x^4\right ) \log \left (\frac {25}{x^3}\right )}{4 x} \, dx=x^{3} + x + \left (\frac {x^{4}}{4} + 5 x^{3} + \frac {75 x^{2}}{2} + 125 x\right ) \log {\left (\frac {25}{x^{3}} \right )} - \frac {1875 \log {\left (x \right )}}{4} \] Input:
integrate(1/4*((4*x**4+60*x**3+300*x**2+500*x)*ln(25/x**3)-3*x**4-48*x**3- 450*x**2-1496*x-1875)/x,x)
Output:
x**3 + x + (x**4/4 + 5*x**3 + 75*x**2/2 + 125*x)*log(25/x**3) - 1875*log(x )/4
Leaf count of result is larger than twice the leaf count of optimal. 51 vs. \(2 (18) = 36\).
Time = 0.03 (sec) , antiderivative size = 51, normalized size of antiderivative = 2.55 \[ \int \frac {-1875-1496 x-450 x^2-48 x^3-3 x^4+\left (500 x+300 x^2+60 x^3+4 x^4\right ) \log \left (\frac {25}{x^3}\right )}{4 x} \, dx=\frac {1}{4} \, x^{4} \log \left (\frac {25}{x^{3}}\right ) + 5 \, x^{3} \log \left (\frac {25}{x^{3}}\right ) + x^{3} + \frac {75}{2} \, x^{2} \log \left (\frac {25}{x^{3}}\right ) + 125 \, x \log \left (\frac {25}{x^{3}}\right ) + x - \frac {1875}{4} \, \log \left (x\right ) \] Input:
integrate(1/4*((4*x^4+60*x^3+300*x^2+500*x)*log(25/x^3)-3*x^4-48*x^3-450*x ^2-1496*x-1875)/x,x, algorithm="maxima")
Output:
1/4*x^4*log(25/x^3) + 5*x^3*log(25/x^3) + x^3 + 75/2*x^2*log(25/x^3) + 125 *x*log(25/x^3) + x - 1875/4*log(x)
Time = 0.11 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.70 \[ \int \frac {-1875-1496 x-450 x^2-48 x^3-3 x^4+\left (500 x+300 x^2+60 x^3+4 x^4\right ) \log \left (\frac {25}{x^3}\right )}{4 x} \, dx=x^{3} + \frac {1}{4} \, {\left (x^{4} + 20 \, x^{3} + 150 \, x^{2} + 500 \, x\right )} \log \left (\frac {25}{x^{3}}\right ) + x - \frac {1875}{4} \, \log \left (x\right ) \] Input:
integrate(1/4*((4*x^4+60*x^3+300*x^2+500*x)*log(25/x^3)-3*x^4-48*x^3-450*x ^2-1496*x-1875)/x,x, algorithm="giac")
Output:
x^3 + 1/4*(x^4 + 20*x^3 + 150*x^2 + 500*x)*log(25/x^3) + x - 1875/4*log(x)
Time = 2.96 (sec) , antiderivative size = 55, normalized size of antiderivative = 2.75 \[ \int \frac {-1875-1496 x-450 x^2-48 x^3-3 x^4+\left (500 x+300 x^2+60 x^3+4 x^4\right ) \log \left (\frac {25}{x^3}\right )}{4 x} \, dx=\frac {625\,\ln \left (\frac {1}{x^3}\right )}{4}+x\,\left (125\,\ln \left (\frac {25}{x^3}\right )+1\right )+x^3\,\left (5\,\ln \left (\frac {25}{x^3}\right )+1\right )+\frac {75\,x^2\,\ln \left (\frac {25}{x^3}\right )}{2}+\frac {x^4\,\ln \left (\frac {25}{x^3}\right )}{4} \] Input:
int(-(374*x + (225*x^2)/2 + 12*x^3 + (3*x^4)/4 - (log(25/x^3)*(500*x + 300 *x^2 + 60*x^3 + 4*x^4))/4 + 1875/4)/x,x)
Output:
(625*log(1/x^3))/4 + x*(125*log(25/x^3) + 1) + x^3*(5*log(25/x^3) + 1) + ( 75*x^2*log(25/x^3))/2 + (x^4*log(25/x^3))/4
Time = 0.19 (sec) , antiderivative size = 51, normalized size of antiderivative = 2.55 \[ \int \frac {-1875-1496 x-450 x^2-48 x^3-3 x^4+\left (500 x+300 x^2+60 x^3+4 x^4\right ) \log \left (\frac {25}{x^3}\right )}{4 x} \, dx=\frac {\mathrm {log}\left (\frac {25}{x^{3}}\right ) x^{4}}{4}+5 \,\mathrm {log}\left (\frac {25}{x^{3}}\right ) x^{3}+\frac {75 \,\mathrm {log}\left (\frac {25}{x^{3}}\right ) x^{2}}{2}+125 \,\mathrm {log}\left (\frac {25}{x^{3}}\right ) x -\frac {1875 \,\mathrm {log}\left (x \right )}{4}+x^{3}+x \] Input:
int(1/4*((4*x^4+60*x^3+300*x^2+500*x)*log(25/x^3)-3*x^4-48*x^3-450*x^2-149 6*x-1875)/x,x)
Output:
(log(25/x**3)*x**4 + 20*log(25/x**3)*x**3 + 150*log(25/x**3)*x**2 + 500*lo g(25/x**3)*x - 1875*log(x) + 4*x**3 + 4*x)/4