Integrand size = 54, antiderivative size = 29 \[ \int \frac {15 x-25 x \log (x)+\left (-4+9 x+2 x^2+3 x^3\right ) \log ^2(x)+(5 x-5 x \log (x)) \log \left (x^2\right )}{4 x \log ^2(x)} \, dx=x-\log (x)+\frac {1}{4} x \left (5+x+x^2-\frac {5 \left (3+\log \left (x^2\right )\right )}{\log (x)}\right ) \] Output:
x-ln(x)+1/4*(x-5*(ln(x^2)+3)/ln(x)+x^2+5)*x
Time = 0.05 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.03 \[ \int \frac {15 x-25 x \log (x)+\left (-4+9 x+2 x^2+3 x^3\right ) \log ^2(x)+(5 x-5 x \log (x)) \log \left (x^2\right )}{4 x \log ^2(x)} \, dx=\frac {1}{4} \left (x \left (9+x+x^2\right )-4 \log (x)-\frac {5 x \left (3+\log \left (x^2\right )\right )}{\log (x)}\right ) \] Input:
Integrate[(15*x - 25*x*Log[x] + (-4 + 9*x + 2*x^2 + 3*x^3)*Log[x]^2 + (5*x - 5*x*Log[x])*Log[x^2])/(4*x*Log[x]^2),x]
Output:
(x*(9 + x + x^2) - 4*Log[x] - (5*x*(3 + Log[x^2]))/Log[x])/4
Time = 0.53 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.24, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {27, 7293, 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.
\(\displaystyle \int \frac {(5 x-5 x \log (x)) \log \left (x^2\right )+\left (3 x^3+2 x^2+9 x-4\right ) \log ^2(x)+15 x-25 x \log (x)}{4 x \log ^2(x)} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{4} \int \frac {-\left (\left (-3 x^3-2 x^2-9 x+4\right ) \log ^2(x)\right )-25 x \log (x)+15 x+5 (x-x \log (x)) \log \left (x^2\right )}{x \log ^2(x)}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {1}{4} \int \left (\frac {3 \log ^2(x) x^3+2 \log ^2(x) x^2+9 \log ^2(x) x-25 \log (x) x+15 x-4 \log ^2(x)}{x \log ^2(x)}-\frac {5 (\log (x)-1) \log \left (x^2\right )}{\log ^2(x)}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{4} \left (x^3+x^2-\frac {5 x \log \left (x^2\right )}{\log (x)}+9 x-\frac {15 x}{\log (x)}-4 \log (x)\right )\) |
Input:
Int[(15*x - 25*x*Log[x] + (-4 + 9*x + 2*x^2 + 3*x^3)*Log[x]^2 + (5*x - 5*x *Log[x])*Log[x^2])/(4*x*Log[x]^2),x]
Output:
(9*x + x^2 + x^3 - (15*x)/Log[x] - 4*Log[x] - (5*x*Log[x^2])/Log[x])/4
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Time = 0.34 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.48
method | result | size |
parallelrisch | \(-\frac {-24 x^{3} \ln \left (x \right )-24 x^{2} \ln \left (x \right )+96 \ln \left (x \right )^{2}-216 x \ln \left (x \right )+120 x \ln \left (x^{2}\right )+360 x}{96 \ln \left (x \right )}\) | \(43\) |
default | \(\frac {5 \left (\ln \left (x^{2}\right )-2 \ln \left (x \right )\right ) \operatorname {expIntegral}_{1}\left (-\ln \left (x \right )\right )}{4}-\frac {x}{4}+\frac {5 \left (\ln \left (x^{2}\right )-2 \ln \left (x \right )\right ) \left (-\frac {x}{\ln \left (x \right )}-\operatorname {expIntegral}_{1}\left (-\ln \left (x \right )\right )\right )}{4}-\frac {15 x}{4 \ln \left (x \right )}+\frac {x^{3}}{4}+\frac {x^{2}}{4}-\ln \left (x \right )\) | \(70\) |
parts | \(\frac {5 \left (\ln \left (x^{2}\right )-2 \ln \left (x \right )\right ) \operatorname {expIntegral}_{1}\left (-\ln \left (x \right )\right )}{4}-\frac {x}{4}+\frac {5 \left (\ln \left (x^{2}\right )-2 \ln \left (x \right )\right ) \left (-\frac {x}{\ln \left (x \right )}-\operatorname {expIntegral}_{1}\left (-\ln \left (x \right )\right )\right )}{4}-\frac {15 x}{4 \ln \left (x \right )}+\frac {x^{3}}{4}+\frac {x^{2}}{4}-\ln \left (x \right )\) | \(70\) |
risch | \(\frac {x^{3}}{4}+\frac {x^{2}}{4}-\frac {x}{4}-\ln \left (x \right )+\frac {5 i x \left (\pi \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )-2 \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2}+\pi \operatorname {csgn}\left (i x^{2}\right )^{3}+6 i\right )}{8 \ln \left (x \right )}\) | \(74\) |
Input:
int(1/4*((-5*x*ln(x)+5*x)*ln(x^2)+(3*x^3+2*x^2+9*x-4)*ln(x)^2-25*x*ln(x)+1 5*x)/x/ln(x)^2,x,method=_RETURNVERBOSE)
Output:
-1/96*(-24*x^3*ln(x)-24*x^2*ln(x)+96*ln(x)^2-216*x*ln(x)+120*x*ln(x^2)+360 *x)/ln(x)
Time = 0.09 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00 \[ \int \frac {15 x-25 x \log (x)+\left (-4+9 x+2 x^2+3 x^3\right ) \log ^2(x)+(5 x-5 x \log (x)) \log \left (x^2\right )}{4 x \log ^2(x)} \, dx=\frac {{\left (x^{3} + x^{2} - x\right )} \log \left (x\right ) - 4 \, \log \left (x\right )^{2} - 15 \, x}{4 \, \log \left (x\right )} \] Input:
integrate(1/4*((-5*x*log(x)+5*x)*log(x^2)+(3*x^3+2*x^2+9*x-4)*log(x)^2-25* x*log(x)+15*x)/x/log(x)^2,x, algorithm="fricas")
Output:
1/4*((x^3 + x^2 - x)*log(x) - 4*log(x)^2 - 15*x)/log(x)
Time = 0.06 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.83 \[ \int \frac {15 x-25 x \log (x)+\left (-4+9 x+2 x^2+3 x^3\right ) \log ^2(x)+(5 x-5 x \log (x)) \log \left (x^2\right )}{4 x \log ^2(x)} \, dx=\frac {x^{3}}{4} + \frac {x^{2}}{4} - \frac {x}{4} - \frac {15 x}{4 \log {\left (x \right )}} - \log {\left (x \right )} \] Input:
integrate(1/4*((-5*x*ln(x)+5*x)*ln(x**2)+(3*x**3+2*x**2+9*x-4)*ln(x)**2-25 *x*ln(x)+15*x)/x/ln(x)**2,x)
Output:
x**3/4 + x**2/4 - x/4 - 15*x/(4*log(x)) - log(x)
Result contains higher order function than in optimal. Order 4 vs. order 3.
Time = 0.07 (sec) , antiderivative size = 69, normalized size of antiderivative = 2.38 \[ \int \frac {15 x-25 x \log (x)+\left (-4+9 x+2 x^2+3 x^3\right ) \log ^2(x)+(5 x-5 x \log (x)) \log \left (x^2\right )}{4 x \log ^2(x)} \, dx=\frac {1}{4} \, x^{3} + \frac {1}{4} \, x^{2} - \frac {5}{4} \, {\rm Ei}\left (\log \left (x\right )\right ) \log \left (x^{2}\right ) + \frac {5}{4} \, \Gamma \left (-1, -\log \left (x\right )\right ) \log \left (x^{2}\right ) + \frac {5}{2} \, {\rm Ei}\left (\log \left (x\right )\right ) \log \left (x\right ) - \frac {5}{2} \, \Gamma \left (-1, -\log \left (x\right )\right ) \log \left (x\right ) - \frac {1}{4} \, x - \frac {15}{4} \, {\rm Ei}\left (\log \left (x\right )\right ) + \frac {15}{4} \, \Gamma \left (-1, -\log \left (x\right )\right ) - \log \left (x\right ) \] Input:
integrate(1/4*((-5*x*log(x)+5*x)*log(x^2)+(3*x^3+2*x^2+9*x-4)*log(x)^2-25* x*log(x)+15*x)/x/log(x)^2,x, algorithm="maxima")
Output:
1/4*x^3 + 1/4*x^2 - 5/4*Ei(log(x))*log(x^2) + 5/4*gamma(-1, -log(x))*log(x ^2) + 5/2*Ei(log(x))*log(x) - 5/2*gamma(-1, -log(x))*log(x) - 1/4*x - 15/4 *Ei(log(x)) + 15/4*gamma(-1, -log(x)) - log(x)
Time = 0.13 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.86 \[ \int \frac {15 x-25 x \log (x)+\left (-4+9 x+2 x^2+3 x^3\right ) \log ^2(x)+(5 x-5 x \log (x)) \log \left (x^2\right )}{4 x \log ^2(x)} \, dx=\frac {1}{4} \, x^{3} + \frac {1}{4} \, x^{2} - \frac {1}{4} \, x - \frac {15 \, x}{4 \, \log \left (x\right )} - \log \left (x\right ) \] Input:
integrate(1/4*((-5*x*log(x)+5*x)*log(x^2)+(3*x^3+2*x^2+9*x-4)*log(x)^2-25* x*log(x)+15*x)/x/log(x)^2,x, algorithm="giac")
Output:
1/4*x^3 + 1/4*x^2 - 1/4*x - 15/4*x/log(x) - log(x)
Time = 3.70 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.21 \[ \int \frac {15 x-25 x \log (x)+\left (-4+9 x+2 x^2+3 x^3\right ) \log ^2(x)+(5 x-5 x \log (x)) \log \left (x^2\right )}{4 x \log ^2(x)} \, dx=\frac {9\,x}{4}-\ln \left (x\right )+\frac {x^2}{4}+\frac {x^3}{4}-\frac {\frac {15\,x}{4}+\frac {5\,x\,\ln \left (x^2\right )}{4}}{\ln \left (x\right )} \] Input:
int(((15*x)/4 + (log(x)^2*(9*x + 2*x^2 + 3*x^3 - 4))/4 + (log(x^2)*(5*x - 5*x*log(x)))/4 - (25*x*log(x))/4)/(x*log(x)^2),x)
Output:
(9*x)/4 - log(x) + x^2/4 + x^3/4 - ((15*x)/4 + (5*x*log(x^2))/4)/log(x)
Time = 0.18 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.38 \[ \int \frac {15 x-25 x \log (x)+\left (-4+9 x+2 x^2+3 x^3\right ) \log ^2(x)+(5 x-5 x \log (x)) \log \left (x^2\right )}{4 x \log ^2(x)} \, dx=\frac {-5 \,\mathrm {log}\left (x^{2}\right ) x -4 \mathrm {log}\left (x \right )^{2}+\mathrm {log}\left (x \right ) x^{3}+\mathrm {log}\left (x \right ) x^{2}+9 \,\mathrm {log}\left (x \right ) x -15 x}{4 \,\mathrm {log}\left (x \right )} \] Input:
int(1/4*((-5*x*log(x)+5*x)*log(x^2)+(3*x^3+2*x^2+9*x-4)*log(x)^2-25*x*log( x)+15*x)/x/log(x)^2,x)
Output:
( - 5*log(x**2)*x - 4*log(x)**2 + log(x)*x**3 + log(x)*x**2 + 9*log(x)*x - 15*x)/(4*log(x))