Integrand size = 78, antiderivative size = 28 \[ \int \frac {5 e^{2 x/5} x^2 \log (16)+(-60+15 x) \log (16)+e^{2 x/5} \left (-8 x^2+2 x^3\right ) \log (16) \log (4-x)+(20-5 x) \log (16) \log \left (2 x^3\right )}{-20 x^2+5 x^3} \, dx=\log (16) \left (e^{2 x/5} \log (4-x)+\frac {\log \left (2 x^3\right )}{x}\right ) \] Output:
4*(exp(1/5*x)^2*ln(4-x)+ln(2*x^3)/x)*ln(2)
Time = 0.09 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \frac {5 e^{2 x/5} x^2 \log (16)+(-60+15 x) \log (16)+e^{2 x/5} \left (-8 x^2+2 x^3\right ) \log (16) \log (4-x)+(20-5 x) \log (16) \log \left (2 x^3\right )}{-20 x^2+5 x^3} \, dx=\frac {\log (16) \left (e^{2 x/5} x \log (4-x)+\log \left (2 x^3\right )\right )}{x} \] Input:
Integrate[(5*E^((2*x)/5)*x^2*Log[16] + (-60 + 15*x)*Log[16] + E^((2*x)/5)* (-8*x^2 + 2*x^3)*Log[16]*Log[4 - x] + (20 - 5*x)*Log[16]*Log[2*x^3])/(-20* x^2 + 5*x^3),x]
Output:
(Log[16]*(E^((2*x)/5)*x*Log[4 - x] + Log[2*x^3]))/x
Time = 1.09 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.71, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.038, Rules used = {2026, 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.
\(\displaystyle \int \frac {(20-5 x) \log (16) \log \left (2 x^3\right )+5 e^{2 x/5} x^2 \log (16)+e^{2 x/5} \left (2 x^3-8 x^2\right ) \log (16) \log (4-x)+(15 x-60) \log (16)}{5 x^3-20 x^2} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {(20-5 x) \log (16) \log \left (2 x^3\right )+5 e^{2 x/5} x^2 \log (16)+e^{2 x/5} \left (2 x^3-8 x^2\right ) \log (16) \log (4-x)+(15 x-60) \log (16)}{x^2 (5 x-20)}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {e^{2 x/5} \log (16) (2 x \log (4-x)-8 \log (4-x)+5)}{5 (x-4)}-\frac {\log (16) \left (\log \left (2 x^3\right )-3\right )}{x^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\log (16) \log \left (2 x^3\right )}{x}+\frac {e^{2 x/5} \log (16) (4 \log (4-x)-x \log (4-x))}{4-x}\) |
Input:
Int[(5*E^((2*x)/5)*x^2*Log[16] + (-60 + 15*x)*Log[16] + E^((2*x)/5)*(-8*x^ 2 + 2*x^3)*Log[16]*Log[4 - x] + (20 - 5*x)*Log[16]*Log[2*x^3])/(-20*x^2 + 5*x^3),x]
Output:
(E^((2*x)/5)*Log[16]*(4*Log[4 - x] - x*Log[4 - x]))/(4 - x) + (Log[16]*Log [2*x^3])/x
Int[(Fx_.)*(Px_)^(p_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Int[x^(p *r)*ExpandToSum[Px/x^r, x]^p*Fx, x] /; IGtQ[r, 0]] /; PolyQ[Px, x] && Integ erQ[p] && !MonomialQ[Px, x] && (ILtQ[p, 0] || !PolyQ[u, x])
Time = 7.60 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.21
method | result | size |
parallelrisch | \(\frac {60 \ln \left (2\right ) {\mathrm e}^{\frac {2 x}{5}} \ln \left (-x +4\right ) x +60 \ln \left (2\right ) \ln \left (2 x^{3}\right )}{15 x}\) | \(34\) |
default | \(4 \ln \left (2\right ) {\mathrm e}^{\frac {2 x}{5}} \ln \left (-x +4\right )-\frac {12 \ln \left (2\right )}{x}-4 \ln \left (2\right ) \left (-\frac {\ln \left (x^{3}\right )}{x}-\frac {3}{x}-\frac {\ln \left (2\right )}{x}\right )\) | \(51\) |
parts | \(4 \ln \left (2\right ) {\mathrm e}^{\frac {2 x}{5}} \ln \left (-x +4\right )-\frac {12 \ln \left (2\right )}{x}-4 \ln \left (2\right ) \left (-\frac {\ln \left (x^{3}\right )}{x}-\frac {3}{x}-\frac {\ln \left (2\right )}{x}\right )\) | \(51\) |
risch | \(4 \ln \left (2\right ) {\mathrm e}^{\frac {2 x}{5}} \ln \left (-x +4\right )+\frac {2 \ln \left (2\right ) \left (-i \pi \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )+2 i \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2}-i \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i x^{3}\right )+i \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{3}\right )^{2}-i \pi \operatorname {csgn}\left (i x^{2}\right )^{3}+i \pi \,\operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i x^{3}\right )^{2}-i \pi \operatorname {csgn}\left (i x^{3}\right )^{3}+2 \ln \left (2\right )+6 \ln \left (x \right )\right )}{x}\) | \(155\) |
orering | \(\text {Expression too large to display}\) | \(2310\) |
Input:
int((4*(-5*x+20)*ln(2)*ln(2*x^3)+4*(2*x^3-8*x^2)*ln(2)*exp(1/5*x)^2*ln(-x+ 4)+20*x^2*ln(2)*exp(1/5*x)^2+4*(15*x-60)*ln(2))/(5*x^3-20*x^2),x,method=_R ETURNVERBOSE)
Output:
1/15*(60*ln(2)*exp(1/5*x)^2*ln(-x+4)*x+60*ln(2)*ln(2*x^3))/x
Time = 0.11 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.04 \[ \int \frac {5 e^{2 x/5} x^2 \log (16)+(-60+15 x) \log (16)+e^{2 x/5} \left (-8 x^2+2 x^3\right ) \log (16) \log (4-x)+(20-5 x) \log (16) \log \left (2 x^3\right )}{-20 x^2+5 x^3} \, dx=\frac {4 \, {\left (x e^{\left (\frac {2}{5} \, x\right )} \log \left (2\right ) \log \left (-x + 4\right ) + \log \left (2\right ) \log \left (2 \, x^{3}\right )\right )}}{x} \] Input:
integrate((4*(-5*x+20)*log(2)*log(2*x^3)+4*(2*x^3-8*x^2)*log(2)*exp(1/5*x) ^2*log(-x+4)+20*x^2*log(2)*exp(1/5*x)^2+4*(15*x-60)*log(2))/(5*x^3-20*x^2) ,x, algorithm="fricas")
Output:
4*(x*e^(2/5*x)*log(2)*log(-x + 4) + log(2)*log(2*x^3))/x
Time = 0.47 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.04 \[ \int \frac {5 e^{2 x/5} x^2 \log (16)+(-60+15 x) \log (16)+e^{2 x/5} \left (-8 x^2+2 x^3\right ) \log (16) \log (4-x)+(20-5 x) \log (16) \log \left (2 x^3\right )}{-20 x^2+5 x^3} \, dx=4 e^{\frac {2 x}{5}} \log {\left (2 \right )} \log {\left (4 - x \right )} + \frac {4 \log {\left (2 \right )} \log {\left (2 x^{3} \right )}}{x} \] Input:
integrate((4*(-5*x+20)*ln(2)*ln(2*x**3)+4*(2*x**3-8*x**2)*ln(2)*exp(1/5*x) **2*ln(-x+4)+20*x**2*ln(2)*exp(1/5*x)**2+4*(15*x-60)*ln(2))/(5*x**3-20*x** 2),x)
Output:
4*exp(2*x/5)*log(2)*log(4 - x) + 4*log(2)*log(2*x**3)/x
\[ \int \frac {5 e^{2 x/5} x^2 \log (16)+(-60+15 x) \log (16)+e^{2 x/5} \left (-8 x^2+2 x^3\right ) \log (16) \log (4-x)+(20-5 x) \log (16) \log \left (2 x^3\right )}{-20 x^2+5 x^3} \, dx=\int { \frac {4 \, {\left (5 \, x^{2} e^{\left (\frac {2}{5} \, x\right )} \log \left (2\right ) + 2 \, {\left (x^{3} - 4 \, x^{2}\right )} e^{\left (\frac {2}{5} \, x\right )} \log \left (2\right ) \log \left (-x + 4\right ) - 5 \, {\left (x - 4\right )} \log \left (2\right ) \log \left (2 \, x^{3}\right ) + 15 \, {\left (x - 4\right )} \log \left (2\right )\right )}}{5 \, {\left (x^{3} - 4 \, x^{2}\right )}} \,d x } \] Input:
integrate((4*(-5*x+20)*log(2)*log(2*x^3)+4*(2*x^3-8*x^2)*log(2)*exp(1/5*x) ^2*log(-x+4)+20*x^2*log(2)*exp(1/5*x)^2+4*(15*x-60)*log(2))/(5*x^3-20*x^2) ,x, algorithm="maxima")
Output:
-4*e^(8/5)*exp_integral_e(1, -2/5*x + 8/5)*log(2) - 3*(4/x + log(x - 4) - log(x))*log(2) + 3*(log(x - 4) - log(x))*log(2) - 4*integrate(e^(2/5*x)/(x - 4), x)*log(2) + 4*(x*e^(2/5*x)*log(2)*log(-x + 4) + log(2)^2 + 3*log(2) *log(x) + 3*log(2))/x
Time = 11.80 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.75 \[ \int \frac {5 e^{2 x/5} x^2 \log (16)+(-60+15 x) \log (16)+e^{2 x/5} \left (-8 x^2+2 x^3\right ) \log (16) \log (4-x)+(20-5 x) \log (16) \log \left (2 x^3\right )}{-20 x^2+5 x^3} \, dx=4 \, e^{\left (\frac {2}{5} \, x\right )} \log \left (2\right ) \log \left (-x + 4\right ) + \frac {4 \, \log \left (2\right )^{2}}{x} + \frac {4 \, \log \left (2\right ) \log \left ({\left (x - 4\right )}^{3} + 12 \, {\left (x - 4\right )}^{2} + 48 \, x - 128\right )}{x} \] Input:
integrate((4*(-5*x+20)*log(2)*log(2*x^3)+4*(2*x^3-8*x^2)*log(2)*exp(1/5*x) ^2*log(-x+4)+20*x^2*log(2)*exp(1/5*x)^2+4*(15*x-60)*log(2))/(5*x^3-20*x^2) ,x, algorithm="giac")
Output:
4*e^(2/5*x)*log(2)*log(-x + 4) + 4*log(2)^2/x + 4*log(2)*log((x - 4)^3 + 1 2*(x - 4)^2 + 48*x - 128)/x
Time = 3.08 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \frac {5 e^{2 x/5} x^2 \log (16)+(-60+15 x) \log (16)+e^{2 x/5} \left (-8 x^2+2 x^3\right ) \log (16) \log (4-x)+(20-5 x) \log (16) \log \left (2 x^3\right )}{-20 x^2+5 x^3} \, dx=\frac {4\,\ln \left (2\right )\,\ln \left (2\,x^3\right )}{x}+4\,{\mathrm {e}}^{\frac {2\,x}{5}}\,\ln \left (2\right )\,\ln \left (4-x\right ) \] Input:
int(-(4*log(2)*(15*x - 60) - 4*log(2)*log(2*x^3)*(5*x - 20) + 20*x^2*exp(( 2*x)/5)*log(2) - 4*exp((2*x)/5)*log(2)*log(4 - x)*(8*x^2 - 2*x^3))/(20*x^2 - 5*x^3),x)
Output:
(4*log(2)*log(2*x^3))/x + 4*exp((2*x)/5)*log(2)*log(4 - x)
Time = 0.18 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.96 \[ \int \frac {5 e^{2 x/5} x^2 \log (16)+(-60+15 x) \log (16)+e^{2 x/5} \left (-8 x^2+2 x^3\right ) \log (16) \log (4-x)+(20-5 x) \log (16) \log \left (2 x^3\right )}{-20 x^2+5 x^3} \, dx=\frac {4 \,\mathrm {log}\left (2\right ) \left (e^{\frac {2 x}{5}} \mathrm {log}\left (-x +4\right ) x +\mathrm {log}\left (2 x^{3}\right )\right )}{x} \] Input:
int((4*(-5*x+20)*log(2)*log(2*x^3)+4*(2*x^3-8*x^2)*log(2)*exp(1/5*x)^2*log (-x+4)+20*x^2*log(2)*exp(1/5*x)^2+4*(15*x-60)*log(2))/(5*x^3-20*x^2),x)
Output:
(4*log(2)*(e**((2*x)/5)*log( - x + 4)*x + log(2*x**3)))/x