Integrand size = 51, antiderivative size = 25 \[ \int \frac {e^{\frac {1}{5} (-4+2 x-\log (5))} (-180+72 x)-18 e^{\frac {1}{5} (-4+2 x-\log (5))} x \log \left (x^4\right )}{5 x} \, dx=9 e^{\frac {1}{5} (-4+2 x-\log (5))} \left (4-\log \left (x^4\right )\right ) \] Output:
9*(4-ln(x^4))*exp(-1/5*ln(5)+2/5*x-4/5)
Time = 0.02 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {e^{\frac {1}{5} (-4+2 x-\log (5))} (-180+72 x)-18 e^{\frac {1}{5} (-4+2 x-\log (5))} x \log \left (x^4\right )}{5 x} \, dx=-\frac {9 e^{\frac {2}{5} (-2+x)} \left (-4+\log \left (x^4\right )\right )}{\sqrt [5]{5}} \] Input:
Integrate[(E^((-4 + 2*x - Log[5])/5)*(-180 + 72*x) - 18*E^((-4 + 2*x - Log [5])/5)*x*Log[x^4])/(5*x),x]
Output:
(-9*E^((2*(-2 + x))/5)*(-4 + Log[x^4]))/5^(1/5)
Time = 0.27 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.88, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.078, Rules used = {27, 27, 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.
\(\displaystyle \int \frac {(72 x-180) e^{\frac {1}{5} (2 x-4-\log (5))}-18 x e^{\frac {1}{5} (2 x-4-\log (5))} \log \left (x^4\right )}{5 x} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{5} \int -\frac {18 \left (2\ 5^{4/5} e^{-\frac {2}{5} (2-x)} (5-2 x)+5^{4/5} e^{-\frac {2}{5} (2-x)} x \log \left (x^4\right )\right )}{5 x}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {18}{25} \int \frac {2\ 5^{4/5} e^{-\frac {2}{5} (2-x)} (5-2 x)+5^{4/5} e^{-\frac {2}{5} (2-x)} x \log \left (x^4\right )}{x}dx\) |
\(\Big \downarrow \) 2010 |
\(\displaystyle -\frac {18}{25} \int \left (5^{4/5} e^{\frac {2 x}{5}-\frac {4}{5}} \log \left (x^4\right )-4\ 5^{4/5} e^{\frac {2 x}{5}-\frac {4}{5}}+\frac {10\ 5^{4/5} e^{\frac {2 x}{5}-\frac {4}{5}}}{x}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {18}{25} \left (\frac {5}{2} 5^{4/5} e^{\frac {2 x}{5}-\frac {4}{5}} \log \left (x^4\right )-10\ 5^{4/5} e^{\frac {2 x}{5}-\frac {4}{5}}\right )\) |
Input:
Int[(E^((-4 + 2*x - Log[5])/5)*(-180 + 72*x) - 18*E^((-4 + 2*x - Log[5])/5 )*x*Log[x^4])/(5*x),x]
Output:
(-18*(-10*5^(4/5)*E^(-4/5 + (2*x)/5) + (5*5^(4/5)*E^(-4/5 + (2*x)/5)*Log[x ^4])/2))/25
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 = 0.19 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.20
method | result | size |
norman | \(-9 \,{\mathrm e}^{-\frac {\ln \left (5\right )}{5}+\frac {2 x}{5}-\frac {4}{5}} \ln \left (x^{4}\right )+36 \,{\mathrm e}^{-\frac {\ln \left (5\right )}{5}+\frac {2 x}{5}-\frac {4}{5}}\) | \(30\) |
parallelrisch | \(-9 \,{\mathrm e}^{-\frac {\ln \left (5\right )}{5}+\frac {2 x}{5}-\frac {4}{5}} \ln \left (x^{4}\right )+36 \,{\mathrm e}^{-\frac {\ln \left (5\right )}{5}+\frac {2 x}{5}-\frac {4}{5}}\) | \(30\) |
default | \(\frac {\left (180-45 \ln \left (x^{4}\right )+180 \ln \left (x \right )\right ) {\mathrm e}^{-\frac {\ln \left (5\right )}{5}+\frac {2 x}{5}-\frac {4}{5}}}{5}-36 \ln \left (x \right ) {\mathrm e}^{-\frac {\ln \left (5\right )}{5}+\frac {2 x}{5}-\frac {4}{5}}\) | \(40\) |
orering | \(\frac {\left (-5+4 x \right ) \left (-18 x \,{\mathrm e}^{-\frac {\ln \left (5\right )}{5}+\frac {2 x}{5}-\frac {4}{5}} \ln \left (x^{4}\right )+\left (72 x -180\right ) {\mathrm e}^{-\frac {\ln \left (5\right )}{5}+\frac {2 x}{5}-\frac {4}{5}}\right )}{2 \left (-5+2 x \right ) x}-\frac {25 x \left (\frac {-18 \,{\mathrm e}^{-\frac {\ln \left (5\right )}{5}+\frac {2 x}{5}-\frac {4}{5}} \ln \left (x^{4}\right )-\frac {36 x \,{\mathrm e}^{-\frac {\ln \left (5\right )}{5}+\frac {2 x}{5}-\frac {4}{5}} \ln \left (x^{4}\right )}{5}+\frac {2 \left (72 x -180\right ) {\mathrm e}^{-\frac {\ln \left (5\right )}{5}+\frac {2 x}{5}-\frac {4}{5}}}{5}}{5 x}-\frac {-18 x \,{\mathrm e}^{-\frac {\ln \left (5\right )}{5}+\frac {2 x}{5}-\frac {4}{5}} \ln \left (x^{4}\right )+\left (72 x -180\right ) {\mathrm e}^{-\frac {\ln \left (5\right )}{5}+\frac {2 x}{5}-\frac {4}{5}}}{5 x^{2}}\right )}{2 \left (-5+2 x \right )}\) | \(159\) |
risch | \(-\frac {36 \ln \left (x \right ) 5^{\frac {4}{5}} {\mathrm e}^{-\frac {4}{5}+\frac {2 x}{5}}}{5}+\frac {9 i 5^{\frac {4}{5}} {\mathrm e}^{-\frac {4}{5}+\frac {2 x}{5}} \left (\pi \operatorname {csgn}\left (i x^{4}\right )^{3}-\pi \operatorname {csgn}\left (i x^{4}\right )^{2} \operatorname {csgn}\left (i x \right )-\pi \operatorname {csgn}\left (i x^{4}\right )^{2} \operatorname {csgn}\left (i x^{3}\right )+\pi \,\operatorname {csgn}\left (i x^{4}\right ) \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{3}\right )+\pi \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )-2 \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2}+\pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i x^{3}\right )-\pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{3}\right )^{2}+\pi \operatorname {csgn}\left (i x^{2}\right )^{3}-\pi \,\operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i x^{3}\right )^{2}+\pi \operatorname {csgn}\left (i x^{3}\right )^{3}-8 i\right )}{10}\) | \(210\) |
Input:
int(1/5*(-18*x*exp(-1/5*ln(5)+2/5*x-4/5)*ln(x^4)+(72*x-180)*exp(-1/5*ln(5) +2/5*x-4/5))/x,x,method=_RETURNVERBOSE)
Output:
-9*exp(-1/5*ln(5)+2/5*x-4/5)*ln(x^4)+36*exp(-1/5*ln(5)+2/5*x-4/5)
Time = 0.07 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.16 \[ \int \frac {e^{\frac {1}{5} (-4+2 x-\log (5))} (-180+72 x)-18 e^{\frac {1}{5} (-4+2 x-\log (5))} x \log \left (x^4\right )}{5 x} \, dx=-9 \, e^{\left (\frac {2}{5} \, x - \frac {1}{5} \, \log \left (5\right ) - \frac {4}{5}\right )} \log \left (x^{4}\right ) + 36 \, e^{\left (\frac {2}{5} \, x - \frac {1}{5} \, \log \left (5\right ) - \frac {4}{5}\right )} \] Input:
integrate(1/5*(-18*x*exp(-1/5*log(5)+2/5*x-4/5)*log(x^4)+(72*x-180)*exp(-1 /5*log(5)+2/5*x-4/5))/x,x, algorithm="fricas")
Output:
-9*e^(2/5*x - 1/5*log(5) - 4/5)*log(x^4) + 36*e^(2/5*x - 1/5*log(5) - 4/5)
Time = 0.32 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.16 \[ \int \frac {e^{\frac {1}{5} (-4+2 x-\log (5))} (-180+72 x)-18 e^{\frac {1}{5} (-4+2 x-\log (5))} x \log \left (x^4\right )}{5 x} \, dx=\frac {\left (- 9 \cdot 5^{\frac {4}{5}} \log {\left (x^{4} \right )} + 36 \cdot 5^{\frac {4}{5}}\right ) e^{\frac {2 x}{5} - \frac {4}{5}}}{5} \] Input:
integrate(1/5*(-18*x*exp(-1/5*ln(5)+2/5*x-4/5)*ln(x**4)+(72*x-180)*exp(-1/ 5*ln(5)+2/5*x-4/5))/x,x)
Output:
(-9*5**(4/5)*log(x**4) + 36*5**(4/5))*exp(2*x/5 - 4/5)/5
Time = 0.16 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.16 \[ \int \frac {e^{\frac {1}{5} (-4+2 x-\log (5))} (-180+72 x)-18 e^{\frac {1}{5} (-4+2 x-\log (5))} x \log \left (x^4\right )}{5 x} \, dx=-9 \, e^{\left (\frac {2}{5} \, x - \frac {1}{5} \, \log \left (5\right ) - \frac {4}{5}\right )} \log \left (x^{4}\right ) + 36 \, e^{\left (\frac {2}{5} \, x - \frac {1}{5} \, \log \left (5\right ) - \frac {4}{5}\right )} \] Input:
integrate(1/5*(-18*x*exp(-1/5*log(5)+2/5*x-4/5)*log(x^4)+(72*x-180)*exp(-1 /5*log(5)+2/5*x-4/5))/x,x, algorithm="maxima")
Output:
-9*e^(2/5*x - 1/5*log(5) - 4/5)*log(x^4) + 36*e^(2/5*x - 1/5*log(5) - 4/5)
\[ \int \frac {e^{\frac {1}{5} (-4+2 x-\log (5))} (-180+72 x)-18 e^{\frac {1}{5} (-4+2 x-\log (5))} x \log \left (x^4\right )}{5 x} \, dx=\int { -\frac {18 \, {\left (x e^{\left (\frac {2}{5} \, x - \frac {1}{5} \, \log \left (5\right ) - \frac {4}{5}\right )} \log \left (x^{4}\right ) - 2 \, {\left (2 \, x - 5\right )} e^{\left (\frac {2}{5} \, x - \frac {1}{5} \, \log \left (5\right ) - \frac {4}{5}\right )}\right )}}{5 \, x} \,d x } \] Input:
integrate(1/5*(-18*x*exp(-1/5*log(5)+2/5*x-4/5)*log(x^4)+(72*x-180)*exp(-1 /5*log(5)+2/5*x-4/5))/x,x, algorithm="giac")
Output:
integrate(-18/5*(x*e^(2/5*x - 1/5*log(5) - 4/5)*log(x^4) - 2*(2*x - 5)*e^( 2/5*x - 1/5*log(5) - 4/5))/x, x)
Time = 3.91 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.68 \[ \int \frac {e^{\frac {1}{5} (-4+2 x-\log (5))} (-180+72 x)-18 e^{\frac {1}{5} (-4+2 x-\log (5))} x \log \left (x^4\right )}{5 x} \, dx=-\frac {9\,5^{4/5}\,{\mathrm {e}}^{\frac {2\,x}{5}-\frac {4}{5}}\,\left (\ln \left (x^4\right )-4\right )}{5} \] Input:
int(((exp((2*x)/5 - log(5)/5 - 4/5)*(72*x - 180))/5 - (18*x*log(x^4)*exp(( 2*x)/5 - log(5)/5 - 4/5))/5)/x,x)
Output:
-(9*5^(4/5)*exp((2*x)/5 - 4/5)*(log(x^4) - 4))/5
Time = 0.18 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84 \[ \int \frac {e^{\frac {1}{5} (-4+2 x-\log (5))} (-180+72 x)-18 e^{\frac {1}{5} (-4+2 x-\log (5))} x \log \left (x^4\right )}{5 x} \, dx=\frac {9 e^{\frac {2 x}{5}} \left (-\mathrm {log}\left (x^{4}\right )+4\right ) 5^{\frac {4}{5}}}{5 e^{\frac {4}{5}}} \] Input:
int(1/5*(-18*x*exp(-1/5*log(5)+2/5*x-4/5)*log(x^4)+(72*x-180)*exp(-1/5*log (5)+2/5*x-4/5))/x,x)
Output:
(9*e**((2*x)/5)*( - log(x**4) + 4))/(e**(4/5)*5**(1/5))