Integrand size = 51, antiderivative size = 25 \[ \int \frac {e^{\frac {1}{5} (-4+2 x-\log (5))} (-180+72 x)-18 e^{\frac {1}{5} (-4+2 x-\log (5))} x \log \left (x^4\right )}{5 x} \, dx=9 e^{\frac {1}{5} (-4+2 x-\log (5))} \left (4-\log \left (x^4\right )\right ) \]
Time = 0.03 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {e^{\frac {1}{5} (-4+2 x-\log (5))} (-180+72 x)-18 e^{\frac {1}{5} (-4+2 x-\log (5))} x \log \left (x^4\right )}{5 x} \, dx=-\frac {9 e^{\frac {2}{5} (-2+x)} \left (-4+\log \left (x^4\right )\right )}{\sqrt [5]{5}} \]
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]
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 |
\(\displaystyle -\frac {18}{25} \left (\frac {5}{2} 5^{4/5} e^{\frac {2 x}{5}-\frac {4}{5}} \log \left (x^4\right )-10\ 5^{4/5} e^{\frac {2 x}{5}-\frac {4}{5}}\right )\) |
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]
3.20.38.3.1 Defintions of rubi rules used
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 = 0.09 (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 | \(\frac {\left (180-45 \ln \left (x^{4}\right )+180 \ln \left (x \right )\right ) {\mathrm e}^{-\frac {\ln \left (5\right )}{5}+\frac {2 x}{5}-\frac {4}{5}}}{5}-36 \ln \left (x \right ) {\mathrm e}^{-\frac {\ln \left (5\right )}{5}+\frac {2 x}{5}-\frac {4}{5}}\) | \(40\) |
risch | \(-\frac {36 \ln \left (x \right ) 5^{\frac {4}{5}} {\mathrm e}^{-\frac {4}{5}+\frac {2 x}{5}}}{5}+\frac {9 i 5^{\frac {4}{5}} {\mathrm e}^{-\frac {4}{5}+\frac {2 x}{5}} \left (\pi \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )-2 \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2}+\pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i x^{3}\right )-\pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{3}\right )^{2}+\pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{3}\right ) \operatorname {csgn}\left (i x^{4}\right )-\pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{4}\right )^{2}+\pi \operatorname {csgn}\left (i x^{2}\right )^{3}-\pi \,\operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i x^{3}\right )^{2}+\pi \operatorname {csgn}\left (i x^{3}\right )^{3}-\pi \,\operatorname {csgn}\left (i x^{3}\right ) \operatorname {csgn}\left (i x^{4}\right )^{2}+\pi \operatorname {csgn}\left (i x^{4}\right )^{3}-8 i\right )}{10}\) | \(210\) |
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)
Time = 0.25 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.16 \[ \int \frac {e^{\frac {1}{5} (-4+2 x-\log (5))} (-180+72 x)-18 e^{\frac {1}{5} (-4+2 x-\log (5))} x \log \left (x^4\right )}{5 x} \, dx=-9 \, e^{\left (\frac {2}{5} \, x - \frac {1}{5} \, \log \left (5\right ) - \frac {4}{5}\right )} \log \left (x^{4}\right ) + 36 \, e^{\left (\frac {2}{5} \, x - \frac {1}{5} \, \log \left (5\right ) - \frac {4}{5}\right )} \]
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=\
Time = 0.33 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.16 \[ \int \frac {e^{\frac {1}{5} (-4+2 x-\log (5))} (-180+72 x)-18 e^{\frac {1}{5} (-4+2 x-\log (5))} x \log \left (x^4\right )}{5 x} \, dx=\frac {\left (- 9 \cdot 5^{\frac {4}{5}} \log {\left (x^{4} \right )} + 36 \cdot 5^{\frac {4}{5}}\right ) e^{\frac {2 x}{5} - \frac {4}{5}}}{5} \]
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)
Time = 0.31 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.16 \[ \int \frac {e^{\frac {1}{5} (-4+2 x-\log (5))} (-180+72 x)-18 e^{\frac {1}{5} (-4+2 x-\log (5))} x \log \left (x^4\right )}{5 x} \, dx=-9 \, e^{\left (\frac {2}{5} \, x - \frac {1}{5} \, \log \left (5\right ) - \frac {4}{5}\right )} \log \left (x^{4}\right ) + 36 \, e^{\left (\frac {2}{5} \, x - \frac {1}{5} \, \log \left (5\right ) - \frac {4}{5}\right )} \]
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=\
\[ \int \frac {e^{\frac {1}{5} (-4+2 x-\log (5))} (-180+72 x)-18 e^{\frac {1}{5} (-4+2 x-\log (5))} x \log \left (x^4\right )}{5 x} \, dx=\int { -\frac {18 \, {\left (x e^{\left (\frac {2}{5} \, x - \frac {1}{5} \, \log \left (5\right ) - \frac {4}{5}\right )} \log \left (x^{4}\right ) - 2 \, {\left (2 \, x - 5\right )} e^{\left (\frac {2}{5} \, x - \frac {1}{5} \, \log \left (5\right ) - \frac {4}{5}\right )}\right )}}{5 \, x} \,d x } \]
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=\
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 = 10.64 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.68 \[ \int \frac {e^{\frac {1}{5} (-4+2 x-\log (5))} (-180+72 x)-18 e^{\frac {1}{5} (-4+2 x-\log (5))} x \log \left (x^4\right )}{5 x} \, dx=-\frac {9\,5^{4/5}\,{\mathrm {e}}^{\frac {2\,x}{5}-\frac {4}{5}}\,\left (\ln \left (x^4\right )-4\right )}{5} \]