\(\int \frac {e^{\frac {1}{5} (-4+2 x-\log (5))} (-180+72 x)-18 e^{\frac {1}{5} (-4+2 x-\log (5))} x \log (x^4)}{5 x} \, dx\) [4938]

Optimal result

Integrand size = 51, antiderivative size = 25 \[ \int \frac {e^{\frac {1}{5} (-4+2 x-\log (5))} (-180+72 x)-18 e^{\frac {1}{5} (-4+2 x-\log (5))} x \log \left (x^4\right )}{5 x} \, dx=9 e^{\frac {1}{5} (-4+2 x-\log (5))} \left (4-\log \left (x^4\right )\right ) \]

[Out]

9*(4-ln(x^4))*exp(-1/5*ln(5)+2/5*x-4/5)

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.64, number of steps used = 8, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.098, Rules used = {12, 14, 2225, 2209, 2634} \[ \int \frac {e^{\frac {1}{5} (-4+2 x-\log (5))} (-180+72 x)-18 e^{\frac {1}{5} (-4+2 x-\log (5))} x \log \left (x^4\right )}{5 x} \, dx=\frac {36 e^{\frac {2 x}{5}-\frac {4}{5}}}{\sqrt [5]{5}}-\frac {9 e^{\frac {2 x}{5}-\frac {4}{5}} \log \left (x^4\right )}{\sqrt [5]{5}} \]

[In]

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]

[Out]

(36*E^(-4/5 + (2*x)/5))/5^(1/5) - (9*E^(-4/5 + (2*x)/5)*Log[x^4])/5^(1/5)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 14

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]]

Rule 2209

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - c*(f/d)))/d)*ExpInteg
ralEi[f*g*(c + d*x)*(Log[F]/d)], x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !TrueQ[$UseGamma]

Rule 2225

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 2634

Int[Log[u_]*(v_), x_Symbol] :> With[{w = IntHide[v, x]}, Dist[Log[u], w, x] - Int[SimplifyIntegrand[w*(D[u, x]
/u), x], x] /; InverseFunctionFreeQ[w, x]] /; InverseFunctionFreeQ[u, x]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{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 )}{x} \, dx \\ & = \frac {1}{5} \int \left (\frac {72 e^{-\frac {4}{5}+\frac {2 x}{5}}}{\sqrt [5]{5}}-\frac {36\ 5^{4/5} e^{-\frac {4}{5}+\frac {2 x}{5}}}{x}-\frac {18 e^{-\frac {4}{5}+\frac {2 x}{5}} \log \left (x^4\right )}{\sqrt [5]{5}}\right ) \, dx \\ & = -\frac {18 \int e^{-\frac {4}{5}+\frac {2 x}{5}} \log \left (x^4\right ) \, dx}{5 \sqrt [5]{5}}+\frac {72 \int e^{-\frac {4}{5}+\frac {2 x}{5}} \, dx}{5 \sqrt [5]{5}}-\frac {36 \int \frac {e^{-\frac {4}{5}+\frac {2 x}{5}}}{x} \, dx}{\sqrt [5]{5}} \\ & = \frac {36 e^{-\frac {4}{5}+\frac {2 x}{5}}}{\sqrt [5]{5}}-\frac {36 \text {Ei}\left (\frac {2 x}{5}\right )}{\sqrt [5]{5} e^{4/5}}-\frac {9 e^{-\frac {4}{5}+\frac {2 x}{5}} \log \left (x^4\right )}{\sqrt [5]{5}}+\frac {18 \int \frac {10 e^{-\frac {4}{5}+\frac {2 x}{5}}}{x} \, dx}{5 \sqrt [5]{5}} \\ & = \frac {36 e^{-\frac {4}{5}+\frac {2 x}{5}}}{\sqrt [5]{5}}-\frac {36 \text {Ei}\left (\frac {2 x}{5}\right )}{\sqrt [5]{5} e^{4/5}}-\frac {9 e^{-\frac {4}{5}+\frac {2 x}{5}} \log \left (x^4\right )}{\sqrt [5]{5}}+\frac {36 \int \frac {e^{-\frac {4}{5}+\frac {2 x}{5}}}{x} \, dx}{\sqrt [5]{5}} \\ & = \frac {36 e^{-\frac {4}{5}+\frac {2 x}{5}}}{\sqrt [5]{5}}-\frac {9 e^{-\frac {4}{5}+\frac {2 x}{5}} \log \left (x^4\right )}{\sqrt [5]{5}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {e^{\frac {1}{5} (-4+2 x-\log (5))} (-180+72 x)-18 e^{\frac {1}{5} (-4+2 x-\log (5))} x \log \left (x^4\right )}{5 x} \, dx=-\frac {9 e^{\frac {2}{5} (-2+x)} \left (-4+\log \left (x^4\right )\right )}{\sqrt [5]{5}} \]

[In]

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]

[Out]

(-9*E^((2*(-2 + x))/5)*(-4 + Log[x^4]))/5^(1/5)

Maple [A] (verified)

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	\(-\frac {36 \ln \left (x \right ) 5^{\frac {4}{5}} {\mathrm e}^{-\frac {4}{5}+\frac {2 x}{5}}}{5}+\frac {9 i 5^{\frac {4}{5}} {\mathrm e}^{-\frac {4}{5}+\frac {2 x}{5}} \left (\pi \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )-2 \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2}+\pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i x^{3}\right )-\pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{3}\right )^{2}+\pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{3}\right ) \operatorname {csgn}\left (i x^{4}\right )-\pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{4}\right )^{2}+\pi \operatorname {csgn}\left (i x^{2}\right )^{3}-\pi \,\operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i x^{3}\right )^{2}+\pi \operatorname {csgn}\left (i x^{3}\right )^{3}-\pi \,\operatorname {csgn}\left (i x^{3}\right ) \operatorname {csgn}\left (i x^{4}\right )^{2}+\pi \operatorname {csgn}\left (i x^{4}\right )^{3}-8 i\right )}{10}\)	\(210\)

[In]

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=_RETURNVERBO
SE)

[Out]

-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)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.16 \[ \int \frac {e^{\frac {1}{5} (-4+2 x-\log (5))} (-180+72 x)-18 e^{\frac {1}{5} (-4+2 x-\log (5))} x \log \left (x^4\right )}{5 x} \, dx=-9 \, e^{\left (\frac {2}{5} \, x - \frac {1}{5} \, \log \left (5\right ) - \frac {4}{5}\right )} \log \left (x^{4}\right ) + 36 \, e^{\left (\frac {2}{5} \, x - \frac {1}{5} \, \log \left (5\right ) - \frac {4}{5}\right )} \]

[In]

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")

[Out]

-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)

Sympy [A] (verification not implemented)

Time = 0.33 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.16 \[ \int \frac {e^{\frac {1}{5} (-4+2 x-\log (5))} (-180+72 x)-18 e^{\frac {1}{5} (-4+2 x-\log (5))} x \log \left (x^4\right )}{5 x} \, dx=\frac {\left (- 9 \cdot 5^{\frac {4}{5}} \log {\left (x^{4} \right )} + 36 \cdot 5^{\frac {4}{5}}\right ) e^{\frac {2 x}{5} - \frac {4}{5}}}{5} \]

[In]

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)

[Out]

(-9*5**(4/5)*log(x**4) + 36*5**(4/5))*exp(2*x/5 - 4/5)/5

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.16 \[ \int \frac {e^{\frac {1}{5} (-4+2 x-\log (5))} (-180+72 x)-18 e^{\frac {1}{5} (-4+2 x-\log (5))} x \log \left (x^4\right )}{5 x} \, dx=-9 \, e^{\left (\frac {2}{5} \, x - \frac {1}{5} \, \log \left (5\right ) - \frac {4}{5}\right )} \log \left (x^{4}\right ) + 36 \, e^{\left (\frac {2}{5} \, x - \frac {1}{5} \, \log \left (5\right ) - \frac {4}{5}\right )} \]

[In]

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")

[Out]

-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)

Giac [F]

\[ \int \frac {e^{\frac {1}{5} (-4+2 x-\log (5))} (-180+72 x)-18 e^{\frac {1}{5} (-4+2 x-\log (5))} x \log \left (x^4\right )}{5 x} \, dx=\int { -\frac {18 \, {\left (x e^{\left (\frac {2}{5} \, x - \frac {1}{5} \, \log \left (5\right ) - \frac {4}{5}\right )} \log \left (x^{4}\right ) - 2 \, {\left (2 \, x - 5\right )} e^{\left (\frac {2}{5} \, x - \frac {1}{5} \, \log \left (5\right ) - \frac {4}{5}\right )}\right )}}{5 \, x} \,d x } \]

[In]

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")

[Out]

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)

Mupad [B] (verification not implemented)

Time = 10.64 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.68 \[ \int \frac {e^{\frac {1}{5} (-4+2 x-\log (5))} (-180+72 x)-18 e^{\frac {1}{5} (-4+2 x-\log (5))} x \log \left (x^4\right )}{5 x} \, dx=-\frac {9\,5^{4/5}\,{\mathrm {e}}^{\frac {2\,x}{5}-\frac {4}{5}}\,\left (\ln \left (x^4\right )-4\right )}{5} \]

[In]

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)

[Out]

-(9*5^(4/5)*exp((2*x)/5 - 4/5)*(log(x^4) - 4))/5