\(\int \frac {-3 x^4+3 x^4 \log (x)+(-20 e^5+6 x^3) \log ^3(x)}{6 x^3 \log ^3(x)} \, dx\) [2528]

Optimal result

Integrand size = 40, antiderivative size = 23 \[ \int \frac {-3 x^4+3 x^4 \log (x)+\left (-20 e^5+6 x^3\right ) \log ^3(x)}{6 x^3 \log ^3(x)} \, dx=\frac {5 e^5}{3 x^2}+x+\frac {x^2}{4 \log ^2(x)} \]

[Out]

5/3*exp(5)/x^2+x+1/4*x^2/ln(x)^2

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {12, 6820, 2343, 2346, 2209} \[ \int \frac {-3 x^4+3 x^4 \log (x)+\left (-20 e^5+6 x^3\right ) \log ^3(x)}{6 x^3 \log ^3(x)} \, dx=\frac {5 e^5}{3 x^2}+\frac {x^2}{4 \log ^2(x)}+x \]

[In]

Int[(-3*x^4 + 3*x^4*Log[x] + (-20*E^5 + 6*x^3)*Log[x]^3)/(6*x^3*Log[x]^3),x]

[Out]

(5*E^5)/(3*x^2) + x + x^2/(4*Log[x]^2)

Rule 12

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

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 2343

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[(d*x)^(m + 1)*((a + b*Log
[c*x^n])^(p + 1)/(b*d*n*(p + 1))), x] - Dist[(m + 1)/(b*n*(p + 1)), Int[(d*x)^m*(a + b*Log[c*x^n])^(p + 1), x]
, x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1] && LtQ[p, -1]

Rule 2346

Int[((a_.) + Log[(c_.)*(x_)]*(b_.))^(p_)*(x_)^(m_.), x_Symbol] :> Dist[1/c^(m + 1), Subst[Int[E^((m + 1)*x)*(a
+ b*x)^p, x], x, Log[c*x]], x] /; FreeQ[{a, b, c, p}, x] && IntegerQ[m]

Rule 6820

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{6} \int \frac {-3 x^4+3 x^4 \log (x)+\left (-20 e^5+6 x^3\right ) \log ^3(x)}{x^3 \log ^3(x)} \, dx \\ & = \frac {1}{6} \int \left (6-\frac {20 e^5}{x^3}-\frac {3 x}{\log ^3(x)}+\frac {3 x}{\log ^2(x)}\right ) \, dx \\ & = \frac {5 e^5}{3 x^2}+x-\frac {1}{2} \int \frac {x}{\log ^3(x)} \, dx+\frac {1}{2} \int \frac {x}{\log ^2(x)} \, dx \\ & = \frac {5 e^5}{3 x^2}+x+\frac {x^2}{4 \log ^2(x)}-\frac {x^2}{2 \log (x)}-\frac {1}{2} \int \frac {x}{\log ^2(x)} \, dx+\int \frac {x}{\log (x)} \, dx \\ & = \frac {5 e^5}{3 x^2}+x+\frac {x^2}{4 \log ^2(x)}-\int \frac {x}{\log (x)} \, dx+\text {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (x)\right ) \\ & = \frac {5 e^5}{3 x^2}+x+\operatorname {ExpIntegralEi}(2 \log (x))+\frac {x^2}{4 \log ^2(x)}-\text {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (x)\right ) \\ & = \frac {5 e^5}{3 x^2}+x+\frac {x^2}{4 \log ^2(x)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {-3 x^4+3 x^4 \log (x)+\left (-20 e^5+6 x^3\right ) \log ^3(x)}{6 x^3 \log ^3(x)} \, dx=\frac {5 e^5}{3 x^2}+x+\frac {x^2}{4 \log ^2(x)} \]

[In]

Integrate[(-3*x^4 + 3*x^4*Log[x] + (-20*E^5 + 6*x^3)*Log[x]^3)/(6*x^3*Log[x]^3),x]

[Out]

(5*E^5)/(3*x^2) + x + x^2/(4*Log[x]^2)

Maple [A] (verified)

Time = 0.11 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83

method	result	size
default	\(\frac {5 \,{\mathrm e}^{5}}{3 x^{2}}+x +\frac {x^{2}}{4 \ln \left (x \right )^{2}}\)	\(19\)
parts	\(\frac {5 \,{\mathrm e}^{5}}{3 x^{2}}+x +\frac {x^{2}}{4 \ln \left (x \right )^{2}}\)	\(19\)
risch	\(\frac {3 x^{3}+5 \,{\mathrm e}^{5}}{3 x^{2}}+\frac {x^{2}}{4 \ln \left (x \right )^{2}}\)	\(26\)
norman	\(\frac {x^{3} \ln \left (x \right )^{2}+\frac {x^{4}}{4}+\frac {5 \,{\mathrm e}^{5} \ln \left (x \right )^{2}}{3}}{x^{2} \ln \left (x \right )^{2}}\)	\(31\)
parallelrisch	\(\frac {12 x^{3} \ln \left (x \right )^{2}+3 x^{4}+20 \,{\mathrm e}^{5} \ln \left (x \right )^{2}}{12 \ln \left (x \right )^{2} x^{2}}\)	\(33\)

[In]

int(1/6*((-20*exp(5)+6*x^3)*ln(x)^3+3*x^4*ln(x)-3*x^4)/x^3/ln(x)^3,x,method=_RETURNVERBOSE)

[Out]

5/3*exp(5)/x^2+x+1/4*x^2/ln(x)^2

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.35 \[ \int \frac {-3 x^4+3 x^4 \log (x)+\left (-20 e^5+6 x^3\right ) \log ^3(x)}{6 x^3 \log ^3(x)} \, dx=\frac {3 \, x^{4} + 4 \, {\left (3 \, x^{3} + 5 \, e^{5}\right )} \log \left (x\right )^{2}}{12 \, x^{2} \log \left (x\right )^{2}} \]

[In]

integrate(1/6*((-20*exp(5)+6*x^3)*log(x)^3+3*x^4*log(x)-3*x^4)/x^3/log(x)^3,x, algorithm="fricas")

[Out]

1/12*(3*x^4 + 4*(3*x^3 + 5*e^5)*log(x)^2)/(x^2*log(x)^2)

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87 \[ \int \frac {-3 x^4+3 x^4 \log (x)+\left (-20 e^5+6 x^3\right ) \log ^3(x)}{6 x^3 \log ^3(x)} \, dx=\frac {x^{2}}{4 \log {\left (x \right )}^{2}} + x + \frac {5 e^{5}}{3 x^{2}} \]

[In]

integrate(1/6*((-20*exp(5)+6*x**3)*ln(x)**3+3*x**4*ln(x)-3*x**4)/x**3/ln(x)**3,x)

[Out]

x**2/(4*log(x)**2) + x + 5*exp(5)/(3*x**2)

Maxima [C] (verification not implemented)

Result contains higher order function than in optimal. Order 4 vs. order 3.

Time = 0.20 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {-3 x^4+3 x^4 \log (x)+\left (-20 e^5+6 x^3\right ) \log ^3(x)}{6 x^3 \log ^3(x)} \, dx=x + \frac {5 \, e^{5}}{3 \, x^{2}} + \Gamma \left (-1, -2 \, \log \left (x\right )\right ) + 2 \, \Gamma \left (-2, -2 \, \log \left (x\right )\right ) \]

[In]

integrate(1/6*((-20*exp(5)+6*x^3)*log(x)^3+3*x^4*log(x)-3*x^4)/x^3/log(x)^3,x, algorithm="maxima")

[Out]

x + 5/3*e^5/x^2 + gamma(-1, -2*log(x)) + 2*gamma(-2, -2*log(x))

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.39 \[ \int \frac {-3 x^4+3 x^4 \log (x)+\left (-20 e^5+6 x^3\right ) \log ^3(x)}{6 x^3 \log ^3(x)} \, dx=\frac {12 \, x^{3} \log \left (x\right )^{2} + 3 \, x^{4} + 20 \, e^{5} \log \left (x\right )^{2}}{12 \, x^{2} \log \left (x\right )^{2}} \]

[In]

integrate(1/6*((-20*exp(5)+6*x^3)*log(x)^3+3*x^4*log(x)-3*x^4)/x^3/log(x)^3,x, algorithm="giac")

[Out]

1/12*(12*x^3*log(x)^2 + 3*x^4 + 20*e^5*log(x)^2)/(x^2*log(x)^2)

Mupad [B] (verification not implemented)

Time = 8.43 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {-3 x^4+3 x^4 \log (x)+\left (-20 e^5+6 x^3\right ) \log ^3(x)}{6 x^3 \log ^3(x)} \, dx=\frac {x^3+\frac {5\,{\mathrm {e}}^5}{3}}{x^2}+\frac {x^2}{4\,{\ln \left (x\right )}^2} \]

[In]

int(-(x^4/2 - (x^4*log(x))/2 + (log(x)^3*(20*exp(5) - 6*x^3))/6)/(x^3*log(x)^3),x)

[Out]

((5*exp(5))/3 + x^3)/x^2 + x^2/(4*log(x)^2)