\(\int \frac {-3 x^3+12 x^3 \log (x)+220 \log ^2(x)}{4 \log ^2(x)} \, dx\) [572]

Optimal result

Integrand size = 27, antiderivative size = 19 \[ \int \frac {-3 x^3+12 x^3 \log (x)+220 \log ^2(x)}{4 \log ^2(x)} \, dx=x+3 \left (18 x+\frac {x^4}{4 \log (x)}\right ) \]

[Out]

55*x+3/4*x^4/ln(x)

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79, number of steps used = 8, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {12, 6874, 2343, 2346, 2209} \[ \int \frac {-3 x^3+12 x^3 \log (x)+220 \log ^2(x)}{4 \log ^2(x)} \, dx=\frac {3 x^4}{4 \log (x)}+55 x \]

[In]

Int[(-3*x^3 + 12*x^3*Log[x] + 220*Log[x]^2)/(4*Log[x]^2),x]

[Out]

55*x + (3*x^4)/(4*Log[x])

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 6874

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

Rubi steps \begin{align*} \text {integral}& = \frac {1}{4} \int \frac {-3 x^3+12 x^3 \log (x)+220 \log ^2(x)}{\log ^2(x)} \, dx \\ & = \frac {1}{4} \int \left (220-\frac {3 x^3}{\log ^2(x)}+\frac {12 x^3}{\log (x)}\right ) \, dx \\ & = 55 x-\frac {3}{4} \int \frac {x^3}{\log ^2(x)} \, dx+3 \int \frac {x^3}{\log (x)} \, dx \\ & = 55 x+\frac {3 x^4}{4 \log (x)}-3 \int \frac {x^3}{\log (x)} \, dx+3 \text {Subst}\left (\int \frac {e^{4 x}}{x} \, dx,x,\log (x)\right ) \\ & = 55 x+3 \operatorname {ExpIntegralEi}(4 \log (x))+\frac {3 x^4}{4 \log (x)}-3 \text {Subst}\left (\int \frac {e^{4 x}}{x} \, dx,x,\log (x)\right ) \\ & = 55 x+\frac {3 x^4}{4 \log (x)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79 \[ \int \frac {-3 x^3+12 x^3 \log (x)+220 \log ^2(x)}{4 \log ^2(x)} \, dx=55 x+\frac {3 x^4}{4 \log (x)} \]

[In]

Integrate[(-3*x^3 + 12*x^3*Log[x] + 220*Log[x]^2)/(4*Log[x]^2),x]

[Out]

55*x + (3*x^4)/(4*Log[x])

Maple [A] (verified)

Time = 0.02 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.74

method	result	size
default	\(55 x +\frac {3 x^{4}}{4 \ln \left (x \right )}\)	\(14\)
risch	\(55 x +\frac {3 x^{4}}{4 \ln \left (x \right )}\)	\(14\)
parts	\(55 x +\frac {3 x^{4}}{4 \ln \left (x \right )}\)	\(14\)
norman	\(\frac {\frac {3 x^{4}}{4}+55 x \ln \left (x \right )}{\ln \left (x \right )}\)	\(17\)
parallelrisch	\(\frac {3 x^{4}+220 x \ln \left (x \right )}{4 \ln \left (x \right )}\)	\(18\)

[In]

int(1/4*(220*ln(x)^2+12*x^3*ln(x)-3*x^3)/ln(x)^2,x,method=_RETURNVERBOSE)

[Out]

55*x+3/4*x^4/ln(x)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89 \[ \int \frac {-3 x^3+12 x^3 \log (x)+220 \log ^2(x)}{4 \log ^2(x)} \, dx=\frac {3 \, x^{4} + 220 \, x \log \left (x\right )}{4 \, \log \left (x\right )} \]

[In]

integrate(1/4*(220*log(x)^2+12*x^3*log(x)-3*x^3)/log(x)^2,x, algorithm="fricas")

[Out]

1/4*(3*x^4 + 220*x*log(x))/log(x)

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.63 \[ \int \frac {-3 x^3+12 x^3 \log (x)+220 \log ^2(x)}{4 \log ^2(x)} \, dx=\frac {3 x^{4}}{4 \log {\left (x \right )}} + 55 x \]

[In]

integrate(1/4*(220*ln(x)**2+12*x**3*ln(x)-3*x**3)/ln(x)**2,x)

[Out]

3*x**4/(4*log(x)) + 55*x

Maxima [C] (verification not implemented)

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

Time = 0.26 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \frac {-3 x^3+12 x^3 \log (x)+220 \log ^2(x)}{4 \log ^2(x)} \, dx=55 \, x + 3 \, {\rm Ei}\left (4 \, \log \left (x\right )\right ) - 3 \, \Gamma \left (-1, -4 \, \log \left (x\right )\right ) \]

[In]

integrate(1/4*(220*log(x)^2+12*x^3*log(x)-3*x^3)/log(x)^2,x, algorithm="maxima")

[Out]

55*x + 3*Ei(4*log(x)) - 3*gamma(-1, -4*log(x))

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.68 \[ \int \frac {-3 x^3+12 x^3 \log (x)+220 \log ^2(x)}{4 \log ^2(x)} \, dx=\frac {3 \, x^{4}}{4 \, \log \left (x\right )} + 55 \, x \]

[In]

integrate(1/4*(220*log(x)^2+12*x^3*log(x)-3*x^3)/log(x)^2,x, algorithm="giac")

[Out]

3/4*x^4/log(x) + 55*x

Mupad [B] (verification not implemented)

Time = 8.17 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.68 \[ \int \frac {-3 x^3+12 x^3 \log (x)+220 \log ^2(x)}{4 \log ^2(x)} \, dx=55\,x+\frac {3\,x^4}{4\,\ln \left (x\right )} \]

[In]

int((3*x^3*log(x) + 55*log(x)^2 - (3*x^3)/4)/log(x)^2,x)

[Out]

55*x + (3*x^4)/(4*log(x))