\(\int \frac {-x^2+x \log (4)+(3 x^2-2 x \log (4)) \log (x)}{5 \log ^2(x)} \, dx\) [7536]

Optimal result

Integrand size = 32, antiderivative size = 19 \[ \int \frac {-x^2+x \log (4)+\left (3 x^2-2 x \log (4)\right ) \log (x)}{5 \log ^2(x)} \, dx=5+\frac {x^2 (x-\log (4))}{5 \log (x)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.32, number of steps used = 17, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {12, 6874, 2395, 2343, 2346, 2209} \[ \int \frac {-x^2+x \log (4)+\left (3 x^2-2 x \log (4)\right ) \log (x)}{5 \log ^2(x)} \, dx=\frac {x^3}{5 \log (x)}-\frac {x^2 \log (4)}{5 \log (x)} \]

[In]

Int[(-x^2 + x*Log[4] + (3*x^2 - 2*x*Log[4])*Log[x])/(5*Log[x]^2),x]

[Out]

x^3/(5*Log[x]) - (x^2*Log[4])/(5*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 2395

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol]
:> With[{u = ExpandIntegrand[(a + b*Log[c*x^n])^p, (f*x)^m*(d + e*x^r)^q, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[
{a, b, c, d, e, f, m, n, p, q, r}, x] && IntegerQ[q] && (GtQ[q, 0] || (IGtQ[p, 0] && IntegerQ[m] && IntegerQ[r
]))

Rule 6874

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

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

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.32 \[ \int \frac {-x^2+x \log (4)+\left (3 x^2-2 x \log (4)\right ) \log (x)}{5 \log ^2(x)} \, dx=\frac {x^3}{5 \log (x)}-\frac {x^2 \log (4)}{5 \log (x)} \]

[In]

Integrate[(-x^2 + x*Log[4] + (3*x^2 - 2*x*Log[4])*Log[x])/(5*Log[x]^2),x]

[Out]

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

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.95

method	result	size
risch	\(-\frac {x^{2} \left (2 \ln \left (2\right )-x \right )}{5 \ln \left (x \right )}\)	\(18\)
norman	\(\frac {\frac {x^{3}}{5}-\frac {2 x^{2} \ln \left (2\right )}{5}}{\ln \left (x \right )}\)	\(19\)
parallelrisch	\(-\frac {2 x^{2} \ln \left (2\right )-x^{3}}{5 \ln \left (x \right )}\)	\(20\)
default	\(\frac {4 \ln \left (2\right ) \operatorname {Ei}_{1}\left (-2 \ln \left (x \right )\right )}{5}+\frac {2 \ln \left (2\right ) \left (-\frac {x^{2}}{\ln \left (x \right )}-2 \,\operatorname {Ei}_{1}\left (-2 \ln \left (x \right )\right )\right )}{5}+\frac {x^{3}}{5 \ln \left (x \right )}\)	\(43\)
parts	\(\frac {4 \ln \left (2\right ) \operatorname {Ei}_{1}\left (-2 \ln \left (x \right )\right )}{5}+\frac {2 \ln \left (2\right ) \left (-\frac {x^{2}}{\ln \left (x \right )}-2 \,\operatorname {Ei}_{1}\left (-2 \ln \left (x \right )\right )\right )}{5}+\frac {x^{3}}{5 \ln \left (x \right )}\)	\(43\)

[In]

int(1/5*((-4*x*ln(2)+3*x^2)*ln(x)+2*x*ln(2)-x^2)/ln(x)^2,x,method=_RETURNVERBOSE)

[Out]

-1/5*x^2*(2*ln(2)-x)/ln(x)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89 \[ \int \frac {-x^2+x \log (4)+\left (3 x^2-2 x \log (4)\right ) \log (x)}{5 \log ^2(x)} \, dx=\frac {x^{3} - 2 \, x^{2} \log \left (2\right )}{5 \, \log \left (x\right )} \]

[In]

integrate(1/5*((-4*x*log(2)+3*x^2)*log(x)+2*x*log(2)-x^2)/log(x)^2,x, algorithm="fricas")

[Out]

1/5*(x^3 - 2*x^2*log(2))/log(x)

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79 \[ \int \frac {-x^2+x \log (4)+\left (3 x^2-2 x \log (4)\right ) \log (x)}{5 \log ^2(x)} \, dx=\frac {x^{3} - 2 x^{2} \log {\left (2 \right )}}{5 \log {\left (x \right )}} \]

[In]

integrate(1/5*((-4*x*ln(2)+3*x**2)*ln(x)+2*x*ln(2)-x**2)/ln(x)**2,x)

[Out]

(x**3 - 2*x**2*log(2))/(5*log(x))

Maxima [C] (verification not implemented)

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

Time = 0.23 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.84 \[ \int \frac {-x^2+x \log (4)+\left (3 x^2-2 x \log (4)\right ) \log (x)}{5 \log ^2(x)} \, dx=-\frac {4}{5} \, {\rm Ei}\left (2 \, \log \left (x\right )\right ) \log \left (2\right ) + \frac {4}{5} \, \Gamma \left (-1, -2 \, \log \left (x\right )\right ) \log \left (2\right ) + \frac {3}{5} \, {\rm Ei}\left (3 \, \log \left (x\right )\right ) - \frac {3}{5} \, \Gamma \left (-1, -3 \, \log \left (x\right )\right ) \]

[In]

integrate(1/5*((-4*x*log(2)+3*x^2)*log(x)+2*x*log(2)-x^2)/log(x)^2,x, algorithm="maxima")

[Out]

-4/5*Ei(2*log(x))*log(2) + 4/5*gamma(-1, -2*log(x))*log(2) + 3/5*Ei(3*log(x)) - 3/5*gamma(-1, -3*log(x))

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.11 \[ \int \frac {-x^2+x \log (4)+\left (3 x^2-2 x \log (4)\right ) \log (x)}{5 \log ^2(x)} \, dx=\frac {x^{3}}{5 \, \log \left (x\right )} - \frac {2 \, x^{2} \log \left (2\right )}{5 \, \log \left (x\right )} \]

[In]

integrate(1/5*((-4*x*log(2)+3*x^2)*log(x)+2*x*log(2)-x^2)/log(x)^2,x, algorithm="giac")

[Out]

1/5*x^3/log(x) - 2/5*x^2*log(2)/log(x)

Mupad [B] (verification not implemented)

Time = 12.69 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.11 \[ \int \frac {-x^2+x \log (4)+\left (3 x^2-2 x \log (4)\right ) \log (x)}{5 \log ^2(x)} \, dx=-\frac {x^3\,\ln \left (4\right )-x^4}{5\,x\,\ln \left (x\right )} \]

[In]

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

[Out]

-(x^3*log(4) - x^4)/(5*x*log(x))