3.29.73 \(\int \frac {15 x-30 x \log (x)+(40 x-20 x^3) \log ^2(x)}{\log ^2(x)} \, dx\)

Optimal. Leaf size=24 \[ \frac {x+5 x^3 \left (4-x^2-\frac {3}{\log (x)}\right )}{x} \]

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Rubi [A]  time = 0.17, antiderivative size = 20, normalized size of antiderivative = 0.83, number of steps used = 12, number of rules used = 7, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {6741, 12, 6688, 14, 2306, 2309, 2178} \begin {gather*} -5 x^4+20 x^2-\frac {15 x^2}{\log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(15*x - 30*x*Log[x] + (40*x - 20*x^3)*Log[x]^2)/Log[x]^2,x]

[Out]

20*x^2 - 5*x^4 - (15*x^2)/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 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 2178

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

Rule 2306

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 2309

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 6688

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

Rule 6741

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {5 x \left (3-6 \log (x)+8 \log ^2(x)-4 x^2 \log ^2(x)\right )}{\log ^2(x)} \, dx\\ &=5 \int \frac {x \left (3-6 \log (x)+8 \log ^2(x)-4 x^2 \log ^2(x)\right )}{\log ^2(x)} \, dx\\ &=5 \int x \left (8-4 x^2+\frac {3}{\log ^2(x)}-\frac {6}{\log (x)}\right ) \, dx\\ &=5 \int \left (-4 x \left (-2+x^2\right )+\frac {3 x}{\log ^2(x)}-\frac {6 x}{\log (x)}\right ) \, dx\\ &=15 \int \frac {x}{\log ^2(x)} \, dx-20 \int x \left (-2+x^2\right ) \, dx-30 \int \frac {x}{\log (x)} \, dx\\ &=-\frac {15 x^2}{\log (x)}-20 \int \left (-2 x+x^3\right ) \, dx+30 \int \frac {x}{\log (x)} \, dx-30 \operatorname {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (x)\right )\\ &=20 x^2-5 x^4-30 \text {Ei}(2 \log (x))-\frac {15 x^2}{\log (x)}+30 \operatorname {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (x)\right )\\ &=20 x^2-5 x^4-\frac {15 x^2}{\log (x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 20, normalized size = 0.83 \begin {gather*} 20 x^2-5 x^4-\frac {15 x^2}{\log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(15*x - 30*x*Log[x] + (40*x - 20*x^3)*Log[x]^2)/Log[x]^2,x]

[Out]

20*x^2 - 5*x^4 - (15*x^2)/Log[x]

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fricas [A]  time = 0.97, size = 24, normalized size = 1.00 \begin {gather*} -\frac {5 \, {\left (3 \, x^{2} + {\left (x^{4} - 4 \, x^{2}\right )} \log \relax (x)\right )}}{\log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-20*x^3+40*x)*log(x)^2-30*x*log(x)+15*x)/log(x)^2,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.25, size = 20, normalized size = 0.83 \begin {gather*} -5 \, x^{4} + 20 \, x^{2} - \frac {15 \, x^{2}}{\log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-20*x^3+40*x)*log(x)^2-30*x*log(x)+15*x)/log(x)^2,x, algorithm="giac")

[Out]

-5*x^4 + 20*x^2 - 15*x^2/log(x)

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maple [A]  time = 0.02, size = 21, normalized size = 0.88

method	result	size
default	\(-5 x^{4}+20 x^{2}-\frac {15 x^{2}}{\ln \relax (x )}\)	\(21\)
risch	\(-5 x^{4}+20 x^{2}-\frac {15 x^{2}}{\ln \relax (x )}\)	\(21\)
norman	\(\frac {-15 x^{2}+20 x^{2} \ln \relax (x )-5 x^{4} \ln \relax (x )}{\ln \relax (x )}\)	\(26\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-20*x^3+40*x)*ln(x)^2-30*x*ln(x)+15*x)/ln(x)^2,x,method=_RETURNVERBOSE)

[Out]

-5*x^4+20*x^2-15*x^2/ln(x)

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maxima [C]  time = 0.70, size = 26, normalized size = 1.08 \begin {gather*} -5 \, x^{4} + 20 \, x^{2} - 30 \, {\rm Ei}\left (2 \, \log \relax (x)\right ) + 30 \, \Gamma \left (-1, -2 \, \log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-20*x^3+40*x)*log(x)^2-30*x*log(x)+15*x)/log(x)^2,x, algorithm="maxima")

[Out]

-5*x^4 + 20*x^2 - 30*Ei(2*log(x)) + 30*gamma(-1, -2*log(x))

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mupad [B]  time = 1.68, size = 20, normalized size = 0.83 \begin {gather*} -5\,x^2\,\left (x^2-4\right )-\frac {15\,x^2}{\ln \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((15*x + log(x)^2*(40*x - 20*x^3) - 30*x*log(x))/log(x)^2,x)

[Out]

- 5*x^2*(x^2 - 4) - (15*x^2)/log(x)

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sympy [A]  time = 0.08, size = 17, normalized size = 0.71 \begin {gather*} - 5 x^{4} + 20 x^{2} - \frac {15 x^{2}}{\log {\relax (x )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-20*x**3+40*x)*ln(x)**2-30*x*ln(x)+15*x)/ln(x)**2,x)

[Out]

-5*x**4 + 20*x**2 - 15*x**2/log(x)

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