3.20.15 \(\int \frac {-8 x-8 x^2-2 x^3+(4 x+6 x^2+2 x^3) \log (x)}{\log ^5(x)} \, dx\)

Optimal. Leaf size=19 \[ \log \left (e^{\frac {x^2 (2+x)^2}{2 \log ^4(x)}}\right ) \]

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Rubi [A]  time = 0.73, antiderivative size = 30, normalized size of antiderivative = 1.58, number of steps used = 41, number of rules used = 8, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {6688, 12, 6742, 2353, 2306, 2309, 2178, 2356} \begin {gather*} \frac {x^4}{2 \log ^4(x)}+\frac {2 x^3}{\log ^4(x)}+\frac {2 x^2}{\log ^4(x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-8*x - 8*x^2 - 2*x^3 + (4*x + 6*x^2 + 2*x^3)*Log[x])/Log[x]^5,x]

[Out]

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

Rule 12

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

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 2353

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 2356

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(Polyx_), x_Symbol] :> Int[ExpandIntegrand[Polyx*(a + b*Log[c*
x^n])^p, x], x] /; FreeQ[{a, b, c, n, p}, x] && PolynomialQ[Polyx, x]

Rule 6688

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

Rule 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {2 x (2+x) (-2-x+(1+x) \log (x))}{\log ^5(x)} \, dx\\ &=2 \int \frac {x (2+x) (-2-x+(1+x) \log (x))}{\log ^5(x)} \, dx\\ &=2 \int \left (-\frac {x (2+x)^2}{\log ^5(x)}+\frac {x (1+x) (2+x)}{\log ^4(x)}\right ) \, dx\\ &=-\left (2 \int \frac {x (2+x)^2}{\log ^5(x)} \, dx\right )+2 \int \frac {x (1+x) (2+x)}{\log ^4(x)} \, dx\\ &=-\left (2 \int \left (\frac {4 x}{\log ^5(x)}+\frac {4 x^2}{\log ^5(x)}+\frac {x^3}{\log ^5(x)}\right ) \, dx\right )+2 \int \left (\frac {2 x}{\log ^4(x)}+\frac {3 x^2}{\log ^4(x)}+\frac {x^3}{\log ^4(x)}\right ) \, dx\\ &=-\left (2 \int \frac {x^3}{\log ^5(x)} \, dx\right )+2 \int \frac {x^3}{\log ^4(x)} \, dx+4 \int \frac {x}{\log ^4(x)} \, dx+6 \int \frac {x^2}{\log ^4(x)} \, dx-8 \int \frac {x}{\log ^5(x)} \, dx-8 \int \frac {x^2}{\log ^5(x)} \, dx\\ &=\frac {2 x^2}{\log ^4(x)}+\frac {2 x^3}{\log ^4(x)}+\frac {x^4}{2 \log ^4(x)}-\frac {4 x^2}{3 \log ^3(x)}-\frac {2 x^3}{\log ^3(x)}-\frac {2 x^4}{3 \log ^3(x)}-2 \int \frac {x^3}{\log ^4(x)} \, dx+\frac {8}{3} \int \frac {x}{\log ^3(x)} \, dx+\frac {8}{3} \int \frac {x^3}{\log ^3(x)} \, dx-4 \int \frac {x}{\log ^4(x)} \, dx-6 \int \frac {x^2}{\log ^4(x)} \, dx+6 \int \frac {x^2}{\log ^3(x)} \, dx\\ &=\frac {2 x^2}{\log ^4(x)}+\frac {2 x^3}{\log ^4(x)}+\frac {x^4}{2 \log ^4(x)}-\frac {4 x^2}{3 \log ^2(x)}-\frac {3 x^3}{\log ^2(x)}-\frac {4 x^4}{3 \log ^2(x)}-\frac {8}{3} \int \frac {x}{\log ^3(x)} \, dx-\frac {8}{3} \int \frac {x^3}{\log ^3(x)} \, dx+\frac {8}{3} \int \frac {x}{\log ^2(x)} \, dx+\frac {16}{3} \int \frac {x^3}{\log ^2(x)} \, dx-6 \int \frac {x^2}{\log ^3(x)} \, dx+9 \int \frac {x^2}{\log ^2(x)} \, dx\\ &=\frac {2 x^2}{\log ^4(x)}+\frac {2 x^3}{\log ^4(x)}+\frac {x^4}{2 \log ^4(x)}-\frac {8 x^2}{3 \log (x)}-\frac {9 x^3}{\log (x)}-\frac {16 x^4}{3 \log (x)}-\frac {8}{3} \int \frac {x}{\log ^2(x)} \, dx-\frac {16}{3} \int \frac {x^3}{\log ^2(x)} \, dx+\frac {16}{3} \int \frac {x}{\log (x)} \, dx-9 \int \frac {x^2}{\log ^2(x)} \, dx+\frac {64}{3} \int \frac {x^3}{\log (x)} \, dx+27 \int \frac {x^2}{\log (x)} \, dx\\ &=\frac {2 x^2}{\log ^4(x)}+\frac {2 x^3}{\log ^4(x)}+\frac {x^4}{2 \log ^4(x)}-\frac {16}{3} \int \frac {x}{\log (x)} \, dx+\frac {16}{3} \operatorname {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (x)\right )-\frac {64}{3} \int \frac {x^3}{\log (x)} \, dx+\frac {64}{3} \operatorname {Subst}\left (\int \frac {e^{4 x}}{x} \, dx,x,\log (x)\right )-27 \int \frac {x^2}{\log (x)} \, dx+27 \operatorname {Subst}\left (\int \frac {e^{3 x}}{x} \, dx,x,\log (x)\right )\\ &=\frac {16}{3} \text {Ei}(2 \log (x))+27 \text {Ei}(3 \log (x))+\frac {64}{3} \text {Ei}(4 \log (x))+\frac {2 x^2}{\log ^4(x)}+\frac {2 x^3}{\log ^4(x)}+\frac {x^4}{2 \log ^4(x)}-\frac {16}{3} \operatorname {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (x)\right )-\frac {64}{3} \operatorname {Subst}\left (\int \frac {e^{4 x}}{x} \, dx,x,\log (x)\right )-27 \operatorname {Subst}\left (\int \frac {e^{3 x}}{x} \, dx,x,\log (x)\right )\\ &=\frac {2 x^2}{\log ^4(x)}+\frac {2 x^3}{\log ^4(x)}+\frac {x^4}{2 \log ^4(x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.08, size = 16, normalized size = 0.84 \begin {gather*} \frac {x^2 (2+x)^2}{2 \log ^4(x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-8*x - 8*x^2 - 2*x^3 + (4*x + 6*x^2 + 2*x^3)*Log[x])/Log[x]^5,x]

[Out]

(x^2*(2 + x)^2)/(2*Log[x]^4)

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fricas [A]  time = 0.58, size = 20, normalized size = 1.05 \begin {gather*} \frac {x^{4} + 4 \, x^{3} + 4 \, x^{2}}{2 \, \log \relax (x)^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^3+6*x^2+4*x)*log(x)-2*x^3-8*x^2-8*x)/log(x)^5,x, algorithm="fricas")

[Out]

1/2*(x^4 + 4*x^3 + 4*x^2)/log(x)^4

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giac [A]  time = 0.54, size = 28, normalized size = 1.47 \begin {gather*} \frac {x^{4}}{2 \, \log \relax (x)^{4}} + \frac {2 \, x^{3}}{\log \relax (x)^{4}} + \frac {2 \, x^{2}}{\log \relax (x)^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^3+6*x^2+4*x)*log(x)-2*x^3-8*x^2-8*x)/log(x)^5,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.02, size = 18, normalized size = 0.95

method	result	size
risch	\(\frac {x^{2} \left (x^{2}+4 x +4\right )}{2 \ln \relax (x )^{4}}\)	\(18\)
norman	\(\frac {2 x^{2}+2 x^{3}+\frac {1}{2} x^{4}}{\ln \relax (x )^{4}}\)	\(22\)
default	\(\frac {x^{4}}{2 \ln \relax (x )^{4}}+\frac {2 x^{3}}{\ln \relax (x )^{4}}+\frac {2 x^{2}}{\ln \relax (x )^{4}}\)	\(29\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*x^3+6*x^2+4*x)*ln(x)-2*x^3-8*x^2-8*x)/ln(x)^5,x,method=_RETURNVERBOSE)

[Out]

1/2*x^2*(x^2+4*x+4)/ln(x)^4

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maxima [C]  time = 0.66, size = 49, normalized size = 2.58 \begin {gather*} 32 \, \Gamma \left (-3, -2 \, \log \relax (x)\right ) + 162 \, \Gamma \left (-3, -3 \, \log \relax (x)\right ) + 128 \, \Gamma \left (-3, -4 \, \log \relax (x)\right ) + 128 \, \Gamma \left (-4, -2 \, \log \relax (x)\right ) + 648 \, \Gamma \left (-4, -3 \, \log \relax (x)\right ) + 512 \, \Gamma \left (-4, -4 \, \log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^3+6*x^2+4*x)*log(x)-2*x^3-8*x^2-8*x)/log(x)^5,x, algorithm="maxima")

[Out]

32*gamma(-3, -2*log(x)) + 162*gamma(-3, -3*log(x)) + 128*gamma(-3, -4*log(x)) + 128*gamma(-4, -2*log(x)) + 648
*gamma(-4, -3*log(x)) + 512*gamma(-4, -4*log(x))

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mupad [B]  time = 1.15, size = 14, normalized size = 0.74 \begin {gather*} \frac {x^2\,{\left (x+2\right )}^2}{2\,{\ln \relax (x)}^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(8*x + 8*x^2 + 2*x^3 - log(x)*(4*x + 6*x^2 + 2*x^3))/log(x)^5,x)

[Out]

(x^2*(x + 2)^2)/(2*log(x)^4)

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sympy [A]  time = 0.10, size = 19, normalized size = 1.00 \begin {gather*} \frac {x^{4} + 4 x^{3} + 4 x^{2}}{2 \log {\relax (x )}^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x**3+6*x**2+4*x)*ln(x)-2*x**3-8*x**2-8*x)/ln(x)**5,x)

[Out]

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

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