3.27.91 \(\int \frac {-32 x^4+(8 x^2+80 x^4) \log (x)-24 x^2 \log ^2(x)+(-4+e^x) \log ^3(x)}{2 \log ^3(x)} \, dx\)

Optimal. Leaf size=32 \[ \frac {1}{2} x \left (-4-\frac {-e^x+x}{x}+\left (-1+\frac {4 x^2}{\log (x)}\right )^2\right ) \]

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Rubi [A]  time = 0.40, antiderivative size = 29, normalized size of antiderivative = 0.91, number of steps used = 20, number of rules used = 7, integrand size = 47, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.149, Rules used = {12, 6742, 2194, 2306, 2309, 2178, 2353} \begin {gather*} \frac {8 x^5}{\log ^2(x)}-\frac {4 x^3}{\log (x)}-2 x+\frac {e^x}{2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-32*x^4 + (8*x^2 + 80*x^4)*Log[x] - 24*x^2*Log[x]^2 + (-4 + E^x)*Log[x]^3)/(2*Log[x]^3),x]

[Out]

E^x/2 - 2*x + (8*x^5)/Log[x]^2 - (4*x^3)/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 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 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

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 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{2} \int \frac {-32 x^4+\left (8 x^2+80 x^4\right ) \log (x)-24 x^2 \log ^2(x)+\left (-4+e^x\right ) \log ^3(x)}{\log ^3(x)} \, dx\\ &=\frac {1}{2} \int \left (e^x-\frac {4 \left (8 x^4-2 x^2 \log (x)-20 x^4 \log (x)+6 x^2 \log ^2(x)+\log ^3(x)\right )}{\log ^3(x)}\right ) \, dx\\ &=\frac {\int e^x \, dx}{2}-2 \int \frac {8 x^4-2 x^2 \log (x)-20 x^4 \log (x)+6 x^2 \log ^2(x)+\log ^3(x)}{\log ^3(x)} \, dx\\ &=\frac {e^x}{2}-2 \int \left (1+\frac {8 x^4}{\log ^3(x)}+\frac {2 x^2 \left (-1-10 x^2\right )}{\log ^2(x)}+\frac {6 x^2}{\log (x)}\right ) \, dx\\ &=\frac {e^x}{2}-2 x-4 \int \frac {x^2 \left (-1-10 x^2\right )}{\log ^2(x)} \, dx-12 \int \frac {x^2}{\log (x)} \, dx-16 \int \frac {x^4}{\log ^3(x)} \, dx\\ &=\frac {e^x}{2}-2 x+\frac {8 x^5}{\log ^2(x)}-4 \int \left (-\frac {x^2}{\log ^2(x)}-\frac {10 x^4}{\log ^2(x)}\right ) \, dx-12 \operatorname {Subst}\left (\int \frac {e^{3 x}}{x} \, dx,x,\log (x)\right )-40 \int \frac {x^4}{\log ^2(x)} \, dx\\ &=\frac {e^x}{2}-2 x-12 \text {Ei}(3 \log (x))+\frac {8 x^5}{\log ^2(x)}+\frac {40 x^5}{\log (x)}+4 \int \frac {x^2}{\log ^2(x)} \, dx+40 \int \frac {x^4}{\log ^2(x)} \, dx-200 \int \frac {x^4}{\log (x)} \, dx\\ &=\frac {e^x}{2}-2 x-12 \text {Ei}(3 \log (x))+\frac {8 x^5}{\log ^2(x)}-\frac {4 x^3}{\log (x)}+12 \int \frac {x^2}{\log (x)} \, dx+200 \int \frac {x^4}{\log (x)} \, dx-200 \operatorname {Subst}\left (\int \frac {e^{5 x}}{x} \, dx,x,\log (x)\right )\\ &=\frac {e^x}{2}-2 x-12 \text {Ei}(3 \log (x))-200 \text {Ei}(5 \log (x))+\frac {8 x^5}{\log ^2(x)}-\frac {4 x^3}{\log (x)}+12 \operatorname {Subst}\left (\int \frac {e^{3 x}}{x} \, dx,x,\log (x)\right )+200 \operatorname {Subst}\left (\int \frac {e^{5 x}}{x} \, dx,x,\log (x)\right )\\ &=\frac {e^x}{2}-2 x+\frac {8 x^5}{\log ^2(x)}-\frac {4 x^3}{\log (x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.08, size = 29, normalized size = 0.91 \begin {gather*} \frac {1}{2} \left (e^x-4 x+\frac {16 x^5}{\log ^2(x)}-\frac {8 x^3}{\log (x)}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-32*x^4 + (8*x^2 + 80*x^4)*Log[x] - 24*x^2*Log[x]^2 + (-4 + E^x)*Log[x]^3)/(2*Log[x]^3),x]

[Out]

(E^x - 4*x + (16*x^5)/Log[x]^2 - (8*x^3)/Log[x])/2

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fricas [A]  time = 0.55, size = 33, normalized size = 1.03 \begin {gather*} \frac {16 \, x^{5} - 8 \, x^{3} \log \relax (x) - {\left (4 \, x - e^{x}\right )} \log \relax (x)^{2}}{2 \, \log \relax (x)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((exp(x)-4)*log(x)^3-24*x^2*log(x)^2+(80*x^4+8*x^2)*log(x)-32*x^4)/log(x)^3,x, algorithm="fricas
")

[Out]

1/2*(16*x^5 - 8*x^3*log(x) - (4*x - e^x)*log(x)^2)/log(x)^2

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giac [A]  time = 0.20, size = 33, normalized size = 1.03 \begin {gather*} \frac {16 \, x^{5} - 8 \, x^{3} \log \relax (x) - 4 \, x \log \relax (x)^{2} + e^{x} \log \relax (x)^{2}}{2 \, \log \relax (x)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((exp(x)-4)*log(x)^3-24*x^2*log(x)^2+(80*x^4+8*x^2)*log(x)-32*x^4)/log(x)^3,x, algorithm="giac")

[Out]

1/2*(16*x^5 - 8*x^3*log(x) - 4*x*log(x)^2 + e^x*log(x)^2)/log(x)^2

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maple [A]  time = 0.02, size = 27, normalized size = 0.84

method	result	size
default	\(-2 x -\frac {4 x^{3}}{\ln \relax (x )}+\frac {8 x^{5}}{\ln \relax (x )^{2}}+\frac {{\mathrm e}^{x}}{2}\)	\(27\)
risch	\(-2 x +\frac {{\mathrm e}^{x}}{2}+\frac {4 x^{3} \left (2 x^{2}-\ln \relax (x )\right )}{\ln \relax (x )^{2}}\)	\(28\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/2*((exp(x)-4)*ln(x)^3-24*x^2*ln(x)^2+(80*x^4+8*x^2)*ln(x)-32*x^4)/ln(x)^3,x,method=_RETURNVERBOSE)

[Out]

-2*x-4*x^3/ln(x)+8*x^5/ln(x)^2+1/2*exp(x)

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maxima [C]  time = 0.45, size = 39, normalized size = 1.22 \begin {gather*} -2 \, x - 12 \, {\rm Ei}\left (3 \, \log \relax (x)\right ) + \frac {1}{2} \, e^{x} + 12 \, \Gamma \left (-1, -3 \, \log \relax (x)\right ) + 200 \, \Gamma \left (-1, -5 \, \log \relax (x)\right ) + 400 \, \Gamma \left (-2, -5 \, \log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((exp(x)-4)*log(x)^3-24*x^2*log(x)^2+(80*x^4+8*x^2)*log(x)-32*x^4)/log(x)^3,x, algorithm="maxima
")

[Out]

-2*x - 12*Ei(3*log(x)) + 1/2*e^x + 12*gamma(-1, -3*log(x)) + 200*gamma(-1, -5*log(x)) + 400*gamma(-2, -5*log(x
))

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mupad [B]  time = 1.55, size = 26, normalized size = 0.81 \begin {gather*} \frac {{\mathrm {e}}^x}{2}-2\,x-\frac {4\,x^3}{\ln \relax (x)}+\frac {8\,x^5}{{\ln \relax (x)}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((log(x)*(8*x^2 + 80*x^4))/2 - 12*x^2*log(x)^2 - 16*x^4 + (log(x)^3*(exp(x) - 4))/2)/log(x)^3,x)

[Out]

exp(x)/2 - 2*x - (4*x^3)/log(x) + (8*x^5)/log(x)^2

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sympy [A]  time = 0.27, size = 26, normalized size = 0.81 \begin {gather*} - 2 x + \frac {8 x^{5} - 4 x^{3} \log {\relax (x )}}{\log {\relax (x )}^{2}} + \frac {e^{x}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((exp(x)-4)*ln(x)**3-24*x**2*ln(x)**2+(80*x**4+8*x**2)*ln(x)-32*x**4)/ln(x)**3,x)

[Out]

-2*x + (8*x**5 - 4*x**3*log(x))/log(x)**2 + exp(x)/2

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