3.101.15 \(\int \frac {-2 x^6+(-8 x^3+2 e^{-4+x} x^4+4 x^6) \log (x)+(8 x^3+e^{-4+x} (-4 x^4-2 x^5)) \log ^2(x)+(-32+2 e^{-8+2 x} x^3+e^{-4+x} (8 x-8 x^2)) \log ^3(x)}{x^3 \log ^3(x)} \, dx\)

Optimal. Leaf size=25 \[ -1+\left (e^{-4+x}-\frac {4+\frac {x^3}{\log (x)}}{x}\right )^2 \]

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Rubi [B]  time = 1.73, antiderivative size = 56, normalized size of antiderivative = 2.24, number of steps used = 18, number of rules used = 10, integrand size = 100, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {6742, 2194, 2288, 6688, 2306, 2309, 2178, 2330, 2297, 2298} \begin {gather*} \frac {x^4}{\log ^2(x)}+\frac {16}{x^2}-\frac {2 e^{x-4} \left (x^4 \log (x)+4 x \log ^2(x)\right )}{x^2 \log ^2(x)}+e^{2 x-8}+\frac {8 x}{\log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-2*x^6 + (-8*x^3 + 2*E^(-4 + x)*x^4 + 4*x^6)*Log[x] + (8*x^3 + E^(-4 + x)*(-4*x^4 - 2*x^5))*Log[x]^2 + (-
32 + 2*E^(-8 + 2*x)*x^3 + E^(-4 + x)*(8*x - 8*x^2))*Log[x]^3)/(x^3*Log[x]^3),x]

[Out]

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

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 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rule 2297

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_), x_Symbol] :> Simp[(x*(a + b*Log[c*x^n])^(p + 1))/(b*n*(p + 1))
, x] - Dist[1/(b*n*(p + 1)), Int[(a + b*Log[c*x^n])^(p + 1), x], x] /; FreeQ[{a, b, c, n}, x] && LtQ[p, -1] &&
IntegerQ[2*p]

Rule 2298

Int[Log[(c_.)*(x_)]^(-1), x_Symbol] :> Simp[LogIntegral[c*x]/c, x] /; FreeQ[c, 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 2330

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

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

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Mathematica [A]  time = 0.21, size = 36, normalized size = 1.44 \begin {gather*} \frac {\left (e^4 x^3+\left (4 e^4-e^x x\right ) \log (x)\right )^2}{e^8 x^2 \log ^2(x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-2*x^6 + (-8*x^3 + 2*E^(-4 + x)*x^4 + 4*x^6)*Log[x] + (8*x^3 + E^(-4 + x)*(-4*x^4 - 2*x^5))*Log[x]^
2 + (-32 + 2*E^(-8 + 2*x)*x^3 + E^(-4 + x)*(8*x - 8*x^2))*Log[x]^3)/(x^3*Log[x]^3),x]

[Out]

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

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fricas [B]  time = 0.93, size = 54, normalized size = 2.16 \begin {gather*} \frac {x^{6} + {\left (x^{2} e^{\left (2 \, x - 8\right )} - 8 \, x e^{\left (x - 4\right )} + 16\right )} \log \relax (x)^{2} - 2 \, {\left (x^{4} e^{\left (x - 4\right )} - 4 \, x^{3}\right )} \log \relax (x)}{x^{2} \log \relax (x)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^3*exp(x-4)^2+(-8*x^2+8*x)*exp(x-4)-32)*log(x)^3+((-2*x^5-4*x^4)*exp(x-4)+8*x^3)*log(x)^2+(2*x^
4*exp(x-4)+4*x^6-8*x^3)*log(x)-2*x^6)/x^3/log(x)^3,x, algorithm="fricas")

[Out]

(x^6 + (x^2*e^(2*x - 8) - 8*x*e^(x - 4) + 16)*log(x)^2 - 2*(x^4*e^(x - 4) - 4*x^3)*log(x))/(x^2*log(x)^2)

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giac [B]  time = 0.15, size = 70, normalized size = 2.80 \begin {gather*} \frac {{\left (x^{6} e^{12} - 2 \, x^{4} e^{\left (x + 8\right )} \log \relax (x) + 8 \, x^{3} e^{12} \log \relax (x) + x^{2} e^{\left (2 \, x + 4\right )} \log \relax (x)^{2} - 8 \, x e^{\left (x + 8\right )} \log \relax (x)^{2} + 16 \, e^{12} \log \relax (x)^{2}\right )} e^{\left (-12\right )}}{x^{2} \log \relax (x)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^3*exp(x-4)^2+(-8*x^2+8*x)*exp(x-4)-32)*log(x)^3+((-2*x^5-4*x^4)*exp(x-4)+8*x^3)*log(x)^2+(2*x^
4*exp(x-4)+4*x^6-8*x^3)*log(x)-2*x^6)/x^3/log(x)^3,x, algorithm="giac")

[Out]

(x^6*e^12 - 2*x^4*e^(x + 8)*log(x) + 8*x^3*e^12*log(x) + x^2*e^(2*x + 4)*log(x)^2 - 8*x*e^(x + 8)*log(x)^2 + 1
6*e^12*log(x)^2)*e^(-12)/(x^2*log(x)^2)

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maple [A]  time = 0.04, size = 48, normalized size = 1.92

method	result	size
risch	\(\frac {x^{2} {\mathrm e}^{2 x -8}-8 x \,{\mathrm e}^{x -4}+16}{x^{2}}+\frac {x \left (x^{3}-2 x \,{\mathrm e}^{x -4} \ln \relax (x )+8 \ln \relax (x )\right )}{\ln \relax (x )^{2}}\)	\(48\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*x^3*exp(x-4)^2+(-8*x^2+8*x)*exp(x-4)-32)*ln(x)^3+((-2*x^5-4*x^4)*exp(x-4)+8*x^3)*ln(x)^2+(2*x^4*exp(x-
4)+4*x^6-8*x^3)*ln(x)-2*x^6)/x^3/ln(x)^3,x,method=_RETURNVERBOSE)

[Out]

(x^2*exp(2*x-8)-8*x*exp(x-4)+16)/x^2+x*(x^3-2*x*exp(x-4)*ln(x)+8*ln(x))/ln(x)^2

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -8 \, {\rm Ei}\relax (x) e^{\left (-4\right )} + 8 \, e^{\left (-4\right )} \Gamma \left (-1, -x\right ) - \frac {2 \, x^{2} e^{\left (x - 4\right )}}{\log \relax (x)} + \frac {16}{x^{2}} + e^{\left (2 \, x - 8\right )} - 8 \, \Gamma \left (-1, -\log \relax (x)\right ) + 16 \, \Gamma \left (-1, -4 \, \log \relax (x)\right ) + 32 \, \Gamma \left (-2, -4 \, \log \relax (x)\right ) + 8 \, \int \frac {1}{\log \relax (x)}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^3*exp(x-4)^2+(-8*x^2+8*x)*exp(x-4)-32)*log(x)^3+((-2*x^5-4*x^4)*exp(x-4)+8*x^3)*log(x)^2+(2*x^
4*exp(x-4)+4*x^6-8*x^3)*log(x)-2*x^6)/x^3/log(x)^3,x, algorithm="maxima")

[Out]

-8*Ei(x)*e^(-4) + 8*e^(-4)*gamma(-1, -x) - 2*x^2*e^(x - 4)/log(x) + 16/x^2 + e^(2*x - 8) - 8*gamma(-1, -log(x)
) + 16*gamma(-1, -4*log(x)) + 32*gamma(-2, -4*log(x)) + 8*integrate(1/log(x), x)

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mupad [B]  time = 7.88, size = 49, normalized size = 1.96 \begin {gather*} {\mathrm {e}}^{2\,x-8}+\frac {8\,x}{\ln \relax (x)}-\frac {8\,{\mathrm {e}}^{x-4}}{x}+\frac {x^4}{{\ln \relax (x)}^2}+\frac {16}{x^2}-\frac {2\,x^2\,{\mathrm {e}}^{x-4}}{\ln \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(x)*(2*x^4*exp(x - 4) - 8*x^3 + 4*x^6) - log(x)^2*(exp(x - 4)*(4*x^4 + 2*x^5) - 8*x^3) - 2*x^6 + log(x
)^3*(exp(x - 4)*(8*x - 8*x^2) + 2*x^3*exp(2*x - 8) - 32))/(x^3*log(x)^3),x)

[Out]

exp(2*x - 8) + (8*x)/log(x) - (8*exp(x - 4))/x + x^4/log(x)^2 + 16/x^2 - (2*x^2*exp(x - 4))/log(x)

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sympy [B]  time = 0.37, size = 53, normalized size = 2.12 \begin {gather*} \frac {x^{4} + 8 x \log {\relax (x )}}{\log {\relax (x )}^{2}} + \frac {x e^{2 x - 8} \log {\relax (x )} + \left (- 2 x^{3} - 8 \log {\relax (x )}\right ) e^{x - 4}}{x \log {\relax (x )}} + \frac {16}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x**3*exp(x-4)**2+(-8*x**2+8*x)*exp(x-4)-32)*ln(x)**3+((-2*x**5-4*x**4)*exp(x-4)+8*x**3)*ln(x)**2
+(2*x**4*exp(x-4)+4*x**6-8*x**3)*ln(x)-2*x**6)/x**3/ln(x)**3,x)

[Out]

(x**4 + 8*x*log(x))/log(x)**2 + (x*exp(2*x - 8)*log(x) + (-2*x**3 - 8*log(x))*exp(x - 4))/(x*log(x)) + 16/x**2

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