3.104.4 \(\int \frac {-2 x^7+8 x^7 \log (x)+(1-2 e^{-7+x^2} x) \log ^3(x)}{\log ^3(x)} \, dx\)

Optimal. Leaf size=19 \[ -e^{-7+x^2}+x+\frac {x^8}{\log ^2(x)} \]

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Rubi [A]  time = 0.29, antiderivative size = 19, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 5, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {6742, 2209, 2306, 2309, 2178} \begin {gather*} \frac {x^8}{\log ^2(x)}-e^{x^2-7}+x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-2*x^7 + 8*x^7*Log[x] + (1 - 2*E^(-7 + x^2)*x)*Log[x]^3)/Log[x]^3,x]

[Out]

-E^(-7 + x^2) + x + x^8/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 2209

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[((e + f*x)^n*
F^(a + b*(c + d*x)^n))/(b*f*n*(c + d*x)^n*Log[F]), x] /; FreeQ[{F, a, b, c, d, e, f, n}, x] && EqQ[m, n - 1] &
& EqQ[d*e - c*f, 0]

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 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^{-7+x^2} x+\frac {-2 x^7+8 x^7 \log (x)+\log ^3(x)}{\log ^3(x)}\right ) \, dx\\ &=-\left (2 \int e^{-7+x^2} x \, dx\right )+\int \frac {-2 x^7+8 x^7 \log (x)+\log ^3(x)}{\log ^3(x)} \, dx\\ &=-e^{-7+x^2}+\int \left (1-\frac {2 x^7}{\log ^3(x)}+\frac {8 x^7}{\log ^2(x)}\right ) \, dx\\ &=-e^{-7+x^2}+x-2 \int \frac {x^7}{\log ^3(x)} \, dx+8 \int \frac {x^7}{\log ^2(x)} \, dx\\ &=-e^{-7+x^2}+x+\frac {x^8}{\log ^2(x)}-\frac {8 x^8}{\log (x)}-8 \int \frac {x^7}{\log ^2(x)} \, dx+64 \int \frac {x^7}{\log (x)} \, dx\\ &=-e^{-7+x^2}+x+\frac {x^8}{\log ^2(x)}-64 \int \frac {x^7}{\log (x)} \, dx+64 \operatorname {Subst}\left (\int \frac {e^{8 x}}{x} \, dx,x,\log (x)\right )\\ &=-e^{-7+x^2}+x+64 \text {Ei}(8 \log (x))+\frac {x^8}{\log ^2(x)}-64 \operatorname {Subst}\left (\int \frac {e^{8 x}}{x} \, dx,x,\log (x)\right )\\ &=-e^{-7+x^2}+x+\frac {x^8}{\log ^2(x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.06, size = 19, normalized size = 1.00 \begin {gather*} -e^{-7+x^2}+x+\frac {x^8}{\log ^2(x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-2*x^7 + 8*x^7*Log[x] + (1 - 2*E^(-7 + x^2)*x)*Log[x]^3)/Log[x]^3,x]

[Out]

-E^(-7 + x^2) + x + x^8/Log[x]^2

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fricas [A]  time = 0.61, size = 24, normalized size = 1.26 \begin {gather*} \frac {x^{8} + {\left (x - e^{\left (x^{2} - 7\right )}\right )} \log \relax (x)^{2}}{\log \relax (x)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x*exp(x^2-7)+1)*log(x)^3+8*x^7*log(x)-2*x^7)/log(x)^3,x, algorithm="fricas")

[Out]

(x^8 + (x - e^(x^2 - 7))*log(x)^2)/log(x)^2

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giac [A]  time = 0.15, size = 32, normalized size = 1.68 \begin {gather*} \frac {{\left (x^{8} e^{7} + x e^{7} \log \relax (x)^{2} - e^{\left (x^{2}\right )} \log \relax (x)^{2}\right )} e^{\left (-7\right )}}{\log \relax (x)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x*exp(x^2-7)+1)*log(x)^3+8*x^7*log(x)-2*x^7)/log(x)^3,x, algorithm="giac")

[Out]

(x^8*e^7 + x*e^7*log(x)^2 - e^(x^2)*log(x)^2)*e^(-7)/log(x)^2

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maple [A]  time = 0.03, size = 19, normalized size = 1.00

method	result	size
default	\(x +\frac {x^{8}}{\ln \relax (x )^{2}}-{\mathrm e}^{x^{2}-7}\)	\(19\)
risch	\(x +\frac {x^{8}}{\ln \relax (x )^{2}}-{\mathrm e}^{x^{2}-7}\)	\(19\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-2*x*exp(x^2-7)+1)*ln(x)^3+8*x^7*ln(x)-2*x^7)/ln(x)^3,x,method=_RETURNVERBOSE)

[Out]

x+x^8/ln(x)^2-exp(x^2-7)

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maxima [C]  time = 0.37, size = 26, normalized size = 1.37 \begin {gather*} x - e^{\left (x^{2} - 7\right )} + 64 \, \Gamma \left (-1, -8 \, \log \relax (x)\right ) + 128 \, \Gamma \left (-2, -8 \, \log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x*exp(x^2-7)+1)*log(x)^3+8*x^7*log(x)-2*x^7)/log(x)^3,x, algorithm="maxima")

[Out]

x - e^(x^2 - 7) + 64*gamma(-1, -8*log(x)) + 128*gamma(-2, -8*log(x))

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mupad [B]  time = 6.80, size = 18, normalized size = 0.95 \begin {gather*} x-{\mathrm {e}}^{x^2-7}+\frac {x^8}{{\ln \relax (x)}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(x)^3*(2*x*exp(x^2 - 7) - 1) - 8*x^7*log(x) + 2*x^7)/log(x)^3,x)

[Out]

x - exp(x^2 - 7) + x^8/log(x)^2

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sympy [A]  time = 0.29, size = 15, normalized size = 0.79 \begin {gather*} \frac {x^{8}}{\log {\relax (x )}^{2}} + x - e^{x^{2} - 7} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x*exp(x**2-7)+1)*ln(x)**3+8*x**7*ln(x)-2*x**7)/ln(x)**3,x)

[Out]

x**8/log(x)**2 + x - exp(x**2 - 7)

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