3.44.79 \(\int \frac {-2 x-2 e^4 x+(2 x+2 e^4 x) \log (x)+9 \log ^3(x)}{\log ^3(x)} \, dx\)

Optimal. Leaf size=26 \[ -\left (\left (-9+\frac {e^3}{x}\right ) x\right )+\frac {\left (1+e^4\right ) x^2}{\log ^2(x)} \]

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Rubi [A]  time = 0.12, antiderivative size = 17, normalized size of antiderivative = 0.65, number of steps used = 10, number of rules used = 5, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.147, Rules used = {6, 6742, 2306, 2309, 2178} \begin {gather*} \frac {\left (1+e^4\right ) x^2}{\log ^2(x)}+9 x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-2*x - 2*E^4*x + (2*x + 2*E^4*x)*Log[x] + 9*Log[x]^3)/Log[x]^3,x]

[Out]

9*x + ((1 + E^4)*x^2)/Log[x]^2

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, 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 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 {\left (-2-2 e^4\right ) x+\left (2 x+2 e^4 x\right ) \log (x)+9 \log ^3(x)}{\log ^3(x)} \, dx\\ &=\int \left (9-\frac {2 \left (1+e^4\right ) x}{\log ^3(x)}+\frac {2 \left (1+e^4\right ) x}{\log ^2(x)}\right ) \, dx\\ &=9 x-\left (2 \left (1+e^4\right )\right ) \int \frac {x}{\log ^3(x)} \, dx+\left (2 \left (1+e^4\right )\right ) \int \frac {x}{\log ^2(x)} \, dx\\ &=9 x+\frac {\left (1+e^4\right ) x^2}{\log ^2(x)}-\frac {2 \left (1+e^4\right ) x^2}{\log (x)}-\left (2 \left (1+e^4\right )\right ) \int \frac {x}{\log ^2(x)} \, dx+\left (4 \left (1+e^4\right )\right ) \int \frac {x}{\log (x)} \, dx\\ &=9 x+\frac {\left (1+e^4\right ) x^2}{\log ^2(x)}-\left (4 \left (1+e^4\right )\right ) \int \frac {x}{\log (x)} \, dx+\left (4 \left (1+e^4\right )\right ) \operatorname {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (x)\right )\\ &=9 x+4 \left (1+e^4\right ) \text {Ei}(2 \log (x))+\frac {\left (1+e^4\right ) x^2}{\log ^2(x)}-\left (4 \left (1+e^4\right )\right ) \operatorname {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (x)\right )\\ &=9 x+\frac {\left (1+e^4\right ) x^2}{\log ^2(x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 23, normalized size = 0.88 \begin {gather*} 9 x+\frac {x^2}{\log ^2(x)}+\frac {e^4 x^2}{\log ^2(x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-2*x - 2*E^4*x + (2*x + 2*E^4*x)*Log[x] + 9*Log[x]^3)/Log[x]^3,x]

[Out]

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

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fricas [A]  time = 0.64, size = 22, normalized size = 0.85 \begin {gather*} \frac {x^{2} e^{4} + 9 \, x \log \relax (x)^{2} + x^{2}}{\log \relax (x)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((9*log(x)^3+(2*x*exp(4)+2*x)*log(x)-2*x*exp(4)-2*x)/log(x)^3,x, algorithm="fricas")

[Out]

(x^2*e^4 + 9*x*log(x)^2 + x^2)/log(x)^2

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giac [A]  time = 0.17, size = 22, normalized size = 0.85 \begin {gather*} 9 \, x + \frac {x^{2} e^{4}}{\log \relax (x)^{2}} + \frac {x^{2}}{\log \relax (x)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((9*log(x)^3+(2*x*exp(4)+2*x)*log(x)-2*x*exp(4)-2*x)/log(x)^3,x, algorithm="giac")

[Out]

9*x + x^2*e^4/log(x)^2 + x^2/log(x)^2

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maple [A]  time = 0.03, size = 17, normalized size = 0.65

method	result	size
risch	\(9 x +\frac {\left ({\mathrm e}^{4}+1\right ) x^{2}}{\ln \relax (x )^{2}}\)	\(17\)
norman	\(\frac {\left ({\mathrm e}^{4}+1\right ) x^{2}+9 x \ln \relax (x )^{2}}{\ln \relax (x )^{2}}\)	\(22\)
default	\(9 x +2 \,{\mathrm e}^{4} \left (-\frac {x^{2}}{\ln \relax (x )}-2 \expIntegralEi \left (1, -2 \ln \relax (x )\right )\right )-2 \,{\mathrm e}^{4} \left (-\frac {x^{2}}{2 \ln \relax (x )^{2}}-\frac {x^{2}}{\ln \relax (x )}-2 \expIntegralEi \left (1, -2 \ln \relax (x )\right )\right )+\frac {x^{2}}{\ln \relax (x )^{2}}\)	\(66\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((9*ln(x)^3+(2*x*exp(4)+2*x)*ln(x)-2*x*exp(4)-2*x)/ln(x)^3,x,method=_RETURNVERBOSE)

[Out]

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

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maxima [C]  time = 0.58, size = 40, normalized size = 1.54 \begin {gather*} 4 \, e^{4} \Gamma \left (-1, -2 \, \log \relax (x)\right ) + 8 \, e^{4} \Gamma \left (-2, -2 \, \log \relax (x)\right ) + 9 \, x + 4 \, \Gamma \left (-1, -2 \, \log \relax (x)\right ) + 8 \, \Gamma \left (-2, -2 \, \log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((9*log(x)^3+(2*x*exp(4)+2*x)*log(x)-2*x*exp(4)-2*x)/log(x)^3,x, algorithm="maxima")

[Out]

4*e^4*gamma(-1, -2*log(x)) + 8*e^4*gamma(-2, -2*log(x)) + 9*x + 4*gamma(-1, -2*log(x)) + 8*gamma(-2, -2*log(x)
)

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mupad [B]  time = 2.95, size = 16, normalized size = 0.62 \begin {gather*} 9\,x+\frac {x^2\,\left ({\mathrm {e}}^4+1\right )}{{\ln \relax (x)}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(2*x + 2*x*exp(4) - 9*log(x)^3 - log(x)*(2*x + 2*x*exp(4)))/log(x)^3,x)

[Out]

9*x + (x^2*(exp(4) + 1))/log(x)^2

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sympy [A]  time = 0.09, size = 17, normalized size = 0.65 \begin {gather*} 9 x + \frac {x^{2} + x^{2} e^{4}}{\log {\relax (x )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((9*ln(x)**3+(2*x*exp(4)+2*x)*ln(x)-2*x*exp(4)-2*x)/ln(x)**3,x)

[Out]

9*x + (x**2 + x**2*exp(4))/log(x)**2

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