3.12.29 \(\int \frac {-2 x+x \log (x)+(2 e^x+8 x) \log ^3(x)+(2 x+2 e^x x) \log ^3(x) \log (10 x)}{x \log ^3(x)} \, dx\)

Optimal. Leaf size=23 \[ 6 (-4+x)+\frac {x}{\log ^2(x)}+2 \left (e^x+x\right ) \log (10 x) \]

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Rubi [A]  time = 0.33, antiderivative size = 26, normalized size of antiderivative = 1.13, number of steps used = 12, number of rules used = 7, integrand size = 49, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {6688, 2178, 2297, 2298, 2194, 2554, 14} \begin {gather*} 6 x+\frac {x}{\log ^2(x)}+2 x \log (10 x)+2 e^x \log (10 x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-2*x + x*Log[x] + (2*E^x + 8*x)*Log[x]^3 + (2*x + 2*E^x*x)*Log[x]^3*Log[10*x])/(x*Log[x]^3),x]

[Out]

6*x + x/Log[x]^2 + 2*E^x*Log[10*x] + 2*x*Log[10*x]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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 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 2554

Int[Log[u_]*(v_), x_Symbol] :> With[{w = IntHide[v, x]}, Dist[Log[u], w, x] - Int[SimplifyIntegrand[(w*D[u, x]
)/u, x], x] /; InverseFunctionFreeQ[w, x]] /; InverseFunctionFreeQ[u, x]

Rule 6688

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (8+\frac {2 e^x}{x}-\frac {2}{\log ^3(x)}+\frac {1}{\log ^2(x)}+2 \left (1+e^x\right ) \log (10 x)\right ) \, dx\\ &=8 x+2 \int \frac {e^x}{x} \, dx-2 \int \frac {1}{\log ^3(x)} \, dx+2 \int \left (1+e^x\right ) \log (10 x) \, dx+\int \frac {1}{\log ^2(x)} \, dx\\ &=8 x+2 \text {Ei}(x)+\frac {x}{\log ^2(x)}-\frac {x}{\log (x)}+2 e^x \log (10 x)+2 x \log (10 x)-2 \int \frac {e^x+x}{x} \, dx-\int \frac {1}{\log ^2(x)} \, dx+\int \frac {1}{\log (x)} \, dx\\ &=8 x+2 \text {Ei}(x)+\frac {x}{\log ^2(x)}+2 e^x \log (10 x)+2 x \log (10 x)+\text {li}(x)-2 \int \left (1+\frac {e^x}{x}\right ) \, dx-\int \frac {1}{\log (x)} \, dx\\ &=6 x+2 \text {Ei}(x)+\frac {x}{\log ^2(x)}+2 e^x \log (10 x)+2 x \log (10 x)-2 \int \frac {e^x}{x} \, dx\\ &=6 x+\frac {x}{\log ^2(x)}+2 e^x \log (10 x)+2 x \log (10 x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.08, size = 27, normalized size = 1.17 \begin {gather*} 6 x+\frac {x}{\log ^2(x)}+\frac {1}{5} \left (10 e^x+10 x\right ) \log (10 x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-2*x + x*Log[x] + (2*E^x + 8*x)*Log[x]^3 + (2*x + 2*E^x*x)*Log[x]^3*Log[10*x])/(x*Log[x]^3),x]

[Out]

6*x + x/Log[x]^2 + ((10*E^x + 10*x)*Log[10*x])/5

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fricas [A]  time = 0.67, size = 36, normalized size = 1.57 \begin {gather*} \frac {2 \, {\left (x + e^{x}\right )} \log \relax (x)^{3} + 2 \, {\left (x \log \left (10\right ) + e^{x} \log \left (10\right ) + 3 \, x\right )} \log \relax (x)^{2} + x}{\log \relax (x)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*exp(x)*x+2*x)*log(x)^3*log(10*x)+(2*exp(x)+8*x)*log(x)^3+x*log(x)-2*x)/x/log(x)^3,x, algorithm="
fricas")

[Out]

(2*(x + e^x)*log(x)^3 + 2*(x*log(10) + e^x*log(10) + 3*x)*log(x)^2 + x)/log(x)^2

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giac [B]  time = 0.38, size = 48, normalized size = 2.09 \begin {gather*} \frac {2 \, x \log \left (10\right ) \log \relax (x)^{2} + 2 \, e^{x} \log \left (10\right ) \log \relax (x)^{2} + 2 \, x \log \relax (x)^{3} + 2 \, e^{x} \log \relax (x)^{3} + 6 \, x \log \relax (x)^{2} + x}{\log \relax (x)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*exp(x)*x+2*x)*log(x)^3*log(10*x)+(2*exp(x)+8*x)*log(x)^3+x*log(x)-2*x)/x/log(x)^3,x, algorithm="
giac")

[Out]

(2*x*log(10)*log(x)^2 + 2*e^x*log(10)*log(x)^2 + 2*x*log(x)^3 + 2*e^x*log(x)^3 + 6*x*log(x)^2 + x)/log(x)^2

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maple [A]  time = 0.06, size = 26, normalized size = 1.13

method	result	size
default	\(6 x +2 \ln \left (10 x \right ) {\mathrm e}^{x}+\frac {x}{\ln \relax (x )^{2}}+2 \ln \left (10 x \right ) x\)	\(26\)
risch	\(\left (2 \,{\mathrm e}^{x}+2 x \right ) \ln \relax (x )+2 x \ln \relax (5)+2 \,{\mathrm e}^{x} \ln \relax (5)+2 x \ln \relax (2)+2 \,{\mathrm e}^{x} \ln \relax (2)+6 x +\frac {x}{\ln \relax (x )^{2}}\)	\(44\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*exp(x)*x+2*x)*ln(x)^3*ln(10*x)+(2*exp(x)+8*x)*ln(x)^3+x*ln(x)-2*x)/x/ln(x)^3,x,method=_RETURNVERBOSE)

[Out]

6*x+2*ln(10*x)*exp(x)+x/ln(x)^2+2*ln(10*x)*x

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maxima [C]  time = 0.56, size = 33, normalized size = 1.43 \begin {gather*} 2 \, x \log \left (10 \, x\right ) + 2 \, e^{x} \log \left (10 \, x\right ) + 6 \, x + \Gamma \left (-1, -\log \relax (x)\right ) + 2 \, \Gamma \left (-2, -\log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*exp(x)*x+2*x)*log(x)^3*log(10*x)+(2*exp(x)+8*x)*log(x)^3+x*log(x)-2*x)/x/log(x)^3,x, algorithm="
maxima")

[Out]

2*x*log(10*x) + 2*e^x*log(10*x) + 6*x + gamma(-1, -log(x)) + 2*gamma(-2, -log(x))

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mupad [B]  time = 0.81, size = 32, normalized size = 1.39 \begin {gather*} 6\,x+\frac {x}{{\ln \relax (x)}^2}+2\,{\mathrm {e}}^x\,\ln \relax (x)+2\,x\,\ln \left (10\right )+2\,{\mathrm {e}}^x\,\ln \left (10\right )+2\,x\,\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(x)^3*(8*x + 2*exp(x)) - 2*x + x*log(x) + log(10*x)*log(x)^3*(2*x + 2*x*exp(x)))/(x*log(x)^3),x)

[Out]

6*x + x/log(x)^2 + 2*exp(x)*log(x) + 2*x*log(10) + 2*exp(x)*log(10) + 2*x*log(x)

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sympy [A]  time = 0.39, size = 34, normalized size = 1.48 \begin {gather*} 2 x \log {\relax (x )} + x \left (2 \log {\left (10 \right )} + 6\right ) + \frac {x}{\log {\relax (x )}^{2}} + \left (2 \log {\relax (x )} + 2 \log {\left (10 \right )}\right ) e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*exp(x)*x+2*x)*ln(x)**3*ln(10*x)+(2*exp(x)+8*x)*ln(x)**3+x*ln(x)-2*x)/x/ln(x)**3,x)

[Out]

2*x*log(x) + x*(2*log(10) + 6) + x/log(x)**2 + (2*log(x) + 2*log(10))*exp(x)

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