3.71.75 \(\int \frac {18-2 x-30 e^x x+(-18+4 x+e^x (60 x+30 x^2)) \log (x)}{3 \log ^2(x)} \, dx\)

Optimal. Leaf size=20 \[ \frac {2 x \left (-3+\frac {x}{3}+5 e^x x\right )}{\log (x)} \]

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Rubi [A]  time = 0.31, antiderivative size = 27, normalized size of antiderivative = 1.35, number of steps used = 18, number of rules used = 8, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {12, 6742, 2288, 2320, 2330, 2298, 2309, 2178} \begin {gather*} \frac {10 e^x x^2}{\log (x)}-\frac {2 (9-x) x}{3 \log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(18 - 2*x - 30*E^x*x + (-18 + 4*x + E^x*(60*x + 30*x^2))*Log[x])/(3*Log[x]^2),x]

[Out]

(-2*(9 - x)*x)/(3*Log[x]) + (10*E^x*x^2)/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 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 2298

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

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 2320

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_)*((d_) + (e_.)*(x_))^(q_.), x_Symbol] :> Simp[(x*(d + e*x)^q*(a
+ b*Log[c*x^n])^(p + 1))/(b*n*(p + 1)), x] + (-Dist[(q + 1)/(b*n*(p + 1)), Int[(d + e*x)^q*(a + b*Log[c*x^n])^
(p + 1), x], x] + Dist[(d*q)/(b*n*(p + 1)), Int[(d + e*x)^(q - 1)*(a + b*Log[c*x^n])^(p + 1), x], x]) /; FreeQ
[{a, b, c, d, e, n}, x] && LtQ[p, -1] && GtQ[q, 0]

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 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}{3} \int \frac {18-2 x-30 e^x x+\left (-18+4 x+e^x \left (60 x+30 x^2\right )\right ) \log (x)}{\log ^2(x)} \, dx\\ &=\frac {1}{3} \int \left (\frac {30 e^x x (-1+2 \log (x)+x \log (x))}{\log ^2(x)}+\frac {2 (9-x-9 \log (x)+2 x \log (x))}{\log ^2(x)}\right ) \, dx\\ &=\frac {2}{3} \int \frac {9-x-9 \log (x)+2 x \log (x)}{\log ^2(x)} \, dx+10 \int \frac {e^x x (-1+2 \log (x)+x \log (x))}{\log ^2(x)} \, dx\\ &=\frac {10 e^x x^2}{\log (x)}+\frac {2}{3} \int \left (\frac {9-x}{\log ^2(x)}+\frac {-9+2 x}{\log (x)}\right ) \, dx\\ &=\frac {10 e^x x^2}{\log (x)}+\frac {2}{3} \int \frac {9-x}{\log ^2(x)} \, dx+\frac {2}{3} \int \frac {-9+2 x}{\log (x)} \, dx\\ &=-\frac {2 (9-x) x}{3 \log (x)}+\frac {10 e^x x^2}{\log (x)}+\frac {2}{3} \int \left (-\frac {9}{\log (x)}+\frac {2 x}{\log (x)}\right ) \, dx+\frac {4}{3} \int \frac {9-x}{\log (x)} \, dx-6 \int \frac {1}{\log (x)} \, dx\\ &=-\frac {2 (9-x) x}{3 \log (x)}+\frac {10 e^x x^2}{\log (x)}-6 \text {li}(x)+\frac {4}{3} \int \left (\frac {9}{\log (x)}-\frac {x}{\log (x)}\right ) \, dx+\frac {4}{3} \int \frac {x}{\log (x)} \, dx-6 \int \frac {1}{\log (x)} \, dx\\ &=-\frac {2 (9-x) x}{3 \log (x)}+\frac {10 e^x x^2}{\log (x)}-12 \text {li}(x)-\frac {4}{3} \int \frac {x}{\log (x)} \, dx+\frac {4}{3} \operatorname {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (x)\right )+12 \int \frac {1}{\log (x)} \, dx\\ &=\frac {4}{3} \text {Ei}(2 \log (x))-\frac {2 (9-x) x}{3 \log (x)}+\frac {10 e^x x^2}{\log (x)}-\frac {4}{3} \operatorname {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (x)\right )\\ &=-\frac {2 (9-x) x}{3 \log (x)}+\frac {10 e^x x^2}{\log (x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.13, size = 18, normalized size = 0.90 \begin {gather*} \frac {2 x \left (-9+x+15 e^x x\right )}{3 \log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(18 - 2*x - 30*E^x*x + (-18 + 4*x + E^x*(60*x + 30*x^2))*Log[x])/(3*Log[x]^2),x]

[Out]

(2*x*(-9 + x + 15*E^x*x))/(3*Log[x])

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fricas [A]  time = 1.06, size = 20, normalized size = 1.00 \begin {gather*} \frac {2 \, {\left (15 \, x^{2} e^{x} + x^{2} - 9 \, x\right )}}{3 \, \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(((30*x^2+60*x)*exp(x)+4*x-18)*log(x)-30*exp(x)*x-2*x+18)/log(x)^2,x, algorithm="fricas")

[Out]

2/3*(15*x^2*e^x + x^2 - 9*x)/log(x)

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giac [A]  time = 0.14, size = 20, normalized size = 1.00 \begin {gather*} \frac {2 \, {\left (15 \, x^{2} e^{x} + x^{2} - 9 \, x\right )}}{3 \, \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(((30*x^2+60*x)*exp(x)+4*x-18)*log(x)-30*exp(x)*x-2*x+18)/log(x)^2,x, algorithm="giac")

[Out]

2/3*(15*x^2*e^x + x^2 - 9*x)/log(x)

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maple [A]  time = 0.04, size = 16, normalized size = 0.80

method	result	size
risch	\(\frac {2 x \left (15 \,{\mathrm e}^{x} x +x -9\right )}{3 \ln \relax (x )}\)	\(16\)
norman	\(\frac {-6 x +\frac {2 x^{2}}{3}+10 \,{\mathrm e}^{x} x^{2}}{\ln \relax (x )}\)	\(22\)
default	\(\frac {2 x^{2}}{3 \ln \relax (x )}-\frac {6 x}{\ln \relax (x )}+\frac {10 \,{\mathrm e}^{x} x^{2}}{\ln \relax (x )}\)	\(29\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/3*(((30*x^2+60*x)*exp(x)+4*x-18)*ln(x)-30*exp(x)*x-2*x+18)/ln(x)^2,x,method=_RETURNVERBOSE)

[Out]

2/3*x*(15*exp(x)*x+x-9)/ln(x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {10 \, x^{2} e^{x}}{\log \relax (x)} + 6 \, \Gamma \left (-1, -\log \relax (x)\right ) - \frac {4}{3} \, \Gamma \left (-1, -2 \, \log \relax (x)\right ) + \frac {2}{3} \, \int \frac {2 \, x - 9}{\log \relax (x)}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(((30*x^2+60*x)*exp(x)+4*x-18)*log(x)-30*exp(x)*x-2*x+18)/log(x)^2,x, algorithm="maxima")

[Out]

10*x^2*e^x/log(x) + 6*gamma(-1, -log(x)) - 4/3*gamma(-1, -2*log(x)) + 2/3*integrate((2*x - 9)/log(x), x)

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mupad [B]  time = 4.19, size = 15, normalized size = 0.75 \begin {gather*} \frac {2\,x\,\left (x+15\,x\,{\mathrm {e}}^x-9\right )}{3\,\ln \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-((2*x)/3 - (log(x)*(4*x + exp(x)*(60*x + 30*x^2) - 18))/3 + 10*x*exp(x) - 6)/log(x)^2,x)

[Out]

(2*x*(x + 15*x*exp(x) - 9))/(3*log(x))

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sympy [A]  time = 0.30, size = 24, normalized size = 1.20 \begin {gather*} \frac {10 x^{2} e^{x}}{\log {\relax (x )}} + \frac {2 x^{2} - 18 x}{3 \log {\relax (x )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(((30*x**2+60*x)*exp(x)+4*x-18)*ln(x)-30*exp(x)*x-2*x+18)/ln(x)**2,x)

[Out]

10*x**2*exp(x)/log(x) + (2*x**2 - 18*x)/(3*log(x))

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