3.48.86 \(\int \frac {126-54 \log (x)+6 \log ^2(x)+e^{\frac {1}{6} (-30+35 x^2+5 x \log (x))} (80+1120 x+(40-560 x) \log (x)+(-35+70 x) \log ^2(x)+5 \log ^3(x))}{96-48 \log (x)+6 \log ^2(x)} \, dx\)

Optimal. Leaf size=29 \[ e^{-5+5 x \left (x+\frac {1}{6} (x+\log (x))\right )}+x+\frac {x}{4-\log (x)} \]

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Rubi [A]  time = 0.66, antiderivative size = 31, normalized size of antiderivative = 1.07, number of steps used = 12, number of rules used = 7, integrand size = 75, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.093, Rules used = {6741, 12, 6742, 2288, 2299, 2178, 2297} \begin {gather*} e^{\frac {35 x^2}{6}-5} x^{5 x/6}+x+\frac {x}{4-\log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(126 - 54*Log[x] + 6*Log[x]^2 + E^((-30 + 35*x^2 + 5*x*Log[x])/6)*(80 + 1120*x + (40 - 560*x)*Log[x] + (-3
5 + 70*x)*Log[x]^2 + 5*Log[x]^3))/(96 - 48*Log[x] + 6*Log[x]^2),x]

[Out]

x + E^(-5 + (35*x^2)/6)*x^((5*x)/6) + x/(4 - 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 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 2299

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

Rule 6741

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

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 {126-54 \log (x)+6 \log ^2(x)+e^{\frac {1}{6} \left (-30+35 x^2+5 x \log (x)\right )} \left (80+1120 x+(40-560 x) \log (x)+(-35+70 x) \log ^2(x)+5 \log ^3(x)\right )}{6 (4-\log (x))^2} \, dx\\ &=\frac {1}{6} \int \frac {126-54 \log (x)+6 \log ^2(x)+e^{\frac {1}{6} \left (-30+35 x^2+5 x \log (x)\right )} \left (80+1120 x+(40-560 x) \log (x)+(-35+70 x) \log ^2(x)+5 \log ^3(x)\right )}{(4-\log (x))^2} \, dx\\ &=\frac {1}{6} \int \left (5 e^{-5+\frac {35 x^2}{6}} x^{5 x/6} (1+14 x+\log (x))+\frac {6 \left (21-9 \log (x)+\log ^2(x)\right )}{(-4+\log (x))^2}\right ) \, dx\\ &=\frac {5}{6} \int e^{-5+\frac {35 x^2}{6}} x^{5 x/6} (1+14 x+\log (x)) \, dx+\int \frac {21-9 \log (x)+\log ^2(x)}{(-4+\log (x))^2} \, dx\\ &=e^{-5+\frac {35 x^2}{6}} x^{5 x/6}+\int \left (1+\frac {1}{4-\log (x)}+\frac {1}{(-4+\log (x))^2}\right ) \, dx\\ &=x+e^{-5+\frac {35 x^2}{6}} x^{5 x/6}+\int \frac {1}{4-\log (x)} \, dx+\int \frac {1}{(-4+\log (x))^2} \, dx\\ &=x+e^{-5+\frac {35 x^2}{6}} x^{5 x/6}+\frac {x}{4-\log (x)}+\int \frac {1}{-4+\log (x)} \, dx+\operatorname {Subst}\left (\int \frac {e^x}{4-x} \, dx,x,\log (x)\right )\\ &=x+e^{-5+\frac {35 x^2}{6}} x^{5 x/6}-e^4 \text {Ei}(-4+\log (x))+\frac {x}{4-\log (x)}+\operatorname {Subst}\left (\int \frac {e^x}{-4+x} \, dx,x,\log (x)\right )\\ &=x+e^{-5+\frac {35 x^2}{6}} x^{5 x/6}+\frac {x}{4-\log (x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.27, size = 37, normalized size = 1.28 \begin {gather*} \frac {1}{6} \left (6 x+6 e^{-5+\frac {35 x^2}{6}} x^{5 x/6}-\frac {6 x}{-4+\log (x)}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(126 - 54*Log[x] + 6*Log[x]^2 + E^((-30 + 35*x^2 + 5*x*Log[x])/6)*(80 + 1120*x + (40 - 560*x)*Log[x]
+ (-35 + 70*x)*Log[x]^2 + 5*Log[x]^3))/(96 - 48*Log[x] + 6*Log[x]^2),x]

[Out]

(6*x + 6*E^(-5 + (35*x^2)/6)*x^((5*x)/6) - (6*x)/(-4 + Log[x]))/6

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fricas [A]  time = 0.55, size = 33, normalized size = 1.14 \begin {gather*} \frac {{\left (\log \relax (x) - 4\right )} e^{\left (\frac {35}{6} \, x^{2} + \frac {5}{6} \, x \log \relax (x) - 5\right )} + x \log \relax (x) - 5 \, x}{\log \relax (x) - 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*log(x)^3+(70*x-35)*log(x)^2+(-560*x+40)*log(x)+1120*x+80)*exp(5/6*x*log(x)+35/6*x^2-5)+6*log(x)^
2-54*log(x)+126)/(6*log(x)^2-48*log(x)+96),x, algorithm="fricas")

[Out]

((log(x) - 4)*e^(35/6*x^2 + 5/6*x*log(x) - 5) + x*log(x) - 5*x)/(log(x) - 4)

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giac [B]  time = 0.18, size = 54, normalized size = 1.86 \begin {gather*} \frac {x e^{5} \log \relax (x) - 5 \, x e^{5} + e^{\left (\frac {35}{6} \, x^{2} + \frac {5}{6} \, x \log \relax (x)\right )} \log \relax (x) - 4 \, e^{\left (\frac {35}{6} \, x^{2} + \frac {5}{6} \, x \log \relax (x)\right )}}{e^{5} \log \relax (x) - 4 \, e^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*log(x)^3+(70*x-35)*log(x)^2+(-560*x+40)*log(x)+1120*x+80)*exp(5/6*x*log(x)+35/6*x^2-5)+6*log(x)^
2-54*log(x)+126)/(6*log(x)^2-48*log(x)+96),x, algorithm="giac")

[Out]

(x*e^5*log(x) - 5*x*e^5 + e^(35/6*x^2 + 5/6*x*log(x))*log(x) - 4*e^(35/6*x^2 + 5/6*x*log(x)))/(e^5*log(x) - 4*
e^5)

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maple [A]  time = 0.20, size = 26, normalized size = 0.90

method	result	size
risch	\(x -\frac {x}{\ln \relax (x )-4}+x^{\frac {5 x}{6}} {\mathrm e}^{-5+\frac {35 x^{2}}{6}}\)	\(26\)
norman	\(\frac {x \ln \relax (x )+\ln \relax (x ) {\mathrm e}^{\frac {5 x \ln \relax (x )}{6}+\frac {35 x^{2}}{6}-5}-5 x -4 \,{\mathrm e}^{\frac {5 x \ln \relax (x )}{6}+\frac {35 x^{2}}{6}-5}}{\ln \relax (x )-4}\)	\(47\)
default	\(\frac {x \left (\ln \relax (x )-5\right )}{\ln \relax (x )-4}+\frac {6 \ln \relax (x ) {\mathrm e}^{\frac {5 x \ln \relax (x )}{6}+\frac {35 x^{2}}{6}-5}-24 \,{\mathrm e}^{\frac {5 x \ln \relax (x )}{6}+\frac {35 x^{2}}{6}-5}}{6 \ln \relax (x )-24}\)	\(55\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((5*ln(x)^3+(70*x-35)*ln(x)^2+(-560*x+40)*ln(x)+1120*x+80)*exp(5/6*x*ln(x)+35/6*x^2-5)+6*ln(x)^2-54*ln(x)+
126)/(6*ln(x)^2-48*ln(x)+96),x,method=_RETURNVERBOSE)

[Out]

x-x/(ln(x)-4)+x^(5/6*x)*exp(-5+35/6*x^2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -\frac {21 \, e^{4} E_{2}\left (-\log \relax (x) + 4\right )}{\log \relax (x) - 4} + \frac {x e^{5} \log \relax (x) + 16 \, x e^{5} + {\left (\log \relax (x) - 4\right )} e^{\left (\frac {35}{6} \, x^{2} + \frac {5}{6} \, x \log \relax (x)\right )}}{e^{5} \log \relax (x) - 4 \, e^{5}} - 21 \, \int \frac {1}{\log \relax (x) - 4}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*log(x)^3+(70*x-35)*log(x)^2+(-560*x+40)*log(x)+1120*x+80)*exp(5/6*x*log(x)+35/6*x^2-5)+6*log(x)^
2-54*log(x)+126)/(6*log(x)^2-48*log(x)+96),x, algorithm="maxima")

[Out]

-21*e^4*exp_integral_e(2, -log(x) + 4)/(log(x) - 4) + (x*e^5*log(x) + 16*x*e^5 + (log(x) - 4)*e^(35/6*x^2 + 5/
6*x*log(x)))/(e^5*log(x) - 4*e^5) - 21*integrate(1/(log(x) - 4), x)

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mupad [B]  time = 3.50, size = 32, normalized size = 1.10 \begin {gather*} x^{\frac {5\,x}{6}}\,{\mathrm {e}}^{\frac {35\,x^2}{6}-5}-\frac {5\,x-x\,\ln \relax (x)}{\ln \relax (x)-4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((6*log(x)^2 - 54*log(x) + exp((5*x*log(x))/6 + (35*x^2)/6 - 5)*(1120*x + 5*log(x)^3 - log(x)*(560*x - 40)
+ log(x)^2*(70*x - 35) + 80) + 126)/(6*log(x)^2 - 48*log(x) + 96),x)

[Out]

x^((5*x)/6)*exp((35*x^2)/6 - 5) - (5*x - x*log(x))/(log(x) - 4)

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sympy [A]  time = 0.41, size = 26, normalized size = 0.90 \begin {gather*} x - \frac {x}{\log {\relax (x )} - 4} + e^{\frac {35 x^{2}}{6} + \frac {5 x \log {\relax (x )}}{6} - 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*ln(x)**3+(70*x-35)*ln(x)**2+(-560*x+40)*ln(x)+1120*x+80)*exp(5/6*x*ln(x)+35/6*x**2-5)+6*ln(x)**2
-54*ln(x)+126)/(6*ln(x)**2-48*ln(x)+96),x)

[Out]

x - x/(log(x) - 4) + exp(35*x**2/6 + 5*x*log(x)/6 - 5)

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