3.50.39 \(\int \frac {-35+180 x^2+36 x^2 \log (3)+(-5+60 x^2+12 x^2 \log (3)) \log (x)+(5 x^2+x^2 \log (3)) \log ^2(x)}{36 x^2+12 x^2 \log (x)+x^2 \log ^2(x)} \, dx\)

Optimal. Leaf size=20 \[ -2+x \log (3)+5 \left (x+\frac {1}{x (6+\log (x))}\right ) \]

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Rubi [A]  time = 0.37, antiderivative size = 18, normalized size of antiderivative = 0.90, number of steps used = 9, number of rules used = 6, integrand size = 72, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {6, 6688, 6742, 2306, 2309, 2178} \begin {gather*} x (5+\log (3))+\frac {5}{x (\log (x)+6)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-35 + 180*x^2 + 36*x^2*Log[3] + (-5 + 60*x^2 + 12*x^2*Log[3])*Log[x] + (5*x^2 + x^2*Log[3])*Log[x]^2)/(36
*x^2 + 12*x^2*Log[x] + x^2*Log[x]^2),x]

[Out]

x*(5 + Log[3]) + 5/(x*(6 + Log[x]))

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 6688

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

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 {-35+x^2 (180+36 \log (3))+\left (-5+60 x^2+12 x^2 \log (3)\right ) \log (x)+\left (5 x^2+x^2 \log (3)\right ) \log ^2(x)}{36 x^2+12 x^2 \log (x)+x^2 \log ^2(x)} \, dx\\ &=\int \frac {-35+36 x^2 (5+\log (3))+\left (-5+12 x^2 (5+\log (3))\right ) \log (x)+x^2 (5+\log (3)) \log ^2(x)}{x^2 (6+\log (x))^2} \, dx\\ &=\int \left (5 \left (1+\frac {\log (3)}{5}\right )-\frac {5}{x^2 (6+\log (x))^2}-\frac {5}{x^2 (6+\log (x))}\right ) \, dx\\ &=x (5+\log (3))-5 \int \frac {1}{x^2 (6+\log (x))^2} \, dx-5 \int \frac {1}{x^2 (6+\log (x))} \, dx\\ &=x (5+\log (3))+\frac {5}{x (6+\log (x))}+5 \int \frac {1}{x^2 (6+\log (x))} \, dx-5 \operatorname {Subst}\left (\int \frac {e^{-x}}{6+x} \, dx,x,\log (x)\right )\\ &=-5 e^6 \text {Ei}(-6-\log (x))+x (5+\log (3))+\frac {5}{x (6+\log (x))}+5 \operatorname {Subst}\left (\int \frac {e^{-x}}{6+x} \, dx,x,\log (x)\right )\\ &=x (5+\log (3))+\frac {5}{x (6+\log (x))}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.10, size = 18, normalized size = 0.90 \begin {gather*} x (5+\log (3))+\frac {5}{x (6+\log (x))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-35 + 180*x^2 + 36*x^2*Log[3] + (-5 + 60*x^2 + 12*x^2*Log[3])*Log[x] + (5*x^2 + x^2*Log[3])*Log[x]^
2)/(36*x^2 + 12*x^2*Log[x] + x^2*Log[x]^2),x]

[Out]

x*(5 + Log[3]) + 5/(x*(6 + Log[x]))

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fricas [A]  time = 0.83, size = 40, normalized size = 2.00 \begin {gather*} \frac {6 \, x^{2} \log \relax (3) + 30 \, x^{2} + {\left (x^{2} \log \relax (3) + 5 \, x^{2}\right )} \log \relax (x) + 5}{x \log \relax (x) + 6 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^2*log(3)+5*x^2)*log(x)^2+(12*x^2*log(3)+60*x^2-5)*log(x)+36*x^2*log(3)+180*x^2-35)/(x^2*log(x)^2
+12*x^2*log(x)+36*x^2),x, algorithm="fricas")

[Out]

(6*x^2*log(3) + 30*x^2 + (x^2*log(3) + 5*x^2)*log(x) + 5)/(x*log(x) + 6*x)

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giac [A]  time = 0.13, size = 19, normalized size = 0.95 \begin {gather*} x {\left (\log \relax (3) + 5\right )} + \frac {5}{x \log \relax (x) + 6 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^2*log(3)+5*x^2)*log(x)^2+(12*x^2*log(3)+60*x^2-5)*log(x)+36*x^2*log(3)+180*x^2-35)/(x^2*log(x)^2
+12*x^2*log(x)+36*x^2),x, algorithm="giac")

[Out]

x*(log(3) + 5) + 5/(x*log(x) + 6*x)

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maple [A]  time = 0.06, size = 20, normalized size = 1.00

method	result	size
risch	\(x \ln \relax (3)+5 x +\frac {5}{\left (\ln \relax (x )+6\right ) x}\)	\(20\)
default	\(\frac {5+30 x^{2}+5 x^{2} \ln \relax (x )}{x \left (\ln \relax (x )+6\right )}+x \ln \relax (3)\)	\(30\)
norman	\(\frac {5+\left (30+6 \ln \relax (3)\right ) x^{2}+\left (5+\ln \relax (3)\right ) x^{2} \ln \relax (x )}{x \left (\ln \relax (x )+6\right )}\)	\(33\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x^2*ln(3)+5*x^2)*ln(x)^2+(12*x^2*ln(3)+60*x^2-5)*ln(x)+36*x^2*ln(3)+180*x^2-35)/(x^2*ln(x)^2+12*x^2*ln(x
)+36*x^2),x,method=_RETURNVERBOSE)

[Out]

x*ln(3)+5*x+5/(ln(x)+6)/x

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maxima [A]  time = 0.48, size = 32, normalized size = 1.60 \begin {gather*} \frac {x^{2} {\left (\log \relax (3) + 5\right )} \log \relax (x) + 6 \, x^{2} {\left (\log \relax (3) + 5\right )} + 5}{x \log \relax (x) + 6 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^2*log(3)+5*x^2)*log(x)^2+(12*x^2*log(3)+60*x^2-5)*log(x)+36*x^2*log(3)+180*x^2-35)/(x^2*log(x)^2
+12*x^2*log(x)+36*x^2),x, algorithm="maxima")

[Out]

(x^2*(log(3) + 5)*log(x) + 6*x^2*(log(3) + 5) + 5)/(x*log(x) + 6*x)

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mupad [B]  time = 4.16, size = 25, normalized size = 1.25 \begin {gather*} 5\,x+x\,\ln \relax (3)+\frac {5\,x}{x^2\,\ln \relax (x)+6\,x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(x)*(12*x^2*log(3) + 60*x^2 - 5) + 36*x^2*log(3) + 180*x^2 + log(x)^2*(x^2*log(3) + 5*x^2) - 35)/(12*x
^2*log(x) + x^2*log(x)^2 + 36*x^2),x)

[Out]

5*x + x*log(3) + (5*x)/(x^2*log(x) + 6*x^2)

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sympy [A]  time = 0.13, size = 15, normalized size = 0.75 \begin {gather*} x \left (\log {\relax (3 )} + 5\right ) + \frac {5}{x \log {\relax (x )} + 6 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x**2*ln(3)+5*x**2)*ln(x)**2+(12*x**2*ln(3)+60*x**2-5)*ln(x)+36*x**2*ln(3)+180*x**2-35)/(x**2*ln(x)
**2+12*x**2*ln(x)+36*x**2),x)

[Out]

x*(log(3) + 5) + 5/(x*log(x) + 6*x)

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