3.61.59 \(\int \frac {110+16 x+(55+16 x+x^2) \log (\frac {3 x^2}{110+32 x+2 x^2})}{55+16 x+x^2} \, dx\)

Optimal. Leaf size=25 \[ -4+x \log \left (\frac {3 x^2}{(11+x) \log \left (e^{10+2 x}\right )}\right ) \]

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Rubi [A]  time = 0.09, antiderivative size = 20, normalized size of antiderivative = 0.80, number of steps used = 9, number of rules used = 6, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.140, Rules used = {6728, 72, 2523, 12, 632, 31} \begin {gather*} x \log \left (\frac {3 x^2}{2 \left (x^2+16 x+55\right )}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(110 + 16*x + (55 + 16*x + x^2)*Log[(3*x^2)/(110 + 32*x + 2*x^2)])/(55 + 16*x + x^2),x]

[Out]

x*Log[(3*x^2)/(2*(55 + 16*x + x^2))]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 72

Int[((e_.) + (f_.)*(x_))^(p_.)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Int[ExpandIntegrand[(
e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[p]

Rule 632

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Dist[
(c*d - e*(b/2 - q/2))/q, Int[1/(b/2 - q/2 + c*x), x], x] - Dist[(c*d - e*(b/2 + q/2))/q, Int[1/(b/2 + q/2 + c*
x), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] && NiceSqrtQ[b^2 - 4*a*
c]

Rule 2523

Int[((a_.) + Log[(c_.)*(RFx_)^(p_.)]*(b_.))^(n_.), x_Symbol] :> Simp[x*(a + b*Log[c*RFx^p])^n, x] - Dist[b*n*p
, Int[SimplifyIntegrand[(x*(a + b*Log[c*RFx^p])^(n - 1)*D[RFx, x])/RFx, x], x], x] /; FreeQ[{a, b, c, p}, x] &
& RationalFunctionQ[RFx, x] && IGtQ[n, 0]

Rule 6728

Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a +
b*x^n + c*x^(2*n)), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {2 (55+8 x)}{(5+x) (11+x)}+\log \left (\frac {3 x^2}{2 \left (55+16 x+x^2\right )}\right )\right ) \, dx\\ &=2 \int \frac {55+8 x}{(5+x) (11+x)} \, dx+\int \log \left (\frac {3 x^2}{2 \left (55+16 x+x^2\right )}\right ) \, dx\\ &=x \log \left (\frac {3 x^2}{2 \left (55+16 x+x^2\right )}\right )+2 \int \left (\frac {5}{2 (5+x)}+\frac {11}{2 (11+x)}\right ) \, dx-\int \frac {2 (55+8 x)}{55+16 x+x^2} \, dx\\ &=5 \log (5+x)+11 \log (11+x)+x \log \left (\frac {3 x^2}{2 \left (55+16 x+x^2\right )}\right )-2 \int \frac {55+8 x}{55+16 x+x^2} \, dx\\ &=5 \log (5+x)+11 \log (11+x)+x \log \left (\frac {3 x^2}{2 \left (55+16 x+x^2\right )}\right )-5 \int \frac {1}{5+x} \, dx-11 \int \frac {1}{11+x} \, dx\\ &=x \log \left (\frac {3 x^2}{2 \left (55+16 x+x^2\right )}\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.04, size = 20, normalized size = 0.80 \begin {gather*} x \log \left (\frac {3 x^2}{2 \left (55+16 x+x^2\right )}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(110 + 16*x + (55 + 16*x + x^2)*Log[(3*x^2)/(110 + 32*x + 2*x^2)])/(55 + 16*x + x^2),x]

[Out]

x*Log[(3*x^2)/(2*(55 + 16*x + x^2))]

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fricas [A]  time = 0.91, size = 18, normalized size = 0.72 \begin {gather*} x \log \left (\frac {3 \, x^{2}}{2 \, {\left (x^{2} + 16 \, x + 55\right )}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^2+16*x+55)*log(3*x^2/(2*x^2+32*x+110))+16*x+110)/(x^2+16*x+55),x, algorithm="fricas")

[Out]

x*log(3/2*x^2/(x^2 + 16*x + 55))

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giac [A]  time = 0.14, size = 18, normalized size = 0.72 \begin {gather*} x \log \left (\frac {3 \, x^{2}}{2 \, {\left (x^{2} + 16 \, x + 55\right )}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^2+16*x+55)*log(3*x^2/(2*x^2+32*x+110))+16*x+110)/(x^2+16*x+55),x, algorithm="giac")

[Out]

x*log(3/2*x^2/(x^2 + 16*x + 55))

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maple [A]  time = 0.45, size = 21, normalized size = 0.84

method	result	size
norman	\(x \ln \left (\frac {3 x^{2}}{2 x^{2}+32 x +110}\right )\)	\(21\)
risch	\(x \ln \left (\frac {3 x^{2}}{2 x^{2}+32 x +110}\right )\)	\(21\)
default	\(x \ln \relax (3)-x \ln \relax (2)+x \ln \left (\frac {x^{2}}{x^{2}+16 x +55}\right )\)	\(28\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x^2+16*x+55)*ln(3*x^2/(2*x^2+32*x+110))+16*x+110)/(x^2+16*x+55),x,method=_RETURNVERBOSE)

[Out]

x*ln(3*x^2/(2*x^2+32*x+110))

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maxima [A]  time = 0.48, size = 45, normalized size = 1.80 \begin {gather*} x {\left (\log \relax (3) - \log \relax (2)\right )} - {\left (x + 11\right )} \log \left (x + 11\right ) - {\left (x + 5\right )} \log \left (x + 5\right ) + 2 \, x \log \relax (x) + 11 \, \log \left (x + 11\right ) + 5 \, \log \left (x + 5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^2+16*x+55)*log(3*x^2/(2*x^2+32*x+110))+16*x+110)/(x^2+16*x+55),x, algorithm="maxima")

[Out]

x*(log(3) - log(2)) - (x + 11)*log(x + 11) - (x + 5)*log(x + 5) + 2*x*log(x) + 11*log(x + 11) + 5*log(x + 5)

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mupad [B]  time = 4.54, size = 20, normalized size = 0.80 \begin {gather*} x\,\ln \left (\frac {3\,x^2}{2\,x^2+32\,x+110}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((16*x + log((3*x^2)/(32*x + 2*x^2 + 110))*(16*x + x^2 + 55) + 110)/(16*x + x^2 + 55),x)

[Out]

x*log((3*x^2)/(32*x + 2*x^2 + 110))

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sympy [A]  time = 0.16, size = 17, normalized size = 0.68 \begin {gather*} x \log {\left (\frac {3 x^{2}}{2 x^{2} + 32 x + 110} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x**2+16*x+55)*ln(3*x**2/(2*x**2+32*x+110))+16*x+110)/(x**2+16*x+55),x)

[Out]

x*log(3*x**2/(2*x**2 + 32*x + 110))

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