3.5.55 \(\int \frac {7680+2974 x+66 x^2-11 x \log (x)+(-30-11 x) \log (30+11 x)}{270 x+99 x^2} \, dx\)

Optimal. Leaf size=28 \[ \frac {1}{3} \left (2 x+\frac {1}{3} (2+\log (x)) (256-\log (x+10 (3+x)))\right ) \]

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Rubi [A]  time = 0.36, antiderivative size = 44, normalized size of antiderivative = 1.57, number of steps used = 11, number of rules used = 6, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {1593, 6742, 893, 2317, 2391, 2392} \begin {gather*} \frac {2 x}{3}-\frac {1}{9} \log \left (\frac {11 x}{30}+1\right ) \log (x)-\frac {1}{9} \log (30) \log (x)+\frac {256 \log (x)}{9}-\frac {2}{9} \log (11 x+30) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(7680 + 2974*x + 66*x^2 - 11*x*Log[x] + (-30 - 11*x)*Log[30 + 11*x])/(270*x + 99*x^2),x]

[Out]

(2*x)/3 + (256*Log[x])/9 - (Log[30]*Log[x])/9 - (Log[1 + (11*x)/30]*Log[x])/9 - (2*Log[30 + 11*x])/9

Rule 893

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> Int[ExpandIntegrand[(d + e*x)^m*(f + g*x)^n*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] &
& NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && I
ntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 0]))

Rule 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 2317

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

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 2392

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

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 {7680+2974 x+66 x^2-11 x \log (x)+(-30-11 x) \log (30+11 x)}{x (270+99 x)} \, dx\\ &=\int \left (\frac {7680+2974 x+66 x^2-11 x \log (x)}{9 x (30+11 x)}-\frac {\log (30+11 x)}{9 x}\right ) \, dx\\ &=\frac {1}{9} \int \frac {7680+2974 x+66 x^2-11 x \log (x)}{x (30+11 x)} \, dx-\frac {1}{9} \int \frac {\log (30+11 x)}{x} \, dx\\ &=-\frac {1}{9} \log (30) \log (x)-\frac {1}{9} \int \frac {\log \left (1+\frac {11 x}{30}\right )}{x} \, dx+\frac {1}{9} \int \left (\frac {2 \left (3840+1487 x+33 x^2\right )}{x (30+11 x)}-\frac {11 \log (x)}{30+11 x}\right ) \, dx\\ &=-\frac {1}{9} \log (30) \log (x)+\frac {1}{9} \text {Li}_2\left (-\frac {11 x}{30}\right )+\frac {2}{9} \int \frac {3840+1487 x+33 x^2}{x (30+11 x)} \, dx-\frac {11}{9} \int \frac {\log (x)}{30+11 x} \, dx\\ &=-\frac {1}{9} \log (30) \log (x)-\frac {1}{9} \log \left (1+\frac {11 x}{30}\right ) \log (x)+\frac {1}{9} \text {Li}_2\left (-\frac {11 x}{30}\right )+\frac {1}{9} \int \frac {\log \left (1+\frac {11 x}{30}\right )}{x} \, dx+\frac {2}{9} \int \left (3+\frac {128}{x}-\frac {11}{30+11 x}\right ) \, dx\\ &=\frac {2 x}{3}+\frac {256 \log (x)}{9}-\frac {1}{9} \log (30) \log (x)-\frac {1}{9} \log \left (1+\frac {11 x}{30}\right ) \log (x)-\frac {2}{9} \log (30+11 x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.05, size = 30, normalized size = 1.07 \begin {gather*} \frac {1}{9} \left (6 x-(-256+\log (30)) \log (x)-\log \left (1+\frac {11 x}{30}\right ) (2+\log (x))\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(7680 + 2974*x + 66*x^2 - 11*x*Log[x] + (-30 - 11*x)*Log[30 + 11*x])/(270*x + 99*x^2),x]

[Out]

(6*x - (-256 + Log[30])*Log[x] - Log[1 + (11*x)/30]*(2 + Log[x]))/9

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fricas [A]  time = 0.50, size = 20, normalized size = 0.71 \begin {gather*} -\frac {1}{9} \, {\left (\log \relax (x) + 2\right )} \log \left (11 \, x + 30\right ) + \frac {2}{3} \, x + \frac {256}{9} \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-11*x-30)*log(11*x+30)-11*x*log(x)+66*x^2+2974*x+7680)/(99*x^2+270*x),x, algorithm="fricas")

[Out]

-1/9*(log(x) + 2)*log(11*x + 30) + 2/3*x + 256/9*log(x)

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giac [A]  time = 0.31, size = 26, normalized size = 0.93 \begin {gather*} -\frac {1}{9} \, \log \left (11 \, x + 30\right ) \log \relax (x) + \frac {2}{3} \, x - \frac {2}{9} \, \log \left (11 \, x + 30\right ) + \frac {256}{9} \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-11*x-30)*log(11*x+30)-11*x*log(x)+66*x^2+2974*x+7680)/(99*x^2+270*x),x, algorithm="giac")

[Out]

-1/9*log(11*x + 30)*log(x) + 2/3*x - 2/9*log(11*x + 30) + 256/9*log(x)

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maple [A]  time = 0.28, size = 27, normalized size = 0.96

method	result	size
norman	\(\frac {2 x}{3}-\frac {\ln \relax (x ) \ln \left (11 x +30\right )}{9}-\frac {2 \ln \left (11 x +30\right )}{9}+\frac {256 \ln \relax (x )}{9}\)	\(27\)
risch	\(\frac {2 x}{3}-\frac {\ln \relax (x ) \ln \left (11 x +30\right )}{9}-\frac {2 \ln \left (11 x +30\right )}{9}+\frac {256 \ln \relax (x )}{9}\)	\(27\)
default	\(-\frac {\left (\ln \left (11 x +30\right )-\ln \left (\frac {11 x}{30}+1\right )\right ) \ln \left (-\frac {11 x}{30}\right )}{9}-\frac {\ln \relax (x ) \ln \left (\frac {11 x}{30}+1\right )}{9}+\frac {2 x}{3}+\frac {256 \ln \relax (x )}{9}-\frac {2 \ln \left (11 x +30\right )}{9}\)	\(48\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-11*x-30)*ln(11*x+30)-11*x*ln(x)+66*x^2+2974*x+7680)/(99*x^2+270*x),x,method=_RETURNVERBOSE)

[Out]

2/3*x-1/9*ln(x)*ln(11*x+30)-2/9*ln(11*x+30)+256/9*ln(x)

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maxima [A]  time = 0.60, size = 26, normalized size = 0.93 \begin {gather*} -\frac {1}{9} \, \log \left (11 \, x + 30\right ) \log \relax (x) + \frac {2}{3} \, x - \frac {2}{9} \, \log \left (11 \, x + 30\right ) + \frac {256}{9} \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-11*x-30)*log(11*x+30)-11*x*log(x)+66*x^2+2974*x+7680)/(99*x^2+270*x),x, algorithm="maxima")

[Out]

-1/9*log(11*x + 30)*log(x) + 2/3*x - 2/9*log(11*x + 30) + 256/9*log(x)

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mupad [B]  time = 0.60, size = 24, normalized size = 0.86 \begin {gather*} \frac {2\,x}{3}-\frac {2\,\ln \left (x+\frac {30}{11}\right )}{9}+\frac {256\,\ln \relax (x)}{9}-\frac {\ln \left (11\,x+30\right )\,\ln \relax (x)}{9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2974*x - log(11*x + 30)*(11*x + 30) - 11*x*log(x) + 66*x^2 + 7680)/(270*x + 99*x^2),x)

[Out]

(2*x)/3 - (2*log(x + 30/11))/9 + (256*log(x))/9 - (log(11*x + 30)*log(x))/9

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sympy [A]  time = 0.35, size = 32, normalized size = 1.14 \begin {gather*} \frac {2 x}{3} - \frac {\log {\relax (x )} \log {\left (11 x + 30 \right )}}{9} + \frac {256 \log {\relax (x )}}{9} - \frac {2 \log {\left (x + \frac {30}{11} \right )}}{9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-11*x-30)*ln(11*x+30)-11*x*ln(x)+66*x**2+2974*x+7680)/(99*x**2+270*x),x)

[Out]

2*x/3 - log(x)*log(11*x + 30)/9 + 256*log(x)/9 - 2*log(x + 30/11)/9

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