3.83.89 \(\int \frac {-18+33 x^3-15 x^6+(1-2 x^3+x^6) \log (6)-9 x^3 \log (x)}{(x-2 x^4+x^7) \log (6)} \, dx\)

Optimal. Leaf size=24 \[ \log (x)-\frac {3 \left (5+\frac {x}{x-x^4}\right ) \log (x)}{\log (6)} \]

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Rubi [A]  time = 0.38, antiderivative size = 31, normalized size of antiderivative = 1.29, number of steps used = 14, number of rules used = 10, integrand size = 49, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.204, Rules used = {12, 1594, 28, 6742, 266, 44, 261, 43, 2335, 260} \begin {gather*} -\frac {3 x^3 \log (x)}{\left (1-x^3\right ) \log (6)}-\frac {18 \log (x)}{\log (6)}+\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-18 + 33*x^3 - 15*x^6 + (1 - 2*x^3 + x^6)*Log[6] - 9*x^3*Log[x])/((x - 2*x^4 + x^7)*Log[6]),x]

[Out]

Log[x] - (18*Log[x])/Log[6] - (3*x^3*Log[x])/((1 - x^3)*Log[6])

Rule 12

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

Rule 28

Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/c^p, Int[u*(b/2 + c*x^n)^(2*
p), x], x] /; FreeQ[{a, b, c, n}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rule 260

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

Rule 261

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

Rule 266

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

Rule 1594

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

Rule 2335

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_), x_Symbol] :> Simp
[((f*x)^(m + 1)*(d + e*x^r)^(q + 1)*(a + b*Log[c*x^n]))/(d*f*(m + 1)), x] - Dist[(b*n)/(d*(m + 1)), Int[(f*x)^
m*(d + e*x^r)^(q + 1), x], x] /; FreeQ[{a, b, c, d, e, f, m, n, q, r}, x] && EqQ[m + r*(q + 1) + 1, 0] && NeQ[
m, -1]

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 {\int \frac {-18+33 x^3-15 x^6+\left (1-2 x^3+x^6\right ) \log (6)-9 x^3 \log (x)}{x-2 x^4+x^7} \, dx}{\log (6)}\\ &=\frac {\int \frac {-18+33 x^3-15 x^6+\left (1-2 x^3+x^6\right ) \log (6)-9 x^3 \log (x)}{x \left (1-2 x^3+x^6\right )} \, dx}{\log (6)}\\ &=\frac {\int \frac {-18+33 x^3-15 x^6+\left (1-2 x^3+x^6\right ) \log (6)-9 x^3 \log (x)}{x \left (-1+x^3\right )^2} \, dx}{\log (6)}\\ &=\frac {\int \left (-\frac {18}{x \left (-1+x^3\right )^2}+\frac {33 x^2}{\left (-1+x^3\right )^2}-\frac {15 x^5}{\left (-1+x^3\right )^2}+\frac {\log (6)}{x}-\frac {9 x^2 \log (x)}{\left (-1+x^3\right )^2}\right ) \, dx}{\log (6)}\\ &=\log (x)-\frac {9 \int \frac {x^2 \log (x)}{\left (-1+x^3\right )^2} \, dx}{\log (6)}-\frac {15 \int \frac {x^5}{\left (-1+x^3\right )^2} \, dx}{\log (6)}-\frac {18 \int \frac {1}{x \left (-1+x^3\right )^2} \, dx}{\log (6)}+\frac {33 \int \frac {x^2}{\left (-1+x^3\right )^2} \, dx}{\log (6)}\\ &=\frac {11}{\left (1-x^3\right ) \log (6)}+\log (x)-\frac {3 x^3 \log (x)}{\left (1-x^3\right ) \log (6)}-\frac {3 \int \frac {x^2}{-1+x^3} \, dx}{\log (6)}-\frac {5 \operatorname {Subst}\left (\int \frac {x}{(-1+x)^2} \, dx,x,x^3\right )}{\log (6)}-\frac {6 \operatorname {Subst}\left (\int \frac {1}{(-1+x)^2 x} \, dx,x,x^3\right )}{\log (6)}\\ &=\frac {11}{\left (1-x^3\right ) \log (6)}+\log (x)-\frac {3 x^3 \log (x)}{\left (1-x^3\right ) \log (6)}-\frac {\log \left (1-x^3\right )}{\log (6)}-\frac {5 \operatorname {Subst}\left (\int \left (\frac {1}{(-1+x)^2}+\frac {1}{-1+x}\right ) \, dx,x,x^3\right )}{\log (6)}-\frac {6 \operatorname {Subst}\left (\int \left (\frac {1}{1-x}+\frac {1}{(-1+x)^2}+\frac {1}{x}\right ) \, dx,x,x^3\right )}{\log (6)}\\ &=\log (x)-\frac {18 \log (x)}{\log (6)}-\frac {3 x^3 \log (x)}{\left (1-x^3\right ) \log (6)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.18, size = 24, normalized size = 1.00 \begin {gather*} \frac {\frac {3 \log (x)}{-1+x^3}+(-15+\log (6)) \log (x)}{\log (6)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-18 + 33*x^3 - 15*x^6 + (1 - 2*x^3 + x^6)*Log[6] - 9*x^3*Log[x])/((x - 2*x^4 + x^7)*Log[6]),x]

[Out]

((3*Log[x])/(-1 + x^3) + (-15 + Log[6])*Log[x])/Log[6]

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fricas [A]  time = 0.72, size = 31, normalized size = 1.29 \begin {gather*} -\frac {{\left (15 \, x^{3} - {\left (x^{3} - 1\right )} \log \relax (6) - 18\right )} \log \relax (x)}{{\left (x^{3} - 1\right )} \log \relax (6)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-9*x^3*log(x)+(x^6-2*x^3+1)*log(6)-15*x^6+33*x^3-18)/(x^7-2*x^4+x)/log(6),x, algorithm="fricas")

[Out]

-(15*x^3 - (x^3 - 1)*log(6) - 18)*log(x)/((x^3 - 1)*log(6))

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giac [A]  time = 0.18, size = 24, normalized size = 1.00 \begin {gather*} \frac {{\left (\log \relax (6) - 15\right )} \log \relax (x) + \frac {3 \, \log \relax (x)}{x^{3} - 1}}{\log \relax (6)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-9*x^3*log(x)+(x^6-2*x^3+1)*log(6)-15*x^6+33*x^3-18)/(x^7-2*x^4+x)/log(6),x, algorithm="giac")

[Out]

((log(6) - 15)*log(x) + 3*log(x)/(x^3 - 1))/log(6)

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maple [A]  time = 0.16, size = 40, normalized size = 1.67

method	result	size
norman	\(\frac {\frac {\left (18-\ln \relax (6)\right ) \ln \relax (x )}{\ln \relax (6)}-\frac {\left (-\ln \relax (6)+15\right ) x^{3} \ln \relax (x )}{\ln \relax (6)}}{x^{3}-1}\)	\(40\)
risch	\(\frac {3 \ln \relax (x )}{\left (\ln \relax (2)+\ln \relax (3)\right ) \left (x^{3}-1\right )}+\frac {\ln \relax (x ) \ln \relax (3)}{\ln \relax (2)+\ln \relax (3)}+\frac {\ln \relax (x ) \ln \relax (2)}{\ln \relax (2)+\ln \relax (3)}-\frac {15 \ln \relax (x )}{\ln \relax (2)+\ln \relax (3)}\)	\(55\)
default	\(\frac {\ln \relax (6) \ln \relax (x )-\frac {i \sqrt {3}\, \ln \relax (x ) \ln \left (\frac {i \sqrt {3}-2 x -1}{-1+i \sqrt {3}}\right )}{3}+\frac {i \sqrt {3}\, \ln \relax (x ) \ln \left (\frac {i \sqrt {3}+2 x +1}{1+i \sqrt {3}}\right )}{3}+\frac {\ln \relax (x ) x}{x -1}+\frac {\ln \relax (x ) \left (i \ln \left (\frac {i \sqrt {3}-2 x -1}{-1+i \sqrt {3}}\right ) \sqrt {3}\, x^{2}-i \ln \left (\frac {i \sqrt {3}+2 x +1}{1+i \sqrt {3}}\right ) \sqrt {3}\, x^{2}+i \ln \left (\frac {i \sqrt {3}-2 x -1}{-1+i \sqrt {3}}\right ) \sqrt {3}\, x -i \ln \left (\frac {i \sqrt {3}+2 x +1}{1+i \sqrt {3}}\right ) \sqrt {3}\, x +i \sqrt {3}\, \ln \left (\frac {i \sqrt {3}-2 x -1}{-1+i \sqrt {3}}\right )-i \sqrt {3}\, \ln \left (\frac {i \sqrt {3}+2 x +1}{1+i \sqrt {3}}\right )+6 x^{2}+3 x \right )}{3 x^{2}+3 x +3}-18 \ln \relax (x )}{\ln \relax (6)}\)	\(290\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-9*x^3*ln(x)+(x^6-2*x^3+1)*ln(6)-15*x^6+33*x^3-18)/(x^7-2*x^4+x)/ln(6),x,method=_RETURNVERBOSE)

[Out]

((18-ln(6))/ln(6)*ln(x)-(-ln(6)+15)/ln(6)*x^3*ln(x))/(x^3-1)

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maxima [B]  time = 0.44, size = 114, normalized size = 4.75 \begin {gather*} -\frac {{\left (\frac {1}{x^{3} - 1} + \log \left (x^{2} + x + 1\right ) + \log \left (x - 1\right ) - 3 \, \log \relax (x)\right )} \log \relax (6) + {\left (\frac {1}{x^{3} - 1} - \log \left (x^{2} + x + 1\right ) - \log \left (x - 1\right )\right )} \log \relax (6) - \frac {2 \, \log \relax (6)}{x^{3} - 1} - \frac {9 \, \log \relax (x)}{x^{3} - 1} + 3 \, \log \left (x^{3} - 1\right ) - 3 \, \log \left (x^{3}\right ) - 3 \, \log \left (x^{2} + x + 1\right ) - 3 \, \log \left (x - 1\right ) + 54 \, \log \relax (x)}{3 \, \log \relax (6)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-9*x^3*log(x)+(x^6-2*x^3+1)*log(6)-15*x^6+33*x^3-18)/(x^7-2*x^4+x)/log(6),x, algorithm="maxima")

[Out]

-1/3*((1/(x^3 - 1) + log(x^2 + x + 1) + log(x - 1) - 3*log(x))*log(6) + (1/(x^3 - 1) - log(x^2 + x + 1) - log(
x - 1))*log(6) - 2*log(6)/(x^3 - 1) - 9*log(x)/(x^3 - 1) + 3*log(x^3 - 1) - 3*log(x^3) - 3*log(x^2 + x + 1) -
3*log(x - 1) + 54*log(x))/log(6)

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mupad [B]  time = 5.58, size = 31, normalized size = 1.29 \begin {gather*} -\frac {\ln \relax (x)\,\left (\ln \relax (6)-x^3\,\ln \relax (6)+15\,x^3-18\right )}{\ln \relax (6)\,\left (x^3-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(9*x^3*log(x) - log(6)*(x^6 - 2*x^3 + 1) - 33*x^3 + 15*x^6 + 18)/(log(6)*(x - 2*x^4 + x^7)),x)

[Out]

-(log(x)*(log(6) - x^3*log(6) + 15*x^3 - 18))/(log(6)*(x^3 - 1))

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sympy [A]  time = 0.17, size = 26, normalized size = 1.08 \begin {gather*} \frac {\left (-15 + \log {\relax (6 )}\right ) \log {\relax (x )}}{\log {\relax (6 )}} + \frac {3 \log {\relax (x )}}{x^{3} \log {\relax (6 )} - \log {\relax (6 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-9*x**3*ln(x)+(x**6-2*x**3+1)*ln(6)-15*x**6+33*x**3-18)/(x**7-2*x**4+x)/ln(6),x)

[Out]

(-15 + log(6))*log(x)/log(6) + 3*log(x)/(x**3*log(6) - log(6))

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