3.71.38 \(\int \frac {-14 x+84 x^3+14 x^4+(1-12 x^2-2 x^3) \log (3)+(7 x-\log (3)) \log (-7 x+\log (3))}{-7 x^3+x^2 \log (3)} \, dx\)

Optimal. Leaf size=24 \[ 3-x-x (11+x)+\frac {-1+\log (-7 x+\log (3))}{x} \]

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Rubi [B]  time = 0.37, antiderivative size = 72, normalized size of antiderivative = 3.00, number of steps used = 9, number of rules used = 7, integrand size = 60, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.117, Rules used = {1593, 6742, 1620, 2395, 36, 29, 31} \begin {gather*} -x^2-12 x-\frac {1}{x}+\frac {\left (343-2 \log ^3(3)+\log ^2(3) \log (9)\right ) \log (7 x-\log (3))}{49 \log (3)}-\frac {7 \log (7 x-\log (3))}{\log (3)}+\frac {\log (\log (3)-7 x)}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-14*x + 84*x^3 + 14*x^4 + (1 - 12*x^2 - 2*x^3)*Log[3] + (7*x - Log[3])*Log[-7*x + Log[3]])/(-7*x^3 + x^2*
Log[3]),x]

[Out]

-x^(-1) - 12*x - x^2 - (7*Log[7*x - Log[3]])/Log[3] + ((343 - 2*Log[3]^3 + Log[3]^2*Log[9])*Log[7*x - Log[3]])
/(49*Log[3]) + Log[-7*x + Log[3]]/x

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

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

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 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 1620

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[Px*(a + b*x)
^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && PolyQ[Px, x] && (IntegersQ[m, n] || IGtQ[m, -2]) &&
GtQ[Expon[Px, x], 2]

Rule 2395

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))*((f_.) + (g_.)*(x_))^(q_.), x_Symbol] :> Simp[((f + g
*x)^(q + 1)*(a + b*Log[c*(d + e*x)^n]))/(g*(q + 1)), x] - Dist[(b*e*n)/(g*(q + 1)), Int[(f + g*x)^(q + 1)/(d +
e*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, n, q}, x] && NeQ[e*f - d*g, 0] && NeQ[q, -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} &=\int \frac {-14 x+84 x^3+14 x^4+\left (1-12 x^2-2 x^3\right ) \log (3)+(7 x-\log (3)) \log (-7 x+\log (3))}{x^2 (-7 x+\log (3))} \, dx\\ &=\int \left (\frac {14 x-14 x^4-\log (3)+12 x^2 \log (3)-x^3 (84-\log (9))}{x^2 (7 x-\log (3))}-\frac {\log (-7 x+\log (3))}{x^2}\right ) \, dx\\ &=\int \frac {14 x-14 x^4-\log (3)+12 x^2 \log (3)-x^3 (84-\log (9))}{x^2 (7 x-\log (3))} \, dx-\int \frac {\log (-7 x+\log (3))}{x^2} \, dx\\ &=\frac {\log (-7 x+\log (3))}{x}+7 \int \frac {1}{x (-7 x+\log (3))} \, dx+\int \left (-12+\frac {1}{x^2}-2 x-\frac {7}{x \log (3)}+\frac {343-2 \log ^3(3)+\log ^2(3) \log (9)}{7 (7 x-\log (3)) \log (3)}\right ) \, dx\\ &=-\frac {1}{x}-12 x-x^2-\frac {7 \log (x)}{\log (3)}+\frac {\left (343-2 \log ^3(3)+\log ^2(3) \log (9)\right ) \log (7 x-\log (3))}{49 \log (3)}+\frac {\log (-7 x+\log (3))}{x}+\frac {7 \int \frac {1}{x} \, dx}{\log (3)}+\frac {49 \int \frac {1}{-7 x+\log (3)} \, dx}{\log (3)}\\ &=-\frac {1}{x}-12 x-x^2-\frac {7 \log (7 x-\log (3))}{\log (3)}+\frac {\left (343-2 \log ^3(3)+\log ^2(3) \log (9)\right ) \log (7 x-\log (3))}{49 \log (3)}+\frac {\log (-7 x+\log (3))}{x}\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.10, size = 59, normalized size = 2.46 \begin {gather*} -\frac {1}{x}-12 x-x^2-\frac {2}{49} \log ^2(3) \log (7 x-\log (3))+\frac {1}{49} \log (3) \log (9) \log (7 x-\log (3))+\frac {\log (-7 x+\log (3))}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-14*x + 84*x^3 + 14*x^4 + (1 - 12*x^2 - 2*x^3)*Log[3] + (7*x - Log[3])*Log[-7*x + Log[3]])/(-7*x^3
+ x^2*Log[3]),x]

[Out]

-x^(-1) - 12*x - x^2 - (2*Log[3]^2*Log[7*x - Log[3]])/49 + (Log[3]*Log[9]*Log[7*x - Log[3]])/49 + Log[-7*x + L
og[3]]/x

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fricas [A]  time = 0.53, size = 24, normalized size = 1.00 \begin {gather*} -\frac {x^{3} + 12 \, x^{2} - \log \left (-7 \, x + \log \relax (3)\right ) + 1}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-log(3)+7*x)*log(log(3)-7*x)+(-2*x^3-12*x^2+1)*log(3)+14*x^4+84*x^3-14*x)/(x^2*log(3)-7*x^3),x, al
gorithm="fricas")

[Out]

-(x^3 + 12*x^2 - log(-7*x + log(3)) + 1)/x

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giac [A]  time = 0.14, size = 25, normalized size = 1.04 \begin {gather*} -x^{2} - 12 \, x + \frac {\log \left (-7 \, x + \log \relax (3)\right )}{x} - \frac {1}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-log(3)+7*x)*log(log(3)-7*x)+(-2*x^3-12*x^2+1)*log(3)+14*x^4+84*x^3-14*x)/(x^2*log(3)-7*x^3),x, al
gorithm="giac")

[Out]

-x^2 - 12*x + log(-7*x + log(3))/x - 1/x

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maple [A]  time = 0.14, size = 24, normalized size = 1.00

method	result	size
norman	\(\frac {-1-12 x^{2}-x^{3}+\ln \left (\ln \relax (3)-7 x \right )}{x}\)	\(24\)
risch	\(\frac {\ln \left (\ln \relax (3)-7 x \right )}{x}-\frac {x^{3}+12 x^{2}+1}{x}\)	\(28\)
derivativedivides	\(\frac {7 \ln \left (7 x \right )}{\ln \relax (3)}+\frac {\ln \left (\ln \relax (3)-7 x \right ) \left (\ln \relax (3)-7 x \right )}{\ln \relax (3) x}-\frac {\left (\ln \relax (3)-7 x \right )^{2}}{49}+\frac {2 \ln \relax (3) \left (\ln \relax (3)-7 x \right )}{49}+\frac {12 \ln \relax (3)}{7}-12 x -\frac {1}{x}-\frac {7 \ln \left (-7 x \right )}{\ln \relax (3)}+\frac {7 \ln \left (\ln \relax (3)-7 x \right )}{\ln \relax (3)}\)	\(88\)
default	\(\frac {7 \ln \left (7 x \right )}{\ln \relax (3)}+\frac {\ln \left (\ln \relax (3)-7 x \right ) \left (\ln \relax (3)-7 x \right )}{\ln \relax (3) x}-\frac {\left (\ln \relax (3)-7 x \right )^{2}}{49}+\frac {2 \ln \relax (3) \left (\ln \relax (3)-7 x \right )}{49}+\frac {12 \ln \relax (3)}{7}-12 x -\frac {1}{x}-\frac {7 \ln \left (-7 x \right )}{\ln \relax (3)}+\frac {7 \ln \left (\ln \relax (3)-7 x \right )}{\ln \relax (3)}\)	\(88\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-ln(3)+7*x)*ln(ln(3)-7*x)+(-2*x^3-12*x^2+1)*ln(3)+14*x^4+84*x^3-14*x)/(x^2*ln(3)-7*x^3),x,method=_RETURN
VERBOSE)

[Out]

(-1-12*x^2-x^3+ln(ln(3)-7*x))/x

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maxima [B]  time = 0.52, size = 83, normalized size = 3.46 \begin {gather*} -{\left (\frac {7 \, \log \left (7 \, x - \log \relax (3)\right )}{\log \relax (3)^{2}} - \frac {7 \, \log \relax (x)}{\log \relax (3)^{2}} + \frac {1}{x \log \relax (3)}\right )} \log \relax (3) - \frac {7 \, \log \relax (x)}{\log \relax (3)} - \frac {x^{3} \log \relax (3) + 12 \, x^{2} \log \relax (3) - {\left (7 \, x + \log \relax (3)\right )} \log \left (-7 \, x + \log \relax (3)\right )}{x \log \relax (3)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-log(3)+7*x)*log(log(3)-7*x)+(-2*x^3-12*x^2+1)*log(3)+14*x^4+84*x^3-14*x)/(x^2*log(3)-7*x^3),x, al
gorithm="maxima")

[Out]

-(7*log(7*x - log(3))/log(3)^2 - 7*log(x)/log(3)^2 + 1/(x*log(3)))*log(3) - 7*log(x)/log(3) - (x^3*log(3) + 12
*x^2*log(3) - (7*x + log(3))*log(-7*x + log(3)))/(x*log(3))

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mupad [B]  time = 4.24, size = 25, normalized size = 1.04 \begin {gather*} \frac {\ln \left (\ln \relax (3)-7\,x\right )}{x}-12\,x-\frac {1}{x}-x^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(log(3) - 7*x)*(7*x - log(3)) - 14*x - log(3)*(12*x^2 + 2*x^3 - 1) + 84*x^3 + 14*x^4)/(x^2*log(3) - 7*
x^3),x)

[Out]

log(log(3) - 7*x)/x - 12*x - 1/x - x^2

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sympy [A]  time = 0.18, size = 19, normalized size = 0.79 \begin {gather*} - x^{2} - 12 x + \frac {\log {\left (- 7 x + \log {\relax (3 )} \right )}}{x} - \frac {1}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-ln(3)+7*x)*ln(ln(3)-7*x)+(-2*x**3-12*x**2+1)*ln(3)+14*x**4+84*x**3-14*x)/(x**2*ln(3)-7*x**3),x)

[Out]

-x**2 - 12*x + log(-7*x + log(3))/x - 1/x

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