3.76.21 \(\int \frac {-2 x+(3-x-30 x^2+10 x^3+(-6 x^2+2 x^3) \log (2)) \log (9-6 x+x^2)}{(-3 x+x^2) \log (9-6 x+x^2)} \, dx\)

Optimal. Leaf size=32 \[ 2 x^2-x^2 (-3-\log (2))-\log (x)-\log \left (\log \left ((3-x)^2\right )\right ) \]

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Rubi [A]  time = 0.30, antiderivative size = 25, normalized size of antiderivative = 0.78, number of steps used = 6, number of rules used = 5, integrand size = 64, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.078, Rules used = {1593, 6688, 2390, 2302, 29} \begin {gather*} \frac {1}{2} x^2 (10+\log (4))-\log (x)-\log \left (\log \left ((x-3)^2\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-2*x + (3 - x - 30*x^2 + 10*x^3 + (-6*x^2 + 2*x^3)*Log[2])*Log[9 - 6*x + x^2])/((-3*x + x^2)*Log[9 - 6*x
+ x^2]),x]

[Out]

(x^2*(10 + Log[4]))/2 - Log[x] - Log[Log[(-3 + x)^2]]

Rule 29

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

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 2302

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/(x_), x_Symbol] :> Dist[1/(b*n), Subst[Int[x^p, x], x, a + b*L
og[c*x^n]], x] /; FreeQ[{a, b, c, n, p}, x]

Rule 2390

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

Rule 6688

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-2 x+\left (3-x-30 x^2+10 x^3+\left (-6 x^2+2 x^3\right ) \log (2)\right ) \log \left (9-6 x+x^2\right )}{(-3+x) x \log \left (9-6 x+x^2\right )} \, dx\\ &=\int \left (-\frac {1}{x}+x (10+\log (4))-\frac {2}{(-3+x) \log \left ((-3+x)^2\right )}\right ) \, dx\\ &=\frac {1}{2} x^2 (10+\log (4))-\log (x)-2 \int \frac {1}{(-3+x) \log \left ((-3+x)^2\right )} \, dx\\ &=\frac {1}{2} x^2 (10+\log (4))-\log (x)-2 \operatorname {Subst}\left (\int \frac {1}{x \log \left (x^2\right )} \, dx,x,-3+x\right )\\ &=\frac {1}{2} x^2 (10+\log (4))-\log (x)-\operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,\log \left ((-3+x)^2\right )\right )\\ &=\frac {1}{2} x^2 (10+\log (4))-\log (x)-\log \left (\log \left ((-3+x)^2\right )\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 28, normalized size = 0.88 \begin {gather*} 5 x^2+\frac {1}{2} x^2 \log (4)-\log (x)-\log \left (\log \left ((-3+x)^2\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-2*x + (3 - x - 30*x^2 + 10*x^3 + (-6*x^2 + 2*x^3)*Log[2])*Log[9 - 6*x + x^2])/((-3*x + x^2)*Log[9
- 6*x + x^2]),x]

[Out]

5*x^2 + (x^2*Log[4])/2 - Log[x] - Log[Log[(-3 + x)^2]]

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fricas [A]  time = 0.54, size = 28, normalized size = 0.88 \begin {gather*} x^{2} \log \relax (2) + 5 \, x^{2} - \log \relax (x) - \log \left (\log \left (x^{2} - 6 \, x + 9\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x^3-6*x^2)*log(2)+10*x^3-30*x^2-x+3)*log(x^2-6*x+9)-2*x)/(x^2-3*x)/log(x^2-6*x+9),x, algorithm=
"fricas")

[Out]

x^2*log(2) + 5*x^2 - log(x) - log(log(x^2 - 6*x + 9))

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giac [A]  time = 0.19, size = 25, normalized size = 0.78 \begin {gather*} x^{2} {\left (\log \relax (2) + 5\right )} - \log \relax (x) - \log \left (\log \left (x^{2} - 6 \, x + 9\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x^3-6*x^2)*log(2)+10*x^3-30*x^2-x+3)*log(x^2-6*x+9)-2*x)/(x^2-3*x)/log(x^2-6*x+9),x, algorithm=
"giac")

[Out]

x^2*(log(2) + 5) - log(x) - log(log(x^2 - 6*x + 9))

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maple [A]  time = 0.26, size = 26, normalized size = 0.81

method	result	size
norman	\(\left (\ln \relax (2)+5\right ) x^{2}-\ln \relax (x )-\ln \left (\ln \left (x^{2}-6 x +9\right )\right )\)	\(26\)
default	\(5 x^{2}-\ln \relax (x )+x^{2} \ln \relax (2)-\ln \left (\ln \left (x^{2}-6 x +9\right )\right )\)	\(29\)
risch	\(5 x^{2}-\ln \relax (x )+x^{2} \ln \relax (2)-\ln \left (\ln \left (x^{2}-6 x +9\right )\right )\)	\(29\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((2*x^3-6*x^2)*ln(2)+10*x^3-30*x^2-x+3)*ln(x^2-6*x+9)-2*x)/(x^2-3*x)/ln(x^2-6*x+9),x,method=_RETURNVERBOS
E)

[Out]

(ln(2)+5)*x^2-ln(x)-ln(ln(x^2-6*x+9))

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maxima [B]  time = 0.48, size = 71, normalized size = 2.22 \begin {gather*} 5 \, x^{2} + {\left (x^{2} + 6 \, x + 18 \, \log \left (x - 3\right )\right )} \log \relax (2) - 6 \, {\left (x + 3 \, \log \left (x - 3\right )\right )} \log \relax (2) - \frac {1}{2} \, \log \left (x^{2} - 6 \, x + 9\right ) \log \left (\log \left (x - 3\right )\right ) + \log \left (x - 3\right ) \log \left (\log \left (x - 3\right )\right ) - \log \relax (x) - \log \left (\log \left (x - 3\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x^3-6*x^2)*log(2)+10*x^3-30*x^2-x+3)*log(x^2-6*x+9)-2*x)/(x^2-3*x)/log(x^2-6*x+9),x, algorithm=
"maxima")

[Out]

5*x^2 + (x^2 + 6*x + 18*log(x - 3))*log(2) - 6*(x + 3*log(x - 3))*log(2) - 1/2*log(x^2 - 6*x + 9)*log(log(x -
3)) + log(x - 3)*log(log(x - 3)) - log(x) - log(log(x - 3))

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mupad [B]  time = 0.33, size = 54, normalized size = 1.69 \begin {gather*} \frac {\left (\ln \relax (2)+5\right )\,x^4+\left (-\ln \left (64\right )-30\right )\,x^3+\left (\ln \left (512\right )+45\right )\,x^2}{x^2-6\,x+9}-\ln \left (\ln \left (x^2-6\,x+9\right )\right )-\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x + log(x^2 - 6*x + 9)*(x + log(2)*(6*x^2 - 2*x^3) + 30*x^2 - 10*x^3 - 3))/(log(x^2 - 6*x + 9)*(3*x - x
^2)),x)

[Out]

(x^4*(log(2) + 5) - x^3*(log(64) + 30) + x^2*(log(512) + 45))/(x^2 - 6*x + 9) - log(log(x^2 - 6*x + 9)) - log(
x)

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sympy [A]  time = 0.16, size = 22, normalized size = 0.69 \begin {gather*} x^{2} \left (\log {\relax (2 )} + 5\right ) - \log {\relax (x )} - \log {\left (\log {\left (x^{2} - 6 x + 9 \right )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x**3-6*x**2)*ln(2)+10*x**3-30*x**2-x+3)*ln(x**2-6*x+9)-2*x)/(x**2-3*x)/ln(x**2-6*x+9),x)

[Out]

x**2*(log(2) + 5) - log(x) - log(log(x**2 - 6*x + 9))

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