3.51.86 \(\int \frac {2-33 x+16 x^2-2 x^3+(16 x-8 x^2+x^3) \log (3 e^{\frac {-1+(-8 x+2 x^2) \log (x)}{-8 x+2 x^2}} x)}{(16 x-8 x^2+x^3) \log ^2(3 e^{\frac {-1+(-8 x+2 x^2) \log (x)}{-8 x+2 x^2}} x)} \, dx\)

Optimal. Leaf size=26 \[ \frac {x}{\log \left (3 e^{\frac {1}{2 (4-x) x}} x^2\right )} \]

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Rubi [A]  time = 0.55, antiderivative size = 26, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 109, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.037, Rules used = {1594, 27, 6688, 6687} \begin {gather*} \frac {x}{\log \left (3 e^{\frac {1}{2 (4-x) x}} x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(2 - 33*x + 16*x^2 - 2*x^3 + (16*x - 8*x^2 + x^3)*Log[3*E^((-1 + (-8*x + 2*x^2)*Log[x])/(-8*x + 2*x^2))*x]
)/((16*x - 8*x^2 + x^3)*Log[3*E^((-1 + (-8*x + 2*x^2)*Log[x])/(-8*x + 2*x^2))*x]^2),x]

[Out]

x/Log[3*E^(1/(2*(4 - x)*x))*x^2]

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

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 6687

Int[(u_)*(y_)^(m_.)*(z_)^(n_.), x_Symbol] :> With[{q = DerivativeDivides[y*z, u*z^(n - m), x]}, Simp[(q*y^(m +
1)*z^(m + 1))/(m + 1), x] /;  !FalseQ[q]] /; FreeQ[{m, n}, x] && NeQ[m, -1]

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 {2-33 x+16 x^2-2 x^3+\left (16 x-8 x^2+x^3\right ) \log \left (3 \exp \left (\frac {-1+\left (-8 x+2 x^2\right ) \log (x)}{-8 x+2 x^2}\right ) x\right )}{x \left (16-8 x+x^2\right ) \log ^2\left (3 \exp \left (\frac {-1+\left (-8 x+2 x^2\right ) \log (x)}{-8 x+2 x^2}\right ) x\right )} \, dx\\ &=\int \frac {2-33 x+16 x^2-2 x^3+\left (16 x-8 x^2+x^3\right ) \log \left (3 \exp \left (\frac {-1+\left (-8 x+2 x^2\right ) \log (x)}{-8 x+2 x^2}\right ) x\right )}{(-4+x)^2 x \log ^2\left (3 \exp \left (\frac {-1+\left (-8 x+2 x^2\right ) \log (x)}{-8 x+2 x^2}\right ) x\right )} \, dx\\ &=\int \frac {2-33 x+16 x^2-2 x^3+(-4+x)^2 x \log \left (3 e^{\frac {1}{8 x-2 x^2}} x^2\right )}{(4-x)^2 x \log ^2\left (3 e^{\frac {1}{(8-2 x) x}} x^2\right )} \, dx\\ &=\frac {x}{\log \left (3 e^{\frac {1}{2 (4-x) x}} x^2\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.26, size = 23, normalized size = 0.88 \begin {gather*} \frac {x}{\log \left (3 e^{\frac {1}{8 x-2 x^2}} x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2 - 33*x + 16*x^2 - 2*x^3 + (16*x - 8*x^2 + x^3)*Log[3*E^((-1 + (-8*x + 2*x^2)*Log[x])/(-8*x + 2*x^
2))*x])/((16*x - 8*x^2 + x^3)*Log[3*E^((-1 + (-8*x + 2*x^2)*Log[x])/(-8*x + 2*x^2))*x]^2),x]

[Out]

x/Log[3*E^(8*x - 2*x^2)^(-1)*x^2]

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fricas [A]  time = 1.21, size = 37, normalized size = 1.42 \begin {gather*} \frac {2 \, {\left (x^{3} - 4 \, x^{2}\right )}}{2 \, {\left (x^{2} - 4 \, x\right )} \log \relax (3) + 4 \, {\left (x^{2} - 4 \, x\right )} \log \relax (x) - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^3-8*x^2+16*x)*log(3*x*exp(((2*x^2-8*x)*log(x)-1)/(2*x^2-8*x)))-2*x^3+16*x^2-33*x+2)/(x^3-8*x^2+1
6*x)/log(3*x*exp(((2*x^2-8*x)*log(x)-1)/(2*x^2-8*x)))^2,x, algorithm="fricas")

[Out]

2*(x^3 - 4*x^2)/(2*(x^2 - 4*x)*log(3) + 4*(x^2 - 4*x)*log(x) - 1)

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giac [A]  time = 3.09, size = 39, normalized size = 1.50 \begin {gather*} \frac {2 \, {\left (x^{3} - 4 \, x^{2}\right )}}{2 \, x^{2} \log \relax (3) + 4 \, x^{2} \log \relax (x) - 8 \, x \log \relax (3) - 16 \, x \log \relax (x) - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^3-8*x^2+16*x)*log(3*x*exp(((2*x^2-8*x)*log(x)-1)/(2*x^2-8*x)))-2*x^3+16*x^2-33*x+2)/(x^3-8*x^2+1
6*x)/log(3*x*exp(((2*x^2-8*x)*log(x)-1)/(2*x^2-8*x)))^2,x, algorithm="giac")

[Out]

2*(x^3 - 4*x^2)/(2*x^2*log(3) + 4*x^2*log(x) - 8*x*log(3) - 16*x*log(x) - 1)

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maple [C]  time = 0.22, size = 251, normalized size = 9.65

method	result	size
risch	\(\frac {2 i x}{\pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i {\mathrm e}^{\frac {2 x^{2} \ln \relax (x )-8 x \ln \relax (x )-1}{2 \left (x -4\right ) x}}\right ) \mathrm {csgn}\left (i x \,{\mathrm e}^{\frac {2 x^{2} \ln \relax (x )-8 x \ln \relax (x )-1}{2 \left (x -4\right ) x}}\right )-\pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x \,{\mathrm e}^{\frac {2 x^{2} \ln \relax (x )-8 x \ln \relax (x )-1}{2 \left (x -4\right ) x}}\right )^{2}-\pi \,\mathrm {csgn}\left (i {\mathrm e}^{\frac {2 x^{2} \ln \relax (x )-8 x \ln \relax (x )-1}{2 \left (x -4\right ) x}}\right ) \mathrm {csgn}\left (i x \,{\mathrm e}^{\frac {2 x^{2} \ln \relax (x )-8 x \ln \relax (x )-1}{2 \left (x -4\right ) x}}\right )^{2}+\pi \mathrm {csgn}\left (i x \,{\mathrm e}^{\frac {2 x^{2} \ln \relax (x )-8 x \ln \relax (x )-1}{2 \left (x -4\right ) x}}\right )^{3}+2 i \ln \relax (3)+2 i \ln \relax (x )+2 i \ln \left ({\mathrm e}^{\frac {2 x^{2} \ln \relax (x )-8 x \ln \relax (x )-1}{2 \left (x -4\right ) x}}\right )}\)	\(251\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x^3-8*x^2+16*x)*ln(3*x*exp(((2*x^2-8*x)*ln(x)-1)/(2*x^2-8*x)))-2*x^3+16*x^2-33*x+2)/(x^3-8*x^2+16*x)/ln(
3*x*exp(((2*x^2-8*x)*ln(x)-1)/(2*x^2-8*x)))^2,x,method=_RETURNVERBOSE)

[Out]

2*I*x/(Pi*csgn(I*x)*csgn(I*exp(1/2*(2*x^2*ln(x)-8*x*ln(x)-1)/(x-4)/x))*csgn(I*x*exp(1/2*(2*x^2*ln(x)-8*x*ln(x)
-1)/(x-4)/x))-Pi*csgn(I*x)*csgn(I*x*exp(1/2*(2*x^2*ln(x)-8*x*ln(x)-1)/(x-4)/x))^2-Pi*csgn(I*exp(1/2*(2*x^2*ln(
x)-8*x*ln(x)-1)/(x-4)/x))*csgn(I*x*exp(1/2*(2*x^2*ln(x)-8*x*ln(x)-1)/(x-4)/x))^2+Pi*csgn(I*x*exp(1/2*(2*x^2*ln
(x)-8*x*ln(x)-1)/(x-4)/x))^3+2*I*ln(3)+2*I*ln(x)+2*I*ln(exp(1/2*(2*x^2*ln(x)-8*x*ln(x)-1)/(x-4)/x)))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^3-8*x^2+16*x)*log(3*x*exp(((2*x^2-8*x)*log(x)-1)/(2*x^2-8*x)))-2*x^3+16*x^2-33*x+2)/(x^3-8*x^2+1
6*x)/log(3*x*exp(((2*x^2-8*x)*log(x)-1)/(2*x^2-8*x)))^2,x, algorithm="maxima")

[Out]

-2*(2*x^6 - 24*x^5 - 2*(x^6 - 12*x^5 + 48*x^4 - 64*x^3)*x^3 + 96*x^4 + 16*(x^6 - 12*x^5 + 48*x^4 - 64*x^3)*x^2
- 128*x^3 - 33*(x^6 - 12*x^5 + 48*x^4 - 64*x^3)*x)/((4*x^5*log(3) - 48*x^4*log(3) + 2*x^3*(97*log(3) - 1) - 4
*x^2*(67*log(3) - 4) + x*(16*log(3) - 33) + 2)*x^3 - 8*(4*x^5*log(3) - 48*x^4*log(3) + 2*x^3*(97*log(3) - 1) -
4*x^2*(67*log(3) - 4) + x*(16*log(3) - 33) + 2)*x^2 + 16*(4*x^5*log(3) - 48*x^4*log(3) + 2*x^3*(97*log(3) - 1
) - 4*x^2*(67*log(3) - 4) + x*(16*log(3) - 33) + 2)*x + 4*((2*x^5 - 24*x^4 + 97*x^3 - 134*x^2 + 8*x)*x^3 - 8*(
2*x^5 - 24*x^4 + 97*x^3 - 134*x^2 + 8*x)*x^2 + 16*(2*x^5 - 24*x^4 + 97*x^3 - 134*x^2 + 8*x)*x)*log(x)) - integ
rate(2*(16*x^6 + (2*x^6 - 16*x^5 + 31*x^4 + 8*x^3 - 20*x^2 + 16*x)*x^4 - 176*x^5 - 12*(2*x^6 - 16*x^5 + 31*x^4
+ 8*x^3 - 20*x^2 + 16*x)*x^3 + 704*x^4 - (2*x^8 - 40*x^7 + 223*x^6 - 504*x^5 + 1056*x^4 - 2432*x^3 + 960*x^2
- 768*x)*x^2 - 1280*x^3 - 4*(35*x^6 - 289*x^5 + 628*x^4 - 112*x^3 - 128*x^2 + 256*x)*x + 1024*x^2)/((8*x^8*log
(3) - 160*x^7*log(3) + 4*x^6*(322*log(3) - 1) - 16*x^5*(327*log(3) - 4) + 2*x^4*(5409*log(3) - 194) - 8*x^3*(1
186*log(3) - 133) + x^2*(1064*log(3) - 1153) - 4*x*(8*log(3) - 33) - 4)*x^4 - 12*(8*x^8*log(3) - 160*x^7*log(3
) + 4*x^6*(322*log(3) - 1) - 16*x^5*(327*log(3) - 4) + 2*x^4*(5409*log(3) - 194) - 8*x^3*(1186*log(3) - 133) +
x^2*(1064*log(3) - 1153) - 4*x*(8*log(3) - 33) - 4)*x^3 + 48*(8*x^8*log(3) - 160*x^7*log(3) + 4*x^6*(322*log(
3) - 1) - 16*x^5*(327*log(3) - 4) + 2*x^4*(5409*log(3) - 194) - 8*x^3*(1186*log(3) - 133) + x^2*(1064*log(3) -
1153) - 4*x*(8*log(3) - 33) - 4)*x^2 - 64*(8*x^8*log(3) - 160*x^7*log(3) + 4*x^6*(322*log(3) - 1) - 16*x^5*(3
27*log(3) - 4) + 2*x^4*(5409*log(3) - 194) - 8*x^3*(1186*log(3) - 133) + x^2*(1064*log(3) - 1153) - 4*x*(8*log
(3) - 33) - 4)*x + 4*((4*x^8 - 80*x^7 + 644*x^6 - 2616*x^5 + 5409*x^4 - 4744*x^3 + 532*x^2 - 16*x)*x^4 - 12*(4
*x^8 - 80*x^7 + 644*x^6 - 2616*x^5 + 5409*x^4 - 4744*x^3 + 532*x^2 - 16*x)*x^3 + 48*(4*x^8 - 80*x^7 + 644*x^6
- 2616*x^5 + 5409*x^4 - 4744*x^3 + 532*x^2 - 16*x)*x^2 - 64*(4*x^8 - 80*x^7 + 644*x^6 - 2616*x^5 + 5409*x^4 -
4744*x^3 + 532*x^2 - 16*x)*x)*log(x)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int \frac {\ln \left (3\,x\,{\mathrm {e}}^{\frac {\ln \relax (x)\,\left (8\,x-2\,x^2\right )+1}{8\,x-2\,x^2}}\right )\,\left (x^3-8\,x^2+16\,x\right )-33\,x+16\,x^2-2\,x^3+2}{{\ln \left (3\,x\,{\mathrm {e}}^{\frac {\ln \relax (x)\,\left (8\,x-2\,x^2\right )+1}{8\,x-2\,x^2}}\right )}^2\,\left (x^3-8\,x^2+16\,x\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(3*x*exp((log(x)*(8*x - 2*x^2) + 1)/(8*x - 2*x^2)))*(16*x - 8*x^2 + x^3) - 33*x + 16*x^2 - 2*x^3 + 2)/
(log(3*x*exp((log(x)*(8*x - 2*x^2) + 1)/(8*x - 2*x^2)))^2*(16*x - 8*x^2 + x^3)),x)

[Out]

int((log(3*x*exp((log(x)*(8*x - 2*x^2) + 1)/(8*x - 2*x^2)))*(16*x - 8*x^2 + x^3) - 33*x + 16*x^2 - 2*x^3 + 2)/
(log(3*x*exp((log(x)*(8*x - 2*x^2) + 1)/(8*x - 2*x^2)))^2*(16*x - 8*x^2 + x^3)), x)

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sympy [B]  time = 0.81, size = 37, normalized size = 1.42 \begin {gather*} \frac {2 x^{3} - 8 x^{2}}{2 x^{2} \log {\relax (3 )} - 8 x \log {\relax (3 )} + \left (4 x^{2} - 16 x\right ) \log {\relax (x )} - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x**3-8*x**2+16*x)*ln(3*x*exp(((2*x**2-8*x)*ln(x)-1)/(2*x**2-8*x)))-2*x**3+16*x**2-33*x+2)/(x**3-8*
x**2+16*x)/ln(3*x*exp(((2*x**2-8*x)*ln(x)-1)/(2*x**2-8*x)))**2,x)

[Out]

(2*x**3 - 8*x**2)/(2*x**2*log(3) - 8*x*log(3) + (4*x**2 - 16*x)*log(x) - 1)

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