3.89.29 \(\int \frac {(-256 x^5+e^{\frac {-1+x-2 x^2+2 \log (5)}{2 x}} (2-4 x^2-4 \log (5))) \log ^3(e^{\frac {-1+x-2 x^2+2 \log (5)}{2 x}}-16 x^4)}{e^{\frac {-1+x-2 x^2+2 \log (5)}{2 x}} x^2-16 x^6} \, dx\)

Optimal. Leaf size=29 \[ \log ^4\left (e^{-x+\frac {-1+x+2 \log (5)}{2 x}}-16 x^4\right ) \]

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Rubi [A]  time = 0.54, antiderivative size = 34, normalized size of antiderivative = 1.17, number of steps used = 1, number of rules used = 1, integrand size = 103, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.010, Rules used = {6686} \begin {gather*} \log ^4\left (5^{\frac {1}{x}} e^{-\frac {2 x^2-x+1}{2 x}}-16 x^4\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((-256*x^5 + E^((-1 + x - 2*x^2 + 2*Log[5])/(2*x))*(2 - 4*x^2 - 4*Log[5]))*Log[E^((-1 + x - 2*x^2 + 2*Log[
5])/(2*x)) - 16*x^4]^3)/(E^((-1 + x - 2*x^2 + 2*Log[5])/(2*x))*x^2 - 16*x^6),x]

[Out]

Log[5^x^(-1)/E^((1 - x + 2*x^2)/(2*x)) - 16*x^4]^4

Rule 6686

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\log ^4\left (5^{\frac {1}{x}} e^{-\frac {1-x+2 x^2}{2 x}}-16 x^4\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.10, size = 28, normalized size = 0.97 \begin {gather*} \log ^4\left (e^{\frac {-1+x-2 x^2+\log (25)}{2 x}}-16 x^4\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((-256*x^5 + E^((-1 + x - 2*x^2 + 2*Log[5])/(2*x))*(2 - 4*x^2 - 4*Log[5]))*Log[E^((-1 + x - 2*x^2 +
2*Log[5])/(2*x)) - 16*x^4]^3)/(E^((-1 + x - 2*x^2 + 2*Log[5])/(2*x))*x^2 - 16*x^6),x]

[Out]

Log[E^((-1 + x - 2*x^2 + Log[25])/(2*x)) - 16*x^4]^4

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fricas [A]  time = 0.56, size = 29, normalized size = 1.00 \begin {gather*} \log \left (-16 \, x^{4} + e^{\left (-\frac {2 \, x^{2} - x - 2 \, \log \relax (5) + 1}{2 \, x}\right )}\right )^{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*log(5)-4*x^2+2)*exp(1/2*(2*log(5)-2*x^2+x-1)/x)-256*x^5)*log(exp(1/2*(2*log(5)-2*x^2+x-1)/x)-16
*x^4)^3/(x^2*exp(1/2*(2*log(5)-2*x^2+x-1)/x)-16*x^6),x, algorithm="fricas")

[Out]

log(-16*x^4 + e^(-1/2*(2*x^2 - x - 2*log(5) + 1)/x))^4

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {2 \, {\left (128 \, x^{5} + {\left (2 \, x^{2} + 2 \, \log \relax (5) - 1\right )} e^{\left (-\frac {2 \, x^{2} - x - 2 \, \log \relax (5) + 1}{2 \, x}\right )}\right )} \log \left (-16 \, x^{4} + e^{\left (-\frac {2 \, x^{2} - x - 2 \, \log \relax (5) + 1}{2 \, x}\right )}\right )^{3}}{16 \, x^{6} - x^{2} e^{\left (-\frac {2 \, x^{2} - x - 2 \, \log \relax (5) + 1}{2 \, x}\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*log(5)-4*x^2+2)*exp(1/2*(2*log(5)-2*x^2+x-1)/x)-256*x^5)*log(exp(1/2*(2*log(5)-2*x^2+x-1)/x)-16
*x^4)^3/(x^2*exp(1/2*(2*log(5)-2*x^2+x-1)/x)-16*x^6),x, algorithm="giac")

[Out]

integrate(2*(128*x^5 + (2*x^2 + 2*log(5) - 1)*e^(-1/2*(2*x^2 - x - 2*log(5) + 1)/x))*log(-16*x^4 + e^(-1/2*(2*
x^2 - x - 2*log(5) + 1)/x))^3/(16*x^6 - x^2*e^(-1/2*(2*x^2 - x - 2*log(5) + 1)/x)), x)

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maple [A]  time = 0.06, size = 30, normalized size = 1.03

method	result	size
risch	\(\ln \left ({\mathrm e}^{\frac {2 \ln \relax (5)-2 x^{2}+x -1}{2 x}}-16 x^{4}\right )^{4}-\frac {3}{2}\)	\(30\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-4*ln(5)-4*x^2+2)*exp(1/2*(2*ln(5)-2*x^2+x-1)/x)-256*x^5)*ln(exp(1/2*(2*ln(5)-2*x^2+x-1)/x)-16*x^4)^3/(x
^2*exp(1/2*(2*ln(5)-2*x^2+x-1)/x)-16*x^6),x,method=_RETURNVERBOSE)

[Out]

ln(exp(1/2*(2*ln(5)-2*x^2+x-1)/x)-16*x^4)^4-3/2

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maxima [B]  time = 0.77, size = 574, normalized size = 19.79 \begin {gather*} -2 \, {\left (\frac {2 \, x^{2} + 1}{x} - 2 \, \log \left (-{\left (16 \, x^{4} e^{\left (x + \frac {1}{2 \, x}\right )} - e^{\left (\frac {\log \relax (5)}{x} + \frac {1}{2}\right )}\right )} e^{\left (-\frac {1}{2}\right )}\right )\right )} \log \left (-16 \, x^{4} + e^{\left (-\frac {2 \, x^{2} - x - 2 \, \log \relax (5) + 1}{2 \, x}\right )}\right )^{3} - \frac {3 \, {\left (4 \, x^{4} + 4 \, x^{2} \log \left (-16 \, x^{4} e^{\left (x + \frac {1}{2 \, x}\right )} + e^{\left (\frac {\log \relax (5)}{x} + \frac {1}{2}\right )}\right )^{2} + 4 \, x^{3} - 4 \, {\left (2 \, x^{3} + x^{2} + x\right )} \log \left (-16 \, x^{4} e^{\left (x + \frac {1}{2 \, x}\right )} + e^{\left (\frac {\log \relax (5)}{x} + \frac {1}{2}\right )}\right ) + 2 \, x + 1\right )} \log \left (-16 \, x^{4} + e^{\left (-\frac {2 \, x^{2} - x - 2 \, \log \relax (5) + 1}{2 \, x}\right )}\right )^{2}}{2 \, x^{2}} - \frac {{\left (8 \, x^{6} - 8 \, x^{3} \log \left (-16 \, x^{4} e^{\left (x + \frac {1}{2 \, x}\right )} + e^{\left (\frac {\log \relax (5)}{x} + \frac {1}{2}\right )}\right )^{3} + 12 \, x^{5} - 12 \, x^{4} + 12 \, {\left (2 \, x^{4} + x^{3} + x^{2}\right )} \log \left (-16 \, x^{4} e^{\left (x + \frac {1}{2 \, x}\right )} + e^{\left (\frac {\log \relax (5)}{x} + \frac {1}{2}\right )}\right )^{2} - 6 \, x^{2} - 6 \, {\left (4 \, x^{5} + 4 \, x^{4} + 2 \, x^{2} + x\right )} \log \left (-16 \, x^{4} e^{\left (x + \frac {1}{2 \, x}\right )} + e^{\left (\frac {\log \relax (5)}{x} + \frac {1}{2}\right )}\right ) + 3 \, x + 1\right )} \log \left (-16 \, x^{4} + e^{\left (-\frac {2 \, x^{2} - x - 2 \, \log \relax (5) + 1}{2 \, x}\right )}\right )}{2 \, x^{3}} - \frac {16 \, x^{8} + 16 \, x^{4} \log \left (-16 \, x^{4} e^{\left (x + \frac {1}{2 \, x}\right )} + e^{\left (\frac {\log \relax (5)}{x} + \frac {1}{2}\right )}\right )^{4} + 32 \, x^{7} - 64 \, x^{6} - 48 \, x^{5} - 32 \, {\left (2 \, x^{5} + x^{4} + x^{3}\right )} \log \left (-16 \, x^{4} e^{\left (x + \frac {1}{2 \, x}\right )} + e^{\left (\frac {\log \relax (5)}{x} + \frac {1}{2}\right )}\right )^{3} - 24 \, x^{3} + 24 \, {\left (4 \, x^{6} + 4 \, x^{5} + 2 \, x^{3} + x^{2}\right )} \log \left (-16 \, x^{4} e^{\left (x + \frac {1}{2 \, x}\right )} + e^{\left (\frac {\log \relax (5)}{x} + \frac {1}{2}\right )}\right )^{2} - 16 \, x^{2} - 8 \, {\left (8 \, x^{7} + 12 \, x^{6} - 12 \, x^{5} - 6 \, x^{3} + 3 \, x^{2} + x\right )} \log \left (-16 \, x^{4} e^{\left (x + \frac {1}{2 \, x}\right )} + e^{\left (\frac {\log \relax (5)}{x} + \frac {1}{2}\right )}\right ) + 4 \, x + 1}{16 \, x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*log(5)-4*x^2+2)*exp(1/2*(2*log(5)-2*x^2+x-1)/x)-256*x^5)*log(exp(1/2*(2*log(5)-2*x^2+x-1)/x)-16
*x^4)^3/(x^2*exp(1/2*(2*log(5)-2*x^2+x-1)/x)-16*x^6),x, algorithm="maxima")

[Out]

-2*((2*x^2 + 1)/x - 2*log(-(16*x^4*e^(x + 1/2/x) - e^(log(5)/x + 1/2))*e^(-1/2)))*log(-16*x^4 + e^(-1/2*(2*x^2
- x - 2*log(5) + 1)/x))^3 - 3/2*(4*x^4 + 4*x^2*log(-16*x^4*e^(x + 1/2/x) + e^(log(5)/x + 1/2))^2 + 4*x^3 - 4*
(2*x^3 + x^2 + x)*log(-16*x^4*e^(x + 1/2/x) + e^(log(5)/x + 1/2)) + 2*x + 1)*log(-16*x^4 + e^(-1/2*(2*x^2 - x
- 2*log(5) + 1)/x))^2/x^2 - 1/2*(8*x^6 - 8*x^3*log(-16*x^4*e^(x + 1/2/x) + e^(log(5)/x + 1/2))^3 + 12*x^5 - 12
*x^4 + 12*(2*x^4 + x^3 + x^2)*log(-16*x^4*e^(x + 1/2/x) + e^(log(5)/x + 1/2))^2 - 6*x^2 - 6*(4*x^5 + 4*x^4 + 2
*x^2 + x)*log(-16*x^4*e^(x + 1/2/x) + e^(log(5)/x + 1/2)) + 3*x + 1)*log(-16*x^4 + e^(-1/2*(2*x^2 - x - 2*log(
5) + 1)/x))/x^3 - 1/16*(16*x^8 + 16*x^4*log(-16*x^4*e^(x + 1/2/x) + e^(log(5)/x + 1/2))^4 + 32*x^7 - 64*x^6 -
48*x^5 - 32*(2*x^5 + x^4 + x^3)*log(-16*x^4*e^(x + 1/2/x) + e^(log(5)/x + 1/2))^3 - 24*x^3 + 24*(4*x^6 + 4*x^5
+ 2*x^3 + x^2)*log(-16*x^4*e^(x + 1/2/x) + e^(log(5)/x + 1/2))^2 - 16*x^2 - 8*(8*x^7 + 12*x^6 - 12*x^5 - 6*x^
3 + 3*x^2 + x)*log(-16*x^4*e^(x + 1/2/x) + e^(log(5)/x + 1/2)) + 4*x + 1)/x^4

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int -\frac {{\ln \left ({\mathrm {e}}^{\frac {-x^2+\frac {x}{2}+\ln \relax (5)-\frac {1}{2}}{x}}-16\,x^4\right )}^3\,\left ({\mathrm {e}}^{\frac {-x^2+\frac {x}{2}+\ln \relax (5)-\frac {1}{2}}{x}}\,\left (4\,x^2+4\,\ln \relax (5)-2\right )+256\,x^5\right )}{x^2\,{\mathrm {e}}^{\frac {-x^2+\frac {x}{2}+\ln \relax (5)-\frac {1}{2}}{x}}-16\,x^6} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(exp((x/2 + log(5) - x^2 - 1/2)/x) - 16*x^4)^3*(exp((x/2 + log(5) - x^2 - 1/2)/x)*(4*log(5) + 4*x^2 -
2) + 256*x^5))/(x^2*exp((x/2 + log(5) - x^2 - 1/2)/x) - 16*x^6),x)

[Out]

int(-(log(exp((x/2 + log(5) - x^2 - 1/2)/x) - 16*x^4)^3*(exp((x/2 + log(5) - x^2 - 1/2)/x)*(4*log(5) + 4*x^2 -
2) + 256*x^5))/(x^2*exp((x/2 + log(5) - x^2 - 1/2)/x) - 16*x^6), x)

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sympy [A]  time = 0.63, size = 24, normalized size = 0.83 \begin {gather*} \log {\left (- 16 x^{4} + e^{\frac {- x^{2} + \frac {x}{2} - \frac {1}{2} + \log {\relax (5 )}}{x}} \right )}^{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*ln(5)-4*x**2+2)*exp(1/2*(2*ln(5)-2*x**2+x-1)/x)-256*x**5)*ln(exp(1/2*(2*ln(5)-2*x**2+x-1)/x)-16
*x**4)**3/(x**2*exp(1/2*(2*ln(5)-2*x**2+x-1)/x)-16*x**6),x)

[Out]

log(-16*x**4 + exp((-x**2 + x/2 - 1/2 + log(5))/x))**4

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