3.34.75 \(\int \frac {(-2 x^3+10 x^4) \log ^2(\frac {4}{x})-8 x^3 \log ^2(\frac {4}{x}) \log (x)+e^{\frac {1}{2} (-10+e^{\frac {3}{\log (\frac {4}{x})}})} ((-2 x+6 x^2) \log ^2(\frac {4}{x})-4 x \log ^2(\frac {4}{x}) \log (x)+e^{\frac {3}{\log (\frac {4}{x})}} (3 x^2-3 x \log (x)))}{2 \log ^2(\frac {4}{x})} \, dx\)

Optimal. Leaf size=34 \[ x^2 \left (e^{-5+\frac {1}{2} e^{\frac {3}{\log \left (\frac {4}{x}\right )}}}+x^2\right ) (x-\log (x)) \]

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Rubi [A]  time = 0.68, antiderivative size = 41, normalized size of antiderivative = 1.21, number of steps used = 5, number of rules used = 4, integrand size = 125, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.032, Rules used = {12, 6688, 2288, 2304} \begin {gather*} x^5-x^4 \log (x)+x^2 e^{\frac {1}{2} \left (e^{\frac {3}{\log \left (\frac {4}{x}\right )}}-10\right )} (x-\log (x)) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((-2*x^3 + 10*x^4)*Log[4/x]^2 - 8*x^3*Log[4/x]^2*Log[x] + E^((-10 + E^(3/Log[4/x]))/2)*((-2*x + 6*x^2)*Log
[4/x]^2 - 4*x*Log[4/x]^2*Log[x] + E^(3/Log[4/x])*(3*x^2 - 3*x*Log[x])))/(2*Log[4/x]^2),x]

[Out]

x^5 + E^((-10 + E^(3/Log[4/x]))/2)*x^2*(x - Log[x]) - x^4*Log[x]

Rule 12

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

Rule 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rule 2304

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Log[c*x^
n]))/(d*(m + 1)), x] - Simp[(b*n*(d*x)^(m + 1))/(d*(m + 1)^2), x] /; FreeQ[{a, b, c, d, 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} &=\frac {1}{2} \int \frac {\left (-2 x^3+10 x^4\right ) \log ^2\left (\frac {4}{x}\right )-8 x^3 \log ^2\left (\frac {4}{x}\right ) \log (x)+e^{\frac {1}{2} \left (-10+e^{\frac {3}{\log \left (\frac {4}{x}\right )}}\right )} \left (\left (-2 x+6 x^2\right ) \log ^2\left (\frac {4}{x}\right )-4 x \log ^2\left (\frac {4}{x}\right ) \log (x)+e^{\frac {3}{\log \left (\frac {4}{x}\right )}} \left (3 x^2-3 x \log (x)\right )\right )}{\log ^2\left (\frac {4}{x}\right )} \, dx\\ &=\frac {1}{2} \int \left (-2 x^3+10 x^4+\frac {e^{\frac {1}{2} \left (-10+e^{\frac {3}{\log \left (\frac {4}{x}\right )}}\right )} x \left (\log ^2\left (\frac {4}{x}\right ) (-2+6 x-4 \log (x))+3 e^{\frac {3}{\log \left (\frac {4}{x}\right )}} (x-\log (x))\right )}{\log ^2\left (\frac {4}{x}\right )}-8 x^3 \log (x)\right ) \, dx\\ &=-\frac {x^4}{4}+x^5+\frac {1}{2} \int \frac {e^{\frac {1}{2} \left (-10+e^{\frac {3}{\log \left (\frac {4}{x}\right )}}\right )} x \left (\log ^2\left (\frac {4}{x}\right ) (-2+6 x-4 \log (x))+3 e^{\frac {3}{\log \left (\frac {4}{x}\right )}} (x-\log (x))\right )}{\log ^2\left (\frac {4}{x}\right )} \, dx-4 \int x^3 \log (x) \, dx\\ &=x^5+e^{\frac {1}{2} \left (-10+e^{\frac {3}{\log \left (\frac {4}{x}\right )}}\right )} x^2 (x-\log (x))-x^4 \log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.25, size = 41, normalized size = 1.21 \begin {gather*} x^5+e^{-5+\frac {1}{2} e^{\frac {3}{\log \left (\frac {4}{x}\right )}}} x^2 (x-\log (x))-x^4 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((-2*x^3 + 10*x^4)*Log[4/x]^2 - 8*x^3*Log[4/x]^2*Log[x] + E^((-10 + E^(3/Log[4/x]))/2)*((-2*x + 6*x^
2)*Log[4/x]^2 - 4*x*Log[4/x]^2*Log[x] + E^(3/Log[4/x])*(3*x^2 - 3*x*Log[x])))/(2*Log[4/x]^2),x]

[Out]

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

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fricas [A]  time = 1.04, size = 59, normalized size = 1.74 \begin {gather*} x^{5} - 2 \, x^{4} \log \relax (2) + x^{4} \log \left (\frac {4}{x}\right ) + {\left (x^{3} - 2 \, x^{2} \log \relax (2) + x^{2} \log \left (\frac {4}{x}\right )\right )} e^{\left (\frac {1}{2} \, e^{\frac {3}{\log \left (\frac {4}{x}\right )}} - 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(((-3*x*log(x)+3*x^2)*exp(3/log(4/x))-4*x*log(4/x)^2*log(x)+(6*x^2-2*x)*log(4/x)^2)*exp(1/2*exp(
3/log(4/x))-5)-8*x^3*log(4/x)^2*log(x)+(10*x^4-2*x^3)*log(4/x)^2)/log(4/x)^2,x, algorithm="fricas")

[Out]

x^5 - 2*x^4*log(2) + x^4*log(4/x) + (x^3 - 2*x^2*log(2) + x^2*log(4/x))*e^(1/2*e^(3/log(4/x)) - 5)

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giac [A]  time = 0.75, size = 60, normalized size = 1.76 \begin {gather*} x^{5} - x^{4} \log \relax (x) + x^{3} e^{\left (\frac {1}{2} \, e^{\left (\frac {3}{2 \, \log \relax (2) - \log \relax (x)}\right )} - 5\right )} - x^{2} e^{\left (\frac {1}{2} \, e^{\left (\frac {3}{2 \, \log \relax (2) - \log \relax (x)}\right )} - 5\right )} \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(((-3*x*log(x)+3*x^2)*exp(3/log(4/x))-4*x*log(4/x)^2*log(x)+(6*x^2-2*x)*log(4/x)^2)*exp(1/2*exp(
3/log(4/x))-5)-8*x^3*log(4/x)^2*log(x)+(10*x^4-2*x^3)*log(4/x)^2)/log(4/x)^2,x, algorithm="giac")

[Out]

x^5 - x^4*log(x) + x^3*e^(1/2*e^(3/(2*log(2) - log(x))) - 5) - x^2*e^(1/2*e^(3/(2*log(2) - log(x))) - 5)*log(x
)

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maple [A]  time = 0.66, size = 41, normalized size = 1.21

method	result	size
risch	\(x^{5}-x^{4} \ln \relax (x )+x^{2} \left (x -\ln \relax (x )\right ) {\mathrm e}^{\frac {{\mathrm e}^{\frac {3}{2 \ln \relax (2)-\ln \relax (x )}}}{2}-5}\)	\(41\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/2*(((-3*x*ln(x)+3*x^2)*exp(3/ln(4/x))-4*x*ln(4/x)^2*ln(x)+(6*x^2-2*x)*ln(4/x)^2)*exp(1/2*exp(3/ln(4/x))-
5)-8*x^3*ln(4/x)^2*ln(x)+(10*x^4-2*x^3)*ln(4/x)^2)/ln(4/x)^2,x,method=_RETURNVERBOSE)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} x^{5} - x^{4} \log \relax (x) - \frac {1}{2} \, \int -\frac {{\left (3 \, x^{2} + 2 \, {\left (12 \, x^{2} \log \relax (2)^{2} - 2 \, x \log \relax (x)^{3} - 4 \, x \log \relax (2)^{2} + {\left (3 \, x^{2} + x {\left (8 \, \log \relax (2) - 1\right )}\right )} \log \relax (x)^{2} - 4 \, {\left (3 \, x^{2} \log \relax (2) + {\left (2 \, \log \relax (2)^{2} - \log \relax (2)\right )} x\right )} \log \relax (x)\right )} e^{\left (-\frac {3}{2 \, \log \relax (2) - \log \relax (x)}\right )} - 3 \, x \log \relax (x)\right )} e^{\left (\frac {3}{2 \, \log \relax (2) - \log \relax (x)} + \frac {1}{2} \, e^{\left (\frac {3}{2 \, \log \relax (2) - \log \relax (x)}\right )}\right )}}{4 \, e^{5} \log \relax (2)^{2} - 4 \, e^{5} \log \relax (2) \log \relax (x) + e^{5} \log \relax (x)^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(((-3*x*log(x)+3*x^2)*exp(3/log(4/x))-4*x*log(4/x)^2*log(x)+(6*x^2-2*x)*log(4/x)^2)*exp(1/2*exp(
3/log(4/x))-5)-8*x^3*log(4/x)^2*log(x)+(10*x^4-2*x^3)*log(4/x)^2)/log(4/x)^2,x, algorithm="maxima")

[Out]

x^5 - x^4*log(x) - 1/2*integrate(-(3*x^2 + 2*(12*x^2*log(2)^2 - 2*x*log(x)^3 - 4*x*log(2)^2 + (3*x^2 + x*(8*lo
g(2) - 1))*log(x)^2 - 4*(3*x^2*log(2) + (2*log(2)^2 - log(2))*x)*log(x))*e^(-3/(2*log(2) - log(x))) - 3*x*log(
x))*e^(3/(2*log(2) - log(x)) + 1/2*e^(3/(2*log(2) - log(x))))/(4*e^5*log(2)^2 - 4*e^5*log(2)*log(x) + e^5*log(
x)^2), x)

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mupad [B]  time = 2.79, size = 41, normalized size = 1.21 \begin {gather*} x^5-x^4\,\ln \relax (x)-{\mathrm {e}}^{\frac {{\mathrm {e}}^{\frac {3}{\ln \left (\frac {4}{x}\right )}}}{2}-5}\,\left (x^2\,\ln \relax (x)-x^3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-((log(4/x)^2*(2*x^3 - 10*x^4))/2 + (exp(exp(3/log(4/x))/2 - 5)*(exp(3/log(4/x))*(3*x*log(x) - 3*x^2) + lo
g(4/x)^2*(2*x - 6*x^2) + 4*x*log(4/x)^2*log(x)))/2 + 4*x^3*log(4/x)^2*log(x))/log(4/x)^2,x)

[Out]

x^5 - x^4*log(x) - exp(exp(3/log(4/x))/2 - 5)*(x^2*log(x) - x^3)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(((-3*x*ln(x)+3*x**2)*exp(3/ln(4/x))-4*x*ln(4/x)**2*ln(x)+(6*x**2-2*x)*ln(4/x)**2)*exp(1/2*exp(3
/ln(4/x))-5)-8*x**3*ln(4/x)**2*ln(x)+(10*x**4-2*x**3)*ln(4/x)**2)/ln(4/x)**2,x)

[Out]

Timed out

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