3.103.23 \(\int \frac {-2-4 x-5 x^2-10 x^3-\log (2)+(2+\log (2)) \log (x)}{5 x^2} \, dx\)

Optimal. Leaf size=30 \[ -4-x-x^2-\log (x)+\frac {(-2+x-\log (2)) \log (x)}{5 x} \]

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Rubi [A]  time = 0.04, antiderivative size = 28, normalized size of antiderivative = 0.93, number of steps used = 6, number of rules used = 3, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {12, 14, 2304} \begin {gather*} -x^2-x-\frac {4 \log (x)}{5}-\frac {(2+\log (2)) \log (x)}{5 x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-x - x^2 - (4*Log[x])/5 - ((2 + Log[2])*Log[x])/(5*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 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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
]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{5} \int \frac {-2-4 x-5 x^2-10 x^3-\log (2)+(2+\log (2)) \log (x)}{x^2} \, dx\\ &=\frac {1}{5} \int \left (\frac {-2-4 x-5 x^2-10 x^3-\log (2)}{x^2}+\frac {(2+\log (2)) \log (x)}{x^2}\right ) \, dx\\ &=\frac {1}{5} \int \frac {-2-4 x-5 x^2-10 x^3-\log (2)}{x^2} \, dx+\frac {1}{5} (2+\log (2)) \int \frac {\log (x)}{x^2} \, dx\\ &=-\frac {2+\log (2)}{5 x}-\frac {(2+\log (2)) \log (x)}{5 x}+\frac {1}{5} \int \left (-5-\frac {4}{x}-10 x+\frac {-2-\log (2)}{x^2}\right ) \, dx\\ &=-x-x^2-\frac {4 \log (x)}{5}-\frac {(2+\log (2)) \log (x)}{5 x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 35, normalized size = 1.17 \begin {gather*} -x-x^2-\frac {4 \log (x)}{5}-\frac {2 \log (x)}{5 x}-\frac {\log (2) \log (x)}{5 x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.55, size = 26, normalized size = 0.87 \begin {gather*} -\frac {5 \, x^{3} + 5 \, x^{2} + {\left (4 \, x + \log \relax (2) + 2\right )} \log \relax (x)}{5 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.24, size = 24, normalized size = 0.80 \begin {gather*} -x^{2} - x - \frac {{\left (\log \relax (2) + 2\right )} \log \relax (x)}{5 \, x} - \frac {4}{5} \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.03, size = 25, normalized size = 0.83

method	result	size
risch	\(-\frac {\left (\ln \relax (2)+2\right ) \ln \relax (x )}{5 x}-x^{2}-x -\frac {4 \ln \relax (x )}{5}\)	\(25\)
norman	\(\frac {\left (-\frac {2}{5}-\frac {\ln \relax (2)}{5}\right ) \ln \relax (x )-\frac {4 x \ln \relax (x )}{5}-x^{2}-x^{3}}{x}\)	\(30\)
default	\(-x^{2}+\frac {\ln \relax (2) \left (-\frac {\ln \relax (x )}{x}-\frac {1}{x}\right )}{5}-x -\frac {2 \ln \relax (x )}{5 x}+\frac {\ln \relax (2)}{5 x}-\frac {4 \ln \relax (x )}{5}\)	\(45\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.41, size = 41, normalized size = 1.37 \begin {gather*} -x^{2} - \frac {1}{5} \, {\left (\frac {\log \relax (x)}{x} + \frac {1}{x}\right )} \log \relax (2) - x + \frac {\log \relax (2)}{5 \, x} - \frac {2 \, \log \relax (x)}{5 \, x} - \frac {4}{5} \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 7.10, size = 24, normalized size = 0.80 \begin {gather*} -x-\frac {4\,\ln \relax (x)}{5}-x^2-\frac {\ln \relax (x)\,\left (\ln \relax (2)+2\right )}{5\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-((4*x)/5 + log(2)/5 - (log(x)*(log(2) + 2))/5 + x^2 + 2*x^3 + 2/5)/x^2,x)

[Out]

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

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sympy [A]  time = 0.16, size = 24, normalized size = 0.80 \begin {gather*} - x^{2} - x - \frac {4 \log {\relax (x )}}{5} + \frac {\left (-2 - \log {\relax (2 )}\right ) \log {\relax (x )}}{5 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x**2 - x - 4*log(x)/5 + (-2 - log(2))*log(x)/(5*x)

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