\(\int \frac {-2-4 x-5 x^2-10 x^3-\log (2)+(2+\log (2)) \log (x)}{5 x^2} \, dx\) [10223]

Optimal result

Integrand size = 33, antiderivative size = 30 \[ \int \frac {-2-4 x-5 x^2-10 x^3-\log (2)+(2+\log (2)) \log (x)}{5 x^2} \, dx=-4-x-x^2-\log (x)+\frac {(-2+x-\log (2)) \log (x)}{5 x} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.93, number of steps used = 6, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {12, 14, 2341} \[ \int \frac {-2-4 x-5 x^2-10 x^3-\log (2)+(2+\log (2)) \log (x)}{5 x^2} \, dx=-x^2-x-\frac {4 \log (x)}{5}-\frac {(2+\log (2)) \log (x)}{5 x} \]

[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 2341

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{align*} \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{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.17 \[ \int \frac {-2-4 x-5 x^2-10 x^3-\log (2)+(2+\log (2)) \log (x)}{5 x^2} \, dx=-x-x^2-\frac {4 \log (x)}{5}-\frac {2 \log (x)}{5 x}-\frac {\log (2) \log (x)}{5 x} \]

[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)

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.83

method	result	size
risch	\(-\frac {\left (\ln \left (2\right )+2\right ) \ln \left (x \right )}{5 x}-x^{2}-x -\frac {4 \ln \left (x \right )}{5}\)	\(25\)
norman	\(\frac {\left (-\frac {2}{5}-\frac {\ln \left (2\right )}{5}\right ) \ln \left (x \right )-\frac {4 x \ln \left (x \right )}{5}-x^{2}-x^{3}}{x}\)	\(30\)
parallelrisch	\(-\frac {5 x^{3}+\ln \left (2\right ) \ln \left (x \right )+5 x^{2}+4 x \ln \left (x \right )+2 \ln \left (x \right )}{5 x}\)	\(31\)
parts	\(-x -x^{2}+\frac {\ln \left (2\right )+2}{5 x}-\frac {4 \ln \left (x \right )}{5}+\left (\frac {2}{5}+\frac {\ln \left (2\right )}{5}\right ) \left (-\frac {\ln \left (x \right )}{x}-\frac {1}{x}\right )\)	\(43\)
default	\(-x^{2}+\frac {\ln \left (2\right ) \left (-\frac {\ln \left (x \right )}{x}-\frac {1}{x}\right )}{5}-x -\frac {2 \ln \left (x \right )}{5 x}+\frac {\ln \left (2\right )}{5 x}-\frac {4 \ln \left (x \right )}{5}\)	\(45\)

[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)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.87 \[ \int \frac {-2-4 x-5 x^2-10 x^3-\log (2)+(2+\log (2)) \log (x)}{5 x^2} \, dx=-\frac {5 \, x^{3} + 5 \, x^{2} + {\left (4 \, x + \log \left (2\right ) + 2\right )} \log \left (x\right )}{5 \, x} \]

[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

Sympy [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.80 \[ \int \frac {-2-4 x-5 x^2-10 x^3-\log (2)+(2+\log (2)) \log (x)}{5 x^2} \, dx=- x^{2} - x - \frac {4 \log {\left (x \right )}}{5} + \frac {\left (-2 - \log {\left (2 \right )}\right ) \log {\left (x \right )}}{5 x} \]

[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)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.37 \[ \int \frac {-2-4 x-5 x^2-10 x^3-\log (2)+(2+\log (2)) \log (x)}{5 x^2} \, dx=-x^{2} - \frac {1}{5} \, {\left (\frac {\log \left (x\right )}{x} + \frac {1}{x}\right )} \log \left (2\right ) - x + \frac {\log \left (2\right )}{5 \, x} - \frac {2 \, \log \left (x\right )}{5 \, x} - \frac {4}{5} \, \log \left (x\right ) \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.80 \[ \int \frac {-2-4 x-5 x^2-10 x^3-\log (2)+(2+\log (2)) \log (x)}{5 x^2} \, dx=-x^{2} - x - \frac {{\left (\log \left (2\right ) + 2\right )} \log \left (x\right )}{5 \, x} - \frac {4}{5} \, \log \left (x\right ) \]

[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)

Mupad [B] (verification not implemented)

Time = 15.42 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.80 \[ \int \frac {-2-4 x-5 x^2-10 x^3-\log (2)+(2+\log (2)) \log (x)}{5 x^2} \, dx=-x-\frac {4\,\ln \left (x\right )}{5}-x^2-\frac {\ln \left (x\right )\,\left (\ln \left (2\right )+2\right )}{5\,x} \]

[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)