\(\int \frac {-100 x^4+50 x^5-6 x^6+(1-2 x^2) \log (2)}{x} \, dx\) [7423]

Optimal result

Integrand size = 30, antiderivative size = 27 \[ \int \frac {-100 x^4+50 x^5-6 x^6+\left (1-2 x^2\right ) \log (2)}{x} \, dx=e^5-(5-x)^2 x^4+\log (2) \left (-x^2+\log (x)\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.04, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.033, Rules used = {14} \[ \int \frac {-100 x^4+50 x^5-6 x^6+\left (1-2 x^2\right ) \log (2)}{x} \, dx=-x^6+10 x^5-25 x^4-x^2 \log (2)+\log (2) \log (x) \]

[In]

Int[(-100*x^4 + 50*x^5 - 6*x^6 + (1 - 2*x^2)*Log[2])/x,x]

[Out]

-25*x^4 + 10*x^5 - x^6 - x^2*Log[2] + Log[2]*Log[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]]

Rubi steps \begin{align*} \text {integral}& = \int \left (-100 x^3+50 x^4-6 x^5+\frac {\log (2)}{x}-2 x \log (2)\right ) \, dx \\ & = -25 x^4+10 x^5-x^6-x^2 \log (2)+\log (2) \log (x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.11 \[ \int \frac {-100 x^4+50 x^5-6 x^6+\left (1-2 x^2\right ) \log (2)}{x} \, dx=-25 x^4+10 x^5-x^6-\frac {1}{2} x^2 \log (4)+\log (2) \log (x) \]

[In]

Integrate[(-100*x^4 + 50*x^5 - 6*x^6 + (1 - 2*x^2)*Log[2])/x,x]

[Out]

-25*x^4 + 10*x^5 - x^6 - (x^2*Log[4])/2 + Log[2]*Log[x]

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.07

method	result	size
default	\(-x^{6}+10 x^{5}-25 x^{4}-x^{2} \ln \left (2\right )+\ln \left (2\right ) \ln \left (x \right )\)	\(29\)
norman	\(-x^{6}+10 x^{5}-25 x^{4}-x^{2} \ln \left (2\right )+\ln \left (2\right ) \ln \left (x \right )\)	\(29\)
risch	\(-x^{6}+10 x^{5}-25 x^{4}-x^{2} \ln \left (2\right )+\ln \left (2\right ) \ln \left (x \right )\)	\(29\)
parallelrisch	\(-x^{6}+10 x^{5}-25 x^{4}-x^{2} \ln \left (2\right )+\ln \left (2\right ) \ln \left (x \right )\)	\(29\)

[In]

int(((-2*x^2+1)*ln(2)-6*x^6+50*x^5-100*x^4)/x,x,method=_RETURNVERBOSE)

[Out]

-x^6+10*x^5-25*x^4-x^2*ln(2)+ln(2)*ln(x)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.04 \[ \int \frac {-100 x^4+50 x^5-6 x^6+\left (1-2 x^2\right ) \log (2)}{x} \, dx=-x^{6} + 10 \, x^{5} - 25 \, x^{4} - x^{2} \log \left (2\right ) + \log \left (2\right ) \log \left (x\right ) \]

[In]

integrate(((-2*x^2+1)*log(2)-6*x^6+50*x^5-100*x^4)/x,x, algorithm="fricas")

[Out]

-x^6 + 10*x^5 - 25*x^4 - x^2*log(2) + log(2)*log(x)

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \[ \int \frac {-100 x^4+50 x^5-6 x^6+\left (1-2 x^2\right ) \log (2)}{x} \, dx=- x^{6} + 10 x^{5} - 25 x^{4} - x^{2} \log {\left (2 \right )} + \log {\left (2 \right )} \log {\left (x \right )} \]

[In]

integrate(((-2*x**2+1)*ln(2)-6*x**6+50*x**5-100*x**4)/x,x)

[Out]

-x**6 + 10*x**5 - 25*x**4 - x**2*log(2) + log(2)*log(x)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.04 \[ \int \frac {-100 x^4+50 x^5-6 x^6+\left (1-2 x^2\right ) \log (2)}{x} \, dx=-x^{6} + 10 \, x^{5} - 25 \, x^{4} - x^{2} \log \left (2\right ) + \log \left (2\right ) \log \left (x\right ) \]

[In]

integrate(((-2*x^2+1)*log(2)-6*x^6+50*x^5-100*x^4)/x,x, algorithm="maxima")

[Out]

-x^6 + 10*x^5 - 25*x^4 - x^2*log(2) + log(2)*log(x)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.07 \[ \int \frac {-100 x^4+50 x^5-6 x^6+\left (1-2 x^2\right ) \log (2)}{x} \, dx=-x^{6} + 10 \, x^{5} - 25 \, x^{4} - x^{2} \log \left (2\right ) + \log \left (2\right ) \log \left ({\left | x \right |}\right ) \]

[In]

integrate(((-2*x^2+1)*log(2)-6*x^6+50*x^5-100*x^4)/x,x, algorithm="giac")

[Out]

-x^6 + 10*x^5 - 25*x^4 - x^2*log(2) + log(2)*log(abs(x))

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.04 \[ \int \frac {-100 x^4+50 x^5-6 x^6+\left (1-2 x^2\right ) \log (2)}{x} \, dx=\ln \left (2\right )\,\ln \left (x\right )-x^2\,\ln \left (2\right )-25\,x^4+10\,x^5-x^6 \]

[In]

int(-(log(2)*(2*x^2 - 1) + 100*x^4 - 50*x^5 + 6*x^6)/x,x)

[Out]

log(2)*log(x) - x^2*log(2) - 25*x^4 + 10*x^5 - x^6