\(\int \frac {2-5 x+20 x^4 \log ^2(4)}{x} \, dx\) [8662]

Optimal result

Integrand size = 18, antiderivative size = 20 \[ \int \frac {2-5 x+20 x^4 \log ^2(4)}{x} \, dx=5 \left (4-x+x^4 \log ^2(4)\right )+\log \left (x^2\right ) \]

[Out]

ln(x^2)+20*x^4*ln(2)^2-5*x+20

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.85, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {14} \[ \int \frac {2-5 x+20 x^4 \log ^2(4)}{x} \, dx=5 x^4 \log ^2(4)-5 x+2 \log (x) \]

[In]

Int[(2 - 5*x + 20*x^4*Log[4]^2)/x,x]

[Out]

-5*x + 5*x^4*Log[4]^2 + 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 (-5+\frac {2}{x}+20 x^3 \log ^2(4)\right ) \, dx \\ & = -5 x+5 x^4 \log ^2(4)+2 \log (x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.85 \[ \int \frac {2-5 x+20 x^4 \log ^2(4)}{x} \, dx=-5 x+5 x^4 \log ^2(4)+2 \log (x) \]

[In]

Integrate[(2 - 5*x + 20*x^4*Log[4]^2)/x,x]

[Out]

-5*x + 5*x^4*Log[4]^2 + 2*Log[x]

Maple [A] (verified)

Time = 0.09 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.90

method	result	size
default	\(-5 x +20 x^{4} \ln \left (2\right )^{2}+2 \ln \left (x \right )\)	\(18\)
norman	\(-5 x +20 x^{4} \ln \left (2\right )^{2}+2 \ln \left (x \right )\)	\(18\)
risch	\(-5 x +20 x^{4} \ln \left (2\right )^{2}+2 \ln \left (x \right )\)	\(18\)
parallelrisch	\(-5 x +20 x^{4} \ln \left (2\right )^{2}+2 \ln \left (x \right )\)	\(18\)

[In]

int((80*x^4*ln(2)^2-5*x+2)/x,x,method=_RETURNVERBOSE)

[Out]

-5*x+20*x^4*ln(2)^2+2*ln(x)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.85 \[ \int \frac {2-5 x+20 x^4 \log ^2(4)}{x} \, dx=20 \, x^{4} \log \left (2\right )^{2} - 5 \, x + 2 \, \log \left (x\right ) \]

[In]

integrate((80*x^4*log(2)^2-5*x+2)/x,x, algorithm="fricas")

[Out]

20*x^4*log(2)^2 - 5*x + 2*log(x)

Sympy [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.85 \[ \int \frac {2-5 x+20 x^4 \log ^2(4)}{x} \, dx=20 x^{4} \log {\left (2 \right )}^{2} - 5 x + 2 \log {\left (x \right )} \]

[In]

integrate((80*x**4*ln(2)**2-5*x+2)/x,x)

[Out]

20*x**4*log(2)**2 - 5*x + 2*log(x)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.85 \[ \int \frac {2-5 x+20 x^4 \log ^2(4)}{x} \, dx=20 \, x^{4} \log \left (2\right )^{2} - 5 \, x + 2 \, \log \left (x\right ) \]

[In]

integrate((80*x^4*log(2)^2-5*x+2)/x,x, algorithm="maxima")

[Out]

20*x^4*log(2)^2 - 5*x + 2*log(x)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.90 \[ \int \frac {2-5 x+20 x^4 \log ^2(4)}{x} \, dx=20 \, x^{4} \log \left (2\right )^{2} - 5 \, x + 2 \, \log \left ({\left | x \right |}\right ) \]

[In]

integrate((80*x^4*log(2)^2-5*x+2)/x,x, algorithm="giac")

[Out]

20*x^4*log(2)^2 - 5*x + 2*log(abs(x))

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.85 \[ \int \frac {2-5 x+20 x^4 \log ^2(4)}{x} \, dx=2\,\ln \left (x\right )-5\,x+20\,x^4\,{\ln \left (2\right )}^2 \]

[In]

int((80*x^4*log(2)^2 - 5*x + 2)/x,x)

[Out]

2*log(x) - 5*x + 20*x^4*log(2)^2