\(\int \frac {-4+4 x^5-\log (x)}{x^2} \, dx\) [3963]

Optimal result

Integrand size = 15, antiderivative size = 18 \[ \int \frac {-4+4 x^5-\log (x)}{x^2} \, dx=\frac {5-2 x+x \left (6+x^4\right )+\log (x)}{x} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.83, number of steps used = 5, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {14, 2341} \[ \int \frac {-4+4 x^5-\log (x)}{x^2} \, dx=x^4+\frac {5}{x}+\frac {\log (x)}{x} \]

[In]

Int[(-4 + 4*x^5 - Log[x])/x^2,x]

[Out]

5/x + x^4 + Log[x]/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}& = \int \left (\frac {4 \left (-1+x^5\right )}{x^2}-\frac {\log (x)}{x^2}\right ) \, dx \\ & = 4 \int \frac {-1+x^5}{x^2} \, dx-\int \frac {\log (x)}{x^2} \, dx \\ & = \frac {1}{x}+\frac {\log (x)}{x}+4 \int \left (-\frac {1}{x^2}+x^3\right ) \, dx \\ & = \frac {5}{x}+x^4+\frac {\log (x)}{x} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.83 \[ \int \frac {-4+4 x^5-\log (x)}{x^2} \, dx=\frac {5}{x}+x^4+\frac {\log (x)}{x} \]

[In]

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

[Out]

5/x + x^4 + Log[x]/x

Maple [A] (verified)

Time = 0.25 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.67

method	result	size
norman	\(\frac {5+x^{5}+\ln \left (x \right )}{x}\)	\(12\)
parallelrisch	\(\frac {5+x^{5}+\ln \left (x \right )}{x}\)	\(12\)
default	\(x^{4}+\frac {\ln \left (x \right )}{x}+\frac {5}{x}\)	\(16\)
parts	\(x^{4}+\frac {\ln \left (x \right )}{x}+\frac {5}{x}\)	\(16\)
risch	\(\frac {\ln \left (x \right )}{x}+\frac {x^{5}+5}{x}\)	\(17\)

[In]

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

[Out]

(5+x^5+ln(x))/x

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.61 \[ \int \frac {-4+4 x^5-\log (x)}{x^2} \, dx=\frac {x^{5} + \log \left (x\right ) + 5}{x} \]

[In]

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

[Out]

(x^5 + log(x) + 5)/x

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.56 \[ \int \frac {-4+4 x^5-\log (x)}{x^2} \, dx=x^{4} + \frac {\log {\left (x \right )}}{x} + \frac {5}{x} \]

[In]

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

[Out]

x**4 + log(x)/x + 5/x

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.83 \[ \int \frac {-4+4 x^5-\log (x)}{x^2} \, dx=x^{4} + \frac {\log \left (x\right )}{x} + \frac {5}{x} \]

[In]

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

[Out]

x^4 + log(x)/x + 5/x

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.83 \[ \int \frac {-4+4 x^5-\log (x)}{x^2} \, dx=x^{4} + \frac {\log \left (x\right )}{x} + \frac {5}{x} \]

[In]

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

[Out]

x^4 + log(x)/x + 5/x

Mupad [B] (verification not implemented)

Time = 9.03 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.67 \[ \int \frac {-4+4 x^5-\log (x)}{x^2} \, dx=\frac {\ln \left (x\right )+5}{x}+x^4 \]

[In]

int(-(log(x) - 4*x^5 + 4)/x^2,x)

[Out]

(log(x) + 5)/x + x^4