3.59.14 \(\int \frac {9-x-10 x^4+2 \log (x)}{5 x^3} \, dx\)

Optimal. Leaf size=21 \[ 5-x^2-\frac {5-x+\log (x)}{5 x^2} \]

________________________________________________________________________________________

Rubi [A]  time = 0.02, antiderivative size = 27, normalized size of antiderivative = 1.29, number of steps used = 6, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {12, 14, 2304} \begin {gather*} -x^2-\frac {1}{x^2}-\frac {\log (x)}{5 x^2}+\frac {1}{5 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(9 - x - 10*x^4 + 2*Log[x])/(5*x^3),x]

[Out]

-x^(-2) + 1/(5*x) - x^2 - Log[x]/(5*x^2)

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 {9-x-10 x^4+2 \log (x)}{x^3} \, dx\\ &=\frac {1}{5} \int \left (\frac {9-x-10 x^4}{x^3}+\frac {2 \log (x)}{x^3}\right ) \, dx\\ &=\frac {1}{5} \int \frac {9-x-10 x^4}{x^3} \, dx+\frac {2}{5} \int \frac {\log (x)}{x^3} \, dx\\ &=-\frac {1}{10 x^2}-\frac {\log (x)}{5 x^2}+\frac {1}{5} \int \left (\frac {9}{x^3}-\frac {1}{x^2}-10 x\right ) \, dx\\ &=-\frac {1}{x^2}+\frac {1}{5 x}-x^2-\frac {\log (x)}{5 x^2}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.00, size = 27, normalized size = 1.29 \begin {gather*} -\frac {1}{x^2}+\frac {1}{5 x}-x^2-\frac {\log (x)}{5 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(9 - x - 10*x^4 + 2*Log[x])/(5*x^3),x]

[Out]

-x^(-2) + 1/(5*x) - x^2 - Log[x]/(5*x^2)

________________________________________________________________________________________

fricas [A]  time = 0.51, size = 17, normalized size = 0.81 \begin {gather*} -\frac {5 \, x^{4} - x + \log \relax (x) + 5}{5 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(2*log(x)-10*x^4-x+9)/x^3,x, algorithm="fricas")

[Out]

-1/5*(5*x^4 - x + log(x) + 5)/x^2

________________________________________________________________________________________

giac [A]  time = 0.24, size = 21, normalized size = 1.00 \begin {gather*} -x^{2} + \frac {x - 5}{5 \, x^{2}} - \frac {\log \relax (x)}{5 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(2*log(x)-10*x^4-x+9)/x^3,x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.03, size = 19, normalized size = 0.90

method	result	size
norman	\(\frac {-1+\frac {x}{5}-x^{4}-\frac {\ln \relax (x )}{5}}{x^{2}}\)	\(19\)
default	\(-\frac {\ln \relax (x )}{5 x^{2}}-\frac {1}{x^{2}}-x^{2}+\frac {1}{5 x}\)	\(24\)
risch	\(-\frac {\ln \relax (x )}{5 x^{2}}-\frac {5 x^{4}-x +5}{5 x^{2}}\)	\(24\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/5*(2*ln(x)-10*x^4-x+9)/x^3,x,method=_RETURNVERBOSE)

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.35, size = 23, normalized size = 1.10 \begin {gather*} -x^{2} + \frac {1}{5 \, x} - \frac {\log \relax (x)}{5 \, x^{2}} - \frac {1}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(2*log(x)-10*x^4-x+9)/x^3,x, algorithm="maxima")

[Out]

-x^2 + 1/5/x - 1/5*log(x)/x^2 - 1/x^2

________________________________________________________________________________________

mupad [B]  time = 4.00, size = 20, normalized size = 0.95 \begin {gather*} -\frac {\frac {\ln \relax (x)}{5}-\frac {x}{5}+1}{x^2}-x^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(x/5 - (2*log(x))/5 + 2*x^4 - 9/5)/x^3,x)

[Out]

- (log(x)/5 - x/5 + 1)/x^2 - x^2

________________________________________________________________________________________

sympy [A]  time = 0.11, size = 20, normalized size = 0.95 \begin {gather*} - x^{2} - \frac {5 - x}{5 x^{2}} - \frac {\log {\relax (x )}}{5 x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(2*ln(x)-10*x**4-x+9)/x**3,x)

[Out]

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

________________________________________________________________________________________