3.41.28 $\int \frac{31-30x-30x^2-30\log (x)}{30x^2}dx$

Optimal. Leaf size=21 $$-\frac{1}{30x}-x-\log (x)+\frac{\log (x)}{x}$$

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Rubi [A]  time = 0.02, antiderivative size = 21, normalized size of antiderivative = 1.00, 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} $$\begin{array}{c}-x-\frac{1}{30x}-\log (x)+\frac{\log (x)}{x}\end{array}$$

Antiderivative was successfully verified.

[In]

Int[(31 - 30*x - 30*x^2 - 30*Log[x])/(30*x^2),x]

[Out]

-1/30*1/x - x - Log[x] + Log[x]/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 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

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\frac{1}{30}\int \frac{31-30x-30x^2-30\log (x)}{x^2}dx\\ & =\frac{1}{30}\int (\frac{31-30x-30x^2}{x^2}-\frac{30\log (x)}{x^2})dx\\ & =\frac{1}{30}\int \frac{31-30x-30x^2}{x^2}dx-\int \frac{\log (x)}{x^2}dx\\ & =\frac{1}{x}+\frac{\log (x)}{x}+\frac{1}{30}\int (-30+\frac{31}{x^2}-\frac{30}{x})dx\\ & =-\frac{1}{30x}-x-\log (x)+\frac{\log (x)}{x}\end{array}\end{array}$$

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Mathematica [A]  time = 0.00, size = 21, normalized size = 1.00 $$\begin{array}{c}-\frac{1}{30x}-x-\log (x)+\frac{\log (x)}{x}\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(31 - 30*x - 30*x^2 - 30*Log[x])/(30*x^2),x]

[Out]

-1/30*1/x - x - Log[x] + Log[x]/x

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fricas [A]  time = 0.88, size = 19, normalized size = 0.90 $$\begin{array}{c}-\frac{30x^2+30(x-1)\log (x)+1}{30x}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/30*(-30*log(x)-30*x^2-30*x+31)/x^2,x, algorithm="fricas")

[Out]

-1/30*(30*x^2 + 30*(x - 1)*log(x) + 1)/x

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giac [A]  time = 0.13, size = 19, normalized size = 0.90 $$\begin{array}{c}-x+\frac{\log (x)}{x}-\frac{1}{30x}-\log (x)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/30*(-30*log(x)-30*x^2-30*x+31)/x^2,x, algorithm="giac")

[Out]

-x + log(x)/x - 1/30/x - log(x)

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maple [A]  time = 0.02, size = 19, normalized size = 0.90

method	result	size
norman	$\frac{-\frac{1}{30}-x\ln (x)-x^2+\ln (x)}{x}$	$19$
default	$\frac{\ln (x)}{x}-x-\frac{1}{30x}-\ln (x)$	$20$
risch	$\frac{\ln (x)}{x}-\frac{30x\ln (x)+30x^2+1}{30x}$	$25$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/30*(-30*ln(x)-30*x^2-30*x+31)/x^2,x,method=_RETURNVERBOSE)

[Out]

(-1/30-x*ln(x)-x^2+ln(x))/x

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maxima [A]  time = 0.35, size = 19, normalized size = 0.90 $$\begin{array}{c}-x+\frac{\log (x)}{x}-\frac{1}{30x}-\log (x)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/30*(-30*log(x)-30*x^2-30*x+31)/x^2,x, algorithm="maxima")

[Out]

-x + log(x)/x - 1/30/x - log(x)

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mupad [B]  time = 3.18, size = 16, normalized size = 0.76 $$\begin{array}{c}\frac{\ln (x)-\frac{1}{30}}{x}-\ln (x)-x\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(x + log(x) + x^2 - 31/30)/x^2,x)

[Out]

(log(x) - 1/30)/x - log(x) - x

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sympy [A]  time = 0.11, size = 14, normalized size = 0.67 $$\begin{array}{c}-x-\log (x)+\frac{\log (x)}{x}-\frac{1}{30x}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/30*(-30*ln(x)-30*x**2-30*x+31)/x**2,x)

[Out]

-x - log(x) + log(x)/x - 1/(30*x)

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