3.33.88 $\int \frac{25-7x-18x^2-27x^3+18x^4}{-9x^2+9x^3}dx$

Optimal. Leaf size=26 $$\frac{25}{9x}-x+x^2-\log ((1-x)x^2)$$

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Rubi [A]  time = 0.05, antiderivative size = 26, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 34, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.059, Rules used = {1593, 1620} $$\begin{array}{c}{x}^2-x+\frac{25}{9x}-\log (1-x)-2\log (x)\end{array}$$

Antiderivative was successfully verified.

[In]

Int[(25 - 7*x - 18*x^2 - 27*x^3 + 18*x^4)/(-9*x^2 + 9*x^3),x]

[Out]

25/(9*x) - x + x^2 - Log[1 - x] - 2*Log[x]

Rule 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 1620

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[Px*(a + b*x)
^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && PolyQ[Px, x] && (IntegersQ[m, n] || IGtQ[m, -2]) &&
GtQ[Expon[Px, x], 2]

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int \frac{25-7x-18x^2-27x^3+18x^4}{x^2(-9+9x)}dx\\ & =\int (-1+\frac{1}{1-x}-\frac{25}{9x^2}-\frac{2}{x}+2x)dx\\ & =\frac{25}{9x}-x+x^2-\log (1-x)-2\log (x)\end{array}\end{array}$$

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Mathematica [A]  time = 0.01, size = 30, normalized size = 1.15 $$\begin{array}{c}\frac{1}{9}(\frac{25}{x}-9x+9x^2-9\log (1-x)-18\log (x))\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(25 - 7*x - 18*x^2 - 27*x^3 + 18*x^4)/(-9*x^2 + 9*x^3),x]

[Out]

(25/x - 9*x + 9*x^2 - 9*Log[1 - x] - 18*Log[x])/9

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fricas [A]  time = 0.58, size = 29, normalized size = 1.12 $$\begin{array}{c}\frac{9x^3-9x^2-9x\log (x-1)-18x\log (x)+25}{9x}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((18*x^4-27*x^3-18*x^2-7*x+25)/(9*x^3-9*x^2),x, algorithm="fricas")

[Out]

1/9*(9*x^3 - 9*x^2 - 9*x*log(x - 1) - 18*x*log(x) + 25)/x

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giac [A]  time = 0.22, size = 24, normalized size = 0.92 $$\begin{array}{c}{x}^2-x+\frac{25}{9x}-\log \left(|x-1|\right)-2\log \left(\left|x\right|\right)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((18*x^4-27*x^3-18*x^2-7*x+25)/(9*x^3-9*x^2),x, algorithm="giac")

[Out]

x^2 - x + 25/9/x - log(abs(x - 1)) - 2*log(abs(x))

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maple [A]  time = 0.51, size = 23, normalized size = 0.88

method	result	size
default	$x^2-x-\ln (x-1)+\frac{25}{9x}-2\ln (x)$	$23$
risch	$x^2-x-\ln (x-1)+\frac{25}{9x}-2\ln (x)$	$23$
norman	$\frac{\frac{25}{9}+x^3-x^2}{x}-2\ln (x)-\ln (x-1)$	$26$
meijerg	$\frac{25}{9x}-2\ln (x)-2i\pi -\ln (1-x)+\frac{x(6+3x)}{3}-3x$	$34$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((18*x^4-27*x^3-18*x^2-7*x+25)/(9*x^3-9*x^2),x,method=_RETURNVERBOSE)

[Out]

x^2-x-ln(x-1)+25/9/x-2*ln(x)

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maxima [A]  time = 0.46, size = 22, normalized size = 0.85 $$\begin{array}{c}{x}^2-x+\frac{25}{9x}-\log (x-1)-2\log (x)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((18*x^4-27*x^3-18*x^2-7*x+25)/(9*x^3-9*x^2),x, algorithm="maxima")

[Out]

x^2 - x + 25/9/x - log(x - 1) - 2*log(x)

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mupad [B]  time = 0.05, size = 22, normalized size = 0.85 $$\begin{array}{c}\frac{25}{9x}-\ln (x-1)-2\ln (x)-x+x^2\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((7*x + 18*x^2 + 27*x^3 - 18*x^4 - 25)/(9*x^2 - 9*x^3),x)

[Out]

25/(9*x) - log(x - 1) - 2*log(x) - x + x^2

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sympy [A]  time = 0.14, size = 19, normalized size = 0.73 $$\begin{array}{c}{x}^2-x-2\log (x)-\log (x-1)+\frac{25}{9x}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((18*x**4-27*x**3-18*x**2-7*x+25)/(9*x**3-9*x**2),x)

[Out]

x**2 - x - 2*log(x) - log(x - 1) + 25/(9*x)

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