3.4.9 \(\int \frac {1+x^3}{-x^2+x^3} \, dx\) [309]

Optimal. Leaf size=17 \[ \frac {1}{x}+x+2 \log (1-x)-\log (x) \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 17, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {1607, 1634} \begin {gather*} x+\frac {1}{x}+2 \log (1-x)-\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 + x^3)/(-x^2 + x^3),x]

[Out]

x^(-1) + x + 2*Log[1 - x] - Log[x]

Rule 1607

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 1634

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

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Mathematica [A]
time = 0.00, size = 17, normalized size = 1.00 \begin {gather*} \frac {1}{x}+x+2 \log (1-x)-\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 + x^3)/(-x^2 + x^3),x]

[Out]

x^(-1) + x + 2*Log[1 - x] - Log[x]

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Mathics [A]
time = 1.82, size = 15, normalized size = 0.88 \begin {gather*} \frac {1}{x}+x-\text {Log}\left [x\right ]+2 \text {Log}\left [-1+x\right ] \end {gather*}

Antiderivative was successfully verified.

[In]

mathics('Integrate[(1 + x^3)/(-x^2 + x^3),x]')

[Out]

1 / x + x - Log[x] + 2 Log[-1 + x]

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Maple [A]
time = 0.06, size = 16, normalized size = 0.94

method	result	size
default	\(x +\frac {1}{x}-\ln \left (x \right )+2 \ln \left (-1+x \right )\)	\(16\)
risch	\(x +\frac {1}{x}-\ln \left (x \right )+2 \ln \left (-1+x \right )\)	\(16\)
norman	\(\frac {x^{2}+1}{x}-\ln \left (x \right )+2 \ln \left (-1+x \right )\)	\(21\)
meijerg	\(\frac {1}{x}-\ln \left (x \right )-i \pi +2 \ln \left (1-x \right )+x\)	\(22\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3+1)/(x^3-x^2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.26, size = 15, normalized size = 0.88 \begin {gather*} x + \frac {1}{x} + 2 \, \log \left (x - 1\right ) - \log \left (x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^3+1)/(x^3-x^2),x, algorithm="maxima")

[Out]

x + 1/x + 2*log(x - 1) - log(x)

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Fricas [A]
time = 0.36, size = 21, normalized size = 1.24 \begin {gather*} \frac {x^{2} + 2 \, x \log \left (x - 1\right ) - x \log \left (x\right ) + 1}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^3+1)/(x^3-x^2),x, algorithm="fricas")

[Out]

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

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Sympy [A]
time = 0.05, size = 14, normalized size = 0.82 \begin {gather*} x - \log {\left (x \right )} + 2 \log {\left (x - 1 \right )} + \frac {1}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**3+1)/(x**3-x**2),x)

[Out]

x - log(x) + 2*log(x - 1) + 1/x

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Giac [A]
time = 0.00, size = 15, normalized size = 0.88 \begin {gather*} x+\frac 1{x}+2 \ln \left |x-1\right |-\ln \left |x\right | \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^3+1)/(x^3-x^2),x)

[Out]

x + 1/x + 2*log(abs(x - 1)) - log(abs(x))

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Mupad [B]
time = 0.04, size = 15, normalized size = 0.88 \begin {gather*} x+2\,\ln \left (x-1\right )-\ln \left (x\right )+\frac {1}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(x^3 + 1)/(x^2 - x^3),x)

[Out]

x + 2*log(x - 1) - log(x) + 1/x

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