3.2.76 \(\int \frac {x^2}{1+x} \, dx\) [176]

Optimal. Leaf size=15 \[ -x+\frac {x^2}{2}+\log (1+x) \]

[Out]

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

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Rubi [A]
time = 0.00, antiderivative size = 15, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {45} \begin {gather*} \frac {x^2}{2}-x+\log (x+1) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2/(1 + x),x]

[Out]

-x + x^2/2 + Log[1 + x]

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

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

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Mathematica [A]
time = 0.00, size = 19, normalized size = 1.27 \begin {gather*} -2 (1+x)+\frac {1}{2} (1+x)^2+\log (1+x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^2/(1 + x),x]

[Out]

-2*(1 + x) + (1 + x)^2/2 + Log[1 + x]

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Mathics [A]
time = 1.77, size = 13, normalized size = 0.87 \begin {gather*} -x+\frac {x^2}{2}+\text {Log}\left [1+x\right ] \end {gather*}

Antiderivative was successfully verified.

[In]

mathics('Integrate[x^2/(1 + x),x]')

[Out]

-x + x ^ 2 / 2 + Log[1 + x]

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Maple [A]
time = 0.04, size = 14, normalized size = 0.93

method	result	size
default	\(-x +\frac {x^{2}}{2}+\ln \left (1+x \right )\)	\(14\)
norman	\(-x +\frac {x^{2}}{2}+\ln \left (1+x \right )\)	\(14\)
meijerg	\(-\frac {x \left (-3 x +6\right )}{6}+\ln \left (1+x \right )\)	\(14\)
risch	\(-x +\frac {x^{2}}{2}+\ln \left (1+x \right )\)	\(14\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.27, size = 13, normalized size = 0.87 \begin {gather*} \frac {1}{2} \, x^{2} - x + \log \left (x + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x^2 - x + log(x + 1)

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Fricas [A]
time = 0.32, size = 13, normalized size = 0.87 \begin {gather*} \frac {1}{2} \, x^{2} - x + \log \left (x + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x^2 - x + log(x + 1)

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Sympy [A]
time = 0.04, size = 10, normalized size = 0.67 \begin {gather*} \frac {x^{2}}{2} - x + \log {\left (x + 1 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x**2/2 - x + log(x + 1)

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Giac [A]
time = 0.00, size = 15, normalized size = 1.00 \begin {gather*} \frac {1}{2} x^{2}-x+\ln \left |x+1\right | \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x^2 - x + log(abs(x + 1))

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Mupad [B]
time = 0.02, size = 13, normalized size = 0.87 \begin {gather*} \ln \left (x+1\right )-x+\frac {x^2}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(x + 1),x)

[Out]

log(x + 1) - x + x^2/2

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