\(\int \frac {x^2}{(1+x)^3} \, dx\) [194]

Optimal result

Integrand size = 9, antiderivative size = 21 \[ \int \frac {x^2}{(1+x)^3} \, dx=-\frac {1}{2 (1+x)^2}+\frac {2}{1+x}+\log (1+x) \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {45} \[ \int \frac {x^2}{(1+x)^3} \, dx=\frac {2}{x+1}-\frac {1}{2 (x+1)^2}+\log (x+1) \]

[In]

Int[x^2/(1 + x)^3,x]

[Out]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int \frac {x^2}{(1+x)^3} \, dx=-\frac {1}{2 (1+x)^2}+\frac {2}{1+x}+\log (1+x) \]

[In]

Integrate[x^2/(1 + x)^3,x]

[Out]

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

Maple [A] (verified)

Time = 0.13 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81

method	result	size
norman	\(\frac {2 x +\frac {3}{2}}{\left (1+x \right )^{2}}+\ln \left (1+x \right )\)	\(17\)
risch	\(\frac {2 x +\frac {3}{2}}{\left (1+x \right )^{2}}+\ln \left (1+x \right )\)	\(17\)
meijerg	\(-\frac {x \left (9 x +6\right )}{6 \left (1+x \right )^{2}}+\ln \left (1+x \right )\)	\(19\)
default	\(-\frac {1}{2 \left (1+x \right )^{2}}+\frac {2}{1+x}+\ln \left (1+x \right )\)	\(20\)
parallelrisch	\(\frac {2 \ln \left (1+x \right ) x^{2}+3+4 \ln \left (1+x \right ) x +2 \ln \left (1+x \right )+4 x}{2 \left (1+x \right )^{2}}\)	\(35\)

[In]

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

[Out]

(2*x+3/2)/(1+x)^2+ln(1+x)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.48 \[ \int \frac {x^2}{(1+x)^3} \, dx=\frac {2 \, {\left (x^{2} + 2 \, x + 1\right )} \log \left (x + 1\right ) + 4 \, x + 3}{2 \, {\left (x^{2} + 2 \, x + 1\right )}} \]

[In]

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

[Out]

1/2*(2*(x^2 + 2*x + 1)*log(x + 1) + 4*x + 3)/(x^2 + 2*x + 1)

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {x^2}{(1+x)^3} \, dx=\frac {4 x + 3}{2 x^{2} + 4 x + 2} + \log {\left (x + 1 \right )} \]

[In]

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

[Out]

(4*x + 3)/(2*x**2 + 4*x + 2) + log(x + 1)

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.05 \[ \int \frac {x^2}{(1+x)^3} \, dx=\frac {4 \, x + 3}{2 \, {\left (x^{2} + 2 \, x + 1\right )}} + \log \left (x + 1\right ) \]

[In]

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

[Out]

1/2*(4*x + 3)/(x^2 + 2*x + 1) + log(x + 1)

Giac [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.86 \[ \int \frac {x^2}{(1+x)^3} \, dx=\frac {4 \, x + 3}{2 \, {\left (x + 1\right )}^{2}} + \log \left ({\left | x + 1 \right |}\right ) \]

[In]

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

[Out]

1/2*(4*x + 3)/(x + 1)^2 + log(abs(x + 1))

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int \frac {x^2}{(1+x)^3} \, dx=\ln \left (x+1\right )+\frac {2\,x+\frac {3}{2}}{x^2+2\,x+1} \]

[In]

int(x^2/(x + 1)^3,x)

[Out]

log(x + 1) + (2*x + 3/2)/(2*x + x^2 + 1)