3.22.78 \(\int \frac {3+3 x+2 x^2}{(1+x)^3} \, dx\) [2178]

Optimal. Leaf size=19 \[ -\frac {1}{(1+x)^2}+\frac {1}{1+x}+2 \log (1+x) \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 19, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {712} \begin {gather*} \frac {1}{x+1}-\frac {1}{(x+1)^2}+2 \log (x+1) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(3 + 3*x + 2*x^2)/(1 + x)^3,x]

[Out]

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

Rule 712

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d +
e*x)^m*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*
e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] && IntegerQ[p] && (GtQ[p, 0] || (EqQ[a, 0] && IntegerQ[m]))

Rubi steps

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

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Mathematica [A]
time = 0.01, size = 19, normalized size = 1.00 \begin {gather*} -\frac {1}{(1+x)^2}+\frac {1}{1+x}+2 \log (1+x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(3 + 3*x + 2*x^2)/(1 + x)^3,x]

[Out]

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

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Maple [A]
time = 0.46, size = 20, normalized size = 1.05

method	result	size
norman	\(\frac {x}{\left (x +1\right )^{2}}+2 \ln \left (x +1\right )\)	\(15\)
risch	\(\frac {x}{\left (x +1\right )^{2}}+2 \ln \left (x +1\right )\)	\(15\)
default	\(-\frac {1}{\left (x +1\right )^{2}}+\frac {1}{x +1}+2 \ln \left (x +1\right )\)	\(20\)
meijerg	\(\frac {3 x \left (2+x \right )}{2 \left (x +1\right )^{2}}-\frac {x \left (6+9 x \right )}{3 \left (x +1\right )^{2}}+2 \ln \left (x +1\right )+\frac {3 x^{2}}{2 \left (x +1\right )^{2}}\)	\(42\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.27, size = 19, normalized size = 1.00 \begin {gather*} \frac {x}{x^{2} + 2 \, x + 1} + 2 \, \log \left (x + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x/(x^2 + 2*x + 1) + 2*log(x + 1)

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Fricas [A]
time = 3.47, size = 27, normalized size = 1.42 \begin {gather*} \frac {2 \, {\left (x^{2} + 2 \, x + 1\right )} \log \left (x + 1\right ) + x}{x^{2} + 2 \, x + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 0.03, size = 15, normalized size = 0.79 \begin {gather*} \frac {x}{x^{2} + 2 x + 1} + 2 \log {\left (x + 1 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x/(x**2 + 2*x + 1) + 2*log(x + 1)

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Giac [A]
time = 1.21, size = 15, normalized size = 0.79 \begin {gather*} \frac {x}{{\left (x + 1\right )}^{2}} + 2 \, \log \left ({\left | x + 1 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x/(x + 1)^2 + 2*log(abs(x + 1))

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Mupad [B]
time = 0.03, size = 14, normalized size = 0.74 \begin {gather*} 2\,\ln \left (x+1\right )+\frac {x}{{\left (x+1\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((3*x + 2*x^2 + 3)/(x + 1)^3,x)

[Out]

2*log(x + 1) + x/(x + 1)^2

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