3.484 \(\int \frac{4+3 x^2+2 x^3}{(1+x)^4} \, dx\)

Optimal. Leaf size=23 \[ \frac{3}{x+1}-\frac{5}{3 (x+1)^3}+2 \log (x+1) \]

[Out]

-5/(3*(1 + x)^3) + 3/(1 + x) + 2*Log[1 + x]

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Rubi [A]  time = 0.0177378, antiderivative size = 23, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.056, Rules used = {1850} \[ \frac{3}{x+1}-\frac{5}{3 (x+1)^3}+2 \log (x+1) \]

Antiderivative was successfully verified.

[In]

Int[(4 + 3*x^2 + 2*x^3)/(1 + x)^4,x]

[Out]

-5/(3*(1 + x)^3) + 3/(1 + x) + 2*Log[1 + x]

Rule 1850

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq*(a + b*x^n)^p, x], x] /; FreeQ[
{a, b, n}, x] && PolyQ[Pq, x] && (IGtQ[p, 0] || EqQ[n, 1])

Rubi steps

\begin{align*} \int \frac{4+3 x^2+2 x^3}{(1+x)^4} \, dx &=\int \left (\frac{5}{(1+x)^4}-\frac{3}{(1+x)^2}+\frac{2}{1+x}\right ) \, dx\\ &=-\frac{5}{3 (1+x)^3}+\frac{3}{1+x}+2 \log (1+x)\\ \end{align*}

Mathematica [A]  time = 0.0098225, size = 23, normalized size = 1. \[ \frac{3}{x+1}-\frac{5}{3 (x+1)^3}+2 \log (x+1) \]

Antiderivative was successfully verified.

[In]

Integrate[(4 + 3*x^2 + 2*x^3)/(1 + x)^4,x]

[Out]

-5/(3*(1 + x)^3) + 3/(1 + x) + 2*Log[1 + x]

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Maple [A]  time = 0.005, size = 22, normalized size = 1. \begin{align*} -{\frac{5}{3\, \left ( 1+x \right ) ^{3}}}+3\, \left ( 1+x \right ) ^{-1}+2\,\ln \left ( 1+x \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x^3+3*x^2+4)/(1+x)^4,x)

[Out]

-5/3/(1+x)^3+3/(1+x)+2*ln(1+x)

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Maxima [A]  time = 1.0971, size = 46, normalized size = 2. \begin{align*} \frac{9 \, x^{2} + 18 \, x + 4}{3 \,{\left (x^{3} + 3 \, x^{2} + 3 \, x + 1\right )}} + 2 \, \log \left (x + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^3+3*x^2+4)/(1+x)^4,x, algorithm="maxima")

[Out]

1/3*(9*x^2 + 18*x + 4)/(x^3 + 3*x^2 + 3*x + 1) + 2*log(x + 1)

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Fricas [B]  time = 1.19698, size = 117, normalized size = 5.09 \begin{align*} \frac{9 \, x^{2} + 6 \,{\left (x^{3} + 3 \, x^{2} + 3 \, x + 1\right )} \log \left (x + 1\right ) + 18 \, x + 4}{3 \,{\left (x^{3} + 3 \, x^{2} + 3 \, x + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^3+3*x^2+4)/(1+x)^4,x, algorithm="fricas")

[Out]

1/3*(9*x^2 + 6*(x^3 + 3*x^2 + 3*x + 1)*log(x + 1) + 18*x + 4)/(x^3 + 3*x^2 + 3*x + 1)

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Sympy [A]  time = 0.104956, size = 31, normalized size = 1.35 \begin{align*} \frac{9 x^{2} + 18 x + 4}{3 x^{3} + 9 x^{2} + 9 x + 3} + 2 \log{\left (x + 1 \right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x**3+3*x**2+4)/(1+x)**4,x)

[Out]

(9*x**2 + 18*x + 4)/(3*x**3 + 9*x**2 + 9*x + 3) + 2*log(x + 1)

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Giac [A]  time = 1.13363, size = 34, normalized size = 1.48 \begin{align*} \frac{9 \, x^{2} + 18 \, x + 4}{3 \,{\left (x + 1\right )}^{3}} + 2 \, \log \left ({\left | x + 1 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^3+3*x^2+4)/(1+x)^4,x, algorithm="giac")

[Out]

1/3*(9*x^2 + 18*x + 4)/(x + 1)^3 + 2*log(abs(x + 1))