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

Optimal. Leaf size=46 \[ \frac{1}{x+2}+\frac{5}{4 (x+3)}+\frac{1}{4 (x+3)^2}+\frac{1}{8} \log (x+1)+2 \log (x+2)-\frac{17}{8} \log (x+3) \]

[Out]

(2 + x)^(-1) + 1/(4*(3 + x)^2) + 5/(4*(3 + x)) + Log[1 + x]/8 + 2*Log[2 + x] - (17*Log[3 + x])/8

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Rubi [A]  time = 0.0234159, antiderivative size = 46, normalized size of antiderivative = 1., 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 = {88} \[ \frac{1}{x+2}+\frac{5}{4 (x+3)}+\frac{1}{4 (x+3)^2}+\frac{1}{8} \log (x+1)+2 \log (x+2)-\frac{17}{8} \log (x+3) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2 + x)^(-1) + 1/(4*(3 + x)^2) + 5/(4*(3 + x)) + Log[1 + x]/8 + 2*Log[2 + x] - (17*Log[3 + x])/8

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rubi steps

\begin{align*} \int \frac{1}{(1+x) (2+x)^2 (3+x)^3} \, dx &=\int \left (\frac{1}{8 (1+x)}-\frac{1}{(2+x)^2}+\frac{2}{2+x}-\frac{1}{2 (3+x)^3}-\frac{5}{4 (3+x)^2}-\frac{17}{8 (3+x)}\right ) \, dx\\ &=\frac{1}{2+x}+\frac{1}{4 (3+x)^2}+\frac{5}{4 (3+x)}+\frac{1}{8} \log (1+x)+2 \log (2+x)-\frac{17}{8} \log (3+x)\\ \end{align*}

Mathematica [A]  time = 0.016478, size = 44, normalized size = 0.96 \[ \frac{1}{8} \left (\frac{8}{x+2}+\frac{10}{x+3}+\frac{2}{(x+3)^2}+\log (-x-1)+16 \log (x+2)-17 \log (x+3)\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(8/(2 + x) + 2/(3 + x)^2 + 10/(3 + x) + Log[-1 - x] + 16*Log[2 + x] - 17*Log[3 + x])/8

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Maple [A]  time = 0.01, size = 39, normalized size = 0.9 \begin{align*} \left ( 2+x \right ) ^{-1}+{\frac{1}{4\, \left ( 3+x \right ) ^{2}}}+{\frac{5}{12+4\,x}}+{\frac{\ln \left ( 1+x \right ) }{8}}+2\,\ln \left ( 2+x \right ) -{\frac{17\,\ln \left ( 3+x \right ) }{8}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/(2+x)+1/4/(3+x)^2+5/4/(3+x)+1/8*ln(1+x)+2*ln(2+x)-17/8*ln(3+x)

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Maxima [A]  time = 1.02433, size = 62, normalized size = 1.35 \begin{align*} \frac{9 \, x^{2} + 50 \, x + 68}{4 \,{\left (x^{3} + 8 \, x^{2} + 21 \, x + 18\right )}} - \frac{17}{8} \, \log \left (x + 3\right ) + 2 \, \log \left (x + 2\right ) + \frac{1}{8} \, \log \left (x + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(9*x^2 + 50*x + 68)/(x^3 + 8*x^2 + 21*x + 18) - 17/8*log(x + 3) + 2*log(x + 2) + 1/8*log(x + 1)

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Fricas [B]  time = 0.988787, size = 239, normalized size = 5.2 \begin{align*} \frac{18 \, x^{2} - 17 \,{\left (x^{3} + 8 \, x^{2} + 21 \, x + 18\right )} \log \left (x + 3\right ) + 16 \,{\left (x^{3} + 8 \, x^{2} + 21 \, x + 18\right )} \log \left (x + 2\right ) +{\left (x^{3} + 8 \, x^{2} + 21 \, x + 18\right )} \log \left (x + 1\right ) + 100 \, x + 136}{8 \,{\left (x^{3} + 8 \, x^{2} + 21 \, x + 18\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(18*x^2 - 17*(x^3 + 8*x^2 + 21*x + 18)*log(x + 3) + 16*(x^3 + 8*x^2 + 21*x + 18)*log(x + 2) + (x^3 + 8*x^2
+ 21*x + 18)*log(x + 1) + 100*x + 136)/(x^3 + 8*x^2 + 21*x + 18)

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Sympy [A]  time = 0.17561, size = 46, normalized size = 1. \begin{align*} \frac{9 x^{2} + 50 x + 68}{4 x^{3} + 32 x^{2} + 84 x + 72} + \frac{\log{\left (x + 1 \right )}}{8} + 2 \log{\left (x + 2 \right )} - \frac{17 \log{\left (x + 3 \right )}}{8} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(9*x**2 + 50*x + 68)/(4*x**3 + 32*x**2 + 84*x + 72) + log(x + 1)/8 + 2*log(x + 2) - 17*log(x + 3)/8

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Giac [A]  time = 1.1537, size = 70, normalized size = 1.52 \begin{align*} \frac{1}{x + 2} - \frac{\frac{7}{x + 2} + 6}{4 \,{\left (\frac{1}{x + 2} + 1\right )}^{2}} + \frac{1}{8} \, \log \left ({\left | -\frac{1}{x + 2} + 1 \right |}\right ) - \frac{17}{8} \, \log \left ({\left | -\frac{1}{x + 2} - 1 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/(x + 2) - 1/4*(7/(x + 2) + 6)/(1/(x + 2) + 1)^2 + 1/8*log(abs(-1/(x + 2) + 1)) - 17/8*log(abs(-1/(x + 2) - 1
))