3.5 \(\int \frac{(1+x)^3}{(-1+x)^4} \, dx\)

Optimal. Leaf size=36 \[ \frac{6}{1-x}-\frac{6}{(1-x)^2}+\frac{8}{3 (1-x)^3}+\log (1-x) \]

[Out]

8/(3*(1 - x)^3) - 6/(1 - x)^2 + 6/(1 - x) + Log[1 - x]

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Rubi [A]  time = 0.0130342, antiderivative size = 36, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.091, Rules used = {43} \[ \frac{6}{1-x}-\frac{6}{(1-x)^2}+\frac{8}{3 (1-x)^3}+\log (1-x) \]

Antiderivative was successfully verified.

[In]

Int[(1 + x)^3/(-1 + x)^4,x]

[Out]

8/(3*(1 - x)^3) - 6/(1 - x)^2 + 6/(1 - x) + Log[1 - x]

Rule 43

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{(1+x)^3}{(-1+x)^4} \, dx &=\int \left (\frac{8}{(-1+x)^4}+\frac{12}{(-1+x)^3}+\frac{6}{(-1+x)^2}+\frac{1}{-1+x}\right ) \, dx\\ &=\frac{8}{3 (1-x)^3}-\frac{6}{(1-x)^2}+\frac{6}{1-x}+\log (1-x)\\ \end{align*}

Mathematica [A]  time = 0.0113566, size = 24, normalized size = 0.67 \[ \log (x-1)-\frac{2 \left (9 x^2-9 x+4\right )}{3 (x-1)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(1 + x)^3/(-1 + x)^4,x]

[Out]

(-2*(4 - 9*x + 9*x^2))/(3*(-1 + x)^3) + Log[-1 + x]

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Maple [A]  time = 0.006, size = 27, normalized size = 0.8 \begin{align*} -{\frac{8}{3\, \left ( -1+x \right ) ^{3}}}-6\, \left ( -1+x \right ) ^{-2}+\ln \left ( -1+x \right ) -6\, \left ( -1+x \right ) ^{-1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1+x)^3/(-1+x)^4,x)

[Out]

-8/3/(-1+x)^3-6/(-1+x)^2+ln(-1+x)-6/(-1+x)

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Maxima [A]  time = 0.948822, size = 43, normalized size = 1.19 \begin{align*} -\frac{2 \,{\left (9 \, x^{2} - 9 \, x + 4\right )}}{3 \,{\left (x^{3} - 3 \, x^{2} + 3 \, x - 1\right )}} + \log \left (x - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.84015, size = 120, normalized size = 3.33 \begin{align*} -\frac{18 \, x^{2} - 3 \,{\left (x^{3} - 3 \, x^{2} + 3 \, x - 1\right )} \log \left (x - 1\right ) - 18 \, x + 8}{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((1+x)^3/(-1+x)^4,x, algorithm="fricas")

[Out]

-1/3*(18*x^2 - 3*(x^3 - 3*x^2 + 3*x - 1)*log(x - 1) - 18*x + 8)/(x^3 - 3*x^2 + 3*x - 1)

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Sympy [A]  time = 0.109386, size = 29, normalized size = 0.81 \begin{align*} - \frac{18 x^{2} - 18 x + 8}{3 x^{3} - 9 x^{2} + 9 x - 3} + \log{\left (x - 1 \right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(18*x**2 - 18*x + 8)/(3*x**3 - 9*x**2 + 9*x - 3) + log(x - 1)

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Giac [A]  time = 1.06742, size = 31, normalized size = 0.86 \begin{align*} -\frac{2 \,{\left (9 \, x^{2} - 9 \, x + 4\right )}}{3 \,{\left (x - 1\right )}^{3}} + \log \left ({\left | x - 1 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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