3.2.88 \(\int \frac {1}{(-1+x)^2 (4+x)} \, dx\) [188]

Optimal. Leaf size=30 \[ \frac {1}{5 (1-x)}-\frac {1}{25} \log (1-x)+\frac {1}{25} \log (4+x) \]

[Out]

1/5/(1-x)-1/25*ln(1-x)+1/25*ln(4+x)

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Rubi [A]
time = 0.01, antiderivative size = 30, normalized size of antiderivative = 1.00, 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 = {46} \begin {gather*} \frac {1}{5 (1-x)}-\frac {1}{25} \log (1-x)+\frac {1}{25} \log (x+4) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((-1 + x)^2*(4 + x)),x]

[Out]

1/(5*(1 - x)) - Log[1 - x]/25 + Log[4 + x]/25

Rule 46

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && Lt
Q[m + n + 2, 0])

Rubi steps

\begin {align*} \int \frac {1}{(-1+x)^2 (4+x)} \, dx &=\int \left (\frac {1}{5 (-1+x)^2}-\frac {1}{25 (-1+x)}+\frac {1}{25 (4+x)}\right ) \, dx\\ &=\frac {1}{5 (1-x)}-\frac {1}{25} \log (1-x)+\frac {1}{25} \log (4+x)\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 22, normalized size = 0.73 \begin {gather*} \frac {1}{25} \left (-\frac {5}{-1+x}-\log (-1+x)+\log (4+x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((-1 + x)^2*(4 + x)),x]

[Out]

(-5/(-1 + x) - Log[-1 + x] + Log[4 + x])/25

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Maple [A]
time = 0.05, size = 21, normalized size = 0.70

method	result	size
default	\(-\frac {1}{5 \left (-1+x \right )}-\frac {\ln \left (-1+x \right )}{25}+\frac {\ln \left (4+x \right )}{25}\)	\(21\)
norman	\(-\frac {1}{5 \left (-1+x \right )}-\frac {\ln \left (-1+x \right )}{25}+\frac {\ln \left (4+x \right )}{25}\)	\(21\)
risch	\(-\frac {1}{5 \left (-1+x \right )}-\frac {\ln \left (-1+x \right )}{25}+\frac {\ln \left (4+x \right )}{25}\)	\(21\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(-1+x)^2/(4+x),x,method=_RETURNVERBOSE)

[Out]

-1/5/(-1+x)-1/25*ln(-1+x)+1/25*ln(4+x)

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Maxima [A]
time = 2.20, size = 20, normalized size = 0.67 \begin {gather*} -\frac {1}{5 \, {\left (x - 1\right )}} + \frac {1}{25} \, \log \left (x + 4\right ) - \frac {1}{25} \, \log \left (x - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/5/(x - 1) + 1/25*log(x + 4) - 1/25*log(x - 1)

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Fricas [A]
time = 0.42, size = 26, normalized size = 0.87 \begin {gather*} \frac {{\left (x - 1\right )} \log \left (x + 4\right ) - {\left (x - 1\right )} \log \left (x - 1\right ) - 5}{25 \, {\left (x - 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/25*((x - 1)*log(x + 4) - (x - 1)*log(x - 1) - 5)/(x - 1)

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Sympy [A]
time = 0.04, size = 19, normalized size = 0.63 \begin {gather*} - \frac {\log {\left (x - 1 \right )}}{25} + \frac {\log {\left (x + 4 \right )}}{25} - \frac {1}{5 x - 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-log(x - 1)/25 + log(x + 4)/25 - 1/(5*x - 5)

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Giac [A]
time = 0.46, size = 21, normalized size = 0.70 \begin {gather*} -\frac {1}{5 \, {\left (x - 1\right )}} + \frac {1}{25} \, \log \left ({\left | -\frac {5}{x - 1} - 1 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/5/(x - 1) + 1/25*log(abs(-5/(x - 1) - 1))

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Mupad [B]
time = 0.06, size = 22, normalized size = 0.73 \begin {gather*} -\frac {\ln \left (\frac {x-1}{x+4}\right )}{25}-\frac {1}{5\,\left (x-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((x - 1)^2*(x + 4)),x)

[Out]

- log((x - 1)/(x + 4))/25 - 1/(5*(x - 1))

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