3.3.21 \(\int \frac {1}{x (1+x)} \, dx\) [221]

Optimal. Leaf size=9 \[ \log (x)-\log (1+x) \]

[Out]

ln(x)-ln(1+x)

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Rubi [A]
time = 0.00, antiderivative size = 9, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {36, 29, 31} \begin {gather*} \log (x)-\log (x+1) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x*(1 + x)),x]

[Out]

Log[x] - Log[1 + x]

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rubi steps

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

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Mathematica [A]
time = 0.00, size = 9, normalized size = 1.00 \begin {gather*} \log (x)-\log (1+x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x*(1 + x)),x]

[Out]

Log[x] - Log[1 + x]

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Mathics [A]
time = 1.77, size = 9, normalized size = 1.00 \begin {gather*} \text {Log}\left [x\right ]-\text {Log}\left [1+x\right ] \end {gather*}

Antiderivative was successfully verified.

[In]

mathics('Integrate[x^(-1)*(x+1)^(-1),x]')

[Out]

Log[x] - Log[1 + x]

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Maple [A]
time = 0.03, size = 10, normalized size = 1.11

method	result	size
default	\(\ln \left (x \right )-\ln \left (1+x \right )\)	\(10\)
norman	\(\ln \left (x \right )-\ln \left (1+x \right )\)	\(10\)
meijerg	\(\ln \left (x \right )-\ln \left (1+x \right )\)	\(10\)
risch	\(\ln \left (x \right )-\ln \left (1+x \right )\)	\(10\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x/(1+x),x,method=_RETURNVERBOSE)

[Out]

ln(x)-ln(1+x)

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Maxima [A]
time = 0.27, size = 9, normalized size = 1.00 \begin {gather*} -\log \left (x + 1\right ) + \log \left (x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-log(x + 1) + log(x)

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Fricas [A]
time = 0.33, size = 9, normalized size = 1.00 \begin {gather*} -\log \left (x + 1\right ) + \log \left (x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-log(x + 1) + log(x)

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Sympy [A]
time = 0.04, size = 7, normalized size = 0.78 \begin {gather*} \log {\left (x \right )} - \log {\left (x + 1 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(1+x),x)

[Out]

log(x) - log(x + 1)

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Giac [A]
time = 0.00, size = 10, normalized size = 1.11 \begin {gather*} \ln \left |x\right |-\ln \left |x+1\right | \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(1+x),x)

[Out]

-log(abs(x + 1)) + log(abs(x))

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Mupad [B]
time = 0.20, size = 8, normalized size = 0.89 \begin {gather*} -\ln \left (\frac {1}{x}+1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x*(x + 1)),x)

[Out]

-log(1/x + 1)

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