3.83.55 \(\int \frac {5-2 x-x^2}{5 x+x^2} \, dx\)

Optimal. Leaf size=19 \[ -x+\log \left (4 e^5 x\right )+\log \left (4 (5+x)^2\right ) \]

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Rubi [A]  time = 0.02, antiderivative size = 12, normalized size of antiderivative = 0.63, number of steps used = 3, number of rules used = 2, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {1593, 893} \begin {gather*} -x+\log (x)+2 \log (x+5) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(5 - 2*x - x^2)/(5*x + x^2),x]

[Out]

-x + Log[x] + 2*Log[5 + x]

Rule 893

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> Int[ExpandIntegrand[(d + e*x)^m*(f + g*x)^n*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] &
& NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && I
ntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 0]))

Rule 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {5-2 x-x^2}{x (5+x)} \, dx\\ &=\int \left (-1+\frac {1}{x}+\frac {2}{5+x}\right ) \, dx\\ &=-x+\log (x)+2 \log (5+x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 12, normalized size = 0.63 \begin {gather*} -x+\log (x)+2 \log (5+x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(5 - 2*x - x^2)/(5*x + x^2),x]

[Out]

-x + Log[x] + 2*Log[5 + x]

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fricas [A]  time = 0.65, size = 12, normalized size = 0.63 \begin {gather*} -x + 2 \, \log \left (x + 5\right ) + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x^2-2*x+5)/(x^2+5*x),x, algorithm="fricas")

[Out]

-x + 2*log(x + 5) + log(x)

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giac [A]  time = 0.15, size = 14, normalized size = 0.74 \begin {gather*} -x + 2 \, \log \left ({\left | x + 5 \right |}\right ) + \log \left ({\left | x \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x^2-2*x+5)/(x^2+5*x),x, algorithm="giac")

[Out]

-x + 2*log(abs(x + 5)) + log(abs(x))

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maple [A]  time = 0.40, size = 13, normalized size = 0.68

method	result	size
default	\(-x +\ln \relax (x )+2 \ln \left (5+x \right )\)	\(13\)
norman	\(-x +\ln \relax (x )+2 \ln \left (5+x \right )\)	\(13\)
risch	\(-x +\ln \relax (x )+2 \ln \left (5+x \right )\)	\(13\)
meijerg	\(-\ln \relax (5)+\ln \relax (x )+2 \ln \left (1+\frac {x}{5}\right )-x\)	\(19\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-x^2-2*x+5)/(x^2+5*x),x,method=_RETURNVERBOSE)

[Out]

-x+ln(x)+2*ln(5+x)

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maxima [A]  time = 0.37, size = 12, normalized size = 0.63 \begin {gather*} -x + 2 \, \log \left (x + 5\right ) + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x^2-2*x+5)/(x^2+5*x),x, algorithm="maxima")

[Out]

-x + 2*log(x + 5) + log(x)

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mupad [B]  time = 4.77, size = 12, normalized size = 0.63 \begin {gather*} 2\,\ln \left (x+5\right )-x+\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(2*x + x^2 - 5)/(5*x + x^2),x)

[Out]

2*log(x + 5) - x + log(x)

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sympy [A]  time = 0.10, size = 10, normalized size = 0.53 \begin {gather*} - x + \log {\relax (x )} + 2 \log {\left (x + 5 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x**2-2*x+5)/(x**2+5*x),x)

[Out]

-x + log(x) + 2*log(x + 5)

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