3.26.14 \(\int \frac {10+10 x+2 x^2}{-25-10 x+5 x^2+2 x^3} \, dx\)

Optimal. Leaf size=22 \[ \log \left (\frac {5-x^2}{\frac {3}{4}+\frac {1+x}{2}}\right ) \]

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Rubi [A]  time = 0.03, antiderivative size = 17, normalized size of antiderivative = 0.77, number of steps used = 3, number of rules used = 2, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {2074, 260} \begin {gather*} \log \left (5-x^2\right )-\log (2 x+5) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(10 + 10*x + 2*x^2)/(-25 - 10*x + 5*x^2 + 2*x^3),x]

[Out]

-Log[5 + 2*x] + Log[5 - x^2]

Rule 260

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

Rule 2074

Int[(P_)^(p_)*(Q_)^(q_.), x_Symbol] :> With[{PP = Factor[P]}, Int[ExpandIntegrand[PP^p*Q^q, x], x] /;  !SumQ[N
onfreeFactors[PP, x]]] /; FreeQ[q, x] && PolyQ[P, x] && PolyQ[Q, x] && IntegerQ[p] && NeQ[P, x]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 25, normalized size = 1.14 \begin {gather*} 2 \left (-\frac {1}{2} \log (5+2 x)+\frac {1}{2} \log \left (5-x^2\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(10 + 10*x + 2*x^2)/(-25 - 10*x + 5*x^2 + 2*x^3),x]

[Out]

2*(-1/2*Log[5 + 2*x] + Log[5 - x^2]/2)

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fricas [A]  time = 0.86, size = 15, normalized size = 0.68 \begin {gather*} \log \left (x^{2} - 5\right ) - \log \left (2 \, x + 5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^2+10*x+10)/(2*x^3+5*x^2-10*x-25),x, algorithm="fricas")

[Out]

log(x^2 - 5) - log(2*x + 5)

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giac [A]  time = 0.20, size = 17, normalized size = 0.77 \begin {gather*} \log \left ({\left | x^{2} - 5 \right |}\right ) - \log \left ({\left | 2 \, x + 5 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^2+10*x+10)/(2*x^3+5*x^2-10*x-25),x, algorithm="giac")

[Out]

log(abs(x^2 - 5)) - log(abs(2*x + 5))

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maple [A]  time = 0.02, size = 16, normalized size = 0.73

method	result	size
default	\(\ln \left (x^{2}-5\right )-\ln \left (5+2 x \right )\)	\(16\)
norman	\(\ln \left (x^{2}-5\right )-\ln \left (5+2 x \right )\)	\(16\)
risch	\(\ln \left (x^{2}-5\right )-\ln \left (5+2 x \right )\)	\(16\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x^2+10*x+10)/(2*x^3+5*x^2-10*x-25),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.51, size = 15, normalized size = 0.68 \begin {gather*} \log \left (x^{2} - 5\right ) - \log \left (2 \, x + 5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^2+10*x+10)/(2*x^3+5*x^2-10*x-25),x, algorithm="maxima")

[Out]

log(x^2 - 5) - log(2*x + 5)

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mupad [B]  time = 0.07, size = 13, normalized size = 0.59 \begin {gather*} \ln \left (x^2-5\right )-\ln \left (x+\frac {5}{2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(10*x + 2*x^2 + 10)/(10*x - 5*x^2 - 2*x^3 + 25),x)

[Out]

log(x^2 - 5) - log(x + 5/2)

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sympy [A]  time = 0.09, size = 12, normalized size = 0.55 \begin {gather*} - \log {\left (2 x + 5 \right )} + \log {\left (x^{2} - 5 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x**2+10*x+10)/(2*x**3+5*x**2-10*x-25),x)

[Out]

-log(2*x + 5) + log(x**2 - 5)

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