3.59.82 \(\int \frac {10+4 x^2+5 x^3}{5 x+x^3+x^4} \, dx\) [5882]

Optimal. Leaf size=17 \[ \log \left (x \left (5+x^2+x^3\right ) (x+x \log (4))\right ) \]

[Out]

ln((x^3+x^2+5)*x*(x+2*x*ln(2)))

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Rubi [A]
time = 0.09, antiderivative size = 14, normalized size of antiderivative = 0.82, number of steps used = 4, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {1608, 6874, 1601} \begin {gather*} \log \left (x^3+x^2+5\right )+2 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(10 + 4*x^2 + 5*x^3)/(5*x + x^3 + x^4),x]

[Out]

2*Log[x] + Log[5 + x^2 + x^3]

Rule 1601

Int[(Pp_)/(Qq_), x_Symbol] :> With[{p = Expon[Pp, x], q = Expon[Qq, x]}, Simp[Coeff[Pp, x, p]*(Log[RemoveConte
nt[Qq, x]]/(q*Coeff[Qq, x, q])), x] /; EqQ[p, q - 1] && EqQ[Pp, Simplify[(Coeff[Pp, x, p]/(q*Coeff[Qq, x, q]))
*D[Qq, x]]]] /; PolyQ[Pp, x] && PolyQ[Qq, x]

Rule 1608

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

Rule 6874

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {10+4 x^2+5 x^3}{x \left (5+x^2+x^3\right )} \, dx\\ &=\int \left (\frac {2}{x}+\frac {x (2+3 x)}{5+x^2+x^3}\right ) \, dx\\ &=2 \log (x)+\int \frac {x (2+3 x)}{5+x^2+x^3} \, dx\\ &=2 \log (x)+\log \left (5+x^2+x^3\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]
time = 0.01, size = 14, normalized size = 0.82 \begin {gather*} 2 \log (x)+\log \left (5+x^2+x^3\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(10 + 4*x^2 + 5*x^3)/(5*x + x^3 + x^4),x]

[Out]

2*Log[x] + Log[5 + x^2 + x^3]

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Maple [A]
time = 0.24, size = 15, normalized size = 0.88

method	result	size
default	\(\ln \left (x^{3}+x^{2}+5\right )+2 \ln \left (x \right )\)	\(15\)
norman	\(\ln \left (x^{3}+x^{2}+5\right )+2 \ln \left (x \right )\)	\(15\)
risch	\(\ln \left (x^{3}+x^{2}+5\right )+2 \ln \left (x \right )\)	\(15\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((5*x^3+4*x^2+10)/(x^4+x^3+5*x),x,method=_RETURNVERBOSE)

[Out]

ln(x^3+x^2+5)+2*ln(x)

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Maxima [A]
time = 0.36, size = 14, normalized size = 0.82 \begin {gather*} \log \left (x^{3} + x^{2} + 5\right ) + 2 \, \log \left (x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x^3+4*x^2+10)/(x^4+x^3+5*x),x, algorithm="maxima")

[Out]

log(x^3 + x^2 + 5) + 2*log(x)

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Fricas [A]
time = 0.35, size = 14, normalized size = 0.82 \begin {gather*} \log \left (x^{3} + x^{2} + 5\right ) + 2 \, \log \left (x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x^3+4*x^2+10)/(x^4+x^3+5*x),x, algorithm="fricas")

[Out]

log(x^3 + x^2 + 5) + 2*log(x)

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Sympy [A]
time = 0.04, size = 14, normalized size = 0.82 \begin {gather*} 2 \log {\left (x \right )} + \log {\left (x^{3} + x^{2} + 5 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x**3+4*x**2+10)/(x**4+x**3+5*x),x)

[Out]

2*log(x) + log(x**3 + x**2 + 5)

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Giac [A]
time = 0.37, size = 16, normalized size = 0.94 \begin {gather*} \log \left ({\left | x^{3} + x^{2} + 5 \right |}\right ) + 2 \, \log \left ({\left | x \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x^3+4*x^2+10)/(x^4+x^3+5*x),x, algorithm="giac")

[Out]

log(abs(x^3 + x^2 + 5)) + 2*log(abs(x))

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Mupad [B]
time = 4.15, size = 14, normalized size = 0.82 \begin {gather*} \ln \left (x^3+x^2+5\right )+2\,\ln \left (x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*x^2 + 5*x^3 + 10)/(5*x + x^3 + x^4),x)

[Out]

log(x^2 + x^3 + 5) + 2*log(x)

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