3.57.22 \(\int \frac {50+19 x+2 x^2}{25 x+10 x^2+x^3} \, dx\)

Optimal. Leaf size=20 \[ -\frac {x}{5 (5+x)}+\log \left (\frac {2 x^2}{e}\right ) \]

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Rubi [A]  time = 0.03, antiderivative size = 10, normalized size of antiderivative = 0.50, 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 = {1594, 27, 893} \begin {gather*} \frac {1}{x+5}+2 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(50 + 19*x + 2*x^2)/(25*x + 10*x^2 + x^3),x]

[Out]

(5 + x)^(-1) + 2*Log[x]

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

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 1594

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]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {50+19 x+2 x^2}{x \left (25+10 x+x^2\right )} \, dx\\ &=\int \frac {50+19 x+2 x^2}{x (5+x)^2} \, dx\\ &=\int \left (\frac {2}{x}-\frac {1}{(5+x)^2}\right ) \, dx\\ &=\frac {1}{5+x}+2 \log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 10, normalized size = 0.50 \begin {gather*} \frac {1}{5+x}+2 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(50 + 19*x + 2*x^2)/(25*x + 10*x^2 + x^3),x]

[Out]

(5 + x)^(-1) + 2*Log[x]

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fricas [A]  time = 1.19, size = 15, normalized size = 0.75 \begin {gather*} \frac {2 \, {\left (x + 5\right )} \log \relax (x) + 1}{x + 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^2+19*x+50)/(x^3+10*x^2+25*x),x, algorithm="fricas")

[Out]

(2*(x + 5)*log(x) + 1)/(x + 5)

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giac [A]  time = 0.18, size = 11, normalized size = 0.55 \begin {gather*} \frac {1}{x + 5} + 2 \, \log \left ({\left | x \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^2+19*x+50)/(x^3+10*x^2+25*x),x, algorithm="giac")

[Out]

1/(x + 5) + 2*log(abs(x))

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maple [A]  time = 0.03, size = 11, normalized size = 0.55

method	result	size
default	\(\frac {1}{5+x}+2 \ln \relax (x )\)	\(11\)
norman	\(\frac {1}{5+x}+2 \ln \relax (x )\)	\(11\)
risch	\(\frac {1}{5+x}+2 \ln \relax (x )\)	\(11\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x^2+19*x+50)/(x^3+10*x^2+25*x),x,method=_RETURNVERBOSE)

[Out]

1/(5+x)+2*ln(x)

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maxima [A]  time = 0.37, size = 10, normalized size = 0.50 \begin {gather*} \frac {1}{x + 5} + 2 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^2+19*x+50)/(x^3+10*x^2+25*x),x, algorithm="maxima")

[Out]

1/(x + 5) + 2*log(x)

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mupad [B]  time = 3.45, size = 10, normalized size = 0.50 \begin {gather*} 2\,\ln \relax (x)+\frac {1}{x+5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((19*x + 2*x^2 + 50)/(25*x + 10*x^2 + x^3),x)

[Out]

2*log(x) + 1/(x + 5)

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sympy [A]  time = 0.08, size = 8, normalized size = 0.40 \begin {gather*} 2 \log {\relax (x )} + \frac {1}{x + 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x**2+19*x+50)/(x**3+10*x**2+25*x),x)

[Out]

2*log(x) + 1/(x + 5)

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