3.54.53 \(\int \frac {15+84 x+63 x^2-12 x^3-20 x^4}{15 x+65 x^2+85 x^3+40 x^4+20 x^5} \, dx\)

Optimal. Leaf size=23 \[ -\frac {4}{5 (1+2 x)}+\log \left (\frac {1}{1+\frac {3}{x}+x}\right ) \]

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Rubi [A]  time = 0.06, antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 47, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.043, Rules used = {2074, 628} \begin {gather*} -\log \left (x^2+x+3\right )-\frac {4}{5 (2 x+1)}+\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(15 + 84*x + 63*x^2 - 12*x^3 - 20*x^4)/(15*x + 65*x^2 + 85*x^3 + 40*x^4 + 20*x^5),x]

[Out]

-4/(5*(1 + 2*x)) + Log[x] - Log[3 + x + x^2]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

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 {1}{x}+\frac {8}{5 (1+2 x)^2}+\frac {-1-2 x}{3+x+x^2}\right ) \, dx\\ &=-\frac {4}{5 (1+2 x)}+\log (x)+\int \frac {-1-2 x}{3+x+x^2} \, dx\\ &=-\frac {4}{5 (1+2 x)}+\log (x)-\log \left (3+x+x^2\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 27, normalized size = 1.17 \begin {gather*} \frac {1}{5} \left (-\frac {4}{1+2 x}+5 \log (x)-5 \log \left (3+x+x^2\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(15 + 84*x + 63*x^2 - 12*x^3 - 20*x^4)/(15*x + 65*x^2 + 85*x^3 + 40*x^4 + 20*x^5),x]

[Out]

(-4/(1 + 2*x) + 5*Log[x] - 5*Log[3 + x + x^2])/5

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fricas [A]  time = 0.62, size = 34, normalized size = 1.48 \begin {gather*} -\frac {5 \, {\left (2 \, x + 1\right )} \log \left (x^{2} + x + 3\right ) - 5 \, {\left (2 \, x + 1\right )} \log \relax (x) + 4}{5 \, {\left (2 \, x + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-20*x^4-12*x^3+63*x^2+84*x+15)/(20*x^5+40*x^4+85*x^3+65*x^2+15*x),x, algorithm="fricas")

[Out]

-1/5*(5*(2*x + 1)*log(x^2 + x + 3) - 5*(2*x + 1)*log(x) + 4)/(2*x + 1)

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giac [A]  time = 0.20, size = 22, normalized size = 0.96 \begin {gather*} -\frac {4}{5 \, {\left (2 \, x + 1\right )}} - \log \left (x^{2} + x + 3\right ) + \log \left ({\left | x \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-20*x^4-12*x^3+63*x^2+84*x+15)/(20*x^5+40*x^4+85*x^3+65*x^2+15*x),x, algorithm="giac")

[Out]

-4/5/(2*x + 1) - log(x^2 + x + 3) + log(abs(x))

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maple [A]  time = 0.04, size = 20, normalized size = 0.87

method	result	size
risch	\(-\frac {2}{5 \left (\frac {1}{2}+x \right )}-\ln \left (x^{2}+x +3\right )+\ln \relax (x )\)	\(20\)
default	\(-\frac {4}{5 \left (2 x +1\right )}+\ln \relax (x )-\ln \left (x^{2}+x +3\right )\)	\(22\)
norman	\(\frac {8 x}{5 \left (2 x +1\right )}-\ln \left (x^{2}+x +3\right )+\ln \relax (x )\)	\(23\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-20*x^4-12*x^3+63*x^2+84*x+15)/(20*x^5+40*x^4+85*x^3+65*x^2+15*x),x,method=_RETURNVERBOSE)

[Out]

-2/5/(1/2+x)-ln(x^2+x+3)+ln(x)

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maxima [A]  time = 0.41, size = 21, normalized size = 0.91 \begin {gather*} -\frac {4}{5 \, {\left (2 \, x + 1\right )}} - \log \left (x^{2} + x + 3\right ) + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-20*x^4-12*x^3+63*x^2+84*x+15)/(20*x^5+40*x^4+85*x^3+65*x^2+15*x),x, algorithm="maxima")

[Out]

-4/5/(2*x + 1) - log(x^2 + x + 3) + log(x)

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mupad [B]  time = 0.08, size = 21, normalized size = 0.91 \begin {gather*} \ln \relax (x)-\ln \left (x^2+x+3\right )-\frac {2}{5\,\left (x+\frac {1}{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((84*x + 63*x^2 - 12*x^3 - 20*x^4 + 15)/(15*x + 65*x^2 + 85*x^3 + 40*x^4 + 20*x^5),x)

[Out]

log(x) - log(x + x^2 + 3) - 2/(5*(x + 1/2))

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sympy [A]  time = 0.12, size = 17, normalized size = 0.74 \begin {gather*} \log {\relax (x )} - \log {\left (x^{2} + x + 3 \right )} - \frac {4}{10 x + 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-20*x**4-12*x**3+63*x**2+84*x+15)/(20*x**5+40*x**4+85*x**3+65*x**2+15*x),x)

[Out]

log(x) - log(x**2 + x + 3) - 4/(10*x + 5)

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