### 3.2238 $$\int \frac{x^2}{2+13 x+15 x^2} \, dx$$

Optimal. Leaf size=26 $\frac{x}{15}-\frac{4}{63} \log (3 x+2)+\frac{1}{175} \log (5 x+1)$

[Out]

x/15 - (4*Log[2 + 3*x])/63 + Log[1 + 5*x]/175

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Rubi [A]  time = 0.0124871, antiderivative size = 26, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 16, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.188, Rules used = {703, 632, 31} $\frac{x}{15}-\frac{4}{63} \log (3 x+2)+\frac{1}{175} \log (5 x+1)$

Antiderivative was successfully veriﬁed.

[In]

Int[x^2/(2 + 13*x + 15*x^2),x]

[Out]

x/15 - (4*Log[2 + 3*x])/63 + Log[1 + 5*x]/175

Rule 703

Int[((d_.) + (e_.)*(x_))^(m_)/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(e*(d + e*x)^(m - 1))/(c*
(m - 1)), x] + Dist[1/c, Int[((d + e*x)^(m - 2)*Simp[c*d^2 - a*e^2 + e*(2*c*d - b*e)*x, x])/(a + b*x + c*x^2),
x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[2*c*d - b*
e, 0] && GtQ[m, 1]

Rule 632

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Dist[
(c*d - e*(b/2 - q/2))/q, Int[1/(b/2 - q/2 + c*x), x], x] - Dist[(c*d - e*(b/2 + q/2))/q, Int[1/(b/2 + q/2 + c*
x), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] && NiceSqrtQ[b^2 - 4*a*
c]

Rule 31

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

Rubi steps

\begin{align*} \int \frac{x^2}{2+13 x+15 x^2} \, dx &=\frac{x}{15}+\frac{1}{15} \int \frac{-2-13 x}{2+13 x+15 x^2} \, dx\\ &=\frac{x}{15}+\frac{3}{35} \int \frac{1}{3+15 x} \, dx-\frac{20}{21} \int \frac{1}{10+15 x} \, dx\\ &=\frac{x}{15}-\frac{4}{63} \log (2+3 x)+\frac{1}{175} \log (1+5 x)\\ \end{align*}

Mathematica [A]  time = 0.0034883, size = 26, normalized size = 1. $\frac{x}{15}-\frac{4}{63} \log (3 x+2)+\frac{1}{175} \log (5 x+1)$

Antiderivative was successfully veriﬁed.

[In]

Integrate[x^2/(2 + 13*x + 15*x^2),x]

[Out]

x/15 - (4*Log[2 + 3*x])/63 + Log[1 + 5*x]/175

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Maple [A]  time = 0.044, size = 21, normalized size = 0.8 \begin{align*}{\frac{x}{15}}-{\frac{4\,\ln \left ( 2+3\,x \right ) }{63}}+{\frac{\ln \left ( 1+5\,x \right ) }{175}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(15*x^2+13*x+2),x)

[Out]

1/15*x-4/63*ln(2+3*x)+1/175*ln(1+5*x)

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Maxima [A]  time = 0.985239, size = 27, normalized size = 1.04 \begin{align*} \frac{1}{15} \, x + \frac{1}{175} \, \log \left (5 \, x + 1\right ) - \frac{4}{63} \, \log \left (3 \, x + 2\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(15*x^2+13*x+2),x, algorithm="maxima")

[Out]

1/15*x + 1/175*log(5*x + 1) - 4/63*log(3*x + 2)

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Fricas [A]  time = 2.2354, size = 66, normalized size = 2.54 \begin{align*} \frac{1}{15} \, x + \frac{1}{175} \, \log \left (5 \, x + 1\right ) - \frac{4}{63} \, \log \left (3 \, x + 2\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(15*x^2+13*x+2),x, algorithm="fricas")

[Out]

1/15*x + 1/175*log(5*x + 1) - 4/63*log(3*x + 2)

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Sympy [A]  time = 0.102507, size = 20, normalized size = 0.77 \begin{align*} \frac{x}{15} + \frac{\log{\left (x + \frac{1}{5} \right )}}{175} - \frac{4 \log{\left (x + \frac{2}{3} \right )}}{63} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2/(15*x**2+13*x+2),x)

[Out]

x/15 + log(x + 1/5)/175 - 4*log(x + 2/3)/63

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Giac [A]  time = 1.10693, size = 30, normalized size = 1.15 \begin{align*} \frac{1}{15} \, x + \frac{1}{175} \, \log \left ({\left | 5 \, x + 1 \right |}\right ) - \frac{4}{63} \, \log \left ({\left | 3 \, x + 2 \right |}\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(15*x^2+13*x+2),x, algorithm="giac")

[Out]

1/15*x + 1/175*log(abs(5*x + 1)) - 4/63*log(abs(3*x + 2))