### 3.2249 $$\int \frac{x}{2 x+13 x^2+15 x^3} \, dx$$

Optimal. Leaf size=21 $\frac{1}{7} \log (5 x+1)-\frac{1}{7} \log (3 x+2)$

[Out]

-Log[2 + 3*x]/7 + Log[1 + 5*x]/7

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Rubi [A]  time = 0.0106777, antiderivative size = 21, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 18, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.167, Rules used = {1585, 616, 31} $\frac{1}{7} \log (5 x+1)-\frac{1}{7} \log (3 x+2)$

Antiderivative was successfully veriﬁed.

[In]

Int[x/(2*x + 13*x^2 + 15*x^3),x]

[Out]

-Log[2 + 3*x]/7 + Log[1 + 5*x]/7

Rule 1585

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

Rule 616

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Dist[c/q, Int[1/Simp
[b/2 - q/2 + c*x, x], x], x] - Dist[c/q, Int[1/Simp[b/2 + q/2 + c*x, x], x], x]] /; FreeQ[{a, b, c}, x] && NeQ
[b^2 - 4*a*c, 0] && PosQ[b^2 - 4*a*c] && PerfectSquareQ[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 x+13 x^2+15 x^3} \, dx &=\int \frac{1}{2+13 x+15 x^2} \, dx\\ &=\frac{15}{7} \int \frac{1}{3+15 x} \, dx-\frac{15}{7} \int \frac{1}{10+15 x} \, dx\\ &=-\frac{1}{7} \log (2+3 x)+\frac{1}{7} \log (1+5 x)\\ \end{align*}

Mathematica [A]  time = 0.0025279, size = 21, normalized size = 1. $\frac{1}{7} \log (5 x+1)-\frac{1}{7} \log (3 x+2)$

Antiderivative was successfully veriﬁed.

[In]

Integrate[x/(2*x + 13*x^2 + 15*x^3),x]

[Out]

-Log[2 + 3*x]/7 + Log[1 + 5*x]/7

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Maple [A]  time = 0.044, size = 18, normalized size = 0.9 \begin{align*} -{\frac{\ln \left ( 2+3\,x \right ) }{7}}+{\frac{\ln \left ( 1+5\,x \right ) }{7}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(x/(15*x^3+13*x^2+2*x),x)

[Out]

-1/7*ln(2+3*x)+1/7*ln(1+5*x)

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Maxima [A]  time = 0.981315, size = 23, normalized size = 1.1 \begin{align*} \frac{1}{7} \, \log \left (5 \, x + 1\right ) - \frac{1}{7} \, \log \left (3 \, x + 2\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(15*x^3+13*x^2+2*x),x, algorithm="maxima")

[Out]

1/7*log(5*x + 1) - 1/7*log(3*x + 2)

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Fricas [A]  time = 2.37545, size = 50, normalized size = 2.38 \begin{align*} \frac{1}{7} \, \log \left (5 \, x + 1\right ) - \frac{1}{7} \, \log \left (3 \, x + 2\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(15*x^3+13*x^2+2*x),x, algorithm="fricas")

[Out]

1/7*log(5*x + 1) - 1/7*log(3*x + 2)

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Sympy [A]  time = 0.101533, size = 15, normalized size = 0.71 \begin{align*} \frac{\log{\left (x + \frac{1}{5} \right )}}{7} - \frac{\log{\left (x + \frac{2}{3} \right )}}{7} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(15*x**3+13*x**2+2*x),x)

[Out]

log(x + 1/5)/7 - log(x + 2/3)/7

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Giac [A]  time = 1.11805, size = 26, normalized size = 1.24 \begin{align*} \frac{1}{7} \, \log \left ({\left | 5 \, x + 1 \right |}\right ) - \frac{1}{7} \, \log \left ({\left | 3 \, x + 2 \right |}\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(15*x^3+13*x^2+2*x),x, algorithm="giac")

[Out]

1/7*log(abs(5*x + 1)) - 1/7*log(abs(3*x + 2))