3.220 \(\int \frac{1}{x (a x^2+b x^3)} \, dx\)

Optimal. Leaf size=42 \[ \frac{b^2 \log (x)}{a^3}-\frac{b^2 \log (a+b x)}{a^3}+\frac{b}{a^2 x}-\frac{1}{2 a x^2} \]

[Out]

-1/(2*a*x^2) + b/(a^2*x) + (b^2*Log[x])/a^3 - (b^2*Log[a + b*x])/a^3

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Rubi [A]  time = 0.0212892, antiderivative size = 42, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.118, Rules used = {1584, 44} \[ \frac{b^2 \log (x)}{a^3}-\frac{b^2 \log (a+b x)}{a^3}+\frac{b}{a^2 x}-\frac{1}{2 a x^2} \]

Antiderivative was successfully verified.

[In]

Int[1/(x*(a*x^2 + b*x^3)),x]

[Out]

-1/(2*a*x^2) + b/(a^2*x) + (b^2*Log[x])/a^3 - (b^2*Log[a + b*x])/a^3

Rule 1584

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

Rule 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \frac{1}{x \left (a x^2+b x^3\right )} \, dx &=\int \frac{1}{x^3 (a+b x)} \, dx\\ &=\int \left (\frac{1}{a x^3}-\frac{b}{a^2 x^2}+\frac{b^2}{a^3 x}-\frac{b^3}{a^3 (a+b x)}\right ) \, dx\\ &=-\frac{1}{2 a x^2}+\frac{b}{a^2 x}+\frac{b^2 \log (x)}{a^3}-\frac{b^2 \log (a+b x)}{a^3}\\ \end{align*}

Mathematica [A]  time = 0.0043671, size = 42, normalized size = 1. \[ \frac{b^2 \log (x)}{a^3}-\frac{b^2 \log (a+b x)}{a^3}+\frac{b}{a^2 x}-\frac{1}{2 a x^2} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x*(a*x^2 + b*x^3)),x]

[Out]

-1/(2*a*x^2) + b/(a^2*x) + (b^2*Log[x])/a^3 - (b^2*Log[a + b*x])/a^3

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Maple [A]  time = 0.007, size = 41, normalized size = 1. \begin{align*} -{\frac{1}{2\,a{x}^{2}}}+{\frac{b}{{a}^{2}x}}+{\frac{{b}^{2}\ln \left ( x \right ) }{{a}^{3}}}-{\frac{{b}^{2}\ln \left ( bx+a \right ) }{{a}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x/(b*x^3+a*x^2),x)

[Out]

-1/2/a/x^2+b/a^2/x+b^2*ln(x)/a^3-b^2*ln(b*x+a)/a^3

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Maxima [A]  time = 1.04069, size = 54, normalized size = 1.29 \begin{align*} -\frac{b^{2} \log \left (b x + a\right )}{a^{3}} + \frac{b^{2} \log \left (x\right )}{a^{3}} + \frac{2 \, b x - a}{2 \, a^{2} x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(b*x^3+a*x^2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 0.780534, size = 103, normalized size = 2.45 \begin{align*} -\frac{2 \, b^{2} x^{2} \log \left (b x + a\right ) - 2 \, b^{2} x^{2} \log \left (x\right ) - 2 \, a b x + a^{2}}{2 \, a^{3} x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(b*x^3+a*x^2),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 1.03191, size = 31, normalized size = 0.74 \begin{align*} \frac{- a + 2 b x}{2 a^{2} x^{2}} + \frac{b^{2} \left (\log{\left (x \right )} - \log{\left (\frac{a}{b} + x \right )}\right )}{a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(b*x**3+a*x**2),x)

[Out]

(-a + 2*b*x)/(2*a**2*x**2) + b**2*(log(x) - log(a/b + x))/a**3

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Giac [A]  time = 1.15355, size = 61, normalized size = 1.45 \begin{align*} -\frac{b^{2} \log \left ({\left | b x + a \right |}\right )}{a^{3}} + \frac{b^{2} \log \left ({\left | x \right |}\right )}{a^{3}} + \frac{2 \, a b x - a^{2}}{2 \, a^{3} x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(b*x^3+a*x^2),x, algorithm="giac")

[Out]

-b^2*log(abs(b*x + a))/a^3 + b^2*log(abs(x))/a^3 + 1/2*(2*a*b*x - a^2)/(a^3*x^2)