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

Optimal. Leaf size=35 \[ \frac {b \log \left (a+b x^2\right )}{2 a^2}-\frac {b \log (x)}{a^2}-\frac {1}{2 a x^2} \]

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Rubi [A]  time = 0.03, antiderivative size = 35, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {1584, 266, 44} \begin {gather*} \frac {b \log \left (a+b x^2\right )}{2 a^2}-\frac {b \log (x)}{a^2}-\frac {1}{2 a x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^2*(a*x + b*x^3)),x]

[Out]

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

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])

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

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]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 35, normalized size = 1.00 \begin {gather*} \frac {b \log \left (a+b x^2\right )}{2 a^2}-\frac {b \log (x)}{a^2}-\frac {1}{2 a x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^2*(a*x + b*x^3)),x]

[Out]

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

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{x^2 \left (a x+b x^3\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[1/(x^2*(a*x + b*x^3)),x]

[Out]

IntegrateAlgebraic[1/(x^2*(a*x + b*x^3)), x]

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fricas [A]  time = 0.40, size = 33, normalized size = 0.94 \begin {gather*} \frac {b x^{2} \log \left (b x^{2} + a\right ) - 2 \, b x^{2} \log \relax (x) - a}{2 \, a^{2} x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.17, size = 43, normalized size = 1.23 \begin {gather*} -\frac {b \log \left (x^{2}\right )}{2 \, a^{2}} + \frac {b \log \left ({\left | b x^{2} + a \right |}\right )}{2 \, a^{2}} + \frac {b x^{2} - a}{2 \, a^{2} x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.05, size = 32, normalized size = 0.91 \begin {gather*} -\frac {b \ln \relax (x )}{a^{2}}+\frac {b \ln \left (b \,x^{2}+a \right )}{2 a^{2}}-\frac {1}{2 a \,x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.27, size = 31, normalized size = 0.89 \begin {gather*} \frac {b \log \left (b x^{2} + a\right )}{2 \, a^{2}} - \frac {b \log \relax (x)}{a^{2}} - \frac {1}{2 \, a x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.06, size = 31, normalized size = 0.89 \begin {gather*} \frac {b\,\ln \left (b\,x^2+a\right )}{2\,a^2}-\frac {1}{2\,a\,x^2}-\frac {b\,\ln \relax (x)}{a^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^2*(a*x + b*x^3)),x)

[Out]

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

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sympy [A]  time = 0.29, size = 31, normalized size = 0.89 \begin {gather*} - \frac {1}{2 a x^{2}} - \frac {b \log {\relax (x )}}{a^{2}} + \frac {b \log {\left (\frac {a}{b} + x^{2} \right )}}{2 a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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