3.1.15 \(\int \frac {x}{a x+b x^3} \, dx\)

Optimal. Leaf size=24 \[ \frac {\tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {b}} \]

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Rubi [A]  time = 0.01, antiderivative size = 24, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {1584, 205} \begin {gather*} \frac {\tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {b}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x/(a*x + b*x^3),x]

[Out]

ArcTan[(Sqrt[b]*x)/Sqrt[a]]/(Sqrt[a]*Sqrt[b])

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

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

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Mathematica [A]  time = 0.00, size = 24, normalized size = 1.00 \begin {gather*} \frac {\tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {b}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x/(a*x + b*x^3),x]

[Out]

ArcTan[(Sqrt[b]*x)/Sqrt[a]]/(Sqrt[a]*Sqrt[b])

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

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[x/(a*x + b*x^3),x]

[Out]

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

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fricas [A]  time = 0.41, size = 67, normalized size = 2.79 \begin {gather*} \left [-\frac {\sqrt {-a b} \log \left (\frac {b x^{2} - 2 \, \sqrt {-a b} x - a}{b x^{2} + a}\right )}{2 \, a b}, \frac {\sqrt {a b} \arctan \left (\frac {\sqrt {a b} x}{a}\right )}{a b}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2*sqrt(-a*b)*log((b*x^2 - 2*sqrt(-a*b)*x - a)/(b*x^2 + a))/(a*b), sqrt(a*b)*arctan(sqrt(a*b)*x/a)/(a*b)]

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giac [A]  time = 0.15, size = 15, normalized size = 0.62 \begin {gather*} \frac {\arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

arctan(b*x/sqrt(a*b))/sqrt(a*b)

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maple [A]  time = 0.04, size = 16, normalized size = 0.67 \begin {gather*} \frac {\arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/(a*b)^(1/2)*arctan(1/(a*b)^(1/2)*b*x)

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maxima [A]  time = 2.95, size = 15, normalized size = 0.62 \begin {gather*} \frac {\arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

arctan(b*x/sqrt(a*b))/sqrt(a*b)

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mupad [B]  time = 0.04, size = 16, normalized size = 0.67 \begin {gather*} \frac {\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )}{\sqrt {a}\,\sqrt {b}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(a*x + b*x^3),x)

[Out]

atan((b^(1/2)*x)/a^(1/2))/(a^(1/2)*b^(1/2))

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sympy [B]  time = 0.15, size = 53, normalized size = 2.21 \begin {gather*} - \frac {\sqrt {- \frac {1}{a b}} \log {\left (- a \sqrt {- \frac {1}{a b}} + x \right )}}{2} + \frac {\sqrt {- \frac {1}{a b}} \log {\left (a \sqrt {- \frac {1}{a b}} + x \right )}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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