3.182 \(\int \frac {x^2}{b x^2+c x^4} \, dx\)

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

[Out]

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

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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 = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {1584, 205} \[ \frac {\tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{\sqrt {b} \sqrt {c}} \]

Antiderivative was successfully verified.

[In]

Int[x^2/(b*x^2 + c*x^4),x]

[Out]

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

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^2}{b x^2+c x^4} \, dx &=\int \frac {1}{b+c x^2} \, dx\\ &=\frac {\tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{\sqrt {b} \sqrt {c}}\\ \end {align*}

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

Antiderivative was successfully verified.

[In]

Integrate[x^2/(b*x^2 + c*x^4),x]

[Out]

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

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fricas [A]  time = 0.70, size = 67, normalized size = 2.79 \[ \left [-\frac {\sqrt {-b c} \log \left (\frac {c x^{2} - 2 \, \sqrt {-b c} x - b}{c x^{2} + b}\right )}{2 \, b c}, \frac {\sqrt {b c} \arctan \left (\frac {\sqrt {b c} x}{b}\right )}{b c}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(c*x^4+b*x^2),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(c*x^4+b*x^2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.00, size = 16, normalized size = 0.67 \[ \frac {\arctan \left (\frac {c x}{\sqrt {b c}}\right )}{\sqrt {b c}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(c*x^4+b*x^2),x)

[Out]

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

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maxima [A]  time = 2.97, size = 15, normalized size = 0.62 \[ \frac {\arctan \left (\frac {c x}{\sqrt {b c}}\right )}{\sqrt {b c}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(c*x^4+b*x^2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 4.20, size = 16, normalized size = 0.67 \[ \frac {\mathrm {atan}\left (\frac {\sqrt {c}\,x}{\sqrt {b}}\right )}{\sqrt {b}\,\sqrt {c}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(b*x^2 + c*x^4),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2/(c*x**4+b*x**2),x)

[Out]

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

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