3.2.86 \(\int \frac {x^2}{(a+b x^2)^3} \, dx\) [186]

Optimal. Leaf size=65 \[ -\frac {x}{4 b \left (a+b x^2\right )^2}+\frac {x}{8 a b \left (a+b x^2\right )}+\frac {\tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{3/2} b^{3/2}} \]

[Out]

-1/4*x/b/(b*x^2+a)^2+1/8*x/a/b/(b*x^2+a)+1/8*arctan(x*b^(1/2)/a^(1/2))/a^(3/2)/b^(3/2)

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Rubi [A]
time = 0.01, antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {294, 205, 211} \begin {gather*} \frac {\text {ArcTan}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{3/2} b^{3/2}}+\frac {x}{8 a b \left (a+b x^2\right )}-\frac {x}{4 b \left (a+b x^2\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2/(a + b*x^2)^3,x]

[Out]

-1/4*x/(b*(a + b*x^2)^2) + x/(8*a*b*(a + b*x^2)) + ArcTan[(Sqrt[b]*x)/Sqrt[a]]/(8*a^(3/2)*b^(3/2))

Rule 205

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^n)^(p + 1)/(a*n*(p + 1))), x] + Dist[(n*(p
+ 1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (
IntegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[
p])

Rule 211

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

Rule 294

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^
n)^(p + 1)/(b*n*(p + 1))), x] - Dist[c^n*((m - n + 1)/(b*n*(p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rubi steps

\begin {align*} \int \frac {x^2}{\left (a+b x^2\right )^3} \, dx &=-\frac {x}{4 b \left (a+b x^2\right )^2}+\frac {\int \frac {1}{\left (a+b x^2\right )^2} \, dx}{4 b}\\ &=-\frac {x}{4 b \left (a+b x^2\right )^2}+\frac {x}{8 a b \left (a+b x^2\right )}+\frac {\int \frac {1}{a+b x^2} \, dx}{8 a b}\\ &=-\frac {x}{4 b \left (a+b x^2\right )^2}+\frac {x}{8 a b \left (a+b x^2\right )}+\frac {\tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{3/2} b^{3/2}}\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 58, normalized size = 0.89 \begin {gather*} \frac {\frac {\sqrt {a} \sqrt {b} x \left (-a+b x^2\right )}{\left (a+b x^2\right )^2}+\tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{3/2} b^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^2/(a + b*x^2)^3,x]

[Out]

((Sqrt[a]*Sqrt[b]*x*(-a + b*x^2))/(a + b*x^2)^2 + ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(8*a^(3/2)*b^(3/2))

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Maple [A]
time = 0.04, size = 49, normalized size = 0.75

method	result	size
default	\(\frac {\frac {x^{3}}{8 a}-\frac {x}{8 b}}{\left (b \,x^{2}+a \right )^{2}}+\frac {\arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 b a \sqrt {a b}}\)	\(49\)
risch	\(\frac {\frac {x^{3}}{8 a}-\frac {x}{8 b}}{\left (b \,x^{2}+a \right )^{2}}-\frac {\ln \left (b x +\sqrt {-a b}\right )}{16 \sqrt {-a b}\, b a}+\frac {\ln \left (-b x +\sqrt {-a b}\right )}{16 \sqrt {-a b}\, b a}\)	\(78\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(b*x^2+a)^3,x,method=_RETURNVERBOSE)

[Out]

(1/8/a*x^3-1/8*x/b)/(b*x^2+a)^2+1/8/b/a/(a*b)^(1/2)*arctan(b*x/(a*b)^(1/2))

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Maxima [A]
time = 0.48, size = 62, normalized size = 0.95 \begin {gather*} \frac {b x^{3} - a x}{8 \, {\left (a b^{3} x^{4} + 2 \, a^{2} b^{2} x^{2} + a^{3} b\right )}} + \frac {\arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(b*x^3 - a*x)/(a*b^3*x^4 + 2*a^2*b^2*x^2 + a^3*b) + 1/8*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*a*b)

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Fricas [A]
time = 0.94, size = 190, normalized size = 2.92 \begin {gather*} \left [\frac {2 \, a b^{2} x^{3} - 2 \, a^{2} b x - {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )} \sqrt {-a b} \log \left (\frac {b x^{2} - 2 \, \sqrt {-a b} x - a}{b x^{2} + a}\right )}{16 \, {\left (a^{2} b^{4} x^{4} + 2 \, a^{3} b^{3} x^{2} + a^{4} b^{2}\right )}}, \frac {a b^{2} x^{3} - a^{2} b x + {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b} x}{a}\right )}{8 \, {\left (a^{2} b^{4} x^{4} + 2 \, a^{3} b^{3} x^{2} + a^{4} b^{2}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/16*(2*a*b^2*x^3 - 2*a^2*b*x - (b^2*x^4 + 2*a*b*x^2 + a^2)*sqrt(-a*b)*log((b*x^2 - 2*sqrt(-a*b)*x - a)/(b*x^
2 + a)))/(a^2*b^4*x^4 + 2*a^3*b^3*x^2 + a^4*b^2), 1/8*(a*b^2*x^3 - a^2*b*x + (b^2*x^4 + 2*a*b*x^2 + a^2)*sqrt(
a*b)*arctan(sqrt(a*b)*x/a))/(a^2*b^4*x^4 + 2*a^3*b^3*x^2 + a^4*b^2)]

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 110 vs. \(2 (51) = 102\).
time = 0.13, size = 110, normalized size = 1.69 \begin {gather*} - \frac {\sqrt {- \frac {1}{a^{3} b^{3}}} \log {\left (- a^{2} b \sqrt {- \frac {1}{a^{3} b^{3}}} + x \right )}}{16} + \frac {\sqrt {- \frac {1}{a^{3} b^{3}}} \log {\left (a^{2} b \sqrt {- \frac {1}{a^{3} b^{3}}} + x \right )}}{16} + \frac {- a x + b x^{3}}{8 a^{3} b + 16 a^{2} b^{2} x^{2} + 8 a b^{3} x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-sqrt(-1/(a**3*b**3))*log(-a**2*b*sqrt(-1/(a**3*b**3)) + x)/16 + sqrt(-1/(a**3*b**3))*log(a**2*b*sqrt(-1/(a**3
*b**3)) + x)/16 + (-a*x + b*x**3)/(8*a**3*b + 16*a**2*b**2*x**2 + 8*a*b**3*x**4)

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Giac [A]
time = 1.35, size = 50, normalized size = 0.77 \begin {gather*} \frac {\arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a b} + \frac {b x^{3} - a x}{8 \, {\left (b x^{2} + a\right )}^{2} a b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*a*b) + 1/8*(b*x^3 - a*x)/((b*x^2 + a)^2*a*b)

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Mupad [B]
time = 4.74, size = 55, normalized size = 0.85 \begin {gather*} \frac {\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )}{8\,a^{3/2}\,b^{3/2}}-\frac {\frac {x}{8\,b}-\frac {x^3}{8\,a}}{a^2+2\,a\,b\,x^2+b^2\,x^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(a + b*x^2)^3,x)

[Out]

atan((b^(1/2)*x)/a^(1/2))/(8*a^(3/2)*b^(3/2)) - (x/(8*b) - x^3/(8*a))/(a^2 + b^2*x^4 + 2*a*b*x^2)

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