3.1877 \(\int \frac {x^2}{(a+\frac {b}{x^2})^3} \, dx\)

Optimal. Leaf size=85 \[ \frac {35 b^{3/2} \tan ^{-1}\left (\frac {\sqrt {a} x}{\sqrt {b}}\right )}{8 a^{9/2}}-\frac {35 b x}{8 a^4}+\frac {35 x^3}{24 a^3}-\frac {7 x^5}{8 a^2 \left (a x^2+b\right )}-\frac {x^7}{4 a \left (a x^2+b\right )^2} \]

[Out]

-35/8*b*x/a^4+35/24*x^3/a^3-1/4*x^7/a/(a*x^2+b)^2-7/8*x^5/a^2/(a*x^2+b)+35/8*b^(3/2)*arctan(x*a^(1/2)/b^(1/2))
/a^(9/2)

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Rubi [A]  time = 0.04, antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {263, 288, 302, 205} \[ \frac {35 b^{3/2} \tan ^{-1}\left (\frac {\sqrt {a} x}{\sqrt {b}}\right )}{8 a^{9/2}}-\frac {7 x^5}{8 a^2 \left (a x^2+b\right )}-\frac {35 b x}{8 a^4}+\frac {35 x^3}{24 a^3}-\frac {x^7}{4 a \left (a x^2+b\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[x^2/(a + b/x^2)^3,x]

[Out]

(-35*b*x)/(8*a^4) + (35*x^3)/(24*a^3) - x^7/(4*a*(b + a*x^2)^2) - (7*x^5)/(8*a^2*(b + a*x^2)) + (35*b^(3/2)*Ar
cTan[(Sqrt[a]*x)/Sqrt[b]])/(8*a^(9/2))

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 263

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

Rule 288

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]

Rule 302

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

Rubi steps

\begin {align*} \int \frac {x^2}{\left (a+\frac {b}{x^2}\right )^3} \, dx &=\int \frac {x^8}{\left (b+a x^2\right )^3} \, dx\\ &=-\frac {x^7}{4 a \left (b+a x^2\right )^2}+\frac {7 \int \frac {x^6}{\left (b+a x^2\right )^2} \, dx}{4 a}\\ &=-\frac {x^7}{4 a \left (b+a x^2\right )^2}-\frac {7 x^5}{8 a^2 \left (b+a x^2\right )}+\frac {35 \int \frac {x^4}{b+a x^2} \, dx}{8 a^2}\\ &=-\frac {x^7}{4 a \left (b+a x^2\right )^2}-\frac {7 x^5}{8 a^2 \left (b+a x^2\right )}+\frac {35 \int \left (-\frac {b}{a^2}+\frac {x^2}{a}+\frac {b^2}{a^2 \left (b+a x^2\right )}\right ) \, dx}{8 a^2}\\ &=-\frac {35 b x}{8 a^4}+\frac {35 x^3}{24 a^3}-\frac {x^7}{4 a \left (b+a x^2\right )^2}-\frac {7 x^5}{8 a^2 \left (b+a x^2\right )}+\frac {\left (35 b^2\right ) \int \frac {1}{b+a x^2} \, dx}{8 a^4}\\ &=-\frac {35 b x}{8 a^4}+\frac {35 x^3}{24 a^3}-\frac {x^7}{4 a \left (b+a x^2\right )^2}-\frac {7 x^5}{8 a^2 \left (b+a x^2\right )}+\frac {35 b^{3/2} \tan ^{-1}\left (\frac {\sqrt {a} x}{\sqrt {b}}\right )}{8 a^{9/2}}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 77, normalized size = 0.91 \[ \frac {35 b^{3/2} \tan ^{-1}\left (\frac {\sqrt {a} x}{\sqrt {b}}\right )}{8 a^{9/2}}-\frac {-8 a^3 x^7+56 a^2 b x^5+175 a b^2 x^3+105 b^3 x}{24 a^4 \left (a x^2+b\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2/(a + b/x^2)^3,x]

[Out]

-1/24*(105*b^3*x + 175*a*b^2*x^3 + 56*a^2*b*x^5 - 8*a^3*x^7)/(a^4*(b + a*x^2)^2) + (35*b^(3/2)*ArcTan[(Sqrt[a]
*x)/Sqrt[b]])/(8*a^(9/2))

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fricas [A]  time = 0.83, size = 230, normalized size = 2.71 \[ \left [\frac {16 \, a^{3} x^{7} - 112 \, a^{2} b x^{5} - 350 \, a b^{2} x^{3} - 210 \, b^{3} x + 105 \, {\left (a^{2} b x^{4} + 2 \, a b^{2} x^{2} + b^{3}\right )} \sqrt {-\frac {b}{a}} \log \left (\frac {a x^{2} + 2 \, a x \sqrt {-\frac {b}{a}} - b}{a x^{2} + b}\right )}{48 \, {\left (a^{6} x^{4} + 2 \, a^{5} b x^{2} + a^{4} b^{2}\right )}}, \frac {8 \, a^{3} x^{7} - 56 \, a^{2} b x^{5} - 175 \, a b^{2} x^{3} - 105 \, b^{3} x + 105 \, {\left (a^{2} b x^{4} + 2 \, a b^{2} x^{2} + b^{3}\right )} \sqrt {\frac {b}{a}} \arctan \left (\frac {a x \sqrt {\frac {b}{a}}}{b}\right )}{24 \, {\left (a^{6} x^{4} + 2 \, a^{5} b x^{2} + a^{4} b^{2}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/48*(16*a^3*x^7 - 112*a^2*b*x^5 - 350*a*b^2*x^3 - 210*b^3*x + 105*(a^2*b*x^4 + 2*a*b^2*x^2 + b^3)*sqrt(-b/a)
*log((a*x^2 + 2*a*x*sqrt(-b/a) - b)/(a*x^2 + b)))/(a^6*x^4 + 2*a^5*b*x^2 + a^4*b^2), 1/24*(8*a^3*x^7 - 56*a^2*
b*x^5 - 175*a*b^2*x^3 - 105*b^3*x + 105*(a^2*b*x^4 + 2*a*b^2*x^2 + b^3)*sqrt(b/a)*arctan(a*x*sqrt(b/a)/b))/(a^
6*x^4 + 2*a^5*b*x^2 + a^4*b^2)]

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giac [A]  time = 0.16, size = 73, normalized size = 0.86 \[ \frac {35 \, b^{2} \arctan \left (\frac {a x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a^{4}} - \frac {13 \, a b^{2} x^{3} + 11 \, b^{3} x}{8 \, {\left (a x^{2} + b\right )}^{2} a^{4}} + \frac {a^{6} x^{3} - 9 \, a^{5} b x}{3 \, a^{9}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

35/8*b^2*arctan(a*x/sqrt(a*b))/(sqrt(a*b)*a^4) - 1/8*(13*a*b^2*x^3 + 11*b^3*x)/((a*x^2 + b)^2*a^4) + 1/3*(a^6*
x^3 - 9*a^5*b*x)/a^9

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maple [A]  time = 0.01, size = 77, normalized size = 0.91 \[ -\frac {13 b^{2} x^{3}}{8 \left (a \,x^{2}+b \right )^{2} a^{3}}+\frac {x^{3}}{3 a^{3}}-\frac {11 b^{3} x}{8 \left (a \,x^{2}+b \right )^{2} a^{4}}+\frac {35 b^{2} \arctan \left (\frac {a x}{\sqrt {a b}}\right )}{8 \sqrt {a b}\, a^{4}}-\frac {3 b x}{a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3/a^3*x^3-3/a^4*b*x-13/8/a^3*b^2/(a*x^2+b)^2*x^3-11/8/a^4*b^3/(a*x^2+b)^2*x+35/8/a^4*b^2/(a*b)^(1/2)*arctan(
1/(a*b)^(1/2)*a*x)

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maxima [A]  time = 1.76, size = 82, normalized size = 0.96 \[ -\frac {13 \, a b^{2} x^{3} + 11 \, b^{3} x}{8 \, {\left (a^{6} x^{4} + 2 \, a^{5} b x^{2} + a^{4} b^{2}\right )}} + \frac {35 \, b^{2} \arctan \left (\frac {a x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a^{4}} + \frac {a x^{3} - 9 \, b x}{3 \, a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(13*a*b^2*x^3 + 11*b^3*x)/(a^6*x^4 + 2*a^5*b*x^2 + a^4*b^2) + 35/8*b^2*arctan(a*x/sqrt(a*b))/(sqrt(a*b)*a
^4) + 1/3*(a*x^3 - 9*b*x)/a^4

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mupad [B]  time = 1.14, size = 77, normalized size = 0.91 \[ \frac {x^3}{3\,a^3}-\frac {\frac {11\,b^3\,x}{8}+\frac {13\,a\,b^2\,x^3}{8}}{a^6\,x^4+2\,a^5\,b\,x^2+a^4\,b^2}+\frac {35\,b^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {a}\,x}{\sqrt {b}}\right )}{8\,a^{9/2}}-\frac {3\,b\,x}{a^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(a + b/x^2)^3,x)

[Out]

x^3/(3*a^3) - ((11*b^3*x)/8 + (13*a*b^2*x^3)/8)/(a^4*b^2 + a^6*x^4 + 2*a^5*b*x^2) + (35*b^(3/2)*atan((a^(1/2)*
x)/b^(1/2)))/(8*a^(9/2)) - (3*b*x)/a^4

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sympy [A]  time = 0.48, size = 133, normalized size = 1.56 \[ - \frac {35 \sqrt {- \frac {b^{3}}{a^{9}}} \log {\left (- \frac {a^{4} \sqrt {- \frac {b^{3}}{a^{9}}}}{b} + x \right )}}{16} + \frac {35 \sqrt {- \frac {b^{3}}{a^{9}}} \log {\left (\frac {a^{4} \sqrt {- \frac {b^{3}}{a^{9}}}}{b} + x \right )}}{16} + \frac {- 13 a b^{2} x^{3} - 11 b^{3} x}{8 a^{6} x^{4} + 16 a^{5} b x^{2} + 8 a^{4} b^{2}} + \frac {x^{3}}{3 a^{3}} - \frac {3 b x}{a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-35*sqrt(-b**3/a**9)*log(-a**4*sqrt(-b**3/a**9)/b + x)/16 + 35*sqrt(-b**3/a**9)*log(a**4*sqrt(-b**3/a**9)/b +
x)/16 + (-13*a*b**2*x**3 - 11*b**3*x)/(8*a**6*x**4 + 16*a**5*b*x**2 + 8*a**4*b**2) + x**3/(3*a**3) - 3*b*x/a**
4

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