\(\int \frac {x^4}{(a+\frac {b}{x^2})^2} \, dx\) [1860]

Optimal result

Integrand size = 13, antiderivative size = 79 \[ \int \frac {x^4}{\left (a+\frac {b}{x^2}\right )^2} \, dx=\frac {7 b^2 x}{2 a^4}-\frac {7 b x^3}{6 a^3}+\frac {7 x^5}{10 a^2}-\frac {x^7}{2 a \left (b+a x^2\right )}-\frac {7 b^{5/2} \arctan \left (\frac {\sqrt {a} x}{\sqrt {b}}\right )}{2 a^{9/2}} \]

[Out]

7/2*b^2*x/a^4-7/6*b*x^3/a^3+7/10*x^5/a^2-1/2*x^7/a/(a*x^2+b)-7/2*b^(5/2)*arctan(x*a^(1/2)/b^(1/2))/a^(9/2)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {269, 294, 308, 211} \[ \int \frac {x^4}{\left (a+\frac {b}{x^2}\right )^2} \, dx=-\frac {7 b^{5/2} \arctan \left (\frac {\sqrt {a} x}{\sqrt {b}}\right )}{2 a^{9/2}}+\frac {7 b^2 x}{2 a^4}-\frac {7 b x^3}{6 a^3}+\frac {7 x^5}{10 a^2}-\frac {x^7}{2 a \left (a x^2+b\right )} \]

[In]

Int[x^4/(a + b/x^2)^2,x]

[Out]

(7*b^2*x)/(2*a^4) - (7*b*x^3)/(6*a^3) + (7*x^5)/(10*a^2) - x^7/(2*a*(b + a*x^2)) - (7*b^(5/2)*ArcTan[(Sqrt[a]*
x)/Sqrt[b]])/(2*a^(9/2))

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 269

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 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]

Rule 308

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*} \text {integral}& = \int \frac {x^8}{\left (b+a x^2\right )^2} \, dx \\ & = -\frac {x^7}{2 a \left (b+a x^2\right )}+\frac {7 \int \frac {x^6}{b+a x^2} \, dx}{2 a} \\ & = -\frac {x^7}{2 a \left (b+a x^2\right )}+\frac {7 \int \left (\frac {b^2}{a^3}-\frac {b x^2}{a^2}+\frac {x^4}{a}-\frac {b^3}{a^3 \left (b+a x^2\right )}\right ) \, dx}{2 a} \\ & = \frac {7 b^2 x}{2 a^4}-\frac {7 b x^3}{6 a^3}+\frac {7 x^5}{10 a^2}-\frac {x^7}{2 a \left (b+a x^2\right )}-\frac {\left (7 b^3\right ) \int \frac {1}{b+a x^2} \, dx}{2 a^4} \\ & = \frac {7 b^2 x}{2 a^4}-\frac {7 b x^3}{6 a^3}+\frac {7 x^5}{10 a^2}-\frac {x^7}{2 a \left (b+a x^2\right )}-\frac {7 b^{5/2} \tan ^{-1}\left (\frac {\sqrt {a} x}{\sqrt {b}}\right )}{2 a^{9/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.90 \[ \int \frac {x^4}{\left (a+\frac {b}{x^2}\right )^2} \, dx=\frac {x \left (90 b^2-20 a b x^2+6 a^2 x^4+\frac {15 b^3}{b+a x^2}\right )}{30 a^4}-\frac {7 b^{5/2} \arctan \left (\frac {\sqrt {a} x}{\sqrt {b}}\right )}{2 a^{9/2}} \]

[In]

Integrate[x^4/(a + b/x^2)^2,x]

[Out]

(x*(90*b^2 - 20*a*b*x^2 + 6*a^2*x^4 + (15*b^3)/(b + a*x^2)))/(30*a^4) - (7*b^(5/2)*ArcTan[(Sqrt[a]*x)/Sqrt[b]]
)/(2*a^(9/2))

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.82

method	result	size
default	\(\frac {\frac {1}{5} x^{5} a^{2}-\frac {2}{3} a b \,x^{3}+3 b^{2} x}{a^{4}}-\frac {b^{3} \left (-\frac {x}{2 \left (a \,x^{2}+b \right )}+\frac {7 \arctan \left (\frac {a x}{\sqrt {a b}}\right )}{2 \sqrt {a b}}\right )}{a^{4}}\)	\(65\)
risch	\(\frac {x^{5}}{5 a^{2}}-\frac {2 b \,x^{3}}{3 a^{3}}+\frac {3 b^{2} x}{a^{4}}+\frac {b^{3} x}{2 \left (a \,x^{2}+b \right ) a^{4}}+\frac {7 \sqrt {-a b}\, b^{2} \ln \left (-\sqrt {-a b}\, x -b \right )}{4 a^{5}}-\frac {7 \sqrt {-a b}\, b^{2} \ln \left (\sqrt {-a b}\, x -b \right )}{4 a^{5}}\)	\(101\)

[In]

int(x^4/(a+b/x^2)^2,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 190, normalized size of antiderivative = 2.41 \[ \int \frac {x^4}{\left (a+\frac {b}{x^2}\right )^2} \, dx=\left [\frac {12 \, a^{3} x^{7} - 28 \, a^{2} b x^{5} + 140 \, a b^{2} x^{3} + 210 \, b^{3} x + 105 \, {\left (a b^{2} x^{2} + b^{3}\right )} \sqrt {-\frac {b}{a}} \log \left (\frac {a x^{2} - 2 \, a x \sqrt {-\frac {b}{a}} - b}{a x^{2} + b}\right )}{60 \, {\left (a^{5} x^{2} + a^{4} b\right )}}, \frac {6 \, a^{3} x^{7} - 14 \, a^{2} b x^{5} + 70 \, a b^{2} x^{3} + 105 \, b^{3} x - 105 \, {\left (a b^{2} x^{2} + b^{3}\right )} \sqrt {\frac {b}{a}} \arctan \left (\frac {a x \sqrt {\frac {b}{a}}}{b}\right )}{30 \, {\left (a^{5} x^{2} + a^{4} b\right )}}\right ] \]

[In]

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

[Out]

[1/60*(12*a^3*x^7 - 28*a^2*b*x^5 + 140*a*b^2*x^3 + 210*b^3*x + 105*(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^5*x^2 + a^4*b), 1/30*(6*a^3*x^7 - 14*a^2*b*x^5 + 70*a*b^2*x^3 + 105*b^3*
x - 105*(a*b^2*x^2 + b^3)*sqrt(b/a)*arctan(a*x*sqrt(b/a)/b))/(a^5*x^2 + a^4*b)]

Sympy [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.57 \[ \int \frac {x^4}{\left (a+\frac {b}{x^2}\right )^2} \, dx=\frac {b^{3} x}{2 a^{5} x^{2} + 2 a^{4} b} + \frac {7 \sqrt {- \frac {b^{5}}{a^{9}}} \log {\left (- \frac {a^{4} \sqrt {- \frac {b^{5}}{a^{9}}}}{b^{2}} + x \right )}}{4} - \frac {7 \sqrt {- \frac {b^{5}}{a^{9}}} \log {\left (\frac {a^{4} \sqrt {- \frac {b^{5}}{a^{9}}}}{b^{2}} + x \right )}}{4} + \frac {x^{5}}{5 a^{2}} - \frac {2 b x^{3}}{3 a^{3}} + \frac {3 b^{2} x}{a^{4}} \]

[In]

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

[Out]

b**3*x/(2*a**5*x**2 + 2*a**4*b) + 7*sqrt(-b**5/a**9)*log(-a**4*sqrt(-b**5/a**9)/b**2 + x)/4 - 7*sqrt(-b**5/a**
9)*log(a**4*sqrt(-b**5/a**9)/b**2 + x)/4 + x**5/(5*a**2) - 2*b*x**3/(3*a**3) + 3*b**2*x/a**4

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.90 \[ \int \frac {x^4}{\left (a+\frac {b}{x^2}\right )^2} \, dx=\frac {b^{3} x}{2 \, {\left (a^{5} x^{2} + a^{4} b\right )}} - \frac {7 \, b^{3} \arctan \left (\frac {a x}{\sqrt {a b}}\right )}{2 \, \sqrt {a b} a^{4}} + \frac {3 \, a^{2} x^{5} - 10 \, a b x^{3} + 45 \, b^{2} x}{15 \, a^{4}} \]

[In]

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

[Out]

1/2*b^3*x/(a^5*x^2 + a^4*b) - 7/2*b^3*arctan(a*x/sqrt(a*b))/(sqrt(a*b)*a^4) + 1/15*(3*a^2*x^5 - 10*a*b*x^3 + 4
5*b^2*x)/a^4

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.92 \[ \int \frac {x^4}{\left (a+\frac {b}{x^2}\right )^2} \, dx=-\frac {7 \, b^{3} \arctan \left (\frac {a x}{\sqrt {a b}}\right )}{2 \, \sqrt {a b} a^{4}} + \frac {b^{3} x}{2 \, {\left (a x^{2} + b\right )} a^{4}} + \frac {3 \, a^{8} x^{5} - 10 \, a^{7} b x^{3} + 45 \, a^{6} b^{2} x}{15 \, a^{10}} \]

[In]

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

[Out]

-7/2*b^3*arctan(a*x/sqrt(a*b))/(sqrt(a*b)*a^4) + 1/2*b^3*x/((a*x^2 + b)*a^4) + 1/15*(3*a^8*x^5 - 10*a^7*b*x^3
+ 45*a^6*b^2*x)/a^10

Mupad [B] (verification not implemented)

Time = 6.07 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.84 \[ \int \frac {x^4}{\left (a+\frac {b}{x^2}\right )^2} \, dx=\frac {x^5}{5\,a^2}-\frac {2\,b\,x^3}{3\,a^3}+\frac {3\,b^2\,x}{a^4}-\frac {7\,b^{5/2}\,\mathrm {atan}\left (\frac {\sqrt {a}\,x}{\sqrt {b}}\right )}{2\,a^{9/2}}+\frac {b^3\,x}{2\,\left (a^5\,x^2+b\,a^4\right )} \]

[In]

int(x^4/(a + b/x^2)^2,x)

[Out]

x^5/(5*a^2) - (2*b*x^3)/(3*a^3) + (3*b^2*x)/a^4 - (7*b^(5/2)*atan((a^(1/2)*x)/b^(1/2)))/(2*a^(9/2)) + (b^3*x)/
(2*(a^4*b + a^5*x^2))