\(\int \frac {1}{(a+\frac {b}{x^2})^3 x^{10}} \, dx\) [1889]

Optimal result

Integrand size = 13, antiderivative size = 87 \[ \int \frac {1}{\left (a+\frac {b}{x^2}\right )^3 x^{10}} \, dx=-\frac {35}{24 b^3 x^3}+\frac {35 a}{8 b^4 x}+\frac {1}{4 b x^3 \left (b+a x^2\right )^2}+\frac {7}{8 b^2 x^3 \left (b+a x^2\right )}+\frac {35 a^{3/2} \arctan \left (\frac {\sqrt {a} x}{\sqrt {b}}\right )}{8 b^{9/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {269, 296, 331, 211} \[ \int \frac {1}{\left (a+\frac {b}{x^2}\right )^3 x^{10}} \, dx=\frac {35 a^{3/2} \arctan \left (\frac {\sqrt {a} x}{\sqrt {b}}\right )}{8 b^{9/2}}+\frac {35 a}{8 b^4 x}+\frac {7}{8 b^2 x^3 \left (a x^2+b\right )}+\frac {1}{4 b x^3 \left (a x^2+b\right )^2}-\frac {35}{24 b^3 x^3} \]

[In]

Int[1/((a + b/x^2)^3*x^10),x]

[Out]

-35/(24*b^3*x^3) + (35*a)/(8*b^4*x) + 1/(4*b*x^3*(b + a*x^2)^2) + 7/(8*b^2*x^3*(b + a*x^2)) + (35*a^(3/2)*ArcT
an[(Sqrt[a]*x)/Sqrt[b]])/(8*b^(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 296

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

Rule 331

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

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

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.91 \[ \int \frac {1}{\left (a+\frac {b}{x^2}\right )^3 x^{10}} \, dx=\frac {-8 b^3+56 a b^2 x^2+175 a^2 b x^4+105 a^3 x^6}{24 b^4 x^3 \left (b+a x^2\right )^2}+\frac {35 a^{3/2} \arctan \left (\frac {\sqrt {a} x}{\sqrt {b}}\right )}{8 b^{9/2}} \]

[In]

Integrate[1/((a + b/x^2)^3*x^10),x]

[Out]

(-8*b^3 + 56*a*b^2*x^2 + 175*a^2*b*x^4 + 105*a^3*x^6)/(24*b^4*x^3*(b + a*x^2)^2) + (35*a^(3/2)*ArcTan[(Sqrt[a]
*x)/Sqrt[b]])/(8*b^(9/2))

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.74

method	result	size
default	\(-\frac {1}{3 b^{3} x^{3}}+\frac {3 a}{b^{4} x}+\frac {a^{2} \left (\frac {\frac {11}{8} a \,x^{3}+\frac {13}{8} b x}{\left (a \,x^{2}+b \right )^{2}}+\frac {35 \arctan \left (\frac {a x}{\sqrt {a b}}\right )}{8 \sqrt {a b}}\right )}{b^{4}}\)	\(64\)
risch	\(\frac {\frac {35 a^{3} x^{6}}{8 b^{4}}+\frac {175 a^{2} x^{4}}{24 b^{3}}+\frac {7 a \,x^{2}}{3 b^{2}}-\frac {1}{3 b}}{x^{3} \left (a \,x^{2}+b \right )^{2}}+\frac {35 \sqrt {-a b}\, a \ln \left (-a x -\sqrt {-a b}\right )}{16 b^{5}}-\frac {35 \sqrt {-a b}\, a \ln \left (-a x +\sqrt {-a b}\right )}{16 b^{5}}\)	\(102\)

[In]

int(1/(a+b/x^2)^3/x^10,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 238, normalized size of antiderivative = 2.74 \[ \int \frac {1}{\left (a+\frac {b}{x^2}\right )^3 x^{10}} \, dx=\left [\frac {210 \, a^{3} x^{6} + 350 \, a^{2} b x^{4} + 112 \, a b^{2} x^{2} - 16 \, b^{3} + 105 \, {\left (a^{3} x^{7} + 2 \, a^{2} b x^{5} + a b^{2} x^{3}\right )} \sqrt {-\frac {a}{b}} \log \left (\frac {a x^{2} + 2 \, b x \sqrt {-\frac {a}{b}} - b}{a x^{2} + b}\right )}{48 \, {\left (a^{2} b^{4} x^{7} + 2 \, a b^{5} x^{5} + b^{6} x^{3}\right )}}, \frac {105 \, a^{3} x^{6} + 175 \, a^{2} b x^{4} + 56 \, a b^{2} x^{2} - 8 \, b^{3} + 105 \, {\left (a^{3} x^{7} + 2 \, a^{2} b x^{5} + a b^{2} x^{3}\right )} \sqrt {\frac {a}{b}} \arctan \left (x \sqrt {\frac {a}{b}}\right )}{24 \, {\left (a^{2} b^{4} x^{7} + 2 \, a b^{5} x^{5} + b^{6} x^{3}\right )}}\right ] \]

[In]

integrate(1/(a+b/x^2)^3/x^10,x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.59 \[ \int \frac {1}{\left (a+\frac {b}{x^2}\right )^3 x^{10}} \, dx=- \frac {35 \sqrt {- \frac {a^{3}}{b^{9}}} \log {\left (x - \frac {b^{5} \sqrt {- \frac {a^{3}}{b^{9}}}}{a^{2}} \right )}}{16} + \frac {35 \sqrt {- \frac {a^{3}}{b^{9}}} \log {\left (x + \frac {b^{5} \sqrt {- \frac {a^{3}}{b^{9}}}}{a^{2}} \right )}}{16} + \frac {105 a^{3} x^{6} + 175 a^{2} b x^{4} + 56 a b^{2} x^{2} - 8 b^{3}}{24 a^{2} b^{4} x^{7} + 48 a b^{5} x^{5} + 24 b^{6} x^{3}} \]

[In]

integrate(1/(a+b/x**2)**3/x**10,x)

[Out]

-35*sqrt(-a**3/b**9)*log(x - b**5*sqrt(-a**3/b**9)/a**2)/16 + 35*sqrt(-a**3/b**9)*log(x + b**5*sqrt(-a**3/b**9
)/a**2)/16 + (105*a**3*x**6 + 175*a**2*b*x**4 + 56*a*b**2*x**2 - 8*b**3)/(24*a**2*b**4*x**7 + 48*a*b**5*x**5 +
24*b**6*x**3)

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.99 \[ \int \frac {1}{\left (a+\frac {b}{x^2}\right )^3 x^{10}} \, dx=\frac {105 \, a^{3} x^{6} + 175 \, a^{2} b x^{4} + 56 \, a b^{2} x^{2} - 8 \, b^{3}}{24 \, {\left (a^{2} b^{4} x^{7} + 2 \, a b^{5} x^{5} + b^{6} x^{3}\right )}} + \frac {35 \, a^{2} \arctan \left (\frac {a x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} b^{4}} \]

[In]

integrate(1/(a+b/x^2)^3/x^10,x, algorithm="maxima")

[Out]

1/24*(105*a^3*x^6 + 175*a^2*b*x^4 + 56*a*b^2*x^2 - 8*b^3)/(a^2*b^4*x^7 + 2*a*b^5*x^5 + b^6*x^3) + 35/8*a^2*arc
tan(a*x/sqrt(a*b))/(sqrt(a*b)*b^4)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.82 \[ \int \frac {1}{\left (a+\frac {b}{x^2}\right )^3 x^{10}} \, dx=\frac {35 \, a^{2} \arctan \left (\frac {a x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} b^{4}} + \frac {11 \, a^{3} x^{3} + 13 \, a^{2} b x}{8 \, {\left (a x^{2} + b\right )}^{2} b^{4}} + \frac {9 \, a x^{2} - b}{3 \, b^{4} x^{3}} \]

[In]

integrate(1/(a+b/x^2)^3/x^10,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 5.87 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.92 \[ \int \frac {1}{\left (a+\frac {b}{x^2}\right )^3 x^{10}} \, dx=\frac {\frac {7\,a\,x^2}{3\,b^2}-\frac {1}{3\,b}+\frac {175\,a^2\,x^4}{24\,b^3}+\frac {35\,a^3\,x^6}{8\,b^4}}{a^2\,x^7+2\,a\,b\,x^5+b^2\,x^3}+\frac {35\,a^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {a}\,x}{\sqrt {b}}\right )}{8\,b^{9/2}} \]

[In]

int(1/(x^10*(a + b/x^2)^3),x)

[Out]

((7*a*x^2)/(3*b^2) - 1/(3*b) + (175*a^2*x^4)/(24*b^3) + (35*a^3*x^6)/(8*b^4))/(a^2*x^7 + b^2*x^3 + 2*a*b*x^5)
+ (35*a^(3/2)*atan((a^(1/2)*x)/b^(1/2)))/(8*b^(9/2))