3.1889 \(\int \frac{1}{(a+\frac{b}{x^2})^3 x^{10}} \, dx\)

Optimal. Leaf size=87 \[ \frac{35 a^{3/2} \tan ^{-1}\left (\frac{\sqrt{a} x}{\sqrt{b}}\right )}{8 b^{9/2}}+\frac{7}{8 b^2 x^3 \left (a x^2+b\right )}+\frac{35 a}{8 b^4 x}+\frac{1}{4 b x^3 \left (a x^2+b\right )^2}-\frac{35}{24 b^3 x^3} \]

[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))

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Rubi [A]  time = 0.0346164, antiderivative size = 87, normalized size of antiderivative = 1., 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, 290, 325, 205} \[ \frac{35 a^{3/2} \tan ^{-1}\left (\frac{\sqrt{a} x}{\sqrt{b}}\right )}{8 b^{9/2}}+\frac{7}{8 b^2 x^3 \left (a x^2+b\right )}+\frac{35 a}{8 b^4 x}+\frac{1}{4 b x^3 \left (a x^2+b\right )^2}-\frac{35}{24 b^3 x^3} \]

Antiderivative was successfully verified.

[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 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 290

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] /; FreeQ[
{a, b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 325

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]

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]

Rubi steps

\begin{align*} \int \frac{1}{\left (a+\frac{b}{x^2}\right )^3 x^{10}} \, dx &=\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]  time = 0.0414718, size = 79, normalized size = 0.91 \[ \frac{175 a^2 b x^4+105 a^3 x^6+56 a b^2 x^2-8 b^3}{24 b^4 x^3 \left (a x^2+b\right )^2}+\frac{35 a^{3/2} \tan ^{-1}\left (\frac{\sqrt{a} x}{\sqrt{b}}\right )}{8 b^{9/2}} \]

Antiderivative was successfully verified.

[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))

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Maple [A]  time = 0.012, size = 79, normalized size = 0.9 \begin{align*}{\frac{11\,{a}^{3}{x}^{3}}{8\,{b}^{4} \left ( a{x}^{2}+b \right ) ^{2}}}+{\frac{13\,{a}^{2}x}{8\,{b}^{3} \left ( a{x}^{2}+b \right ) ^{2}}}+{\frac{35\,{a}^{2}}{8\,{b}^{4}}\arctan \left ({ax{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}-{\frac{1}{3\,{b}^{3}{x}^{3}}}+3\,{\frac{a}{{b}^{4}x}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.45662, size = 504, normalized size = 5.79 \begin{align*} \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 ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)]

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Sympy [A]  time = 1.09174, size = 138, normalized size = 1.59 \begin{align*} - \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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Giac [A]  time = 1.20722, size = 96, normalized size = 1.1 \begin{align*} \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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)