3.1887 \(\int \frac {1}{(a+\frac {b}{x^2})^3 x^8} \, dx\)

Optimal. Leaf size=76 \[ -\frac {15 \sqrt {a} \tan ^{-1}\left (\frac {\sqrt {a} x}{\sqrt {b}}\right )}{8 b^{7/2}}+\frac {5}{8 b^2 x \left (a x^2+b\right )}+\frac {1}{4 b x \left (a x^2+b\right )^2}-\frac {15}{8 b^3 x} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {263, 290, 325, 205} \[ \frac {5}{8 b^2 x \left (a x^2+b\right )}-\frac {15 \sqrt {a} \tan ^{-1}\left (\frac {\sqrt {a} x}{\sqrt {b}}\right )}{8 b^{7/2}}+\frac {1}{4 b x \left (a x^2+b\right )^2}-\frac {15}{8 b^3 x} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b/x^2)^3*x^8),x]

[Out]

-15/(8*b^3*x) + 1/(4*b*x*(b + a*x^2)^2) + 5/(8*b^2*x*(b + a*x^2)) - (15*Sqrt[a]*ArcTan[(Sqrt[a]*x)/Sqrt[b]])/(
8*b^(7/2))

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 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]

Rubi steps

\begin {align*} \int \frac {1}{\left (a+\frac {b}{x^2}\right )^3 x^8} \, dx &=\int \frac {1}{x^2 \left (b+a x^2\right )^3} \, dx\\ &=\frac {1}{4 b x \left (b+a x^2\right )^2}+\frac {5 \int \frac {1}{x^2 \left (b+a x^2\right )^2} \, dx}{4 b}\\ &=\frac {1}{4 b x \left (b+a x^2\right )^2}+\frac {5}{8 b^2 x \left (b+a x^2\right )}+\frac {15 \int \frac {1}{x^2 \left (b+a x^2\right )} \, dx}{8 b^2}\\ &=-\frac {15}{8 b^3 x}+\frac {1}{4 b x \left (b+a x^2\right )^2}+\frac {5}{8 b^2 x \left (b+a x^2\right )}-\frac {(15 a) \int \frac {1}{b+a x^2} \, dx}{8 b^3}\\ &=-\frac {15}{8 b^3 x}+\frac {1}{4 b x \left (b+a x^2\right )^2}+\frac {5}{8 b^2 x \left (b+a x^2\right )}-\frac {15 \sqrt {a} \tan ^{-1}\left (\frac {\sqrt {a} x}{\sqrt {b}}\right )}{8 b^{7/2}}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 68, normalized size = 0.89 \[ -\frac {15 a^2 x^4+25 a b x^2+8 b^2}{8 b^3 x \left (a x^2+b\right )^2}-\frac {15 \sqrt {a} \tan ^{-1}\left (\frac {\sqrt {a} x}{\sqrt {b}}\right )}{8 b^{7/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + b/x^2)^3*x^8),x]

[Out]

-1/8*(8*b^2 + 25*a*b*x^2 + 15*a^2*x^4)/(b^3*x*(b + a*x^2)^2) - (15*Sqrt[a]*ArcTan[(Sqrt[a]*x)/Sqrt[b]])/(8*b^(
7/2))

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fricas [A]  time = 0.81, size = 202, normalized size = 2.66 \[ \left [-\frac {30 \, a^{2} x^{4} + 50 \, a b x^{2} - 15 \, {\left (a^{2} x^{5} + 2 \, a b x^{3} + b^{2} x\right )} \sqrt {-\frac {a}{b}} \log \left (\frac {a x^{2} - 2 \, b x \sqrt {-\frac {a}{b}} - b}{a x^{2} + b}\right ) + 16 \, b^{2}}{16 \, {\left (a^{2} b^{3} x^{5} + 2 \, a b^{4} x^{3} + b^{5} x\right )}}, -\frac {15 \, a^{2} x^{4} + 25 \, a b x^{2} + 15 \, {\left (a^{2} x^{5} + 2 \, a b x^{3} + b^{2} x\right )} \sqrt {\frac {a}{b}} \arctan \left (x \sqrt {\frac {a}{b}}\right ) + 8 \, b^{2}}{8 \, {\left (a^{2} b^{3} x^{5} + 2 \, a b^{4} x^{3} + b^{5} x\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x^2)^3/x^8,x, algorithm="fricas")

[Out]

[-1/16*(30*a^2*x^4 + 50*a*b*x^2 - 15*(a^2*x^5 + 2*a*b*x^3 + b^2*x)*sqrt(-a/b)*log((a*x^2 - 2*b*x*sqrt(-a/b) -
b)/(a*x^2 + b)) + 16*b^2)/(a^2*b^3*x^5 + 2*a*b^4*x^3 + b^5*x), -1/8*(15*a^2*x^4 + 25*a*b*x^2 + 15*(a^2*x^5 + 2
*a*b*x^3 + b^2*x)*sqrt(a/b)*arctan(x*sqrt(a/b)) + 8*b^2)/(a^2*b^3*x^5 + 2*a*b^4*x^3 + b^5*x)]

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giac [A]  time = 0.16, size = 57, normalized size = 0.75 \[ -\frac {15 \, a \arctan \left (\frac {a x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} b^{3}} - \frac {7 \, a^{2} x^{3} + 9 \, a b x}{8 \, {\left (a x^{2} + b\right )}^{2} b^{3}} - \frac {1}{b^{3} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x^2)^3/x^8,x, algorithm="giac")

[Out]

-15/8*a*arctan(a*x/sqrt(a*b))/(sqrt(a*b)*b^3) - 1/8*(7*a^2*x^3 + 9*a*b*x)/((a*x^2 + b)^2*b^3) - 1/(b^3*x)

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maple [A]  time = 0.01, size = 66, normalized size = 0.87 \[ -\frac {7 a^{2} x^{3}}{8 \left (a \,x^{2}+b \right )^{2} b^{3}}-\frac {9 a x}{8 \left (a \,x^{2}+b \right )^{2} b^{2}}-\frac {15 a \arctan \left (\frac {a x}{\sqrt {a b}}\right )}{8 \sqrt {a b}\, b^{3}}-\frac {1}{b^{3} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b/x^2)^3/x^8,x)

[Out]

-7/8*a^2/b^3/(a*x^2+b)^2*x^3-9/8*a/b^2/(a*x^2+b)^2*x-15/8*a/b^3/(a*b)^(1/2)*arctan(1/(a*b)^(1/2)*a*x)-1/b^3/x

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maxima [A]  time = 1.96, size = 71, normalized size = 0.93 \[ -\frac {15 \, a^{2} x^{4} + 25 \, a b x^{2} + 8 \, b^{2}}{8 \, {\left (a^{2} b^{3} x^{5} + 2 \, a b^{4} x^{3} + b^{5} x\right )}} - \frac {15 \, a \arctan \left (\frac {a x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x^2)^3/x^8,x, algorithm="maxima")

[Out]

-1/8*(15*a^2*x^4 + 25*a*b*x^2 + 8*b^2)/(a^2*b^3*x^5 + 2*a*b^4*x^3 + b^5*x) - 15/8*a*arctan(a*x/sqrt(a*b))/(sqr
t(a*b)*b^3)

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mupad [B]  time = 1.19, size = 66, normalized size = 0.87 \[ -\frac {\frac {1}{b}+\frac {25\,a\,x^2}{8\,b^2}+\frac {15\,a^2\,x^4}{8\,b^3}}{a^2\,x^5+2\,a\,b\,x^3+b^2\,x}-\frac {15\,\sqrt {a}\,\mathrm {atan}\left (\frac {\sqrt {a}\,x}{\sqrt {b}}\right )}{8\,b^{7/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^8*(a + b/x^2)^3),x)

[Out]

- (1/b + (25*a*x^2)/(8*b^2) + (15*a^2*x^4)/(8*b^3))/(b^2*x + a^2*x^5 + 2*a*b*x^3) - (15*a^(1/2)*atan((a^(1/2)*
x)/b^(1/2)))/(8*b^(7/2))

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sympy [A]  time = 0.43, size = 116, normalized size = 1.53 \[ \frac {15 \sqrt {- \frac {a}{b^{7}}} \log {\left (x - \frac {b^{4} \sqrt {- \frac {a}{b^{7}}}}{a} \right )}}{16} - \frac {15 \sqrt {- \frac {a}{b^{7}}} \log {\left (x + \frac {b^{4} \sqrt {- \frac {a}{b^{7}}}}{a} \right )}}{16} + \frac {- 15 a^{2} x^{4} - 25 a b x^{2} - 8 b^{2}}{8 a^{2} b^{3} x^{5} + 16 a b^{4} x^{3} + 8 b^{5} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x**2)**3/x**8,x)

[Out]

15*sqrt(-a/b**7)*log(x - b**4*sqrt(-a/b**7)/a)/16 - 15*sqrt(-a/b**7)*log(x + b**4*sqrt(-a/b**7)/a)/16 + (-15*a
**2*x**4 - 25*a*b*x**2 - 8*b**2)/(8*a**2*b**3*x**5 + 16*a*b**4*x**3 + 8*b**5*x)

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