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

Optimal. Leaf size=113 \[ \frac {99 b^{7/2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{13/2}}+\frac {99 b^3}{8 a^6 x}-\frac {33 b^2}{8 a^5 x^3}+\frac {99 b}{40 a^4 x^5}-\frac {99}{56 a^3 x^7}+\frac {11}{8 a^2 x^7 \left (a+b x^2\right )}+\frac {1}{4 a x^7 \left (a+b x^2\right )^2} \]

[Out]

-99/56/a^3/x^7+99/40*b/a^4/x^5-33/8*b^2/a^5/x^3+99/8*b^3/a^6/x+1/4/a/x^7/(b*x^2+a)^2+11/8/a^2/x^7/(b*x^2+a)+99
/8*b^(7/2)*arctan(x*b^(1/2)/a^(1/2))/a^(13/2)

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Rubi [A]  time = 0.06, antiderivative size = 113, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 3, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {290, 325, 205} \[ -\frac {33 b^2}{8 a^5 x^3}+\frac {99 b^3}{8 a^6 x}+\frac {99 b^{7/2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{13/2}}+\frac {99 b}{40 a^4 x^5}+\frac {11}{8 a^2 x^7 \left (a+b x^2\right )}-\frac {99}{56 a^3 x^7}+\frac {1}{4 a x^7 \left (a+b x^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^8*(a + b*x^2)^3),x]

[Out]

-99/(56*a^3*x^7) + (99*b)/(40*a^4*x^5) - (33*b^2)/(8*a^5*x^3) + (99*b^3)/(8*a^6*x) + 1/(4*a*x^7*(a + b*x^2)^2)
+ 11/(8*a^2*x^7*(a + b*x^2)) + (99*b^(7/2)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(8*a^(13/2))

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 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}{x^8 \left (a+b x^2\right )^3} \, dx &=\frac {1}{4 a x^7 \left (a+b x^2\right )^2}+\frac {11 \int \frac {1}{x^8 \left (a+b x^2\right )^2} \, dx}{4 a}\\ &=\frac {1}{4 a x^7 \left (a+b x^2\right )^2}+\frac {11}{8 a^2 x^7 \left (a+b x^2\right )}+\frac {99 \int \frac {1}{x^8 \left (a+b x^2\right )} \, dx}{8 a^2}\\ &=-\frac {99}{56 a^3 x^7}+\frac {1}{4 a x^7 \left (a+b x^2\right )^2}+\frac {11}{8 a^2 x^7 \left (a+b x^2\right )}-\frac {(99 b) \int \frac {1}{x^6 \left (a+b x^2\right )} \, dx}{8 a^3}\\ &=-\frac {99}{56 a^3 x^7}+\frac {99 b}{40 a^4 x^5}+\frac {1}{4 a x^7 \left (a+b x^2\right )^2}+\frac {11}{8 a^2 x^7 \left (a+b x^2\right )}+\frac {\left (99 b^2\right ) \int \frac {1}{x^4 \left (a+b x^2\right )} \, dx}{8 a^4}\\ &=-\frac {99}{56 a^3 x^7}+\frac {99 b}{40 a^4 x^5}-\frac {33 b^2}{8 a^5 x^3}+\frac {1}{4 a x^7 \left (a+b x^2\right )^2}+\frac {11}{8 a^2 x^7 \left (a+b x^2\right )}-\frac {\left (99 b^3\right ) \int \frac {1}{x^2 \left (a+b x^2\right )} \, dx}{8 a^5}\\ &=-\frac {99}{56 a^3 x^7}+\frac {99 b}{40 a^4 x^5}-\frac {33 b^2}{8 a^5 x^3}+\frac {99 b^3}{8 a^6 x}+\frac {1}{4 a x^7 \left (a+b x^2\right )^2}+\frac {11}{8 a^2 x^7 \left (a+b x^2\right )}+\frac {\left (99 b^4\right ) \int \frac {1}{a+b x^2} \, dx}{8 a^6}\\ &=-\frac {99}{56 a^3 x^7}+\frac {99 b}{40 a^4 x^5}-\frac {33 b^2}{8 a^5 x^3}+\frac {99 b^3}{8 a^6 x}+\frac {1}{4 a x^7 \left (a+b x^2\right )^2}+\frac {11}{8 a^2 x^7 \left (a+b x^2\right )}+\frac {99 b^{7/2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{13/2}}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 101, normalized size = 0.89 \[ \frac {99 b^{7/2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{13/2}}+\frac {-40 a^5+88 a^4 b x^2-264 a^3 b^2 x^4+1848 a^2 b^3 x^6+5775 a b^4 x^8+3465 b^5 x^{10}}{280 a^6 x^7 \left (a+b x^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^8*(a + b*x^2)^3),x]

[Out]

(-40*a^5 + 88*a^4*b*x^2 - 264*a^3*b^2*x^4 + 1848*a^2*b^3*x^6 + 5775*a*b^4*x^8 + 3465*b^5*x^10)/(280*a^6*x^7*(a
+ b*x^2)^2) + (99*b^(7/2)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(8*a^(13/2))

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fricas [A]  time = 0.95, size = 286, normalized size = 2.53 \[ \left [\frac {6930 \, b^{5} x^{10} + 11550 \, a b^{4} x^{8} + 3696 \, a^{2} b^{3} x^{6} - 528 \, a^{3} b^{2} x^{4} + 176 \, a^{4} b x^{2} - 80 \, a^{5} + 3465 \, {\left (b^{5} x^{11} + 2 \, a b^{4} x^{9} + a^{2} b^{3} x^{7}\right )} \sqrt {-\frac {b}{a}} \log \left (\frac {b x^{2} + 2 \, a x \sqrt {-\frac {b}{a}} - a}{b x^{2} + a}\right )}{560 \, {\left (a^{6} b^{2} x^{11} + 2 \, a^{7} b x^{9} + a^{8} x^{7}\right )}}, \frac {3465 \, b^{5} x^{10} + 5775 \, a b^{4} x^{8} + 1848 \, a^{2} b^{3} x^{6} - 264 \, a^{3} b^{2} x^{4} + 88 \, a^{4} b x^{2} - 40 \, a^{5} + 3465 \, {\left (b^{5} x^{11} + 2 \, a b^{4} x^{9} + a^{2} b^{3} x^{7}\right )} \sqrt {\frac {b}{a}} \arctan \left (x \sqrt {\frac {b}{a}}\right )}{280 \, {\left (a^{6} b^{2} x^{11} + 2 \, a^{7} b x^{9} + a^{8} x^{7}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/560*(6930*b^5*x^10 + 11550*a*b^4*x^8 + 3696*a^2*b^3*x^6 - 528*a^3*b^2*x^4 + 176*a^4*b*x^2 - 80*a^5 + 3465*(
b^5*x^11 + 2*a*b^4*x^9 + a^2*b^3*x^7)*sqrt(-b/a)*log((b*x^2 + 2*a*x*sqrt(-b/a) - a)/(b*x^2 + a)))/(a^6*b^2*x^1
1 + 2*a^7*b*x^9 + a^8*x^7), 1/280*(3465*b^5*x^10 + 5775*a*b^4*x^8 + 1848*a^2*b^3*x^6 - 264*a^3*b^2*x^4 + 88*a^
4*b*x^2 - 40*a^5 + 3465*(b^5*x^11 + 2*a*b^4*x^9 + a^2*b^3*x^7)*sqrt(b/a)*arctan(x*sqrt(b/a)))/(a^6*b^2*x^11 +
2*a^7*b*x^9 + a^8*x^7)]

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giac [A]  time = 0.61, size = 93, normalized size = 0.82 \[ \frac {99 \, b^{4} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a^{6}} + \frac {19 \, b^{5} x^{3} + 21 \, a b^{4} x}{8 \, {\left (b x^{2} + a\right )}^{2} a^{6}} + \frac {350 \, b^{3} x^{6} - 70 \, a b^{2} x^{4} + 21 \, a^{2} b x^{2} - 5 \, a^{3}}{35 \, a^{6} x^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

99/8*b^4*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*a^6) + 1/8*(19*b^5*x^3 + 21*a*b^4*x)/((b*x^2 + a)^2*a^6) + 1/35*(350
*b^3*x^6 - 70*a*b^2*x^4 + 21*a^2*b*x^2 - 5*a^3)/(a^6*x^7)

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maple [A]  time = 0.01, size = 101, normalized size = 0.89 \[ \frac {19 b^{5} x^{3}}{8 \left (b \,x^{2}+a \right )^{2} a^{6}}+\frac {21 b^{4} x}{8 \left (b \,x^{2}+a \right )^{2} a^{5}}+\frac {99 b^{4} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \sqrt {a b}\, a^{6}}+\frac {10 b^{3}}{a^{6} x}-\frac {2 b^{2}}{a^{5} x^{3}}+\frac {3 b}{5 a^{4} x^{5}}-\frac {1}{7 a^{3} x^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/7/a^3/x^7+10*b^3/a^6/x-2*b^2/a^5/x^3+3/5*b/a^4/x^5+19/8/a^6*b^5/(b*x^2+a)^2*x^3+21/8/a^5*b^4/(b*x^2+a)^2*x+
99/8/a^6*b^4/(a*b)^(1/2)*arctan(1/(a*b)^(1/2)*b*x)

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maxima [A]  time = 3.02, size = 108, normalized size = 0.96 \[ \frac {3465 \, b^{5} x^{10} + 5775 \, a b^{4} x^{8} + 1848 \, a^{2} b^{3} x^{6} - 264 \, a^{3} b^{2} x^{4} + 88 \, a^{4} b x^{2} - 40 \, a^{5}}{280 \, {\left (a^{6} b^{2} x^{11} + 2 \, a^{7} b x^{9} + a^{8} x^{7}\right )}} + \frac {99 \, b^{4} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/280*(3465*b^5*x^10 + 5775*a*b^4*x^8 + 1848*a^2*b^3*x^6 - 264*a^3*b^2*x^4 + 88*a^4*b*x^2 - 40*a^5)/(a^6*b^2*x
^11 + 2*a^7*b*x^9 + a^8*x^7) + 99/8*b^4*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*a^6)

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mupad [B]  time = 4.98, size = 102, normalized size = 0.90 \[ \frac {\frac {11\,b\,x^2}{35\,a^2}-\frac {1}{7\,a}-\frac {33\,b^2\,x^4}{35\,a^3}+\frac {33\,b^3\,x^6}{5\,a^4}+\frac {165\,b^4\,x^8}{8\,a^5}+\frac {99\,b^5\,x^{10}}{8\,a^6}}{a^2\,x^7+2\,a\,b\,x^9+b^2\,x^{11}}+\frac {99\,b^{7/2}\,\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )}{8\,a^{13/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^8*(a + b*x^2)^3),x)

[Out]

((11*b*x^2)/(35*a^2) - 1/(7*a) - (33*b^2*x^4)/(35*a^3) + (33*b^3*x^6)/(5*a^4) + (165*b^4*x^8)/(8*a^5) + (99*b^
5*x^10)/(8*a^6))/(a^2*x^7 + b^2*x^11 + 2*a*b*x^9) + (99*b^(7/2)*atan((b^(1/2)*x)/a^(1/2)))/(8*a^(13/2))

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sympy [A]  time = 0.58, size = 162, normalized size = 1.43 \[ - \frac {99 \sqrt {- \frac {b^{7}}{a^{13}}} \log {\left (- \frac {a^{7} \sqrt {- \frac {b^{7}}{a^{13}}}}{b^{4}} + x \right )}}{16} + \frac {99 \sqrt {- \frac {b^{7}}{a^{13}}} \log {\left (\frac {a^{7} \sqrt {- \frac {b^{7}}{a^{13}}}}{b^{4}} + x \right )}}{16} + \frac {- 40 a^{5} + 88 a^{4} b x^{2} - 264 a^{3} b^{2} x^{4} + 1848 a^{2} b^{3} x^{6} + 5775 a b^{4} x^{8} + 3465 b^{5} x^{10}}{280 a^{8} x^{7} + 560 a^{7} b x^{9} + 280 a^{6} b^{2} x^{11}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-99*sqrt(-b**7/a**13)*log(-a**7*sqrt(-b**7/a**13)/b**4 + x)/16 + 99*sqrt(-b**7/a**13)*log(a**7*sqrt(-b**7/a**1
3)/b**4 + x)/16 + (-40*a**5 + 88*a**4*b*x**2 - 264*a**3*b**2*x**4 + 1848*a**2*b**3*x**6 + 5775*a*b**4*x**8 + 3
465*b**5*x**10)/(280*a**8*x**7 + 560*a**7*b*x**9 + 280*a**6*b**2*x**11)

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