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

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

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {290, 325, 205} \[ \frac {5}{8 a^2 x \left (a+b x^2\right )}-\frac {15 \sqrt {b} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{7/2}}-\frac {15}{8 a^3 x}+\frac {1}{4 a x \left (a+b x^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^2*(a + b*x^2)^3),x]

[Out]

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

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

Antiderivative was successfully verified.

[In]

Integrate[1/(x^2*(a + b*x^2)^3),x]

[Out]

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

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fricas [A]  time = 0.94, size = 202, normalized size = 2.66 \[ \left [-\frac {30 \, b^{2} x^{4} + 50 \, a b x^{2} - 15 \, {\left (b^{2} x^{5} + 2 \, a b x^{3} + a^{2} x\right )} \sqrt {-\frac {b}{a}} \log \left (\frac {b x^{2} - 2 \, a x \sqrt {-\frac {b}{a}} - a}{b x^{2} + a}\right ) + 16 \, a^{2}}{16 \, {\left (a^{3} b^{2} x^{5} + 2 \, a^{4} b x^{3} + a^{5} x\right )}}, -\frac {15 \, b^{2} x^{4} + 25 \, a b x^{2} + 15 \, {\left (b^{2} x^{5} + 2 \, a b x^{3} + a^{2} x\right )} \sqrt {\frac {b}{a}} \arctan \left (x \sqrt {\frac {b}{a}}\right ) + 8 \, a^{2}}{8 \, {\left (a^{3} b^{2} x^{5} + 2 \, a^{4} b x^{3} + a^{5} x\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.64, size = 57, normalized size = 0.75 \[ -\frac {15 \, b \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a^{3}} - \frac {7 \, b^{2} x^{3} + 9 \, a b x}{8 \, {\left (b x^{2} + a\right )}^{2} a^{3}} - \frac {1}{a^{3} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 2.98, size = 71, normalized size = 0.93 \[ -\frac {15 \, b^{2} x^{4} + 25 \, a b x^{2} + 8 \, a^{2}}{8 \, {\left (a^{3} b^{2} x^{5} + 2 \, a^{4} b x^{3} + a^{5} x\right )}} - \frac {15 \, b \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 4.67, size = 66, normalized size = 0.87 \[ -\frac {\frac {1}{a}+\frac {25\,b\,x^2}{8\,a^2}+\frac {15\,b^2\,x^4}{8\,a^3}}{a^2\,x+2\,a\,b\,x^3+b^2\,x^5}-\frac {15\,\sqrt {b}\,\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )}{8\,a^{7/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^2*(a + b*x^2)^3),x)

[Out]

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

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sympy [A]  time = 0.42, size = 116, normalized size = 1.53 \[ \frac {15 \sqrt {- \frac {b}{a^{7}}} \log {\left (- \frac {a^{4} \sqrt {- \frac {b}{a^{7}}}}{b} + x \right )}}{16} - \frac {15 \sqrt {- \frac {b}{a^{7}}} \log {\left (\frac {a^{4} \sqrt {- \frac {b}{a^{7}}}}{b} + x \right )}}{16} + \frac {- 8 a^{2} - 25 a b x^{2} - 15 b^{2} x^{4}}{8 a^{5} x + 16 a^{4} b x^{3} + 8 a^{3} b^{2} x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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