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

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

[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))

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Rubi [A]  time = 0.0258144, antiderivative size = 76, normalized size of antiderivative = 1., 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 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}{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*}

Mathematica [A]  time = 0.0387667, size = 68, normalized size = 0.89 \[ -\frac{8 a^2+25 a b x^2+15 b^2 x^4}{8 a^3 x \left (a+b x^2\right )^2}-\frac{15 \sqrt{b} \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{8 a^{7/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^2*(a + b*x^2)^3),x]

[Out]

-(8*a^2 + 25*a*b*x^2 + 15*b^2*x^4)/(8*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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Maple [A]  time = 0.01, size = 66, normalized size = 0.9 \begin{align*} -{\frac{1}{{a}^{3}x}}-{\frac{7\,{b}^{2}{x}^{3}}{8\,{a}^{3} \left ( b{x}^{2}+a \right ) ^{2}}}-{\frac{9\,bx}{8\,{a}^{2} \left ( b{x}^{2}+a \right ) ^{2}}}-{\frac{15\,b}{8\,{a}^{3}}\arctan \left ({bx{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}} \end{align*}

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(b*x/(a*b)^(1/2))

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.32791, size = 428, normalized size = 5.63 \begin{align*} \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 ] \end{align*}

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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Sympy [A]  time = 0.636164, size = 114, normalized size = 1.5 \begin{align*} \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}} \end{align*}

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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Giac [A]  time = 2.10063, size = 77, normalized size = 1.01 \begin{align*} -\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} \end{align*}

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)