3.643 \(\int \frac{1}{x^2 (a^2+2 a b x^2+b^2 x^4)^{3/2}} \, dx\)

Optimal. Leaf size=169 \[ -\frac{15 \left (a+b x^2\right )}{8 a^3 x \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{5}{8 a^2 x \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{1}{4 a x \sqrt{a^2+2 a b x^2+b^2 x^4} \left (a+b x^2\right )}-\frac{15 \sqrt{b} \left (a+b x^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{8 a^{7/2} \sqrt{a^2+2 a b x^2+b^2 x^4}} \]

[Out]

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

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Rubi [A]  time = 0.0653457, antiderivative size = 169, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {1112, 290, 325, 205} \[ -\frac{15 \left (a+b x^2\right )}{8 a^3 x \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{5}{8 a^2 x \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{1}{4 a x \sqrt{a^2+2 a b x^2+b^2 x^4} \left (a+b x^2\right )}-\frac{15 \sqrt{b} \left (a+b x^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{8 a^{7/2} \sqrt{a^2+2 a b x^2+b^2 x^4}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^2*(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2)),x]

[Out]

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

Rule 1112

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Dist[(a + b*x^2 + c*x^4)^FracPa
rt[p]/(c^IntPart[p]*(b/2 + c*x^2)^(2*FracPart[p])), Int[(d*x)^m*(b/2 + c*x^2)^(2*p), x], x] /; FreeQ[{a, b, c,
d, m, p}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p - 1/2]

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

Mathematica [A]  time = 0.0322544, size = 93, normalized size = 0.55 \[ \frac{-\sqrt{a} \left (8 a^2+25 a b x^2+15 b^2 x^4\right )-15 \sqrt{b} x \left (a+b x^2\right )^2 \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{8 a^{7/2} x \left (a+b x^2\right ) \sqrt{\left (a+b x^2\right )^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^2*(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2)),x]

[Out]

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

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Maple [A]  time = 0.239, size = 119, normalized size = 0.7 \begin{align*} -{\frac{b{x}^{2}+a}{8\,x{a}^{3}} \left ( 15\,\arctan \left ({\frac{bx}{\sqrt{ab}}} \right ){x}^{5}{b}^{3}+15\,\sqrt{ab}{x}^{4}{b}^{2}+30\,\arctan \left ({\frac{bx}{\sqrt{ab}}} \right ){x}^{3}a{b}^{2}+25\,\sqrt{ab}{x}^{2}ab+15\,\arctan \left ({\frac{bx}{\sqrt{ab}}} \right ) x{a}^{2}b+8\,\sqrt{ab}{a}^{2} \right ){\frac{1}{\sqrt{ab}}} \left ( \left ( b{x}^{2}+a \right ) ^{2} \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^2/(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x)

[Out]

-1/8*(15*arctan(b*x/(a*b)^(1/2))*x^5*b^3+15*(a*b)^(1/2)*x^4*b^2+30*arctan(b*x/(a*b)^(1/2))*x^3*a*b^2+25*(a*b)^
(1/2)*x^2*a*b+15*arctan(b*x/(a*b)^(1/2))*x*a^2*b+8*(a*b)^(1/2)*a^2)*(b*x^2+a)/(a*b)^(1/2)/x/a^3/((b*x^2+a)^2)^
(3/2)

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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^2*x^4+2*a*b*x^2+a^2)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.32713, size = 428, normalized size = 2.53 \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^2*x^4+2*a*b*x^2+a^2)^(3/2),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 [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x^{2} \left (\left (a + b x^{2}\right )^{2}\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**2/(b**2*x**4+2*a*b*x**2+a**2)**(3/2),x)

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{sage}_{0} x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x, algorithm="giac")

[Out]

sage0*x