3.657 \(\int \frac{1}{x^2 (a^2+2 a b x^2+b^2 x^4)^{5/2}} \, dx\)

Optimal. Leaf size=251 \[ -\frac{315 \left (a+b x^2\right )}{128 a^5 x \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{105}{128 a^4 x \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{21}{64 a^3 x \sqrt{a^2+2 a b x^2+b^2 x^4} \left (a+b x^2\right )}+\frac{3}{16 a^2 x \sqrt{a^2+2 a b x^2+b^2 x^4} \left (a+b x^2\right )^2}+\frac{1}{8 a x \sqrt{a^2+2 a b x^2+b^2 x^4} \left (a+b x^2\right )^3}-\frac{315 \sqrt{b} \left (a+b x^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{128 a^{11/2} \sqrt{a^2+2 a b x^2+b^2 x^4}} \]

[Out]

105/(128*a^4*x*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) + 1/(8*a*x*(a + b*x^2)^3*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) + 3/
(16*a^2*x*(a + b*x^2)^2*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) + 21/(64*a^3*x*(a + b*x^2)*Sqrt[a^2 + 2*a*b*x^2 + b^2
*x^4]) - (315*(a + b*x^2))/(128*a^5*x*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) - (315*Sqrt[b]*(a + b*x^2)*ArcTan[(Sqrt
[b]*x)/Sqrt[a]])/(128*a^(11/2)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])

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Rubi [A]  time = 0.109376, antiderivative size = 251, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {1112, 290, 325, 205} \[ -\frac{315 \left (a+b x^2\right )}{128 a^5 x \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{105}{128 a^4 x \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{21}{64 a^3 x \sqrt{a^2+2 a b x^2+b^2 x^4} \left (a+b x^2\right )}+\frac{3}{16 a^2 x \sqrt{a^2+2 a b x^2+b^2 x^4} \left (a+b x^2\right )^2}+\frac{1}{8 a x \sqrt{a^2+2 a b x^2+b^2 x^4} \left (a+b x^2\right )^3}-\frac{315 \sqrt{b} \left (a+b x^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{128 a^{11/2} \sqrt{a^2+2 a b x^2+b^2 x^4}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^2*(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2)),x]

[Out]

105/(128*a^4*x*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) + 1/(8*a*x*(a + b*x^2)^3*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) + 3/
(16*a^2*x*(a + b*x^2)^2*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) + 21/(64*a^3*x*(a + b*x^2)*Sqrt[a^2 + 2*a*b*x^2 + b^2
*x^4]) - (315*(a + b*x^2))/(128*a^5*x*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) - (315*Sqrt[b]*(a + b*x^2)*ArcTan[(Sqrt
[b]*x)/Sqrt[a]])/(128*a^(11/2)*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 )^{5/2}} \, dx &=\frac{\left (b^4 \left (a b+b^2 x^2\right )\right ) \int \frac{1}{x^2 \left (a b+b^2 x^2\right )^5} \, dx}{\sqrt{a^2+2 a b x^2+b^2 x^4}}\\ &=\frac{1}{8 a x \left (a+b x^2\right )^3 \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{\left (9 b^3 \left (a b+b^2 x^2\right )\right ) \int \frac{1}{x^2 \left (a b+b^2 x^2\right )^4} \, dx}{8 a \sqrt{a^2+2 a b x^2+b^2 x^4}}\\ &=\frac{1}{8 a x \left (a+b x^2\right )^3 \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{3}{16 a^2 x \left (a+b x^2\right )^2 \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{\left (21 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}{16 a^2 \sqrt{a^2+2 a b x^2+b^2 x^4}}\\ &=\frac{1}{8 a x \left (a+b x^2\right )^3 \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{3}{16 a^2 x \left (a+b x^2\right )^2 \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{21}{64 a^3 x \left (a+b x^2\right ) \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{\left (105 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}{64 a^3 \sqrt{a^2+2 a b x^2+b^2 x^4}}\\ &=\frac{105}{128 a^4 x \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{1}{8 a x \left (a+b x^2\right )^3 \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{3}{16 a^2 x \left (a+b x^2\right )^2 \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{21}{64 a^3 x \left (a+b x^2\right ) \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{\left (315 \left (a b+b^2 x^2\right )\right ) \int \frac{1}{x^2 \left (a b+b^2 x^2\right )} \, dx}{128 a^4 \sqrt{a^2+2 a b x^2+b^2 x^4}}\\ &=\frac{105}{128 a^4 x \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{1}{8 a x \left (a+b x^2\right )^3 \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{3}{16 a^2 x \left (a+b x^2\right )^2 \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{21}{64 a^3 x \left (a+b x^2\right ) \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{315 \left (a+b x^2\right )}{128 a^5 x \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{\left (315 b \left (a b+b^2 x^2\right )\right ) \int \frac{1}{a b+b^2 x^2} \, dx}{128 a^5 \sqrt{a^2+2 a b x^2+b^2 x^4}}\\ &=\frac{105}{128 a^4 x \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{1}{8 a x \left (a+b x^2\right )^3 \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{3}{16 a^2 x \left (a+b x^2\right )^2 \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{21}{64 a^3 x \left (a+b x^2\right ) \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{315 \left (a+b x^2\right )}{128 a^5 x \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{315 \sqrt{b} \left (a+b x^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{128 a^{11/2} \sqrt{a^2+2 a b x^2+b^2 x^4}}\\ \end{align*}

Mathematica [A]  time = 0.0440012, size = 115, normalized size = 0.46 \[ \frac{-\sqrt{a} \left (1533 a^2 b^2 x^4+837 a^3 b x^2+128 a^4+1155 a b^3 x^6+315 b^4 x^8\right )-315 \sqrt{b} x \left (a+b x^2\right )^4 \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{128 a^{11/2} x \left (a+b x^2\right )^3 \sqrt{\left (a+b x^2\right )^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^2*(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2)),x]

[Out]

(-(Sqrt[a]*(128*a^4 + 837*a^3*b*x^2 + 1533*a^2*b^2*x^4 + 1155*a*b^3*x^6 + 315*b^4*x^8)) - 315*Sqrt[b]*x*(a + b
*x^2)^4*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(128*a^(11/2)*x*(a + b*x^2)^3*Sqrt[(a + b*x^2)^2])

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Maple [A]  time = 0.233, size = 191, normalized size = 0.8 \begin{align*} -{\frac{b{x}^{2}+a}{128\,x{a}^{5}} \left ( 315\,\arctan \left ({\frac{bx}{\sqrt{ab}}} \right ){x}^{9}{b}^{5}+315\,\sqrt{ab}{x}^{8}{b}^{4}+1260\,\arctan \left ({\frac{bx}{\sqrt{ab}}} \right ){x}^{7}a{b}^{4}+1155\,\sqrt{ab}{x}^{6}a{b}^{3}+1890\,\arctan \left ({\frac{bx}{\sqrt{ab}}} \right ){x}^{5}{a}^{2}{b}^{3}+1533\,\sqrt{ab}{x}^{4}{a}^{2}{b}^{2}+1260\,\arctan \left ({\frac{bx}{\sqrt{ab}}} \right ){x}^{3}{a}^{3}{b}^{2}+837\,\sqrt{ab}{x}^{2}{a}^{3}b+315\,\arctan \left ({\frac{bx}{\sqrt{ab}}} \right ) x{a}^{4}b+128\,\sqrt{ab}{a}^{4} \right ){\frac{1}{\sqrt{ab}}} \left ( \left ( b{x}^{2}+a \right ) ^{2} \right ) ^{-{\frac{5}{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)^(5/2),x)

[Out]

-1/128*(315*arctan(b*x/(a*b)^(1/2))*x^9*b^5+315*(a*b)^(1/2)*x^8*b^4+1260*arctan(b*x/(a*b)^(1/2))*x^7*a*b^4+115
5*(a*b)^(1/2)*x^6*a*b^3+1890*arctan(b*x/(a*b)^(1/2))*x^5*a^2*b^3+1533*(a*b)^(1/2)*x^4*a^2*b^2+1260*arctan(b*x/
(a*b)^(1/2))*x^3*a^3*b^2+837*(a*b)^(1/2)*x^2*a^3*b+315*arctan(b*x/(a*b)^(1/2))*x*a^4*b+128*(a*b)^(1/2)*a^4)*(b
*x^2+a)/(a*b)^(1/2)/x/a^5/((b*x^2+a)^2)^(5/2)

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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)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.40395, size = 721, normalized size = 2.87 \begin{align*} \left [-\frac{630 \, b^{4} x^{8} + 2310 \, a b^{3} x^{6} + 3066 \, a^{2} b^{2} x^{4} + 1674 \, a^{3} b x^{2} + 256 \, a^{4} - 315 \,{\left (b^{4} x^{9} + 4 \, a b^{3} x^{7} + 6 \, a^{2} b^{2} x^{5} + 4 \, a^{3} b x^{3} + a^{4} 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 )}{256 \,{\left (a^{5} b^{4} x^{9} + 4 \, a^{6} b^{3} x^{7} + 6 \, a^{7} b^{2} x^{5} + 4 \, a^{8} b x^{3} + a^{9} x\right )}}, -\frac{315 \, b^{4} x^{8} + 1155 \, a b^{3} x^{6} + 1533 \, a^{2} b^{2} x^{4} + 837 \, a^{3} b x^{2} + 128 \, a^{4} + 315 \,{\left (b^{4} x^{9} + 4 \, a b^{3} x^{7} + 6 \, a^{2} b^{2} x^{5} + 4 \, a^{3} b x^{3} + a^{4} x\right )} \sqrt{\frac{b}{a}} \arctan \left (x \sqrt{\frac{b}{a}}\right )}{128 \,{\left (a^{5} b^{4} x^{9} + 4 \, a^{6} b^{3} x^{7} + 6 \, a^{7} b^{2} x^{5} + 4 \, a^{8} b x^{3} + a^{9} 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)^(5/2),x, algorithm="fricas")

[Out]

[-1/256*(630*b^4*x^8 + 2310*a*b^3*x^6 + 3066*a^2*b^2*x^4 + 1674*a^3*b*x^2 + 256*a^4 - 315*(b^4*x^9 + 4*a*b^3*x
^7 + 6*a^2*b^2*x^5 + 4*a^3*b*x^3 + a^4*x)*sqrt(-b/a)*log((b*x^2 - 2*a*x*sqrt(-b/a) - a)/(b*x^2 + a)))/(a^5*b^4
*x^9 + 4*a^6*b^3*x^7 + 6*a^7*b^2*x^5 + 4*a^8*b*x^3 + a^9*x), -1/128*(315*b^4*x^8 + 1155*a*b^3*x^6 + 1533*a^2*b
^2*x^4 + 837*a^3*b*x^2 + 128*a^4 + 315*(b^4*x^9 + 4*a*b^3*x^7 + 6*a^2*b^2*x^5 + 4*a^3*b*x^3 + a^4*x)*sqrt(b/a)
*arctan(x*sqrt(b/a)))/(a^5*b^4*x^9 + 4*a^6*b^3*x^7 + 6*a^7*b^2*x^5 + 4*a^8*b*x^3 + a^9*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{5}{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)**(5/2),x)

[Out]

Integral(1/(x**2*((a + b*x**2)**2)**(5/2)), 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)^(5/2),x, algorithm="giac")

[Out]

sage0*x