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

Optimal. Leaf size=209 \[ \frac{35 b \left (a+b x^2\right )}{8 a^4 x \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{35 \left (a+b x^2\right )}{24 a^3 x^3 \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{1}{4 a x^3 \left (a+b x^2\right ) \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{7}{8 a^2 x^3 \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{35 b^{3/2} \left (a+b x^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{8 a^{9/2} \sqrt{a^2+2 a b x^2+b^2 x^4}} \]

[Out]

7/(8*a^2*x^3*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) + 1/(4*a*x^3*(a + b*x^2)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) - (35*
(a + b*x^2))/(24*a^3*x^3*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) + (35*b*(a + b*x^2))/(8*a^4*x*Sqrt[a^2 + 2*a*b*x^2 +
b^2*x^4]) + (35*b^(3/2)*(a + b*x^2)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(8*a^(9/2)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])

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Rubi [A]  time = 0.0791235, antiderivative size = 209, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {1112, 290, 325, 205} \[ \frac{35 b \left (a+b x^2\right )}{8 a^4 x \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{35 \left (a+b x^2\right )}{24 a^3 x^3 \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{1}{4 a x^3 \left (a+b x^2\right ) \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{7}{8 a^2 x^3 \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{35 b^{3/2} \left (a+b x^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{8 a^{9/2} \sqrt{a^2+2 a b x^2+b^2 x^4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

7/(8*a^2*x^3*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) + 1/(4*a*x^3*(a + b*x^2)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) - (35*
(a + b*x^2))/(24*a^3*x^3*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) + (35*b*(a + b*x^2))/(8*a^4*x*Sqrt[a^2 + 2*a*b*x^2 +
b^2*x^4]) + (35*b^(3/2)*(a + b*x^2)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(8*a^(9/2)*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^4 \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^4 \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^3 \left (a+b x^2\right ) \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{\left (7 b \left (a b+b^2 x^2\right )\right ) \int \frac{1}{x^4 \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{7}{8 a^2 x^3 \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{1}{4 a x^3 \left (a+b x^2\right ) \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{\left (35 \left (a b+b^2 x^2\right )\right ) \int \frac{1}{x^4 \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{7}{8 a^2 x^3 \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{1}{4 a x^3 \left (a+b x^2\right ) \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{35 \left (a+b x^2\right )}{24 a^3 x^3 \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{\left (35 b \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^3 \sqrt{a^2+2 a b x^2+b^2 x^4}}\\ &=\frac{7}{8 a^2 x^3 \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{1}{4 a x^3 \left (a+b x^2\right ) \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{35 \left (a+b x^2\right )}{24 a^3 x^3 \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{35 b \left (a+b x^2\right )}{8 a^4 x \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{\left (35 b^2 \left (a b+b^2 x^2\right )\right ) \int \frac{1}{a b+b^2 x^2} \, dx}{8 a^4 \sqrt{a^2+2 a b x^2+b^2 x^4}}\\ &=\frac{7}{8 a^2 x^3 \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{1}{4 a x^3 \left (a+b x^2\right ) \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{35 \left (a+b x^2\right )}{24 a^3 x^3 \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{35 b \left (a+b x^2\right )}{8 a^4 x \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{35 b^{3/2} \left (a+b x^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{8 a^{9/2} \sqrt{a^2+2 a b x^2+b^2 x^4}}\\ \end{align*}

Mathematica [A]  time = 0.0380871, size = 105, normalized size = 0.5 \[ \frac{\sqrt{a} \left (56 a^2 b x^2-8 a^3+175 a b^2 x^4+105 b^3 x^6\right )+105 b^{3/2} x^3 \left (a+b x^2\right )^2 \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{24 a^{9/2} x^3 \left (a+b x^2\right ) \sqrt{\left (a+b x^2\right )^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a]*(-8*a^3 + 56*a^2*b*x^2 + 175*a*b^2*x^4 + 105*b^3*x^6) + 105*b^(3/2)*x^3*(a + b*x^2)^2*ArcTan[(Sqrt[b]
*x)/Sqrt[a]])/(24*a^(9/2)*x^3*(a + b*x^2)*Sqrt[(a + b*x^2)^2])

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Maple [A]  time = 0.235, size = 139, normalized size = 0.7 \begin{align*}{\frac{b{x}^{2}+a}{24\,{a}^{4}{x}^{3}} \left ( 105\,\arctan \left ({\frac{bx}{\sqrt{ab}}} \right ){x}^{7}{b}^{4}+105\,\sqrt{ab}{x}^{6}{b}^{3}+210\,\arctan \left ({\frac{bx}{\sqrt{ab}}} \right ){x}^{5}a{b}^{3}+175\,\sqrt{ab}{x}^{4}a{b}^{2}+105\,\arctan \left ({\frac{bx}{\sqrt{ab}}} \right ){x}^{3}{a}^{2}{b}^{2}+56\,\sqrt{ab}{x}^{2}{a}^{2}b-8\,\sqrt{ab}{a}^{3} \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^4/(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x)

[Out]

1/24*(105*arctan(b*x/(a*b)^(1/2))*x^7*b^4+105*(a*b)^(1/2)*x^6*b^3+210*arctan(b*x/(a*b)^(1/2))*x^5*a*b^3+175*(a
*b)^(1/2)*x^4*a*b^2+105*arctan(b*x/(a*b)^(1/2))*x^3*a^2*b^2+56*(a*b)^(1/2)*x^2*a^2*b-8*(a*b)^(1/2)*a^3)*(b*x^2
+a)/(a*b)^(1/2)/x^3/a^4/((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^4/(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 = 1.23424, size = 504, normalized size = 2.41 \begin{align*} \left [\frac{210 \, b^{3} x^{6} + 350 \, a b^{2} x^{4} + 112 \, a^{2} b x^{2} - 16 \, a^{3} + 105 \,{\left (b^{3} x^{7} + 2 \, a b^{2} x^{5} + a^{2} b x^{3}\right )} \sqrt{-\frac{b}{a}} \log \left (\frac{b x^{2} + 2 \, a x \sqrt{-\frac{b}{a}} - a}{b x^{2} + a}\right )}{48 \,{\left (a^{4} b^{2} x^{7} + 2 \, a^{5} b x^{5} + a^{6} x^{3}\right )}}, \frac{105 \, b^{3} x^{6} + 175 \, a b^{2} x^{4} + 56 \, a^{2} b x^{2} - 8 \, a^{3} + 105 \,{\left (b^{3} x^{7} + 2 \, a b^{2} x^{5} + a^{2} b x^{3}\right )} \sqrt{\frac{b}{a}} \arctan \left (x \sqrt{\frac{b}{a}}\right )}{24 \,{\left (a^{4} b^{2} x^{7} + 2 \, a^{5} b x^{5} + a^{6} x^{3}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/48*(210*b^3*x^6 + 350*a*b^2*x^4 + 112*a^2*b*x^2 - 16*a^3 + 105*(b^3*x^7 + 2*a*b^2*x^5 + a^2*b*x^3)*sqrt(-b/
a)*log((b*x^2 + 2*a*x*sqrt(-b/a) - a)/(b*x^2 + a)))/(a^4*b^2*x^7 + 2*a^5*b*x^5 + a^6*x^3), 1/24*(105*b^3*x^6 +
175*a*b^2*x^4 + 56*a^2*b*x^2 - 8*a^3 + 105*(b^3*x^7 + 2*a*b^2*x^5 + a^2*b*x^3)*sqrt(b/a)*arctan(x*sqrt(b/a)))
/(a^4*b^2*x^7 + 2*a^5*b*x^5 + a^6*x^3)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x^{4} \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**4/(b**2*x**4+2*a*b*x**2+a**2)**(3/2),x)

[Out]

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

[Out]

sage0*x