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

Optimal. Leaf size=147 \[ \frac{1}{4 a \left (a+b x^2\right ) \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{1}{2 a^2 \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{\log (x) \left (a+b x^2\right )}{a^3 \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{\left (a+b x^2\right ) \log \left (a+b x^2\right )}{2 a^3 \sqrt{a^2+2 a b x^2+b^2 x^4}} \]

[Out]

1/(2*a^2*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) + 1/(4*a*(a + b*x^2)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) + ((a + b*x^2)
*Log[x])/(a^3*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) - ((a + b*x^2)*Log[a + b*x^2])/(2*a^3*Sqrt[a^2 + 2*a*b*x^2 + b^
2*x^4])

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Rubi [A]  time = 0.0826822, antiderivative size = 147, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.115, Rules used = {1112, 266, 44} \[ \frac{1}{4 a \left (a+b x^2\right ) \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{1}{2 a^2 \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{\log (x) \left (a+b x^2\right )}{a^3 \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{\left (a+b x^2\right ) \log \left (a+b x^2\right )}{2 a^3 \sqrt{a^2+2 a b x^2+b^2 x^4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

1/(2*a^2*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) + 1/(4*a*(a + b*x^2)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) + ((a + b*x^2)
*Log[x])/(a^3*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) - ((a + b*x^2)*Log[a + b*x^2])/(2*a^3*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 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \frac{1}{x \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 \left (a b+b^2 x^2\right )^3} \, dx}{\sqrt{a^2+2 a b x^2+b^2 x^4}}\\ &=\frac{\left (b^2 \left (a b+b^2 x^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{x \left (a b+b^2 x\right )^3} \, dx,x,x^2\right )}{2 \sqrt{a^2+2 a b x^2+b^2 x^4}}\\ &=\frac{\left (b^2 \left (a b+b^2 x^2\right )\right ) \operatorname{Subst}\left (\int \left (\frac{1}{a^3 b^3 x}-\frac{1}{a b^2 (a+b x)^3}-\frac{1}{a^2 b^2 (a+b x)^2}-\frac{1}{a^3 b^2 (a+b x)}\right ) \, dx,x,x^2\right )}{2 \sqrt{a^2+2 a b x^2+b^2 x^4}}\\ &=\frac{1}{2 a^2 \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{1}{4 a \left (a+b x^2\right ) \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{\left (a+b x^2\right ) \log (x)}{a^3 \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{\left (a+b x^2\right ) \log \left (a+b x^2\right )}{2 a^3 \sqrt{a^2+2 a b x^2+b^2 x^4}}\\ \end{align*}

Mathematica [A]  time = 0.0263441, size = 74, normalized size = 0.5 \[ \frac{a \left (3 a+2 b x^2\right )+4 \log (x) \left (a+b x^2\right )^2-2 \left (a+b x^2\right )^2 \log \left (a+b x^2\right )}{4 a^3 \left (a+b x^2\right ) \sqrt{\left (a+b x^2\right )^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(3*a + 2*b*x^2) + 4*(a + b*x^2)^2*Log[x] - 2*(a + b*x^2)^2*Log[a + b*x^2])/(4*a^3*(a + b*x^2)*Sqrt[(a + b*x
^2)^2])

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Maple [A]  time = 0.231, size = 107, normalized size = 0.7 \begin{align*} -{\frac{ \left ( 2\,\ln \left ( b{x}^{2}+a \right ){x}^{4}{b}^{2}-4\,\ln \left ( x \right ){x}^{4}{b}^{2}+4\,\ln \left ( b{x}^{2}+a \right ){x}^{2}ab-8\,\ln \left ( x \right ){x}^{2}ab-2\,ab{x}^{2}+2\,{a}^{2}\ln \left ( b{x}^{2}+a \right ) -4\,{a}^{2}\ln \left ( x \right ) -3\,{a}^{2} \right ) \left ( b{x}^{2}+a \right ) }{4\,{a}^{3}} \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/(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x)

[Out]

-1/4*(2*ln(b*x^2+a)*x^4*b^2-4*ln(x)*x^4*b^2+4*ln(b*x^2+a)*x^2*a*b-8*ln(x)*x^2*a*b-2*a*b*x^2+2*a^2*ln(b*x^2+a)-
4*a^2*ln(x)-3*a^2)*(b*x^2+a)/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/(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 = 1.26835, size = 196, normalized size = 1.33 \begin{align*} \frac{2 \, a b x^{2} + 3 \, a^{2} - 2 \,{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )} \log \left (b x^{2} + a\right ) + 4 \,{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )} \log \left (x\right )}{4 \,{\left (a^{3} b^{2} x^{4} + 2 \, a^{4} b x^{2} + a^{5}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(2*a*b*x^2 + 3*a^2 - 2*(b^2*x^4 + 2*a*b*x^2 + a^2)*log(b*x^2 + a) + 4*(b^2*x^4 + 2*a*b*x^2 + a^2)*log(x))/
(a^3*b^2*x^4 + 2*a^4*b*x^2 + a^5)

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

[Out]

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

[Out]

sage0*x