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

Optimal. Leaf size=122 \[ -\frac{a+b x^2}{2 a x^2 \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{b \log (x) \left (a+b x^2\right )}{a^2 \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{b \left (a+b x^2\right ) \log \left (a+b x^2\right )}{2 a^2 \sqrt{a^2+2 a b x^2+b^2 x^4}} \]

[Out]

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

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Rubi [A]  time = 0.0496018, antiderivative size = 122, 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} \[ -\frac{a+b x^2}{2 a x^2 \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{b \log (x) \left (a+b x^2\right )}{a^2 \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{b \left (a+b x^2\right ) \log \left (a+b x^2\right )}{2 a^2 \sqrt{a^2+2 a b x^2+b^2 x^4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.0153979, size = 54, normalized size = 0.44 \[ -\frac{\left (a+b x^2\right ) \left (-b x^2 \log \left (a+b x^2\right )+a+2 b x^2 \log (x)\right )}{2 a^2 x^2 \sqrt{\left (a+b x^2\right )^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.214, size = 51, normalized size = 0.4 \begin{align*} -{\frac{ \left ( b{x}^{2}+a \right ) \left ( 2\,b\ln \left ( x \right ){x}^{2}-b\ln \left ( b{x}^{2}+a \right ){x}^{2}+a \right ) }{2\,{a}^{2}{x}^{2}}{\frac{1}{\sqrt{ \left ( b{x}^{2}+a \right ) ^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(b*x^2+a)*(2*b*ln(x)*x^2-b*ln(b*x^2+a)*x^2+a)/((b*x^2+a)^2)^(1/2)/x^2/a^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^3/((b*x^2+a)^2)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.29492, size = 80, normalized size = 0.66 \begin{align*} \frac{b x^{2} \log \left (b x^{2} + a\right ) - 2 \, b x^{2} \log \left (x\right ) - a}{2 \, a^{2} x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.464759, size = 31, normalized size = 0.25 \begin{align*} - \frac{1}{2 a x^{2}} - \frac{b \log{\left (x \right )}}{a^{2}} + \frac{b \log{\left (\frac{a}{b} + x^{2} \right )}}{2 a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/(2*a*x**2) - b*log(x)/a**2 + b*log(a/b + x**2)/(2*a**2)

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Giac [A]  time = 1.10707, size = 70, normalized size = 0.57 \begin{align*} -\frac{1}{2} \,{\left (\frac{b \log \left (x^{2}\right )}{a^{2}} - \frac{b \log \left ({\left | b x^{2} + a \right |}\right )}{a^{2}} - \frac{b x^{2} - a}{a^{2} x^{2}}\right )} \mathrm{sgn}\left (b x^{2} + a\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(b*log(x^2)/a^2 - b*log(abs(b*x^2 + a))/a^2 - (b*x^2 - a)/(a^2*x^2))*sgn(b*x^2 + a)