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

Optimal. Leaf size=44 \[ \frac{\left (a+b x^2\right ) \log \left (a+b x^2\right )}{2 b \sqrt{a^2+2 a b x^2+b^2 x^4}} \]

[Out]

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

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Rubi [A]  time = 0.0316934, antiderivative size = 44, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {1107, 608, 31} \[ \frac{\left (a+b x^2\right ) \log \left (a+b x^2\right )}{2 b \sqrt{a^2+2 a b x^2+b^2 x^4}} \]

Antiderivative was successfully verified.

[In]

Int[x/Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4],x]

[Out]

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

Rule 1107

Int[(x_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[Int[(a + b*x + c*x^2)^p, x],
x, x^2], x] /; FreeQ[{a, b, c, p}, x]

Rule 608

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[(b/2 + c*x)/Sqrt[a + b*x + c*x^2], Int[1/(b/2
+ c*x), x], x] /; FreeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rubi steps

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

Mathematica [A]  time = 0.0073811, size = 35, normalized size = 0.8 \[ \frac{\left (a+b x^2\right ) \log \left (a+b x^2\right )}{2 b \sqrt{\left (a+b x^2\right )^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x/Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4],x]

[Out]

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

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Maple [A]  time = 0.207, size = 32, normalized size = 0.7 \begin{align*}{\frac{ \left ( b{x}^{2}+a \right ) \ln \left ( b{x}^{2}+a \right ) }{2\,b}{\frac{1}{\sqrt{ \left ( b{x}^{2}+a \right ) ^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(b*x^2+a)*ln(b*x^2+a)/b/((b*x^2+a)^2)^(1/2)

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Maxima [A]  time = 0.997002, size = 23, normalized size = 0.52 \begin{align*} \frac{1}{2} \, \sqrt{\frac{1}{b^{2}}} \log \left (x^{2} + \frac{a}{b}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*sqrt(b^(-2))*log(x^2 + a/b)

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Fricas [A]  time = 1.18735, size = 30, normalized size = 0.68 \begin{align*} \frac{\log \left (b x^{2} + a\right )}{2 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*log(b*x^2 + a)/b

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Sympy [A]  time = 0.136152, size = 10, normalized size = 0.23 \begin{align*} \frac{\log{\left (a + b x^{2} \right )}}{2 b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(a + b*x**2)/(2*b)

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Giac [A]  time = 1.14841, size = 30, normalized size = 0.68 \begin{align*} \frac{\log \left ({\left | b x^{2} + a \right |}\right ) \mathrm{sgn}\left (b x^{2} + a\right )}{2 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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