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

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

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Rubi [A]  time = 0.03, antiderivative size = 44, normalized size of antiderivative = 1.00, 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} \begin {gather*} \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 {gather*}

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 31

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

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*}

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Mathematica [A]  time = 0.01, size = 35, normalized size = 0.80 \begin {gather*} \frac {\left (a+b x^2\right ) \log \left (a+b x^2\right )}{2 b \sqrt {\left (a+b x^2\right )^2}} \end {gather*}

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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IntegrateAlgebraic [B]  time = 0.25, size = 149, normalized size = 3.39 \begin {gather*} -\frac {\log \left (\sqrt {a^2+2 a b x^2+b^2 x^4}-a-\sqrt {b^2} x^2\right )}{4 \sqrt {b^2}}-\frac {\log \left (\sqrt {a^2+2 a b x^2+b^2 x^4}+a-\sqrt {b^2} x^2\right )}{4 \sqrt {b^2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {b^2} x^2}{a}-\frac {\sqrt {a^2+2 a b x^2+b^2 x^4}}{a}\right )}{2 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.63, size = 13, normalized size = 0.30 \begin {gather*} \frac {\log \left (b x^{2} + a\right )}{2 \, b} \end {gather*}

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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giac [A]  time = 0.15, size = 22, normalized size = 0.50 \begin {gather*} \frac {\log \left ({\left | b x^{2} + a \right |}\right ) \mathrm {sgn}\left (b x^{2} + a\right )}{2 \, b} \end {gather*}

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

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maple [A]  time = 0.00, size = 32, normalized size = 0.73 \begin {gather*} \frac {\left (b \,x^{2}+a \right ) \ln \left (b \,x^{2}+a \right )}{2 \sqrt {\left (b \,x^{2}+a \right )^{2}}\, b} \end {gather*}

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 = 1.35, size = 13, normalized size = 0.30 \begin {gather*} \frac {\log \left (b x^{2} + a\right )}{2 \, b} \end {gather*}

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*log(b*x^2 + a)/b

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mupad [B]  time = 4.42, size = 33, normalized size = 0.75 \begin {gather*} \frac {\ln \left (b^2\,x^2+a\,b\right )\,\mathrm {sign}\left (2\,b^2\,x^2+2\,a\,b\right )}{2\,\sqrt {b^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/((a + b*x^2)^2)^(1/2),x)

[Out]

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

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sympy [A]  time = 0.15, size = 10, normalized size = 0.23 \begin {gather*} \frac {\log {\left (a + b x^{2} \right )}}{2 b} \end {gather*}

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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