\(\int \frac {x}{\sqrt {a^2+2 a b x^2+b^2 x^4}} \, dx\) [626]

Optimal result

Integrand size = 24, antiderivative size = 44 \[ \int \frac {x}{\sqrt {a^2+2 a b x^2+b^2 x^4}} \, dx=\frac {\left (a+b x^2\right ) \log \left (a+b x^2\right )}{2 b \sqrt {a^2+2 a b x^2+b^2 x^4}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {1121, 622, 31} \[ \int \frac {x}{\sqrt {a^2+2 a b x^2+b^2 x^4}} \, dx=\frac {\left (a+b x^2\right ) \log \left (a+b x^2\right )}{2 b \sqrt {a^2+2 a b x^2+b^2 x^4}} \]

[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 622

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 1121

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*} \text {integral}& = \frac {1}{2} \text {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 ) \text {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] (verified)

Time = 0.10 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.11 \[ \int \frac {x}{\sqrt {a^2+2 a b x^2+b^2 x^4}} \, dx=-\frac {\text {arctanh}\left (\frac {\frac {\sqrt {a^2}}{b}-\frac {\sqrt {a^2+2 a b x^2+b^2 x^4}}{b}}{x^2}\right )}{b} \]

[In]

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

[Out]

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

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 0.12 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.50

method	result	size
pseudoelliptic	\(\frac {\ln \left (b \,x^{2}+a \right ) \operatorname {csgn}\left (b \,x^{2}+a \right )}{2 b}\)	\(22\)
default	\(\frac {\left (b \,x^{2}+a \right ) \ln \left (b \,x^{2}+a \right )}{2 b \sqrt {\left (b \,x^{2}+a \right )^{2}}}\)	\(32\)
risch	\(\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \ln \left (b \,x^{2}+a \right )}{2 \left (b \,x^{2}+a \right ) b}\)	\(34\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.30 \[ \int \frac {x}{\sqrt {a^2+2 a b x^2+b^2 x^4}} \, dx=\frac {\log \left (b x^{2} + a\right )}{2 \, b} \]

[In]

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

[Out]

1/2*log(b*x^2 + a)/b

Sympy [F]

\[ \int \frac {x}{\sqrt {a^2+2 a b x^2+b^2 x^4}} \, dx=\int \frac {x}{\sqrt {\left (a + b x^{2}\right )^{2}}}\, dx \]

[In]

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

[Out]

Integral(x/sqrt((a + b*x**2)**2), x)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.30 \[ \int \frac {x}{\sqrt {a^2+2 a b x^2+b^2 x^4}} \, dx=\frac {\log \left (b x^{2} + a\right )}{2 \, b} \]

[In]

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

[Out]

1/2*log(b*x^2 + a)/b

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.50 \[ \int \frac {x}{\sqrt {a^2+2 a b x^2+b^2 x^4}} \, dx=\frac {\log \left ({\left | b x^{2} + a \right |}\right ) \mathrm {sgn}\left (b x^{2} + a\right )}{2 \, b} \]

[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

Mupad [B] (verification not implemented)

Time = 13.55 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.75 \[ \int \frac {x}{\sqrt {a^2+2 a b x^2+b^2 x^4}} \, dx=\frac {\ln \left (b^2\,x^2+a\,b\right )\,\mathrm {sign}\left (2\,b^2\,x^2+2\,a\,b\right )}{2\,\sqrt {b^2}} \]

[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))