\(\int \frac {x}{a+b x^4} \, dx\) [78]

Optimal result

Integrand size = 11, antiderivative size = 25 \[ \int \frac {x}{a+b x^4} \, dx=\frac {\arctan \left (\sqrt {\frac {b}{a}} x^2\right )}{2 \sqrt {a b}} \]

[Out]

1/2/(a*b)^(1/2)*arctan(x^2*(b/a)^(1/2))

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.16, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {281, 211} \[ \int \frac {x}{a+b x^4} \, dx=\frac {\arctan \left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )}{2 \sqrt {a} \sqrt {b}} \]

[In]

Int[x/(a + b*x^4),x]

[Out]

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

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 281

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {1}{a+b x^2} \, dx,x,x^2\right ) \\ & = \frac {\arctan \left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )}{2 \sqrt {a} \sqrt {b}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.16 \[ \int \frac {x}{a+b x^4} \, dx=\frac {\arctan \left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )}{2 \sqrt {a} \sqrt {b}} \]

[In]

Integrate[x/(a + b*x^4),x]

[Out]

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

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.76

method	result	size
default	\(\frac {\arctan \left (\frac {b \,x^{2}}{\sqrt {a b}}\right )}{2 \sqrt {a b}}\)	\(19\)
risch	\(-\frac {\ln \left (x^{2} \sqrt {-a b}-a \right )}{4 \sqrt {-a b}}+\frac {\ln \left (x^{2} \sqrt {-a b}+a \right )}{4 \sqrt {-a b}}\)	\(46\)

[In]

int(x/(b*x^4+a),x,method=_RETURNVERBOSE)

[Out]

1/2/(a*b)^(1/2)*arctan(b*x^2/(a*b)^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 72, normalized size of antiderivative = 2.88 \[ \int \frac {x}{a+b x^4} \, dx=\left [-\frac {\sqrt {-a b} \log \left (\frac {b x^{4} - 2 \, \sqrt {-a b} x^{2} - a}{b x^{4} + a}\right )}{4 \, a b}, -\frac {\sqrt {a b} \arctan \left (\frac {\sqrt {a b}}{b x^{2}}\right )}{2 \, a b}\right ] \]

[In]

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

[Out]

[-1/4*sqrt(-a*b)*log((b*x^4 - 2*sqrt(-a*b)*x^2 - a)/(b*x^4 + a))/(a*b), -1/2*sqrt(a*b)*arctan(sqrt(a*b)/(b*x^2
))/(a*b)]

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 56 vs. \(2 (19) = 38\).

Time = 0.08 (sec) , antiderivative size = 56, normalized size of antiderivative = 2.24 \[ \int \frac {x}{a+b x^4} \, dx=- \frac {\sqrt {- \frac {1}{a b}} \log {\left (- a \sqrt {- \frac {1}{a b}} + x^{2} \right )}}{4} + \frac {\sqrt {- \frac {1}{a b}} \log {\left (a \sqrt {- \frac {1}{a b}} + x^{2} \right )}}{4} \]

[In]

integrate(x/(b*x**4+a),x)

[Out]

-sqrt(-1/(a*b))*log(-a*sqrt(-1/(a*b)) + x**2)/4 + sqrt(-1/(a*b))*log(a*sqrt(-1/(a*b)) + x**2)/4

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.72 \[ \int \frac {x}{a+b x^4} \, dx=\frac {\arctan \left (\frac {b x^{2}}{\sqrt {a b}}\right )}{2 \, \sqrt {a b}} \]

[In]

integrate(x/(b*x^4+a),x, algorithm="maxima")

[Out]

1/2*arctan(b*x^2/sqrt(a*b))/sqrt(a*b)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.72 \[ \int \frac {x}{a+b x^4} \, dx=\frac {\arctan \left (\frac {b x^{2}}{\sqrt {a b}}\right )}{2 \, \sqrt {a b}} \]

[In]

integrate(x/(b*x^4+a),x, algorithm="giac")

[Out]

1/2*arctan(b*x^2/sqrt(a*b))/sqrt(a*b)

Mupad [B] (verification not implemented)

Time = 13.95 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.76 \[ \int \frac {x}{a+b x^4} \, dx=\frac {\mathrm {atan}\left (\frac {\sqrt {b}\,x^2}{\sqrt {a}}\right )}{2\,\sqrt {a}\,\sqrt {b}} \]

[In]

int(x/(a + b*x^4),x)

[Out]

atan((b^(1/2)*x^2)/a^(1/2))/(2*a^(1/2)*b^(1/2))